From 9bcf44644b875d2b0efb1cba4e8d58178181791a Mon Sep 17 00:00:00 2001
From: Mohsen Tabibian <36014419+mohsentabibian@users.noreply.github.com>
Date: Tue, 14 Nov 2023 16:42:34 +0000
Subject: [PATCH 1/7] Update documentation

---
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diff --git a/_build/html/09. Heap - Priority Queue/README.html b/09. Heap - Priority Queue/README.html
similarity index 100%
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diff --git a/_build/html/10. Backtracking/39. Combination Sum.html b/10. Backtracking/39. Combination Sum.html
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diff --git a/_build/html/10. Backtracking/79. Word Search.html b/10. Backtracking/79. Word Search.html
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diff --git a/_build/html/11. Graphs/133. Clone Graph.html b/11. Graphs/133. Clone Graph.html
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diff --git a/_build/html/11. Graphs/178. Graph Valid Tree.html b/11. Graphs/178. Graph Valid Tree.html
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diff --git a/_build/html/11. Graphs/3651. Number of Connected Components in an Undirected Graph.html b/11. Graphs/3651. Number of Connected Components in an Undirected Graph.html
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diff --git a/_build/html/13. One-D Dynamic Programming/198. House Robber.html b/13. One-D Dynamic Programming/198. House Robber.html
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diff --git a/_build/html/Book.html b/Book.html
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diff --git a/Book.ipynb b/Book.ipynb
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index e7b4f46..0000000
--- a/Book.ipynb
+++ /dev/null
@@ -1,400 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!python -m pip install --upgrade pip\n",
-    "%pip install jupyter-book"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions\n"
-     ]
-    }
-   ],
-   "source": [
-    "%cd /workspaces/LeetCode-Solutions"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book toc migrate /workspaces/LeetCode-Solutions/_toc.yml -o /workspaces/LeetCode-Solutions/_toc.yml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "\u001b[01mmaking output directory... \u001b[39;49;00mdone\n",
-      "[etoc] Changing master_doc to 'README'\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 77 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 77 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]  \n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.26 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.00 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]     \n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.92 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 1.28 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executed notebook in 0.91 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executing notebook using local CWD [mystnb]      \n",
-      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executed notebook in 0.91 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executed notebook in 1.17 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executed notebook in 1.16 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executing notebook using local CWD [mystnb] \n",
-      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executed notebook in 1.27 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executing notebook using local CWD [mystnb]\n",
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-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executed notebook in 4.03 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executing notebook using local CWD [mystnb]   \n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/Book.ipynb: Executing notebook using local CWD [mystnb]                          \n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mSECURITY\u001b[39;49;00m                                              \n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb:30008: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Executing notebook failed: CellTimeoutError [mystnb.exec]\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Notebook exception traceback saved in: /workspaces/LeetCode-Solutions/_build/html/reports/Book.err.log [mystnb.exec]\u001b[39;49;00m\n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CODE_OF_CONDUCT.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/SECURITY.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mSECURITY\u001b[39;49;00m                                               raph\u001b[39;49;00m0m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:8: WARNING: 'myst' reference target not found: #ways-to-contribute\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:9: WARNING: 'myst' reference target not found: #getting-started\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:10: WARNING: 'myst' reference target not found: #contributing-process\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:11: WARNING: 'myst' reference target not found: #code-of-conduct\u001b[39;49;00m\n",
-      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
-      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
-      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
-      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
-      "[etoc] missing index.html written as redirect to 'README.html'\n",
-      "\u001b[01mbuild succeeded, 11 warnings.\u001b[39;49;00m\n",
-      "\n",
-      "The HTML pages are in _build/html.\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Finished generating HTML for book.\n",
-      "Your book's HTML pages are here:\n",
-      "    _build/html/\n",
-      "You can look at your book by opening this file in a browser:\n",
-      "    _build/html/index.html\n",
-      "Or paste this line directly into your browser bar:\n",
-      "    file:///workspaces/LeetCode-Solutions/_build/html/index.html            \n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build ."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book build /workspaces/LeetCode-Solutions/ --builder pdfhtml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/html": [
-       "<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>"
-      ],
-      "text/plain": [
-       "<IPython.core.display.HTML object>"
-      ]
-     },
-     "execution_count": 5,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "from IPython.display import HTML\n",
-    "HTML('<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>')\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Your book template can be found at\n",
-      "\n",
-      "    mybook/\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book create mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions/mybook\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_build/html\n",
-      "[sphinxcontrib-bibtex] Beware that docutils versions 0.18 and 0.19 (you are running 0.18.1) are known to generate invalid html for citations. If this issue affects you, please use docutils<0.18 (or >=0.20 once released) instead. For more details, see https://sourceforge.net/p/docutils/patches/195/\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "[etoc] Changing master_doc to 'intro'\n",
-      "checking bibtex cache... out of date\n",
-      "parsing bibtex file /workspaces/LeetCode-Solutions/mybook/references.bib... parsed 5 entries\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='force', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/mybook/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 14 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 14 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.35 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.97 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 0.83 seconds [mystnb]\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mintro\u001b[39;49;00m                                                 \n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mintro\u001b[39;49;00m                                                  \n",
-      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
-      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
-      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
-      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
-      "[etoc] missing index.html written as redirect to 'intro.html'\n",
-      "\u001b[01mbuild succeeded, 2 warnings.\u001b[39;49;00m\n",
-      "\n",
-      "The HTML pages are in mybook/_build/html.\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Finished generating HTML for book.\n",
-      "Your book's HTML pages are here:\n",
-      "    mybook/_build/html/\n",
-      "You can look at your book by opening this file in a browser:\n",
-      "    mybook/_build/html/index.html\n",
-      "Or paste this line directly into your browser bar:\n",
-      "    file:///workspaces/LeetCode-Solutions/mybook/_build/html/index.html            \n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "format: jb-book\n",
-      "root: intro\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book toc from-project /workspaces/LeetCode-Solutions/mybook/ -f jb-book"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 2
-}
diff --git a/_build/html/CODE_OF_CONDUCT.html b/CODE_OF_CONDUCT.html
similarity index 100%
rename from _build/html/CODE_OF_CONDUCT.html
rename to CODE_OF_CONDUCT.html
diff --git a/_build/html/CONTRIBUTING.html b/CONTRIBUTING.html
similarity index 100%
rename from _build/html/CONTRIBUTING.html
rename to CONTRIBUTING.html
diff --git a/LICENSE b/LICENSE
deleted file mode 100644
index 8818d1f..0000000
--- a/LICENSE
+++ /dev/null
@@ -1,21 +0,0 @@
-MIT License
-
-Copyright (c) 2023 Mohsen Tabibian
-
-Permission is hereby granted, free of charge, to any person obtaining a copy
-of this software and associated documentation files (the "Software"), to deal
-in the Software without restriction, including without limitation the rights
-to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
-copies of the Software, and to permit persons to whom the Software is
-furnished to do so, subject to the following conditions:
-
-The above copyright notice and this permission notice shall be included in all
-copies or substantial portions of the Software.
-
-THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
-IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
-FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
-AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
-LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
-OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
-SOFTWARE.
diff --git a/_build/html/README.html b/README.html
similarity index 100%
rename from _build/html/README.html
rename to README.html
diff --git a/_build/html/SECURITY.html b/SECURITY.html
similarity index 100%
rename from _build/html/SECURITY.html
rename to SECURITY.html
diff --git a/_build/.doctrees/01. Array Hashing/1. Two Sum.doctree b/_build/.doctrees/01. Array Hashing/1. Two Sum.doctree
deleted file mode 100644
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diff --git a/_build/.doctrees/03. Sliding Window/424. Longest Repeating Character Replacement.doctree b/_build/.doctrees/03. Sliding Window/424. Longest Repeating Character Replacement.doctree
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diff --git a/_build/.doctrees/07. Trees/297. Serialize and Deserialize Binary Tree.doctree b/_build/.doctrees/07. Trees/297. Serialize and Deserialize Binary Tree.doctree
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diff --git a/_build/.doctrees/11. Graphs/3651. Number of Connected Components in an Undirected Graph.doctree b/_build/.doctrees/11. Graphs/3651. Number of Connected Components in an Undirected Graph.doctree
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diff --git a/_build/html/_sources/01. Array Hashing/1. Two Sum.ipynb b/_build/html/_sources/01. Array Hashing/1. Two Sum.ipynb
deleted file mode 100644
index b3b517f..0000000
--- a/_build/html/_sources/01. Array Hashing/1. Two Sum.ipynb	
+++ /dev/null
@@ -1,211 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 100: Same Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-        "\n",
-        "---\n",
-        "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-        "\n",
-        "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-        "\n",
-        "**Constraints**\n",
-        "\n",
-        "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-        "2. $-10^4$ <= `Node.val` <= $10^4$"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def isSameTree(p, q):\n",
-        "    # Base case: If both p and q are None, the trees are the same.\n",
-        "    if not p and not q:\n",
-        "        return True\n",
-        "    \n",
-        "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-        "    if not p or not q:\n",
-        "        return False\n",
-        "    \n",
-        "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-        "    if p.val != q.val:\n",
-        "        return False\n",
-        "    \n",
-        "    # Recursively check the left and right subtrees of p and q.\n",
-        "    # If both subtrees are the same, the entire trees are the same.\n",
-        "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-        "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-        "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-        "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-        "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-        "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "\n",
-        "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-        "\n",
-        "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "print(isSameTree(p1, q1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "\n",
-        "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-        "\n",
-        "p2 = TreeNode(1, TreeNode(2), None)\n",
-        "q2 = TreeNode(1, None, TreeNode(2))\n",
-        "print(isSameTree(p2, q2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "\n",
-        "#Input: p = [1,2,1], q = [1,1,2]\n",
-        "\n",
-        "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-        "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-        "print(isSameTree(p3, q3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "**Time Complexity**\n",
-        "\n",
-        "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-        "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-        "\n",
-        "**Space Complexity**\n",
-        "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-        "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-        "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-        "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "+ Time Complexity: $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-        "+ Space Complexity: $O(n)$ due to the recursive call stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb b/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb
deleted file mode 100644
index a77a03d..0000000
--- a/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 128. Longest Consecutive Sequence\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Consecutive Sequence problem on LeetCode, click here!](https://leetcode.com/problems/longest-consecutive-sequence/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an unsorted array of integers `nums`, return *the length of the longest consecutive elements sequence.*\n",
-        "\n",
-        "You must write an algorithm that runs in `O(n)` time.\n",
-        "\n",
-        "*Constraints:**\n",
-        "- 0 <= `nums.length` <= $10^5$\n",
-        "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 58,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def longestConsecutive(nums):\n",
-        "    if not nums:\n",
-        "        return 0\n",
-        "\n",
-        "    num_set = set(nums)\n",
-        "    max_length = 0\n",
-        "\n",
-        "    for num in num_set:\n",
-        "        if num - 1 not in num_set:\n",
-        "            current_num = num\n",
-        "            current_length = 1\n",
-        "\n",
-        "            while current_num + 1 in num_set:\n",
-        "                current_num += 1\n",
-        "                current_length += 1\n",
-        "\n",
-        "            max_length = max(max_length, current_length)\n",
-        "\n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining the `longestConsecutive` function that takes the `nums` array as input.\n",
-        "\n",
-        "2. We check if the `nums` array is empty. If it's empty, there are no consecutive elements, so we return 0.\n",
-        "\n",
-        "3. We create a Python set called `num_set` and insert all elements from the `nums` array into it. Using a set allows us to efficiently check for the existence of elements in O(1) time.\n",
-        "\n",
-        "4. We initialize a variable `max_length` to 0. This variable will keep track of the maximum length of consecutive elements sequence found.\n",
-        "\n",
-        "5. We iterate through the elements of the `num_set`. For each element `num`, we check if `num - 1` exists in the `num_set`. If it doesn't exist, it means `num` is the starting element of a potential consecutive sequence.\n",
-        "\n",
-        "6. Inside the loop, we initialize two variables: `current_num` to the current element `num` and `current_length` to 1. We start with a length of 1 because `num` itself is part of the sequence.\n",
-        "\n",
-        "7. We then enter a while loop that continues as long as `current_num + 1` exists in the `num_set`. This means we are incrementing the consecutive sequence.\n",
-        "\n",
-        "8. Inside the while loop, we increment `current_num` by 1 and also increment `current_length` by 1 to account for the next consecutive element.\n",
-        "\n",
-        "9. We compare `current_length` with the `max_length` and update `max_length` if the current sequence is longer.\n",
-        "\n",
-        "10. After the while loop, we move to the next element in the outer loop and repeat the process.\n",
-        "\n",
-        "11. Finally, we return the `max_length` as the result, which represents the length of the longest consecutive elements sequence in the `nums` array.\n",
-        "\n",
-        "The key idea here is to use a set to efficiently check for the existence of elements and to iterate through the elements, considering each element as the potential start of a consecutive sequence. By doing this, we can find the longest consecutive sequence in O(n) time complexity, where n is the number of elements in the array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 59,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "nums1 = [100, 4, 200, 1, 3, 2]\n",
-        "print(longestConsecutive(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 60,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "9\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "nums2 = [0, 3, 7, 2, 5, 8, 4, 6, 0, 1]\n",
-        "print(longestConsecutive(nums2)) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The code is designed to run in O(n) time complexity, where n is the number of elements in the `nums` array. Here's the breakdown:\n",
-        "\n",
-        "1. Constructing the `num_set` by inserting all elements from `nums` into it takes O(n) time because we perform an insertion operation for each element in `nums`.\n",
-        "\n",
-        "2. The main loop iterates through the elements in `num_set`. In the worst case, each element is visited only once, so the loop itself takes O(n) time.\n",
-        "\n",
-        "3. Within the loop, we have a while loop that may also take O(n) time in the worst case. However, this while loop is nested inside the main loop, so its overall time complexity remains O(n).\n",
-        "\n",
-        "Therefore, the overall time complexity of the code is O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of the code is determined by the space used by the `num_set` and a few additional variables. Here's the breakdown:\n",
-        "\n",
-        "1. `num_set` is a set that stores the unique elements from `nums`. In the worst case, it can store all n elements from `nums`, so the space complexity is O(n).\n",
-        "\n",
-        "2. The additional variables used, such as `max_length`, `current_num`, and `current_length`, have constant space requirements and do not depend on the size of the input array.\n",
-        "\n",
-        "Therefore, the overall space complexity of the code is O(n) due to the space used by the `num_set`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. Write an algorithm that not only returns the length of the longest consecutive elements sequence but also returns the actual consecutive sequence itself.\n",
-        "\n",
-        "2. Extend the problem to allow elements to be considered consecutive if they are within a certain absolute difference (e.g., less than or equal to k) instead of exactly 1. Write an algorithm that finds the longest such sequence and runs in O(n) time."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb b/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb
deleted file mode 100644
index 57bb55d..0000000
--- a/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb	
+++ /dev/null
@@ -1,218 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 217. Contains Duplicate\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Contains Duplicate problem on LeetCode, click here!](https://leetcode.com/problems/contains-duplicate/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `nums.length` <= $10^5$\n",
-        "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 19,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def containsDuplicate(nums):\n",
-        "    # Create an empty set to store unique elements\n",
-        "    unique_elements = set()\n",
-        "    \n",
-        "    # Iterate through the array\n",
-        "    for num in nums:\n",
-        "        # If the element is already in the set, return True\n",
-        "        if num in unique_elements:\n",
-        "            return True\n",
-        "        # Otherwise, add it to the set\n",
-        "        else:\n",
-        "            unique_elements.add(num)\n",
-        "    \n",
-        "    # If the loop completes without finding duplicates, return False\n",
-        "    return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `containsDuplicate` function takes a single argument, `nums`, which is the input integer array.\n",
-        "\n",
-        "2. Inside the function, an empty set called `unique_elements` is created. This set will be used to keep track of unique elements in the input array.\n",
-        "\n",
-        "3. The function then iterates through the input array `nums` using a `for` loop.\n",
-        "\n",
-        "4. For each element `num` in the array, it checks whether `num` is already in the `unique_elements` set using the `if num in unique_elements:` condition.\n",
-        "\n",
-        "5. If `num` is already in the set, it means there is a duplicate element in the array, and the function immediately returns `True`.\n",
-        "\n",
-        "6. If `num` is not in the set, it is added to the `unique_elements` set using `unique_elements.add(num)`.\n",
-        "\n",
-        "7. The loop continues to the next element, and the process repeats.\n",
-        "\n",
-        "8. If the loop completes without finding any duplicates, it means that all elements in the array are distinct, and the function returns `False`.\n",
-        "\n",
-        "The code efficiently utilizes a set data structure to keep track of unique elements while iterating through the array, allowing it to quickly detect duplicate elements. This code meets the problem's requirements and constraints, as explained earlier."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "nums1 = [1, 2, 3, 1]\n",
-        "print(containsDuplicate(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "nums2 = [1, 2, 3, 4]\n",
-        "print(containsDuplicate(nums2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 22,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "nums3 = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]\n",
-        "print(containsDuplicate(nums3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for the \"Contains Duplicate\" problem:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The primary operation that affects the time complexity is the `for` loop that iterates through the input array `nums`. In the worst case, the loop will iterate through all `n` elements of the array, where `n` is the length of the input array.\n",
-        "\n",
-        "- Iterating through the array: O(n)\n",
-        "\n",
-        "Inside the loop, we perform two operations:\n",
-        "\n",
-        "1. Checking whether an element exists in the `unique_elements` set (`if num in unique_elements`). This operation has an average time complexity of O(1) for a set.\n",
-        "2. Adding an element to the `unique_elements` set (`unique_elements.add(num)`), which also has an average time complexity of O(1) for a set.\n",
-        "\n",
-        "Since these operations are performed for each element in the array, the overall time complexity remains O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity is determined by the additional data structures used in the code, which are the `unique_elements` set.\n",
-        "\n",
-        "- `unique_elements` set: This set stores unique elements from the input array. In the worst case, if all elements in the input array are distinct, the set will store all `n` elements.\n",
-        "\n",
-        "Therefore, the space complexity is O(n) because, in the worst case, the set's size will grow linearly with the input array's size.\n",
-        "\n",
-        "**In summary**, the time complexity of the code is O(n), where `n` is the length of the input array, and the space complexity is also O(n) in the worst case. This code is efficient and meets the constraints of the problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K Duplicates:** Modify the problem to return `true` if there are exactly K duplicate elements in the array. Write a function that takes an additional integer parameter K and returns `true` if there are exactly K duplicates.\n",
-        "2. **Frequency Count:** Write a function that returns a list of all the elements that appear more than once in the array, along with their frequencies."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb b/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb
deleted file mode 100644
index 1ff9a9c..0000000
--- a/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb	
+++ /dev/null
@@ -1,197 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 238. Product of Array Except Self\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Product of Array Except Self problem on LeetCode, click here!](https://leetcode.com/problems/product-of-array-except-self/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array `nums`, return *an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.*\n",
-        "\n",
-        "The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-        "\n",
-        "You must write an algorithm that runs in O(n) time and without using the division operation.\n",
-        "\n",
-        "Constraints:\n",
-        "- 2 <= `nums.length` <= $10^5$\n",
-        "- `-30 <= nums[i] <= 30`\n",
-        "- The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-        "\n",
-        "**Follow-up:** Can you solve the problem in O(1) extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 50,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def productExceptSelf(nums):\n",
-        "    n = len(nums)\n",
-        "    \n",
-        "    # Initialize two arrays to store products to the left and right of each element\n",
-        "    left_products = [1] * n\n",
-        "    right_products = [1] * n\n",
-        "    \n",
-        "    # Calculate products to the left of each element\n",
-        "    left_product = 1\n",
-        "    for i in range(n):\n",
-        "        left_products[i] = left_product\n",
-        "        left_product *= nums[i]\n",
-        "    \n",
-        "    # Calculate products to the right of each element\n",
-        "    right_product = 1\n",
-        "    for i in range(n - 1, -1, -1):\n",
-        "        right_products[i] = right_product\n",
-        "        right_product *= nums[i]\n",
-        "    \n",
-        "    # Calculate the final answer using left and right products\n",
-        "    answer = [left_products[i] * right_products[i] for i in range(n)]\n",
-        "    \n",
-        "    return answer"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `n` is initialized as the length of the `nums` array.\n",
-        "2. Two arrays, `left_products` and `right_products`, are created, each with a length of `n` and filled with ones initially.\n",
-        "3. `left_product` is initialized to 1.\n",
-        "4. A loop iterates through the `nums` array, from left to right. For each element at index `i`, it stores the product of all elements to its left (including itself) in the `left_products` array and updates `left_product` accordingly.\n",
-        "5. `right_product` is initialized to 1.\n",
-        "6. Another loop iterates through the `nums` array in reverse order, from right to left. For each element at index `i`, it stores the product of all elements to its right (including itself) in the `right_products` array and updates `right_product`.\n",
-        "7. The final `answer` list is constructed by multiplying corresponding elements from the `left_products` and `right_products` arrays for each index `i`.\n",
-        "8. The function returns the `answer` list, which contains the desired results.\n",
-        "\n",
-        "Overall, this algorithm efficiently computes the product of all elements except the current element, as required. It runs in O(n) time complexity and uses O(n) extra space, which is within the problem's constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 52,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[24, 12, 8, 6]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "nums = [1,2,3,4]\n",
-        "result1 = productExceptSelf(nums)\n",
-        "print(result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 53,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[0, 0, 9, 0, 0]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums = [-1,1,0,-3,3]\n",
-        "result2 = productExceptSelf(nums)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The first loop iterates through the `nums` array to calculate the products to the left of each element. This loop runs in O(n) time, where n is the length of the `nums` array.\n",
-        "2. The second loop also iterates through the `nums` array in reverse order to calculate the products to the right of each element. This loop also runs in O(n) time.\n",
-        "3. The final loop constructs the `answer` list by multiplying elements from the `left_products` and `right_products` arrays. This loop runs in O(n) time.\n",
-        "\n",
-        "Since all the loops are independent and sequential, the overall time complexity is O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. Two additional arrays, `left_products` and `right_products`, are created with a length of `n`, where `n` is the length of the `nums` array. Therefore, these arrays consume O(n) extra space.\n",
-        "2. The `left_product` and `right_product` variables are used to keep track of the product to the left and right of each element, respectively. These variables occupy O(1) extra space.\n",
-        "3. The `answer` list, which contains the final results, also consumes O(n) space, but it's not counted toward the extra space complexity analysis as per the problem's constraints.\n",
-        "\n",
-        "Overall, the space complexity is O(n) for this solution."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Handle Zeros in the Input**: Modify the solution to handle cases where the input array contains zeros. For example, if `nums = [0, 1, 2, 3, 0]`, the output should be `[0, 0, 0, 0, 0]` because any element multiplied by zero results in zero.\n",
-        "\n",
-        "2. **In-Place Modification**: Attempt to solve the problem with in-place modification of the `nums` array, where the output is stored directly in the input array without using any additional space."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb b/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb
deleted file mode 100644
index d165899..0000000
--- a/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb	
+++ /dev/null
@@ -1,198 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 242. Valid Anagram\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Anagram problem on LeetCode, click here!](https://leetcode.com/problems/valid-anagram/)\n",
-        "\n",
-        "---\n",
-        "Given two strings `s` and `t`, return `true` if `t` is an anagram of `s`, and `false` otherwise.\n",
-        "\n",
-        "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `s.length`, `t.length` <= $5 * 10^4$\n",
-        "- `s` and `t` consist of lowercase English letters.\n",
-        "\n",
-        "**Follow-up:** What if the inputs contain Unicode characters? How would you adapt your solution to such a case?\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 29,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isAnagram(s: str, t: str) -> bool:\n",
-        "    # Check if the lengths of s and t are not equal\n",
-        "    if len(s) != len(t):\n",
-        "        return False\n",
-        "\n",
-        "    # Create dictionaries to store character frequencies for s and t\n",
-        "    countS, countT = {}, {}\n",
-        "\n",
-        "    # Count character frequencies in s\n",
-        "    for i in range(len(s)):\n",
-        "        countS[s[i]] = 1 + countS.get(s[i], 0)\n",
-        "\n",
-        "    # Count character frequencies in t\n",
-        "    for i in range(len(t)):\n",
-        "        countT[t[i]] = 1 + countT.get(t[i], 0)\n",
-        "\n",
-        "    # Check if the character frequencies in s and t are the same\n",
-        "    return countS == countT"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The code defines a class called `Solution` with a method `isAnagram` that takes two input strings, `s` and `t`, and returns a boolean (`True` or `False`) based on whether `t` is an anagram of `s`.\n",
-        "\n",
-        "2. The first check performed in the method is whether the lengths of `s` and `t` are equal. If they are not equal, it immediately returns `False` because strings of different lengths cannot be anagrams of each other.\n",
-        "\n",
-        "3. Two dictionaries, `countS` and `countT`, are created to store the frequency of characters in `s` and `t`, respectively. These dictionaries will be used to count the occurrences of each character in the strings.\n",
-        "\n",
-        "4. The code then enters a loop to iterate through each character in string `s` using a `for` loop. Inside the loop, it updates the `countS` dictionary. The line `countS[s[i]] = 1 + countS.get(s[i], 0)` increments the count of character `s[i]` in `countS` by 1. If the character is not already in the dictionary, it initializes the count to 1.\n",
-        "\n",
-        "5. Similarly, the code iterates through each character in string `t` and updates the `countT` dictionary in the same way.\n",
-        "\n",
-        "6. After counting the character frequencies in both strings, the code compares the two dictionaries using `countS == countT`. If the dictionaries are equal, it means that both strings have the same character frequencies, and therefore, `t` is an anagram of `s`. In this case, the method returns `True`.\n",
-        "\n",
-        "7. If the dictionaries are not equal, the method returns `False`, indicating that `t` is not an anagram of `s`.\n",
-        "\n",
-        "In summary, this code efficiently determines whether two input strings are anagrams by counting the character frequencies in each string using dictionaries and then comparing these counts. If the character frequencies match, the strings are considered anagrams, and the method returns `True`; otherwise, it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 30,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "s = \"anagram\"\n",
-        "t = \"nagaram\"\n",
-        "print(isAnagram(s, t))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "s = \"rat\"\n",
-        "t = \"car\"\n",
-        "print(isAnagram(s, t))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. The code first checks whether the lengths of the input strings `s` and `t` are equal, which takes constant time. This check has a time complexity of O(1).\n",
-        "\n",
-        "2. The code then iterates through both strings, `s` and `t`, once to count the character frequencies. Since both strings have a maximum length of 5 * 10^4, the worst-case scenario is that the code iterates through 5 * 10^4 characters in each string. This results in a linear time complexity of O(n), where n is the length of the longer of the two input strings (`s` or `t`).\n",
-        "\n",
-        "3. Finally, the code compares the two dictionaries `countS` and `countT` to check if the character frequencies match. This comparison takes O(n) time in the worst case, where n is the length of the longer string.\n",
-        "\n",
-        "Overall, the time complexity of the code is O(n), where n is the length of the longer input string.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The code uses two dictionaries, `countS` and `countT`, to store the character frequencies of the input strings `s` and `t`. In the worst case, both dictionaries can contain all unique characters from the input strings. Since the input strings can have a maximum length of 5 * 10^4, the space complexity for these dictionaries is O(5 * 10^4) or simply O(n), where n is the length of the longer input string.\n",
-        "\n",
-        "2. The code uses a few additional variables for bookkeeping, but these variables have constant space requirements and do not depend on the input size. Therefore, they do not significantly impact the overall space complexity.\n",
-        "\n",
-        "**In summary**, the space complexity of the code is O(n), where n is the length of the longer input string."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Anagram Chains**: Given a list of words, find the longest chain of anagrams, where each word in the chain is an anagram of the previous word. For example, given [\"bat\", \"tab\", \"cat\", \"tac\", \"dog\"], the longest anagram chain is [\"bat\", \"tab\", \"cat\", \"tac\"].\n",
-        "\n",
-        "2. **Minimum Deletions to Make Anagrams**: Given two strings `s` and `t`, find the minimum number of character deletions required in both strings to make them anagrams of each other. For example, if `s = \"hello\"` and `t = \"billion\"`, you can remove the characters \"heo\" from `s` and \"billi\" from `t` to make them anagrams (\"lo\" and \"on\" are left in the two strings)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb b/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb
deleted file mode 100644
index ecfa494..0000000
--- a/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 347. Top K Frequent Elements\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Top K Frequent Elements problem on LeetCode, click here!](https://leetcode.com/problems/top-k-frequent-elements/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array `nums` and an integer `k`, return the `k` most frequent elements. You may return the answer in **any order**.\n",
-        "\n",
-        "Constraints:\n",
-        "- 1 <= `nums.length` <= $10^5$\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- `k` is in the range `[1, the number of unique elements in the array]`.\n",
-        "- It is **guaranteed** that the answer is **unique**.\n",
-        "\n",
-        "**Follow-up:** Your algorithm's time complexity must be better than $O(n\\ log\\ n)$, where n is the array's size.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 46,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def topKFrequent(nums, k):\n",
-        "    # Create a dictionary to store the frequency of each element\n",
-        "    count = {}\n",
-        "    # Create a list of lists to store elements with the same frequency\n",
-        "    frequency = [[] for i in range(len(nums) + 1)]\n",
-        "\n",
-        "    # Count the frequency of each element in nums\n",
-        "    for n in nums:\n",
-        "        count[n] = 1 + count.get(n, 0)\n",
-        "\n",
-        "    # Place elements in the freq list according to their frequency\n",
-        "    for n, c in count.items():\n",
-        "        frequency[c].append(n)\n",
-        "\n",
-        "    res = []\n",
-        "    # Traverse freq list from the end (higher frequencies)\n",
-        "    for i in range(len(frequency) - 1, 0, -1):\n",
-        "        for n in frequency[i]:\n",
-        "            res.append(n)\n",
-        "            if len(res) == k:\n",
-        "                return res"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `def topKFrequent(nums, k):`: This is a function that takes two arguments: `nums`, which is the input array of integers, and `k`, which is the number of most frequent elements to return.\n",
-        "\n",
-        "2. `count = {}`: This dictionary `count` will be used to store the frequency of each unique element in the `nums` array. The keys are elements from the input array, and the values are their corresponding frequencies.\n",
-        "\n",
-        "3. `frequency = [[] for i in range(len(nums) + 1)]`: This creates a list of empty lists called `frequency`. It's used to store elements based on their frequencies, similar to the `freq` list in the previous code. The size of this list is set to be one greater than the length of the input `nums`.\n",
-        "\n",
-        "4. `for n in nums:`: This loop iterates through each element `n` in the input `nums` array.\n",
-        "\n",
-        "5. `count[n] = 1 + count.get(n, 0)`: This line counts the frequency of each element `n` in the `nums` array. It uses the `count.get(n, 0)` method to retrieve the current count of `n` from the `count` dictionary. If `n` is not in the dictionary, it defaults to 0. It then increments the count by 1.\n",
-        "\n",
-        "6. After the above loop, the `count` dictionary will contain counts of each unique element in the `nums` array.\n",
-        "\n",
-        "7. `for n, c in count.items():`: This loop iterates through the items (key-value pairs) of the `count` dictionary. `n` represents the element, and `c` represents its frequency.\n",
-        "\n",
-        "8. `frequency[c].append(n)`: This line places the element `n` into the bucket corresponding to its frequency `c`. Buckets are represented by the `frequency` list. For example, if an element `n` has a frequency of 3, it will be added to `frequency[3]`.\n",
-        "\n",
-        "9. After this loop, the `frequency` list will contain buckets of elements grouped by their frequencies.\n",
-        "\n",
-        "10. `res = []`: This list `res` will be used to store the k most frequent elements.\n",
-        "\n",
-        "11. `for i in range(len(freq) - 1, 0, -1):`: This loop iterates in reverse order through the `frequency` list, starting from the highest frequency and going down to 1. Note that there is a typo here; it should be `frequency` instead of `freq`.\n",
-        "\n",
-        "12. `for n in frequency[i]:`: For each element `n` in the current bucket (i.e., elements with frequency `i`), it appends `n` to the result list `res`.\n",
-        "\n",
-        "13. `if len(res) == k:`: Once `res` contains k elements, the code exits and returns the result.\n",
-        "\n",
-        "The code efficiently finds the k most frequent elements in the input array without using sorting, similar to the previous explanation, with the only difference being the variable names (e.g., `frequency` instead of `freq`)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 48,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 2]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "nums = [1,1,1,2,2,3]\n",
-        "k = 2\n",
-        "result1 = topKFrequent(nums, k)\n",
-        "print(result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 49,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums = [1]\n",
-        "k = 1\n",
-        "result2 = topKFrequent(nums, k)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The first loop that counts the frequency of elements by iterating through `nums` has a time complexity of O(n), where n is the number of elements in the input array `nums`.\n",
-        "\n",
-        "2. The second loop iterates through the keys in the `count` dictionary, which has at most `k` unique elements (as per the constraints). Therefore, the second loop also has a time complexity of O(k).\n",
-        "\n",
-        "3. The third loop iterates through the `frequency` list, which has a length equal to the maximum frequency of elements in `nums`. In the worst case, this loop can have a time complexity of O(n), where n is the number of elements in `nums`.\n",
-        "\n",
-        "Overall, the dominant factor in terms of time complexity is the loop that iterates through the `count` dictionary. So, the total time complexity of the code is O(n + k).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The `count` dictionary stores the frequency of each unique element in `nums`. In the worst case, it can have at most `k` unique elements, so the space complexity for `count` is O(k).\n",
-        "\n",
-        "2. The `frequency` list is used to store elements grouped by their frequencies. It has a length of `len(nums) + 1`, which can be at most 105 based on the constraints. Therefore, the space complexity for `frequency` is O(105).\n",
-        "\n",
-        "3. The `res` list stores the k most frequent elements, which can be at most `k` elements, so the space complexity for `res` is O(k).\n",
-        "\n",
-        "**In summary**, the space complexity is dominated by the `count` dictionary and the `frequency` list, both of which have a space complexity of O(k + 105)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimized k Most Frequent Elements:** Modify the code to find the k most frequent elements in an array while ensuring that the time complexity is O(n + klogk). You can use a priority queue (heap) to achieve this.\n",
-        "\n",
-        "2. **Handling Duplicate Frequencies:** Extend the code to handle cases where multiple elements have the same frequency and are among the k most frequent elements. Ensure that the output contains exactly k elements."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb b/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb
deleted file mode 100644
index afe89ec..0000000
--- a/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb	
+++ /dev/null
@@ -1,229 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 49. Group Anagrams\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Group Anagrams problem on LeetCode, click here!](https://leetcode.com/problems/group-anagrams/)\n",
-        "\n",
-        "---\n",
-        "Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.\n",
-        "\n",
-        "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `strs.length <= $10^4$\n",
-        "- `0 <= strs[i].length <= 100`\n",
-        "- `strs[i]` consists of lowercase English letters.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 36,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict\n",
-        "\n",
-        "\n",
-        "def groupAnagrams(strs):\n",
-        "    ans = defaultdict(list)\n",
-        "\n",
-        "    # Iterate through the list of input strings\n",
-        "    for s in strs:\n",
-        "        # Initialize a list to represent character counts for each character (a-z)\n",
-        "        count = [0] * 26\n",
-        "\n",
-        "        # Count the occurrences of each character in the current word\n",
-        "        for c in s:\n",
-        "            count[ord(c) - ord(\"a\")] += 1\n",
-        "\n",
-        "        # Use a tuple of character counts as the key and append the word to the anagram group\n",
-        "        ans[tuple(count)].append(s)\n",
-        "\n",
-        "    # Convert the values (lists of anagrams) to a list of lists\n",
-        "    return list(ans.values())"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. Inside the method, a `defaultdict` called `ans` is created to store anagram groups. This dictionary will have a list as its default value, meaning that each key in the dictionary will be associated with an empty list by default.\n",
-        "\n",
-        "2. The code then iterates through the list of input strings, `strs`, using a `for` loop. For each word `s` in `strs`, it performs the following steps:\n",
-        "\n",
-        "   - It initializes a list `count` of length 26, where each element represents the count of a specific character (a-z) in the word `s`.\n",
-        "\n",
-        "   - The code then iterates through the characters of `s`. For each character `c`, it increments the corresponding count in the `count` list based on its ASCII value.\n",
-        "\n",
-        "   - After counting the characters in `s`, it converts the `count` list into a tuple `tuple(count)` to use as a key for the `ans` dictionary.\n",
-        "\n",
-        "   - It appends the word `s` to the list associated with the key (tuple) in the `ans` dictionary. This groups all anagrams of the same word together under the same key.\n",
-        "\n",
-        "3. After processing all words in `strs`, the code converts the values of the `ans` dictionary (which are lists of anagrams) to a list of lists using the `list()` constructor.\n",
-        "\n",
-        "4. Finally, the code returns the list of anagram groups, which is the result of grouping the anagrams in the input list.\n",
-        "\n",
-        "In summary, the code efficiently groups anagrams together by counting the characters in each word and using a dictionary to store them under the same key. The result is a list of lists, where each inner list represents a group of anagrams."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 37,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: Group anagrams from a list of words\n",
-        "strs1 = [\"eat\", \"tea\", \"tan\", \"ate\", \"nat\", \"bat\"]\n",
-        "result1 = groupAnagrams(strs1)\n",
-        "print(result1)\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 38,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2: Group anagrams from a list with an empty string\n",
-        "strs2 = [\"\"]\n",
-        "result2 = groupAnagrams(strs2)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 40,
-      "id": "c8367e2a-e7d5-4e18-8235-2870e341a04b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['a']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3: Group anagrams from a list with a single word\n",
-        "strs3 = [\"a\"]\n",
-        "result3 = groupAnagrams(strs3)\n",
-        "print(result3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `groupAnagrams` method:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. First, we create an instance of the `Solution` class, which is an O(1) operation.\n",
-        "\n",
-        "2. Inside the `groupAnagrams` method, we iterate through the list of input strings `strs` once in a loop:\n",
-        "   - For each word in `strs`, we perform character counting, which is done in O(K) time, where K is the maximum length of a word in the list.\n",
-        "\n",
-        "3. The overall time complexity of the method is O(N * K), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. We use a `defaultdict` called `ans` to store the anagram groups. In the worst case, each word belongs to a distinct anagram group, so the space complexity for `ans` is O(N).\n",
-        "\n",
-        "2. Within the loop, we create a `count` list of length 26 to store character counts for each word. This is a fixed-size list and does not depend on the input size, so it has a constant space complexity of O(26), which is effectively O(1).\n",
-        "\n",
-        "3. The space used for other variables is also constant and does not depend on the input size.\n",
-        "\n",
-        "4. The final result, `result`, is a list of lists that contains the grouped anagrams. In the worst case, each word is a distinct anagram group, so the space complexity for `result` is O(N).\n",
-        "\n",
-        "5. The overall space complexity of the method is O(N) due to the space used by `ans` and `result`. The space used by the `count` list and other variables is constant and does not significantly affect the overall space complexity.\n",
-        "\n",
-        "**In summary,** the time complexity of the `groupAnagrams` method is O(N * K), and the space complexity is O(N), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Anagramic Palindromes**: Given a list of strings, find all the groups of anagrams where each group contains words that can be rearranged into a palindrome. For example, in the list [\"bat\", \"tab\", \"act\", \"tac\", \"dog\"], there are two groups: [\"bat\", \"tab\", \"act\", \"tac\"] and [\"dog\"].\n",
-        "\n",
-        "2. **Largest Anagram Group**: Given a list of strings, find the largest group of anagrams. Return the list of anagrams in that group. If there are multiple largest groups, return all of them."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb b/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb
deleted file mode 100644
index 037c476..0000000
--- a/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb	
+++ /dev/null
@@ -1,204 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 659· Encode and Decode Strings\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Encode and Decode Strings problem on LintCode, click here!](https://www.lintcode.com/problem/659/)\n",
-        "\n",
-        "---\n",
-        "Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.\n",
-        "\n",
-        "Please implement `encode` and `decode`.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 54,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class Codec:\n",
-        "\n",
-        "    def encode(self, strs):\n",
-        "        \"\"\"Encodes a list of strings to a single string.\n",
-        "        \n",
-        "        :type strs: List[str]\n",
-        "        :rtype: str\n",
-        "        \"\"\"\n",
-        "        encoded = \"\"\n",
-        "        for s in strs:\n",
-        "            encoded += s.replace(':', '::') + ':;'\n",
-        "        return encoded\n",
-        "\n",
-        "    def decode(self, s):\n",
-        "        \"\"\"Decodes a single string to a list of strings.\n",
-        "        \n",
-        "        :type s: str\n",
-        "        :rtype: List[str]\n",
-        "        \"\"\"\n",
-        "        decoded = []\n",
-        "        i = 0\n",
-        "        while i < len(s):\n",
-        "            end = s.find(':;', i)\n",
-        "            if end == -1:\n",
-        "                end = len(s)\n",
-        "            decoded.append(s[i:end].replace('::', ':'))\n",
-        "            i = end + 2\n",
-        "        return decoded"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a class called `Codec` to encapsulate the encoding and decoding operations for a list of strings.\n",
-        "\n",
-        "2. In the `encode` method, we take a list of strings (`strs`) as input and return a single encoded string. The encoding process involves concatenating the strings together with a delimiter `:;`. We iterate through each string in the input list, replace any occurrence of `:` with `::` (to avoid conflicts with the delimiter), and then append `:;` to indicate the end of that string.\n",
-        "\n",
-        "3. In the `decode` method, we take an encoded string (`s`) as input and return a list of strings. The decoding process involves splitting the encoded string using the `:;` delimiter. We start from the beginning of the encoded string (`i = 0`) and repeatedly find the next occurrence of `:;`. We extract the substring between the current position (`i`) and the next delimiter (`end`). Before adding it to the result list, we replace any `::` with `:` to revert the encoding. We then update the current position `i` to be after the delimiter, so we can find the next substring in the next iteration.\n",
-        "\n",
-        "4. Finally, we create an instance of the `Codec` class and test it with two examples:\n",
-        "\n",
-        "By using the `:;` delimiter and handling the `::` escaping, this code can encode and decode lists of strings, preserving the original content even if it contains colons or the delimiter itself."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 55,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['lint', 'code', 'love', 'you']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "codec = Codec()\n",
-        "\n",
-        "input1 = [\"lint\", \"code\", \"love\", \"you\"]\n",
-        "encoded1 = codec.encode(input1)\n",
-        "decoded1 = codec.decode(encoded1)\n",
-        "print(decoded1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 57,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['we', 'say', ':', 'yes']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "codec = Codec() \n",
-        "\n",
-        "input2 = [\"we\", \"say\", \":\", \"yes\"]\n",
-        "encoded2 = codec.encode(input2)\n",
-        "decoded2 = codec.decode(encoded2)\n",
-        "print(decoded2)  "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `encode` and `decode` methods in the provided code.\n",
-        "\n",
-        "**Encode Method (`encode`):**\n",
-        "\n",
-        "- Time Complexity: O(n * m)\n",
-        "  - Here, `n` is the number of strings in the input list `strs`, and `m` is the average length of these strings.\n",
-        "  - In the worst case, for each string in the list, we iterate over its characters once to replace any colons (`:`) with double colons (`::`) and then append `:;`. This is done for each string in the list.\n",
-        "\n",
-        "- Space Complexity: O(n * m)\n",
-        "  - The space complexity is also O(n * m) because we build the encoded string by concatenating the modified strings. In the worst case, the encoded string can be of the same length as the original strings.\n",
-        "\n",
-        "**Decode Method (`decode`):**\n",
-        "\n",
-        "- Time Complexity: O(n * m)\n",
-        "  - Similar to the encode method, we iterate through the encoded string in the decode method. In the worst case, we may have to scan each character in the encoded string to find the `:;` delimiter.\n",
-        "\n",
-        "- Space Complexity: O(n * m)\n",
-        "  - The space complexity of the decode method is also O(n * m) because we build the list of decoded strings, which can be of the same length as the encoded string.\n",
-        "\n",
-        "Overall, both the `encode` and `decode` methods have time and space complexities of O(n * m), where `n` is the number of strings in the input list, and `m` is the average length of these strings. The space complexity arises from storing the encoded or decoded strings, and the time complexity arises from iterating through the characters in these strings."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Delimiters**: Modify the code to allow for multiple delimiters, not just one. Each string could have its own delimiter, and the code should be able to handle this complexity.\n",
-        "\n",
-        "2. **Optimize Encoding**: Modify the `encode` method to achieve a more efficient encoding algorithm. Try to minimize the length of the encoded string while ensuring that it can be correctly decoded."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/README.md b/_build/html/_sources/01. Array Hashing/README.md
deleted file mode 100644
index a331dab..0000000
--- a/_build/html/_sources/01. Array Hashing/README.md	
+++ /dev/null
@@ -1,27 +0,0 @@
-# Array & Hashing Solutions - Blind 75 LeetCode Problems
-![Array & Hashing Solutions](https://img.shields.io/badge/Array_&_Hashing_Solutions-blue)![Blind 75 LeetCode](https://img.shields.io/badge/Blind_75_LeetCode-Problems-blue?labelColor=red)
-
-This repository contains my solutions to Array & Hashing Problems from Blind 75 LeetCode problems. I've organized the solutions by categories for easier navigation and reference. Each problem solution is presented in Jupyter Notebook format (`.ipynb`).
-
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [217. Contains Duplicate](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/217.%20Contains%20Duplicate.ipynb)                            | Easy       |
-| [242. Valid Anagram](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/242.%20Valid%20Anagram.ipynb)                                      | Easy       |
-| [1. Two Sum](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/1.%20Two%20Sum.ipynb)                                                      | Easy       |
-| [49. Group Anagrams](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/49.%20Group%20Anagrams.ipynb)                                      | Medium     |
-| [347. Top K Frequent Elements](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.ipynb)              | Medium     |
-| [238. Product of Array Except Self](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.ipynb)  | Medium     |
-| [659. Encode and Decode Strings](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.ipynb)          | Medium     |
-| [128. Longest Consecutive Sequence](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.ipynb)      | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb b/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb
deleted file mode 100644
index 284c832..0000000
--- a/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 11. Container With Most Water\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Container With Most Water problem on LeetCode, click here!](https://leetcode.com/problems/container-with-most-water/)\n",
-        "\n",
-        "---\n",
-        "You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the $i^{th}$ line are `(i, 0)` and `(i, height[i])`.\n",
-        "\n",
-        "Find two lines that together with the x-axis form a container, such that the container contains the most water.\n",
-        "\n",
-        "Return *the maximum amount of water a container can store.*\n",
-        "\n",
-        "**Notice** that you may not slant the container.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `n == height.length`\n",
-        "- 2 <= `n` <= $10^5$\n",
-        "- 0 <= `height[i]` <= $10^4$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def maxArea(height):\n",
-        "    # Initialize the maximum area to 0.\n",
-        "    max_area = 0\n",
-        "    # Initialize two pointers, one at the beginning (left) and one at the end (right) of the array.\n",
-        "    left = 0\n",
-        "    right = len(height) - 1\n",
-        "\n",
-        "    # Iterate until the left pointer is less than the right pointer.\n",
-        "    while left < right:\n",
-        "        # Calculate the height of the container, which is the minimum height of the two lines.\n",
-        "        h = min(height[left], height[right])\n",
-        "        # Calculate the width of the container, which is the distance between the two pointers.\n",
-        "        width = right - left\n",
-        "        # Calculate the area of the container using height and width and update max_area if it's greater.\n",
-        "        max_area = max(max_area, h * width)\n",
-        "\n",
-        "        # Move the pointer with the shorter line inward.\n",
-        "        if height[left] < height[right]:\n",
-        "            left += 1\n",
-        "        else:\n",
-        "            right -= 1\n",
-        "\n",
-        "    # Continue this process until left < right, and then return the maximum area.\n",
-        "    return max_area"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start with initializing `max_area` to 0, which will store the maximum area of water that can be contained by the lines.\n",
-        "\n",
-        "2. Two pointers, `left` and `right`, are initialized to the beginning and end of the input `height` array.\n",
-        "\n",
-        "3. The main loop runs while `left` is less than `right`. This ensures that we are checking all possible pairs of lines in a systematic way.\n",
-        "\n",
-        "4. Inside the loop:\n",
-        "   - We calculate the height of the container `h` as the minimum of the heights at the `left` and `right` pointers. This represents the height of the water the container can hold.\n",
-        "   - We calculate the width of the container as the difference between `right` and `left`. This represents the distance between the two lines.\n",
-        "   - We calculate the area of the container using `h * width` and update `max_area` if the calculated area is greater than the current `max_area`.\n",
-        "\n",
-        "5. After calculating the area and updating `max_area`, we move one of the pointers. We move the pointer pointing to the shorter line inward because moving the pointer pointing to the taller line won't increase the height and will only decrease the width, which will result in a smaller area.\n",
-        "\n",
-        "6. We continue this process until `left` is no longer less than `right`, indicating that we have checked all possible pairs of lines.\n",
-        "\n",
-        "7. Finally, we return the `max_area` which contains the maximum area of water that can be contained by two lines from the input array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 49\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "height1 = [1, 8, 6, 2, 5, 4, 8, 3, 7]\n",
-        "result1 = maxArea(height1)\n",
-        "print(\"Output:\", result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "height2 = [1, 1]\n",
-        "result2 = maxArea(height2)\n",
-        "print(\"Output:\", result2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 45\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Additional Example\n",
-        "height3 = [3, 9, 3, 4, 7, 2, 12, 6]\n",
-        "result3 = maxArea(height3)\n",
-        "print(\"Output:\", result3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `maxArea` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The function uses a two-pointer approach to find the maximum area of water. In the worst case, both the left and right pointers traverse the entire input array once. Therefore, the time complexity is O(n), where n is the number of elements in the `height` array.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The function uses a constant amount of extra space for variables like `max_area`, `left`, `right`, `h`, and `width`. It does not use any data structures whose space consumption depends on the input size. As a result, the space complexity is O(1), which means it has constant space complexity.\n",
-        "\n",
-        "**In summary,** the `maxArea` function has a time complexity of O(n) and a space complexity of O(1), making it an efficient algorithm for finding the maximum area of water that can be contained by two lines in the input array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the maxArea function to not only return the maximum area but also the indices of the two lines that form the container with the maximum area.\n",
-        "2. **Variant Problem:** Instead of finding the maximum area of water, find the minimum amount of water required to fill all the gaps between the lines. You'll need to return the total water volume needed."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb b/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb
deleted file mode 100644
index 8520b3b..0000000
--- a/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb	
+++ /dev/null
@@ -1,218 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 125. Valid Palindrome\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Palindrome problem on LeetCode, click here!](https://leetcode.com/problems/valid-palindrome/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.\n",
-        "\n",
-        "Given a string `s`, *return `true` if it is a **palindrome**, or `false` otherwise.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $2*10^5$\n",
-        "- `s` consists only of printable ASCII characters.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isPalindrome(s):\n",
-        "    # Step 1: Convert the input string to lowercase\n",
-        "    s = s.lower()\n",
-        "    \n",
-        "    # Step 2: Remove all non-alphanumeric characters from the string\n",
-        "    cleaned_s = ''.join(c for c in s if c.isalnum())\n",
-        "    \n",
-        "    # Step 3: Initialize two pointers, one at the beginning and one at the end of the cleaned string\n",
-        "    left, right = 0, len(cleaned_s) - 1\n",
-        "    \n",
-        "    # Step 4: Compare characters using the two pointers while moving them towards each other\n",
-        "    while left < right:\n",
-        "        if cleaned_s[left] != cleaned_s[right]:\n",
-        "            return False  # If characters don't match, it's not a palindrome\n",
-        "        left += 1\n",
-        "        right -= 1\n",
-        "    \n",
-        "    # Step 5: If the loop completes without returning False, it's a palindrome\n",
-        "    return True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `isPalindrome` function takes an input string `s`.\n",
-        "\n",
-        "2. It starts by converting the input string to lowercase using `s.lower()`. This step ensures that the function is case-insensitive when checking for palindromes.\n",
-        "\n",
-        "3. Next, it removes all non-alphanumeric characters from the string and stores the result in the `cleaned_s` variable. It does this by iterating through each character in the input string and only keeping characters that are alphanumeric (letters and digits). This step removes spaces, punctuation, and other non-alphanumeric characters.\n",
-        "\n",
-        "4. The function then initializes two pointers, `left` and `right`, which will be used to compare characters from the beginning and end of the cleaned string.\n",
-        "\n",
-        "5. In a while loop, it compares characters using the two pointers while moving them towards each other. The loop continues until the `left` pointer is less than the `right` pointer. This is done to check all characters in the cleaned string.\n",
-        "\n",
-        "6. Inside the loop, it checks if the characters at the `left` and `right` pointers don't match. If they don't match, it means the input is not a palindrome, so the function returns `False`.\n",
-        "\n",
-        "7. If the characters at the `left` and `right` pointers match, the pointers are moved closer to each other by incrementing `left` and decrementing `right`.\n",
-        "\n",
-        "8. The loop continues to compare characters until it either finds a mismatch (returning `False`) or until the `left` pointer is greater than or equal to the `right` pointer. If the loop completes without finding a mismatch, it means the input is a palindrome, so the function returns `True`.\n",
-        "\n",
-        "In summary, the code converts the input string to lowercase, removes non-alphanumeric characters, and then uses two pointers to compare characters from the cleaned string's beginning and end. If all characters match during the comparison, the function returns `True`, indicating that the input is a palindrome. If any characters do not match, it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "input1 = \"A man, a plan, a canal: Panama\"\n",
-        "output1 = isPalindrome(input1)\n",
-        "print(output1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "input2 = \"race a car\"\n",
-        "output2 = isPalindrome(input2)\n",
-        "print(output2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "input3 = \"\"\n",
-        "output3 = isPalindrome(input3)\n",
-        "print(output3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `isPalindrome` function that uses two pointers:\n",
-        "\n",
-        "1. **Time Complexity:**\n",
-        "   - Converting the input string to lowercase using `s.lower()` takes O(n) time, where 'n' is the length of the input string.\n",
-        "   - Removing non-alphanumeric characters using the list comprehension `''.join(c for c in s if c.isalnum())` also takes O(n) time because it iterates through each character in the string once.\n",
-        "   - The comparison of characters using the two pointers (`left` and `right`) is done in a loop that runs until `left` is less than `right`. In the worst case, the loop iterates through half of the string, so it takes O(n/2) time, which is still considered O(n).\n",
-        "\n",
-        "   Therefore, the overall time complexity of the `isPalindrome` function is O(n), where 'n' is the length of the input string.\n",
-        "\n",
-        "2. **Space Complexity:**\n",
-        "   - The space complexity is primarily determined by the additional string `cleaned_s` created to store the cleaned version of the input string. In the worst case, if all characters in the input string are alphanumeric, this cleaned string can be as large as the original input string.\n",
-        "   - Additionally, the function uses two pointers, `left` and `right`, which consume constant space and do not depend on the size of the input.\n",
-        "\n",
-        "**In summary,** the overall space complexity of the `isPalindrome` function is O(n) in the worst case, where 'n' is the length of the input string. The space complexity is dominated by the space required for `cleaned_s`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Longest Palindromic Substring:** Given a string `s`, find and return the longest palindromic substring within `s`. For example, if `s = \"babad\"`, the function should return `\"bab\"` or `\"aba\"`.\n",
-        "2. **Palindrome with Limited Characters:** Given a string `s` and a list of characters `allowed_chars`, check if `s` is a palindrome considering only the characters from `allowed_chars`. Ignore all other characters. For example, if `s = \"A man, a plan, a canal: Panama\"` and `allowed_chars = ['a', 'm', 'n', 'p']`, the function should return `True`."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb b/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb
deleted file mode 100644
index 06de549..0000000
--- a/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb	
+++ /dev/null
@@ -1,250 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 15. 3Sum\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the 3Sum problem on LeetCode, click here!](https://leetcode.com/problems/3sum/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array nums, *return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.*\n",
-        "\n",
-        "**Notice** that the solution set must not contain duplicate triplets.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `3 <= nums.length <= 3000`\n",
-        "- $-10^5$ <= `nums[i]` <= $10^5$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def threeSum(nums):\n",
-        "    # Sort the input array in ascending order\n",
-        "    nums.sort()\n",
-        "    triplets = []\n",
-        "    \n",
-        "    for i in range(len(nums) - 2):\n",
-        "        # Skip duplicates to avoid duplicate triplets\n",
-        "        if i > 0 and nums[i] == nums[i - 1]:\n",
-        "            continue\n",
-        "        \n",
-        "        left, right = i + 1, len(nums) - 1\n",
-        "        \n",
-        "        while left < right:\n",
-        "            # Calculate the total sum of the current triplet\n",
-        "            total = nums[i] + nums[left] + nums[right]\n",
-        "            \n",
-        "            if total == 0:\n",
-        "                # If the total is zero, we found a valid triplet\n",
-        "                triplets.append([nums[i], nums[left], nums[right]])\n",
-        "                \n",
-        "                # Skip duplicates of left and right pointers\n",
-        "                while left < right and nums[left] == nums[left + 1]:\n",
-        "                    left += 1\n",
-        "                while left < right and nums[right] == nums[right - 1]:\n",
-        "                    right -= 1\n",
-        "                \n",
-        "                # Move the pointers to the next unique elements\n",
-        "                left += 1\n",
-        "                right -= 1\n",
-        "            elif total < 0:\n",
-        "                # If the total is negative, we need to increase the sum by moving the left pointer to the right\n",
-        "                left += 1\n",
-        "            else:\n",
-        "                # If the total is positive, we need to decrease the sum by moving the right pointer to the left\n",
-        "                right -= 1\n",
-        "    \n",
-        "    return triplets"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. It starts by sorting the input array `nums` in ascending order. Sorting helps in efficiently finding triplets that sum up to zero.\n",
-        "\n",
-        "2. It initializes an empty list `triplets` to store the unique triplets that meet the given conditions.\n",
-        "\n",
-        "3. The code then iterates through the `nums` array using a loop with index `i`. This loop will consider each element of the array as a potential starting point for a triplet.\n",
-        "\n",
-        "4. Inside the loop, it checks for duplicates and skips them. This is done to ensure that the solution set does not contain duplicate triplets. If `nums[i]` is the same as the previous element `nums[i-1]`, it continues to the next iteration of the loop.\n",
-        "\n",
-        "5. Two pointers, `left` and `right`, are initialized. `left` starts just after the current element `nums[i]`, and `right` starts at the end of the array.\n",
-        "\n",
-        "6. The code enters another loop with `left` and `right` pointers, trying to find a triplet that sums up to zero.\n",
-        "\n",
-        "7. It calculates the `total` sum of the current triplet as `nums[i] + nums[left] + nums[right]`.\n",
-        "\n",
-        "8. If `total` is zero, it means a valid triplet is found, so it appends `[nums[i], nums[left], nums[right]]` to the `triplets` list. Then, it moves both the `left` and `right` pointers to their next unique elements while skipping duplicates.\n",
-        "\n",
-        "9. If `total` is less than zero, it means the sum is negative, so it increments the `left` pointer to move towards a larger sum.\n",
-        "\n",
-        "10. If `total` is greater than zero, it means the sum is positive, so it decrements the `right` pointer to move towards a smaller sum.\n",
-        "\n",
-        "11. This process continues until all possible triplets have been considered.\n",
-        "\n",
-        "12. Finally, the function returns the `triplets` list, which contains all the unique triplets that sum up to zero in the sorted `nums` array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[-1, -1, 2], [-1, 0, 1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "nums = [-1, 0, 1, 2, -1, -4]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "nums = [0, 1, 1]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 0, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "nums = [0, 0, 0]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `threeSum` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. Sorting the input array `nums` takes O(n log n) time, where n is the length of the array.\n",
-        "\n",
-        "2. The outer loop runs for each element of the array, so it iterates O(n) times.\n",
-        "\n",
-        "3. Inside the outer loop, we have a while loop with two pointers (`left` and `right`). In the worst case, the while loop can iterate O(n) times (when all elements are unique).\n",
-        "\n",
-        "4. Inside the while loop, we have constant time operations (additions, comparisons, and list appends).\n",
-        "\n",
-        "Overall, the time complexity of the `threeSum` function is dominated by the sorting step, so it is O(n log n) due to the sorting. The other operations inside the loops contribute linearly but are dominated by the sorting step.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The space used by the `triplets` list to store the output triplets is O(k), where k is the number of unique triplets that sum up to zero. In the worst case, this can be O(n^2/3) because there can be roughly O(n^2) possible triplets, and many of them may sum up to zero.\n",
-        "\n",
-        "2. Other than the `triplets` list, the function uses only a constant amount of additional space for variables like `i`, `left`, `right`, and `total`.\n",
-        "\n",
-        "Overall, the space complexity of the `threeSum` function is O(k), where k is the number of unique triplets that meet the criteria. It's important to note that the space complexity does not depend on the size of the input array but rather on the number of valid triplets found.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: O(n log n) due to sorting (dominant factor) and O(n^2) in the worst case for the nested loops.\n",
-        "- Space Complexity: O(k) for storing the output triplets, where k is the number of unique triplets that meet the criteria."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K-Sum Problem:** Generalize the problem to the K-Sum problem, where instead of finding triplets that sum to zero, you need to find K elements that sum to a target value. Implement a function for this generalization.\n",
-        "2. **Count the Number of Unique Triplets:** Modify the algorithm to count the number of unique triplets that satisfy the conditions instead of returning the triplets themselves. This will require some modifications to handle duplicates efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/02. Two Pointers/README.md b/_build/html/_sources/02. Two Pointers/README.md
deleted file mode 100644
index 0d4c7cf..0000000
--- a/_build/html/_sources/02. Two Pointers/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Two Pointers Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)                            | Easy       |
-| [15. 3Sum](https://leetcode.com/problems/3sum/)                                                       | Medium     |
-| [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)           | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb b/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb
deleted file mode 100644
index 554c940..0000000
--- a/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb	
+++ /dev/null
@@ -1,191 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 121. Best Time to Buy and Sell Stock\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Best Time to Buy and Sell Stock problem on LeetCode, click here!](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given an array `prices` where `prices[i]` is the price of a given stock on the $i^{th}$ day.\n",
-        "\n",
-        "You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.\n",
-        "\n",
-        "*Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `prices.length` <= $10^5$\n",
-        "- 0 <= `prices[i]` <= $10^4$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 65,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def maxProfit(prices):\n",
-        "    if not prices:\n",
-        "        return 0\n",
-        "    \n",
-        "    min_price = prices[0]\n",
-        "    max_profit = 0\n",
-        "    \n",
-        "    for price in prices:\n",
-        "        # Update the minimum price if a lower price is encountered\n",
-        "        min_price = min(min_price, price)\n",
-        "        # Calculate the profit if selling at the current price\n",
-        "        max_profit = max(max_profit, price - min_price)\n",
-        "    \n",
-        "    return max_profit"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- The `maxProfit` function takes a single argument, `prices`, which is a list of stock prices.\n",
-        "- The first line of the function checks if the `prices` list is empty. If it's empty, there are no prices to analyze, so the function returns 0 (no profit can be made).\n",
-        "\n",
-        "- Then, we initialize two variables:\n",
-        "  - `min_price`: This variable will keep track of the minimum price seen so far. We initialize it with the price of the first day in the list (`prices[0]`).\n",
-        "  - `max_profit`: This variable will keep track of the maximum profit we can achieve. We initialize it to 0 because initially, we haven't made any profit.\n",
-        "\n",
-        "- We start iterating through the `prices` list, where `price` represents the stock price for the current day.\n",
-        "\n",
-        "- In each iteration, we check if the current `price` is lower than the `min_price` we've seen so far. If it is, we update `min_price` to the current `price`. This is because we want to buy the stock at the lowest possible price.\n",
-        "\n",
-        "- Next, we calculate the profit we would make if we sold the stock at the current `price` (assuming we bought it at `min_price`). We do this by subtracting `min_price` from the current `price`. This represents the profit for the current day.\n",
-        "\n",
-        "- We also use the `max` function to keep track of the maximum profit seen so far. If the profit calculated for the current day is greater than the `max_profit` we've seen previously, we update `max_profit` with this higher value.\n",
-        "\n",
-        "- Finally, after iterating through all the days in the `prices` list, we return `max_profit`, which represents the maximum profit that can be achieved by buying and selling a single stock."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 66,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "5\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "prices1 = [7, 1, 5, 3, 6, 4]\n",
-        "print(maxProfit(prices1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 67,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "0\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "prices2 = [7, 6, 4, 3, 1]\n",
-        "print(maxProfit(prices2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code iterates through the `prices` array once, examining each price exactly once.\n",
-        "- Within the loop, there are constant-time operations such as comparisons, additions, and subtractions.\n",
-        "- Therefore, the time complexity of this code is O(n), where n is the length of the `prices` array. It performs a linear number of operations relative to the size of the input.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the code is constant, O(1).\n",
-        "- Regardless of the size of the input `prices` array, the code only uses a fixed amount of additional memory to store two variables (`min_price` and `max_profit`) and a few loop variables.\n",
-        "- The memory usage does not depend on the size of the input, so it is considered constant space complexity.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: O(n) where n is the length of the `prices` array.\n",
-        "- Space Complexity: O(1) (constant space usage)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Transactions Allowed:** Modify the problem to allow multiple transactions (buy and sell multiple times), but you must sell before buying again. Find the maximum profit in this scenario.\n",
-        "2. **K Transactions Allowed:** Extend the problem to allow at most k transactions. Find the maximum profit considering this constraint.\n",
-        "3. **With Transaction Fee:** Introduce a transaction fee for each buy/sell operation. Modify the code to maximize profit while considering the transaction fee."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb b/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb
deleted file mode 100644
index 4a516de..0000000
--- a/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb	
+++ /dev/null
@@ -1,214 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 3. Longest Substring Without Repeating Characters\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Substring Without Repeating Characters problem on LeetCode, click here!](https://leetcode.com/problems/longest-substring-without-repeating-characters/)\n",
-        "\n",
-        "---\n",
-        "Given a string `s`, find the length of the **longest substring** without repeating characters.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 0 <= `s.length` <= $5 * 10^4$\n",
-        "- `s` consists of English letters, digits, symbols and spaces.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 61,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def length_of_longest_substring(s):\n",
-        "    # Create a dictionary to store the index of each character's last occurrence.\n",
-        "    char_index = {}\n",
-        "    \n",
-        "    # Initialize variables to keep track of the start and end of the current substring.\n",
-        "    start = 0\n",
-        "    max_length = 0\n",
-        "    \n",
-        "    for end in range(len(s)):\n",
-        "        # If the character is in the dictionary and its last occurrence is after the start of the current substring,\n",
-        "        # update the start of the substring to the next character after its last occurrence.\n",
-        "        if s[end] in char_index and char_index[s[end]] >= start:\n",
-        "            start = char_index[s[end]] + 1\n",
-        "        \n",
-        "        # Update the last occurrence index of the current character.\n",
-        "        char_index[s[end]] = end\n",
-        "        \n",
-        "        # Calculate the length of the current substring and update the maximum length if necessary.\n",
-        "        current_length = end - start + 1\n",
-        "        max_length = max(max_length, current_length)\n",
-        "    \n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a function called `length_of_longest_substring` that takes a single argument `s`, which is the input string for which we want to find the length of the longest substring without repeating characters.\n",
-        "\n",
-        "2. We create a dictionary called `char_index` to store the index of the last occurrence of each character in the input string. This dictionary will help us keep track of where each character was last seen.\n",
-        "\n",
-        "3. We initialize two variables, `start` and `max_length`. \n",
-        "   - `start` represents the start index of the current substring without repeating characters. It starts at 0.\n",
-        "   - `max_length` is used to keep track of the maximum length found so far and is initially set to 0.\n",
-        "\n",
-        "4. We then iterate through the input string `s` using a for loop, where the loop variable `end` represents the current end index of the substring we are considering.\n",
-        "\n",
-        "5. Inside the loop:\n",
-        "   - We check if the character `s[end]` is already in the `char_index` dictionary and if its last occurrence is within or after the current substring. If so, it means that we've encountered a repeating character, and we need to update the `start` of the substring to the next character after the last occurrence of `s[end]`. This ensures that we have a new valid substring without repeating characters.\n",
-        "   - We then update the `char_index` dictionary by storing the current index `end` as the last occurrence index of the character `s[end]`.\n",
-        "\n",
-        "6. Next, we calculate the length of the current substring without repeating characters, which is `current_length = end - start + 1`. We add 1 to account for the fact that the indices are zero-based.\n",
-        "\n",
-        "7. We update the `max_length` with the maximum of its current value and the `current_length`. This step ensures that we keep track of the longest valid substring we have encountered so far.\n",
-        "\n",
-        "8. The loop continues to iterate through the input string, and at the end, we return the `max_length`, which represents the length of the longest substring without repeating characters."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 62,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(length_of_longest_substring(\"abcabcbb\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 63,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(length_of_longest_substring(\"bbbbb\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 64,
-      "id": "cf04eba0-0ead-4e8b-ba33-4316ea64a9d3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(length_of_longest_substring(\"pwwkew\"))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "Let's analyze the time and space complexity of the `length_of_longest_substring` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The function iterates through the input string `s` using a single for loop. The loop runs from the beginning of the string to the end once.\n",
-        "- Inside the loop, we perform constant-time operations such as dictionary lookups and updates, comparisons, and arithmetic operations.\n",
-        "- Therefore, the time complexity of the function is O(n), where n is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The primary data structure that consumes space in this function is the `char_index` dictionary.\n",
-        "- In the worst case, if there are no repeating characters in the input string, the dictionary can store all unique characters in the string.\n",
-        "- Therefore, the space complexity is O(min(n, m)), where n is the length of the input string `s`, and m is the number of unique characters in the string. In the worst case, when all characters are unique, m is equal to n, so the space complexity is O(n).\n",
-        "- Additionally, there are a few integer variables used for indices and lengths, which consume constant space.\n",
-        "- Overall, the space complexity of the function is O(min(n, m)) or simply O(n) in the worst case.\n",
-        "\n",
-        "**In summary,** the time complexity of the `length_of_longest_substring` function is O(n), and the space complexity is O(n) in the worst case due to the `char_index` dictionary. This algorithm provides an efficient way to find the length of the longest substring without repeating characters in linear time."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Longest Substring with K Distinct Characters:** Modify the problem to find the length of the longest substring with exactly K distinct characters. For example, given the input string \"abcabcbb\" and K = 2, the answer would be 4 because the longest substring with two distinct characters is \"abca.\"\n",
-        "2. **Longest Substring with Unique Characters:** Write a function to find the length of the longest substring in a given string where all characters are unique. For example, given the input string \"abcabcbb,\" the answer would be 4 because \"abcd\" is the longest substring with unique characters."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb b/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb
deleted file mode 100644
index 07e96a2..0000000
--- a/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb	
+++ /dev/null
@@ -1,195 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 424. Longest Repeating Character Replacement\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Repeating Character Replacement problem on LeetCode, click here!](https://leetcode.com/problems/longest-repeating-character-replacement/)\n",
-        "\n",
-        "---\n",
-        "You are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.\n",
-        "\n",
-        "*Return the length of the longest substring containing the same letter you can get after performing the above operations.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $10^5$\n",
-        "- `s` consists of only uppercase English letters.\n",
-        "- `0 <= k <= s.length`"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 58,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def characterReplacement(s, k):\n",
-        "    max_length = 0  # Initialize the maximum length\n",
-        "    max_count = 0   # Initialize the maximum count of repeating characters\n",
-        "    start = 0       # Initialize the start of the sliding window\n",
-        "    char_count = {} # Dictionary to store the count of each character\n",
-        "\n",
-        "    for end in range(len(s)):\n",
-        "        # Update the count of the current character in the dictionary\n",
-        "        char_count[s[end]] = char_count.get(s[end], 0) + 1\n",
-        "        \n",
-        "        # Update the maximum count of repeating characters\n",
-        "        max_count = max(max_count, char_count[s[end]])\n",
-        "        \n",
-        "        # Check if the current window size is greater than k\n",
-        "        if (end - start + 1) - max_count > k:\n",
-        "            # Move the start of the window to the right\n",
-        "            char_count[s[start]] -= 1\n",
-        "            start += 1\n",
-        "        \n",
-        "        # Update the maximum length\n",
-        "        max_length = max(max_length, end - start + 1)\n",
-        "\n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We initialize some variables:\n",
-        "   - `max_length` to keep track of the maximum length of the substring containing the same letter.\n",
-        "   - `max_count` to keep track of the maximum count of repeating characters within the current window.\n",
-        "   - `start` to represent the start index of the sliding window.\n",
-        "   - `char_count` is a dictionary that will store the count of each character within the current window.\n",
-        "\n",
-        "2. We use a `for` loop to iterate through the characters of the string `s`.\n",
-        "\n",
-        "3. Inside the loop, we do the following for each character at position `end`:\n",
-        "   - Update the count of the current character in the `char_count` dictionary.\n",
-        "   - Update the `max_count` to be the maximum of the current `max_count` and the count of the current character. This keeps track of the maximum count of repeating characters within the current window.\n",
-        "\n",
-        "4. We check if the current window size (the difference between `end` and `start` plus one) minus the `max_count` is greater than `k`. This condition checks whether we have exceeded the allowed number of replacements (`k`) within the current window.\n",
-        "\n",
-        "   - If we have exceeded the allowed replacements, it means we need to shrink the window from the left side. We do this by moving the `start` of the window to the right and decrementing the count of the character at `s[start]` in the `char_count` dictionary. This effectively removes characters from the left side of the window until we have a valid window again.\n",
-        "\n",
-        "   - By doing this, we ensure that the difference between the current window size and the `max_count` is always less than or equal to `k`.\n",
-        "\n",
-        "5. After each character, we update the `max_length` to be the maximum of the current `max_length` and the size of the current window (`end - start + 1`).\n",
-        "\n",
-        "6. Finally, we return `max_length`, which holds the length of the longest substring containing the same letter with at most `k` replacements."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 59,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "s1 = \"ABAB\"\n",
-        "k1 = 2\n",
-        "print(characterReplacement(s1, k1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 60,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "s2 = \"AABABBA\"\n",
-        "k2 = 1\n",
-        "print(characterReplacement(s2, k2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code uses a single `for` loop to iterate through the characters of the input string `s`. The loop runs from the beginning to the end of the string.\n",
-        "- Inside the loop, we perform constant-time operations such as updating the `char_count` dictionary and updating variables like `max_count` and `max_length`.\n",
-        "- The code's time complexity is primarily determined by the loop, which iterates through each character in the string once. Therefore, the time complexity is O(n), where n is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The code uses several variables to store information, but the space they consume is constant and does not depend on the size of the input string. These variables include `max_length`, `max_count`, `start`, and `char_count`. Therefore, the space complexity is O(1) or constant space.\n",
-        "\n",
-        "**In summary**, the time complexity of the code is O(n), where n is the length of the input string, and the space complexity is O(1), as it uses a constant amount of additional space regardless of the input size. This algorithm efficiently solves the problem with a linear time complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the code to find the longest substring containing the same letter with at most k replacements in O(n) time complexity and O(1) space complexity. Hint: You may need to update the approach to achieve this.\n",
-        "2. **Variation with Lowercase Letters:** Extend the problem to include both uppercase and lowercase English letters in the input string s. Write a function that can handle this extended input and still find the longest substring with at most k replacements efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb b/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb
deleted file mode 100644
index 33c2a44..0000000
--- a/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb	
+++ /dev/null
@@ -1,273 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 76. Minimum Window Substring\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Minimum Window Substring problem on LeetCode, click here!](https://leetcode.com/problems/minimum-window-substring/)\n",
-        "\n",
-        "---\n",
-        "Given two strings `s` and `t` of lengths `m` and `n` respectively, return *the **minimum window substring** of `s` such that every character in `t` (**including duplicates**) is included in the window. If there is no such substring, return the empty string `\"\"`.*\n",
-        "\n",
-        "The testcases will be generated such that the answer is **unique**.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `m == s.length`\n",
-        "- `n == t.length`\n",
-        "- 1 <= `m, n` <= $10^5$\n",
-        "- `s` and `t` consist of uppercase and lowercase English letters.\n",
-        "\n",
-        "**Follow up**: Could you find an algorithm that runs in $O(m + n)$ time?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 54,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def min_window_substring(s, t):\n",
-        "    if not s or not t:\n",
-        "        return \"\"\n",
-        "\n",
-        "    # Initialize dictionaries to keep track of character counts for t and the current window in s.\n",
-        "    t_dict = {}\n",
-        "    current_window_dict = {}\n",
-        "    \n",
-        "    # Populate t_dict with character counts for string t.\n",
-        "    for char in t:\n",
-        "        t_dict[char] = t_dict.get(char, 0) + 1\n",
-        "\n",
-        "    # Initialize pointers for the sliding window.\n",
-        "    left = 0\n",
-        "    min_len = float('inf')\n",
-        "    min_window = \"\"\n",
-        "    required_chars = len(t_dict)\n",
-        "\n",
-        "    # Initialize a variable to keep track of how many required characters have been found in the current window.\n",
-        "    found_chars = 0\n",
-        "\n",
-        "    # Iterate over the string s using the right pointer.\n",
-        "    for right in range(len(s)):\n",
-        "        char = s[right]\n",
-        "\n",
-        "        # Update the current_window_dict.\n",
-        "        current_window_dict[char] = current_window_dict.get(char, 0) + 1\n",
-        "\n",
-        "        # Check if the current character is a required character and if its count matches the required count.\n",
-        "        if char in t_dict and current_window_dict[char] == t_dict[char]:\n",
-        "            found_chars += 1\n",
-        "\n",
-        "        # Try to minimize the window by moving the left pointer.\n",
-        "        while left <= right and found_chars == required_chars:\n",
-        "            # Calculate the current window size.\n",
-        "            window_size = right - left + 1\n",
-        "\n",
-        "            # If the current window is smaller than the minimum found so far, update min_len and min_window.\n",
-        "            if window_size < min_len:\n",
-        "                min_len = window_size\n",
-        "                min_window = s[left:right+1]\n",
-        "\n",
-        "            # Move the left pointer to the right to shrink the window.\n",
-        "            left_char = s[left]\n",
-        "            current_window_dict[left_char] -= 1\n",
-        "\n",
-        "            # Check if the character removed from the window was a required character.\n",
-        "            if left_char in t_dict and current_window_dict[left_char] < t_dict[left_char]:\n",
-        "                found_chars -= 1\n",
-        "\n",
-        "            # Move the left pointer further to continue shrinking the window.\n",
-        "            left += 1\n",
-        "\n",
-        "    return min_window"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by checking if either of the input strings `s` or `t` is empty. If either of them is empty, we return an empty string since there can't be any valid substring in this case.\n",
-        "\n",
-        "2. We initialize two dictionaries: `t_dict` and `current_window_dict`. These dictionaries will be used to keep track of character counts in the string `t` and the current window in string `s`, respectively.\n",
-        "\n",
-        "3. We populate the `t_dict` dictionary by iterating through string `t`. For each character, we increment its count in the dictionary using `t_dict.get(char, 0) + 1`. This allows us to count the occurrences of each character in `t`.\n",
-        "\n",
-        "4. We initialize two pointers: `left` and `right`. The `left` pointer will represent the start of the current window, and the `right` pointer will represent the end of the current window. We also initialize `min_len` to store the length of the minimum window found so far, and `min_window` to store the actual minimum window substring.\n",
-        "\n",
-        "5. We determine the number of required characters in `t` (i.e., the number of distinct characters in `t`) and store it in the variable `required_chars`.\n",
-        "\n",
-        "6. We initialize a variable `found_chars` to keep track of how many required characters have been found in the current window. Initially, it is set to 0.\n",
-        "\n",
-        "7. We iterate over the string `s` using the `right` pointer. In each iteration, we do the following:\n",
-        "\n",
-        "   - Update the `current_window_dict` by incrementing the count of the current character.\n",
-        "   \n",
-        "   - Check if the current character is a required character (present in `t_dict`) and if its count in the `current_window_dict` matches the required count from `t_dict`. If so, we increment `found_chars`.\n",
-        "\n",
-        "8. After updating the window, we attempt to minimize the window size by moving the `left` pointer to the right. In this step, we:\n",
-        "\n",
-        "   - Calculate the size of the current window.\n",
-        "   \n",
-        "   - If the current window size is smaller than the minimum found so far (`min_len`), we update `min_len` and `min_window` to store the current window substring.\n",
-        "   \n",
-        "   - Move the `left` pointer to the right to shrink the window.\n",
-        "   \n",
-        "   - Check if the character being removed from the window was a required character. If it was, we decrement `found_chars`.\n",
-        "   \n",
-        "   - Continue moving the `left` pointer further to continue shrinking the window if the window still contains all required characters.\n",
-        "\n",
-        "9. Finally, we return `min_window`, which will contain the minimum window substring that contains all characters from `t`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 55,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "BANC\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(min_window_substring(\"ADOBECODEBANC\", \"ABC\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 56,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "a\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(min_window_substring(\"a\", \"a\")) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 57,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": []
-        },
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(min_window_substring(\"a\", \"aa\")) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `min_window_substring` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The main loop iterates through the string `s` from left to right using the `right` pointer. This loop runs in O(m) time, where 'm' is the length of string `s`.\n",
-        "\n",
-        "- Inside the loop, we have a while loop that moves the `left` pointer to the right to shrink the window as needed. In the worst case, this while loop can also run in O(m) time because in the worst case, we may have to move the `left` pointer all the way to the end of the string.\n",
-        "\n",
-        "- Within each iteration of the while loop, we perform constant time operations, such as updating dictionaries and comparing character counts.\n",
-        "\n",
-        "- The initialization of the `t_dict` dictionary takes O(n) time, where 'n' is the length of string `t`.\n",
-        "\n",
-        "Therefore, the overall time complexity of the function is O(m + n) because the dominant factor is the length of string `s`, and the length of string `t` has a smaller impact.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity is determined by the space used by the `t_dict` dictionary, the `current_window_dict` dictionary, and a few variables.\n",
-        "\n",
-        "- The `t_dict` dictionary stores character counts for string `t`. In the worst case, when all characters in `t` are distinct, this dictionary can have a maximum of 'n' key-value pairs. So, the space complexity for `t_dict` is O(n).\n",
-        "\n",
-        "- The `current_window_dict` dictionary stores character counts for the current window. In the worst case, it can have a maximum of 'm' key-value pairs. So, the space complexity for `current_window_dict` is also O(m).\n",
-        "\n",
-        "- Other variables used in the function, such as `left`, `right`, `min_len`, `min_window`, `required_chars`, and `found_chars`, are all of constant size and do not depend on the input sizes.\n",
-        "\n",
-        "**In summary**, the overall space complexity of the function is O(max(m, n)), which means it is determined by the larger of the two input strings' lengths. This is because the space used for `t_dict` and `current_window_dict` dominates the space complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Smallest Distinct Substring:** Instead of finding the minimum window substring containing all characters, find the smallest distinct (unique) substring of s that contains all characters from t. This variation adds complexity because you must find a substring with distinct characters.\n",
-        "2. **No Extra Space:** Solve the problem without using any extra space, such as dictionaries or arrays, other than a constant amount of space. This is a significant optimization challenge."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/03. Sliding Window/README.md b/_build/html/_sources/03. Sliding Window/README.md
deleted file mode 100644
index e1294ff..0000000
--- a/_build/html/_sources/03. Sliding Window/README.md	
+++ /dev/null
@@ -1,19 +0,0 @@
-# Sliding Window Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [121. Best Time to Buy And Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | Easy       |
-| [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) | Medium |
-| [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) | Medium |
-| [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)             | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb b/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb
deleted file mode 100644
index c47292f..0000000
--- a/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 20. Valid Parentheses\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Parentheses problem on LeetCode, click here!](https://leetcode.com/problems/valid-parentheses/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given a string s containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.\n",
-        "\n",
-        "An input string is valid if:\n",
-        "\n",
-        "1. Open brackets must be closed by the same type of brackets.\n",
-        "2. Open brackets must be closed in the correct order.\n",
-        "3. Every close bracket has a corresponding open bracket of the same type.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $10^4$\n",
-        "- `s` consists of parentheses only `'()[]{}'`."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 49,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isValid(s):\n",
-        "    # Create an empty stack to store opening brackets\n",
-        "    stack = []\n",
-        "    \n",
-        "    # Define a mapping for matching brackets\n",
-        "    bracket_mapping = {')': '(', '}': '{', ']': '['}\n",
-        "    \n",
-        "    # Iterate through the characters in the input string\n",
-        "    for char in s:\n",
-        "        # If the character is an opening bracket, push it onto the stack\n",
-        "        if char in bracket_mapping.values():\n",
-        "            stack.append(char)\n",
-        "        # If the character is a closing bracket\n",
-        "        elif char in bracket_mapping.keys():\n",
-        "            # Pop the top element from the stack if it exists, or use a dummy value '#'\n",
-        "            top_element = stack.pop() if stack else '#'\n",
-        "            # If the popped element does not match the corresponding opening bracket, return False\n",
-        "            if bracket_mapping[char] != top_element:\n",
-        "                return False\n",
-        "        # If the character is not a bracket, ignore it\n",
-        "        \n",
-        "    # After processing the entire string, if there are any unmatched opening brackets left in the stack, return False\n",
-        "    return len(stack) == 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining the `isValid` function that takes a single argument `s`, which is the input string containing parentheses and brackets.\n",
-        "\n",
-        "2. Inside the function, we create an empty stack, which is a list in Python, to store opening brackets as we encounter them in the input string. The stack will help us keep track of the brackets and their order.\n",
-        "\n",
-        "3. We define a `bracket_mapping` dictionary that maps each closing bracket to its corresponding opening bracket. This mapping will be used to check if a closing bracket matches the most recent opening bracket in the stack.\n",
-        "\n",
-        "4. We iterate through each character `char` in the input string `s` using a `for` loop.\n",
-        "\n",
-        "5. If the character `char` is an opening bracket (i.e., it exists in the `values()` of `bracket_mapping`), we push it onto the stack using the `append()` method.\n",
-        "\n",
-        "6. If the character `char` is a closing bracket (i.e., it exists in the `keys()` of `bracket_mapping`), we need to check if it matches the corresponding opening bracket. To do this, we pop the top element from the stack (if it exists) and store it in `top_element`. We use a dummy value `'#'` if the stack is empty to avoid errors.\n",
-        "\n",
-        "7. We then compare `top_element` with the corresponding opening bracket using `bracket_mapping[char]`. If they do not match, it means the string is not valid, and we return `False`.\n",
-        "\n",
-        "8. If the character `char` is not a bracket, we simply ignore it and continue the loop.\n",
-        "\n",
-        "9. After processing the entire string, we check if there are any unmatched opening brackets left in the stack. If the stack is empty, it means all opening brackets have been properly matched and closed, and we return `True`. Otherwise, we return `False`.\n",
-        "\n",
-        "10. Finally, we provide some test cases at the bottom of the code to demonstrate how the function works for different input strings."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 50,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(isValid(\"()\"))    "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 51,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(isValid(\"()[]{}\")) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 52,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(isValid(\"(]\"))     "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `isValid` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The function iterates through each character in the input string `s` once, performing constant-time operations for each character.\n",
-        "- Pushing and popping elements from the stack (list) also takes constant time in most cases.\n",
-        "- Therefore, the overall time complexity of the function is $O(n)$, where $n$ is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the function is determined by the space used by the stack and the `bracket_mapping` dictionary.\n",
-        "- In the worst case, if the input string `s` consists entirely of opening brackets, the stack can potentially contain all of these opening brackets, resulting in a space complexity of $O(n)$ in terms of the stack.\n",
-        "- The `bracket_mapping` dictionary has a constant number of key-value pairs (3 pairs in this case).\n",
-        "- Therefore, the overall space complexity of the function is $O(n)$, where $n$ is the length of the input string `s`, mainly due to the stack space.\n",
-        "\n",
-        "**In summary:**\n",
-        "- The time complexity is $O(n)$ because we iterate through the string once.\n",
-        "- The space complexity is $O(n)$ due to the stack used to keep track of opening brackets."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Valid Expressions:** Modify the problem to validate not only brackets but also arithmetic expressions containing parentheses, such as \"$2 * (3 + 5) / (4 - 2)$\".\n",
-        "2. **Valid Parentheses Combinations:**\n",
-        "Write a function to generate all valid combinations of n pairs of parentheses, where $n$ is a positive integer. For example, for $n = 3$, the valid combinations are [\"((()))\", \"(()())\", \"(())()\", \"()(())\", \"()()()\"]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/04. Stack/README.md b/_build/html/_sources/04. Stack/README.md
deleted file mode 100644
index 2ed115b..0000000
--- a/_build/html/_sources/04. Stack/README.md	
+++ /dev/null
@@ -1,16 +0,0 @@
-# Stack Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)                            | Easy       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb b/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb
deleted file mode 100644
index 98f2331..0000000
--- a/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,228 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 153. Find Minimum in Rotated Sorted Array\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Find Minimum in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:\n",
-        "\n",
-        "- `[4,5,6,7,0,1,2]` if it was rotated `4` times.\n",
-        "- `[0,1,2,4,5,6,7]` if it was rotated `7` times.\n",
-        "\n",
-        "Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.\n",
-        "\n",
-        "Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array.*\n",
-        "\n",
-        "You must write an algorithm that runs in $O(log\\ n)$ time.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `n == nums.length`\n",
-        "- `1 <= n <= 5000`\n",
-        "- `-5000 <= nums[i] <= 5000`\n",
-        "- All the integers of `nums` are **unique**.\n",
-        "- `nums` is sorted and rotated between `1` and `n` times."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 39,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def findMin(nums):\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "    \n",
-        "    while left < right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "        \n",
-        "        # If the mid element is greater than the rightmost element,\n",
-        "        # it means the minimum element is in the right half.\n",
-        "        if nums[mid] > nums[right]:\n",
-        "            left = mid + 1\n",
-        "        # If the mid element is less than or equal to the rightmost element,\n",
-        "        # it means the minimum element is in the left half or at the mid itself.\n",
-        "        else:\n",
-        "            right = mid\n",
-        "    \n",
-        "    # The loop will break when left and right converge to the minimum element.\n",
-        "    # At this point, left (or right) will be pointing to the minimum element.\n",
-        "    return nums[left]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The function `findMin(nums)` takes an input list `nums`, which represents the rotated sorted array.\n",
-        "\n",
-        "2. Initialize two pointers `left` and `right` to track the current search range. Initially, `left` is set to 0 (the beginning of the array), and `right` is set to `len(nums) - 1` (the end of the array).\n",
-        "\n",
-        "3. Enter a `while` loop with the condition `left < right`. This loop continues until `left` and `right` converge to a single element, which will be the minimum element in the rotated array.\n",
-        "\n",
-        "4. Inside the loop, calculate the middle index `mid` using integer division. This helps in finding the middle element of the current search range.\n",
-        "\n",
-        "5. Check if the element at the middle index (`nums[mid]`) is greater than the element at the rightmost index (`nums[right]`). If this condition is true, it means that the minimum element must be in the right half of the current search range. So, update `left` to `mid + 1`, effectively eliminating the left half of the search range.\n",
-        "\n",
-        "6. If the condition from step 5 is not met, it implies that the minimum element can be in the left half of the current search range or could be the element at the `mid` index itself. In this case, update `right` to `mid`, effectively eliminating the right half of the search range.\n",
-        "\n",
-        "7. Repeat steps 4-6 until `left` and `right` converge to the minimum element.\n",
-        "\n",
-        "8. When the loop exits, it means that `left` (or `right`) points to the minimum element in the array.\n",
-        "\n",
-        "9. Return `nums[left]` (or `nums[right]`) as the result, which is the minimum element in the rotated sorted array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 43,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1:\n",
-        "nums1 = [3, 4, 5, 1, 2]\n",
-        "print(findMin(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 41,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "0\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "print(findMin(nums2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 42,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "11\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "nums3 = [11, 13, 15, 17]\n",
-        "print(findMin(nums3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this code is $O(log\\ n)$, where '$n$' is the length of the input array `nums`. This is because the binary search algorithm is employed, which continually divides the search range in half with each iteration.\n",
-        "\n",
-        "In each iteration of the `while` loop, the search range is halved, and this process continues until the `left` and `right` pointers converge to the minimum element. In the worst case, it will take logarithmic time to reduce the search range to a single element.\n",
-        "\n",
-        "Therefore, the binary search used in this code runs in $O(log\\ n)$ time complexity.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of this code is $O(1)$, which means it uses a constant amount of additional space that does not depend on the size of the input array.\n",
-        "\n",
-        "The algorithm uses a fixed number of variables (`left`, `right`, `mid`) to keep track of the search range and indices, but the number of these variables does not grow with the size of the input array. Hence, the space complexity is constant.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: $O(log\\ n)$\n",
-        "- Space Complexity: $O(1)$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. Solve the problem with a variation where the input array may contain duplicates.\n",
-        "\n",
-        "2. Modify the problem to return the index of the minimum element in the rotated sorted array, rather than the element itself. The algorithm should still run in $O(log\\ n)$ time.\n",
-        "\n",
-        "3. Implement a function that finds the maximum element in a rotated sorted array. How would you adapt the binary search algorithm to solve this problem efficiently in $O(log\\ n)$ time?\n",
-        "\n",
-        "4. Implement a function that can find the kth smallest element in a rotated sorted array. This is an extension of the original problem, and the algorithm should still run in $O(log\\ n)$ time."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb b/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb
deleted file mode 100644
index d8d4202..0000000
--- a/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 33. Search in Rotated Sorted Array\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Search in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/search-in-rotated-sorted-array/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "There is an integer array `nums` sorted in ascending order (with **distinct** values).\n",
-        "\n",
-        "Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k (1 <= k < nums.length)` such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index 3 and become `[4,5,6,7,0,1,2]`.\n",
-        "\n",
-        "Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.*\n",
-        "\n",
-        "You must write an algorithm with $O(log\\ n)$ runtime complexity.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `1 <= nums.length <= 5000`\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- All values of `nums` are **unique**.\n",
-        "- `nums` is an ascending array that is possibly rotated.\n",
-        "- $-10^4$ <= `target` <= $10^4$"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 44,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def search(nums, target):\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "\n",
-        "    # Find the pivot index using binary search\n",
-        "    while left < right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "\n",
-        "        if nums[mid] > nums[right]:\n",
-        "            left = mid + 1\n",
-        "        else:\n",
-        "            right = mid\n",
-        "\n",
-        "    pivot = left\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "\n",
-        "    # Determine which part of the array to search in\n",
-        "    if target >= nums[pivot] and target <= nums[right]:\n",
-        "        left = pivot\n",
-        "    else:\n",
-        "        right = pivot\n",
-        "\n",
-        "    # Perform binary search to find the target\n",
-        "    while left <= right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "\n",
-        "        if nums[mid] == target:\n",
-        "            return mid\n",
-        "        elif nums[mid] < target:\n",
-        "            left = mid + 1\n",
-        "        else:\n",
-        "            right = mid - 1\n",
-        "\n",
-        "    return -1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `search` function takes two arguments: `nums`, which is the rotated sorted array, and `target`, which is the value we want to find in the array.\n",
-        "\n",
-        "2. We initialize two pointers, `left` and `right`, which represent the range in which we are going to perform binary search. Initially, `left` is set to 0, and `right` is set to the index of the last element in the array (`len(nums) - 1`).\n",
-        "\n",
-        "3. We start by finding the pivot index using binary search. The pivot index is the index at which the array is rotated. We use a `while` loop to continue the search until `left` is less than `right`.\n",
-        "\n",
-        "4. Inside the loop, we calculate the middle index `mid` using integer division. We then compare the value at `mid` with the value at `right`. If `nums[mid]` is greater than `nums[right]`, it means the pivot point lies to the right of `mid`, so we update `left` to `mid + 1`. Otherwise, the pivot point lies to the left of `mid`, so we update `right` to `mid`. This process continues until we find the pivot index.\n",
-        "\n",
-        "5. After finding the pivot index, we have divided the array into two parts: one part is sorted in ascending order, and the other part is also sorted in ascending order but rotated. We need to determine which part of the array contains the target value.\n",
-        "\n",
-        "6. We reset `left` and `right` pointers. If the target value is within the range `[nums[pivot], nums[right]]`, we set `left` to `pivot` (the start of the rotated part), indicating that we should search in the rotated part. Otherwise, we set `right` to `pivot` (the end of the sorted part), indicating that we should search in the sorted part.\n",
-        "\n",
-        "7. We then perform binary search again within the chosen range (`left` to `right`) to find the target element. The binary search continues until `left` is less than or equal to `right`.\n",
-        "\n",
-        "8. Inside the binary search loop, we calculate the middle index `mid` and compare `nums[mid]` with the target value. If they are equal, we return `mid` as the index of the target element. If `nums[mid]` is less than the target, we update `left` to `mid + 1` to search in the right half. If `nums[mid]` is greater than the target, we update `right` to `mid - 1` to search in the left half.\n",
-        "\n",
-        "9. If we exit the binary search loop without finding the target, we return -1 to indicate that the target is not present in the array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 45,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "nums1 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "target1 = 0\n",
-        "result1 = search(nums1, target1)\n",
-        "print(result1) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 46,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "-1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "target2 = 3\n",
-        "result2 = search(nums2, target2)\n",
-        "print(result2) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 47,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "-1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "nums3 = [1]\n",
-        "target3 = 0\n",
-        "result3 = search(nums3, target3)\n",
-        "print(result3) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. Finding the pivot index using binary search takes $O(log\\ n)$ time, where '$n$' is the number of elements in the array.\n",
-        "2. After finding the pivot index, performing binary search to find the target element also takes $O(log\\ n)$ time in the worst case.\n",
-        "\n",
-        "The dominant factor in the time complexity is the binary search, and since we perform two binary searches sequentially, the overall time complexity of the code is $O(log\\ n)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is very minimal:\n",
-        "\n",
-        "1. We use a constant amount of additional space for variables such as `left`, `right`, `pivot`, and `mid`. These variables do not depend on the input size, so they contribute $O(1)$ space complexity.\n",
-        "2. The function does not use any additional data structures that scale with the input size.\n",
-        "\n",
-        "Therefore, the space complexity of the code is $O(1)$, which means it has a constant space complexity and does not depend on the size of the input array.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: $O(log\\ n)$\n",
-        "- Space Complexity: $O(1)$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Find Minimum Element in Rotated Sorted Array**: Write an algorithm to find the minimum element in a rotated sorted array. This is a variation of the problem where you don't need to search for a target value but instead find the smallest element.\n",
-        "\n",
-        "2. **Search in a Circularly Sorted Array**: Consider an array that is sorted in a circular manner (e.g., [4, 5, 6, 7, 0, 1, 2, 3]). Adapt the search algorithm to work for circularly sorted arrays while maintaining $O(log\\ n)$ complexity."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/05. Binary Search/README.md b/_build/html/_sources/05. Binary Search/README.md
deleted file mode 100644
index 6086b9b..0000000
--- a/_build/html/_sources/05. Binary Search/README.md	
+++ /dev/null
@@ -1,17 +0,0 @@
-# Binary search Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [153. Find Minimum In Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) | Medium |
-| [33. Search In Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)   | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb b/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb
deleted file mode 100644
index 0336bc8..0000000
--- a/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb	
+++ /dev/null
@@ -1,271 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 141. Linked List Cycle\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Linked List Cycle problem on LeetCode, click here!](https://leetcode.com/problems/linked-list-cycle/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given `head`, the head of a linked list, determine if the linked list has a cycle in it.\n",
-        "\n",
-        "There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is connected to. **Note that `pos` is not passed as a parameter.**\n",
-        "\n",
-        "Return `true` if there is a cycle in the linked list. Otherwise, return `false`.\n",
-        "\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of the nodes in the list is in the range $[0, 10^4]$.\n",
-        "- $-10^5$ <= Node.val <= $10^5$\n",
-        "- `pos` is `-1` or a **valid index** in the linked-list.\n",
-        "\n",
-        "**Follow up:** Can you solve it using $O(1)$ (i.e. constant) memory?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "    def __init__(self, val=0):\n",
-        "        self.val = val\n",
-        "        self.next = None\n",
-        "\n",
-        "class Solution:\n",
-        "    def hasCycle(self, head: ListNode) -> bool:\n",
-        "        if not head:\n",
-        "            return False\n",
-        "        \n",
-        "        slow = head\n",
-        "        fast = head\n",
-        "        \n",
-        "        while fast and fast.next:\n",
-        "            slow = slow.next\n",
-        "            fast = fast.next.next\n",
-        "            \n",
-        "            if slow == fast:\n",
-        "                return True\n",
-        "        \n",
-        "        return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The given code defines a class `Solution` with a method `hasCycle` for determining if a linked list has a cycle.\n",
-        "\n",
-        "The `ListNode` class represents nodes in the linked list, where each node has a `val` (a value associated with the node) and a `next` attribute (a reference to the next node in the list).\n",
-        "\n",
-        "The `hasCycle` method takes the `head` of a linked list as input and returns `True` if there is a cycle in the linked list, and `False` otherwise. It uses two pointers, `slow` and `fast`, to traverse the linked list. If there is a cycle, the `fast` pointer will eventually catch up to the `slow` pointer.\n",
-        "\n",
-        "The algorithm works as follows:\n",
-        "1. It checks if the input `head` is `None` (an empty list). If it is, the method returns `False` because an empty list can't have a cycle.\n",
-        "2. Two pointers, `slow` and `fast`, initially point to the `head` of the list.\n",
-        "3. The code enters a `while` loop where `fast` moves two steps at a time, and `slow` moves one step at a time.\n",
-        "4. If there is a cycle, `fast` will eventually catch up to `slow`, and the method returns `True`.\n",
-        "5. If the loop completes without finding a cycle, the method returns `False`."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list with a cycle\n",
-        "def create_linked_list_with_cycle(values, pos):\n",
-        "    dummy = ListNode()\n",
-        "    current = dummy\n",
-        "    cycle_node = None\n",
-        "    \n",
-        "    for i, val in enumerate(values):\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "        \n",
-        "        if i == pos:\n",
-        "            cycle_node = current\n",
-        "    \n",
-        "    if cycle_node:\n",
-        "        current.next = cycle_node\n",
-        "    \n",
-        "    return dummy.next"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "This helper function is designed to create a linked list with a cycle for testing purposes. It takes two arguments:\n",
-        "\n",
-        "1. `values`: A list of values representing the nodes in the linked list.\n",
-        "2. `pos`: An integer representing the index of the node where the cycle begins.\n",
-        "\n",
-        "Here's how the function works:\n",
-        "\n",
-        "- It starts by creating an empty `ListNode` called `dummy`. This `dummy` node is used to simplify the creation of the linked list.\n",
-        "- It initializes a `current` pointer to the `dummy` node. This pointer will be used to traverse the linked list.\n",
-        "- It initializes a `cycle_node` variable to `None`. This variable will hold the reference to the node where the cycle begins, if any.\n",
-        "- It then iterates through the `values` list, creating a new `ListNode` for each value and appending it to the linked list.\n",
-        "- Inside the loop, it checks if the current index `i` is equal to the specified `pos`. If they match, it sets `cycle_node` to the current node. This simulates the creation of a cycle in the linked list.\n",
-        "- After the loop, if `cycle_node` is not `None` (indicating a cycle should be created), it connects the last node in the linked list to `cycle_node`, effectively creating a cycle.\n",
-        "- Finally, it returns the reference to the first node of the linked list (not the `dummy` node).\n",
-        "\n",
-        "This helper function allows you to easily create test cases where you can specify the values for the linked list and the position where the cycle starts. It's particularly useful for testing the code's ability to detect cycles in linked lists."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "head = [3,2,0,-4]\n",
-        "pos = 1\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "head = [1,2]\n",
-        "pos = 0\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 22,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "head = [1]\n",
-        "pos = -1\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's discuss the time and space complexity of the code for detecting a cycle in a linked list:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this code is $O(n)$, where \"$n$\" is the number of nodes in the linked list.\n",
-        "\n",
-        "The reason for this is that both the `slow` and `fast` pointers traverse the linked list, and they move at different speeds. In the worst case, when there is no cycle, the `fast` pointer will reach the end of the list after going through approximately $n/2$ nodes. In the case of a cycle, it may take some extra iterations for the `fast` pointer to catch up to the `slow` pointer. However, the total number of iterations is still proportional to the number of nodes in the linked list, making it $O(n)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this code is $O(1)$, which means it uses constant space.\n",
-        "\n",
-        "The reason for this is that regardless of the size of the linked list, the code only uses two additional pointers (`slow` and `fast`) to traverse the list. These pointers do not depend on the size of the input linked list, so the space complexity remains constant.\n",
-        "\n",
-        "**In summary**, this code efficiently detects cycles in a linked list with a time complexity of $O(n)$ and a space complexity of $O(1)$. It is an example of an algorithm that solves a complex problem with minimal memory usage."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/143. Reorder List.ipynb b/_build/html/_sources/06. Linked List/143. Reorder List.ipynb
deleted file mode 100644
index a83545b..0000000
--- a/_build/html/_sources/06. Linked List/143. Reorder List.ipynb	
+++ /dev/null
@@ -1,285 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 143. Reorder List\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Reorder List problem on LeetCode, click here!](https://leetcode.com/problems/reorder-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given the head of a singly linked-list. The list can be represented as:\n",
-        "\n",
-        "$L_0 → L_1 → … → L_{n - 1} → L_n$\n",
-        "\n",
-        "Reorder the list to be on the following form:\n",
-        "\n",
-        "$L_0 → L_n → L_1 → L_{n - 1} → L_2 → L_{n - 2} → …$\n",
-        "\n",
-        "You may not modify the values in the list's nodes. Only nodes themselves may be changed.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is in the range $[1, 5 * 10^4]$.\n",
-        "- `1 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 28,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def reorderList(head):\n",
-        "    if not head or not head.next:\n",
-        "        return head\n",
-        "\n",
-        "    # Step 1: Find the middle of the linked list\n",
-        "    slow, fast = head, head\n",
-        "    while fast.next and fast.next.next:\n",
-        "        slow = slow.next\n",
-        "        fast = fast.next.next\n",
-        "\n",
-        "    # Split the list into two halves\n",
-        "    first_half = head\n",
-        "    second_half = slow.next\n",
-        "    slow.next = None\n",
-        "\n",
-        "    # Step 2: Reverse the second half of the linked list\n",
-        "    prev = None\n",
-        "    current = second_half\n",
-        "    while current:\n",
-        "        next_node = current.next\n",
-        "        current.next = prev\n",
-        "        prev = current\n",
-        "        current = next_node\n",
-        "    second_half = prev\n",
-        "\n",
-        "    # Step 3: Merge the first and reversed second halves alternately\n",
-        "    p1, p2 = first_half, second_half\n",
-        "    while p2:\n",
-        "        next_p1 = p1.next\n",
-        "        next_p2 = p2.next\n",
-        "        p1.next = p2\n",
-        "        p2.next = next_p1\n",
-        "        p1 = next_p1\n",
-        "        p2 = next_p2\n",
-        "\n",
-        "    return head"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `reorderList` function is designed to reorder a singly linked list following the specific pattern described in the problem statement. Here's a detailed explanation of how the function works:\n",
-        "\n",
-        "1. **Base Cases Handling**:\n",
-        "   - The function first checks if the input linked list is empty (`not head`) or contains only one element (`not head.next`). If either of these conditions is met, the list cannot be reordered, so the function returns the original list as-is.\n",
-        "\n",
-        "2. **Finding the Middle of the Linked List**:\n",
-        "   - To reorder the list, we first need to find the middle point so that we can split the list into two halves. This is done using two pointers, `slow` and `fast`, initialized to the head of the list.\n",
-        "   - `slow` moves one step at a time while `fast` moves two steps at a time. When `fast` reaches the end of the list or the second-to-last node, `slow` will be at the middle node.\n",
-        "\n",
-        "3. **Splitting the List**:\n",
-        "   - After finding the middle node, we split the list into two halves:\n",
-        "     - The first half, `first_half`, contains nodes from the beginning up to the middle.\n",
-        "     - The second half, `second_half`, contains nodes from the middle to the end.\n",
-        "   - To split the list, we set the `next` pointer of the node before the middle node to `None`.\n",
-        "\n",
-        "4. **Reversing the Second Half**:\n",
-        "   - We reverse the second half of the linked list using the `prev`, `current`, and `next_node` pointers.\n",
-        "   - `prev` is initially `None`, and we iterate through the second half. For each node, we:\n",
-        "     - Set the `next` of the current node to `prev`, effectively reversing the next pointer direction.\n",
-        "     - Update `prev` to the current node.\n",
-        "     - Move to the next node using the `next_node`.\n",
-        "\n",
-        "5. **Merging the Two Halves Alternately**:\n",
-        "   - We now have two linked lists: `first_half` and the reversed `second_half`.\n",
-        "   - We merge these two lists alternately by adjusting the `next` pointers of the nodes.\n",
-        "   - `p1` and `p2` are pointers to the current nodes in `first_half` and `second_half`, respectively.\n",
-        "   - We iterate through both lists while reordering nodes:\n",
-        "     - Set the `next` of `p1` to `p2` to link a node from the first half to a node from the reversed second half.\n",
-        "     - Update `p1` and `p2` to their respective next nodes.\n",
-        "     - Repeat this process until we have processed all nodes in both halves.\n",
-        "\n",
-        "6. **Returning the Reordered List**:\n",
-        "   - After the merging process is complete, the linked list is reordered as specified.\n",
-        "   - The function returns the `head` of the reordered list.\n",
-        "\n",
-        "The overall result is a singly linked list that has been reordered according to the given pattern. The time complexity of this algorithm is O(N), where N is the number of nodes in the linked list, as we traverse the list once to find the middle, once to reverse the second half, and once to merge the two halves."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 29,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to print the linked list\n",
-        "def printLinkedList(head):\n",
-        "    result = []\n",
-        "    while head:\n",
-        "        result.append(head.val)\n",
-        "        head = head.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the helper functions:\n",
-        "**printLinkedList Function**:\n",
-        "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-        "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-        "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-        "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-        "\n",
-        "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 30,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 4, 2, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "head1 = ListNode(1, ListNode(2, ListNode(3, ListNode(4))))\n",
-        "reorderList(head1)\n",
-        "print(printLinkedList(head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 5, 2, 4, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "head2 = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))\n",
-        "reorderList(head2)\n",
-        "print(printLinkedList(head2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `reorderList` function:\n",
-        "\n",
-        "**Time Complexity**:\n",
-        "1. **Finding the Middle of the Linked List**: In this step, we use two pointers, one moving one step at a time (`slow`) and the other moving two steps at a time (`fast`). This process takes $O(N/2)$ time, where $N$ is the number of nodes in the linked list.\n",
-        "\n",
-        "2. **Splitting the List**: After finding the middle, we split the list into two halves by setting the `next` pointer of the node before the middle node to `None`. This step takes $O(1)$ time.\n",
-        "\n",
-        "3. **Reversing the Second Half**: We reverse the second half of the linked list using a while loop that iterates through the second half of the list once. Therefore, this step also takes $O(N/2)$ time.\n",
-        "\n",
-        "4. **Merging the Two Halves Alternately**: In this step, we merge the two halves alternately by adjusting the `next` pointers of the nodes. We iterate through both halves once, so this step takes $O(N/2)$ time.\n",
-        "\n",
-        "Overall, the time complexity of the `reorderList` function is dominated by the steps involving reversing and merging the two halves, both of which take $O(N/2)$ time. Therefore, the total time complexity is $O(N)$.\n",
-        "\n",
-        "**Space Complexity**:\n",
-        "The space complexity of the `reorderList` function is primarily determined by the variables and data structures used within the function. Let's break down the space complexity components:\n",
-        "\n",
-        "1. **Constant Space Variables**: The variables `slow`, `fast`, `prev`, `current`, `next_node`, `p1`, and `p2` are used to traverse and manipulate the linked list. These variables occupy constant space, regardless of the input size. Therefore, they contribute $O(1)$ to the space complexity.\n",
-        "\n",
-        "2. **Split Linked Lists**: In the splitting step, we create two new linked list segments (`first_half` and `second_half`) that store references to nodes from the original linked list. These segments occupy space proportional to half of the input list, $O(N/2)$. \n",
-        "\n",
-        "3. **Reversed Second Half**: During the reversal step, we reverse the second half of the linked list in-place without creating any additional data structures. Therefore, it doesn't contribute to additional space complexity.\n",
-        "\n",
-        "4. **Overall**: Combining the above components, the total space complexity is $O(N/2)$, which simplifies to $O(N)$ in terms of space complexity.\n",
-        "\n",
-        "**In summary**, the `reorderList` function has a time complexity of $O(N)$ and a space complexity of $O(N)$ due to the creation of two new linked list segments while splitting the list. The constant space variables used for traversal do not significantly impact the overall space complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Reorder by K Elements:** Generalize the reorderList function to reorder the list in a different pattern, where you reorder the list by $K$ elements at a time. For example, for $K=3$, you would reorder it as ($L_0 → L_1 → L_2 → L_n → L_{n-1} → L_{n-2} → ...$).\n",
-        "You may need to handle cases where the number of nodes is not a multiple of $K$.\n",
-        "\n",
-        "2. **Reorder by Odd and Even Nodes:** Modify the reorderList function to reorder the list in a pattern where odd-indexed nodes (1-based index) come before even-indexed nodes. For example, ($L_0 → L_n → L_1 → L_{n-1} → L_2 → L_{n-2} → ...$)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb b/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb
deleted file mode 100644
index 263124e..0000000
--- a/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb	
+++ /dev/null
@@ -1,278 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 19. Remove Nth Node From End of List\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Remove Nth Node From End of List problem on LeetCode, click here!](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the head of a linked list, remove the $n^{th}$ node from the end of the list and return its head.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is `sz`.\n",
-        "- `1 <= sz <= 30`\n",
-        "- `0 <= Node.val <= 100`\n",
-        "- `1 <= n <= sz`\n",
-        "\n",
-        "**Follow up:** Could you do this in one pass?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 23,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def removeNthFromEnd(head, n):\n",
-        "    # Create a dummy node to handle the case of removing the head node.\n",
-        "    dummy = ListNode(0)\n",
-        "    dummy.next = head\n",
-        "    first = dummy\n",
-        "    second = dummy\n",
-        "    \n",
-        "    # Advance the first pointer by n+1 nodes.\n",
-        "    for _ in range(n + 1):\n",
-        "        first = first.next\n",
-        "    \n",
-        "    # Move both pointers simultaneously until first reaches the end.\n",
-        "    while first is not None:\n",
-        "        first = first.next\n",
-        "        second = second.next\n",
-        "    \n",
-        "    # Remove the nth node by updating the next pointer of the previous node.\n",
-        "    second.next = second.next.next\n",
-        "    \n",
-        "    return dummy.next  # Return the modified head of the linked list"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a `ListNode` class to represent individual nodes in a linked list. Each node has two attributes: `val` (the value of the node) and `next` (a reference to the next node in the list).\n",
-        "\n",
-        "2. The `removeNthFromEnd` function takes the head of a linked list (`head`) and the value `n`, which represents the position from the end of the list of the node to be removed. It returns the modified head of the linked list after the removal.\n",
-        "\n",
-        "3. We create a dummy node (`dummy`) at the beginning of the list. This dummy node simplifies the code by handling cases where the head of the list needs to be removed.\n",
-        "\n",
-        "4. We initialize two pointers, `first` and `second`, both initially pointing to the dummy node.\n",
-        "\n",
-        "5. To advance the `first` pointer by `n+1` nodes, we use a `for` loop. This positions the `first` pointer `n` nodes ahead of the `second` pointer.\n",
-        "\n",
-        "6. We then move both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This ensures that the `second` pointer ends up pointing to the node that needs to be removed.\n",
-        "\n",
-        "7. To remove the nth node from the end, we update the `next` pointer of the node pointed to by the `second` pointer to skip over the node to be removed.\n",
-        "\n",
-        "8. Finally, we return `dummy.next`, which is the modified head of the linked list without the removed node.\n",
-        "\n",
-        "9. Additionally, there are two helper functions:\n",
-        "   - `createLinkedList(values)` creates a linked list from a list of values.\n",
-        "   - `convertToList(head)` converts a linked list back into a list of values."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list from a list of values.\n",
-        "def createLinkedList(values):\n",
-        "    if not values:\n",
-        "        return None\n",
-        "    head = ListNode(values[0])\n",
-        "    current = head\n",
-        "    for val in values[1:]:\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "    return head\n",
-        "\n",
-        "# Helper function to convert a linked list to a list of values.\n",
-        "def convertToList(head):\n",
-        "    result = []\n",
-        "    current = head\n",
-        "    while current:\n",
-        "        result.append(current.val)\n",
-        "        current = current.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the two helper functions:\n",
-        "\n",
-        "1. `createLinkedList(values)`: This function creates a linked list from a list of values.\n",
-        "\n",
-        "   - Input: `values` is a list containing values that you want to insert into the linked list.\n",
-        "   - Output: The function returns the head of the created linked list.\n",
-        "\n",
-        "   Explanation:\n",
-        "   - The function starts by checking if the `values` list is empty. If it is, it returns `None` to indicate an empty linked list.\n",
-        "   - If the `values` list is not empty, it creates the first node of the linked list with the value from the first element of `values`.\n",
-        "   - It then iterates through the remaining elements of `values` and creates new nodes for each value, connecting them together using the `next` attribute to form a linked list.\n",
-        "   - Finally, it returns the head of the linked list, which is the first node created.\n",
-        "   \n",
-        "\n",
-        "2. `convertToList(head)`: This function converts a linked list into a list of values.\n",
-        "\n",
-        "   - Input: `head` is the head node of the linked list that you want to convert.\n",
-        "   - Output: The function returns a list containing the values of the nodes in the linked list in order.\n",
-        "\n",
-        "   Explanation:\n",
-        "   - The function starts by initializing an empty list, `result`, to store the values of the linked list nodes.\n",
-        "   - It then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - The iteration continues until it reaches the end of the linked list (i.e., the `next` attribute of the current node becomes `None`).\n",
-        "   - After the iteration is complete, the function returns the `result` list, which now contains the values of all the nodes in the linked list in the same order as they appear in the linked list.\n",
-        "\n",
-        "These helper functions are useful for creating linked lists from lists of values and converting linked lists back into lists of values, making it easier to work with linked list examples and test cases in a more familiar list format."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 2, 3, 5]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "head1 = createLinkedList([1, 2, 3, 4, 5])\n",
-        "n1 = 2\n",
-        "new_head1 = removeNthFromEnd(head1, n1)\n",
-        "print(convertToList(new_head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 26,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "head2 = createLinkedList([1])\n",
-        "n2 = 1\n",
-        "new_head2 = removeNthFromEnd(head2, n2)\n",
-        "print(convertToList(new_head2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 27,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "head3 = createLinkedList([1, 2])\n",
-        "n3 = 1\n",
-        "new_head3 = removeNthFromEnd(head3, n3)\n",
-        "print(convertToList(new_head3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Certainly! Let's analyze the time and space complexity of the `removeNthFromEnd` function in isolation:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The `removeNthFromEnd` function performs two passes through the linked list. In the worst case, it needs to traverse the entire linked list of length $N$.\n",
-        "- The first pass positions the `first` pointer $n$ nodes ahead of the `second` pointer. This pass takes $O(N)$ time because it involves iterating through all N nodes.\n",
-        "- The second pass moves both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This also takes $O(N)$ time in the worst case.\n",
-        "- Therefore, the overall time complexity of the `removeNthFromEnd` function is $O(N)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The `removeNthFromEnd` function uses a constant amount of extra space for its variables (e.g., `first`, `second`, and `dummy`). This space usage does not depend on the size of the linked list.\n",
-        "- The space complexity is $O(1)$, indicating that the space used by the function is constant and independent of the size of the input linked list.\n",
-        "\n",
-        "**In summary**, the `removeNthFromEnd` function has a time complexity of $O(N)$ and a space complexity of $O(1)$. It efficiently removes the nth node from the end of the linked list with a single pass through the list and constant extra space usage."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb b/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb
deleted file mode 100644
index c97de2d..0000000
--- a/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb	
+++ /dev/null
@@ -1,316 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 206. Reverse Linked List\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Reverse Linked List problem on LeetCode, click here!](https://leetcode.com/problems/reverse-linked-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the `head` of a singly linked list, reverse the list, and return the reversed list.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is the range `[0, 5000]`.\n",
-        "- `-5000 <= Node.val <= 5000`\n",
-        "\n",
-        "**Follow up:** A linked list can be reversed either iteratively or recursively. Could you implement both?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 33,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "# Function to reverse a linked list iteratively\n",
-        "def reverseListIterative(head):\n",
-        "    prev = None\n",
-        "    current = head\n",
-        "    \n",
-        "    while current:\n",
-        "        next_node = current.next\n",
-        "        current.next = prev\n",
-        "        prev = current\n",
-        "        current = next_node\n",
-        "    \n",
-        "    return prev\n",
-        "\n",
-        "# Function to reverse a linked list recursively\n",
-        "def reverseListRecursive(head):\n",
-        "    if not head or not head.next:\n",
-        "        return head\n",
-        "\n",
-        "    new_head = reverseListRecursive(head.next)\n",
-        "    head.next.next = head\n",
-        "    head.next = None\n",
-        "\n",
-        "    return new_head"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `reverseListIterative(head)`: This function reverses a linked list iteratively using a loop. Here's a step-by-step explanation:\n",
-        "\n",
-        "   - It takes the `head` of the input linked list as a parameter.\n",
-        "   - It initializes two pointers, `prev` and `current`, initially set to `None` and the `head` of the list, respectively.\n",
-        "   - It enters a `while` loop that continues until `current` reaches the end of the list (i.e., becomes `None`).\n",
-        "   - Inside the loop:\n",
-        "     - It stores the `next_node` which is the next node after `current`.\n",
-        "     - It updates the `next` pointer of `current` to point to the `prev` node. This effectively reverses the link direction.\n",
-        "     - It moves `prev` to `current` and `current` to `next_node`, advancing the pointers one step further in the list.\n",
-        "   - Once the loop completes, `prev` will be pointing to the new head of the reversed list (which was the last node of the original list), so it returns `prev`.\n",
-        "\n",
-        "2. `reverseListRecursive(head)`: This function reverses a linked list recursively. Here's how it works:\n",
-        "\n",
-        "   - It takes the `head` of the input linked list as a parameter.\n",
-        "   - The base case is checked: if `head` is `None` or `head.next` is `None`, meaning the list is empty or has only one node, it returns `head` as there's no need to reverse a list with zero or one element.\n",
-        "   - In the recursive case:\n",
-        "     - It calls itself with `head.next`, effectively moving down the list until it reaches the end.\n",
-        "     - Once it reaches the end, it starts reversing the links:\n",
-        "       - `head.next.next` is set to `head`, reversing the link between `head` and the next node.\n",
-        "       - `head.next` is set to `None` to avoid cycles.\n",
-        "   - Finally, it returns the new head of the reversed list, which will be the last node of the original list.\n",
-        "\n",
-        "**In summary,** `reverseListIterative` reverses the linked list by iterating through it and changing the next pointers, while `reverseListRecursive` reverses the linked list by recursively reaching the end and then reversing the links on the way back up the recursion stack. Both functions achieve the same result: reversing the linked list."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 36,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list from a list of values\n",
-        "def createLinkedList(values):\n",
-        "    if not values:\n",
-        "        return None\n",
-        "    head = ListNode(values[0])\n",
-        "    current = head\n",
-        "    for val in values[1:]:\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "    return head\n",
-        "\n",
-        "# Helper function to convert a linked list to a list for testing\n",
-        "def linkedListToList(head):\n",
-        "    result = []\n",
-        "    current = head\n",
-        "    while current:\n",
-        "        result.append(current.val)\n",
-        "        current = current.next\n",
-        "    return result\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the two helper functions used in the provided code:\n",
-        "\n",
-        "1. `createLinkedList(values)`:\n",
-        "\n",
-        "   - Purpose: This function is used to create a linked list from a list of values. It takes a list of values as input and returns the head of the linked list.\n",
-        "\n",
-        "   - How it works:\n",
-        "     - It first checks if the input list `values` is empty. If it's empty, it returns `None` to indicate an empty linked list.\n",
-        "     - If `values` is not empty, it initializes a `head` node with the value of the first element in `values`.\n",
-        "     - It then iterates through the remaining values in `values`, creating new nodes for each value and linking them together to form a linked list.\n",
-        "     - Finally, it returns the `head` node of the newly created linked list.\n",
-        "\n",
-        "2. `linkedListToList(head)`:\n",
-        "\n",
-        "   - Purpose: This function is used to convert a linked list back into a Python list for testing and output purposes.\n",
-        "\n",
-        "   - How it works:\n",
-        "     - It takes the `head` of the linked list as input.\n",
-        "     - It initializes an empty list called `result`.\n",
-        "     - It then iterates through the linked list, starting from `head`, and appends each node's `val` (value) to the `result` list.\n",
-        "     - This process continues until the end of the linked list is reached.\n",
-        "     - Finally, it returns the `result` list, which contains the values from the linked list in the same order.\n",
-        "\n",
-        "These helper functions are used to simplify the process of creating linked lists from lists of values and converting linked lists back into lists, making it easier to work with linked lists in the provided examples and test cases."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 35,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[1, 1, 2, 3, 4, 4]"
-            ]
-          },
-          "execution_count": 35,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 1: Iterative\n",
-        "head1 = [1,2,3,4,5]\n",
-        "head1 = createLinkedList(head1)\n",
-        "reversed_head1 = reverseListIterative(head1)\n",
-        "print(linkedListToList(reversed_head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[]"
-            ]
-          },
-          "execution_count": 37,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 2: Recursive\n",
-        "head2 = [1,2]\n",
-        "head2 = createLinkedList(head2)\n",
-        "reversed_head2 = reverseListRecursive(head2)\n",
-        "print(linkedListToList(reversed_head2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 38,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[0]"
-            ]
-          },
-          "execution_count": 38,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 3: Empty list\n",
-        "head3 = []\n",
-        "head3 = createLinkedList(head3)\n",
-        "reversed_head3 = reverseListIterative(head3)\n",
-        "print(linkedListToList(reversed_head3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the two functions:\n",
-        "\n",
-        "1. `reverseListIterative(head)`:\n",
-        "\n",
-        "   - Time Complexity: $O(n)$\n",
-        "     - The iterative function visits each node in the linked list exactly once in a single pass, where \"$n$\" is the number of nodes in the list. Therefore, the time complexity is linear in the number of nodes.\n",
-        "\n",
-        "   - Space Complexity: $O(1)$\n",
-        "     - This function uses a constant amount of extra space for the `prev`, `current`, and `next_node` pointers. Regardless of the size of the input list, the amount of additional memory used remains the same, so the space complexity is constant.\n",
-        "\n",
-        "2. `reverseListRecursive(head)`:\n",
-        "\n",
-        "   - Time Complexity: $O(n)$\n",
-        "     - The recursive function also visits each node in the linked list exactly once. It recursively traverses the list from the head to the tail, reversing the links on the way back. Therefore, like the iterative approach, the time complexity is $O(n)$, where \"$n$\" is the number of nodes.\n",
-        "\n",
-        "   - Space Complexity: $O(n)$\n",
-        "     - The recursive function uses space on the call stack for each recursive call. In the worst case, when the list has \"$n$\" nodes, it will create \"$n$\" recursive function calls on the stack, resulting in a space complexity of $O(n)$. This is because each function call stores information about its state, including the `head` pointer, until it reaches the base case and starts returning.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "- The time complexity of both functions is $O(n)$ because they both visit each node once.\n",
-        "- The space complexity of `reverseListIterative` is $O(1)$ because it uses a constant amount of extra space.\n",
-        "- The space complexity of `reverseListRecursive` is $O(n)$ due to the recursive function calls and the associated call stack space.\n",
-        "\n",
-        "Both functions are efficient in terms of time complexity, but `reverseListIterative` is more memory-efficient because it uses a constant amount of additional space, while `reverseListRecursive` uses space on the call stack proportional to the number of nodes in the list."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K-Group Reverse:** Modify the reverseListIterative function to reverse the linked list in groups of $K$ elements. For example, if the input linked list is [1, 2, 3, 4, 5, 6], and $K$ is 3, the output should be [3, 2, 1, 6, 5, 4].\n",
-        "\n",
-        "2. **Rotate Linked List:** Write a function to rotate a linked list to the right by $K$ places. For example, if the input is [1, 2, 3, 4, 5] and $K$ is 2, the output should be [4, 5, 1, 2, 3]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb b/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb
deleted file mode 100644
index 876d633..0000000
--- a/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 21. Merge Two Sorted Lists\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Merge Two Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-two-sorted-lists/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given the heads of two sorted linked lists `list1` and `list2`.\n",
-        "\n",
-        "Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.\n",
-        "\n",
-        "*Return the head of the merged linked list.*\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in both lists is in the range `[0, 50]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "- Both `list1` and `list2` are sorted in non-decreasing order."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def mergeTwoLists(list1, list2):\n",
-        "    # Create a dummy node to simplify the code.\n",
-        "    dummy = ListNode()\n",
-        "    current = dummy\n",
-        "\n",
-        "    while list1 and list2:\n",
-        "        if list1.val < list2.val:\n",
-        "            current.next = list1\n",
-        "            list1 = list1.next\n",
-        "        else:\n",
-        "            current.next = list2\n",
-        "            list2 = list2.next\n",
-        "        current = current.next\n",
-        "\n",
-        "    # Append the remaining elements from either list.\n",
-        "    if list1:\n",
-        "        current.next = list1\n",
-        "    else:\n",
-        "        current.next = list2\n",
-        "\n",
-        "    return dummy.next"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a class `ListNode` to represent the nodes of a singly-linked list. Each node has a `val` attribute to store the value of the node and a `next` attribute to point to the next node in the list.\n",
-        "\n",
-        "2. We define the `mergeTwoLists` function that takes two linked list heads, `list1` and `list2`, as input.\n",
-        "\n",
-        "3. We create a `dummy` node at the beginning of the merged list. The `dummy` node simplifies the code by serving as a placeholder, and its `next` attribute will point to the actual merged list.\n",
-        "\n",
-        "4. We initialize a `current` pointer to point to the `dummy` node initially. This `current` pointer helps us traverse and build the merged list.\n",
-        "\n",
-        "5. We enter a `while` loop that continues until either `list1` or `list2` becomes empty. Inside the loop, we compare the values of the nodes at the current positions of `list1` and `list2`.\n",
-        "\n",
-        "6. If the `val` of the node in `list1` is smaller than the `val` of the node in `list2`, we attach the node from `list1` to the merged list by updating the `next` attribute of the `current` node to point to the node in `list1`. We then move the `list1` pointer to the next node.\n",
-        "\n",
-        "7. If the `val` of the node in `list2` is smaller than or equal to the `val` of the node in `list1`, we attach the node from `list2` to the merged list in a similar manner. We move the `list2` pointer to the next node.\n",
-        "\n",
-        "8. After attaching a node to the merged list, we move the `current` pointer to the newly added node. This step is essential for keeping track of the end of the merged list.\n",
-        "\n",
-        "9. The loop continues until either `list1` or `list2` becomes empty.\n",
-        "\n",
-        "10. Once the loop exits, we check if there are any remaining elements in `list1` or `list2`. If `list1` is not empty, we attach the remaining elements of `list1` to the merged list. If `list2` is not empty, we attach the remaining elements of `list2`.\n",
-        "\n",
-        "11. Finally, we return the `next` attribute of the `dummy` node as the head of the merged list, which represents the sorted merged list."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to print the linked list\n",
-        "def printLinkedList(head):\n",
-        "    result = []\n",
-        "    while head:\n",
-        "        result.append(head.val)\n",
-        "        head = head.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the helper functions:\n",
-        "**printLinkedList Function**:\n",
-        "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-        "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-        "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-        "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-        "\n",
-        "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 1:\n",
-        "# Input: list1 = [1,2,4], list2 = [1,3,4]\n",
-        "list1 = ListNode(1, ListNode(2, ListNode(4)))\n",
-        "list2 = ListNode(1, ListNode(3, ListNode(4)))\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 2:\n",
-        "# Input: list1 = [], list2 = []\n",
-        "list1 = None\n",
-        "list2 = None\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 3:\n",
-        "# list1 = [], list2 = [0]\n",
-        "list1 = None\n",
-        "list2 = ListNode(0)\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `mergeTwoLists` function.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this function is $O(N + M$), where $N$ and $M$ are the lengths of `list1` and `list2`, respectively. Here's why:\n",
-        "\n",
-        "1. In the worst case, we need to traverse both `list1` and `list2` completely. This involves iterating through all the nodes in both lists once.\n",
-        "\n",
-        "2. The while loop runs until either `list1` or `list2` becomes empty. The number of iterations depends on the total number of nodes in both lists, which is N + M in the worst case.\n",
-        "\n",
-        "3. Inside the loop, we perform constant-time operations for each iteration, such as comparisons and updating pointers.\n",
-        "\n",
-        "Therefore, the dominant factor in the time complexity is the combined length of both input lists, resulting in a linear time complexity of $O(N + M)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this function is $O(1)$, which means it uses a constant amount of extra space regardless of the input sizes. Here's why:\n",
-        "\n",
-        "1. We create a few extra pointers (`dummy`, `current`) and temporary variables (`val`) to manage the merging process. These consume a fixed amount of memory, regardless of the input sizes. The number of these extra variables is independent of the lengths of `list1` and `list2`.\n",
-        "\n",
-        "2. We do not create a new data structure or allocate memory for the merged list. Instead, we rearrange the existing nodes from `list1` and `list2` to form the merged list. This operation does not consume additional memory proportional to the input sizes.\n",
-        "\n",
-        "**In summary**, the `mergeTwoLists` function has a time complexity of $O(N + M)$ and a space complexity of $O(1)$, making it an efficient solution for merging two sorted linked lists."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Merge K Sorted Lists:** Extend the problem to merge $K$ sorted linked lists instead of just two. You need to efficiently merge $K$ lists into one sorted list.\n",
-        "\n",
-        "2. **Merge Lists in a Zigzag Pattern:** Merge two sorted lists in a zigzag pattern. For example, given [1, 2, 4] and [1, 3, 5], the merged list should be [1, 1, 2, 3, 4, 5]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": "sql",
-      "file_extension": "",
-      "mimetype": "",
-      "name": "sql",
-      "version": "3.32.3"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb b/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb
deleted file mode 100644
index a850734..0000000
--- a/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb	
+++ /dev/null
@@ -1,285 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 23. Merge k Sorted Lists\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Merge k Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-k-sorted-lists/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given an array of $k$ linked-lists lists, each linked-list is sorted in ascending order.\n",
-        "\n",
-        "Merge all the linked-lists into one sorted linked-list and return it.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `k == lists.length`\n",
-        "- 0 <= k <= $10^4$\n",
-        "- `0 <= lists[i].length <= 500`\n",
-        "- $-10^4$ <= lists[i][j] <= $10^4$\n",
-        "- `lists[i]` is sorted in ascending order.\n",
-        "- The sum of `lists[i].length` will not exceed $10^4$.\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "     def __init__(self, val=0, next=None):\n",
-        "         self.val = val\n",
-        "         self.next = next\n",
-        "\n",
-        "class Solution:\n",
-        "    def mergeKLists(self, lists: [[ListNode]]) -> ListNode:\n",
-        "        # Check if the input list of linked lists is empty\n",
-        "        if not lists or len(lists) == 0:\n",
-        "            return None\n",
-        "\n",
-        "        # Loop until there is only one merged list left\n",
-        "        while len(lists) > 1:\n",
-        "            mergedLists = []\n",
-        "\n",
-        "            # Merge pairs of lists\n",
-        "            for i in range(0, len(lists), 2):\n",
-        "                l1 = lists[i]\n",
-        "                l2 = lists[i + 1] if (i + 1) < len(lists) else None\n",
-        "                mergedLists.append(self.mergeList(l1, l2))\n",
-        "\n",
-        "            # Update the list of lists with merged results\n",
-        "            lists = mergedLists\n",
-        "        \n",
-        "        # Return the final merged list\n",
-        "        return lists[0]\n",
-        "\n",
-        "    def mergeList(self, l1, l2):\n",
-        "        # Create a dummy node to simplify merging\n",
-        "        dummy = ListNode()\n",
-        "        tail = dummy\n",
-        "\n",
-        "        # Merge the two sorted lists\n",
-        "        while l1 and l2:\n",
-        "            if l1.val < l2.val:\n",
-        "                tail.next = l1\n",
-        "                l1 = l1.next\n",
-        "            else:\n",
-        "                tail.next = l2\n",
-        "                l2 = l2.next\n",
-        "            tail = tail.next\n",
-        "\n",
-        "        # Append any remaining elements from l1 or l2 (if any)\n",
-        "        if l1:\n",
-        "            tail.next = l1\n",
-        "        if l2:\n",
-        "            tail.next = l2\n",
-        "\n",
-        "        # Return the merged result starting from the next of the dummy node\n",
-        "        return dummy.next\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided code defines a Python class `Solution` with two methods for merging k sorted linked lists:\n",
-        "\n",
-        "1. `mergeKLists(self, lists: List[ListNode]) -> ListNode`: This method takes a list of k sorted linked lists as input and returns a single merged sorted linked list. It uses a divide-and-conquer approach to repeatedly merge pairs of lists until only one merged list remains.\n",
-        "\n",
-        "2. `mergeList(self, l1, l2)`: This method takes two sorted linked lists, `l1` and `l2`, as input and merges them into a single sorted linked list. It uses a dummy node to simplify the merging process.\n",
-        "\n",
-        "Here's a high-level overview of how the code works:\n",
-        "\n",
-        "- The `mergeKLists` method checks if the input list of linked lists is empty or contains no lists. If there are no lists, it returns `None`.\n",
-        "\n",
-        "- Inside a `while` loop, the code repeatedly merges pairs of linked lists until only one merged list remains in the `lists` array. It does this by iterating through the input lists in pairs and calling the `mergeList` method to merge each pair.\n",
-        "\n",
-        "- The `mergeList` method takes two sorted linked lists, `l1` and `l2`, and merges them into a single sorted linked list. It uses a dummy node (`dummy`) and a `tail` pointer to keep track of the merged list while comparing and merging elements from `l1` and `l2`.\n",
-        "\n",
-        "- After merging all pairs of lists and updating the `lists` array with the merged results, the loop continues until only one merged list remains in the `lists` array.\n",
-        "\n",
-        "- Finally, the `mergeKLists` method returns the merged list.\n",
-        "\n",
-        "Overall, this code efficiently merges k sorted linked lists using a divide-and-conquer strategy, resulting in a single merged sorted linked list as the output."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> "
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: lists = [[1,4,5],[1,3,4],[2,6]]\n",
-        "\n",
-        "lists1 = [\n",
-        "    ListNode(1, ListNode(4, ListNode(5))),\n",
-        "    ListNode(1, ListNode(3, ListNode(4))),\n",
-        "    ListNode(2, ListNode(6))\n",
-        "]\n",
-        "solution = Solution()\n",
-        "result1 = solution.mergeKLists(lists1)\n",
-        "\n",
-        "# Print the result\n",
-        "if result1:\n",
-        "    current = result1\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: lists = []\n",
-        "\n",
-        "lists2 = []\n",
-        "solution = Solution()\n",
-        "result2 = solution.mergeKLists(lists2)\n",
-        "\n",
-        "# Print the result\n",
-        "if result2:\n",
-        "    current = result2\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "# Input: lists = [[]]\n",
-        "\n",
-        "lists3 = [None]\n",
-        "solution = Solution()\n",
-        "result3 = solution.mergeKLists(lists3)\n",
-        "\n",
-        "# Print the result\n",
-        "if result3:\n",
-        "    current = result3\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. **Heap Initialization:** The code does not use a heap. Instead, it uses a divide-and-conquer approach. The initial check for empty input lists takes $O(1)$ time.\n",
-        "\n",
-        "2. **Merging:** The merging operation is performed in a divide-and-conquer fashion. In each iteration of the `while` loop, we merge pairs of linked lists, and the number of comparisons made is proportional to the total number of nodes across all the linked lists ($n$). In each merge step, we effectively process each node once. The number of iterations required to reduce k lists to 1 is $O(log\\ k)$.\n",
-        "\n",
-        "   Therefore, the overall time complexity of the code is $O(n * log\\ k)$, where n is the total number of nodes across all lists, and $k$ is the number of linked lists.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. **Heap Space:** The code doesn't use a heap data structure, so there's no additional space complexity due to a heap.\n",
-        "\n",
-        "2. **Merged Lists:** In the `mergeList` method, we create a new merged list. However, this list is not stored in memory for all lists; it's replaced with each merged pair. The space used for these merged lists is proportional to the size of the largest merged list, which is $O(n)$ in the worst case.\n",
-        "\n",
-        "3. **Additional Variables:** The code uses a few additional variables, such as `dummy` and `tail`, but these occupy a constant amount of space and don't depend on the input size.\n",
-        "\n",
-        "   Therefore, the overall space complexity of the code is $O(n)$, where n is the total number of nodes across all lists.\n",
-        "\n",
-        "In summary, the code's time complexity is $O(n * log(k))$, and its space complexity is $O(n)$. This code efficiently merges $k$ sorted linked lists using a divide-and-conquer approach with a relatively low space overhead."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/README.md b/_build/html/_sources/06. Linked List/README.md
deleted file mode 100644
index fb7b3ee..0000000
--- a/_build/html/_sources/06. Linked List/README.md	
+++ /dev/null
@@ -1,21 +0,0 @@
-# Linked List Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)                       | Easy       |
-| [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)                   | Easy       |
-| [143. Reorder List](https://leetcode.com/problems/reorder-list/)                                     | Medium     |
-| [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | Medium |
-| [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)                           | Medium     |
-| [23. Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)                       | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/07. Trees/100. Same Tree.ipynb b/_build/html/_sources/07. Trees/100. Same Tree.ipynb
deleted file mode 100644
index a270fab..0000000
--- a/_build/html/_sources/07. Trees/100. Same Tree.ipynb	
+++ /dev/null
@@ -1,211 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 100: Same Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-        "\n",
-        "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-        "\n",
-        "**Constraints**\n",
-        "\n",
-        "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-        "2. $-10^4$ <= Node.val <= $10^4$\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def isSameTree(p, q):\n",
-        "    # Base case: If both p and q are None, the trees are the same.\n",
-        "    if not p and not q:\n",
-        "        return True\n",
-        "    \n",
-        "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-        "    if not p or not q:\n",
-        "        return False\n",
-        "    \n",
-        "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-        "    if p.val != q.val:\n",
-        "        return False\n",
-        "    \n",
-        "    # Recursively check the left and right subtrees of p and q.\n",
-        "    # If both subtrees are the same, the entire trees are the same.\n",
-        "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-        "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-        "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-        "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-        "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-        "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "\n",
-        "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-        "\n",
-        "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "print(isSameTree(p1, q1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "\n",
-        "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-        "\n",
-        "p2 = TreeNode(1, TreeNode(2), None)\n",
-        "q2 = TreeNode(1, None, TreeNode(2))\n",
-        "print(isSameTree(p2, q2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "\n",
-        "#Input: p = [1,2,1], q = [1,1,2]\n",
-        "\n",
-        "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-        "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-        "print(isSameTree(p3, q3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time Complexity\n",
-        "\n",
-        "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-        "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-        "\n",
-        "## Space Complexity\n",
-        "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-        "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-        "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-        "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-        "\n",
-        "## In summary:\n",
-        "\n",
-        "+ **Time Complexity:** $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-        "+ **Space Complexity:** $O(n)$ due to the recursive call stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb b/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb
deleted file mode 100644
index 8cb08b7..0000000
--- a/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb	
+++ /dev/null
@@ -1,271 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 102. Binary Tree Level Order Traversal\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Level Order Traversal problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-level-order-traversal/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, return *the level order traversal of its nodes' values*. (i.e., from left to right, level by level).\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-        "- `-1000 <= Node.val <= 1000`\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def levelOrder(root):\n",
-        "    # Check if the tree is empty, if so, return an empty list.\n",
-        "    if not root:\n",
-        "        return []\n",
-        "\n",
-        "    # Initialize an empty list to store the result.\n",
-        "    result = []\n",
-        "    \n",
-        "    # Initialize a queue with the root node to perform BFS.\n",
-        "    queue = [root]\n",
-        "\n",
-        "    while queue:\n",
-        "        # Initialize an empty list to store the values of nodes at the current level.\n",
-        "        level_values = []\n",
-        "        \n",
-        "        # Get the number of nodes at the current level.\n",
-        "        level_size = len(queue)\n",
-        "\n",
-        "        # Iterate through the nodes at the current level.\n",
-        "        for _ in range(level_size):\n",
-        "            # Dequeue the front node from the queue.\n",
-        "            node = queue.pop(0)\n",
-        "            \n",
-        "            # Append the value of the current node to the level_values list.\n",
-        "            level_values.append(node.val)\n",
-        "\n",
-        "            # Enqueue the left and right children of the current node if they exist.\n",
-        "            if node.left:\n",
-        "                queue.append(node.left)\n",
-        "            if node.right:\n",
-        "                queue.append(node.right)\n",
-        "\n",
-        "        # Append the level_values list (values at the current level) to the result list.\n",
-        "        result.append(level_values)\n",
-        "\n",
-        "    # Return the final result, which is a list of lists representing level order traversal.\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a `TreeNode` class that represents a node in the binary tree. Each `TreeNode` object has a value (`val`) and two child nodes: `left` and `right`.\n",
-        "\n",
-        "2. The `levelOrder` function takes the root of the binary tree as its input and returns the level order traversal of the tree as a list of lists.\n",
-        "\n",
-        "3. We start by checking if the input `root` is `None`, which indicates an empty tree. If the tree is empty, we return an empty list because there are no nodes to traverse.\n",
-        "\n",
-        "4. We initialize an empty list called `result` to store the final result, which will be a list of lists containing node values at each level.\n",
-        "\n",
-        "5. We initialize a queue called `queue` with the root node. This queue will be used for breadth-first traversal of the tree.\n",
-        "\n",
-        "6. We enter a while loop that continues until the `queue` is empty. Inside the loop, we perform the following steps:\n",
-        "\n",
-        "   - We initialize an empty list called `level_values` to store the values of nodes at the current level.\n",
-        "\n",
-        "   - We determine the number of nodes at the current level by getting the length of the `queue`. This is done to process nodes level by level.\n",
-        "\n",
-        "   - We iterate through the nodes at the current level using a for loop. For each node in the current level:\n",
-        "\n",
-        "     - We dequeue (remove) the front node from the `queue`.\n",
-        "\n",
-        "     - We append the value of the dequeued node to the `level_values` list, effectively collecting the values of nodes at the current level.\n",
-        "\n",
-        "     - If the dequeued node has a left child, we enqueue the left child to the `queue`.\n",
-        "\n",
-        "     - If the dequeued node has a right child, we enqueue the right child to the `queue`.\n",
-        "\n",
-        "   - After processing all nodes at the current level, we append the `level_values` list to the `result` list. This represents the values at the current level.\n",
-        "\n",
-        "7. The loop continues until all levels have been traversed, and the `queue` becomes empty.\n",
-        "\n",
-        "8. Finally, we return the `result` list, which contains lists of node values at each level, representing the level order traversal of the binary tree.\n",
-        "\n",
-        "The code effectively performs a breadth-first traversal of the binary tree, processing nodes level by level, and constructs the result list that represents the level order traversal."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[3], [9, 20], [15, 7]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(9)\n",
-        "root.right = TreeNode(20)\n",
-        "root.right.left = TreeNode(15)\n",
-        "root.right.right = TreeNode(7)\n",
-        "\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Using the same tree as before\n",
-        "root = TreeNode(1)\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root = None\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `levelOrder` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this function is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the breadth-first traversal.\n",
-        "\n",
-        "In the worst case, we have to enqueue and dequeue all nodes in the binary tree, which is proportional to the number of nodes (N). In each level, we process all nodes in that level, and since there are a total of log(N) levels in a balanced binary tree, the time complexity can also be approximated as O(N) for unbalanced trees.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this function is O(N), where N is the number of nodes in the binary tree. Here's how the space complexity breaks down:\n",
-        "\n",
-        "1. The `result` list stores the level order traversal, and in the worst case, it contains N/2 levels (for a completely unbalanced binary tree). So, the space used by `result` is O(N).\n",
-        "\n",
-        "2. The `queue` data structure is used for BFS traversal. In the worst case, it can store all nodes at the last level of the tree. In a balanced binary tree, the maximum number of nodes at any level is 2^(log(N)), which is still O(N). In the case of an unbalanced tree, it can be even worse. So, the space used by `queue` is O(N).\n",
-        "\n",
-        "Overall, the dominant factor in terms of space complexity is the `queue`, and the space complexity is O(N).\n",
-        "\n",
-        "In summary, the function's time complexity is O(N), and its space complexity is also O(N). It is an efficient and optimal solution for performing a level order traversal of a binary tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Reverse Level Order Traversal:** Modify the `levelOrder` function to return the level order traversal in reverse order (from bottom to top). For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[15, 7], [9, 20], [3]]`.\n",
-        "\n",
-        "2. **Zigzag Level Order Traversal:** Write a function that performs a level order traversal of a binary tree in a zigzag pattern. In a zigzag traversal, the nodes at even levels are traversed from left to right, and nodes at odd levels are traversed from right to left. For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[3], [20, 9], [15, 7]]`."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb b/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb
deleted file mode 100644
index 213989d..0000000
--- a/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 104. Maximum Depth of Binary Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the `root` of a binary tree, return its maximum depth.\n",
-        "\n",
-        "A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 19,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def maxDepth(root):\n",
-        "    # Base case: If the root is None, the depth is 0.\n",
-        "    if root is None:\n",
-        "        return 0\n",
-        "    \n",
-        "    # Recursively calculate the maximum depth of the left and right subtrees.\n",
-        "    left_depth = maxDepth(root.left)\n",
-        "    right_depth = maxDepth(root.right)\n",
-        "    \n",
-        "    # Return the maximum depth of the tree by adding 1 to the maximum depth of the subtrees.\n",
-        "    return max(left_depth, right_depth) + 1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's go through the code step by step to understand how it works:\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class to represent the nodes of the binary tree. Each node has three attributes:\n",
-        "   - `val`: the value stored in the node.\n",
-        "   - `left`: a reference to the left child node.\n",
-        "   - `right`: a reference to the right child node.\n",
-        "\n",
-        "2. The `maxDepth` function is the main function that calculates the maximum depth of the binary tree. It takes a single argument, `root`, which is the root node of the binary tree.\n",
-        "\n",
-        "3. In the `maxDepth` function, we have a base case:\n",
-        "   - If the `root` is `None`, it means we have reached the end of a branch of the tree (a leaf node or an empty subtree). In this case, the depth is 0 because there are no nodes to count.\n",
-        "\n",
-        "4. If the `root` is not `None`, we recursively calculate the maximum depth of the left and right subtrees:\n",
-        "   - We call `maxDepth` on the `root.left` to calculate the maximum depth of the left subtree and store it in the variable `left_depth`.\n",
-        "   - We call `maxDepth` on the `root.right` to calculate the maximum depth of the right subtree and store it in the variable `right_depth`.\n",
-        "\n",
-        "5. Finally, we return the maximum depth of the tree by taking the maximum of `left_depth` and `right_depth` and adding 1 to it. This is because we are counting the current level as well."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: root = [3,9,20,null,null,15,7]\n",
-        "\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(9)\n",
-        "root.right = TreeNode(20)\n",
-        "root.right.left = TreeNode(15)\n",
-        "root.right.right = TreeNode(7)\n",
-        "\n",
-        "print(maxDepth(root))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a230be4f-1f47-4919-87e3-62fc3f072d97",
-      "metadata": {},
-      "source": [
-        "In this example, we create a binary tree based on the given input `[3,9,20,null,null,15,7]` and calculate its maximum depth, which is `3`. The tree structure is as follows:\n",
-        "   ```\n",
-        "       3\n",
-        "      / \\\n",
-        "     9  20\n",
-        "       /  \\\n",
-        "      15   7\n",
-        "   ```\n",
-        "\n",
-        "So, the maximum depth is the length of the longest path from the root to a leaf node, which is 3 in this case."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: root = [1,null,2]\n",
-        "\n",
-        "root = TreeNode(1)\n",
-        "root.right = TreeNode(2)\n",
-        "\n",
-        "# Calculate the maximum depth of the tree and print the result\n",
-        "print(maxDepth(root))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The time complexity of the `maxDepth` function is $O(N)$, where $N$ is the number of nodes in the binary tree.\n",
-        "- This is because in the worst case, the function visits every node exactly once in a depth-first manner.\n",
-        "- The recursion explores all nodes of the tree, so the time complexity is linear with respect to the number of nodes.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the `maxDepth` function is $O(H)$, where H is the height of the binary tree.\n",
-        "- In the worst case, if the binary tree is completely unbalanced (skewed), the recursion stack can go as deep as the height of the tree.\n",
-        "- In the best case, if the binary tree is perfectly balanced, the height is $O(log\\ N)$, where $N$ is the number of nodes.\n",
-        "- Therefore, the space complexity can vary from $O(log\\ N)$ to $O(N)$ depending on the shape of the tree.\n",
-        "- In addition to the recursion stack, there is a small constant amount of space used for variables and function call overhead.\n",
-        "\n",
-        "**In summary:**\n",
-        "- The time complexity of the code is $O(N)$ as it visits each node once.\n",
-        "- The space complexity is $O(H)$, where H is the height of the tree, which can vary from $O(log\\ N)$ to $O(N)$ depending on the tree's shape."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb b/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb
deleted file mode 100644
index 31c9e69..0000000
--- a/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb	
+++ /dev/null
@@ -1,313 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 105. Construct Binary Tree from Preorder and Inorder Traversal\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Construct Binary Tree from Preorder and Inorder Traversal problem on LeetCode, click here!](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)\n",
-        "\n",
-        "---\n",
-        "Given two integer arrays `preorder` and `inorder` where `preorder` is the preorder traversal of a binary tree and `inorder` is the inorder traversal of the same tree, construct and return *the binary tree.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= preorder.length <= 3000`\n",
-        "- `inorder.length == preorder.length`\n",
-        "- `-3000 <= preorder[i], inorder[i] <= 3000`\n",
-        "- `preorder` and `inorder` consist of **unique** values.\n",
-        "- Each value of `inorder` also appears in `preorder`.\n",
-        "- `preorder` is **guaranteed** to be the preorder traversal of the tree.\n",
-        "- `inorder` is **guaranteed** to be the inorder traversal of the tree.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def buildTree(preorder, inorder):\n",
-        "    if not preorder:\n",
-        "        return None\n",
-        "\n",
-        "    # The first element in the preorder traversal is the root of the current subtree\n",
-        "    root_val = preorder[0]\n",
-        "    root = TreeNode(root_val)\n",
-        "\n",
-        "    # Find the index of the root value in the inorder traversal\n",
-        "    root_idx_inorder = inorder.index(root_val)\n",
-        "\n",
-        "    # Recursively build left and right subtrees\n",
-        "    root.left = buildTree(preorder[1:1 + root_idx_inorder], inorder[:root_idx_inorder])\n",
-        "    root.right = buildTree(preorder[1 + root_idx_inorder:], inorder[root_idx_inorder + 1:])\n",
-        "\n",
-        "    return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `class TreeNode:`: This is a class definition for a binary tree node. It has three attributes:\n",
-        "   - `val`: The value of the node.\n",
-        "   - `left`: A reference to the left child node.\n",
-        "   - `right`: A reference to the right child node.\n",
-        "\n",
-        "   This class is used to create instances of binary tree nodes, which will be used to build the binary tree.\n",
-        "\n",
-        "2. `def buildTree(preorder, inorder):`: This is the main function that constructs a binary tree from its preorder and inorder traversals.\n",
-        "\n",
-        "   - `preorder`: A list representing the preorder traversal of the binary tree.\n",
-        "   - `inorder`: A list representing the inorder traversal of the binary tree.\n",
-        "\n",
-        "3. `if not preorder: return None`: This line checks if the `preorder` list is empty. If it is, it means there are no nodes to construct, so the function returns `None`.\n",
-        "\n",
-        "4. `root_val = preorder[0]`: The value of the root node is extracted from the first element of the `preorder` list.\n",
-        "\n",
-        "5. `root = TreeNode(root_val)`: A new `TreeNode` object is created with the `root_val` as its value. This represents the root of the current subtree.\n",
-        "\n",
-        "6. `root_idx_inorder = inorder.index(root_val)`: The index of the `root_val` in the `inorder` list is found. This index indicates the position of the root node in the inorder traversal.\n",
-        "\n",
-        "7. `root.left`: The left subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The left subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists up to the `root_idx_inorder`.\n",
-        "\n",
-        "8. `root.right`: The right subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The right subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists starting from `root_idx_inorder + 1`.\n",
-        "\n",
-        "9. Finally, the function returns the `root` of the subtree it just constructed.\n",
-        "\n",
-        "The recursive approach used here divides the problem of constructing the entire binary tree into smaller subproblems, starting with the root node and then recursively building the left and right subtrees until the entire tree is constructed."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "2d82857b-4fbe-474f-bf33-17b55c20147b",
-      "metadata": {},
-      "source": [
-        "## Helper Function \n",
-        "\n",
-        "Here is a helper function to represent the binary tree elements."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "13e6f3e1-659d-479d-9521-75629b8c782f",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def treeToList(root):\n",
-        "    # Initialize an empty list to store the elements of the binary tree\n",
-        "    result = []\n",
-        "    \n",
-        "    # Initialize a queue for level-order traversal starting with the root\n",
-        "    queue = [root]\n",
-        "\n",
-        "    while queue:\n",
-        "        # Pop the front node from the queue\n",
-        "        node = queue.pop(0)\n",
-        "        \n",
-        "        # If the node is not None (i.e., a valid node), add its value to the result list\n",
-        "        if node:\n",
-        "            result.append(node.val)\n",
-        "            \n",
-        "            # Add the left and right children of the node to the queue\n",
-        "            queue.append(node.left)\n",
-        "            queue.append(node.right)\n",
-        "        else:\n",
-        "            # If the node is None, add None to the result to represent an empty node\n",
-        "            result.append(None)\n",
-        "\n",
-        "    # Remove any trailing None values from the result list\n",
-        "    while result and result[-1] is None:\n",
-        "        result.pop()\n",
-        "\n",
-        "    # Return the resulting list representing the binary tree elements\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "86cd4c9d-a120-4dd1-a1bd-530436012574",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's break down the `treeToList` helper function step by step:\n",
-        "\n",
-        "1. `def treeToList(root):`: This function takes the root of a binary tree as input and converts it into a list.\n",
-        "\n",
-        "2. `result = []`: Initialize an empty list called `result` to store the elements of the binary tree in list format.\n",
-        "\n",
-        "3. `queue = [root]`: Initialize a queue data structure for a level-order traversal, starting with the root node.\n",
-        "\n",
-        "4. `while queue:`: This starts a loop that continues until the queue is empty, indicating that all nodes have been processed.\n",
-        "\n",
-        "5. `node = queue.pop(0)`: Dequeue the first node from the queue, effectively processing it.\n",
-        "\n",
-        "6. `if node:`: Check if the `node` is not None, which means it's a valid node in the binary tree.\n",
-        "\n",
-        "    a. `result.append(node.val)`: If the node is valid, append its value (`node.val`) to the `result` list. This represents the value of the current node in the binary tree.\n",
-        "\n",
-        "    b. `queue.append(node.left)`: Enqueue the left child of the current node if it exists. This adds the left child to the queue for processing in the next iteration.\n",
-        "\n",
-        "    c. `queue.append(node.right)`: Enqueue the right child of the current node if it exists. This adds the right child to the queue for processing in the next iteration.\n",
-        "\n",
-        "7. `else:`: If the `node` is None, it represents an empty node in the binary tree.\n",
-        "\n",
-        "    a. `result.append(None)`: Append `None` to the `result` list to indicate an empty node.\n",
-        "\n",
-        "8. `while result and result[-1] is None:`: After the traversal is complete, there might be trailing `None` values in the `result` list. This loop removes any such trailing `None` values to ensure a clean representation of the tree's elements.\n",
-        "\n",
-        "9. `return result`: Return the `result` list, which now contains the elements of the binary tree in a format where `None` represents empty nodes and the order of elements reflects a level-order traversal of the tree.\n",
-        "\n",
-        "In summary, the `treeToList` function performs a level-order traversal of the binary tree using a queue, constructing a list where each element corresponds to a node's value or represents an empty node with `None`. This list represents the binary tree's elements in a structured format."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[3, 9, 20, None, None, 15, 7]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "preorder = [3, 9, 20, 15, 7]\n",
-        "inorder = [9, 3, 15, 20, 7]\n",
-        "result = buildTree(preorder, inorder)\n",
-        "\n",
-        "print(treeToList(result))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[-1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "preorder = [-1]\n",
-        "inorder = [-1]\n",
-        "result = buildTree(preorder, inorder)\n",
-        "\n",
-        "print(treeToList(result))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `buildTree` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of the `buildTree` function can be analyzed in terms of the number of nodes in the binary tree. Let's assume there are `n` nodes in the binary tree.\n",
-        "\n",
-        "1. Finding the root value in the `preorder` traversal takes O(1) time as it's just extracting the first element from the list.\n",
-        "2. Finding the index of the root value in the `inorder` traversal using `inorder.index(root_val)` takes O(n) time in the worst case because in the worst case, it might have to search through all `n` elements in the `inorder` list to find the index.\n",
-        "3. The recursive calls to `buildTree` for the left and right subtrees are made once for each node in the tree. Therefore, the recursive calls have a combined time complexity of O(n) as they process each node once.\n",
-        "\n",
-        "The total time complexity of the `buildTree` function is O(n) due to the recursive calls and finding the root's index in the `inorder` list.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity is determined by the space used by the function's call stack during recursion and any additional data structures used. \n",
-        "\n",
-        "1. The space used by the function call stack during recursion depends on the height of the binary tree. In the worst case, where the tree is highly unbalanced (e.g., a skewed tree), the space complexity for the call stack is O(n) as it can go as deep as the number of nodes in the tree.\n",
-        "\n",
-        "2. Additionally, the function creates TreeNode objects for each node in the binary tree. Therefore, the space complexity for these objects is also O(n).\n",
-        "\n",
-        "Overall, the space complexity of the `buildTree` function is O(n) due to the space used by the call stack and the TreeNode objects.\n",
-        "\n",
-        "In summary, the time complexity of `buildTree` is O(n), and the space complexity is O(n), where 'n' is the number of nodes in the binary tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Construct Binary Tree from Postorder and Inorder Traversal:**\n",
-        "   Modify the problem to construct a binary tree from its postorder and inorder traversals. Implement a function similar to `buildTree` but for postorder and inorder traversals.\n",
-        "\n",
-        "2. **Reconstruct Binary Tree with Duplicate Values:**\n",
-        "   Extend the problem to handle binary trees with duplicate values. Ensure that your solution correctly handles scenarios where there are duplicate values in the preorder and inorder traversals."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb b/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb
deleted file mode 100644
index 59d8e7d..0000000
--- a/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 124. Binary Tree Maximum Path Sum\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Maximum Path Sum problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-maximum-path-sum/)\n",
-        "\n",
-        "---\n",
-        "A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.\n",
-        "\n",
-        "The **path sum** of a path is the sum of the node's values in the path.\n",
-        "\n",
-        "Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[1, 3*10^4]$.\n",
-        "- `-1000 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to find the maximum path sum in a binary tree. A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. The path sum of a path is the sum of the node values in that path.\n",
-        "\n",
-        "In this problem, you are given the root of a binary tree, and you need to find the maximum path sum among all possible paths in the tree. Note that the path does not need to start at the root node or end at a leaf node; it can start and end at any nodes in the tree.\n",
-        "\n",
-        "To find the maximum path sum, you need to consider both left and right subtrees of each node while traversing the tree. At each node, you have two choices:\n",
-        "\n",
-        "1. Include the current node in the path: In this case, you add the value of the current node to the sum and continue exploring both the left and right subtrees for possible extensions of the path.\n",
-        "\n",
-        "2. Start a new path from the current node: In this case, you do not include the current node in the path sum, and you choose either the left subtree or the right subtree to start a new path.\n",
-        "\n",
-        "To solve this problem, you can use a recursive approach to traverse the binary tree. For each node, you calculate the maximum path sum that can pass through that node (option 1) and also return the maximum path sum that can be extended from that node (option 2). You keep track of the global maximum path sum encountered during the traversal.\n",
-        "\n",
-        "Ultimately, the maximum path sum among all paths in the tree will be the maximum of the global maximum path sum and the maximum path sum calculated for each node.\n",
-        "\n",
-        "The constraints for this problem include:\n",
-        "\n",
-        "- The number of nodes in the tree is in the range $[1, 3 * 10^4]$.\n",
-        "- Node values are in the range [-1000, 1000].\n",
-        "\n",
-        "By considering all possible paths through the tree, the algorithm aims to find the most significant sum of node values in any path in the binary tree.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "class Solution:\n",
-        "    def maxPathSum(self, root):\n",
-        "        def max_path_sum_helper(node):\n",
-        "            if not node:\n",
-        "                return 0\n",
-        "            \n",
-        "            # Recursively calculate the maximum path sum for left and right subtrees\n",
-        "            left_max = max(0, max_path_sum_helper(node.left))\n",
-        "            right_max = max(0, max_path_sum_helper(node.right))\n",
-        "            \n",
-        "            # Update the global maximum path sum\n",
-        "            self.max_sum = max(self.max_sum, left_max + right_max + node.val)\n",
-        "            \n",
-        "            # Return the maximum path sum starting from the current node\n",
-        "            return max(left_max, right_max) + node.val\n",
-        "        \n",
-        "        self.max_sum = float('-inf')  # Initialize the global maximum to negative infinity\n",
-        "        max_path_sum_helper(root)  # Start the recursive traversal from the root\n",
-        "        return self.max_sum"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The code starts by defining a `TreeNode` class to represent nodes in the binary tree. Each `TreeNode` has a `val` attribute representing its value, a `left` attribute pointing to the left child, and a `right` attribute pointing to the right child.\n",
-        "\n",
-        "2. Next, a `Solution` class is defined to contain the `maxPathSum` method, which will be used to find the maximum path sum in the binary tree.\n",
-        "\n",
-        "3. Within the `maxPathSum` method, a helper function called `max_path_sum_helper` is defined. This recursive function takes a single argument, `node`, which is the current node being considered during traversal.\n",
-        "\n",
-        "4. Inside the `max_path_sum_helper` function:\n",
-        "   - If `node` is `None` (i.e., the current node is a null node), it returns 0. This is the base case of the recursion.\n",
-        "   - It calculates the maximum path sum starting from the left subtree (`left_max`) and the right subtree (`right_max`). Importantly, it uses `max(0, ...)` to ensure that negative values (which would make the path sum smaller) are not considered. If a subtree's path sum is negative, it's better to not include it in the path.\n",
-        "   - The code then updates the global maximum path sum (`self.max_sum`) by checking if the sum of `left_max`, `right_max`, and the current node's value is greater than the current maximum.\n",
-        "   - Finally, it returns the maximum path sum starting from the current node, which is the maximum of `left_max` and `right_max` plus the current node's value.\n",
-        "\n",
-        "5. Before calling the `max_path_sum_helper` function, the `max_sum` attribute of the `Solution` class is initialized to negative infinity (`float('-inf')`). This is done to ensure that any valid path sum encountered during traversal will be greater than the initial value of `max_sum`.\n",
-        "\n",
-        "6. The `max_path_sum_helper` function is called with the root of the binary tree, effectively starting the traversal from the root node.\n",
-        "\n",
-        "7. Once the traversal is complete, the method returns the value of `self.max_sum`, which contains the maximum path sum found in the binary tree.\n",
-        "\n",
-        "8. Example usage at the bottom of the code demonstrates how to create binary trees and use the `maxPathSum` method to find the maximum path sum for two different tree structures.\n",
-        "\n",
-        "In summary, the code uses a recursive approach to explore all possible paths in the binary tree, ensuring that negative path sums are not considered. It keeps track of the maximum path sum encountered and returns that maximum value as the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "6\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "root1 = TreeNode(1)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(3)\n",
-        "solution = Solution()\n",
-        "print(solution.maxPathSum(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "42\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "root2 = TreeNode(-10)\n",
-        "root2.left = TreeNode(9)\n",
-        "root2.right = TreeNode(20)\n",
-        "root2.right.left = TreeNode(15)\n",
-        "root2.right.right = TreeNode(7)\n",
-        "print(solution.maxPathSum(root2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "Time Complexity:\n",
-        "- The code uses a recursive depth-first traversal to explore each node in the binary tree exactly once.\n",
-        "- During the traversal, for each node, we perform constant time operations, such as updating the `max_sum` variable and calculating the maximum path sum starting from that node.\n",
-        "- Therefore, the time complexity of the code is O(N), where N is the number of nodes in the binary tree.\n",
-        "\n",
-        "Space Complexity:\n",
-        "- The space complexity of the code is determined by the space used in the call stack during the recursive traversal.\n",
-        "- In the worst case, the depth of the call stack can be equal to the height of the binary tree. In an unbalanced tree, this could be O(N), but in a balanced binary tree, the height is O(log N).\n",
-        "- Additionally, the code uses a constant amount of space for variables like `left_max`, `right_max`, and `max_sum`.\n",
-        "- Therefore, the overall space complexity is O(H), where H is the height of the binary tree. In the worst case, it's O(N), and in the best case (a balanced binary tree), it's O(log N).\n",
-        "\n",
-        "In summary:\n",
-        "- Time Complexity: O(N)\n",
-        "- Space Complexity: O(H), where H is the height of the binary tree, ranging from O(log N) to O(N) depending on the tree's balance."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Print the Maximum Path:**\n",
-        "   Modify the code to not only find the maximum path sum but also print the nodes involved in the maximum path. You can print the nodes in the correct order from the root to the leaf.\n",
-        "\n",
-        "2. **Path Sum Count:**\n",
-        "   Instead of finding the maximum path sum, find the count of unique paths that sum to a given target value. Each path can start and end anywhere in the tree."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb b/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb
deleted file mode 100644
index 55dc944..0000000
--- a/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb	
+++ /dev/null
@@ -1,257 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 199. Binary Tree Right Side View\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Right Side View problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-right-side-view/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, imagine yourself standing on the **right side** of it, return *the values of the nodes you can see ordered from top to bottom*.\n",
-        "\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def rightSideView(root):\n",
-        "    if not root:\n",
-        "        return []  # If the tree is empty, return an empty list.\n",
-        "\n",
-        "    result = []  # Initialize a list to store the right side view values.\n",
-        "    queue = [root]  # Initialize a queue for level-order traversal with the root node.\n",
-        "\n",
-        "    while queue:\n",
-        "        # Get the number of nodes at the current level.\n",
-        "        level_size = len(queue)\n",
-        "\n",
-        "        # Traverse all nodes at the current level and add the rightmost node to the result.\n",
-        "        for i in range(level_size):\n",
-        "            node = queue.pop(0)  # Dequeue the first node from the queue.\n",
-        "\n",
-        "            # If it's the rightmost node at this level, add its value to the result.\n",
-        "            if i == level_size - 1:\n",
-        "                result.append(node.val)\n",
-        "\n",
-        "            # Add the children of the current node to the queue.\n",
-        "            if node.left:\n",
-        "                queue.append(node.left)\n",
-        "            if node.right:\n",
-        "                queue.append(node.right)\n",
-        "\n",
-        "    return result  # Return the list of right side view values.\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start with a binary tree represented using a TreeNode class. Each node has a value (`val`) and can have a left child (`left`) and a right child (`right`).\n",
-        "\n",
-        "2. The `rightSideView` function takes the root node of the binary tree as input and returns a list of values representing the nodes you can see when standing on the right side of the tree.\n",
-        "\n",
-        "3. We check if the `root` is `None`, which means the tree is empty. If it's empty, we return an empty list because there are no nodes to see.\n",
-        "\n",
-        "4. We initialize an empty list called `result` to store the right side view values.\n",
-        "\n",
-        "5. We initialize a queue called `queue` with the root node. This queue is used for a level-order traversal of the tree.\n",
-        "\n",
-        "6. We enter a while loop that continues until the `queue` is empty. Inside this loop, we perform the following steps for each level of the tree:\n",
-        "\n",
-        "   a. We determine the number of nodes at the current level by getting the length of the `queue`. This is important for processing nodes at the same level together.\n",
-        "\n",
-        "   b. We then iterate through the nodes at the current level using a for loop. For each node in the level, we dequeue it from the `queue` using `queue.pop(0)`.\n",
-        "\n",
-        "   c. If the node we dequeue is the rightmost node at the current level (determined by `i == level_size - 1`), we add its `val` to the `result` list. This is because, when standing on the right side of the tree, you can see the rightmost node at each level.\n",
-        "\n",
-        "   d. We enqueue the left and right children of the current node (if they exist) to the `queue`. This ensures that we process the next level in the subsequent iterations of the while loop.\n",
-        "\n",
-        "7. After processing all levels of the tree, the `result` list contains the values of the rightmost nodes in each level, ordered from top to bottom.\n",
-        "\n",
-        "8. Finally, we return the `result` list as the output of the function, which represents the right side view of the binary tree.\n",
-        "\n",
-        "The code effectively performs a level-order traversal of the binary tree while keeping track of the rightmost nodes at each level and adding them to the result list. This ensures that we obtain the correct order of nodes visible from the right side of the tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 3, 4]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root1 = TreeNode(1)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(3)\n",
-        "root1.left.right = TreeNode(5)\n",
-        "root1.right.right = TreeNode(4)\n",
-        "print(rightSideView(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "root2 = TreeNode(1)\n",
-        "root2.right = TreeNode(3)\n",
-        "print(rightSideView(root2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root = None\n",
-        "result = rightSideView(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of the code is O(N), where N is the number of nodes in the binary tree. Here's why:\n",
-        "\n",
-        "1. We perform a level-order traversal of the tree using a queue. In the worst case, we visit all nodes once, which takes O(N) time since we visit each node exactly once.\n",
-        "\n",
-        "2. For each node, we perform constant-time operations like dequeuing it from the queue, checking if it's the rightmost node at the current level, and enqueuing its children (if they exist). These operations do not depend on the size of the tree, so they do not contribute to the overall time complexity.\n",
-        "\n",
-        "Therefore, the dominant factor in the time complexity is the level-order traversal, making it O(N).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is also O(N). Here's why:\n",
-        "\n",
-        "1. We use a queue to perform level-order traversal. In the worst case, the queue can contain all nodes at the maximum width of the tree, which can be up to N/2 nodes for a completely unbalanced tree. Therefore, the space required for the queue is O(N).\n",
-        "\n",
-        "2. The `result` list stores the values of the rightmost nodes. In the worst case, when the binary tree is a complete binary tree, it can have roughly N/2 rightmost nodes. Therefore, the space required for the `result` list is also O(N).\n",
-        "\n",
-        "3. Other auxiliary variables like `level_size`, `i`, and `node` require only constant space and do not depend on the size of the tree.\n",
-        "\n",
-        "Combining the space used by the queue and the `result` list, we get a space complexity of O(N).\n",
-        "\n",
-        "In summary, the code has a time complexity of O(N) and a space complexity of O(N), where N is the number of nodes in the binary tree. These complexities are efficient and scale linearly with the size of the input tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Variant with Left Side View**: Modify the code to find and return the values of nodes visible from the left side of the binary tree instead of the right side.\n",
-        "\n",
-        "2. **Zigzag Right Side View**: Extend the code to return the right side view values in a zigzag order. In a zigzag order, you alternate between starting from the rightmost node at level 0, then the leftmost at level 1, rightmost at level 2, and so on."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb b/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb
deleted file mode 100644
index 8ea4609..0000000
--- a/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb	
+++ /dev/null
@@ -1,223 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 226: Invert Binary Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the root of a binary tree, invert the tree, and return its root.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the tree is in the range `[0, 100]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def invertTree(root):\n",
-        "    # Base case: If the root is None or the tree is empty, return None.\n",
-        "    if not root:\n",
-        "        return None\n",
-        "\n",
-        "    # Swap the left and right subtrees of the current node.\n",
-        "    root.left, root.right = root.right, root.left\n",
-        "\n",
-        "    # Recursively invert the left and right subtrees.\n",
-        "    invertTree(root.left)\n",
-        "    invertTree(root.right)\n",
-        "\n",
-        "    return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a TreeNode class, which represents a node in a binary tree. Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-        "\n",
-        "2. The `invertTree` function is defined to invert the binary tree. It takes the root node of the tree as an argument.\n",
-        "\n",
-        "3. In the `invertTree` function, we have a base case to handle the scenario when the root is `None` (empty tree). In such cases, we return `None` because there's nothing to invert.\n",
-        "\n",
-        "4. For non-empty trees, we swap the left and right subtrees of the current node. This effectively inverts the tree at the current node.\n",
-        "\n",
-        "5. We then recursively call `invertTree` on the left and right subtrees to invert them.\n",
-        "\n",
-        "6. Finally, we return the root of the inverted tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inverted Tree (Example 1): <__main__.TreeNode object at 0x270c9c0>\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: root = [4,2,7,1,3,6,9]\n",
-        "\n",
-        "root1 = TreeNode(4)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(7)\n",
-        "root1.left.left = TreeNode(1)\n",
-        "root1.left.right = TreeNode(3)\n",
-        "root1.right.left = TreeNode(6)\n",
-        "root1.right.right = TreeNode(9)\n",
-        "\n",
-        "inverted_root1 = invertTree(root1)\n",
-        "print(\"Inverted Tree (Example 1):\", inverted_root1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inorder Traversal of Inverted Tree (Example 2): <__main__.TreeNode object at 0x2255fc0>\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: root = [2,1,3]\n",
-        "\n",
-        "root2 = TreeNode(2)\n",
-        "root2.left = TreeNode(1)\n",
-        "root2.right = TreeNode(3)\n",
-        "\n",
-        "inverted_root2 = invertTree(root2)\n",
-        "print(\"Inverted Tree (Example 2):\", inverted_root2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inverted Tree (Example 3): None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "# Input: root = []\n",
-        "\n",
-        "root3 = None  # Empty tree\n",
-        "\n",
-        "inverted_root3 = invertTree(root3)\n",
-        "print(\"Inverted Tree (Example 3):\", inverted_root3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's discuss the time and space complexity of the provided Python code to invert a binary tree.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of the `invertTree` function is $O(n)$, where $n$ is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal of the tree. In the worst case, we have to visit all nodes in the tree.\n",
-        "\n",
-        "The reason for this time complexity is the depth-first traversal of the tree, where we recursively process the left and right subtrees of each node.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is determined by the function call stack during the recursion. In the worst case, the space complexity is $O(h)$, where $h$ is the height of the binary tree. \n",
-        "\n",
-        "In a completely unbalanced binary tree (essentially a linked list), the height $h$ is equal to the number of nodes $n$, resulting in a space complexity of $O(n)$. This occurs when the tree is skewed to one side.\n",
-        "\n",
-        "In a balanced binary tree, such as a full binary tree, the height $h$ is $O(log\\ n)$, and the space complexity is $O(log\\ n)$.\n",
-        "\n",
-        "The space complexity depends on how balanced the tree is. In practical scenarios, binary trees are often approximately balanced, so the space complexity is typically closer to $O(log\\ n)$.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "- Time Complexity: $O(n)$ where $n$ is the number of nodes.\n",
-        "- Space Complexity: $O(h)$, where $h$ is the height of the binary tree. In the worst case, it can be $O(n)$, and in a balanced tree, it is $O(log\\ n)$.\n",
-        "\n",
-        "Keep in mind that these complexities are based on the recursive implementation provided. Iterative solutions can achieve the same task with $O(1)$ space complexity, using auxiliary data structures like stacks or queues to mimic the recursion stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb b/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb
deleted file mode 100644
index 673ced2..0000000
--- a/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 230. Kth Smallest Element in a BST\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Kth Smallest Element in a BST problem on LeetCode, click here!](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary search tree, and an integer `k`, return *the $k^{th}$ smallest value (**1-indexed**) of all the values of the nodes in the tree.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is `n`.\n",
-        "- 1 <= `k <= n` <= $10^4$\n",
-        "- 0 <= `Node.val` <= $10^4$\n",
-        "\n",
-        "**Follow up**: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def kthSmallest(root, k):\n",
-        "    stack = []  # Initialize an empty stack to simulate the traversal\n",
-        "    while root or stack:\n",
-        "        while root:\n",
-        "            stack.append(root)  # Push the current node onto the stack\n",
-        "            root = root.left  # Move to the left child\n",
-        "        root = stack.pop()  # Pop a node from the stack\n",
-        "        k -= 1  # Decrement k since we've visited a node\n",
-        "        if k == 0:\n",
-        "            return root.val  # If k becomes 0, return the current node's value as the kth smallest\n",
-        "        root = root.right  # Move to the right child to continue the traversal"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in the binary search tree (BST). Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-        "\n",
-        "2. The `kthSmallest` function takes two parameters: `root`, which is the root of the BST, and `k`, which represents the kth smallest element we want to find.\n",
-        "\n",
-        "3. We initialize an empty stack (`stack`) to simulate the traversal of the BST. The stack will be used to keep track of nodes as we traverse the tree in an iterative manner.\n",
-        "\n",
-        "4. We enter a while loop that continues until either `root` becomes `None` (indicating we have traversed the entire tree) or the `stack` is empty.\n",
-        "\n",
-        "5. Within the loop, we start another while loop to traverse as far left as possible in the BST. We repeatedly push nodes onto the `stack` and move to their left children until we reach the leftmost leaf node. This is the smallest node in the BST.\n",
-        "\n",
-        "6. Once we have reached the leftmost leaf node (the smallest node), we pop nodes from the `stack` one by one. As we pop each node, we decrement `k` by 1 to keep track of the number of nodes we have visited.\n",
-        "\n",
-        "7. If `k` becomes 0 after decrementing, it means we have found the kth smallest element. In this case, we return the `val` of the current node as the result.\n",
-        "\n",
-        "8. If `k` is still greater than 0, it means we haven't found the kth smallest element yet. In this case, we move to the right child of the current node to continue the traversal, as the kth smallest element, if it exists, will be in the right subtree of the current node.\n",
-        "\n",
-        "9. The process continues until we find the kth smallest element or traverse the entire tree.\n",
-        "\n",
-        "10. Finally, we return the kth smallest element found.\n",
-        "\n",
-        "The key idea in this code is to perform an in-order traversal of the BST while keeping track of the kth smallest element. By visiting nodes in ascending order, we can efficiently find the kth smallest element in O(h + k) time, where h is the height of the BST and k is the desired kth element. This approach is both concise and efficient for this problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Create the tree for the first example: root = [3,1,4,null,2], k = 1\n",
-        "root1 = TreeNode(3)\n",
-        "root1.left = TreeNode(1)\n",
-        "root1.right = TreeNode(4)\n",
-        "root1.left.right = TreeNode(2)\n",
-        "k1 = 1\n",
-        "print(kthSmallest(root1, k1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Create the tree for the second example: root = [5,3,6,2,4,null,null,1], k = 3\n",
-        "root2 = TreeNode(5)\n",
-        "root2.left = TreeNode(3)\n",
-        "root2.right = TreeNode(6)\n",
-        "root2.left.left = TreeNode(2)\n",
-        "root2.left.right = TreeNode(4)\n",
-        "root2.left.left.left = TreeNode(1)\n",
-        "k2 = 3\n",
-        "print(kthSmallest(root2, k2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this code is O(h + k), where:\n",
-        "- h is the height of the binary search tree (BST).\n",
-        "- k is the desired kth smallest element we want to find.\n",
-        "\n",
-        "1. In the worst case, where the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the time complexity is O(n + k).\n",
-        "\n",
-        "2. In the best case, where the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the time complexity is O(log(n) + k).\n",
-        "\n",
-        "So, the time complexity can vary from O(log(n) + k) in the best-case scenario to O(n + k) in the worst-case scenario. Typically, for balanced BSTs, the time complexity is closer to O(log(n) + k).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this code is O(h) due to the stack used for the iterative in-order traversal, where h is the height of the BST.\n",
-        "\n",
-        "1. In the worst case, when the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the space complexity is O(n) because the stack can potentially store all n nodes.\n",
-        "\n",
-        "2. In the best case, when the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the space complexity is O(log(n)) because the stack will have at most log(n) nodes.\n",
-        "\n",
-        "So, the space complexity depends on the height of the BST and can vary from O(log(n)) in the best-case scenario to O(n) in the worst-case scenario.\n",
-        "\n",
-        "In practical terms, for balanced BSTs or moderately unbalanced BSTs, the space complexity is usually close to O(log(n)), and the code is efficient in terms of space usage."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Kth Largest Element**: Modify the code to find the kth largest element in the BST instead of the kth smallest element.\n",
-        "\n",
-        "2. **Kth Smallest Element in Two BSTs**: Given two BSTs, find the kth smallest element when considering both BSTs as a single sorted list. This involves merging the two BSTs while finding the kth element efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb b/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb
deleted file mode 100644
index 666bcd9..0000000
--- a/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 235. Lowest Common Ancestor of a Binary Search Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Lowest Common Ancestor of a Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)\n",
-        "\n",
-        "---\n",
-        "Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.\n",
-        "\n",
-        "According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[2, 10^5]$.\n",
-        "- $-10^9$ <= `Node.val` <= $10^9$\n",
-        "- All `Node.val` are **unique**.\n",
-        "- `p != q`\n",
-        "- `p` and `q` will exist in the BST.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def lowestCommonAncestor(root, p, q):\n",
-        "    # Ensure p is less than q\n",
-        "    if p.val > q.val:\n",
-        "        p, q = q, p\n",
-        "\n",
-        "    while root:\n",
-        "        # If the current node value is greater than both p and q, move left\n",
-        "        if root.val > q.val:\n",
-        "            root = root.left\n",
-        "        # If the current node value is less than both p and q, move right\n",
-        "        elif root.val < p.val:\n",
-        "            root = root.right\n",
-        "        # If the current node value is between p and q (inclusive), or it matches either p or q, it's the LCA\n",
-        "        else:\n",
-        "            return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `class TreeNode:` defines a simple class for representing nodes in a binary tree. Each `TreeNode` has three attributes:\n",
-        "   - `val`: The value of the node.\n",
-        "   - `left`: A reference to the left child node.\n",
-        "   - `right`: A reference to the right child node.\n",
-        "\n",
-        "2. `def lowestCommonAncestor(root, p, q):` is a function that takes three arguments:\n",
-        "   - `root`: The root node of the BST.\n",
-        "   - `p`: A TreeNode representing one of the nodes for which we want to find the LCA.\n",
-        "   - `q`: A TreeNode representing the other node for which we want to find the LCA.\n",
-        "\n",
-        "3. `if p.val > q.val:` checks if the value of `p` is greater than the value of `q`. In a BST, it's essential to ensure that `p` represents the smaller value, and `q` represents the larger value. If `p` is greater than `q`, the code swaps their values.\n",
-        "\n",
-        "4. The main logic is inside the `while` loop, which runs until `root` becomes `None`. It performs the following steps to find the LCA:\n",
-        "   - If the current `root` node's value is greater than the value of `q`, it means both `p` and `q` are on the left subtree of the current node. So, we move to the left child of the current node by setting `root = root.left`.\n",
-        "   - If the current `root` node's value is less than the value of `p`, it means both `p` and `q` are on the right subtree of the current node. So, we move to the right child of the current node by setting `root = root.right`.\n",
-        "   - If neither of the above conditions is met, it means the current `root` node's value is between `p` and `q`, or it matches either `p` or `q`. In this case, the current node is the lowest common ancestor (LCA), so we return `root`.\n",
-        "\n",
-        "The algorithm is efficient because it takes advantage of the properties of a BST. It eliminates subtrees that cannot contain the LCA by comparing the values of `p`, `q`, and the current `root` node. Eventually, it reaches the LCA node, and that node is returned as the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "6\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root = TreeNode(6)\n",
-        "root.left = TreeNode(2)\n",
-        "root.right = TreeNode(8)\n",
-        "root.left.left = TreeNode(0)\n",
-        "root.left.right = TreeNode(4)\n",
-        "root.right.left = TreeNode(7)\n",
-        "root.right.right = TreeNode(9)\n",
-        "root.left.right.left = TreeNode(3)\n",
-        "root.left.right.right = TreeNode(5)\n",
-        "\n",
-        "p = root.left  # Node 2\n",
-        "q = root.right  # Node 8\n",
-        "\n",
-        "result = lowestCommonAncestor(root, p, q)\n",
-        "print(result.val)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Using the same tree as before\n",
-        "p = root.left  # Node 2\n",
-        "q = root.left.right  # Node 4\n",
-        "\n",
-        "result = lowestCommonAncestor(root, p, q)\n",
-        "print(result.val) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root2 = TreeNode(2)\n",
-        "root2.left = TreeNode(1)\n",
-        "\n",
-        "p = root2  # Node 2\n",
-        "q = root2.left  # Node 1\n",
-        "\n",
-        "result = lowestCommonAncestor(root2, p, q)\n",
-        "print(result.val)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided algorithm for finding the lowest common ancestor (LCA) in a Binary Search Tree (BST).\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this algorithm is O(h), where \"h\" is the height of the BST. In the worst case, where the tree is completely unbalanced (essentially a linked list), the height of the tree is O(n), where \"n\" is the number of nodes in the tree. However, in a well-balanced BST, the height is logarithmic, which is O(log n).\n",
-        "\n",
-        "The reason for this time complexity is that the algorithm efficiently narrows down the search space by traversing either left or right subtrees based on the values of the target nodes `p` and `q`. It eliminates entire subtrees that cannot contain the LCA, leading to a relatively quick search.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of the algorithm is O(1). This is because it uses a constant amount of extra space, regardless of the size of the input BST. The only variables used are `root`, `p`, and `q`, and there are no data structures like stacks or queues used for additional space. The algorithm performs a simple traversal without recursion, so it does not consume extra memory as the tree depth increases.\n",
-        "\n",
-        "**In summary**, the provided algorithm for finding the LCA in a BST is both time and space-efficient. Its time complexity is O(h), where \"h\" is the height of the tree, and its space complexity is O(1), making it suitable for practical use even in large BSTs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "### Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple LCAs:** Extend the algorithm to find all the lowest common ancestors of two given nodes `p` and `q` in a BST. In some cases, there can be multiple LCAs.\n",
-        "\n",
-        "2. **LCA with k Nodes:** Given a BST and k nodes, find the lowest common ancestor of these k nodes. This is an extension of the problem where you must find the LCA of more than two nodes."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb b/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb
deleted file mode 100644
index 75c48ea..0000000
--- a/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb	
+++ /dev/null
@@ -1,297 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 297. Serialize and Deserialize Binary Tree\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Serialize and Deserialize Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.\n",
-        "\n",
-        "Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.\n",
-        "\n",
-        "**Clarification:** The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-        "- `-1000 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "TThe problem is to design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of characters or bytes so that it can be easily stored in a file, transmitted over a network, or otherwise persisted. Deserialization is the reverse process of converting the serialized data back into the original data structure.\n",
-        "\n",
-        "In the context of this problem:\n",
-        "\n",
-        "1. **Serialization**: You are given a binary tree, and your task is to convert it into a string representation such that you can later recreate the same binary tree from this string. The format of serialization is flexible, but it should allow you to reconstruct the original binary tree accurately.\n",
-        "\n",
-        "2. **Deserialization**: Given a serialized string, you need to reconstruct the binary tree it represents, ensuring that it is identical to the original tree.\n",
-        "\n",
-        "Here are some key points to consider in solving this problem:\n",
-        "\n",
-        "- The input can include any valid binary tree, including trees with nodes having integer values within the range [-1000, 1000].\n",
-        "- You don't necessarily need to follow a specific format for serialization, but it should be designed in a way that allows unambiguous deserialization.\n",
-        "- The goal is to serialize the tree structure and its values and then deserialize it back into the same structure and values.\n",
-        "\n",
-        "A common approach for serialization is to use a traversal method like preorder traversal, where you visit nodes in the order: root, left subtree, right subtree. This way, you can serialize the tree into a string, and during deserialization, you can reconstruct the tree by parsing the string in the same order.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode(object):\n",
-        "    def __init__(self, x):\n",
-        "        self.val = x\n",
-        "        self.left = None\n",
-        "        self.right = None\n",
-        "\n",
-        "class Codec:\n",
-        "    def serialize(self, root):\n",
-        "        serialized  = []\n",
-        "\n",
-        "        def dfs(node):\n",
-        "            if not node:\n",
-        "                # If the node is None, represent it as \"Null\" in the serialized string\n",
-        "                serialized .append(\"Null\")\n",
-        "                return\n",
-        "            # Convert the node's value to a string and add it to the serialized string\n",
-        "            serialized .append(str(node.val))\n",
-        "            dfs(node.left)\n",
-        "            dfs(node.right)\n",
-        "\n",
-        "        dfs(root)\n",
-        "        # Join the serialized values with \",\"\n",
-        "        return \",\".join(serialized )\n",
-        "\n",
-        "    def deserialize(self, data):\n",
-        "        vals = data.split(\",\")\n",
-        "        self.i = 0\n",
-        "\n",
-        "        def dfs():\n",
-        "            if vals[self.i] == \"Null\":\n",
-        "                self.i += 1\n",
-        "                return None\n",
-        "            # Convert the value to an integer to create a new node\n",
-        "            node = TreeNode(int(vals[self.i]))\n",
-        "            self.i += 1\n",
-        "            node.left = dfs()\n",
-        "            node.right = dfs()\n",
-        "            return node\n",
-        "\n",
-        "        return dfs()"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Here's a step-by-step explanation of the code:\n",
-        "\n",
-        "1. **Definition of TreeNode**:\n",
-        "   - The code defines a simple `TreeNode` class that represents a node in a binary tree. Each `TreeNode` has three attributes:\n",
-        "     - `val`: The integer value stored in the node.\n",
-        "     - `left`: A reference to the left child node (or None if there is no left child).\n",
-        "     - `right`: A reference to the right child node (or None if there is no right child).\n",
-        "   - This class allows you to create binary tree nodes with integer values and left and right child references.\n",
-        "\n",
-        "2. **Serialization (serialize method)**:\n",
-        "   - The `serialize` method takes a `root` node as input, which is the root of the binary tree that needs to be serialized.\n",
-        "   - It initializes an empty list `serialized` to store the serialized elements.\n",
-        "   - The `dfs` (depth-first search) function is defined within the `serialize` method to perform a preorder traversal of the binary tree.\n",
-        "   - In the `dfs` function:\n",
-        "     - If the current `node` is `None` (i.e., a leaf node or a child of a leaf node), it appends the string \"Null\" to the `serialized` list to represent the absence of a node.\n",
-        "     - If the current `node` is not `None`, it appends the string representation of the `node.val` to the `serialized` list and then recursively calls `dfs` on the left and right children.\n",
-        "   - After the `dfs` traversal, the `serialized` list contains the serialized binary tree elements.\n",
-        "   - The method returns a string obtained by joining the elements in the `serialized` list with commas.\n",
-        "\n",
-        "3. **Deserialization (deserialize method)**:\n",
-        "   - The `deserialize` method takes a serialized string `data` as input, which represents a binary tree in the custom format.\n",
-        "   - It splits the `data` string by commas to obtain a list of elements called `vals`.\n",
-        "   - The method initializes an index `self.i` to 0. This index keeps track of the current position in the `vals` list during deserialization.\n",
-        "   - The `dfs` function is defined within the `deserialize` method to perform the deserialization process using a recursive approach.\n",
-        "   - In the `dfs` function:\n",
-        "     - If the current element in `vals` is \"Null,\" it means there is no node at this position in the binary tree, so it returns `None`.\n",
-        "     - If the current element is not \"Null,\" it converts the element to an integer to create a new `TreeNode` with that value.\n",
-        "     - It then recursively calls `dfs` to set the left and right children of the current node.\n",
-        "   - The method returns the root node of the reconstructed binary tree.\n",
-        "\n",
-        "Overall, this code demonstrates a way to serialize a binary tree into a string format and then deserialize it back into the original tree structure, allowing for the representation of both the tree structure and the values stored in the nodes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1,2,Null,Null,3,4,Null,Null,5,Null,Null\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Serialize a binary tree\n",
-        "root = TreeNode(1)\n",
-        "root.left = TreeNode(2)\n",
-        "root.right = TreeNode(3)\n",
-        "root.right.left = TreeNode(4)\n",
-        "root.right.right = TreeNode(5)\n",
-        "\n",
-        "codec = Codec()\n",
-        "serialized_tree = codec.serialize(root)\n",
-        "print(serialized_tree)\n",
-        "\n",
-        "# Deserialize the serialized tree\n",
-        "new_root = codec.deserialize(serialized_tree)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Null\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Example usage:\n",
-        "# Serialize a binary tree\n",
-        "root = None\n",
-        "\n",
-        "codec = Codec()\n",
-        "serialized_tree = codec.serialize(root)\n",
-        "print(serialized_tree)\n",
-        "\n",
-        "# Deserialize the serialized tree\n",
-        "new_root = codec.deserialize(serialized_tree)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for serializing and deserializing a binary tree:\n",
-        "\n",
-        "1. **Serialization (serialize method)**:\n",
-        "\n",
-        "   - Time Complexity:\n",
-        "     - The `serialize` method uses a depth-first traversal (preorder) of the binary tree.\n",
-        "     - It visits each node exactly once.\n",
-        "     - At each node, it performs constant-time operations (string conversion and list append).\n",
-        "     - Therefore, the time complexity for serialization is O(N), where N is the number of nodes in the binary tree.\n",
-        "\n",
-        "   - Space Complexity:\n",
-        "     - The space complexity for serialization includes the space used by the `serialized` list and the recursive call stack.\n",
-        "     - The `serialized` list stores the serialized elements, and its space is proportional to the number of nodes.\n",
-        "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-        "     - Therefore, the space complexity for serialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-        "\n",
-        "2. **Deserialization (deserialize method)**:\n",
-        "\n",
-        "   - Time Complexity:\n",
-        "     - The `deserialize` method splits the serialized string into a list of elements, which takes O(N) time.\n",
-        "     - The `dfs` function is a recursive function that visits each element in the list exactly once.\n",
-        "     - At each element, it performs constant-time operations (string comparison, integer conversion, and recursive function calls).\n",
-        "     - Therefore, the time complexity for deserialization is O(N).\n",
-        "\n",
-        "   - Space Complexity:\n",
-        "     - The space complexity for deserialization includes the space used by the `vals` list and the recursive call stack.\n",
-        "     - The `vals` list stores the elements from the serialized string, and its space is proportional to the number of nodes.\n",
-        "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-        "     - Therefore, the space complexity for deserialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-        "\n",
-        "Overall, the provided code has a time complexity of O(N) for both serialization and deserialization and a space complexity of O(N) in the majority of cases (unless the tree is severely unbalanced, in which case the height H dominates the space complexity). It efficiently serializes and deserializes a binary tree while using a linear amount of memory."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Efficient Serialization**:\n",
-        "   Optimize the serialization process to minimize the length of the serialized string. Consider using techniques like binary encoding to represent node values more efficiently.\n",
-        "\n",
-        "2. **Custom Serialization Format**:\n",
-        "   Modify the serialization and deserialization methods to use a custom format different from the one provided. Ensure that the new format allows for accurate reconstruction of the original binary tree."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb b/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb
deleted file mode 100644
index 8a41c3b..0000000
--- a/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb	
+++ /dev/null
@@ -1,265 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 572: Subtree of Another Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Subtree of Another Tree problem on LeetCode, click here!](https://leetcode.com/problems/subtree-of-another-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "**Problem Description:**\n",
-        "\n",
-        "Given the roots of two binary trees `root` and `subRoot`, return `True` if there is a subtree of `root` with the same structure and node values as `subRoot`, and `False` otherwise.\n",
-        "\n",
-        "A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "1. The number of nodes in the `root` tree is in the range $[1, 2000]$.\n",
-        "2. The number of nodes in the `subRoot` tree is in the range $[1, 1000]$.\n",
-        "3. $-10^4$ <= root.val <= $10^4$\n",
-        "4. $-10^4$ <= subRoot.val <= $10^4$\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "class Solution:\n",
-        "    def isSubtree(self, root, subRoot):\n",
-        "        if not root:\n",
-        "            return False\n",
-        "        \n",
-        "        # Check if the current subtree is equal to the subRoot\n",
-        "        if self.isSameTree(root, subRoot):\n",
-        "            return True\n",
-        "        \n",
-        "        # Recursively check the left and right subtrees\n",
-        "        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)\n",
-        "    \n",
-        "    def isSameTree(self, tree1, tree2):\n",
-        "        if not tree1 and not tree2:\n",
-        "            return True\n",
-        "        if not tree1 or not tree2:\n",
-        "            return False\n",
-        "        \n",
-        "        return (\n",
-        "            tree1.val == tree2.val and\n",
-        "            self.isSameTree(tree1.left, tree2.left) and\n",
-        "            self.isSameTree(tree1.right, tree2.right)\n",
-        "        )"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "This code defines a TreeNode class for the binary tree nodes and a Solution class with two methods:\n",
-        "\n",
-        "1. `isSubtree`: This method checks if there is a subtree of the `root` tree with the same structure and node values as the `subRoot` tree. It uses a helper function `isSameTree` to compare two trees for equality.\n",
-        "\n",
-        "2. `isSameTree`: This helper method recursively compares two trees to check if they are the same in structure and node values.\n",
-        "\n",
-        "### Here's a detailed explanation of the code:\n",
-        "\n",
-        "1. **TreeNode Class:**\n",
-        "\n",
-        "The `TreeNode` class is defined to represent nodes in a binary tree. Each node has a `val` (the node's value) and may have a `left` and `right` child.\n",
-        "\n",
-        "2. **Solution Class:**\n",
-        "\n",
-        "    + The `Solution` class contains the solution for the problem and defines two important methods:\n",
-        "\n",
-        "    + `isSubtree(self, root, subRoot)`:\n",
-        "\n",
-        "        - This method checks whether `subRoot` is a subtree of `root`. It takes two tree nodes, `root` and `subRoot`, as input arguments.\n",
-        "        - If `root` is `None`, it returns `False` because there is no subtree to search.\n",
-        "        - It then checks if the current subtree with `root` as its root is equal to `subRoot` using the `isSameTree` method. If they are the same, it returns `True`.\n",
-        "        - If the current subtree is not the same as `subRoot`, it recursively checks the left and right subtrees of `root` to see if `subRoot` is a subtree of any of them.\n",
-        "        - It returns `True` if `subRoot` is found in either the left or right subtree; otherwise, it returns `False`.\n",
-        "\n",
-        "    + `isSameTree(self, tree1, tree2)`:\n",
-        "\n",
-        "        - This method checks whether two trees, `tree1` and `tree2`, are the same.\n",
-        "        - If both `tree1` and `tree2` are `None`, they are considered the same tree, so it returns `True`.\n",
-        "        - If only one of them is `None` (but not both), they are different trees, so it returns `False`.\n",
-        "        - If both `tree1` and `tree2` have values, it checks if their values are equal and recursively checks if their left and right subtrees are the same.\n",
-        "        - It returns `True` if the trees are the same; otherwise, it returns `False`.\n",
-        "\n",
-        "The code effectively uses recursion to traverse the binary trees and check for subtree equality. The `isSubtree` method starts the recursive search, and the `isSameTree` method is used to compare individual subtrees. The approach is efficient and avoids unnecessary checks when possible.\n",
-        "\n",
-        "This solution allows you to determine if there exists a subtree within the `root` tree that matches the structure and node values of `subRoot`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Is subRoot a subtree of root? True\n"
-          ]
-        }
-      ],
-      "source": [
-        "#root = [3,4,5,1,2]\n",
-        "#subRoot = [4,1,2]\n",
-        "\n",
-        "# Example input\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(4)\n",
-        "root.right = TreeNode(5)\n",
-        "root.left.left = TreeNode(1)\n",
-        "root.left.right = TreeNode(2)\n",
-        "\n",
-        "subRoot = TreeNode(4)\n",
-        "subRoot.left = TreeNode(1)\n",
-        "subRoot.right = TreeNode(2)\n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test the isSubtree method\n",
-        "result = solution.isSubtree(root, subRoot)\n",
-        "print(\"Is subRoot a subtree of root?\", result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Is subRoot a subtree of root? False\n"
-          ]
-        }
-      ],
-      "source": [
-        "#root = [3,4,5,1,2,null,null,null,null,0] \n",
-        "#subRoot = [4,1,2]\n",
-        "\n",
-        "\n",
-        "# Example input\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(4)\n",
-        "root.right = TreeNode(5)\n",
-        "root.left.left = TreeNode(1)\n",
-        "root.left.right = TreeNode(2)\n",
-        "root.left.right.left = TreeNode(0)\n",
-        "\n",
-        "subRoot = TreeNode(4)\n",
-        "subRoot.left = TreeNode(1)\n",
-        "subRoot.right = TreeNode(2)\n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test the isSubtree method\n",
-        "result = solution.isSubtree(root, subRoot)\n",
-        "print(\"Is subRoot a subtree of root?\", result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "86a34f0d-0789-498c-818c-2d30f84754d0",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code for the \"Subtree of Another Tree\" problem:\n",
-        "\n",
-        "**Time Complexity**\n",
-        "\n",
-        "The time complexity of the code primarily depends on the recursive traversal of the binary tree. In both the `isSubtree` and `isSameTree` functions, we visit each node in the binary trees once. Let's break down the time complexity:\n",
-        "\n",
-        "1. The `isSubtree` function:\n",
-        "   - In the worst case, we need to visit every node in the `root` tree.\n",
-        "   - For each node in the `root` tree, we call the `isSameTree` function, which has its own traversal.\n",
-        "   - So, the total time complexity is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree.\n",
-        "\n",
-        "2. The `isSameTree` function:\n",
-        "   - In the worst case, we visit every node in both `tree1` and `tree2`.\n",
-        "   - The number of recursive calls made is proportional to the number of nodes in the trees.\n",
-        "   - So, the time complexity of this function is $O(max(n, m))$, where $n$ and $m$ are the numbers of nodes in `tree1` and `tree2`, respectively.\n",
-        "\n",
-        "Overall, the time complexity of the entire code is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree. In practice, it may be less than $O(n * m)$ if a subtree mismatch is detected early during the traversal.\n",
-        "\n",
-        "**Space Complexity**\n",
-        "\n",
-        "The space complexity of the code is determined by the function call stack during recursion and the space used by the recursive functions. Let's analyze the space complexity:\n",
-        "\n",
-        "1. The `isSubtree` function:\n",
-        "   - It uses the call stack for recursion.\n",
-        "   - The maximum depth of the recursion is equal to the height of the `root` tree, which can be $O(n)$ in the worst case (unbalanced tree).\n",
-        "   - Additionally, the function doesn't use any significant extra space other than the recursion stack.\n",
-        "\n",
-        "2. The `isSameTree` function:\n",
-        "   - It also uses the call stack for recursion.\n",
-        "   - The maximum depth of the recursion is equal to the height of the `tree1` or `tree2`, whichever is greater.\n",
-        "   - So, the maximum space used for the call stack is $O(max(n, m))$.\n",
-        "\n",
-        "In summary, the space complexity of the code is $O(max(n, m))$ due to the function call stack. It scales with the maximum height of the trees being compared.\n",
-        "\n",
-        "Overall, the code is efficient and works well for trees with moderate sizes. However, it's important to keep in mind that the worst-case time complexity is $O(n * m)$, so for very large trees, the performance may degrade."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb b/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb
deleted file mode 100644
index ad86882..0000000
--- a/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 98. Validate Binary Search Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Validate Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/validate-binary-search-tree/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, *determine if it is a valid binary search tree (BST)*.\n",
-        "\n",
-        "A **valid BST** is defined as follows:\n",
-        "\n",
-        "- The left subtree of a node contains only nodes with keys **less than** the node's key.\n",
-        "- The right subtree of a node contains only nodes with keys **greater than** the node's key.\n",
-        "- Both the left and right subtrees must also be binary search trees. \n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[1, 10^4]$.\n",
-        "- $-2^{31}$ <= `Node.val` <= $2^{31} - 1$\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def is_valid_BST(root):\n",
-        "    def is_valid(node, min_val, max_val):\n",
-        "        # Base case: If the node is None, it's a valid BST.\n",
-        "        if node is None:\n",
-        "            return True\n",
-        "        \n",
-        "        # Check if the current node's value is within the valid range.\n",
-        "        if not (min_val < node.val < max_val):\n",
-        "            return False\n",
-        "        \n",
-        "        # Recursively check the left and right subtrees with updated ranges.\n",
-        "        return (is_valid(node.left, min_val, node.val) and\n",
-        "                is_valid(node.right, node.val, max_val))\n",
-        "    \n",
-        "    # Call the helper function starting with a wide range for root node.\n",
-        "    return is_valid(root, float('-inf'), float('inf'))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- It defines a `TreeNode` class to represent nodes in the binary tree.\n",
-        "\n",
-        "- The `is_valid_BST` function takes the root node of the binary tree as input and returns `True` if the tree is a valid BST, and `False` otherwise.\n",
-        "\n",
-        "- Inside the `is_valid_BST` function, there is a helper function called `is_valid` that performs the actual validation using a recursive approach.\n",
-        "\n",
-        "- The `is_valid` function checks each node in the tree to ensure that it satisfies the properties of a BST:\n",
-        "  - The value of the current node must be within a valid range defined by `min_val` and `max_val`.\n",
-        "  - The left subtree of the current node should contain values less than the current node's value.\n",
-        "  - The right subtree of the current node should contain values greater than the current node's value.\n",
-        "\n",
-        "- If any of these conditions are violated, the function returns `False`, indicating that the tree is not a valid BST.\n",
-        "\n",
-        "- If all nodes satisfy these conditions, the function returns `True`, indicating that the tree is a valid BST.\n",
-        "\n",
-        "- The code calls the `is_valid` function with the root node and initial range values of negative infinity to positive infinity to start the validation.\n",
-        "\n",
-        "- The time complexity of the code is O(N), where N is the number of nodes in the tree, as it traverses each node once.\n",
-        "\n",
-        "- The space complexity depends on the height of the tree. In the average case for a balanced BST, the space complexity is O(log N), but in the worst case (skewed tree), it can be O(N) due to the recursive call stack.\n",
-        "\n",
-        "In summary, this code checks whether a given binary tree is a valid BST by recursively validating each node's value and its left and right subtrees while maintaining valid value ranges. If all nodes satisfy the BST properties, the tree is considered a valid BST."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Create the tree for the first example: [2, 1, 3]\n",
-        "root1 = TreeNode(2)\n",
-        "root1.left = TreeNode(1)\n",
-        "root1.right = TreeNode(3)\n",
-        "print(is_valid_BST(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Create the tree for the second example: [5, 1, 4, None, None, 3, 6]\n",
-        "root2 = TreeNode(5)\n",
-        "root2.left = TreeNode(1)\n",
-        "root2.right = TreeNode(4)\n",
-        "root2.right.left = TreeNode(3)\n",
-        "root2.right.right = TreeNode(6)\n",
-        "print(is_valid_BST(root2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `is_valid_BST` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The time complexity of the function primarily depends on the number of nodes in the binary tree.\n",
-        "- In the worst case, we may have to visit every node in the tree once to validate whether it's part of a valid BST.\n",
-        "- Since we are performing a depth-first search (DFS) traversal of the tree, the time complexity is O(N), where N is the number of nodes in the tree.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity is determined by the space used by the recursive call stack during the DFS traversal.\n",
-        "- In the worst case, if the tree is completely unbalanced (a skewed tree), the maximum depth of the call stack would be equal to the height of the tree.\n",
-        "- In a balanced BST, the height is approximately log2(N), where N is the number of nodes.\n",
-        "- Therefore, the space complexity of the call stack is O(log N) for a balanced BST.\n",
-        "- In the worst case (skewed tree), the space complexity can be O(N) as the height of the tree can be equal to N.\n",
-        "\n",
-        "**Overall:**\n",
-        "- Time Complexity: O(N)\n",
-        "- Space Complexity: O(log N) on average for a balanced BST, and O(N) in the worst case for a skewed tree.\n",
-        "\n",
-        "The space complexity is typically dominated by the recursive call stack, and it varies depending on the shape of the binary tree. In practice, for balanced binary trees, the space complexity is often close to O(log N), which is quite efficient."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Complex Constraints**: Modify the problem to have more complex constraints on the tree. For example, consider a binary tree with constraints like \"The left subtree of a node contains only nodes with keys less than or equal to the node's key,\" and \"The right subtree of a node contains only nodes with keys greater than or equal to the node's key.\" How would you adapt the code to handle these constraints?\n",
-        "\n",
-        "2. **Handling Duplicates**: Modify the code to handle binary search trees that allow duplicate values. In a standard BST, each value is unique, but in this case, multiple nodes can have the same value. The tree should still be considered valid if it follows the BST property."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/README.md b/_build/html/_sources/07. Trees/README.md
deleted file mode 100644
index 1eba5d4..0000000
--- a/_build/html/_sources/07. Trees/README.md	
+++ /dev/null
@@ -1,26 +0,0 @@
-# Trees Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)                          | Easy       |
-| [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)     | Easy       |
-| [100. Same Tree](https://leetcode.com/problems/same-tree/)                                           | Easy       |
-| [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)               | Easy       |
-| [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/) | Easy |
-| [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/) | Medium |
-| [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)       | Medium     |
-| [230. Kth Smallest Element In a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)   | Medium     |
-| [105. Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) | Medium |
-| [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)     | Hard       |
-| [297. Serialize And Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/) | Hard |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb b/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb
deleted file mode 100644
index 9da2a17..0000000
--- a/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 208. Implement Trie (Prefix Tree)\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Implement Trie problem on LeetCode, click here!](https://leetcode.com/problems/implement-trie-prefix-tree/)\n",
-        "\n",
-        "---\n",
-        "A **trie** (pronounced as \"try\") or **prefix tree** is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.\n",
-        "\n",
-        "Implement the Trie class:\n",
-        "\n",
-        "- `Trie()` Initializes the trie object.\n",
-        "- `void insert(String word)` Inserts the string `word` into the trie.\n",
-        "- `boolean search(String word)` Returns `true` if the string `word` is in the trie (i.e., was inserted before), and `false` otherwise.\n",
-        "- `boolean startsWith(String prefix)` Returns `true` if there is a previously inserted string `word` that has the prefix `prefix`, and `false` otherwise.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= word.length, prefix.length <= 2000`\n",
-        "- `word` and `prefix` consist only of lowercase English letters.\n",
-        "- At most $3 * 10^4$ calls in total will be made to `insert`, `search`, and `startsWith`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to implement a data structure known as a Trie (pronounced \"try\") or Prefix Tree. A Trie is a tree-like data structure that is used to efficiently store and retrieve a set of strings or words. It's particularly useful for tasks involving string manipulation, such as autocomplete and spell-checking. The problem statement defines the following operations for implementing the Trie class:\n",
-        "\n",
-        "1. `Trie()`: This is the constructor that initializes the Trie object.\n",
-        "\n",
-        "2. `void insert(String word)`: This method inserts the given string `word` into the Trie. It effectively adds the characters of the word to the Trie's structure, creating nodes for each character if they don't already exist. At the end of the word, a special flag is set to indicate that this node represents the end of a valid word.\n",
-        "\n",
-        "3. `boolean search(String word)`: This method checks if the given string `word` is present in the Trie. It starts from the root of the Trie and traverses the Trie by following the characters in the word. If it successfully traverses the Trie and reaches the end of the word, it returns `true`, indicating that the word exists in the Trie; otherwise, it returns `false`.\n",
-        "\n",
-        "4. `boolean startsWith(String prefix)`: This method checks if there is any previously inserted word in the Trie that has the given `prefix`. It's similar to the `search` method but does not require that the prefix be a complete word. If there is any word in the Trie that starts with the given prefix, it returns `true`; otherwise, it returns `false`.\n",
-        "\n",
-        "The problem also provides an example scenario where these operations are called, demonstrating the expected output for each operation.\n",
-        "\n",
-        "In summary, the goal is to create a data structure (Trie) that efficiently stores a set of strings and provides methods to insert new strings, search for complete words, and check if a given prefix exists in the stored set of strings.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        self.children = {}           # A dictionary to store child nodes\n",
-        "        self.is_end_of_word = False  # Indicates if a word ends at this node\n",
-        "\n",
-        "class Trie:\n",
-        "    def __init__(self):\n",
-        "        self.root = TrieNode()       # Initialize the Trie with a root node\n",
-        "\n",
-        "    def insert(self, word):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                node.children[char] = TrieNode()  # Create a new node if the character is not present\n",
-        "            node = node.children[char]           # Move to the child node\n",
-        "        node.is_end_of_word = True                # Mark the end of the inserted word\n",
-        "\n",
-        "    def search(self, word):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                return False         # If the character is not found, the word is not in the Trie\n",
-        "            node = node.children[char]  # Move to the child node\n",
-        "        return node.is_end_of_word    # Check if the node represents the end of a valid word\n",
-        "\n",
-        "    def startsWith(self, prefix):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in prefix:\n",
-        "            if char not in node.children:\n",
-        "                return False         # If the character is not found, no word starts with the prefix\n",
-        "            node = node.children[char]  # Move to the child node\n",
-        "        return True                  # Prefix found in the Trie"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's explain how the code works step by step:\n",
-        "\n",
-        "1. `TrieNode` Class:\n",
-        "   - `TrieNode` is a class that represents nodes in the Trie. Each node has two attributes:\n",
-        "     - `children`: This is a dictionary that stores child nodes. Each key in the dictionary represents a character, and the corresponding value is the child node for that character.\n",
-        "     - `is_end_of_word`: This boolean flag indicates whether the node represents the end of a complete word. Initially, it's set to `False` for all nodes.\n",
-        "\n",
-        "2. `Trie` Class:\n",
-        "   - `Trie` is the main class that implements the Trie data structure. It has the following methods:\n",
-        "\n",
-        "   - `__init__(self)`: The constructor initializes the Trie object by creating a root node, which serves as the starting point for all Trie operations.\n",
-        "\n",
-        "   - `insert(self, word)`: This method inserts a string `word` into the Trie. It starts from the root node and iterates through each character in the word.\n",
-        "     - If a character is not present as a child node, it creates a new node for that character.\n",
-        "     - It then moves to the child node and continues the process until the entire word is inserted.\n",
-        "     - Finally, it sets the `is_end_of_word` flag to `True` for the last node to mark the end of the inserted word.\n",
-        "\n",
-        "   - `search(self, word)`: This method checks if a complete word (string) exists in the Trie. It starts from the root node and iterates through each character in the word.\n",
-        "     - If a character is not found as a child node, it immediately returns `False` because the word is not in the Trie.\n",
-        "     - It continues to move to the child node for each character.\n",
-        "     - After reaching the end of the word, it checks if the `is_end_of_word` flag is `True` for the last node to confirm the presence of the word.\n",
-        "\n",
-        "   - `startsWith(self, prefix)`: This method checks if there is any previously inserted word in the Trie that starts with a given `prefix`. It follows the same logic as the `search` method but does not require the entire word to be present.\n",
-        "     - If the prefix is found in the Trie, it returns `True`; otherwise, it returns `False`.\n",
-        "\n",
-        "3. Example Usage:\n",
-        "   - The code demonstrates how to create a Trie object, insert words (\"apple\" and \"app\"), and perform operations on the Trie:\n",
-        "     - It inserts the word \"apple\" into the Trie.\n",
-        "     - It checks if \"apple\" is in the Trie, which returns `True`.\n",
-        "     - It checks if \"app\" is in the Trie, which returns `False`.\n",
-        "     - It checks if there is any word in the Trie that starts with \"app,\" which returns `True`.\n",
-        "     - It inserts the word \"app\" into the Trie.\n",
-        "     - It checks if \"app\" is in the Trie again, which now returns `True`.\n",
-        "\n",
-        "In summary, the code efficiently implements a Trie data structure with the ability to insert words, search for complete words, and check for the existence of words with a given prefix. The Trie is organized as a tree of nodes, with each node representing a character in the words being stored."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n",
-            "False\n",
-            "True\n",
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "\n",
-        "trie = Trie()\n",
-        "trie.insert(\"apple\")\n",
-        "print(trie.search(\"apple\"))   # Output: True\n",
-        "print(trie.search(\"app\"))     # Output: False\n",
-        "print(trie.startsWith(\"app\")) # Output: True\n",
-        "trie.insert(\"app\")\n",
-        "print(trie.search(\"app\"))     # Output: True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the Trie implementation:\n",
-        "\n",
-        "1. Time Complexity:\n",
-        "   - Insertion (`insert` method): The time complexity of inserting a word into the Trie is O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to insert it into the Trie.\n",
-        "   - Search (`search` method): The time complexity of searching for a word in the Trie is also O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to determine if it exists in the Trie.\n",
-        "   - Starts with (`startsWith` method): The time complexity of checking if a prefix exists in the Trie is O(p), where p is the length of the prefix. In the worst case, you may need to traverse the entire prefix to determine its presence.\n",
-        "\n",
-        "   Therefore, for each operation (insertion, search, or starts with), the time complexity is O(m) or O(p), where m is the length of the word being operated on, and p is the length of the prefix.\n",
-        "\n",
-        "2. Space Complexity:\n",
-        "   - The space complexity of the Trie is determined by the number of nodes and characters stored in it. In the worst case, where none of the inserted words share common prefixes, the space complexity is O(N), where N is the total number of characters across all inserted words.\n",
-        "   - Each node in the Trie represents a character, and the number of nodes is directly related to the total number of characters.\n",
-        "\n",
-        "   In practice, the space complexity can be less than O(N) because common prefixes are shared among words in the Trie, leading to space optimization.\n",
-        "\n",
-        "In summary:\n",
-        "- Time complexity for each operation (insertion, search, starts with) is O(m) or O(p), where m is the length of the word and p is the length of the prefix.\n",
-        "- Space complexity is O(N) in the worst case, where N is the total number of characters across all inserted words. However, space optimization occurs due to common prefix sharing in practice, potentially reducing the actual space used."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Count Prefixes**:\n",
-        "   Implement a method to count the number of words in the Trie that have a specific prefix. This exercise requires you to maintain additional data in the Trie nodes to keep track of the count.\n",
-        "\n",
-        "2. **Auto-Complete Suggestions**:\n",
-        "   Implement an auto-complete feature using the Trie. Given a prefix, the program should return a list of suggested words based on the words previously inserted into the Trie."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb b/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb
deleted file mode 100644
index 2de472c..0000000
--- a/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb	
+++ /dev/null
@@ -1,275 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 211. Design Add and Search Words Data Structure\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Design Add and Search Words Data Structure problem on LeetCode, click here!](https://leetcode.com/problems/design-add-and-search-words-data-structure/)\n",
-        "\n",
-        "---\n",
-        "Design a data structure that supports adding new words and finding if a string matches any previously added string.\n",
-        "\n",
-        "Implement the `WordDictionary` class:\n",
-        "\n",
-        "- `WordDictionary()` Initializes the object.\n",
-        "- `void addWord(word)` Adds `word` to the data structure, it can be matched later.\n",
-        "- `bool search(word)` Returns `true` if there is any string in the data structure that matches `word` or `false` otherwise. `word` may contain dots `'.'` where dots can be matched with any letter.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= word.length <= 25`\n",
-        "- `word` in `addWord` consists of lowercase English letters.\n",
-        "- `word` in `search` consist of `'.'` or lowercase English letters.\n",
-        "- There will be at most `2` dots in `word` for `search` queries.\n",
-        "- At most $10^4$ calls will be made to `addWord` and `search`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to design a data structure called `WordDictionary` that allows you to add words and search for words efficiently. The key feature of this data structure is that it should support wildcard search, where a dot ('.') can match any letter.\n",
-        "\n",
-        "Here are the specific requirements and explanations for the problem:\n",
-        "\n",
-        "1. Initialize the `WordDictionary` object using the constructor `WordDictionary()`.\n",
-        "\n",
-        "2. The `addWord(word)` method allows you to add a word to the data structure. Once added, you should be able to search for this word later. The words consist of lowercase English letters, and each word has a length between 1 and 25 characters.\n",
-        "\n",
-        "3. The `search(word)` method allows you to search for a word in the data structure. This method returns `True` if there is any string in the data structure that matches the given word or `False` otherwise. The word you are searching for may contain dots ('.'), where each dot can match any letter.\n",
-        "\n",
-        "4. The wildcard character ('.') in the `search` method allows for partial or fuzzy searching. For example, if the word dictionary contains the words \"bad,\" \"dad,\" and \"mad,\" then:\n",
-        "   - Searching for \"pad\" should return `False` because there is no exact match.\n",
-        "   - Searching for \"bad\" should return `True` because \"bad\" is in the dictionary.\n",
-        "   - Searching for \".ad\" should return `True` because it can match \"bad,\" \"dad,\" or \"mad.\"\n",
-        "   - Searching for \"b..\" should return `True` because it can match \"bad,\" \"bed,\" \"bet,\" etc.\n",
-        "\n",
-        "5. The problem specifies that there will be at most two dots ('.') in search queries, and there will be at most 10^4 calls to both `addWord` and `search`.\n",
-        "\n",
-        "In summary, you need to implement the `WordDictionary` class that efficiently supports adding words and searching for words, where the search can include wildcard characters ('.') that match any letter.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        # Initialize a TrieNode with children and a flag to indicate the end of a word.\n",
-        "        self.children = {}\n",
-        "        self.is_end_of_word = False\n",
-        "\n",
-        "class WordDictionary:\n",
-        "\n",
-        "    def __init__(self):\n",
-        "        # Initialize the WordDictionary with a root node.\n",
-        "        self.root = TrieNode()\n",
-        "\n",
-        "    def addWord(self, word: str) -> None:\n",
-        "        node = self.root\n",
-        "        # Iterate through the characters in the word.\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                # Create a new node for the character if it doesn't exist.\n",
-        "                node.children[char] = TrieNode()\n",
-        "            # Move to the next node.\n",
-        "            node = node.children[char]\n",
-        "        # Mark the end of the word by setting the flag to True.\n",
-        "        node.is_end_of_word = True\n",
-        "\n",
-        "    def search_helper(self, node: TrieNode, word: str) -> bool:\n",
-        "        if not word:\n",
-        "            # If there are no characters left in the word, check if we reached the end of a word in the trie.\n",
-        "            return node.is_end_of_word\n",
-        "        \n",
-        "        char = word[0]\n",
-        "        if char == '.':\n",
-        "            # If the character is a dot, explore all possible child nodes.\n",
-        "            for child in node.children.values():\n",
-        "                if self.search_helper(child, word[1:]):\n",
-        "                    return True\n",
-        "        elif char in node.children:\n",
-        "            # If the character is in the children of the current node, move to the next node.\n",
-        "            return self.search_helper(node.children[char], word[1:])\n",
-        "        \n",
-        "        # If the character is not found and is not a dot, the word is not in the trie.\n",
-        "        return False\n",
-        "\n",
-        "    def search(self, word: str) -> bool:\n",
-        "        # Start the search from the root of the trie.\n",
-        "        return self.search_helper(self.root, word)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "This code defines a `WordDictionary` class that uses a Trie data structure to efficiently store and search for words. Here's a step-by-step explanation:\n",
-        "\n",
-        "1. `TrieNode` class: This class represents nodes in the Trie data structure. Each node has two attributes: `children` (a dictionary to store child nodes) and `is_end_of_word` (a flag to indicate whether the node marks the end of a word).\n",
-        "\n",
-        "2. `WordDictionary` class: This class initializes with an empty Trie by creating a root node.\n",
-        "\n",
-        "3. `addWord` method: This method adds a word to the Trie. It iterates through each character in the word and creates Trie nodes as necessary. It sets the `is_end_of_word` flag to `True` for the final character of the word to mark the end of the word.\n",
-        "\n",
-        "4. `search_helper` method: This is a recursive helper function for searching words in the Trie. It takes a Trie node (`node`) and a word to search (`word`) as input. If there are no characters left in the word (`not word`), it checks if the current node marks the end of a word. If it does, it returns `True`.\n",
-        "\n",
-        "5. If the first character of the word is a dot ('.'), the function explores all possible child nodes by iterating through `node.children.values()` and recursively calls `search_helper` for each child with the remaining part of the word (`word[1:]`).\n",
-        "\n",
-        "6. If the first character of the word is not a dot and is found in the children of the current node, the function recursively moves to the next node by calling `search_helper` on the child node and the remaining part of the word.\n",
-        "\n",
-        "7. If the character is not found in the children and is not a dot, the word is not in the Trie, so the function returns `False`.\n",
-        "\n",
-        "8. `search` method: This is the public method for searching words. It initiates the search process by calling `search_helper` starting from the root node of the Trie.\n",
-        "\n",
-        "Overall, this code efficiently implements a Trie-based data structure that allows adding and searching for words, including support for wildcard characters represented by dots ('.')."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n",
-            "True\n",
-            "True\n",
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "\n",
-        "wordDictionary = WordDictionary()\n",
-        "wordDictionary.addWord(\"bad\")\n",
-        "wordDictionary.addWord(\"dad\")\n",
-        "wordDictionary.addWord(\"mad\")\n",
-        "print(wordDictionary.search(\"pad\"))  # Output: False\n",
-        "print(wordDictionary.search(\"bad\"))  # Output: True\n",
-        "print(wordDictionary.search(\".ad\"))  # Output: True\n",
-        "print(wordDictionary.search(\"b..\"))  # Output: True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `WordDictionary` class methods:\n",
-        "\n",
-        "1. `addWord` Method:\n",
-        "\n",
-        "   - Time Complexity: O(L)\n",
-        "     - L is the length of the word being added.\n",
-        "     - In the worst case, we need to iterate through each character in the word and create nodes in the Trie. This takes O(L) time.\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for storing the word in the Trie is also O(L) because we create nodes for each character in the word.\n",
-        "\n",
-        "2. `search_helper` Method (called by `search`):\n",
-        "\n",
-        "   - Time Complexity: O(2^M * L)\n",
-        "     - M is the maximum number of dots ('.') in the search word.\n",
-        "     - In the worst case, when there are M dots in the search word, we may explore all possible child nodes at each dot. This results in a branching factor of 26 (for lowercase letters) at each dot.\n",
-        "     - So, the time complexity is exponential in the number of dots and linear in the length of the word being searched (L).\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for the recursive call stack is O(L) in the worst case because we may have to recurse to a depth equal to the length of the search word.\n",
-        "\n",
-        "3. `search` Method:\n",
-        "\n",
-        "   - Time Complexity: O(2^M * L)\n",
-        "     - The `search` method calls the `search_helper` method, which has the same time complexity explained above.\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for the recursive call stack is O(L) in the worst case, as explained earlier.\n",
-        "\n",
-        "Overall, the time complexity of the `search` method is mainly affected by the number of dots ('.') in the search word. In the worst case, when there are multiple dots and many possibilities to explore at each dot, the time complexity can be exponential. However, for most practical cases, it performs reasonably well.\n",
-        "\n",
-        "The space complexity primarily depends on the space required to store the words in the Trie. It is proportional to the total number of characters in all added words and is linear with respect to the length of the words added.\n",
-        "\n",
-        "In summary:\n",
-        "\n",
-        "- Time Complexity for `addWord`: O(L)\n",
-        "- Time Complexity for `search`: O(2^M * L)\n",
-        "- Space Complexity: O(L) for storing words in the Trie, O(L) for the recursive call stack during search.\n",
-        "\n",
-        "Note: The space complexity analysis assumes that the dictionary contains a reasonable number of words and does not account for the overhead of Python objects or system-level memory allocation."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Dictionary Auto-Completion**: Extend the `WordDictionary` class to provide auto-completion suggestions based on the prefix of a word. Implement a method that returns a list of words that match the given prefix.\n",
-        "\n",
-        "2. **Support Word Deletion**: Extend the `WordDictionary` class to support word deletion. Add a method `deleteWord(word)` that removes a word from the data structure. Ensure that the search operation still works correctly after deletion."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/08. Tries/212. Word Search II.ipynb b/_build/html/_sources/08. Tries/212. Word Search II.ipynb
deleted file mode 100644
index 71597b8..0000000
--- a/_build/html/_sources/08. Tries/212. Word Search II.ipynb	
+++ /dev/null
@@ -1,316 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 212. Word Search II\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Word Search II problem on LeetCode, click here!](https://leetcode.com/problems/word-search-ii/)\n",
-        "\n",
-        "---\n",
-        "Given an `m x n` `board` of characters and a list of strings `words`, the task is to return *all words on the board*.\n",
-        "\n",
-        "Each word must be constructed from letters of sequentially **adjacent cells**, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == board.length`\n",
-        "- `n == board[i].length`\n",
-        "- `1 <= m, n <= 12`\n",
-        "- `board[i][j]` is a lowercase English letter.\n",
-        "- 1 <= `words.length` <= $3 * 10^4$\n",
-        "- `1 <= words[i].length <= 10`\n",
-        "- `words[i]` consists of lowercase English letters.\n",
-        "- All the strings of `words` are unique."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "To solve the \"Word Search II\" problem, you can use a Trie data structure and perform a depth-first search (DFS) on the board. Here's a step-by-step approach to solve the problem:\n",
-        "\n",
-        "1. Build a Trie from the given list of words:\n",
-        "   - Create a Trie data structure to store all the words from the input list.\n",
-        "   - For each word in the input list, insert it into the Trie, character by character. Make sure to mark the end of each word in the Trie.\n",
-        "\n",
-        "2. Perform DFS on the board:\n",
-        "   - Iterate through each cell (i, j) on the board.\n",
-        "   - For each cell, start a DFS traversal from that cell to search for words.\n",
-        "   - During the DFS traversal, maintain a current path string.\n",
-        "   - At each cell, check if the current path string, concatenated with the character in the cell, matches any word prefix in the Trie. If it does, continue the DFS.\n",
-        "   - If the current path string matches a word in the Trie, add it to the result.\n",
-        "\n",
-        "3. Implement DFS with backtracking:\n",
-        "   - During the DFS traversal, you will explore neighboring cells (horizontally and vertically) if they are within bounds and haven't been visited before.\n",
-        "   - To prevent revisiting the same cell within the same word, mark the cell as visited (e.g., change its character to a special character like '#') before exploring it and restore its original character after the DFS traversal.\n",
-        "\n",
-        "4. Return the result:\n",
-        "   - After completing the DFS traversal for all cells on the board, you will have collected all the valid words in the result set.\n",
-        "   - Convert the result set to a list and return it as the final output.\n",
-        "\n",
-        "This approach efficiently finds all words on the board by utilizing the Trie data structure to prune unnecessary DFS paths and avoiding duplicate visits to the same cell within the same word. It ensures that each word is constructed from letters of sequentially adjacent cells without using the same letter cell more than once in a word.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Define a TrieNode class to represent nodes in the Trie.\n",
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        self.children = {}  # Dictionary to store child nodes.\n",
-        "        self.is_end_of_word = False  # Flag to mark the end of a word.\n",
-        "\n",
-        "# Main function to find words on the board.\n",
-        "def findWords(board, words):\n",
-        "    # DFS function to search for words starting from a given cell.\n",
-        "    def dfs(node, i, j, path):\n",
-        "        char = board[i][j]  # Get the character at the current cell.\n",
-        "        curr_node = node.children.get(char)  # Find the corresponding Trie node.\n",
-        "\n",
-        "        if not curr_node:\n",
-        "            return  # If no matching Trie node, stop the search.\n",
-        "\n",
-        "        path += char  # Add the character to the current path.\n",
-        "\n",
-        "        if curr_node.is_end_of_word:\n",
-        "            result.add(path)  # If the path matches a word, add it to the result.\n",
-        "\n",
-        "        temp, board[i][j] = board[i][j], \"#\"  # Mark the cell as visited.\n",
-        "\n",
-        "        # Explore neighboring cells (up, down, left, right).\n",
-        "        if i > 0:\n",
-        "            dfs(curr_node, i - 1, j, path)\n",
-        "        if i < m - 1:\n",
-        "            dfs(curr_node, i + 1, j, path)\n",
-        "        if j > 0:\n",
-        "            dfs(curr_node, i, j - 1, path)\n",
-        "        if j < n - 1:\n",
-        "            dfs(curr_node, i, j + 1, path)\n",
-        "\n",
-        "        board[i][j] = temp  # Restore the cell's original character.\n",
-        "\n",
-        "    # Function to build a Trie from the list of words.\n",
-        "    def buildTrie():\n",
-        "        root = TrieNode()  # Create a root node.\n",
-        "        for word in words:\n",
-        "            node = root\n",
-        "            for char in word:\n",
-        "                if char not in node.children:\n",
-        "                    node.children[char] = TrieNode()  # Create a new node if needed.\n",
-        "                node = node.children[char]\n",
-        "            node.is_end_of_word = True  # Mark the end of the word.\n",
-        "        return root\n",
-        "\n",
-        "    m, n = len(board), len(board[0])  # Get the dimensions of the board.\n",
-        "    root = buildTrie()  # Build the Trie from the list of words.\n",
-        "    result = set()  # Use a set to store unique words found on the board.\n",
-        "\n",
-        "    # Iterate through each cell on the board and start DFS from there.\n",
-        "    for i in range(m):\n",
-        "        for j in range(n):\n",
-        "            dfs(root, i, j, \"\")  # Start DFS from each cell on the board.\n",
-        "\n",
-        "    return list(result)  # Convert the set to a list and return the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code solves the \"Word Search II\" problem, where you have a grid of characters and a list of words. The goal is to find all words from the list that can be constructed by traversing adjacent cells (horizontally or vertically) in the grid, without using the same cell more than once for a single word.\n",
-        "\n",
-        "The code does the following:\n",
-        "\n",
-        "1. Defines a `TrieNode` class to represent nodes in a Trie data structure. Each node has children (other characters in the word) and a flag to mark the end of a word.\n",
-        "\n",
-        "2. Defines the `findWords` function, which takes the grid (`board`) and a list of words (`words`) as input.\n",
-        "\n",
-        "3. Inside the `findWords` function, there is a `dfs` (depth-first search) function. This function is used to explore the grid starting from a specific cell. It checks if the current path matches any word in the Trie and adds it to the result if found.\n",
-        "\n",
-        "4. The code also includes a `buildTrie` function that constructs a Trie data structure from the list of words. It iterates through each word, creating nodes for each character in the Trie and marking the end of words.\n",
-        "\n",
-        "5. The dimensions of the grid are obtained (m x n).\n",
-        "\n",
-        "6. A Trie is built using the `buildTrie` function.\n",
-        "\n",
-        "7. A set called `result` is used to store unique words found in the grid.\n",
-        "\n",
-        "8. The code iterates through each cell on the grid (using nested loops) and starts a DFS search from each cell.\n",
-        "\n",
-        "9. During the DFS, it explores neighboring cells (up, down, left, right) if they haven't been visited before and if the current path matches a prefix in the Trie.\n",
-        "\n",
-        "10. When a word is found during the DFS traversal, it is added to the `result` set.\n",
-        "\n",
-        "11. The grid cell is marked as visited during the DFS by changing its character to \"#\", and it is restored to its original character after the DFS traversal.\n",
-        "\n",
-        "12. Finally, the set of unique words found is converted to a list and returned as the output.\n",
-        "\n",
-        "The code effectively uses a Trie data structure to optimize the search and ensures that each word is constructed from adjacent cells without reusing the same cell for the same word."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['eat', 'oath']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "board1 = [[\"o\",\"a\",\"a\",\"n\"],[\"e\",\"t\",\"a\",\"e\"],[\"i\",\"h\",\"k\",\"r\"],[\"i\",\"f\",\"l\",\"v\"]]\n",
-        "words1 = [\"oath\",\"pea\",\"eat\",\"rain\"]\n",
-        "print(findWords(board1, words1))  # Output: [\"eat\", \"oath\"]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "c748b879-a2d8-4298-a42c-c7f61f9c7fc6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "\n",
-        "board2 = [[\"a\",\"b\"],[\"c\",\"d\"]]\n",
-        "words2 = [\"abcb\"]\n",
-        "print(findWords(board2, words2))  # Output: []"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. **Building the Trie (`buildTrie` function):** \n",
-        "   - Suppose there are `N` words in the input list, and the average length of a word is `L`.\n",
-        "   - Building the Trie takes O(N * L) time, as we iterate through each word and insert each character into the Trie.\n",
-        "\n",
-        "2. **DFS Traversal (`dfs` function):** \n",
-        "   - In the worst case, the DFS function explores all possible paths in the grid.\n",
-        "   - The maximum number of recursive calls for each cell is 4 (up, down, left, right).\n",
-        "   - The maximum depth of the DFS is limited by the length of the longest word in the Trie, which is at most `L`.\n",
-        "   - Therefore, the DFS traversal for each cell takes $O(4^L)$ time.\n",
-        "\n",
-        "3. **Iterating Through the Grid (`findWords` function):**\n",
-        "   - In the worst case, we iterate through all `m` rows and `n` columns in the grid.\n",
-        "   - For each cell, we start a DFS traversal.\n",
-        "   - Therefore, iterating through the entire grid takes $O(m * n * 4^L)$ time.\n",
-        "\n",
-        "Overall, the time complexity of the code is $O(N * L + m * n * 4^L)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. **Trie Data Structure:**\n",
-        "   - The space required to store the Trie depends on the number of unique characters in all the words.\n",
-        "   - In the worst case, where all words are unique and have no common prefixes, the space complexity of the Trie is O(N * L).\n",
-        "   - In practice, it may be less if there are common prefixes among words.\n",
-        "\n",
-        "2. **DFS Stack (Recursive Calls):**\n",
-        "   - The depth of the recursive call stack during DFS is at most `L`, where `L` is the length of the longest word in the Trie.\n",
-        "   - Therefore, the space complexity for the recursive call stack is O(L).\n",
-        "\n",
-        "3. **Result Set (`result` set):**\n",
-        "   - The space used to store the result set depends on the number of valid words found in the grid.\n",
-        "   - In the worst case, when all words are found, the space complexity of the result set is O(N * L).\n",
-        "\n",
-        "Overall, the space complexity of the code is O(N * L) for the Trie and O(L) for the recursive call stack.\n",
-        "\n",
-        "In summary, the time complexity is dominated by the DFS traversal and is influenced by the number of words, their lengths, and the size of the grid. The space complexity is mainly determined by the Trie structure and the recursive call stack depth during DFS."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Word Search Paths:**\n",
-        "   Extend the basic word search exercise to find all possible paths (sequences of cells) that spell out a given word on the board. Return a list of paths if the word can be formed, otherwise an empty list.\n",
-        "\n",
-        "\n",
-        "2. **Reverse Word Search:**\n",
-        "   Modify the code to search for words in reverse order on the board. Given a 2D grid of characters and a list of words, find all words from the list that can be formed by traversing the board in reverse (right to left, bottom to top, etc.)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/08. Tries/README.md b/_build/html/_sources/08. Tries/README.md
deleted file mode 100644
index 22c3bcd..0000000
--- a/_build/html/_sources/08. Tries/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Trie Problems - Blind 75 LeetCode
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [208. Implement Trie Prefix Tree](https://leetcode.com/problems/implement-trie-prefix-tree/)         | Medium     |
-| [211. Design Add And Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/) | Medium |
-| [212. Word Search II](https://leetcode.com/problems/word-search-ii/)                                 | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb b/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb
deleted file mode 100644
index cb66b6a..0000000
--- a/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb	
+++ /dev/null
@@ -1,254 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 1046. Last Stone Weight\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Last Stone Weight problem on LeetCode, click here!](https://leetcode.com/problems/last-stone-weight/)\n",
-        "\n",
-        "---\n",
-        "You are given an array of integers `stones` where `stones[i]` is the weight of the `i`th stone.\n",
-        "\n",
-        "We are playing a game with the stones. On each turn, you choose the **heaviest two stones** and smash them together. Suppose the heaviest two stones have weights `x` and `y` with `x <= y`. The result of this smash is:\n",
-        "\n",
-        "1. If `x == y`, both stones are destroyed.\n",
-        "2. If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - x`.\n",
-        "\n",
-        "At the end of the game, there is **at most one** stone left.\n",
-        "\n",
-        "Return *the weight of the last remaining stone*. If there are no stones left, return `0`.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= stones.length <= 30`\n",
-        "- `1 <= stones[i] <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The \"Last Stone Weight\" problem involves a game played with a collection of stones, each stone having a certain weight. The goal is to find the weight of the last remaining stone after applying a specific set of rules.\n",
-        "\n",
-        "Here's how the game works:\n",
-        "\n",
-        "1. You start with an array of stones, where each stone is represented by its weight. The weights of the stones are given in the input as an array.\n",
-        "\n",
-        "2. In each turn of the game, you select the two heaviest stones from the remaining stones. If there is only one stone left, you've found the answer.\n",
-        "\n",
-        "3. If there are two stones with weights `x` and `y`, where `x` is less than or equal to `y`, the following happens:\n",
-        "   - If `x` is equal to `y`, both stones are completely destroyed, and they are removed from the array of stones.\n",
-        "   - If `x` is not equal to `y`, the stone with weight `x` is destroyed, and the stone with weight `y` now has a new weight of `y - x`. The stone with weight `x` is removed from the array, and the modified stone with weight `y - x` remains in the array.\n",
-        "\n",
-        "4. The game continues with the modified array of stones until there is at most one stone left.\n",
-        "\n",
-        "5. The goal is to find the weight of the last remaining stone, or if there are no stones left, return `0`.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "\n",
-        "def lastStoneWeight(stones):\n",
-        "    # Convert the input list to a max heap (negate the values to simulate a max heap)\n",
-        "    max_heap = [-stone for stone in stones]\n",
-        "    heapq.heapify(max_heap)\n",
-        "\n",
-        "    # Continue as long as there are more than 1 stone in the heap\n",
-        "    while len(max_heap) > 1:\n",
-        "        # Get the two heaviest stones\n",
-        "        y = -heapq.heappop(max_heap)  # Get the heaviest stone and negate it back to positive\n",
-        "        x = -heapq.heappop(max_heap)  # Get the second heaviest stone and negate it back to positive\n",
-        "\n",
-        "        # Calculate the new stone's weight after smashing\n",
-        "        if x != y:\n",
-        "            new_stone = y - x\n",
-        "            heapq.heappush(max_heap, -new_stone)  # Negate the value and push it back to the heap\n",
-        "\n",
-        "    # If there is one stone left, return its weight (negate it back to positive)\n",
-        "    if max_heap:\n",
-        "        return -max_heap[0]\n",
-        "    else:\n",
-        "        return 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a Python function called `lastStoneWeight` that takes a list of stone weights as input. It aims to find the weight of the last remaining stone after playing the stone smashing game described in the problem statement.\n",
-        "\n",
-        "Here's a step-by-step explanation of the code:\n",
-        "\n",
-        "1. We import the `heapq` library, which allows us to create and manipulate a heap (priority queue).\n",
-        "\n",
-        "2. Inside the `lastStoneWeight` function:\n",
-        "   - We create a max heap (negating stone weights to simulate a max heap) using the `heapq.heapify` function. This heap will keep track of the heaviest stones.\n",
-        "   \n",
-        "   - We enter a loop that continues as long as there are more than 1 stone in the heap.\n",
-        "   \n",
-        "   - Inside the loop, we:\n",
-        "     - Retrieve and negate the heaviest stone (`y`) from the heap.\n",
-        "     - Retrieve and negate the second heaviest stone (`x`) from the heap.\n",
-        "   \n",
-        "     - Calculate the new stone's weight after smashing (`y - x`), and negate it back to simulate a max heap.\n",
-        "     - Add the new stone back to the heap using `heapq.heappush`.\n",
-        "   \n",
-        "3. After the loop, if there is one stone left in the heap, we retrieve and negate it to get its actual weight, and return it as the result.\n",
-        "   \n",
-        "4. If there are no stones left (the heap is empty), we return `0`, indicating that no stones remain.\n",
-        "\n",
-        "5. Finally, we provide two test cases to demonstrate the function's usage and correctness.\n",
-        "\n",
-        "In summary, the code efficiently implements the stone smashing game by maintaining a max heap, selecting the heaviest stones in each turn, smashing them, and updating the heap until there is at most one stone left. It then returns the weight of the last remaining stone or 0 if there are no stones left."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "stones1 = [2, 7, 4, 1, 8, 1]\n",
-        "print(lastStoneWeight(stones1))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "b7e79514-4821-48d0-bb9d-0f2de8150945",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2: \n",
-        "\n",
-        "stones2 = [1]\n",
-        "print(lastStoneWeight(stones2))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided Python code for the \"Last Stone Weight\" problem:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The `heapq.heapify` operation to create the max heap from the input stone weights has a time complexity of O(n), where n is the number of stones.\n",
-        "\n",
-        "2. The loop that continues until there are more than 1 stone in the heap performs the following operations in each iteration:\n",
-        "   - Retrieving and negating the heaviest stone (`y`) from the heap, which takes O(log n) time.\n",
-        "   - Retrieving and negating the second heaviest stone (`x`) from the heap, also taking O(log n) time.\n",
-        "   - Calculating the new stone's weight and pushing it back into the heap using `heapq.heappush`, which takes O(log n) time.\n",
-        "\n",
-        "3. In the worst case, the loop runs until there is only one stone left, which requires approximately (n-1) iterations.\n",
-        "\n",
-        "Overall, the dominant time complexity is O(n) for the initial heap creation, and within the loop, each iteration takes O(log n) time. Therefore, the overall time complexity of the algorithm is O(n log n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The space complexity is primarily determined by the space required for the max heap. In the worst case, this heap can contain all the stones, so the space complexity for the heap is O(n).\n",
-        "\n",
-        "2. The rest of the variables used in the function (e.g., `x`, `y`, `new_stone`) require only constant space, so they do not significantly contribute to the space complexity.\n",
-        "\n",
-        "Therefore, the overall space complexity of the algorithm is O(n) due to the space used for the max heap.\n",
-        "\n",
-        "In summary:\n",
-        "- Time Complexity: $O(n\\ log\\ n)$\n",
-        "- Space Complexity: $O(n)$\n",
-        "\n",
-        "The algorithm is efficient enough to handle the problem's constraints, as the worst-case time and space complexities are both linearithmic in terms of the number of stones."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Kth Last Stone Weight**: Modify the problem to find the weight of the Kth last remaining stone, where K is an integer input. Your function should work efficiently for large values of K.\n",
-        "\n",
-        "\n",
-        "2. **Multiple Stones Smash**: Modify the problem so that instead of smashing two stones at a time, you can smash up to K stones at each turn, where K is an integer input. Determine the weight of the last remaining stone in this variation."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb b/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb
deleted file mode 100644
index 847dd60..0000000
--- a/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 295. Find Median from Data Stream\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Find Median from Data Stream problem on LeetCode, click here!](https://leetcode.com/problems/find-median-from-data-stream/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.\n",
-        "\n",
-        "Implement the MedianFinder class:\n",
-        "\n",
-        "- `MedianFinder()` initializes the `MedianFinder` object.\n",
-        "- `void addNum(int num)` adds the integer `num` from the data stream to the data structure.\n",
-        "- `double findMedian()` returns the median of all elements so far. Answers within $10^{-5}$ of the actual answer will be accepted.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- $-10^{5}$ <= `num` <= $10^{5}$\n",
-        "- There will be at least one element in the data structure before calling `findMedian`.\n",
-        "- At most $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-        "\n",
-        "**Follow-up:**\n",
-        "\n",
-        "1. If all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?\n",
-        "2. If `99%` of all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem, \"Find Median from Data Stream,\" deals with efficiently computing the median of a sequence of numbers as they are incrementally provided. The median is the middle value in an ordered list of numbers. When the list has an even number of elements, the median is the average of the two middle values.\n",
-        "\n",
-        "Here's a breakdown of the problem:\n",
-        "\n",
-        "1. **Initialization**: You are asked to implement a `MedianFinder` class that supports the following operations:\n",
-        "\n",
-        "    - `MedianFinder()`: Initializes the MedianFinder object.\n",
-        "    - `addNum(int num)`: Adds an integer `num` from the data stream to the data structure.\n",
-        "    - `findMedian()`: Returns the median of all elements added to the data structure.\n",
-        "\n",
-        "2. **Median Calculation**: The `findMedian` operation should efficiently compute the median, and the result should have a precision of at least $10^{-5}$ (i.e., answers within this range will be considered correct).\n",
-        "\n",
-        "3. **Examples**:\n",
-        "   \n",
-        "   - If you add numbers [1, 2], the median should be 1.5 because (1 + 2) / 2 = 1.5.\n",
-        "   - If you add numbers [1, 2, 3], the median should be 2 because it's the middle value.\n",
-        "   \n",
-        "4. **Constraints**:\n",
-        "\n",
-        "   - The integers provided in the data stream are in the range from $-10^5$ to $10^5$.\n",
-        "   - There will be at least one element in the data structure before calling `findMedian`.\n",
-        "   - At most, $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-        "\n",
-        "The problem can be solved by using two priority queues (heaps): one max-heap to store the smaller half of the numbers and one min-heap to store the larger half of the numbers. The max-heap ensures that the largest number in the smaller half is at the top, and the min-heap ensures that the smallest number in the larger half is at the top. These heaps are balanced to efficiently find the median when requested.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "\n",
-        "class MedianFinder:\n",
-        "\n",
-        "    def __init__(self):\n",
-        "        # Initialize two heaps: a max-heap (small_half) for the smaller half of numbers,\n",
-        "        # and a min-heap (large_half) for the larger half of numbers.\n",
-        "        self.small_half = []  # Max-heap for the smaller half of the numbers\n",
-        "        self.large_half = []  # Min-heap for the larger half of the numbers\n",
-        "\n",
-        "    def addNum(self, num: int) -> None:\n",
-        "        # Add the number to the appropriate heap based on its value\n",
-        "        if not self.small_half or num <= -self.small_half[0]:\n",
-        "            # If the number is less than or equal to the current maximum in the smaller half,\n",
-        "            # add it to the smaller half (max-heap)\n",
-        "            heapq.heappush(self.small_half, -num)\n",
-        "        else:\n",
-        "            # Otherwise, add it to the larger half (min-heap)\n",
-        "            heapq.heappush(self.large_half, num)\n",
-        "\n",
-        "        # Balance the heaps if necessary to ensure the size difference is at most 1\n",
-        "        if len(self.small_half) > len(self.large_half) + 1:\n",
-        "            # If the size of the smaller half is more than one greater than the size of the larger half,\n",
-        "            # move the maximum from the smaller half to the larger half to balance them.\n",
-        "            heapq.heappush(self.large_half, -heapq.heappop(self.small_half))\n",
-        "        elif len(self.large_half) > len(self.small_half):\n",
-        "            # If the size of the larger half is greater than the size of the smaller half,\n",
-        "            # move the minimum from the larger half to the smaller half to balance them.\n",
-        "            heapq.heappush(self.small_half, -heapq.heappop(self.large_half))\n",
-        "\n",
-        "    def findMedian(self) -> float:\n",
-        "        if len(self.small_half) == len(self.large_half):\n",
-        "            # If both halves have the same size, there's an even number of elements,\n",
-        "            # so the median is the average of the top elements in both heaps.\n",
-        "            return (-self.small_half[0] + self.large_half[0]) / 2.0\n",
-        "        else:\n",
-        "            # If the smaller half has more elements, it contains the median (odd number of elements).\n",
-        "            return -self.small_half[0]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a `MedianFinder` class for efficiently finding the median of a stream of numbers. It uses two heaps: a max-heap (`small_half`) to store the smaller half of the numbers and a min-heap (`large_half`) to store the larger half. Here's how the code works:\n",
-        "\n",
-        "1. The `__init__` method initializes the two heaps.\n",
-        "\n",
-        "2. The `addNum` method adds a number to the appropriate heap, ensuring that the heaps remain balanced. It does this by comparing the number to the current maximum in the smaller half and adding it to the appropriate heap. Then, it checks the sizes of the two heaps and rebalances them if necessary.\n",
-        "\n",
-        "3. The `findMedian` method calculates and returns the median. If both halves have the same size (even number of elements), it computes the average of the tops of both heaps. If the smaller half has more elements (odd number of elements), it returns the top of the max-heap, which is the median.\n",
-        "\n",
-        "In summary, this code efficiently maintains two heaps to divide the elements into smaller and larger halves, allowing for quick median retrieval. It offers a time complexity of O(log N) for adding elements and O(1) for finding the median, where N is the total number of elements processed. The space complexity is O(1) as the space used by the two heaps is independent of the input size."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1.5\n",
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "medianFinder = MedianFinder()\n",
-        "medianFinder.addNum(1)\n",
-        "medianFinder.addNum(2)\n",
-        "print(medianFinder.findMedian())  # Output: 1.5\n",
-        "medianFinder.addNum(3)\n",
-        "print(medianFinder.findMedian())  # Output: 2.0\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `MedianFinder` class:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. `MedianFinder()` Initialization: This operation has a time complexity of O(1). It simply initializes the two heaps.\n",
-        "\n",
-        "2. `addNum(int num)`: The time complexity for adding a number is O(log N), where N is the total number of elements processed so far. This is because, in the worst case, you might need to perform logarithmic operations on the heaps to maintain their balance.\n",
-        "\n",
-        "3. `findMedian()`: Finding the median has a time complexity of O(1). It simply involves accessing the tops of the two heaps, which takes constant time.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. `MedianFinder()` Initialization: The space complexity for initialization is O(1). It involves creating two empty heaps, which do not depend on the size of the data stream.\n",
-        "\n",
-        "2. `addNum(int num)`: The space complexity for adding a number is also O(1). It involves adding the number to one of the two heaps, which do not significantly affect the overall space complexity.\n",
-        "\n",
-        "3. `findMedian()`: The space complexity for finding the median is O(1). It doesn't require any additional data structures that grow with the input size.\n",
-        "\n",
-        "In summary, the time complexity of the `addNum` operation is O(log N), and the time complexity of the `findMedian` operation is O(1). The space complexity is O(1) as well, as the space used by the two heaps is independent of the size of the data stream. This solution efficiently maintains and retrieves the median, making it suitable for the given problem constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Generalized kth Element**: Extend the `MedianFinder` class to support finding the kth element in the data stream efficiently. This means finding the kth smallest or kth largest element in the stream.\n",
-        "\n",
-        "2. **Sliding Window Median**: Given an array and a sliding window size, find the median within the window as it slides over the array. This is an extension of the problem to a moving window scenario."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb b/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb
deleted file mode 100644
index 2e1522d..0000000
--- a/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb	
+++ /dev/null
@@ -1,239 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 703. Kth Largest Element in a Stream\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Kth Largest Element in a Stream problem on LeetCode, click here!](https://leetcode.com/problems/kth-largest-element-in-a-stream/)\n",
-        "\n",
-        "---\n",
-        "Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.\n",
-        "\n",
-        "Implement the `KthLargest` class with the following methods:\n",
-        "\n",
-        "- `KthLargest(int k, int[] nums)`: Initializes the object with the integer `k` and the stream of integers `nums`.\n",
-        "- `int add(int val)`: Appends the integer `val` to the stream and returns the element representing the `kth` largest element in the stream.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `k` <= $10^4$\n",
-        "- 0 <= `nums.length` <= $10^4$\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- $-10^4$ <= `val` <= $10^4$\n",
-        "- At most $10^4$ calls will be made to `add`.\n",
-        "- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem you're trying to solve is to design a class called `KthLargest` that can efficiently find the kth largest element in a stream of integers. You need to support two main operations:\n",
-        "\n",
-        "To solve this problem efficiently, you can use a min-heap data structure. Here's how the approach works:\n",
-        "\n",
-        "1. Initialize a min-heap to store the k largest elements. Initially, it's empty.\n",
-        "\n",
-        "2. In the `__init__` method, populate the min-heap with the first `k` elements from `nums`. This ensures that you have the k largest elements seen so far.\n",
-        "\n",
-        "3. Whenever you add a new element to the stream using the `add` method, follow these steps:\n",
-        "   - Add the new element to the min-heap.\n",
-        "   - If the size of the min-heap exceeds `k`, remove the smallest element from the min-heap. This ensures that you always have the k largest elements in the min-heap.\n",
-        "   - The smallest element in the min-heap (the root) will always represent the kth largest element in the stream.\n",
-        "\n",
-        "Here's why this approach works efficiently:\n",
-        "\n",
-        "- By using a min-heap, you can quickly maintain the k largest elements, and finding the smallest element in the heap (the root) takes constant time.\n",
-        "\n",
-        "- When you add a new element, the min-heap's size is kept at most `k`, which ensures that you only track the k largest elements and discard the smaller ones.\n",
-        "\n",
-        "- The time complexity for adding an element is O(log k), which is very efficient compared to sorting the entire stream, which would be O(n log n).\n",
-        "\n",
-        "- This approach meets the constraints of the problem, including handling large streams and having a low time complexity for both initialization and adding elements.\n",
-        "\n",
-        "In summary, the min-heap approach efficiently tracks the kth largest element in the stream by maintaining a heap of the k largest elements seen so far, updating it as new elements are added. This approach provides a fast and scalable solution to the problem.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "from typing import List  # Import the List type from the typing module\n",
-        "\n",
-        "class KthLargest:\n",
-        "\n",
-        "    def __init__(self, k: int, nums: List[int]):\n",
-        "        # Initialize the KthLargest object with k and nums.\n",
-        "        self.k = k\n",
-        "        # Create a min-heap to store the k largest elements.\n",
-        "        self.min_heap = []\n",
-        "        # Populate the min-heap with the first k elements from nums.\n",
-        "        for num in nums:\n",
-        "            self.add(num)\n",
-        "\n",
-        "    def add(self, val: int) -> int:\n",
-        "        # Add val to the min-heap.\n",
-        "        heapq.heappush(self.min_heap, val)\n",
-        "        # If the size of the min-heap exceeds k, remove the smallest element.\n",
-        "        if len(self.min_heap) > self.k:\n",
-        "            heapq.heappop(self.min_heap)\n",
-        "        # The root of the min-heap is the kth largest element.\n",
-        "        return self.min_heap[0]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- We start by importing the necessary modules:\n",
-        "  - `heapq`: This module provides functions to create and manipulate heaps.\n",
-        "  - `List` from the `typing` module: This is used to specify the type of the `nums` parameter.\n",
-        "\n",
-        "- We define the `KthLargest` class:\n",
-        "  - The `__init__` method initializes the object with the integer `k` and the list of integers `nums`.\n",
-        "  - It also creates an empty min-heap called `self.min_heap` to store the k largest elements.\n",
-        "  - It populates the min-heap with the first `k` elements from `nums` by calling the `add` method.\n",
-        "\n",
-        "- The `add` method:\n",
-        "  - Adds the new integer `val` to the min-heap using `heapq.heappush`. This maintains the min-heap property.\n",
-        "  - Checks if the size of the min-heap exceeds `k`. If it does, it removes the smallest element (the k+1th largest) using `heapq.heappop`.\n",
-        "  - Finally, it returns the smallest element in the min-heap, which is always the kth largest element.\n",
-        "\n",
-        "Overall, this code implements a class that efficiently finds the kth largest element in a stream of integers by maintaining a min-heap of the k largest elements seen so far.\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n",
-            "5\n",
-            "5\n",
-            "8\n",
-            "8\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "kthLargest = KthLargest(3, [4, 5, 8, 2])\n",
-        "print(kthLargest.add(3))  # Output: 4\n",
-        "print(kthLargest.add(5))  # Output: 5\n",
-        "print(kthLargest.add(10)) # Output: 5\n",
-        "print(kthLargest.add(9))  # Output: 8\n",
-        "print(kthLargest.add(4))  # Output: 8"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `KthLargest` class using the min-heap approach:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. `__init__` method:\n",
-        "   - In the `__init__` method, you iterate over the first `k` elements in `nums` and add them to the min-heap. Each `heapq.heappush` operation takes O(log k) time.\n",
-        "   - Therefore, the time complexity of the `__init__` method is O(k * log k).\n",
-        "\n",
-        "2. `add` method:\n",
-        "   - In the `add` method, you perform the following operations:\n",
-        "     - `heapq.heappush`: O(log k) to add the new element to the min-heap.\n",
-        "     - If the size of the min-heap exceeds `k`, you perform `heapq.heappop`, which is also O(log k).\n",
-        "     - Finally, you return the smallest element from the min-heap, which is O(1) because it's always the root.\n",
-        "   - Overall, the time complexity of the `add` method is O(log k).\n",
-        "\n",
-        "3. Overall, if you make `n` calls to the `add` method, the total time complexity is O(n * log k), where `n` is the total number of elements added to the stream.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The space complexity is determined by the space used to store the min-heap and the instance variables.\n",
-        "2. The min-heap stores at most `k` elements at any given time, so its space complexity is O(k).\n",
-        "3. The instance variables (such as `self.k` and `self.min_heap`) have constant space requirements.\n",
-        "4. Therefore, the overall space complexity of the `KthLargest` class is O(k).\n",
-        "\n",
-        "In summary, the time complexity of the `KthLargest` class is O(n * log k) for `n` add operations, and the space complexity is O(k), where `k` is the parameter passed during initialization. This implementation efficiently maintains the kth largest element in the stream while meeting the problem's constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Dynamic k:** Modify the `KthLargest` class to support dynamic changes in the value of `k`. Implement a method `update_k` that allows changing the value of `k` during the lifetime of the object. Ensure that the object can still correctly find the kth largest element based on the updated value of `k`.\n",
-        "\n",
-        "\n",
-        "2. **Implement kth Smallest:** Create a new class called `KthSmallest` that finds the kth smallest element in a stream instead of the kth largest. You may need to modify the data structure used in the implementation."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/09. Heap - Priority Queue/README.md b/_build/html/_sources/09. Heap - Priority Queue/README.md
deleted file mode 100644
index 2eac70c..0000000
--- a/_build/html/_sources/09. Heap - Priority Queue/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Heap / Priority Queue Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [703. Kth Largest Element in a Stream](https://leetcode.com/problems/kth-largest-element-in-a-stream/)     | Easy       |
-| [1046. Last Stone Weight](https://leetcode.com/problems/last-stone-weight/)     | Easy       |
-| [295. Find Median From Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)     | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb b/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb
deleted file mode 100644
index 228a62e..0000000
--- a/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb	
+++ /dev/null
@@ -1,282 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 39. Combination Sum\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Combination Sum problem on LeetCode, click here!](https://leetcode.com/problems/combination-sum/)\n",
-        "\n",
-        "---\n",
-        "Given an array of **distinct** integers `candidates` and a target integer `target`, return *a list of all **unique combinations** of `candidates` where the chosen numbers sum to the `target`*. You may return the combinations in **any order**. \n",
-        "\n",
-        "The **same** number may be chosen from `candidates` an **unlimited number** of times. Two combinations are unique if the **frequency** of at least one of the chosen numbers is different.\n",
-        "\n",
-        "The test cases are generated such that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= candidates.length <= 30`\n",
-        "- `2 <= candidates[i] <= 40`\n",
-        "- All elements of `candidates` are **distinct**.\n",
-        "- `1 <= target <= 40`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to find all unique combinations of numbers from a given array (candidates) such that their sum equals a target number. Here are the details of the problem:\n",
-        "\n",
-        "- You are given an array of distinct integers called \"candidates.\"\n",
-        "- You are also given a target integer called \"target.\"\n",
-        "\n",
-        "The goal is to find all unique combinations of numbers from the candidates array where the sum of the selected numbers is equal to the target. You can use the same number from the candidates array an unlimited number of times. A combination is considered unique if it has a different frequency (i.e., a different number of occurrences) of at least one chosen number compared to other combinations.\n",
-        "\n",
-        "For example:\n",
-        "\n",
-        "**Example 1:**\n",
-        "```python\n",
-        "Input: candidates = [2, 3, 6, 7], target = 7\n",
-        "Output: [[2, 2, 3], [7]]\n",
-        "```\n",
-        "In this example, there are two unique combinations that sum up to the target:\n",
-        "- 2 + 2 + 3 = 7\n",
-        "- 7 = 7\n",
-        "\n",
-        "**Example 2:**\n",
-        "```python\n",
-        "Input: candidates = [2, 3, 5], target = 8\n",
-        "Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n",
-        "```\n",
-        "Here, there are three unique combinations that sum up to the target.\n",
-        "\n",
-        "**Example 3:**\n",
-        "```python\n",
-        "Input: candidates = [2], target = 1\n",
-        "Output: []\n",
-        "```\n",
-        "In this case, there are no combinations that can be formed from the candidates to reach the target of 1, so the output is an empty list.\n",
-        "\n",
-        "The problem asks you to find and return these combinations in any order.\n",
-        "\n",
-        "The constraints for this problem include the length of the candidates array, the values of candidates, and the target value.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def combinationSum(candidates, target):\n",
-        "    # Define a recursive DFS function to find combinations\n",
-        "    def dfs(remaining, start, path):\n",
-        "        # If the remaining target is 0, we found a valid combination\n",
-        "        if remaining == 0:\n",
-        "            result.append(path)\n",
-        "            return\n",
-        "        # If the remaining target is negative, this path is invalid\n",
-        "        if remaining < 0:\n",
-        "            return\n",
-        "        # Iterate through candidates starting from 'start' to avoid duplicates\n",
-        "        for i in range(start, len(candidates)):\n",
-        "            # Explore the current candidate by subtracting it from the remaining target\n",
-        "            # Add the current candidate to the path\n",
-        "            dfs(remaining - candidates[i], i, path + [candidates[i]])\n",
-        "\n",
-        "    result = []  # Initialize an empty list to store the result\n",
-        "    candidates.sort()  # Sort the candidates for deduplication and early stopping\n",
-        "    dfs(target, 0, [])  # Start the DFS from the target value with an empty path\n",
-        "    return result  # Return the list of unique combinations"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `combinationSum` function takes an array of distinct integers `candidates` and a target integer `target`. It returns a list of all unique combinations of `candidates` where the chosen numbers sum to the `target`. Each number in the `candidates` array can be used an unlimited number of times.\n",
-        "\n",
-        "Here's an overview of how the code works:\n",
-        "\n",
-        "1. The `combinationSum` function is defined, and it takes two arguments: `candidates` and `target`.\n",
-        "\n",
-        "2. Inside the function, there is a nested helper function called `backtrack`. This function is responsible for the actual combination generation using a recursive approach.\n",
-        "\n",
-        "3. The `backtrack` function is called with three arguments: `start`, `target`, and `path`. The `start` variable helps keep track of the current index in the `candidates` array, `target` keeps track of the remaining sum to be achieved, and `path` is a list that stores the current combination.\n",
-        "\n",
-        "4. Within the `backtrack` function, there are three main conditional statements:\n",
-        "   - If `target` becomes zero, it means we have found a combination that sums up to the target, so we add `path` to the `result` list.\n",
-        "   - If `target` becomes negative, it means the current combination doesn't work, so we return without doing anything.\n",
-        "   - Otherwise, we enter a loop that iterates over the `candidates` array, starting from the current index `start`.\n",
-        "\n",
-        "5. In the loop, the `backtrack` function is called recursively with the updated `target` and `path` after including the current candidate. This process explores different combinations by considering the current candidate or moving to the next candidate.\n",
-        "\n",
-        "6. After the loop completes, the `result` list contains all unique combinations that sum to the target.\n",
-        "\n",
-        "7. The `candidates` array is sorted to optimize the search. Sorting helps in avoiding unnecessary recursive branches and reduces the number of explored combinations.\n",
-        "\n",
-        "8. Finally, the `backtrack` function is initially called from the `combinationSum` function with `start` set to 0, `target` set to the original target value, and an empty `path`. The result is returned as a list of unique combinations.\n",
-        "\n",
-        "The code includes a few example cases to demonstrate how the function works with different inputs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[2, 2, 3], [7]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example 1\n",
-        "candidates1 = [2, 3, 6, 7]\n",
-        "target1 = 7\n",
-        "print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "candidates2 = [2, 3, 5]\n",
-        "target2 = 8\n",
-        "print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "candidates3 = [2]\n",
-        "target3 = 1\n",
-        "print(combinationSum(candidates3, target3))  # Output: []"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `combinationSum` function.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this function is influenced by the number of recursive calls made by the `backtrack` function and the amount of work done within each call. In the worst case, the function will explore all possible combinations.\n",
-        "\n",
-        "The worst-case scenario occurs when we have many combinations to reach the target. The time complexity can be expressed as $O(2^n)$, where 'n' is the maximum number of recursive calls. In this context, 'n' corresponds to the number of elements in the `candidates` list, and it may be up to 30 (as per the constraints). However, the actual number of combinations explored is generally much smaller because we eliminate branches when the target becomes negative, and we skip over candidates that cannot lead to a solution. The sorting step also helps to optimize the search.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity is determined by the auxiliary space used for storing the results and the stack space for the recursive calls.\n",
-        "\n",
-        "1. The space required for the `result` list can be up to $O(2^n)$ in the worst case because each combination is stored.\n",
-        "\n",
-        "2. The stack space for the recursive calls can also be up to O(n) in the worst case, where 'n' is the number of elements in the `candidates` list.\n",
-        "\n",
-        "Therefore, the overall space complexity of the function is $O(2^n + n)$. However, in practice, the space used for the results list is often the dominant factor, and the actual space used may be much smaller than 2^n because not all combinations are explored.\n",
-        "\n",
-        "In summary, the time complexity is exponential but typically smaller in practice due to optimization, and the space complexity is influenced by the number of results and the depth of the recursion."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Find the Number of Unique Combinations:** Instead of listing all unique combinations, modify the function to return the count of unique combinations that sum to the target.\n",
-        "\n",
-        "2. **Combinations with a Maximum Value:** Add a maximum value constraint for each combination, so they cannot exceed a certain value. Modify the code to find combinations respecting this constraint."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/10. Backtracking/79. Word Search.ipynb b/_build/html/_sources/10. Backtracking/79. Word Search.ipynb
deleted file mode 100644
index cee509b..0000000
--- a/_build/html/_sources/10. Backtracking/79. Word Search.ipynb	
+++ /dev/null
@@ -1,298 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 79. Word Search\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Word Search problem on LeetCode, click here!](https://leetcode.com/problems/word-search/)\n",
-        "\n",
-        "---\n",
-        "Given an `m x n` grid of characters `board` and a string `word`, return *`true` if `word` exists in the grid.*\n",
-        "\n",
-        "The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == board.length`\n",
-        "- `n = board[i].length`\n",
-        "- `1 <= m, n <= 6`\n",
-        "- `1 <= word.length <= 15`\n",
-        "- `board` and `word` consists of only lowercase and uppercase English letters.\n",
-        "\n",
-        "**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether a given word can be constructed from characters in an `m x n` grid of characters. The word can be formed by moving sequentially through horizontally or vertically neighboring cells in the grid, and you cannot use the same cell more than once.\n",
-        "\n",
-        "Here are the specifics of the problem:\n",
-        "\n",
-        "- You are given an `m x n` grid, which is a 2D board of characters.\n",
-        "- You are also given a string `word` consisting of lowercase and uppercase English letters.\n",
-        "- Your task is to determine if the word can be found in the grid by moving from one cell to an adjacent cell (horizontally or vertically) and collecting the characters sequentially to form the word.\n",
-        "- You can't use the same cell more than once when forming the word.\n",
-        "- If it's possible to form the word in the grid, return `True`; otherwise, return `False`.\n",
-        "\n",
-        "For example, consider the following grid and word:\n",
-        "\n",
-        "```python\n",
-        "board = [[\"A\", \"B\", \"C\", \"E\"],\n",
-        "         [\"S\", \"F\", \"C\", \"S\"],\n",
-        "         [\"A\", \"D\", \"E\", \"E\"]]         \n",
-        "word = \"ABCCED\"\n",
-        "```\n",
-        "\n",
-        "In this case, the word \"ABCCED\" can be formed by moving through the cells (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) in the grid. Therefore, the function should return `True`.\n",
-        "\n",
-        "Conversely, if the word is impossible to construct, such as in the case of the word \"ABCB\" for the same grid, where there's no valid path that forms the word, the function should return `False`.\n",
-        "\n",
-        "The problem involves searching for a path through the grid that matches the characters in the word while respecting the constraints mentioned. It can be solved using depth-first search (DFS) or backtracking.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def exist(self, board: List[List[str]], word: str) -> bool:\n",
-        "        def dfs(i, j, k):\n",
-        "            # Check if the current position is out of bounds or doesn't match the character in the word.\n",
-        "            if not (0 <= i < len(board)) or not (0 <= j < len(board[0])) or board[i][j] != word[k]:\n",
-        "                return False\n",
-        "            \n",
-        "            # If we have matched all characters in the word, return True.\n",
-        "            if k == len(word) - 1:\n",
-        "                return True\n",
-        "\n",
-        "            # Temporarily mark the current cell to prevent revisiting.\n",
-        "            tmp, board[i][j] = board[i][j], \"/\"\n",
-        "\n",
-        "            # Explore adjacent cells by making recursive calls in all four directions.\n",
-        "            found = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)\n",
-        "\n",
-        "            # Restore the original character in the cell.\n",
-        "            board[i][j] = tmp\n",
-        "\n",
-        "            return found\n",
-        "\n",
-        "        # Iterate through all cells in the board.\n",
-        "        for i in range(len(board)):\n",
-        "            for j in range(len(board[0])):\n",
-        "                # Check if the 'dfs' function starting from this cell returns True.\n",
-        "                if dfs(i, j, 0):\n",
-        "                    return True\n",
-        "\n",
-        "        # If no match is found after exploring all cells, return False.\n",
-        "        return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. Import the necessary module:\n",
-        "   - `from typing import List`: This line imports the `List` type from the `typing` module, which is used to specify the type of the `board` parameter in the method signature.\n",
-        "\n",
-        "2. Define the `Solution` class:\n",
-        "   - `class Solution:`: This line defines a class called `Solution` to encapsulate the solution for the word search problem.\n",
-        "\n",
-        "3. Define the `exist` method:\n",
-        "   - `def exist(self, board: List[List[str]], word: str) -> bool:`: This method is a member of the `Solution` class and takes two parameters: `board`, which is a 2D list of characters, and `word`, which is a string. It returns a boolean value, indicating whether the word exists in the board.\n",
-        "\n",
-        "4. Implement the inner `dfs` function:\n",
-        "   - `def dfs(i, j, k):`: This is a helper function used for depth-first search (DFS). It is called recursively to explore the board and search for the word. It takes three parameters: `i` and `j` representing the current position on the board, and `k` representing the current index in the `word` string.\n",
-        "\n",
-        "   - Inside `dfs`, it checks if the current position is out of bounds or if the character at the current position on the board does not match the character at the current index in the `word`. If either of these conditions is met, it returns `False`.\n",
-        "\n",
-        "   - If `k` is equal to the length of the `word` minus one, it means we have successfully matched all characters in the `word`, so it returns `True`.\n",
-        "\n",
-        "   - The function then temporarily replaces the character at the current position with a marker, \"/\", to prevent revisiting the same cell. It then explores adjacent cells by making recursive calls in all four directions: up, down, left, and right. If any of these recursive calls return `True`, it propagates the `True` result up the call stack. After the exploration is complete, it restores the original character in the cell.\n",
-        "\n",
-        "5. Iterate through the board:\n",
-        "   - The outer loop iterates through all cells in the `board` using two nested loops. For each cell, it checks if the `dfs` function starting from that cell (with `k` initially set to 0) returns `True`. If it does, it means the word exists, so the method returns `True`.\n",
-        "\n",
-        "6. Return `False`:\n",
-        "   - If the loop completes without finding the word in the board, the method returns `False`.\n",
-        "\n",
-        "To use this code, you can create an instance of the `Solution` class and call the `exist` method on that instance, passing the `board` and the `word` as arguments to check if the word exists in the board."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test case 1: Word \"ABCCED\" can be found in the grid\n",
-        "board1 = [[\"A\",\"B\",\"C\",\"E\"],\n",
-        "          [\"S\",\"F\",\"C\",\"S\"],\n",
-        "          [\"A\",\"D\",\"E\",\"E\"]]\n",
-        "word1 = \"ABCCED\"\n",
-        "print(solution.exist(board1, word1))  # Expected output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "\n",
-        "# Test case 2: Word \"SEE\" can be found in the grid\n",
-        "word2 = \"SEE\"\n",
-        "print(solution.exist(board1, word2))  # Expected output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "\n",
-        "# Test case 3: Word \"ABCB\" cannot be found in the grid\n",
-        "word3 = \"ABCB\"\n",
-        "print(solution.exist(board1, word3))  # Expected output: False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `exist` method in the given code.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The primary work in this code is done by the `dfs` function, which performs a depth-first search through the grid. The time complexity of the `dfs` function is $O(4^(n))$, where \"n\" is the length of the `word`. This is because, in the worst case, the function explores all possible directions (up, down, left, right) for each character in the `word`.\n",
-        "\n",
-        "The outer loops iterate through all cells in the `board`, which has dimensions `m x n`. So, the total number of iterations of `dfs` is O(m * n). However, since the `dfs` function has a higher time complexity, the overall time complexity is still dominated by the `dfs` function.\n",
-        "\n",
-        "Therefore, the overall time complexity of the `exist` method is $O(m * n * 4^n)$, where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The primary space usage comes from the recursive calls in the `dfs` function. The depth of the recursion can be at most equal to the length of the `word`, which is O(n).\n",
-        "\n",
-        "2. Additionally, the `tmp` variable and the recursive calls on the call stack consume a small amount of additional memory.\n",
-        "\n",
-        "3. The marking of visited cells with the \"/\" character temporarily modifies the `board` in-place but doesn't significantly contribute to the space complexity.\n",
-        "\n",
-        "Therefore, the overall space complexity is O(n) due to the recursive call stack.\n",
-        "\n",
-        "In summary, the time complexity is $O(m * n * 4^n)$, and the space complexity is O(n), where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word. The time complexity is exponential in the worst case due to the depth-first search through all possible paths in the grid."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the code to find all unique words that can be constructed from the grid. Instead of returning `True` or `False`, return a list of words found. Ensure that the same cell is not used more than once in constructing each word.\n",
-        "\n",
-        "2. **Board Generation:** Create a function to generate random grids of characters and then use the `exist` method to solve them. Test the efficiency and correctness of your code by generating larger boards and words."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/10. Backtracking/README.md b/_build/html/_sources/10. Backtracking/README.md
deleted file mode 100644
index 0d1ea13..0000000
--- a/_build/html/_sources/10. Backtracking/README.md	
+++ /dev/null
@@ -1,17 +0,0 @@
-# Backtracking Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [39. Combination Sum](https://leetcode.com/problems/combination-sum/)        | Medium     |
-| [79. Word Search](https://leetcode.com/problems/word-search/)                | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb b/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb
deleted file mode 100644
index 07351b6..0000000
--- a/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb	
+++ /dev/null
@@ -1,317 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 133. Clone Graph\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Clone Graph problem on LeetCode, click here!](https://leetcode.com/problems/clone-graph/)\n",
-        "\n",
-        "---\n",
-        "Given a reference of a node in a **connected** undirected graph.\n",
-        "\n",
-        "Return a **deep copy** (clone) of the graph.\n",
-        "\n",
-        "Each node in the graph contains a value (`int`) and a list (`List[Node]`) of its neighbors.\n",
-        "\n",
-        "```\n",
-        "class Node {\n",
-        "    public int val;\n",
-        "    public List<Node> neighbors;\n",
-        "}\n",
-        "```\n",
-        "\n",
-        "**Test case format:**\n",
-        "\n",
-        "For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with `val == 1`, the second node with `val == 2`, and so on. The graph is represented in the test case using an adjacency list.\n",
-        "\n",
-        "**An adjacency list** is a collection of unordered **lists** used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.\n",
-        "\n",
-        "The given node will always be the first node with `val = 1`. You must return the **copy of the given node** as a reference to the cloned graph.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the graph is in the range `[0, 100]`.\n",
-        "- `1 <= Node.val <= 100`\n",
-        "- `Node.val` is unique for each node.\n",
-        "- There are no repeated edges and no self-loops in the graph.\n",
-        "- The Graph is connected and all nodes can be visited starting from the given node."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to clone a connected undirected graph, which means we need to create a deep copy of the original graph while preserving its structure and relationships between nodes. The graph is represented using adjacency lists, where each node has a value (an integer) and a list of its neighbors.\n",
-        "\n",
-        "Here are the key aspects of the problem:\n",
-        "\n",
-        "1. **Input**: You are given a reference to one of the nodes in the original graph. This reference node serves as the entry point to the graph. The entire graph can be explored starting from this node.\n",
-        "\n",
-        "2. **Output**: The goal is to create a deep copy of the graph and return a reference to the cloned graph. The cloned graph should have the same structure and relationships as the original one but should be a separate instance in memory.\n",
-        "\n",
-        "3. **Graph Structure**: The graph is composed of nodes (vertices) and edges. Nodes have values (integers) and are connected to other nodes through edges (undirected connections). Each node has a list of neighbors, which are other nodes it is connected to.\n",
-        "\n",
-        "4. **Connected Graph**: The problem assumes that the graph is connected, meaning that you can start from the reference node provided and traverse the entire graph by following the edges. This ensures that there are no isolated components in the graph.\n",
-        "\n",
-        "To solve this problem, you typically need to perform a graph traversal (e.g., depth-first search or breadth-first search) to visit all the nodes in the original graph, create corresponding nodes in the clone, and establish the same connections between nodes in the cloned graph as in the original one. To avoid duplication of nodes, you also need to keep track of visited nodes to ensure that the cloning process is efficient and that nodes are not duplicated in the clone.\n",
-        "\n",
-        "The problem can be challenging because you need to create a deep copy of the graph while handling cyclic dependencies between nodes and ensuring that the structure and relationships of the original graph are preserved in the clone. The solution requires recursion or a data structure to maintain the mapping between original nodes and their clones to achieve an efficient and accurate cloning process.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a Node.\n",
-        "class Node:\n",
-        "    def __init__(self, val=0, neighbors=None):\n",
-        "        self.val = val\n",
-        "        self.neighbors = neighbors if neighbors is not None else []\n",
-        "\n",
-        "class Solution:\n",
-        "    def cloneGraph(self, node: 'Node') -> 'Node':\n",
-        "        if not node:\n",
-        "            return None\n",
-        "\n",
-        "        visited = {}  # Dictionary to keep track of visited nodes\n",
-        "\n",
-        "        def dfs(original_node):\n",
-        "            if original_node in visited:\n",
-        "                return visited[original_node]\n",
-        "\n",
-        "            # Create a copy of the original node\n",
-        "            copy_node = Node(original_node.val)\n",
-        "            visited[original_node] = copy_node\n",
-        "\n",
-        "            # Recursively clone the neighbors of the original node\n",
-        "            for neighbor in original_node.neighbors:\n",
-        "                copy_node.neighbors.append(dfs(neighbor))\n",
-        "\n",
-        "            return copy_node\n",
-        "\n",
-        "        return dfs(node)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The given code defines a Python class, `Node`, and a solution class, `Solution`, to clone a connected undirected graph represented by nodes and their neighbors. Here's an explanation of the code:\n",
-        "\n",
-        "1. The `Node` class defines a node in the graph. Each node has a value (`val`) which is an integer and a list of neighbors (`neighbors`) that are other nodes in the graph.\n",
-        "\n",
-        "2. The `Solution` class contains the method `cloneGraph`, which takes a reference to the first node of the graph (`node`) as input and returns a deep copy of the entire graph, preserving the structure and relationships between nodes.\n",
-        "\n",
-        "3. Inside the `cloneGraph` method:\n",
-        "   - It first checks if the input `node` is `None`. If it's `None`, it means there is no graph to clone, and the method returns `None`.\n",
-        "\n",
-        "   - It initializes a `visited` dictionary to keep track of visited nodes. This dictionary is used to ensure that each node is cloned only once to avoid duplication.\n",
-        "\n",
-        "   - The `dfs` (depth-first search) function is defined within the `cloneGraph` method. It takes an original node as input and returns its corresponding clone. This function performs the actual cloning of the graph.\n",
-        "\n",
-        "   - Inside the `dfs` function:\n",
-        "     - It checks if the original node has been visited before by looking it up in the `visited` dictionary. If it has been visited, it returns the corresponding clone from the `visited` dictionary.\n",
-        "\n",
-        "     - If the original node is not visited, it creates a new node, `copy_node`, with the same value as the original node. This is the first step in creating the clone of the original node.\n",
-        "\n",
-        "     - It then recursively clones the neighbors of the original node by iterating through the `neighbors` list of the original node and appending the corresponding cloned neighbors to the `neighbors` list of the `copy_node`. This step ensures that the relationships between nodes are preserved in the clone.\n",
-        "\n",
-        "     - The `copy_node` is added to the `visited` dictionary with the original node as the key and the clone as the value.\n",
-        "\n",
-        "     - Finally, the `copy_node` is returned as the result of the DFS for the original node.\n",
-        "\n",
-        "4. The `cloneGraph` method returns the result of the DFS on the input `node`, which is the clone of the entire graph with its structure and relationships intact.\n",
-        "\n",
-        "In summary, the code uses depth-first search (DFS) to traverse the original graph, creating a clone for each node and its neighbors. It ensures that each node is cloned only once to avoid duplication and returns a reference to the cloned graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1, 2, 4]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "def get_adjacency_list(node):\n",
-        "    visited = {}  # Dictionary to keep track of visited nodes\n",
-        "\n",
-        "    def dfs(node):\n",
-        "        if not node:\n",
-        "            return []\n",
-        "\n",
-        "        if node in visited:\n",
-        "            return visited[node]\n",
-        "\n",
-        "        neighbors = [n.val for n in node.neighbors]\n",
-        "        visited[node] = neighbors\n",
-        "\n",
-        "        for neighbor in node.neighbors:\n",
-        "            dfs(neighbor)\n",
-        "\n",
-        "    dfs(node)\n",
-        "    result = []\n",
-        "\n",
-        "    def get_node_and_neighbors(node):\n",
-        "        result.append([node.val] + visited[node])\n",
-        "\n",
-        "    get_node_and_neighbors(node)\n",
-        "    return result\n",
-        "\n",
-        "\n",
-        "# Testing the code with the provided examples\n",
-        "adjList1 = [[2, 4], [1, 3], [2, 4], [1, 3]]\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Create the graph from adjList\n",
-        "nodes = [Node(i) for i in range(1, len(adjList1) + 1)]\n",
-        "for i, adj in enumerate(adjList1):\n",
-        "    nodes[i].neighbors = [nodes[j - 1] for j in adj]\n",
-        "\n",
-        "# Clone the graph and get its adjacency list\n",
-        "cloned_node1 = solution.cloneGraph(nodes[0])\n",
-        "cloned_adjList1 = get_adjacency_list(cloned_node1)\n",
-        "\n",
-        "# Print the cloned graph as an adjacency list\n",
-        "print(cloned_adjList1)\n",
-        "\n",
-        "# Output should match adjList1\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "# Input: adjList = [[]]\n",
-        "\n",
-        "adjList2 = [[]]\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Create the graph from adjList\n",
-        "node1 = Node(1)\n",
-        "\n",
-        "# Clone the graph and get its adjacency list\n",
-        "cloned_node1 = solution.cloneGraph(node1)\n",
-        "cloned_adjList2 = get_adjacency_list(cloned_node1)\n",
-        "\n",
-        "# Print the cloned graph as an adjacency list\n",
-        "print(cloned_adjList2)\n",
-        "\n",
-        "# Output should be [[]], which represents an empty graph"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for cloning a graph using depth-first search (DFS).\n",
-        "\n",
-        "Time Complexity:\n",
-        "1. Creating the graph: The time complexity to create the original graph from `adjList` is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. We iterate through each node and its neighbors in `adjList`.\n",
-        "\n",
-        "2. Cloning the graph (DFS): The depth-first search (DFS) is used to traverse the original graph and create a clone. The time complexity of DFS is O(V + E), as we visit each node and each edge once.\n",
-        "\n",
-        "Overall, the time complexity is O(V + E) for both creating the original graph and cloning it.\n",
-        "\n",
-        "Space Complexity:\n",
-        "1. Storage for the original graph: We need to store the original graph, which includes all nodes and their adjacency lists. In the worst case, this requires O(V + E) space, where V is the number of nodes, and E is the number of edges.\n",
-        "\n",
-        "2. Storage for the visited dictionary: The `visited` dictionary stores mappings from original nodes to their corresponding clones. In the worst case, this requires O(V) space because each node is visited once.\n",
-        "\n",
-        "3. Recursive stack for DFS: The space used for the recursive call stack during DFS can be up to O(V) in the worst case, where V is the number of nodes.\n",
-        "\n",
-        "Overall, the space complexity is O(V + E) for storing the original graph, and O(V) for auxiliary data structures, including the `visited` dictionary and the DFS call stack. Therefore, the overall space complexity is O(V + E).\n",
-        "\n",
-        "In summary, the time and space complexities of the provided code for cloning a graph are both O(V + E), where V is the number of nodes, and E is the number of edges in the graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Cyclic Graph**: Design a graph with cycles (nodes connected in a cycle). Test if the code can correctly clone graphs with cycles and doesn't fall into an infinite loop during traversal.\n",
-        "\n",
-        "\n",
-        "2. **Partial Graph Cloning**: Modify the code to clone only a subset of the graph starting from a given reference node, rather than the entire graph. Implement this as an extension to the problem."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb b/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb
deleted file mode 100644
index e0235d6..0000000
--- a/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb	
+++ /dev/null
@@ -1,248 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 178. Graph Valid Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Graph Valid Tree problem on LintCode, click here!](https://www.lintcode.com/problem/178/)\n",
-        "\n",
-        "---\n",
-        "Given `n` nodes labeled from `0` to `n - 1` and a list of `undirected` edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- You can assume that no duplicate edges will appear in edges. Since all edges are `undirected`, `[0, 1]` is the same as `[1, 0]` and thus will not appear together in edges."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether a given set of undirected edges can form a valid tree. In this context, a \"tree\" is a specific type of graph with the following properties:\n",
-        "\n",
-        "1. It is a connected graph, meaning there is a path between any pair of nodes in the graph.\n",
-        "2. It has no cycles, which means there are no closed loops or circuits in the graph.\n",
-        "3. It has exactly `n - 1` edges, where `n` is the number of nodes in the tree.\n",
-        "\n",
-        "The problem is typically given with two pieces of information:\n",
-        "\n",
-        "1. The number of nodes (`n`), which are usually labeled from 0 to `n-1`.\n",
-        "2. A list of undirected edges, where each edge is represented as a pair of nodes.\n",
-        "\n",
-        "The task is to write a function or algorithm to determine whether the provided set of edges can be arranged to form a valid tree based on the properties mentioned above.\n",
-        "\n",
-        "To solve this problem, you need to check the following conditions:\n",
-        "\n",
-        "1. There should be exactly `n - 1` edges to ensure that the graph is connected and tree-like.\n",
-        "2. There should be no cycles in the graph. If there are any cycles, the set of edges cannot form a tree.\n",
-        "3. The graph should be connected, which means you can reach any node from any other node through a series of edges.\n",
-        "\n",
-        "The solution typically involves using graph traversal algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the graph and check these conditions. If all conditions are met, the given edges form a valid tree; otherwise, they do not.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def validTree(n, edges):\n",
-        "    if len(edges) != n - 1:\n",
-        "        return False  # A tree with n nodes must have n-1 edges.\n",
-        "\n",
-        "    # Create an adjacency list to represent the graph.\n",
-        "    adj_list = {i: [] for i in range(n)}\n",
-        "    for u, v in edges:\n",
-        "        adj_list[u].append(v)\n",
-        "        adj_list[v].append(u)\n",
-        "\n",
-        "    visited = set()\n",
-        "\n",
-        "    # Define a recursive DFS function.\n",
-        "    def dfs(node, parent):\n",
-        "        if node in visited:\n",
-        "            return False  # If we encounter a visited node, there's a cycle.\n",
-        "        visited.add(node)\n",
-        "\n",
-        "        for neighbor in adj_list[node]:\n",
-        "            if neighbor != parent:\n",
-        "                if not dfs(neighbor, node):\n",
-        "                    return False\n",
-        "\n",
-        "        return True\n",
-        "\n",
-        "    # Start DFS from any node (e.g., node 0).\n",
-        "    if not dfs(0, -1):\n",
-        "        return False  # If there's a cycle, it's not a valid tree.\n",
-        "\n",
-        "    return len(visited) == n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `validTree` function checks if a given set of undirected edges forms a valid tree. A valid tree must satisfy these conditions:\n",
-        "\n",
-        "1. It has exactly `n - 1` edges, where `n` is the number of nodes.\n",
-        "2. There are no cycles in the graph.\n",
-        "3. The graph is connected, meaning all nodes are reachable from any starting node.\n",
-        "\n",
-        "The code accomplishes this as follows:\n",
-        "\n",
-        "- It first checks if the number of edges is equal to `n - 1`. If not, it immediately returns `False` since a tree with `n` nodes must have `n - 1` edges to be connected.\n",
-        "\n",
-        "- It then constructs an adjacency list (`adj_list`) to represent the graph efficiently. This data structure stores the neighbors of each node.\n",
-        "\n",
-        "- The code uses a `visited` set to keep track of nodes visited during depth-first search (DFS).\n",
-        "\n",
-        "- The `dfs` function is a recursive function that checks for cycles in the graph. It does this by visiting nodes and marking them as visited. If it encounters a node that has already been visited (indicating a cycle), it returns `False`.\n",
-        "\n",
-        "- The DFS function explores the neighbors of each node, avoiding revisiting the parent node to prevent cycles.\n",
-        "\n",
-        "- After the DFS is complete, the code checks if the number of visited nodes is equal to `n`, ensuring the graph is connected.\n",
-        "\n",
-        "- Finally, if all conditions are met, the code returns `True`, indicating that the given edges form a valid tree. Otherwise, it returns `False`.\n",
-        "\n",
-        "The code provides examples to demonstrate how to use the `validTree` function to check whether a set of edges forms a valid tree or not."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "n1 = 5\n",
-        "edges1 = [[0, 1], [0, 2], [0, 3], [1, 4]]\n",
-        "print(validTree(n1, edges1))  # Output: true"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "n2 = 5\n",
-        "edges2 = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]\n",
-        "print(validTree(n2, edges2))  # Output: false"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexities of the `validTree` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. Constructing the adjacency list (`adj_list`) takes O(E) time, where E is the number of edges in the input graph (|edges|).\n",
-        "\n",
-        "2. The DFS function, when considering all nodes, has a time complexity of O(V + E), where V is the number of nodes in the graph. This is because, in the worst case, the DFS function visits each node and each edge once.\n",
-        "\n",
-        "3. Checking if the number of visited nodes is equal to `n` takes O(1) time.\n",
-        "\n",
-        "Overall, the time complexity of the `validTree` function is O(E + V) in the worst case.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The adjacency list (`adj_list`) consumes additional space to store the graph structure, taking O(V + E) space. This is because it stores information about each node and its adjacent nodes.\n",
-        "\n",
-        "2. The `visited` set keeps track of visited nodes, and its space complexity is O(V) in the worst case, as there can be at most V unique nodes in the set.\n",
-        "\n",
-        "3. The depth of the function call stack during DFS can be at most V, which contributes O(V) space to the space complexity.\n",
-        "\n",
-        "Overall, the space complexity of the `validTree` function is O(V + E) due to the adjacency list and O(V) due to the `visited` set and DFS call stack.\n",
-        "\n",
-        "In summary, the time complexity is O(E + V), and the space complexity is O(V + E). These complexities are based on the worst-case scenario, where all nodes and edges are considered. In practical cases, the actual complexities may be smaller, depending on the specific structure of the graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Minimum Spanning Tree (MST)**: Implement an algorithm to find the minimum spanning tree of a connected graph using either Kruskal's or Prim's algorithm. Compare the MST to the original graph to determine if the original graph is a tree. This exercise reinforces the concept of tree properties.\n",
-        "\n",
-        "2. **Detecting Cycles**: Modify the code to not only check if the input forms a tree but also to identify and print out any cycles present in the graph if it's not a tree. This exercise enhances your ability to detect and visualize cycles in a graph."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb b/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb
deleted file mode 100644
index a8a7727..0000000
--- a/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 200. Number of Islands\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Number of Islands problem on LeetCode, click here!](https://leetcode.com/problems/number-of-islands/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an `m x n` 2D binary grid `grid` which represents a map of `'1'`s (land) and `'0'`s (water), return *the number of islands*. \n",
-        "\n",
-        "An **island** is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == grid.length`\n",
-        "- `n == grid[i].length`\n",
-        "- `1 <= m, n <= 300`\n",
-        "- `grid[i][j]` is `'0'` or `'1'`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is about counting the number of islands in a 2D binary grid. In this grid, '1' represents land, and '0' represents water. An island is defined as a group of '1' cells that are adjacent to each other either horizontally or vertically. It's important to note that diagonal connections do not count. Additionally, it's assumed that all four edges of the grid are surrounded by water, meaning the grid is finite and doesn't \"wrap around.\"\n",
-        "\n",
-        "The objective is to determine how many distinct islands exist in the grid. To clarify, an island consists of connected '1' cells, and these connections can only occur horizontally or vertically. If there are no '1' cells in the grid, the number of islands would be zero.\n",
-        "\n",
-        "The problem often involves using graph traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to identify and count these islands. You start at one '1' cell and explore its adjacent cells to determine the extent of the island. Once you've visited all the '1' cells in an island, you move on to the next unvisited '1' cell and repeat the process until you've counted all the islands in the grid.\n",
-        "\n",
-        "In essence, the problem asks you to analyze the grid to find groups of connected '1' cells and count them as individual islands. The goal is to return the total count of islands in the given grid.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def numIslands(self, grid: List[List[str]]) -> int:\n",
-        "        if not grid:\n",
-        "            return 0\n",
-        "\n",
-        "        m, n = len(grid), len(grid[0])  # Get the dimensions of the grid\n",
-        "        num_islands = 0  # Initialize the count of islands\n",
-        "\n",
-        "        def dfs(row, col):\n",
-        "            if row < 0 or row >= m or col < 0 or col >= n or grid[row][col] == '0':\n",
-        "                return\n",
-        "\n",
-        "            grid[row][col] = '0'  # Mark the current cell as visited by changing it to '0'\n",
-        "            directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Define four possible directions to move\n",
-        "\n",
-        "            for dr, dc in directions:\n",
-        "                # Explore adjacent cells\n",
-        "                dfs(row + dr, col + dc)\n",
-        "\n",
-        "        for i in range(m):\n",
-        "            for j in range(n):\n",
-        "                if grid[i][j] == '1':\n",
-        "                    num_islands += 1  # Increment the island count\n",
-        "                    dfs(i, j)  # Start a DFS from the current '1' cell to explore the entire island\n",
-        "\n",
-        "        return num_islands\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided code defines a Python class called `Solution` with a method called `numIslands`. This class is designed to count the number of islands in a 2D binary grid.\n",
-        "\n",
-        "The `numIslands` method takes a 2D grid as input, where '1' represents land, and '0' represents water. It uses a depth-first search (DFS) approach to identify and count islands. The key steps in the code include:\n",
-        "\n",
-        "1. Checking if the input grid is empty, and if so, returning 0 (indicating there are no islands).\n",
-        "\n",
-        "2. Determining the dimensions of the grid (number of rows and columns).\n",
-        "\n",
-        "3. Initializing a variable `num_islands` to keep track of the count of islands.\n",
-        "\n",
-        "4. Defining a nested function `dfs` to perform depth-first search. This function recursively explores adjacent land cells connected to the current cell and marks them as visited by changing '1' to '0'.\n",
-        "\n",
-        "5. In the `dfs` function, it checks for boundary conditions and whether the current cell is '0'. If these conditions are met, it returns without further exploration.\n",
-        "\n",
-        "6. The `dfs` function explores neighboring cells in all four directions (up, down, left, right) using a loop.\n",
-        "\n",
-        "7. The main loop iterates through each cell of the grid. When a '1' cell is encountered, it increments the `num_islands` count and starts a DFS from that cell to explore the entire island.\n",
-        "\n",
-        "8. Finally, the method returns the total count of islands found in the grid.\n",
-        "\n",
-        "This code encapsulates the solution in a class, allowing you to create an instance of the `Solution` class and call the `numIslands` method on it to count the number of islands in a given grid."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 1 Output: 1\n"
-          ]
-        }
-      ],
-      "source": [
-        "grid1 = [\n",
-        "  [\"1\", \"1\", \"1\", \"1\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"1\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"0\", \"0\", \"0\"]\n",
-        "]\n",
-        "\n",
-        "solution = Solution()\n",
-        "result1 = solution.numIslands(grid1)\n",
-        "print(\"Example 1 Output:\", result1)  # Expected output: 1\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 2 Output: 3\n"
-          ]
-        }
-      ],
-      "source": [
-        "grid2 = [\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"1\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"0\", \"1\", \"1\"]\n",
-        "]\n",
-        "\n",
-        "result2 = solution.numIslands(grid2)\n",
-        "print(\"Example 2 Output:\", result2)  # Expected output: 3\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "Time Complexity:\n",
-        "- The code uses a depth-first search (DFS) to explore the grid and count the number of islands.\n",
-        "- In the worst case, it visits every cell in the grid exactly once.\n",
-        "- The DFS function explores neighboring cells in four directions (up, down, left, right), so it has a time complexity of $O(4^{max(m, n)})$, where m is the number of rows and n is the number of columns.\n",
-        "- Therefore, the overall time complexity of the code is O(m * n), where m is the number of rows, and n is the number of columns.\n",
-        "\n",
-        "Space Complexity:\n",
-        "- The space complexity is determined by the recursive calls and the depth of the DFS stack.\n",
-        "- In the worst case, when the entire grid consists of '1's, the depth of the DFS recursion can be as deep as max(m, n).\n",
-        "- Therefore, the space complexity of the code is O(max(m, n)), representing the maximum depth of the DFS stack.\n",
-        "\n",
-        "In practical terms, for reasonably sized grids, the time and space complexity is linear, and the code should perform well. However, it's important to keep in mind that the worst-case time complexity is O(m * n), and the space complexity can be O(max(m, n))."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Identify Largest Island:** In addition to counting the number of islands, modify the code to identify the largest island in the grid and return its size.\n",
-        "\n",
-        "2. **Optimize for Space:** Modify the solution to reduce its space complexity to O(1) without modifying the input grid. This is a more memory-efficient version of the problem."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb b/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb
deleted file mode 100644
index 37624f6..0000000
--- a/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 207. Course Schedule\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Course Schedule problem on LeetCode, click here!](https://leetcode.com/problems/course-schedule/)\n",
-        "\n",
-        "---\n",
-        "There are a total of `numCourses` courses to take, labeled from 0 to `numCourses - 1`. You are given an array `prerequisites` where `prerequisites[i] = [ai, bi]` indicates that you **must** take course `bi` first if you want to take course `ai`. \n",
-        "\n",
-        "- For example, the pair `[0, 1]`, indicates that to take course `0` you have to first take course `1`.\n",
-        "\n",
-        "Return `true` if you can finish all courses. Otherwise, return `false`.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `numCourses` <= 2000\n",
-        "- 0 <= `prerequisites.length` <= 5000\n",
-        "- `prerequisites[i].length` == 2\n",
-        "- 0 <= `ai`, `bi` < `numCourses`\n",
-        "- All the pairs `prerequisites[i]` are **unique**."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether it's possible to complete a set of courses given a list of prerequisites. You are given the following:\n",
-        "\n",
-        "- `numCourses`: The total number of courses to be taken, labeled from 0 to `numCourses - 1`.\n",
-        "- `prerequisites`: A list of prerequisite courses, where each element `prerequisites[i]` is a pair `[ai, bi]`, indicating that you must take course `bi` before you can take course `ai`.\n",
-        "\n",
-        "The goal is to check whether it's possible to complete all the courses while respecting the prerequisite requirements. In other words, you need to determine if there are any circular dependencies in the course prerequisites that would prevent you from taking all the courses.\n",
-        "\n",
-        "For example:\n",
-        "\n",
-        "- If `numCourses` is 2, and `prerequisites` is `[[1, 0]]`, it means you have two courses, and you must finish course 0 before you can take course 1. In this case, it's possible to complete all courses, so the function should return `True`.\n",
-        "\n",
-        "- If `numCourses` is 2, and `prerequisites` is `[[1, 0], [0, 1]]`, it means you have two courses, but there's a circular dependency where course 1 requires course 0 and course 0 requires course 1. In this case, it's impossible to complete all courses, so the function should return `False`.\n",
-        "\n",
-        "The problem is essentially about checking for the presence of cycles in a directed graph, where the courses are nodes, and the prerequisites represent directed edges between them. If there are no cycles, it's possible to complete all courses; otherwise, it's not possible.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict, deque\n",
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:\n",
-        "        # Create a graph using an adjacency list to represent the courses and their prerequisites.\n",
-        "        graph = defaultdict(list)  # Initialize an empty adjacency list.\n",
-        "        in_degree = [0] * numCourses  # Initialize an array to store in-degrees of courses.\n",
-        "\n",
-        "        # Populate the graph and calculate in-degrees.\n",
-        "        for course, prerequisite in prerequisites:\n",
-        "            graph[prerequisite].append(course)  # Add the course as a neighbor to its prerequisite.\n",
-        "            in_degree[course] += 1  # Increment the in-degree of the course.\n",
-        "\n",
-        "        # Initialize a queue with courses that have no prerequisites.\n",
-        "        queue = deque()\n",
-        "        for course in range(numCourses):\n",
-        "            if in_degree[course] == 0:\n",
-        "                queue.append(course)\n",
-        "\n",
-        "        # Perform topological sorting.\n",
-        "        while queue:\n",
-        "            course = queue.popleft()  # Take a course with no prerequisites.\n",
-        "            numCourses -= 1  # Decrement the count of remaining courses.\n",
-        "\n",
-        "            for neighbor in graph[course]:\n",
-        "                in_degree[neighbor] -= 1  # Remove the prerequisite relationship.\n",
-        "                if in_degree[neighbor] == 0:\n",
-        "                    queue.append(neighbor)  # If no more prerequisites, add to the queue.\n",
-        "\n",
-        "        # If all courses were successfully taken (numCourses becomes 0), return True.\n",
-        "        return numCourses == 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "This Python code is designed to determine whether it is possible to complete all the required courses given a set of prerequisites. It uses a topological sorting algorithm to accomplish this. Here's a brief explanation of the code:\n",
-        "\n",
-        "1. The code defines a `Solution` class with a `canFinish` method that takes two parameters: `numCourses` (the total number of courses) and `prerequisites` (a list of pairs indicating which course must be taken before another).\n",
-        "\n",
-        "2. It initializes an empty graph as an adjacency list using a `defaultdict` to represent the courses and their prerequisites. It also initializes a list called `in_degree` to keep track of the in-degrees of courses, which is used in the topological sorting process.\n",
-        "\n",
-        "3. The code then populates the graph by iterating through the `prerequisites` list. For each pair `(ai, bi)`, it adds course `ai` as a neighbor to its prerequisite course `bi` in the graph and increments the in-degree of course `ai`.\n",
-        "\n",
-        "4. Next, it initializes a queue with courses that have no prerequisites. It does this by iterating through all the courses and adding those with an in-degree of 0 to the queue.\n",
-        "\n",
-        "5. The code performs a topological sorting of the courses using a while loop. In each iteration of the loop, it dequeues a course from the front of the queue (a course with no prerequisites).\n",
-        "\n",
-        "6. For each neighbor (course that depends on the dequeued course), it decrements the in-degree of the neighbor. If the neighbor's in-degree becomes 0, it means all of its prerequisites have been taken, so the neighbor is added to the queue.\n",
-        "\n",
-        "7. The loop continues until there are no more courses with no prerequisites to dequeue, and the in-degrees are updated accordingly.\n",
-        "\n",
-        "8. After the loop, if all courses were successfully taken (i.e., `numCourses` becomes 0), it means that there were no circular dependencies, and it's possible to finish all courses. In this case, the method returns `True`. Otherwise, if there are remaining courses (indicating a circular dependency), it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "numCourses1 = 2\n",
-        "prerequisites1 = [[1, 0]]\n",
-        "sol = Solution()\n",
-        "print(sol.canFinish(numCourses1, prerequisites1))  # Output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "numCourses2 = 2\n",
-        "prerequisites2 = [[1, 0], [0, 1]]\n",
-        "print(sol.canFinish(numCourses2, prerequisites2))  # Output: False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `canFinish` method:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. Constructing the graph: In the first loop where we iterate through the `prerequisites` list, we populate the graph with prerequisites. This loop has a time complexity of O(E), where E is the number of prerequisites (edges).\n",
-        "\n",
-        "2. Initializing in-degrees and finding courses with no prerequisites: In the worst case, we iterate through all `numCourses` courses to initialize in-degrees and find the courses with no prerequisites. This has a time complexity of O(V), where V is the number of courses (vertices).\n",
-        "\n",
-        "3. Topological Sorting: In the while loop, we perform topological sorting, processing each course once and its outgoing edges. In the worst case, each course is processed once. This has a time complexity of O(V + E).\n",
-        "\n",
-        "The overall time complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. Graph Representation: The space complexity for storing the graph as an adjacency list is O(E), where E is the number of prerequisites.\n",
-        "\n",
-        "2. In-Degree Array: The space complexity for storing the in-degrees of courses is O(V), where V is the number of courses.\n",
-        "\n",
-        "3. Queue: The space complexity for the queue is O(V), as it may contain all the courses.\n",
-        "\n",
-        "The overall space complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-        "\n",
-        "In most practical cases, the number of prerequisites (E) is much smaller than the number of courses (V), so the time and space complexity can often be considered as O(V)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimizing Course Scheduling**: Given a list of courses with their durations and prerequisites, find the most efficient way to schedule these courses to minimize the time it takes to complete all of them.\n",
-        "\n",
-        "2. **Detecting Cycles in Course Dependencies**: Modify the original problem to not just determine if you can finish all courses but also to identify the specific courses that form a cycle in the prerequisite dependencies."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb b/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb
deleted file mode 100644
index fb186f2..0000000
--- a/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb	
+++ /dev/null
@@ -1,236 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 3651. Number of Connected Components in an Undirected Graph\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Number of Connected Components in an Undirected Graph problem on LintCode, click here!](https://www.lintcode.com/problem/3651/)\n",
-        "\n",
-        "---\n",
-        "In this problem, there is an undirected graph with `n` nodes. There is also an `edges` array. Where `edges[i] = [a, b]` means that there is an edge between *node a* and *node b* in the graph.\n",
-        "\n",
-        "You need to return *the number of connected components in that graph*."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to determine the number of connected components in an undirected graph. Here's a more detailed explanation of the problem:\n",
-        "\n",
-        "- You are given an undirected graph, which consists of nodes and edges. The graph may have multiple distinct connected components.\n",
-        "\n",
-        "- A connected component is a group of nodes within the graph where each node can be reached from every other node in the same component. In other words, there is a path between any two nodes in the same connected component, but there are no paths connecting nodes from different components.\n",
-        "\n",
-        "- Your task is to write a function that takes as input the number of nodes `n` and a list of edges `edges`, where each edge is represented as a pair of nodes `[a, b]` indicating an edge between node `a` and node `b`.\n",
-        "\n",
-        "- The goal is to determine the number of distinct connected components in the graph.\n",
-        "\n",
-        "- You need to implement an algorithm that can identify these connected components and count how many there are in the given graph.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def countConnectedComponents(n, edges):\n",
-        "    # Define a helper function for depth-first search (DFS).\n",
-        "    def dfs(node):\n",
-        "        if not visited[node]:  # If the node has not been visited yet:\n",
-        "            visited[node] = True  # Mark it as visited.\n",
-        "            for neighbor in graph[node]:  # Explore all neighbors of the current node.\n",
-        "                dfs(neighbor)  # Recursively call DFS on the neighbor.\n",
-        "\n",
-        "    # Create an empty adjacency list to represent the graph.\n",
-        "    graph = [[] for _ in range(n)]\n",
-        "\n",
-        "    # Initialize a boolean list to keep track of visited nodes.\n",
-        "    visited = [False] * n\n",
-        "\n",
-        "    # Populate the adjacency list with edges from the input.\n",
-        "    for a, b in edges:\n",
-        "        graph[a].append(b)\n",
-        "        graph[b].append(a)  # Since the graph is undirected, we add edges in both directions.\n",
-        "\n",
-        "    # Initialize a counter to keep track of the number of connected components.\n",
-        "    count = 0\n",
-        "\n",
-        "    # Iterate through all nodes in the graph.\n",
-        "    for i in range(n):\n",
-        "        if not visited[i]:  # If the node has not been visited:\n",
-        "            count += 1  # Increment the count since we've found a new connected component.\n",
-        "            dfs(i)  # Start a depth-first search from the unvisited node to explore the component.\n",
-        "\n",
-        "    return count"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided Python code calculates the number of connected components in an undirected graph. Here's a high-level explanation of how the code works:\n",
-        "\n",
-        "1. The `countConnectedComponents` function takes two arguments: `n` (the number of nodes in the graph) and `edges` (a list of edge pairs representing connections between nodes).\n",
-        "\n",
-        "2. Inside the function, there is a helper function called `dfs` (depth-first search), which is used to explore and mark nodes as visited.\n",
-        "\n",
-        "3. An empty adjacency list called `graph` is created to represent the graph. The `graph` is a list of lists where each element represents a node, and the list associated with each node contains its neighboring nodes.\n",
-        "\n",
-        "4. Another list called `visited` is created to keep track of visited nodes. Initially, all nodes are marked as not visited (`False`).\n",
-        "\n",
-        "5. The code populates the `graph` by iterating through the `edges` list and adding each edge to the adjacency list. Since the graph is undirected, both directions of each edge are added.\n",
-        "\n",
-        "6. A counter variable called `count` is initialized to keep track of the number of connected components.\n",
-        "\n",
-        "7. The code iterates through all nodes from `0` to `n-1`. For each node, it checks whether it has been visited yet. If it hasn't been visited, it means a new connected component is found. The counter `count` is incremented, and a depth-first search (DFS) is initiated from this unvisited node to explore the connected component.\n",
-        "\n",
-        "8. The `dfs` function recursively visits all nodes in the connected component, marking them as visited. It does this by checking each neighbor of the current node and recursively calling `dfs` on unvisited neighbors.\n",
-        "\n",
-        "9. After the loop through all nodes, the `count` variable contains the total number of connected components in the graph.\n",
-        "\n",
-        "10. The function returns the value of `count`, which represents the number of connected components.\n",
-        "\n",
-        "11. Two examples are provided at the end to demonstrate how to use the function with different inputs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "n1 = 3\n",
-        "edges1 = [[0, 1], [0, 2]]\n",
-        "print(countConnectedComponents(n1, edges1))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "n2 = 6\n",
-        "edges2 = [[0, 1], [1, 2], [2, 3], [4, 5]]\n",
-        "print(countConnectedComponents(n2, edges2))  # Output: 2"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the code for counting the number of connected components in an undirected graph using depth-first search (DFS).\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code iterates through all nodes in the graph (from 0 to `n-1`).\n",
-        "- For each unvisited node, it initiates a depth-first search (DFS) from that node to explore the connected component.\n",
-        "- In the worst case, every node may belong to a separate connected component, and you perform a DFS for each node.\n",
-        "- The time complexity of the DFS for a single connected component is O(V + E), where V is the number of nodes in the component, and E is the number of edges.\n",
-        "- In the worst case, you perform O(n) DFS calls, each with its own O(V + E) complexity.\n",
-        "\n",
-        "So, the overall time complexity of the code is O(n * (V + E)). In the worst case, if the graph is connected (a single connected component), this simplifies to O(n * (n + m)), where m is the number of edges.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The main data structures used are the `graph` (adjacency list), `visited` (boolean array), and the recursive call stack for DFS.\n",
-        "- The `graph` has a space complexity of O(n + m), where n is the number of nodes, and m is the number of edges.\n",
-        "- The `visited` boolean array also has a space complexity of O(n) to keep track of whether nodes have been visited.\n",
-        "- The depth of the recursive call stack for DFS can go up to O(n) in the worst case, when all nodes are part of a single connected component.\n",
-        "\n",
-        "So, the overall space complexity of the code is O(n + m) for data structures and O(n) for the call stack. In the worst case, this simplifies to O(n + m).\n",
-        "\n",
-        "In summary, the time complexity is O(n * (V + E)), and the space complexity is O(n + m)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Disconnected Components with Multiple Queries:** You are given an undirected graph with `n` nodes and `m` edges. Initially, the graph is disconnected, and you need to process `q` queries. Each query consists of adding an edge to connect two nodes. After each query, you need to output the number of connected components in the updated graph.\n",
-        "\n",
-        "2. **Maximum Size of Connected Components:** Given an undirected graph with `n` nodes and `m` edges, find the size of the largest connected component in the graph. Additionally, identify the nodes that belong to this largest component. You can assume there is at least one connected component in the graph.\n"
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb b/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb
deleted file mode 100644
index eb34497..0000000
--- a/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb	
+++ /dev/null
@@ -1,261 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 417. Pacific Atlantic Water Flow\n",
-        "\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Pacific Atlantic Water Flow problem on LeetCode, click here!](https://leetcode.com/problems/pacific-atlantic-water-flow/)\n",
-        "\n",
-        "---\n",
-        "There is an `m x n` rectangular island that borders both the **Pacific Ocean** and **Atlantic Ocean**. The **Pacific Ocea**n touches the island's left and top edges, and the **Atlantic Ocean** touches the island's right and bottom edges.\n",
-        "\n",
-        "The island is partitioned into a grid of square cells. You are given an `m x n` integer matrix `heights` where `heights[r][c]` represents the **height above sea level** of the cell at coordinate `(r, c)`.\n",
-        "\n",
-        "The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is **less than or equal to** the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.\n",
-        "\n",
-        "Return *a **2D list** of grid coordinates `result` where $result[i] = [r_i, c_i]$ denotes that rain water can flow from cell $(r_i, c_i)$ to **both** the Pacific and Atlantic oceans*.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == heights.length`\n",
-        "- `n == heights[r].length`\n",
-        "- `1 <= m, n <= 200`\n",
-        "- 0 <= `heights[r][c]` <= $10^5$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand involves a scenario with a rectangular island surrounded by both the Pacific Ocean and the Atlantic Ocean. The Pacific Ocean is in contact with the island's left and top edges, while the Atlantic Ocean is in contact with the island's right and bottom edges. The goal is to determine which cells on the island can allow rainwater to flow to both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "Here are the key details and constraints of the problem:\n",
-        "\n",
-        "1. The island is divided into a grid of square cells.\n",
-        "\n",
-        "2. Each cell is represented by an integer value in a matrix `heights`. The value at `heights[r][c]` represents the height above sea level of the cell at coordinates `(r, c)` on the island.\n",
-        "\n",
-        "3. Rainwater can flow from a cell to neighboring cells directly north, south, east, or west only if the neighboring cell's height is less than or equal to the current cell's height.\n",
-        "\n",
-        "4. Water can flow from any cell adjacent to the ocean into the ocean. This means water can flow from cells on the edge of the island to the Pacific Ocean and the Atlantic Ocean.\n",
-        "\n",
-        "The task is to find and return a 2D list of grid coordinates where rainwater can flow from a cell to both the Pacific and Atlantic Oceans. In other words, you need to identify the cells that can send water to both the left and top edges (Pacific) and the right and bottom edges (Atlantic) of the island.\n",
-        "\n",
-        "The problem is solved by performing a depth-first search (DFS) starting from the ocean edges and marking the cells that can flow to each ocean. Then, the algorithm identifies cells that are reachable from both oceans and returns their coordinates as the result.\n",
-        "\n",
-        "The problem involves both traversal of the island and backtracking to find the solution, which makes it a classic graph traversal problem.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:\n",
-        "        # Check if the input matrix is empty\n",
-        "        if not heights:\n",
-        "            return []\n",
-        "        \n",
-        "        m, n = len(heights), len(heights[0])\n",
-        "        \n",
-        "        # Initialize boolean matrices to keep track of cells reachable from each ocean\n",
-        "        pacific_reachable = [[False] * n for _ in range(m)]\n",
-        "        atlantic_reachable = [[False] * n for _ in range(m)]\n",
-        "        \n",
-        "        # Depth-First Search (DFS) function to mark cells that can be reached from an ocean\n",
-        "        def dfs(x, y, reachable):\n",
-        "            if reachable[x][y]:\n",
-        "                return\n",
-        "            reachable[x][y] = True\n",
-        "            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:\n",
-        "                new_x, new_y = x + dx, y + dy\n",
-        "                if 0 <= new_x < m and 0 <= new_y < n and heights[new_x][new_y] >= heights[x][y]:\n",
-        "                    dfs(new_x, new_y, reachable)\n",
-        "        \n",
-        "        # Start DFS from the Pacific Ocean (left and top edges)\n",
-        "        for i in range(m):\n",
-        "            dfs(i, 0, pacific_reachable)\n",
-        "            dfs(i, n - 1, atlantic_reachable)\n",
-        "        \n",
-        "        for j in range(n):\n",
-        "            dfs(0, j, pacific_reachable)\n",
-        "            dfs(m - 1, j, atlantic_reachable)\n",
-        "        \n",
-        "        result = []\n",
-        "        \n",
-        "        # Find cells that can flow to both the Pacific and Atlantic oceans\n",
-        "        for i in range(m):\n",
-        "            for j in range(n):\n",
-        "                if pacific_reachable[i][j] and atlantic_reachable[i][j]:\n",
-        "                    result.append([i, j])\n",
-        "        \n",
-        "        return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a class called `Solution` with a method named `pacificAtlantic`. This method takes a 2D matrix `heights`, where each element represents the height above sea level of a cell on a rectangular island. The goal of the method is to find and return a list of grid coordinates (cells) from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "The code uses a Depth-First Search (DFS) algorithm to traverse the island, starting from the ocean edges, and marks cells that are reachable from either the Pacific or Atlantic Ocean. It does this by creating two boolean matrices, `pacific_reachable` and `atlantic_reachable`, to keep track of cells that can be reached from each ocean.\n",
-        "\n",
-        "The code iterates through the entire island, initiating DFS searches from the edges of the island. If a cell is reachable from an ocean (either Pacific or Atlantic), it is marked as reachable in the corresponding boolean matrix. The DFS function ensures that the traversal follows the rule that water can flow to a neighboring cell if the neighboring cell's height is less than or equal to the current cell's height.\n",
-        "\n",
-        "After marking all reachable cells from both oceans, the code then looks for cells that are reachable from both the Pacific and Atlantic Oceans. If a cell satisfies this condition, it is added to the `result` list.\n",
-        "\n",
-        "Finally, the code returns the `result` list, which contains the grid coordinates of cells from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "The code provides a solution to the problem of finding cells on the island where water can reach both oceans, demonstrating a traversal algorithm that explores the connectivity of cells on the island."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "heights1 = [\n",
-        "    [1,2,2,3,5],\n",
-        "    [3,2,3,4,4],\n",
-        "    [2,4,5,3,1],\n",
-        "    [6,7,1,4,5],\n",
-        "    [5,1,1,2,4]\n",
-        "]\n",
-        "\n",
-        "solution = Solution()\n",
-        "print(solution.pacificAtlantic(heights1)) # Expected Output: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 26,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "heights2 = [[1]]\n",
-        "print(solution.pacificAtlantic(heights2)) # Expected Output: [[0, 0]]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Time Complexity:\n",
-        "The time complexity of this solution is O(m * n), where 'm' is the number of rows in the input matrix and 'n' is the number of columns. Here's a breakdown of the time complexity:\n",
-        "\n",
-        "1. Creating the `pacific_reachable` and `atlantic_reachable` matrices with dimensions m x n takes O(m * n) time.\n",
-        "\n",
-        "2. Initiating Depth-First Search (DFS) from the edges of the island takes O(m * n) time because it processes each cell once.\n",
-        "\n",
-        "3. The DFS function visits each cell at most once, and its time complexity is O(1) for each cell. The number of cells visited by the DFS in the worst case is m * n.\n",
-        "\n",
-        "4. The final step of identifying and appending cells to the result list takes O(m * n) time in the worst case.\n",
-        "\n",
-        "Overall, the time complexity is dominated by the DFS and is O(m * n).\n",
-        "\n",
-        "Space Complexity:\n",
-        "The space complexity of this solution is also O(m * n). Here's how the space is used:\n",
-        "\n",
-        "1. Two boolean matrices, `pacific_reachable` and `atlantic_reachable`, are created with dimensions m x n. Each matrix takes O(m * n) space, resulting in O(2 * m * n) space usage.\n",
-        "\n",
-        "2. The depth-first search (DFS) stack space during the recursive calls can go as deep as the diagonal of the grid, which is at most min(m, n). This additional space for the call stack is relatively small compared to the boolean matrices and can be considered O(min(m, n)).\n",
-        "\n",
-        "3. The `result` list stores the grid coordinates, and its size is determined by the number of cells that can flow to both oceans. In the worst case, this list can contain all m * n cells, so it takes O(m * n) space.\n",
-        "\n",
-        "The overall space complexity is the sum of these components, which is O(m * n) + O(min(m, n)) + O(m * n). In big O notation, we can simplify this to O(m * n)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the algorithm to find the maximum flow of water from any cell to both the Pacific and Atlantic Oceans. In other words, find the cell(s) that can send the most water to both oceans.\n",
-        "\n",
-        "2. **Efficiency Challenge:** Optimize the time and space complexity of the algorithm while maintaining correctness. Aim to reduce both time and space complexity as much as possible."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/README.md b/_build/html/_sources/11. Graphs/README.md
deleted file mode 100644
index aad262c..0000000
--- a/_build/html/_sources/11. Graphs/README.md	
+++ /dev/null
@@ -1,23 +0,0 @@
-# Graphs Problems - Blind 75 LeetCode
-
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)                           | Medium     |
-| [133. Clone Graph](https://leetcode.com/problems/clone-graph/)                                       | Medium     |
-| [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)       | Medium     |
-| [207. Course Schedule](https://leetcode.com/problems/course-schedule/)                                 | Medium     |
-| [323. Number of Connected Components In An Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/) | Medium |
-| [261. Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)                               | Medium     |
-
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb b/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb
deleted file mode 100644
index 81f0e5b..0000000
--- a/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb	
+++ /dev/null
@@ -1,256 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 892. Alien Dictionary\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Alien Dictionary problem on LintCode, click here!](https://www.lintcode.com/problem/892/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of **non-empty** words from the dictionary, where words are **sorted lexicographically by the rules of this new language**. Derive the order of letters in this language.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- You may assume all letters are in lowercase.\n",
-        "- The dictionary is invalid, if string a is prefix of string b and b is appear before a.\n",
-        "- If the order is invalid, return an empty string.\n",
-        "- There may be multiple valid order of letters, return the smallest in normal lexicographical order.\n",
-        "- The letters in **one** string are of the same rank by default and are sorted in Human dictionary order."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand involves determining the order of letters in a new alien language that uses the Latin alphabet. However, the order of letters in this alien language is unknown. To solve this problem, we are given a list of non-empty words from the dictionary, where the words are sorted lexicographically based on the rules of the alien language. The goal is to derive the correct order of letters in this new language.\n",
-        "\n",
-        "Here's a more detailed explanation of the problem:\n",
-        "\n",
-        "1. The alphabet: The new alien language uses the Latin alphabet, which consists of lowercase letters.\n",
-        "\n",
-        "2. Dictionary: We are provided with a list of non-empty words. These words are sorted in lexicographical order based on the rules of the alien language. The order of letters in the words reflects the correct order in the new language.\n",
-        "\n",
-        "3. Invalid order: The problem specifies that the dictionary is considered invalid if there exists a situation where one string 'a' is a prefix of another string 'b,' and 'b' appears before 'a' in the dictionary. This condition ensures that there are no inconsistencies in the order of letters.\n",
-        "\n",
-        "4. Multiple valid orders: It's possible that there are multiple valid orders of letters that satisfy the provided dictionary. However, we are instructed to return the smallest valid order in normal lexicographical order.\n",
-        "\n",
-        "To solve this problem, you need to analyze the given list of words and determine the correct order of letters in the alien language while adhering to the specified constraints. You should return the order as a string in lexicographical order if it's valid, and an empty string if the order is invalid.\n",
-        "\n",
-        "The key to solving this problem is to construct a directed graph of letter relationships based on the order information provided by the words in the dictionary. Then, perform a topological sort on this graph to derive the order of letters, ensuring that the order is consistent and valid according to the given constraints.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict, deque\n",
-        "\n",
-        "def alienOrder(words):\n",
-        "    # Create an adjacency list to represent the graph\n",
-        "    graph = defaultdict(list)\n",
-        "    \n",
-        "    # Create a dictionary to store the in-degrees of each letter\n",
-        "    in_degree = {}\n",
-        "    \n",
-        "    # Initialize in-degrees to 0 for all letters\n",
-        "    for word in words:\n",
-        "        for char in word:\n",
-        "            if char not in in_degree:\n",
-        "                in_degree[char] = 0\n",
-        "    \n",
-        "    # Build the graph and update in-degrees\n",
-        "    for i in range(1, len(words)):\n",
-        "        word1, word2 = words[i-1], words[i]\n",
-        "        min_length = min(len(word1), len(word2))\n",
-        "        \n",
-        "        # Compare characters in the two words\n",
-        "        j = 0\n",
-        "        while j < min_length and word1[j] == word2[j]:\n",
-        "            j += 1\n",
-        "        \n",
-        "        if j < min_length:\n",
-        "            # If word1[j] is lexicographically before word2[j], add an edge\n",
-        "            graph[word1[j]].append(word2[j])\n",
-        "            in_degree[word2[j]] += 1\n",
-        "            \n",
-        "            # Break the loop to avoid further comparisons\n",
-        "            break\n",
-        "    \n",
-        "    # Perform topological sorting using Kahn's algorithm\n",
-        "    result = []\n",
-        "    queue = deque([char for char in in_degree if in_degree[char] == 0])\n",
-        "    \n",
-        "    while queue:\n",
-        "        char = queue.popleft()\n",
-        "        result.append(char)\n",
-        "        \n",
-        "        for neighbor in graph[char]:\n",
-        "            in_degree[neighbor] -= 1\n",
-        "            if in_degree[neighbor] == 0:\n",
-        "                queue.append(neighbor)\n",
-        "    \n",
-        "    # Check if there is a valid order\n",
-        "    if len(result) < len(in_degree):\n",
-        "        return \"\"\n",
-        "    \n",
-        "    return \"\".join(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code is designed to determine the order of letters in an unknown alien language that uses the Latin alphabet. It takes a list of non-empty words from a dictionary as input, where the words are sorted lexicographically based on the rules of this alien language.\n",
-        "\n",
-        "The code follows these main steps:\n",
-        "\n",
-        "1. Create a data structure to represent the graph of letters and their relationships in the alien language. It uses a defaultdict to store a list of letters that come after each letter. Additionally, it uses a dictionary called `in_degree` to keep track of the in-degrees (the number of letters that precede a letter) for each letter.\n",
-        "\n",
-        "2. Initialize the in-degrees of all letters to 0. This is done by iterating through all the words and characters in the words, adding each character to the `in_degree` dictionary with an initial in-degree of 0.\n",
-        "\n",
-        "3. Build the graph by comparing adjacent words in the dictionary. For each pair of adjacent words, the code finds the first differing character between the two words. If such a character exists, it means there's an order relationship between the letters represented by these characters. The code updates the graph and in-degrees accordingly.\n",
-        "\n",
-        "4. After constructing the graph, the code performs a topological sorting of the letters using Kahn's algorithm. It starts by initializing an empty result list and a queue containing letters with in-degrees of 0. The algorithm repeatedly removes a letter with an in-degree of 0 from the queue, adds it to the result list, and updates the in-degrees of its neighbors. This process continues until the queue is empty.\n",
-        "\n",
-        "5. Finally, the code checks if the topological sorting was successful by comparing the length of the result list with the number of unique letters in the input. If the lengths differ, it means there's a cycle in the graph, and the order is invalid. In such a case, an empty string is returned. Otherwise, the result list is joined together to form the smallest order of letters in normal lexicographical order, and it is returned as the output.\n",
-        "\n",
-        "The code provides a way to determine the order of letters in the alien language while handling various edge cases, including checking for invalid order conditions."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "wrtef\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "words1 = [\"wrt\", \"wrf\", \"er\", \"ett\", \"rftt\"]\n",
-        "print(alienOrder(words1))  # Output: \"wertf\""
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "zx\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "words2 = [\"z\", \"x\"]\n",
-        "print(alienOrder(words2))  # Output: \"zx\""
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for determining the order of letters in an alien language.\n",
-        "\n",
-        "Time Complexity:\n",
-        "1. Building the graph: The code iterates through the list of words and compares adjacent words. In the worst case, where all words have the same prefix, it takes O(N) time to build the graph, where N is the total number of characters in all the words.\n",
-        "2. Topological sorting: Performing a topological sort on the graph takes O(V + E) time, where V is the number of unique letters (vertices) and E is the number of relationships between letters (edges) in the graph.\n",
-        "\n",
-        "Overall, the time complexity is O(N + V + E), where N is the total number of characters in the words, V is the number of unique letters, and E is the number of order relationships between letters.\n",
-        "\n",
-        "Space Complexity:\n",
-        "1. Graph and In-degrees: The code uses data structures to represent the graph and in-degrees. The space complexity for these data structures is O(V + E), where V is the number of unique letters, and E is the number of order relationships between letters in the words.\n",
-        "2. Result List: The result list stores the order of letters, which can have a maximum size of V, where V is the number of unique letters.\n",
-        "3. Queue: The space used by the queue during topological sorting is also O(V).\n",
-        "\n",
-        "Overall, the space complexity is O(V + E) due to the data structures used for graph representation and the space needed for storing the result and the queue.\n",
-        "\n",
-        "In the worst case, where there are many unique letters and many order relationships, the space complexity is dominated by the number of unique letters and their relationships.\n",
-        "\n",
-        "In summary, the time complexity of the code is O(N + V + E), and the space complexity is O(V + E). The actual performance will depend on the specific input data, such as the number of unique letters and the relationships between them in the alien language."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Valid Orders**: Extend the problem to handle cases where there are multiple valid orders of letters in the alien language. Modify the code to return all valid orders instead of just one. Be mindful of performance.\n",
-        "\n",
-        "2. **Detect Invalid Dictionary**: Given a dictionary of words, write a function to detect whether the dictionary is valid or not based on the constraints mentioned in the problem statement."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/12. Advanced Graphs/README.md b/_build/html/_sources/12. Advanced Graphs/README.md
deleted file mode 100644
index f362de7..0000000
--- a/_build/html/_sources/12. Advanced Graphs/README.md	
+++ /dev/null
@@ -1,16 +0,0 @@
-# Advanced Graphs Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [269. Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)                               | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb b/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb
deleted file mode 100644
index 8d32e21..0000000
--- a/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb	
+++ /dev/null
@@ -1,103 +0,0 @@
-{
-  "metadata": {
-    "kernelspec": {
-      "name": "python",
-      "display_name": "Python (Pyodide)",
-      "language": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
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-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
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-  },
-  "nbformat_minor": 5,
-  "nbformat": 4,
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "source": "# House Robber\n**Difficulty:** Medium\n\n---\nYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and **it will automatically contact the police if two adjacent houses were broken into on the same night.**\n\nGiven an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight **without alerting the police**.*\n\n**Constraints:**\n- `1 <= nums.length <= 100`\n- `0 <= nums[i] <= 400`",
-      "metadata": {},
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Probelm Explanation:\n\nThe problem is a classic dynamic programming challenge known as the \"House Robber\" problem. The scenario is that you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, but the constraint is that adjacent houses have security systems connected. If two adjacent houses are broken into on the same night, the security systems will automatically contact the police.\n\nThe goal is to determine the maximum amount of money you can rob tonight without alerting the police. You need to find a strategy to maximize your earnings while avoiding robbing two adjacent houses.\n\nThe input to the problem is an array `nums`, where `nums[i]` represents the amount of money in the `i`-th house. The task is to return the maximum amount of money that can be robbed without triggering the security system.\n\nFor example:\n```\nInput: nums = [1,2,3,1]\nOutput: 4\nExplanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).\nTotal amount you can rob = 1 + 3 = 4.\n\nInput: nums = [2,7,9,3,1]\nOutput: 12\nExplanation: Rob house 1 (money = 2), rob house 3 (money = 9), and rob house 5 (money = 1).\nTotal amount you can rob = 2 + 9 + 1 = 12.\n```\n\nThe challenge involves designing an algorithm to efficiently solve this problem and find the optimal strategy for robbing houses to maximize the stolen money while adhering to the constraint of not robbing adjacent houses. The provided solution utilizes dynamic programming to achieve this.\n\n## Solution:\nHere's a Python function to implement this algorithm:",
-      "metadata": {},
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76"
-    },
-    {
-      "cell_type": "code",
-      "source": "from typing import List\n\nclass Solution:\n    def rob(self, nums: List[int]) -> int:\n        # Initialize variables to keep track of the maximum amount of money robbed at the previous and current houses\n        prev, curr = 0, 0\n\n        # Iterate through the houses\n        for n in nums:\n            # Calculate the maximum amount of money that can be robbed at the current house\n            # It is the maximum of either robbing the current house and the money robbed at the house before the previous one (prev),\n            # or skipping the current house and taking the maximum amount from the previous house (curr)\n            temp = max(n + prev, curr)\n\n            # Update the variables for the next iteration\n            prev, curr = curr, temp\n\n        # The final result is the maximum amount of money that can be robbed at the last house\n        return curr",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 4,
-      "outputs": [],
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Explanation:\n\nThe provided code is a Python implementation of the \"House Robber\" problem using a dynamic programming approach. It's part of a class `Solution` and includes a method `rob` that takes a list of integers `nums` as input and returns the maximum amount of money that can be robbed without alerting the police.\n\nThe key idea of the algorithm is to maintain two variables, `prev` and `curr`, which represent the maximum amount of money that can be robbed up to the previous house and the current house, respectively. The algorithm iterates through the houses, updating these variables based on the maximum amount that can be robbed at the current house, considering the constraint of not robbing adjacent houses.\n\nThe loop calculates a temporary variable `temp` that represents the maximum amount of money that can be robbed at the current house. It considers two scenarios: either robbing the current house and adding the money robbed at the house before the previous one (`n + prev`), or skipping the current house and taking the maximum amount from the previous house (`curr`). The maximum of these two scenarios is assigned to `temp`.\n\nThe `prev` and `curr` variables are then updated for the next iteration. `prev` is set to the previous value of `curr`, and `curr` is set to the calculated `temp`.\n\nThe final result is the value stored in the `curr` variable, representing the maximum amount of money that can be robbed at the last house. This value is returned as the result of the `rob` method.\n\nThe provided test cases at the end demonstrate the usage of the `rob` method with different input arrays.",
-      "metadata": {},
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Test cases:\n\nHere's how you can use this solution:",
-      "metadata": {},
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 1:\nsol = Solution()\n\nnums1 = [1, 2, 3, 1]\nprint(sol.rob(nums1))  # Output: 4",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 5,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "4\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums2 = [2, 7, 9, 3, 1]\nprint(sol.rob(nums2))  # Output: 12",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 6,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "12\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Time and Space Complexity Analysis\n\nLet's analyze the time and space complexity of the provided code.\n\n### Time Complexity:\n\nThe time complexity of the code is O(n), where n is the number of houses. The algorithm iterates through the array once, and in each iteration, it performs constant time operations. The loop runs for each house, and the number of operations inside the loop is independent of the input size. Therefore, the overall time complexity is linear, O(n).\n\n### Space Complexity:\n\nThe space complexity is O(1), constant space. The algorithm uses only a constant amount of extra space regardless of the input size. The variables `prev`, `curr`, and `temp` are the only additional variables used, and they do not depend on the size of the input array. Thus, the space complexity is constant, O(1).\n\nIn summary:\n- Time Complexity: O(n)\n- Space Complexity: O(1)",
-      "metadata": {},
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Challenging Exercises:\n\n1. **Variation with Maximum Number of Houses to Rob:** Modify the problem to include an additional constraint: you are allowed to rob at most 'k' houses without triggering the alarm. Find the maximum amount you can rob under this new constraint.\n\n2. **Robbing in a Circular Street:** Extend the problem to a circular street where the first and last houses are adjacent. You cannot rob both the first and last houses simultaneously. Find the maximum amount you can rob in this circular setting.\n\n\n**Link to Problem:** [To see the House Robber problem on LeetCode, click here!](https://leetcode.com/problems/house-robber/description/)\n",
-      "metadata": {},
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6"
-    }
-  ]
-}
\ No newline at end of file
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb b/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb
deleted file mode 100644
index b1080e8..0000000
--- a/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb	
+++ /dev/null
@@ -1,119 +0,0 @@
-{
-  "metadata": {
-    "kernelspec": {
-      "name": "python",
-      "display_name": "Python (Pyodide)",
-      "language": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
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-      "pygments_lexer": "ipython3",
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-  "nbformat_minor": 5,
-  "nbformat": 4,
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "source": "# House Robber II\n**Difficulty:** Medium\n\n---\nYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle**. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.\n\nGiven an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight without alerting the police*.\n\n**Constraints:**\n- `1 <= nums.length <= 100`\n- `0 <= nums[i] <= 1000`",
-      "metadata": {},
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Probelm Explanation:\n\nCertainly! The problem statement describes a scenario where you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. However, there is a security system in place: adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.\n\nThe twist in this problem is that the houses are arranged in a circle. That means the first house is the neighbor of the last one. So, robbing the first house has implications for the last house, and vice versa.\n\nThe goal is to find the maximum amount of money you can rob tonight without alerting the police. You need to implement a function that takes an integer array `nums` representing the amount of money in each house and returns the maximum amount of money that can be robbed without triggering the security system.\n\nHere are a few examples to illustrate the problem:\n\n1. **Example 1:**\n   ```python\n   Input: nums = [2, 3, 2]\n   Output: 3\n   Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2) because they are adjacent houses. The optimal strategy is to rob house 2 (money = 3).\n   ```\n\n2. **Example 2:**\n   ```python\n   Input: nums = [1, 2, 3, 1]\n   Output: 4\n   Explanation: Robbing house 1 (money = 1) and then robbing house 3 (money = 3) results in a total amount of 4. Robbing house 2 is skipped to avoid alerting the police.\n   ```\n\n3. **Example 3:**\n   ```python\n   Input: nums = [1, 2, 3]\n   Output: 3\n   Explanation: Since the houses are in a circle, you can choose either house 1, 2, or 3. The optimal strategy is to rob house 3 for a total amount of 3.\n   ```\n\nThe constraints for the problem are that the length of the `nums` array is between 1 and 100, and the amount of money in each house (`nums[i]`) is between 0 and 1000.\n\n## Solution:\nHere's a Python function to implement this algorithm:",
-      "metadata": {},
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76"
-    },
-    {
-      "cell_type": "code",
-      "source": "from typing import List\n\nclass Solution:\n    def rob(self, nums: List[int]) -> int:\n        def simple_rob(nums):\n            # Helper function for basic house robbery problem\n            prev, curr = 0, 0\n            for num in nums:\n                # Calculate the maximum amount of money that can be robbed\n                # without alerting the police for each house\n                prev, curr = curr, max(prev + num, curr)\n            return curr\n\n        # Check the base cases where the length of nums is 1 or 2\n        if len(nums) == 1:\n            return nums[0]\n        elif len(nums) == 2:\n            return max(nums[0], nums[1])\n\n        # Case 1: First house is robbed, last house is not\n        case1 = simple_rob(nums[:-1])\n\n        # Case 2: First house is not robbed, last house is\n        case2 = simple_rob(nums[1:])\n\n        # Maximum of the two cases\n        return max(case1, case2)",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 7,
-      "outputs": [],
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Explanation:\n\nThe provided code defines a class `Solution` with a method `rob`, which aims to determine the maximum amount of money a professional robber can steal from houses arranged in a circle without alerting the police. Each house contains a certain amount of money, and adjacent houses have a security system that triggers if both are robbed on the same night.\n\nThe code uses dynamic programming to solve the problem efficiently. It contains a helper function `simple_rob` that computes the maximum amount of money that can be robbed from a linear arrangement of houses. The main method `rob` handles the circular arrangement by considering two cases: robbing the first house and not robbing the last, and not robbing the first house and robbing the last. The result is the maximum of these two cases.\n\nThe class is instantiated as `solution`, and the `rob` method is applied to three different scenarios (test cases) to demonstrate its functionality. The final output represents the maximum amount of money the robber can steal without triggering the security system.",
-      "metadata": {},
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Test cases:\n\nHere's how you can use this solution:",
-      "metadata": {},
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 1:\n\n# Create an instance of the Solution class\nsolution = Solution()\n\nnums1 = [2, 3, 2]\nprint(solution.rob(nums1))  # Output: 3",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 8,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "3\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums2 = [1, 2, 3, 1]\nprint(solution.rob(nums2))  # Output: 4",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 9,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "4\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums3 = [1, 2, 3]\nprint(solution.rob(nums3))  # Output: 3",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 10,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "3\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "e6e1554f-ae77-410c-8af2-e59e5428166c"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Time and Space Complexity Analysis\n\nLet's analyze the time and space complexity of the provided code:\n\n### Time Complexity:\nThe time complexity is primarily determined by the `simple_rob` helper function, which has a time complexity of O(N), where N is the number of houses in the input array. The main `rob` function calls `simple_rob` twice, once for the case of robbing the first house and not robbing the last, and once for the case of not robbing the first house and robbing the last. Therefore, the overall time complexity is O(N).\n\n### Space Complexity:\nThe space complexity is O(1) because the space used by the algorithm is constant, regardless of the size of the input array. The only variables used are `prev` and `curr` in the `simple_rob` function, which represent the previous and current maximum amounts of money that can be robbed without alerting the police. These variables are updated iteratively, and no additional data structures are used that scale with the input size. Therefore, the space complexity is constant, or O(1).\n\nIn summary:\n- Time Complexity: O(N)\n- Space Complexity: O(1)",
-      "metadata": {},
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Challenging Exercises:\n\n1. **Optimization Challenge:** Optimize the space complexity of the solution. Can you solve the problem with O(1) space complexity without sacrificing the time complexity?\n\n2. **Circular Robbery Variations:** Consider variations of the circular house robbery problem, such as allowing the robber to skip a certain number of houses or restricting the number of consecutive houses that can be robbed. Modify the solution to accommodate these variations.\n\n**Link to Problem:** [To see the House Robber II problem on LeetCode, click here!](https://leetcode.com/problems/house-robber-ii/)",
-      "metadata": {},
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6"
-    }
-  ]
-}
\ No newline at end of file
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb b/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb
deleted file mode 100644
index b727489..0000000
--- a/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb	
+++ /dev/null
@@ -1,210 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 70. Climbing Stairs\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Climbing Stairs problem on LeetCode, click here!](https://leetcode.com/problems/climbing-stairs/)\n",
-        "\n",
-        "---\n",
-        "You are climbing a staircase. It takes `n` steps to reach the top.\n",
-        "\n",
-        "Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= n <= 45`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The \"Climbing Stairs\" problem is a classic algorithmic problem that asks you to find the number of distinct ways to climb a staircase with a given number of steps. Here's the problem statement:\n",
-        "\n",
-        "You are given a staircase with 'n' steps. Each time you can either climb 1 step or 2 steps. You need to find out how many distinct ways there are to reach the top of the staircase.\n",
-        "\n",
-        "For example, if 'n' is 2, there are two ways to climb to the top:\n",
-        "\n",
-        "1. Climbing two steps at once.\n",
-        "2. Climbing one step, and then another step.\n",
-        "\n",
-        "If 'n' is 3, there are three ways:\n",
-        "\n",
-        "1. Climbing one step, then one step, and finally one step.\n",
-        "2. Climbing one step, then two steps.\n",
-        "3. Climbing two steps, then one step.\n",
-        "\n",
-        "The problem essentially asks you to find a solution that counts the possible combinations of 1-step and 2-step climbs that can be taken to reach the top of the staircase. It's a classic example of a dynamic programming problem where you can break it down into smaller subproblems and build a solution incrementally, as shown in the code I provided earlier.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class Solution:\n",
-        "    def climbStairs(self, n: int) -> int:\n",
-        "        # If there are 1 or 2 steps, there are exactly n ways to reach the top.\n",
-        "        if n <= 2:\n",
-        "            return n\n",
-        "\n",
-        "        # Create an array to store the number of ways to reach each step.\n",
-        "        dp = [0] * (n + 1)\n",
-        "\n",
-        "        # There is one way to reach the first step (base case).\n",
-        "        dp[1] = 1\n",
-        "\n",
-        "        # There are two ways to reach the second step (base case).\n",
-        "        dp[2] = 2\n",
-        "\n",
-        "        # Calculate the number of ways to reach each step from the previous steps.\n",
-        "        for i in range(3, n + 1):\n",
-        "            # The number of ways to reach step 'i' is the sum of the ways to reach\n",
-        "            # the previous two steps, as you can take 1 or 2 steps at a time.\n",
-        "            dp[i] = dp[i - 1] + dp[i - 2]\n",
-        "\n",
-        "        # The result is the number of ways to reach the top, which is stored in dp[n].\n",
-        "        return dp[n]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a class `Solution` with a method `climbStairs` that takes an integer `n` as input and returns the number of distinct ways to climb a staircase with `n` steps.\n",
-        "\n",
-        "1. If `n` is 1 or 2, the function returns `n` directly because there are only 1 or 2 distinct ways to reach the top in those cases.\n",
-        "\n",
-        "2. For `n` greater than 2, the code uses dynamic programming to calculate the number of ways to reach each step. It creates an array `dp` of length `n + 1` to store these values.\n",
-        "\n",
-        "3. The `dp[1]` and `dp[2]` are initialized to 1 and 2, respectively. This is because there is only one way to reach the first step (by taking 1 step), and there are two ways to reach the second step (by taking 2 steps or two 1-step climbs).\n",
-        "\n",
-        "4. The code then uses a loop to fill in the `dp` array for steps 3 and above. For each step `i` from 3 to `n`, it calculates the number of ways to reach that step by adding the number of ways to reach the previous two steps, as you can either take 1 step or 2 steps at a time.\n",
-        "\n",
-        "5. Finally, the function returns the value stored in `dp[n]`, which represents the total number of distinct ways to climb the staircase with `n` steps."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 1 - Input: n = 2, Output: 2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "n = 2\n",
-        "solution = Solution()\n",
-        "print(f\"Example 1 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 2"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 2 - Input: n = 3, Output: 3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "n = 3\n",
-        "print(f\"Example 2 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 3\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Time Complexity:\n",
-        "The time complexity of the solution is O(n) because we use a single for loop that iterates from 3 to n to fill the `dp` array. In each iteration, we perform constant time operations (addition and assignment). Therefore, the time complexity is linear with respect to the input value `n`.\n",
-        "\n",
-        "Space Complexity:\n",
-        "The space complexity of the solution is O(n) as well. We use an array `dp` of length `n + 1` to store the number of ways to reach each step. The size of this array is directly proportional to the input value `n`, which makes the space complexity linear in terms of `n`. The rest of the variables used in the function (e.g., `i`) occupy constant space and do not contribute significantly to the overall space complexity.\n",
-        "\n",
-        "In summary, the time complexity is O(n), and the space complexity is O(n) for this dynamic programming solution to the \"Climbing Stairs\" problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Generalized Step Sizes**: Instead of being limited to 1 step or 2 steps, consider a scenario where you can take steps of sizes `[1, 2, 3]`. Implement a function that calculates the number of ways to climb the staircase with these different step sizes.\n",
-        "\n",
-        "2. **Minimize Steps**: Given an integer `n`, find the minimum number of steps required to reach the top of the staircase. Return both the minimum number of steps and the specific sequence of steps taken."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/README.md b/_build/html/_sources/13. One-D Dynamic Programming/README.md
deleted file mode 100644
index 73f4de3..0000000
--- a/_build/html/_sources/13. One-D Dynamic Programming/README.md	
+++ /dev/null
@@ -1,25 +0,0 @@
-# 1-D Dynamic Programming Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)                                   | Easy       |
-| [198. House Robber](https://leetcode.com/problems/house-robber/)                                       | Medium     |
-| [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)                                 | Medium     |
-| [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)       | Medium     |
-| [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)                   | Medium     |
-| [91. Decode Ways](https://leetcode.com/problems/decode-ways/)                                         | Medium     |
-| [322. Coin Change](https://leetcode.com/problems/coin-change/)                                         | Medium     |
-| [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)             | Medium     |
-| [139. Word Break](https://leetcode.com/problems/word-break/)                                           | Medium     |
-| [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/14. Two-D Dynamic Programming/README.md b/_build/html/_sources/14. Two-D Dynamic Programming/README.md
deleted file mode 100644
index 8bab876..0000000
--- a/_build/html/_sources/14. Two-D Dynamic Programming/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### 2-D Dynamic Programming
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [62. Unique Paths](https://leetcode.com/problems/unique-paths/)                                         | Medium     |
-| [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)       | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/15. Greedy/README.md b/_build/html/_sources/15. Greedy/README.md
deleted file mode 100644
index 82b6543..0000000
--- a/_build/html/_sources/15. Greedy/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Greedy
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)                                   | Easy       |
-| [55. Jump Game](https://leetcode.com/problems/jump-game/)                                                 | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/16. Intervals/README.md b/_build/html/_sources/16. Intervals/README.md
deleted file mode 100644
index a57713e..0000000
--- a/_build/html/_sources/16. Intervals/README.md	
+++ /dev/null
@@ -1,50 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Intervals
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [57. Insert Interval](https://leetcode.com/problems/insert-interval/)                                     | Hard       |
-| [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)                                   | Medium     |
-| [435. Non Overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)             | Medium     |
-| [252. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)                                       | Easy       |
-| [253. Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)                                 | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/17. Math Geometry/README.md b/_build/html/_sources/17. Math Geometry/README.md
deleted file mode 100644
index 8b2f180..0000000
--- a/_build/html/_sources/17. Math Geometry/README.md	
+++ /dev/null
@@ -1,48 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Math & Geometry
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [48. Rotate Image](https://leetcode.com/problems/rotate-image/)                                           | Medium     |
-| [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)                                         | Medium     |
-| [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)                               | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/18. Bit Manipulation/README.md b/_build/html/_sources/18. Bit Manipulation/README.md
deleted file mode 100644
index 05b2fef..0000000
--- a/_build/html/_sources/18. Bit Manipulation/README.md	
+++ /dev/null
@@ -1,51 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Bit Manupulation
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)                                   | Easy       |
-| [338. Counting Bits](https://leetcode.com/problems/counting-bits/)                                         | Medium     |
-| [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)                                           | Easy       |
-| [268. Missing Number](https://leetcode.com/problems/missing-number/)                                       | Easy       |
-| [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)                             | Medium     |
-
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/CODE_OF_CONDUCT.md b/_build/html/_sources/CODE_OF_CONDUCT.md
deleted file mode 100644
index 5fe5041..0000000
--- a/_build/html/_sources/CODE_OF_CONDUCT.md
+++ /dev/null
@@ -1,128 +0,0 @@
-# Contributor Covenant Code of Conduct
-
-## Our Pledge
-
-We as members, contributors, and leaders pledge to make participation in our
-community a harassment-free experience for everyone, regardless of age, body
-size, visible or invisible disability, ethnicity, sex characteristics, gender
-identity and expression, level of experience, education, socio-economic status,
-nationality, personal appearance, race, religion, or sexual identity
-and orientation.
-
-We pledge to act and interact in ways that contribute to an open, welcoming,
-diverse, inclusive, and healthy community.
-
-## Our Standards
-
-Examples of behavior that contributes to a positive environment for our
-community include:
-
-* Demonstrating empathy and kindness toward other people
-* Being respectful of differing opinions, viewpoints, and experiences
-* Giving and gracefully accepting constructive feedback
-* Accepting responsibility and apologizing to those affected by our mistakes,
-  and learning from the experience
-* Focusing on what is best not just for us as individuals, but for the
-  overall community
-
-Examples of unacceptable behavior include:
-
-* The use of sexualized language or imagery, and sexual attention or
-  advances of any kind
-* Trolling, insulting or derogatory comments, and personal or political attacks
-* Public or private harassment
-* Publishing others' private information, such as a physical or email
-  address, without their explicit permission
-* Other conduct which could reasonably be considered inappropriate in a
-  professional setting
-
-## Enforcement Responsibilities
-
-Community leaders are responsible for clarifying and enforcing our standards of
-acceptable behavior and will take appropriate and fair corrective action in
-response to any behavior that they deem inappropriate, threatening, offensive,
-or harmful.
-
-Community leaders have the right and responsibility to remove, edit, or reject
-comments, commits, code, wiki edits, issues, and other contributions that are
-not aligned to this Code of Conduct, and will communicate reasons for moderation
-decisions when appropriate.
-
-## Scope
-
-This Code of Conduct applies within all community spaces, and also applies when
-an individual is officially representing the community in public spaces.
-Examples of representing our community include using an official e-mail address,
-posting via an official social media account, or acting as an appointed
-representative at an online or offline event.
-
-## Enforcement
-
-Instances of abusive, harassing, or otherwise unacceptable behavior may be
-reported to the community leaders responsible for enforcement at
-Email.
-All complaints will be reviewed and investigated promptly and fairly.
-
-All community leaders are obligated to respect the privacy and security of the
-reporter of any incident.
-
-## Enforcement Guidelines
-
-Community leaders will follow these Community Impact Guidelines in determining
-the consequences for any action they deem in violation of this Code of Conduct:
-
-### 1. Correction
-
-**Community Impact**: Use of inappropriate language or other behavior deemed
-unprofessional or unwelcome in the community.
-
-**Consequence**: A private, written warning from community leaders, providing
-clarity around the nature of the violation and an explanation of why the
-behavior was inappropriate. A public apology may be requested.
-
-### 2. Warning
-
-**Community Impact**: A violation through a single incident or series
-of actions.
-
-**Consequence**: A warning with consequences for continued behavior. No
-interaction with the people involved, including unsolicited interaction with
-those enforcing the Code of Conduct, for a specified period of time. This
-includes avoiding interactions in community spaces as well as external channels
-like social media. Violating these terms may lead to a temporary or
-permanent ban.
-
-### 3. Temporary Ban
-
-**Community Impact**: A serious violation of community standards, including
-sustained inappropriate behavior.
-
-**Consequence**: A temporary ban from any sort of interaction or public
-communication with the community for a specified period of time. No public or
-private interaction with the people involved, including unsolicited interaction
-with those enforcing the Code of Conduct, is allowed during this period.
-Violating these terms may lead to a permanent ban.
-
-### 4. Permanent Ban
-
-**Community Impact**: Demonstrating a pattern of violation of community
-standards, including sustained inappropriate behavior,  harassment of an
-individual, or aggression toward or disparagement of classes of individuals.
-
-**Consequence**: A permanent ban from any sort of public interaction within
-the community.
-
-## Attribution
-
-This Code of Conduct is adapted from the [Contributor Covenant][homepage],
-version 2.0, available at
-https://www.contributor-covenant.org/version/2/0/code_of_conduct.html.
-
-Community Impact Guidelines were inspired by [Mozilla's code of conduct
-enforcement ladder](https://github.com/mozilla/diversity).
-
-[homepage]: https://www.contributor-covenant.org
-
-For answers to common questions about this code of conduct, see the FAQ at
-https://www.contributor-covenant.org/faq. Translations are available at
-https://www.contributor-covenant.org/translations.
diff --git a/_build/html/_sources/CONTRIBUTING.md b/_build/html/_sources/CONTRIBUTING.md
deleted file mode 100644
index ca794ef..0000000
--- a/_build/html/_sources/CONTRIBUTING.md
+++ /dev/null
@@ -1,79 +0,0 @@
-# Contributing Guidelines
-
-Thank you for considering contributing to the Blind 75 LeetCode repository! We welcome contributions from the community to help improve this resource and make it even more valuable.
-
-Please take a moment to review this document to understand how you can contribute.
-
-## Table of Contents
-- [Ways to Contribute](#ways-to-contribute)
-- [Getting Started](#getting-started)
-- [Contributing Process](#contributing-process)
-- [Code of Conduct](#code-of-conduct)
-
-## Ways to Contribute
-
-There are several ways you can contribute to this repository:
-
-1. **Suggest Improvements**: If you have ideas for making the Blind 75 LeetCode repository better, please open an issue to discuss your suggestions.
-
-2. **Report Issues**: If you encounter any issues with the existing content or code, please open an issue and provide as much detail as possible.
-
-3. **Contribute Solutions**: You can contribute solutions to the problems listed in the repository. Please follow the guidelines mentioned below when submitting your solutions.
-
-4. **Enhance Documentation**: Improve documentation, including explanations and time/space complexity analysis, for existing solutions.
-
-5. **Fix Bugs**: If you find and fix any bugs in the existing code or documentation, you are welcome to submit a pull request.
-
-## Getting Started
-
-Before you start contributing, please ensure you have the following prerequisites:
-
-- A GitHub account.
-- Familiarity with Git and GitHub.
-- Code editor or IDE for making changes.
-
-## Contributing Process
-
-Here's the general process for contributing:
-
-1. **Fork the Repository**: Click the "Fork" button at the top of this repository to create your copy.
-
-2. **Clone Your Fork**: Clone your fork to your local machine.
-
-    ```
-    git clone https://github.com/your-username/Blind-75-LeetCode.git
-    ```
-
-3. **Create a Branch**: Create a new branch for your contribution.
-
-    ```
-    git checkout -b feature/your-feature-name
-    ```
-
-4. **Make Changes**: Implement your changes, following the guidelines provided.
-
-5. **Test Your Changes**: Ensure that your changes do not introduce new issues and are consistent with the repository's style and structure.
-
-6. **Commit Changes**: Commit your changes with a meaningful commit message.
-
-    ```
-    git commit -m "Add solution to problem X"
-    ```
-
-7. **Push Changes**: Push your changes to your fork on GitHub.
-
-    ```
-    git push origin feature/your-feature-name
-    ```
-
-8. **Create a Pull Request**: Go to the original repository and create a pull request. Please provide a detailed description of your changes.
-
-9. **Review and Collaboration**: Your pull request will be reviewed, and you may need to make further changes. Be responsive to comments and feedback.
-
-10. **Merge**: Once your pull request is approved, it will be merged into the main branch.
-
-## Code of Conduct
-
-Please note that we have a [Code of Conduct](CODE_OF_CONDUCT.md) that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.
-
-We appreciate your contributions to the Blind 75 LeetCode repository and look forward to working together to make it an even better resource for the community!
diff --git a/_build/html/_sources/README.md b/_build/html/_sources/README.md
deleted file mode 100644
index a9b8b2e..0000000
--- a/_build/html/_sources/README.md
+++ /dev/null
@@ -1,216 +0,0 @@
-# Blind 75 LeetCode Solutions
-
-![Blind 75 LeetCode](https://img.shields.io/badge/Blind_75_LeetCode-Solutions-blue?labelColor=red)
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Array & Hashing
-![Array & Hashing](https://img.shields.io/badge/Array_&_Hashing-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [217. Contains Duplicate](https://leetcode.com/problems/contains-duplicate/)                        | Easy       |
-| [242. Valid Anagram](https://leetcode.com/problems/valid-anagram/)                                  | Easy       |
-| [1. Two Sum](https://leetcode.com/problems/two-sum/)                                              | Easy       |
-| [49. Group Anagrams](https://leetcode.com/problems/group-anagrams/)                                | Medium     |
-| [347. Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/)              | Medium     |
-| [238. Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/)    | Medium     |
-| [659. Encode and Decode Strings](https://leetcode.com/problems/encode-and-decode-strings/)          | Medium     |
-| [128. Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/)    | Hard       |
-
-### Two Pointers
-![Array & Hashing](https://img.shields.io/badge/Two_Pointers-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)                            | Easy       |
-| [15. 3Sum](https://leetcode.com/problems/3sum/)                                                       | Medium     |
-| [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)           | Medium     |
-
-### Sliding Window
-![Array & Hashing](https://img.shields.io/badge/Sliding_Window-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [121. Best Time to Buy And Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | Easy       |
-| [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) | Medium |
-| [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) | Medium |
-| [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)             | Hard       |
-
-### Stack
-![Array & Hashing](https://img.shields.io/badge/Stack-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)                            | Easy       |
-
-### Binary Search
-![Array & Hashing](https://img.shields.io/badge/Binary_Search-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [153. Find Minimum In Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) | Medium |
-| [33. Search In Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)   | Medium     |
-
-### Linked List
-![Array & Hashing](https://img.shields.io/badge/Linked_List-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)                       | Easy       |
-| [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)                   | Easy       |
-| [143. Reorder List](https://leetcode.com/problems/reorder-list/)                                     | Medium     |
-| [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | Medium |
-| [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)                           | Medium     |
-| [23. Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)                       | Hard       |
-
-### Trees
-![Array & Hashing](https://img.shields.io/badge/Trees-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)                          | Easy       |
-| [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)     | Easy       |
-| [100. Same Tree](https://leetcode.com/problems/same-tree/)                                           | Easy       |
-| [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)               | Easy       |
-| [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/) | Easy |
-| [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/) | Medium |
-| [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)       | Medium     |
-| [230. Kth Smallest Element In a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)   | Medium     |
-| [105. Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) | Medium |
-| [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)     | Hard       |
-| [297. Serialize And Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/) | Hard |
-
-### Tries
-![Array & Hashing](https://img.shields.io/badge/Tries-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [208. Implement Trie Prefix Tree](https://leetcode.com/problems/implement-trie-prefix-tree/)         | Medium     |
-| [211. Design Add And Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/) | Medium |
-| [212. Word Search II](https://leetcode.com/problems/word-search-ii/)                                 | Hard       |
-
-### Heap / Priority Queue
-![Array & Hashing](https://img.shields.io/badge/Heap_/_Priority_Queue-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [295. Find Median From Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)     | Hard       |
-
-### Backtracking
-![Array & Hashing](https://img.shields.io/badge/Backtracking-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [39. Combination Sum](https://leetcode.com/problems/combination-sum/)                                 | Medium     |
-| [79. Word Search](https://leetcode.com/problems/word-search/)                                         | Medium     |
-
-### Graphs
-![Array & Hashing](https://img.shields.io/badge/Graphs-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)                           | Medium     |
-| [133. Clone Graph](https://leetcode.com/problems/clone-graph/)                                       | Medium     |
-| [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)       | Medium     |
-| [207. Course Schedule](https://leetcode.com/problems/course-schedule/)                                 | Medium     |
-| [323. Number of Connected Components In An Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/) | Medium |
-| [261. Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)                               | Medium     |
-
-### Advanced Graphs
-![Array & Hashing](https://img.shields.io/badge/Advanced_Graphs-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [269. Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)                               | Hard       |
-
-### 1-D Dynamic Programming
-![Array & Hashing](https://img.shields.io/badge/1_D_Dynamic_Programming-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)                                   | Easy       |
-| [198. House Robber](https://leetcode.com/problems/house-robber/)                                       | Easy       |
-| [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)                                 | Medium     |
-| [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)       | Medium     |
-| [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)                   | Medium     |
-| [91. Decode Ways](https://leetcode.com/problems/decode-ways/)                                         | Medium     |
-| [322. Coin Change](https://leetcode.com/problems/coin-change/)                                         | Medium     |
-| [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)             | Medium     |
-| [139. Word Break](https://leetcode.com/problems/word-break/)                                           | Medium     |
-| [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | Medium     |
-
-### 2-D Dynamic Programming
-![Array & Hashing](https://img.shields.io/badge/2_D_Dynamic_Programming-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [62. Unique Paths](https://leetcode.com/problems/unique-paths/)                                         | Medium     |
-| [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)       | Medium     |
-
-### Greedy
-![Array & Hashing](https://img.shields.io/badge/Greedy-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)                                   | Easy       |
-| [55. Jump Game](https://leetcode.com/problems/jump-game/)                                                 | Medium     |
-
-### Intervals
-![Array & Hashing](https://img.shields.io/badge/Intervals-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [57. Insert Interval](https://leetcode.com/problems/insert-interval/)                                     | Hard       |
-| [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)                                   | Medium     |
-| [435. Non Overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)             | Medium     |
-| [252. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)                                       | Easy       |
-| [253. Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)                                 | Medium     |
-
-### Math & Geometry
-![Array & Hashing](https://img.shields.io/badge/Math_&_Geometry-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [48. Rotate Image](https://leetcode.com/problems/rotate-image/)                                           | Medium     |
-| [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)                                         | Medium     |
-| [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)                               | Medium     |
-
-### Bit Manipulation
-![Array & Hashing](https://img.shields.io/badge/Bit_Manipulation-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)                                   | Easy       |
-| [338. Counting Bits](https://leetcode.com/problems/counting-bits/)                                         | Medium     |
-| [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)                                           | Easy       |
-| [268. Missing Number](https://leetcode.com/problems/missing-number/)                                       | Easy       |
-| [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)                             | Medium     |
-
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-## Code of Conduct
-
-We have a [Code of Conduct](CODE_OF_CONDUCT.md) that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.
-
-## Security Policy
-
-We take security seriously. Please review our [Security Policy](SECURITY.md) for information on reporting security vulnerabilities.
-
-Happy coding!
diff --git a/_build/html/_sources/SECURITY.md b/_build/html/_sources/SECURITY.md
deleted file mode 100644
index 7c19993..0000000
--- a/_build/html/_sources/SECURITY.md
+++ /dev/null
@@ -1,42 +0,0 @@
-# Security Policy
-
-## Reporting a Vulnerability
-
-We take the security of the Blind 75 LeetCode repository seriously. If you believe you have found a security vulnerability, please help us by following these steps:
-
-1. **Do Not Disclose Publicly**: Please do not disclose the issue publicly until it has been addressed by our team.
-
-2. **Report Privately**: Create a GitHub issue with the title "Security Vulnerability Report" and provide a detailed description of the vulnerability. You can use a sample report template as follows:
-
-    ```
-    **Description**
-    A clear and concise description of the security vulnerability.
-
-    **Steps to Reproduce**
-    Provide detailed steps on how to reproduce the vulnerability.
-
-    **Expected Behavior**
-    Describe what you expected to happen.
-
-    **Actual Behavior**
-    Describe what actually happened.
-
-    **Additional Information**
-    Any additional information that can help us understand and reproduce the issue, such as environment details, configuration, or screenshots.
-    ```
-
-3. **Secure Communication**: If the vulnerability involves sensitive data, please encrypt your communication with our PGP key. Contact us for our PGP key information.
-
-4. **Response**: We will acknowledge your report within 5 business days and provide an estimated timeline for the resolution.
-
-5. **Fix and Disclosure**: We will work to fix the issue promptly and privately. Once resolved, we will provide a security advisory and give credit to the reporter if desired.
-
-6. **Responsible Disclosure**: We encourage responsible disclosure. Please do not take advantage of the vulnerability or attempt to exploit it for malicious purposes.
-
-## Security Updates
-
-We are committed to promptly addressing and resolving any security issues. We recommend that you stay up-to-date with the latest releases of this repository, as they may include security updates and fixes.
-
-## Contact
-
-If you have any questions or need further assistance regarding security, you can contact us at mohsentabibian@gmail.com.
diff --git a/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb b/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb
deleted file mode 100644
index 5d55ca7..0000000
--- a/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 100: Same Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-    "\n",
-    "---\n",
-    "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-    "\n",
-    "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-    "\n",
-    "**Constraints**\n",
-    "\n",
-    "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-    "2. $-10^4$ <= `Node.val` <= $10^4$"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def isSameTree(p, q):\n",
-    "    # Base case: If both p and q are None, the trees are the same.\n",
-    "    if not p and not q:\n",
-    "        return True\n",
-    "    \n",
-    "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-    "    if not p or not q:\n",
-    "        return False\n",
-    "    \n",
-    "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-    "    if p.val != q.val:\n",
-    "        return False\n",
-    "    \n",
-    "    # Recursively check the left and right subtrees of p and q.\n",
-    "    # If both subtrees are the same, the entire trees are the same.\n",
-    "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-    "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-    "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-    "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-    "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-    "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "\n",
-    "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-    "\n",
-    "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "print(isSameTree(p1, q1)) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "\n",
-    "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-    "\n",
-    "p2 = TreeNode(1, TreeNode(2), None)\n",
-    "q2 = TreeNode(1, None, TreeNode(2))\n",
-    "print(isSameTree(p2, q2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "\n",
-    "#Input: p = [1,2,1], q = [1,1,2]\n",
-    "\n",
-    "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-    "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-    "print(isSameTree(p3, q3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "**Time Complexity**\n",
-    "\n",
-    "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-    "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-    "\n",
-    "**Space Complexity**\n",
-    "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-    "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-    "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-    "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-    "\n",
-    "**In summary:**\n",
-    "\n",
-    "+ Time Complexity: $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-    "+ Space Complexity: $O(n)$ due to the recursive call stack."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb b/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb
deleted file mode 100644
index 9256d0e..0000000
--- a/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 128. Longest Consecutive Sequence\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Longest Consecutive Sequence problem on LeetCode, click here!](https://leetcode.com/problems/longest-consecutive-sequence/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given an unsorted array of integers `nums`, return *the length of the longest consecutive elements sequence.*\n",
-    "\n",
-    "You must write an algorithm that runs in `O(n)` time.\n",
-    "\n",
-    "*Constraints:**\n",
-    "- 0 <= `nums.length` <= $10^5$\n",
-    "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def longestConsecutive(nums):\n",
-    "    if not nums:\n",
-    "        return 0\n",
-    "\n",
-    "    num_set = set(nums)\n",
-    "    max_length = 0\n",
-    "\n",
-    "    for num in num_set:\n",
-    "        if num - 1 not in num_set:\n",
-    "            current_num = num\n",
-    "            current_length = 1\n",
-    "\n",
-    "            while current_num + 1 in num_set:\n",
-    "                current_num += 1\n",
-    "                current_length += 1\n",
-    "\n",
-    "            max_length = max(max_length, current_length)\n",
-    "\n",
-    "    return max_length"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining the `longestConsecutive` function that takes the `nums` array as input.\n",
-    "\n",
-    "2. We check if the `nums` array is empty. If it's empty, there are no consecutive elements, so we return 0.\n",
-    "\n",
-    "3. We create a Python set called `num_set` and insert all elements from the `nums` array into it. Using a set allows us to efficiently check for the existence of elements in O(1) time.\n",
-    "\n",
-    "4. We initialize a variable `max_length` to 0. This variable will keep track of the maximum length of consecutive elements sequence found.\n",
-    "\n",
-    "5. We iterate through the elements of the `num_set`. For each element `num`, we check if `num - 1` exists in the `num_set`. If it doesn't exist, it means `num` is the starting element of a potential consecutive sequence.\n",
-    "\n",
-    "6. Inside the loop, we initialize two variables: `current_num` to the current element `num` and `current_length` to 1. We start with a length of 1 because `num` itself is part of the sequence.\n",
-    "\n",
-    "7. We then enter a while loop that continues as long as `current_num + 1` exists in the `num_set`. This means we are incrementing the consecutive sequence.\n",
-    "\n",
-    "8. Inside the while loop, we increment `current_num` by 1 and also increment `current_length` by 1 to account for the next consecutive element.\n",
-    "\n",
-    "9. We compare `current_length` with the `max_length` and update `max_length` if the current sequence is longer.\n",
-    "\n",
-    "10. After the while loop, we move to the next element in the outer loop and repeat the process.\n",
-    "\n",
-    "11. Finally, we return the `max_length` as the result, which represents the length of the longest consecutive elements sequence in the `nums` array.\n",
-    "\n",
-    "The key idea here is to use a set to efficiently check for the existence of elements and to iterate through the elements, considering each element as the potential start of a consecutive sequence. By doing this, we can find the longest consecutive sequence in O(n) time complexity, where n is the number of elements in the array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "nums1 = [100, 4, 200, 1, 3, 2]\n",
-    "print(longestConsecutive(nums1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "9\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "nums2 = [0, 3, 7, 2, 5, 8, 4, 6, 0, 1]\n",
-    "print(longestConsecutive(nums2)) "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The code is designed to run in O(n) time complexity, where n is the number of elements in the `nums` array. Here's the breakdown:\n",
-    "\n",
-    "1. Constructing the `num_set` by inserting all elements from `nums` into it takes O(n) time because we perform an insertion operation for each element in `nums`.\n",
-    "\n",
-    "2. The main loop iterates through the elements in `num_set`. In the worst case, each element is visited only once, so the loop itself takes O(n) time.\n",
-    "\n",
-    "3. Within the loop, we have a while loop that may also take O(n) time in the worst case. However, this while loop is nested inside the main loop, so its overall time complexity remains O(n).\n",
-    "\n",
-    "Therefore, the overall time complexity of the code is O(n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity of the code is determined by the space used by the `num_set` and a few additional variables. Here's the breakdown:\n",
-    "\n",
-    "1. `num_set` is a set that stores the unique elements from `nums`. In the worst case, it can store all n elements from `nums`, so the space complexity is O(n).\n",
-    "\n",
-    "2. The additional variables used, such as `max_length`, `current_num`, and `current_length`, have constant space requirements and do not depend on the size of the input array.\n",
-    "\n",
-    "Therefore, the overall space complexity of the code is O(n) due to the space used by the `num_set`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. Write an algorithm that not only returns the length of the longest consecutive elements sequence but also returns the actual consecutive sequence itself.\n",
-    "\n",
-    "2. Extend the problem to allow elements to be considered consecutive if they are within a certain absolute difference (e.g., less than or equal to k) instead of exactly 1. Write an algorithm that finds the longest such sequence and runs in O(n) time."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb b/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb
deleted file mode 100644
index f7dbbe9..0000000
--- a/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb	
+++ /dev/null
@@ -1,210 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 217. Contains Duplicate\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Contains Duplicate problem on LeetCode, click here!](https://leetcode.com/problems/contains-duplicate/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `nums.length` <= $10^5$\n",
-    "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def containsDuplicate(nums):\n",
-    "    # Create an empty set to store unique elements\n",
-    "    unique_elements = set()\n",
-    "    \n",
-    "    # Iterate through the array\n",
-    "    for num in nums:\n",
-    "        # If the element is already in the set, return True\n",
-    "        if num in unique_elements:\n",
-    "            return True\n",
-    "        # Otherwise, add it to the set\n",
-    "        else:\n",
-    "            unique_elements.add(num)\n",
-    "    \n",
-    "    # If the loop completes without finding duplicates, return False\n",
-    "    return False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The `containsDuplicate` function takes a single argument, `nums`, which is the input integer array.\n",
-    "\n",
-    "2. Inside the function, an empty set called `unique_elements` is created. This set will be used to keep track of unique elements in the input array.\n",
-    "\n",
-    "3. The function then iterates through the input array `nums` using a `for` loop.\n",
-    "\n",
-    "4. For each element `num` in the array, it checks whether `num` is already in the `unique_elements` set using the `if num in unique_elements:` condition.\n",
-    "\n",
-    "5. If `num` is already in the set, it means there is a duplicate element in the array, and the function immediately returns `True`.\n",
-    "\n",
-    "6. If `num` is not in the set, it is added to the `unique_elements` set using `unique_elements.add(num)`.\n",
-    "\n",
-    "7. The loop continues to the next element, and the process repeats.\n",
-    "\n",
-    "8. If the loop completes without finding any duplicates, it means that all elements in the array are distinct, and the function returns `False`.\n",
-    "\n",
-    "The code efficiently utilizes a set data structure to keep track of unique elements while iterating through the array, allowing it to quickly detect duplicate elements. This code meets the problem's requirements and constraints, as explained earlier."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "nums1 = [1, 2, 3, 1]\n",
-    "print(containsDuplicate(nums1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "nums2 = [1, 2, 3, 4]\n",
-    "print(containsDuplicate(nums2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "nums3 = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]\n",
-    "print(containsDuplicate(nums3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for the \"Contains Duplicate\" problem:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The primary operation that affects the time complexity is the `for` loop that iterates through the input array `nums`. In the worst case, the loop will iterate through all `n` elements of the array, where `n` is the length of the input array.\n",
-    "\n",
-    "- Iterating through the array: O(n)\n",
-    "\n",
-    "Inside the loop, we perform two operations:\n",
-    "\n",
-    "1. Checking whether an element exists in the `unique_elements` set (`if num in unique_elements`). This operation has an average time complexity of O(1) for a set.\n",
-    "2. Adding an element to the `unique_elements` set (`unique_elements.add(num)`), which also has an average time complexity of O(1) for a set.\n",
-    "\n",
-    "Since these operations are performed for each element in the array, the overall time complexity remains O(n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity is determined by the additional data structures used in the code, which are the `unique_elements` set.\n",
-    "\n",
-    "- `unique_elements` set: This set stores unique elements from the input array. In the worst case, if all elements in the input array are distinct, the set will store all `n` elements.\n",
-    "\n",
-    "Therefore, the space complexity is O(n) because, in the worst case, the set's size will grow linearly with the input array's size.\n",
-    "\n",
-    "**In summary**, the time complexity of the code is O(n), where `n` is the length of the input array, and the space complexity is also O(n) in the worst case. This code is efficient and meets the constraints of the problem."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **K Duplicates:** Modify the problem to return `true` if there are exactly K duplicate elements in the array. Write a function that takes an additional integer parameter K and returns `true` if there are exactly K duplicates.\n",
-    "2. **Frequency Count:** Write a function that returns a list of all the elements that appear more than once in the array, along with their frequencies."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb b/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb
deleted file mode 100644
index c5a69c0..0000000
--- a/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb	
+++ /dev/null
@@ -1,191 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 238. Product of Array Except Self\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Product of Array Except Self problem on LeetCode, click here!](https://leetcode.com/problems/product-of-array-except-self/)\n",
-    "\n",
-    "---\n",
-    "Given an integer array `nums`, return *an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.*\n",
-    "\n",
-    "The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-    "\n",
-    "You must write an algorithm that runs in O(n) time and without using the division operation.\n",
-    "\n",
-    "Constraints:\n",
-    "- 2 <= `nums.length` <= $10^5$\n",
-    "- `-30 <= nums[i] <= 30`\n",
-    "- The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-    "\n",
-    "**Follow-up:** Can you solve the problem in O(1) extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def productExceptSelf(nums):\n",
-    "    n = len(nums)\n",
-    "    \n",
-    "    # Initialize two arrays to store products to the left and right of each element\n",
-    "    left_products = [1] * n\n",
-    "    right_products = [1] * n\n",
-    "    \n",
-    "    # Calculate products to the left of each element\n",
-    "    left_product = 1\n",
-    "    for i in range(n):\n",
-    "        left_products[i] = left_product\n",
-    "        left_product *= nums[i]\n",
-    "    \n",
-    "    # Calculate products to the right of each element\n",
-    "    right_product = 1\n",
-    "    for i in range(n - 1, -1, -1):\n",
-    "        right_products[i] = right_product\n",
-    "        right_product *= nums[i]\n",
-    "    \n",
-    "    # Calculate the final answer using left and right products\n",
-    "    answer = [left_products[i] * right_products[i] for i in range(n)]\n",
-    "    \n",
-    "    return answer"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `n` is initialized as the length of the `nums` array.\n",
-    "2. Two arrays, `left_products` and `right_products`, are created, each with a length of `n` and filled with ones initially.\n",
-    "3. `left_product` is initialized to 1.\n",
-    "4. A loop iterates through the `nums` array, from left to right. For each element at index `i`, it stores the product of all elements to its left (including itself) in the `left_products` array and updates `left_product` accordingly.\n",
-    "5. `right_product` is initialized to 1.\n",
-    "6. Another loop iterates through the `nums` array in reverse order, from right to left. For each element at index `i`, it stores the product of all elements to its right (including itself) in the `right_products` array and updates `right_product`.\n",
-    "7. The final `answer` list is constructed by multiplying corresponding elements from the `left_products` and `right_products` arrays for each index `i`.\n",
-    "8. The function returns the `answer` list, which contains the desired results.\n",
-    "\n",
-    "Overall, this algorithm efficiently computes the product of all elements except the current element, as required. It runs in O(n) time complexity and uses O(n) extra space, which is within the problem's constraints."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[24, 12, 8, 6]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "nums = [1,2,3,4]\n",
-    "result1 = productExceptSelf(nums)\n",
-    "print(result1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[0, 0, 9, 0, 0]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums = [-1,1,0,-3,3]\n",
-    "result2 = productExceptSelf(nums)\n",
-    "print(result2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. The first loop iterates through the `nums` array to calculate the products to the left of each element. This loop runs in O(n) time, where n is the length of the `nums` array.\n",
-    "2. The second loop also iterates through the `nums` array in reverse order to calculate the products to the right of each element. This loop also runs in O(n) time.\n",
-    "3. The final loop constructs the `answer` list by multiplying elements from the `left_products` and `right_products` arrays. This loop runs in O(n) time.\n",
-    "\n",
-    "Since all the loops are independent and sequential, the overall time complexity is O(n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. Two additional arrays, `left_products` and `right_products`, are created with a length of `n`, where `n` is the length of the `nums` array. Therefore, these arrays consume O(n) extra space.\n",
-    "2. The `left_product` and `right_product` variables are used to keep track of the product to the left and right of each element, respectively. These variables occupy O(1) extra space.\n",
-    "3. The `answer` list, which contains the final results, also consumes O(n) space, but it's not counted toward the extra space complexity analysis as per the problem's constraints.\n",
-    "\n",
-    "Overall, the space complexity is O(n) for this solution."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Handle Zeros in the Input**: Modify the solution to handle cases where the input array contains zeros. For example, if `nums = [0, 1, 2, 3, 0]`, the output should be `[0, 0, 0, 0, 0]` because any element multiplied by zero results in zero.\n",
-    "\n",
-    "2. **In-Place Modification**: Attempt to solve the problem with in-place modification of the `nums` array, where the output is stored directly in the input array without using any additional space."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb b/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb
deleted file mode 100644
index 5649123..0000000
--- a/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb	
+++ /dev/null
@@ -1,192 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 242. Valid Anagram\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Valid Anagram problem on LeetCode, click here!](https://leetcode.com/problems/valid-anagram/)\n",
-    "\n",
-    "---\n",
-    "Given two strings `s` and `t`, return `true` if `t` is an anagram of `s`, and `false` otherwise.\n",
-    "\n",
-    "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `s.length`, `t.length` <= $5 * 10^4$\n",
-    "- `s` and `t` consist of lowercase English letters.\n",
-    "\n",
-    "**Follow-up:** What if the inputs contain Unicode characters? How would you adapt your solution to such a case?\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def isAnagram(s: str, t: str) -> bool:\n",
-    "    # Check if the lengths of s and t are not equal\n",
-    "    if len(s) != len(t):\n",
-    "        return False\n",
-    "\n",
-    "    # Create dictionaries to store character frequencies for s and t\n",
-    "    countS, countT = {}, {}\n",
-    "\n",
-    "    # Count character frequencies in s\n",
-    "    for i in range(len(s)):\n",
-    "        countS[s[i]] = 1 + countS.get(s[i], 0)\n",
-    "\n",
-    "    # Count character frequencies in t\n",
-    "    for i in range(len(t)):\n",
-    "        countT[t[i]] = 1 + countT.get(t[i], 0)\n",
-    "\n",
-    "    # Check if the character frequencies in s and t are the same\n",
-    "    return countS == countT"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The code defines a class called `Solution` with a method `isAnagram` that takes two input strings, `s` and `t`, and returns a boolean (`True` or `False`) based on whether `t` is an anagram of `s`.\n",
-    "\n",
-    "2. The first check performed in the method is whether the lengths of `s` and `t` are equal. If they are not equal, it immediately returns `False` because strings of different lengths cannot be anagrams of each other.\n",
-    "\n",
-    "3. Two dictionaries, `countS` and `countT`, are created to store the frequency of characters in `s` and `t`, respectively. These dictionaries will be used to count the occurrences of each character in the strings.\n",
-    "\n",
-    "4. The code then enters a loop to iterate through each character in string `s` using a `for` loop. Inside the loop, it updates the `countS` dictionary. The line `countS[s[i]] = 1 + countS.get(s[i], 0)` increments the count of character `s[i]` in `countS` by 1. If the character is not already in the dictionary, it initializes the count to 1.\n",
-    "\n",
-    "5. Similarly, the code iterates through each character in string `t` and updates the `countT` dictionary in the same way.\n",
-    "\n",
-    "6. After counting the character frequencies in both strings, the code compares the two dictionaries using `countS == countT`. If the dictionaries are equal, it means that both strings have the same character frequencies, and therefore, `t` is an anagram of `s`. In this case, the method returns `True`.\n",
-    "\n",
-    "7. If the dictionaries are not equal, the method returns `False`, indicating that `t` is not an anagram of `s`.\n",
-    "\n",
-    "In summary, this code efficiently determines whether two input strings are anagrams by counting the character frequencies in each string using dictionaries and then comparing these counts. If the character frequencies match, the strings are considered anagrams, and the method returns `True`; otherwise, it returns `False`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "s = \"anagram\"\n",
-    "t = \"nagaram\"\n",
-    "print(isAnagram(s, t))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "s = \"rat\"\n",
-    "t = \"car\"\n",
-    "print(isAnagram(s, t))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. The code first checks whether the lengths of the input strings `s` and `t` are equal, which takes constant time. This check has a time complexity of O(1).\n",
-    "\n",
-    "2. The code then iterates through both strings, `s` and `t`, once to count the character frequencies. Since both strings have a maximum length of 5 * 10^4, the worst-case scenario is that the code iterates through 5 * 10^4 characters in each string. This results in a linear time complexity of O(n), where n is the length of the longer of the two input strings (`s` or `t`).\n",
-    "\n",
-    "3. Finally, the code compares the two dictionaries `countS` and `countT` to check if the character frequencies match. This comparison takes O(n) time in the worst case, where n is the length of the longer string.\n",
-    "\n",
-    "Overall, the time complexity of the code is O(n), where n is the length of the longer input string.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. The code uses two dictionaries, `countS` and `countT`, to store the character frequencies of the input strings `s` and `t`. In the worst case, both dictionaries can contain all unique characters from the input strings. Since the input strings can have a maximum length of 5 * 10^4, the space complexity for these dictionaries is O(5 * 10^4) or simply O(n), where n is the length of the longer input string.\n",
-    "\n",
-    "2. The code uses a few additional variables for bookkeeping, but these variables have constant space requirements and do not depend on the input size. Therefore, they do not significantly impact the overall space complexity.\n",
-    "\n",
-    "**In summary**, the space complexity of the code is O(n), where n is the length of the longer input string."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Anagram Chains**: Given a list of words, find the longest chain of anagrams, where each word in the chain is an anagram of the previous word. For example, given [\"bat\", \"tab\", \"cat\", \"tac\", \"dog\"], the longest anagram chain is [\"bat\", \"tab\", \"cat\", \"tac\"].\n",
-    "\n",
-    "2. **Minimum Deletions to Make Anagrams**: Given two strings `s` and `t`, find the minimum number of character deletions required in both strings to make them anagrams of each other. For example, if `s = \"hello\"` and `t = \"billion\"`, you can remove the characters \"heo\" from `s` and \"billi\" from `t` to make them anagrams (\"lo\" and \"on\" are left in the two strings)."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb b/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb
deleted file mode 100644
index 1b11faf..0000000
--- a/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 347. Top K Frequent Elements\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Top K Frequent Elements problem on LeetCode, click here!](https://leetcode.com/problems/top-k-frequent-elements/)\n",
-    "\n",
-    "---\n",
-    "Given an integer array `nums` and an integer `k`, return the `k` most frequent elements. You may return the answer in **any order**.\n",
-    "\n",
-    "Constraints:\n",
-    "- 1 <= `nums.length` <= $10^5$\n",
-    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-    "- `k` is in the range `[1, the number of unique elements in the array]`.\n",
-    "- It is **guaranteed** that the answer is **unique**.\n",
-    "\n",
-    "**Follow-up:** Your algorithm's time complexity must be better than $O(n\\ log\\ n)$, where n is the array's size.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def topKFrequent(nums, k):\n",
-    "    # Create a dictionary to store the frequency of each element\n",
-    "    count = {}\n",
-    "    # Create a list of lists to store elements with the same frequency\n",
-    "    frequency = [[] for i in range(len(nums) + 1)]\n",
-    "\n",
-    "    # Count the frequency of each element in nums\n",
-    "    for n in nums:\n",
-    "        count[n] = 1 + count.get(n, 0)\n",
-    "\n",
-    "    # Place elements in the freq list according to their frequency\n",
-    "    for n, c in count.items():\n",
-    "        frequency[c].append(n)\n",
-    "\n",
-    "    res = []\n",
-    "    # Traverse freq list from the end (higher frequencies)\n",
-    "    for i in range(len(frequency) - 1, 0, -1):\n",
-    "        for n in frequency[i]:\n",
-    "            res.append(n)\n",
-    "            if len(res) == k:\n",
-    "                return res"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `def topKFrequent(nums, k):`: This is a function that takes two arguments: `nums`, which is the input array of integers, and `k`, which is the number of most frequent elements to return.\n",
-    "\n",
-    "2. `count = {}`: This dictionary `count` will be used to store the frequency of each unique element in the `nums` array. The keys are elements from the input array, and the values are their corresponding frequencies.\n",
-    "\n",
-    "3. `frequency = [[] for i in range(len(nums) + 1)]`: This creates a list of empty lists called `frequency`. It's used to store elements based on their frequencies, similar to the `freq` list in the previous code. The size of this list is set to be one greater than the length of the input `nums`.\n",
-    "\n",
-    "4. `for n in nums:`: This loop iterates through each element `n` in the input `nums` array.\n",
-    "\n",
-    "5. `count[n] = 1 + count.get(n, 0)`: This line counts the frequency of each element `n` in the `nums` array. It uses the `count.get(n, 0)` method to retrieve the current count of `n` from the `count` dictionary. If `n` is not in the dictionary, it defaults to 0. It then increments the count by 1.\n",
-    "\n",
-    "6. After the above loop, the `count` dictionary will contain counts of each unique element in the `nums` array.\n",
-    "\n",
-    "7. `for n, c in count.items():`: This loop iterates through the items (key-value pairs) of the `count` dictionary. `n` represents the element, and `c` represents its frequency.\n",
-    "\n",
-    "8. `frequency[c].append(n)`: This line places the element `n` into the bucket corresponding to its frequency `c`. Buckets are represented by the `frequency` list. For example, if an element `n` has a frequency of 3, it will be added to `frequency[3]`.\n",
-    "\n",
-    "9. After this loop, the `frequency` list will contain buckets of elements grouped by their frequencies.\n",
-    "\n",
-    "10. `res = []`: This list `res` will be used to store the k most frequent elements.\n",
-    "\n",
-    "11. `for i in range(len(freq) - 1, 0, -1):`: This loop iterates in reverse order through the `frequency` list, starting from the highest frequency and going down to 1. Note that there is a typo here; it should be `frequency` instead of `freq`.\n",
-    "\n",
-    "12. `for n in frequency[i]:`: For each element `n` in the current bucket (i.e., elements with frequency `i`), it appends `n` to the result list `res`.\n",
-    "\n",
-    "13. `if len(res) == k:`: Once `res` contains k elements, the code exits and returns the result.\n",
-    "\n",
-    "The code efficiently finds the k most frequent elements in the input array without using sorting, similar to the previous explanation, with the only difference being the variable names (e.g., `frequency` instead of `freq`)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 2]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "nums = [1,1,1,2,2,3]\n",
-    "k = 2\n",
-    "result1 = topKFrequent(nums, k)\n",
-    "print(result1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums = [1]\n",
-    "k = 1\n",
-    "result2 = topKFrequent(nums, k)\n",
-    "print(result2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. The first loop that counts the frequency of elements by iterating through `nums` has a time complexity of O(n), where n is the number of elements in the input array `nums`.\n",
-    "\n",
-    "2. The second loop iterates through the keys in the `count` dictionary, which has at most `k` unique elements (as per the constraints). Therefore, the second loop also has a time complexity of O(k).\n",
-    "\n",
-    "3. The third loop iterates through the `frequency` list, which has a length equal to the maximum frequency of elements in `nums`. In the worst case, this loop can have a time complexity of O(n), where n is the number of elements in `nums`.\n",
-    "\n",
-    "Overall, the dominant factor in terms of time complexity is the loop that iterates through the `count` dictionary. So, the total time complexity of the code is O(n + k).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The `count` dictionary stores the frequency of each unique element in `nums`. In the worst case, it can have at most `k` unique elements, so the space complexity for `count` is O(k).\n",
-    "\n",
-    "2. The `frequency` list is used to store elements grouped by their frequencies. It has a length of `len(nums) + 1`, which can be at most 105 based on the constraints. Therefore, the space complexity for `frequency` is O(105).\n",
-    "\n",
-    "3. The `res` list stores the k most frequent elements, which can be at most `k` elements, so the space complexity for `res` is O(k).\n",
-    "\n",
-    "**In summary**, the space complexity is dominated by the `count` dictionary and the `frequency` list, both of which have a space complexity of O(k + 105)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimized k Most Frequent Elements:** Modify the code to find the k most frequent elements in an array while ensuring that the time complexity is O(n + klogk). You can use a priority queue (heap) to achieve this.\n",
-    "\n",
-    "2. **Handling Duplicate Frequencies:** Extend the code to handle cases where multiple elements have the same frequency and are among the k most frequent elements. Ensure that the output contains exactly k elements."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb b/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb
deleted file mode 100644
index 256e8f7..0000000
--- a/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 49. Group Anagrams\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Group Anagrams problem on LeetCode, click here!](https://leetcode.com/problems/group-anagrams/)\n",
-    "\n",
-    "---\n",
-    "Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.\n",
-    "\n",
-    "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `strs.length <= $10^4$\n",
-    "- `0 <= strs[i].length <= 100`\n",
-    "- `strs[i]` consists of lowercase English letters.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from collections import defaultdict\n",
-    "\n",
-    "\n",
-    "def groupAnagrams(strs):\n",
-    "    ans = defaultdict(list)\n",
-    "\n",
-    "    # Iterate through the list of input strings\n",
-    "    for s in strs:\n",
-    "        # Initialize a list to represent character counts for each character (a-z)\n",
-    "        count = [0] * 26\n",
-    "\n",
-    "        # Count the occurrences of each character in the current word\n",
-    "        for c in s:\n",
-    "            count[ord(c) - ord(\"a\")] += 1\n",
-    "\n",
-    "        # Use a tuple of character counts as the key and append the word to the anagram group\n",
-    "        ans[tuple(count)].append(s)\n",
-    "\n",
-    "    # Convert the values (lists of anagrams) to a list of lists\n",
-    "    return list(ans.values())"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. Inside the method, a `defaultdict` called `ans` is created to store anagram groups. This dictionary will have a list as its default value, meaning that each key in the dictionary will be associated with an empty list by default.\n",
-    "\n",
-    "2. The code then iterates through the list of input strings, `strs`, using a `for` loop. For each word `s` in `strs`, it performs the following steps:\n",
-    "\n",
-    "   - It initializes a list `count` of length 26, where each element represents the count of a specific character (a-z) in the word `s`.\n",
-    "\n",
-    "   - The code then iterates through the characters of `s`. For each character `c`, it increments the corresponding count in the `count` list based on its ASCII value.\n",
-    "\n",
-    "   - After counting the characters in `s`, it converts the `count` list into a tuple `tuple(count)` to use as a key for the `ans` dictionary.\n",
-    "\n",
-    "   - It appends the word `s` to the list associated with the key (tuple) in the `ans` dictionary. This groups all anagrams of the same word together under the same key.\n",
-    "\n",
-    "3. After processing all words in `strs`, the code converts the values of the `ans` dictionary (which are lists of anagrams) to a list of lists using the `list()` constructor.\n",
-    "\n",
-    "4. Finally, the code returns the list of anagram groups, which is the result of grouping the anagrams in the input list.\n",
-    "\n",
-    "In summary, the code efficiently groups anagrams together by counting the characters in each word and using a dictionary to store them under the same key. The result is a list of lists, where each inner list represents a group of anagrams."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: Group anagrams from a list of words\n",
-    "strs1 = [\"eat\", \"tea\", \"tan\", \"ate\", \"nat\", \"bat\"]\n",
-    "result1 = groupAnagrams(strs1)\n",
-    "print(result1)\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[['']]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2: Group anagrams from a list with an empty string\n",
-    "strs2 = [\"\"]\n",
-    "result2 = groupAnagrams(strs2)\n",
-    "print(result2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "c8367e2a-e7d5-4e18-8235-2870e341a04b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[['a']]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3: Group anagrams from a list with a single word\n",
-    "strs3 = [\"a\"]\n",
-    "result3 = groupAnagrams(strs3)\n",
-    "print(result3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `groupAnagrams` method:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. First, we create an instance of the `Solution` class, which is an O(1) operation.\n",
-    "\n",
-    "2. Inside the `groupAnagrams` method, we iterate through the list of input strings `strs` once in a loop:\n",
-    "   - For each word in `strs`, we perform character counting, which is done in O(K) time, where K is the maximum length of a word in the list.\n",
-    "\n",
-    "3. The overall time complexity of the method is O(N * K), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. We use a `defaultdict` called `ans` to store the anagram groups. In the worst case, each word belongs to a distinct anagram group, so the space complexity for `ans` is O(N).\n",
-    "\n",
-    "2. Within the loop, we create a `count` list of length 26 to store character counts for each word. This is a fixed-size list and does not depend on the input size, so it has a constant space complexity of O(26), which is effectively O(1).\n",
-    "\n",
-    "3. The space used for other variables is also constant and does not depend on the input size.\n",
-    "\n",
-    "4. The final result, `result`, is a list of lists that contains the grouped anagrams. In the worst case, each word is a distinct anagram group, so the space complexity for `result` is O(N).\n",
-    "\n",
-    "5. The overall space complexity of the method is O(N) due to the space used by `ans` and `result`. The space used by the `count` list and other variables is constant and does not significantly affect the overall space complexity.\n",
-    "\n",
-    "**In summary,** the time complexity of the `groupAnagrams` method is O(N * K), and the space complexity is O(N), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Anagramic Palindromes**: Given a list of strings, find all the groups of anagrams where each group contains words that can be rearranged into a palindrome. For example, in the list [\"bat\", \"tab\", \"act\", \"tac\", \"dog\"], there are two groups: [\"bat\", \"tab\", \"act\", \"tac\"] and [\"dog\"].\n",
-    "\n",
-    "2. **Largest Anagram Group**: Given a list of strings, find the largest group of anagrams. Return the list of anagrams in that group. If there are multiple largest groups, return all of them."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb b/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb
deleted file mode 100644
index 2fae2a5..0000000
--- a/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb	
+++ /dev/null
@@ -1,198 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 659· Encode and Decode Strings\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Encode and Decode Strings problem on LintCode, click here!](https://www.lintcode.com/problem/659/)\n",
-    "\n",
-    "---\n",
-    "Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.\n",
-    "\n",
-    "Please implement `encode` and `decode`.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class Codec:\n",
-    "\n",
-    "    def encode(self, strs):\n",
-    "        \"\"\"Encodes a list of strings to a single string.\n",
-    "        \n",
-    "        :type strs: List[str]\n",
-    "        :rtype: str\n",
-    "        \"\"\"\n",
-    "        encoded = \"\"\n",
-    "        for s in strs:\n",
-    "            encoded += s.replace(':', '::') + ':;'\n",
-    "        return encoded\n",
-    "\n",
-    "    def decode(self, s):\n",
-    "        \"\"\"Decodes a single string to a list of strings.\n",
-    "        \n",
-    "        :type s: str\n",
-    "        :rtype: List[str]\n",
-    "        \"\"\"\n",
-    "        decoded = []\n",
-    "        i = 0\n",
-    "        while i < len(s):\n",
-    "            end = s.find(':;', i)\n",
-    "            if end == -1:\n",
-    "                end = len(s)\n",
-    "            decoded.append(s[i:end].replace('::', ':'))\n",
-    "            i = end + 2\n",
-    "        return decoded"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We define a class called `Codec` to encapsulate the encoding and decoding operations for a list of strings.\n",
-    "\n",
-    "2. In the `encode` method, we take a list of strings (`strs`) as input and return a single encoded string. The encoding process involves concatenating the strings together with a delimiter `:;`. We iterate through each string in the input list, replace any occurrence of `:` with `::` (to avoid conflicts with the delimiter), and then append `:;` to indicate the end of that string.\n",
-    "\n",
-    "3. In the `decode` method, we take an encoded string (`s`) as input and return a list of strings. The decoding process involves splitting the encoded string using the `:;` delimiter. We start from the beginning of the encoded string (`i = 0`) and repeatedly find the next occurrence of `:;`. We extract the substring between the current position (`i`) and the next delimiter (`end`). Before adding it to the result list, we replace any `::` with `:` to revert the encoding. We then update the current position `i` to be after the delimiter, so we can find the next substring in the next iteration.\n",
-    "\n",
-    "4. Finally, we create an instance of the `Codec` class and test it with two examples:\n",
-    "\n",
-    "By using the `:;` delimiter and handling the `::` escaping, this code can encode and decode lists of strings, preserving the original content even if it contains colons or the delimiter itself."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "['lint', 'code', 'love', 'you']\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "codec = Codec()\n",
-    "\n",
-    "input1 = [\"lint\", \"code\", \"love\", \"you\"]\n",
-    "encoded1 = codec.encode(input1)\n",
-    "decoded1 = codec.decode(encoded1)\n",
-    "print(decoded1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "['we', 'say', ':', 'yes']\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "codec = Codec() \n",
-    "\n",
-    "input2 = [\"we\", \"say\", \":\", \"yes\"]\n",
-    "encoded2 = codec.encode(input2)\n",
-    "decoded2 = codec.decode(encoded2)\n",
-    "print(decoded2)  "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `encode` and `decode` methods in the provided code.\n",
-    "\n",
-    "**Encode Method (`encode`):**\n",
-    "\n",
-    "- Time Complexity: O(n * m)\n",
-    "  - Here, `n` is the number of strings in the input list `strs`, and `m` is the average length of these strings.\n",
-    "  - In the worst case, for each string in the list, we iterate over its characters once to replace any colons (`:`) with double colons (`::`) and then append `:;`. This is done for each string in the list.\n",
-    "\n",
-    "- Space Complexity: O(n * m)\n",
-    "  - The space complexity is also O(n * m) because we build the encoded string by concatenating the modified strings. In the worst case, the encoded string can be of the same length as the original strings.\n",
-    "\n",
-    "**Decode Method (`decode`):**\n",
-    "\n",
-    "- Time Complexity: O(n * m)\n",
-    "  - Similar to the encode method, we iterate through the encoded string in the decode method. In the worst case, we may have to scan each character in the encoded string to find the `:;` delimiter.\n",
-    "\n",
-    "- Space Complexity: O(n * m)\n",
-    "  - The space complexity of the decode method is also O(n * m) because we build the list of decoded strings, which can be of the same length as the encoded string.\n",
-    "\n",
-    "Overall, both the `encode` and `decode` methods have time and space complexities of O(n * m), where `n` is the number of strings in the input list, and `m` is the average length of these strings. The space complexity arises from storing the encoded or decoded strings, and the time complexity arises from iterating through the characters in these strings."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple Delimiters**: Modify the code to allow for multiple delimiters, not just one. Each string could have its own delimiter, and the code should be able to handle this complexity.\n",
-    "\n",
-    "2. **Optimize Encoding**: Modify the `encode` method to achieve a more efficient encoding algorithm. Try to minimize the length of the encoded string while ensuring that it can be correctly decoded."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb b/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb
deleted file mode 100644
index e330eb0..0000000
--- a/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb	
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 11. Container With Most Water\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Container With Most Water problem on LeetCode, click here!](https://leetcode.com/problems/container-with-most-water/)\n",
-    "\n",
-    "---\n",
-    "You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the $i^{th}$ line are `(i, 0)` and `(i, height[i])`.\n",
-    "\n",
-    "Find two lines that together with the x-axis form a container, such that the container contains the most water.\n",
-    "\n",
-    "Return *the maximum amount of water a container can store.*\n",
-    "\n",
-    "**Notice** that you may not slant the container.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `n == height.length`\n",
-    "- 2 <= `n` <= $10^5$\n",
-    "- 0 <= `height[i]` <= $10^4$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def maxArea(height):\n",
-    "    # Initialize the maximum area to 0.\n",
-    "    max_area = 0\n",
-    "    # Initialize two pointers, one at the beginning (left) and one at the end (right) of the array.\n",
-    "    left = 0\n",
-    "    right = len(height) - 1\n",
-    "\n",
-    "    # Iterate until the left pointer is less than the right pointer.\n",
-    "    while left < right:\n",
-    "        # Calculate the height of the container, which is the minimum height of the two lines.\n",
-    "        h = min(height[left], height[right])\n",
-    "        # Calculate the width of the container, which is the distance between the two pointers.\n",
-    "        width = right - left\n",
-    "        # Calculate the area of the container using height and width and update max_area if it's greater.\n",
-    "        max_area = max(max_area, h * width)\n",
-    "\n",
-    "        # Move the pointer with the shorter line inward.\n",
-    "        if height[left] < height[right]:\n",
-    "            left += 1\n",
-    "        else:\n",
-    "            right -= 1\n",
-    "\n",
-    "    # Continue this process until left < right, and then return the maximum area.\n",
-    "    return max_area"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start with initializing `max_area` to 0, which will store the maximum area of water that can be contained by the lines.\n",
-    "\n",
-    "2. Two pointers, `left` and `right`, are initialized to the beginning and end of the input `height` array.\n",
-    "\n",
-    "3. The main loop runs while `left` is less than `right`. This ensures that we are checking all possible pairs of lines in a systematic way.\n",
-    "\n",
-    "4. Inside the loop:\n",
-    "   - We calculate the height of the container `h` as the minimum of the heights at the `left` and `right` pointers. This represents the height of the water the container can hold.\n",
-    "   - We calculate the width of the container as the difference between `right` and `left`. This represents the distance between the two lines.\n",
-    "   - We calculate the area of the container using `h * width` and update `max_area` if the calculated area is greater than the current `max_area`.\n",
-    "\n",
-    "5. After calculating the area and updating `max_area`, we move one of the pointers. We move the pointer pointing to the shorter line inward because moving the pointer pointing to the taller line won't increase the height and will only decrease the width, which will result in a smaller area.\n",
-    "\n",
-    "6. We continue this process until `left` is no longer less than `right`, indicating that we have checked all possible pairs of lines.\n",
-    "\n",
-    "7. Finally, we return the `max_area` which contains the maximum area of water that can be contained by two lines from the input array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output: 49\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "height1 = [1, 8, 6, 2, 5, 4, 8, 3, 7]\n",
-    "result1 = maxArea(height1)\n",
-    "print(\"Output:\", result1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output: 1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "height2 = [1, 1]\n",
-    "result2 = maxArea(height2)\n",
-    "print(\"Output:\", result2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output: 45\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Additional Example\n",
-    "height3 = [3, 9, 3, 4, 7, 2, 12, 6]\n",
-    "result3 = maxArea(height3)\n",
-    "print(\"Output:\", result3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `maxArea` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The function uses a two-pointer approach to find the maximum area of water. In the worst case, both the left and right pointers traverse the entire input array once. Therefore, the time complexity is O(n), where n is the number of elements in the `height` array.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The function uses a constant amount of extra space for variables like `max_area`, `left`, `right`, `h`, and `width`. It does not use any data structures whose space consumption depends on the input size. As a result, the space complexity is O(1), which means it has constant space complexity.\n",
-    "\n",
-    "**In summary,** the `maxArea` function has a time complexity of O(n) and a space complexity of O(1), making it an efficient algorithm for finding the maximum area of water that can be contained by two lines in the input array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the maxArea function to not only return the maximum area but also the indices of the two lines that form the container with the maximum area.\n",
-    "2. **Variant Problem:** Instead of finding the maximum area of water, find the minimum amount of water required to fill all the gaps between the lines. You'll need to return the total water volume needed."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb b/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb
deleted file mode 100644
index 12c7daa..0000000
--- a/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb	
+++ /dev/null
@@ -1,210 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 125. Valid Palindrome\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Valid Palindrome problem on LeetCode, click here!](https://leetcode.com/problems/valid-palindrome/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.\n",
-    "\n",
-    "Given a string `s`, *return `true` if it is a **palindrome**, or `false` otherwise.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `s.length` <= $2*10^5$\n",
-    "- `s` consists only of printable ASCII characters.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def isPalindrome(s):\n",
-    "    # Step 1: Convert the input string to lowercase\n",
-    "    s = s.lower()\n",
-    "    \n",
-    "    # Step 2: Remove all non-alphanumeric characters from the string\n",
-    "    cleaned_s = ''.join(c for c in s if c.isalnum())\n",
-    "    \n",
-    "    # Step 3: Initialize two pointers, one at the beginning and one at the end of the cleaned string\n",
-    "    left, right = 0, len(cleaned_s) - 1\n",
-    "    \n",
-    "    # Step 4: Compare characters using the two pointers while moving them towards each other\n",
-    "    while left < right:\n",
-    "        if cleaned_s[left] != cleaned_s[right]:\n",
-    "            return False  # If characters don't match, it's not a palindrome\n",
-    "        left += 1\n",
-    "        right -= 1\n",
-    "    \n",
-    "    # Step 5: If the loop completes without returning False, it's a palindrome\n",
-    "    return True"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The `isPalindrome` function takes an input string `s`.\n",
-    "\n",
-    "2. It starts by converting the input string to lowercase using `s.lower()`. This step ensures that the function is case-insensitive when checking for palindromes.\n",
-    "\n",
-    "3. Next, it removes all non-alphanumeric characters from the string and stores the result in the `cleaned_s` variable. It does this by iterating through each character in the input string and only keeping characters that are alphanumeric (letters and digits). This step removes spaces, punctuation, and other non-alphanumeric characters.\n",
-    "\n",
-    "4. The function then initializes two pointers, `left` and `right`, which will be used to compare characters from the beginning and end of the cleaned string.\n",
-    "\n",
-    "5. In a while loop, it compares characters using the two pointers while moving them towards each other. The loop continues until the `left` pointer is less than the `right` pointer. This is done to check all characters in the cleaned string.\n",
-    "\n",
-    "6. Inside the loop, it checks if the characters at the `left` and `right` pointers don't match. If they don't match, it means the input is not a palindrome, so the function returns `False`.\n",
-    "\n",
-    "7. If the characters at the `left` and `right` pointers match, the pointers are moved closer to each other by incrementing `left` and decrementing `right`.\n",
-    "\n",
-    "8. The loop continues to compare characters until it either finds a mismatch (returning `False`) or until the `left` pointer is greater than or equal to the `right` pointer. If the loop completes without finding a mismatch, it means the input is a palindrome, so the function returns `True`.\n",
-    "\n",
-    "In summary, the code converts the input string to lowercase, removes non-alphanumeric characters, and then uses two pointers to compare characters from the cleaned string's beginning and end. If all characters match during the comparison, the function returns `True`, indicating that the input is a palindrome. If any characters do not match, it returns `False`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "input1 = \"A man, a plan, a canal: Panama\"\n",
-    "output1 = isPalindrome(input1)\n",
-    "print(output1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "input2 = \"race a car\"\n",
-    "output2 = isPalindrome(input2)\n",
-    "print(output2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "input3 = \"\"\n",
-    "output3 = isPalindrome(input3)\n",
-    "print(output3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `isPalindrome` function that uses two pointers:\n",
-    "\n",
-    "1. **Time Complexity:**\n",
-    "   - Converting the input string to lowercase using `s.lower()` takes O(n) time, where 'n' is the length of the input string.\n",
-    "   - Removing non-alphanumeric characters using the list comprehension `''.join(c for c in s if c.isalnum())` also takes O(n) time because it iterates through each character in the string once.\n",
-    "   - The comparison of characters using the two pointers (`left` and `right`) is done in a loop that runs until `left` is less than `right`. In the worst case, the loop iterates through half of the string, so it takes O(n/2) time, which is still considered O(n).\n",
-    "\n",
-    "   Therefore, the overall time complexity of the `isPalindrome` function is O(n), where 'n' is the length of the input string.\n",
-    "\n",
-    "2. **Space Complexity:**\n",
-    "   - The space complexity is primarily determined by the additional string `cleaned_s` created to store the cleaned version of the input string. In the worst case, if all characters in the input string are alphanumeric, this cleaned string can be as large as the original input string.\n",
-    "   - Additionally, the function uses two pointers, `left` and `right`, which consume constant space and do not depend on the size of the input.\n",
-    "\n",
-    "**In summary,** the overall space complexity of the `isPalindrome` function is O(n) in the worst case, where 'n' is the length of the input string. The space complexity is dominated by the space required for `cleaned_s`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Longest Palindromic Substring:** Given a string `s`, find and return the longest palindromic substring within `s`. For example, if `s = \"babad\"`, the function should return `\"bab\"` or `\"aba\"`.\n",
-    "2. **Palindrome with Limited Characters:** Given a string `s` and a list of characters `allowed_chars`, check if `s` is a palindrome considering only the characters from `allowed_chars`. Ignore all other characters. For example, if `s = \"A man, a plan, a canal: Panama\"` and `allowed_chars = ['a', 'm', 'n', 'p']`, the function should return `True`."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb b/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb
deleted file mode 100644
index 15d189a..0000000
--- a/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 15. 3Sum\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the 3Sum problem on LeetCode, click here!](https://leetcode.com/problems/3sum/)\n",
-    "\n",
-    "---\n",
-    "Given an integer array nums, *return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.*\n",
-    "\n",
-    "**Notice** that the solution set must not contain duplicate triplets.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `3 <= nums.length <= 3000`\n",
-    "- $-10^5$ <= `nums[i]` <= $10^5$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def threeSum(nums):\n",
-    "    # Sort the input array in ascending order\n",
-    "    nums.sort()\n",
-    "    triplets = []\n",
-    "    \n",
-    "    for i in range(len(nums) - 2):\n",
-    "        # Skip duplicates to avoid duplicate triplets\n",
-    "        if i > 0 and nums[i] == nums[i - 1]:\n",
-    "            continue\n",
-    "        \n",
-    "        left, right = i + 1, len(nums) - 1\n",
-    "        \n",
-    "        while left < right:\n",
-    "            # Calculate the total sum of the current triplet\n",
-    "            total = nums[i] + nums[left] + nums[right]\n",
-    "            \n",
-    "            if total == 0:\n",
-    "                # If the total is zero, we found a valid triplet\n",
-    "                triplets.append([nums[i], nums[left], nums[right]])\n",
-    "                \n",
-    "                # Skip duplicates of left and right pointers\n",
-    "                while left < right and nums[left] == nums[left + 1]:\n",
-    "                    left += 1\n",
-    "                while left < right and nums[right] == nums[right - 1]:\n",
-    "                    right -= 1\n",
-    "                \n",
-    "                # Move the pointers to the next unique elements\n",
-    "                left += 1\n",
-    "                right -= 1\n",
-    "            elif total < 0:\n",
-    "                # If the total is negative, we need to increase the sum by moving the left pointer to the right\n",
-    "                left += 1\n",
-    "            else:\n",
-    "                # If the total is positive, we need to decrease the sum by moving the right pointer to the left\n",
-    "                right -= 1\n",
-    "    \n",
-    "    return triplets"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. It starts by sorting the input array `nums` in ascending order. Sorting helps in efficiently finding triplets that sum up to zero.\n",
-    "\n",
-    "2. It initializes an empty list `triplets` to store the unique triplets that meet the given conditions.\n",
-    "\n",
-    "3. The code then iterates through the `nums` array using a loop with index `i`. This loop will consider each element of the array as a potential starting point for a triplet.\n",
-    "\n",
-    "4. Inside the loop, it checks for duplicates and skips them. This is done to ensure that the solution set does not contain duplicate triplets. If `nums[i]` is the same as the previous element `nums[i-1]`, it continues to the next iteration of the loop.\n",
-    "\n",
-    "5. Two pointers, `left` and `right`, are initialized. `left` starts just after the current element `nums[i]`, and `right` starts at the end of the array.\n",
-    "\n",
-    "6. The code enters another loop with `left` and `right` pointers, trying to find a triplet that sums up to zero.\n",
-    "\n",
-    "7. It calculates the `total` sum of the current triplet as `nums[i] + nums[left] + nums[right]`.\n",
-    "\n",
-    "8. If `total` is zero, it means a valid triplet is found, so it appends `[nums[i], nums[left], nums[right]]` to the `triplets` list. Then, it moves both the `left` and `right` pointers to their next unique elements while skipping duplicates.\n",
-    "\n",
-    "9. If `total` is less than zero, it means the sum is negative, so it increments the `left` pointer to move towards a larger sum.\n",
-    "\n",
-    "10. If `total` is greater than zero, it means the sum is positive, so it decrements the `right` pointer to move towards a smaller sum.\n",
-    "\n",
-    "11. This process continues until all possible triplets have been considered.\n",
-    "\n",
-    "12. Finally, the function returns the `triplets` list, which contains all the unique triplets that sum up to zero in the sorted `nums` array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[-1, -1, 2], [-1, 0, 1]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "nums = [-1, 0, 1, 2, -1, -4]\n",
-    "result = threeSum(nums)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "nums = [0, 1, 1]\n",
-    "result = threeSum(nums)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[0, 0, 0]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "nums = [0, 0, 0]\n",
-    "result = threeSum(nums)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `threeSum` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. Sorting the input array `nums` takes O(n log n) time, where n is the length of the array.\n",
-    "\n",
-    "2. The outer loop runs for each element of the array, so it iterates O(n) times.\n",
-    "\n",
-    "3. Inside the outer loop, we have a while loop with two pointers (`left` and `right`). In the worst case, the while loop can iterate O(n) times (when all elements are unique).\n",
-    "\n",
-    "4. Inside the while loop, we have constant time operations (additions, comparisons, and list appends).\n",
-    "\n",
-    "Overall, the time complexity of the `threeSum` function is dominated by the sorting step, so it is O(n log n) due to the sorting. The other operations inside the loops contribute linearly but are dominated by the sorting step.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The space used by the `triplets` list to store the output triplets is O(k), where k is the number of unique triplets that sum up to zero. In the worst case, this can be O(n^2/3) because there can be roughly O(n^2) possible triplets, and many of them may sum up to zero.\n",
-    "\n",
-    "2. Other than the `triplets` list, the function uses only a constant amount of additional space for variables like `i`, `left`, `right`, and `total`.\n",
-    "\n",
-    "Overall, the space complexity of the `threeSum` function is O(k), where k is the number of unique triplets that meet the criteria. It's important to note that the space complexity does not depend on the size of the input array but rather on the number of valid triplets found.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: O(n log n) due to sorting (dominant factor) and O(n^2) in the worst case for the nested loops.\n",
-    "- Space Complexity: O(k) for storing the output triplets, where k is the number of unique triplets that meet the criteria."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **K-Sum Problem:** Generalize the problem to the K-Sum problem, where instead of finding triplets that sum to zero, you need to find K elements that sum to a target value. Implement a function for this generalization.\n",
-    "2. **Count the Number of Unique Triplets:** Modify the algorithm to count the number of unique triplets that satisfy the conditions instead of returning the triplets themselves. This will require some modifications to handle duplicates efficiently."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb b/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb
deleted file mode 100644
index 451a21d..0000000
--- a/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb	
+++ /dev/null
@@ -1,185 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 121. Best Time to Buy and Sell Stock\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Best Time to Buy and Sell Stock problem on LeetCode, click here!](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given an array `prices` where `prices[i]` is the price of a given stock on the $i^{th}$ day.\n",
-    "\n",
-    "You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.\n",
-    "\n",
-    "*Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `prices.length` <= $10^5$\n",
-    "- 0 <= `prices[i]` <= $10^4$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def maxProfit(prices):\n",
-    "    if not prices:\n",
-    "        return 0\n",
-    "    \n",
-    "    min_price = prices[0]\n",
-    "    max_profit = 0\n",
-    "    \n",
-    "    for price in prices:\n",
-    "        # Update the minimum price if a lower price is encountered\n",
-    "        min_price = min(min_price, price)\n",
-    "        # Calculate the profit if selling at the current price\n",
-    "        max_profit = max(max_profit, price - min_price)\n",
-    "    \n",
-    "    return max_profit"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "- The `maxProfit` function takes a single argument, `prices`, which is a list of stock prices.\n",
-    "- The first line of the function checks if the `prices` list is empty. If it's empty, there are no prices to analyze, so the function returns 0 (no profit can be made).\n",
-    "\n",
-    "- Then, we initialize two variables:\n",
-    "  - `min_price`: This variable will keep track of the minimum price seen so far. We initialize it with the price of the first day in the list (`prices[0]`).\n",
-    "  - `max_profit`: This variable will keep track of the maximum profit we can achieve. We initialize it to 0 because initially, we haven't made any profit.\n",
-    "\n",
-    "- We start iterating through the `prices` list, where `price` represents the stock price for the current day.\n",
-    "\n",
-    "- In each iteration, we check if the current `price` is lower than the `min_price` we've seen so far. If it is, we update `min_price` to the current `price`. This is because we want to buy the stock at the lowest possible price.\n",
-    "\n",
-    "- Next, we calculate the profit we would make if we sold the stock at the current `price` (assuming we bought it at `min_price`). We do this by subtracting `min_price` from the current `price`. This represents the profit for the current day.\n",
-    "\n",
-    "- We also use the `max` function to keep track of the maximum profit seen so far. If the profit calculated for the current day is greater than the `max_profit` we've seen previously, we update `max_profit` with this higher value.\n",
-    "\n",
-    "- Finally, after iterating through all the days in the `prices` list, we return `max_profit`, which represents the maximum profit that can be achieved by buying and selling a single stock."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "5\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "prices1 = [7, 1, 5, 3, 6, 4]\n",
-    "print(maxProfit(prices1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "0\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "prices2 = [7, 6, 4, 3, 1]\n",
-    "print(maxProfit(prices2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The code iterates through the `prices` array once, examining each price exactly once.\n",
-    "- Within the loop, there are constant-time operations such as comparisons, additions, and subtractions.\n",
-    "- Therefore, the time complexity of this code is O(n), where n is the length of the `prices` array. It performs a linear number of operations relative to the size of the input.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity of the code is constant, O(1).\n",
-    "- Regardless of the size of the input `prices` array, the code only uses a fixed amount of additional memory to store two variables (`min_price` and `max_profit`) and a few loop variables.\n",
-    "- The memory usage does not depend on the size of the input, so it is considered constant space complexity.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: O(n) where n is the length of the `prices` array.\n",
-    "- Space Complexity: O(1) (constant space usage)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple Transactions Allowed:** Modify the problem to allow multiple transactions (buy and sell multiple times), but you must sell before buying again. Find the maximum profit in this scenario.\n",
-    "2. **K Transactions Allowed:** Extend the problem to allow at most k transactions. Find the maximum profit considering this constraint.\n",
-    "3. **With Transaction Fee:** Introduce a transaction fee for each buy/sell operation. Modify the code to maximize profit while considering the transaction fee."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb b/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb
deleted file mode 100644
index ded03f3..0000000
--- a/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb	
+++ /dev/null
@@ -1,206 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 3. Longest Substring Without Repeating Characters\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Longest Substring Without Repeating Characters problem on LeetCode, click here!](https://leetcode.com/problems/longest-substring-without-repeating-characters/)\n",
-    "\n",
-    "---\n",
-    "Given a string `s`, find the length of the **longest substring** without repeating characters.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 0 <= `s.length` <= $5 * 10^4$\n",
-    "- `s` consists of English letters, digits, symbols and spaces.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def length_of_longest_substring(s):\n",
-    "    # Create a dictionary to store the index of each character's last occurrence.\n",
-    "    char_index = {}\n",
-    "    \n",
-    "    # Initialize variables to keep track of the start and end of the current substring.\n",
-    "    start = 0\n",
-    "    max_length = 0\n",
-    "    \n",
-    "    for end in range(len(s)):\n",
-    "        # If the character is in the dictionary and its last occurrence is after the start of the current substring,\n",
-    "        # update the start of the substring to the next character after its last occurrence.\n",
-    "        if s[end] in char_index and char_index[s[end]] >= start:\n",
-    "            start = char_index[s[end]] + 1\n",
-    "        \n",
-    "        # Update the last occurrence index of the current character.\n",
-    "        char_index[s[end]] = end\n",
-    "        \n",
-    "        # Calculate the length of the current substring and update the maximum length if necessary.\n",
-    "        current_length = end - start + 1\n",
-    "        max_length = max(max_length, current_length)\n",
-    "    \n",
-    "    return max_length"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a function called `length_of_longest_substring` that takes a single argument `s`, which is the input string for which we want to find the length of the longest substring without repeating characters.\n",
-    "\n",
-    "2. We create a dictionary called `char_index` to store the index of the last occurrence of each character in the input string. This dictionary will help us keep track of where each character was last seen.\n",
-    "\n",
-    "3. We initialize two variables, `start` and `max_length`. \n",
-    "   - `start` represents the start index of the current substring without repeating characters. It starts at 0.\n",
-    "   - `max_length` is used to keep track of the maximum length found so far and is initially set to 0.\n",
-    "\n",
-    "4. We then iterate through the input string `s` using a for loop, where the loop variable `end` represents the current end index of the substring we are considering.\n",
-    "\n",
-    "5. Inside the loop:\n",
-    "   - We check if the character `s[end]` is already in the `char_index` dictionary and if its last occurrence is within or after the current substring. If so, it means that we've encountered a repeating character, and we need to update the `start` of the substring to the next character after the last occurrence of `s[end]`. This ensures that we have a new valid substring without repeating characters.\n",
-    "   - We then update the `char_index` dictionary by storing the current index `end` as the last occurrence index of the character `s[end]`.\n",
-    "\n",
-    "6. Next, we calculate the length of the current substring without repeating characters, which is `current_length = end - start + 1`. We add 1 to account for the fact that the indices are zero-based.\n",
-    "\n",
-    "7. We update the `max_length` with the maximum of its current value and the `current_length`. This step ensures that we keep track of the longest valid substring we have encountered so far.\n",
-    "\n",
-    "8. The loop continues to iterate through the input string, and at the end, we return the `max_length`, which represents the length of the longest substring without repeating characters."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "print(length_of_longest_substring(\"abcabcbb\"))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "print(length_of_longest_substring(\"bbbbb\"))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "cf04eba0-0ead-4e8b-ba33-4316ea64a9d3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "print(length_of_longest_substring(\"pwwkew\"))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "Let's analyze the time and space complexity of the `length_of_longest_substring` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The function iterates through the input string `s` using a single for loop. The loop runs from the beginning of the string to the end once.\n",
-    "- Inside the loop, we perform constant-time operations such as dictionary lookups and updates, comparisons, and arithmetic operations.\n",
-    "- Therefore, the time complexity of the function is O(n), where n is the length of the input string `s`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The primary data structure that consumes space in this function is the `char_index` dictionary.\n",
-    "- In the worst case, if there are no repeating characters in the input string, the dictionary can store all unique characters in the string.\n",
-    "- Therefore, the space complexity is O(min(n, m)), where n is the length of the input string `s`, and m is the number of unique characters in the string. In the worst case, when all characters are unique, m is equal to n, so the space complexity is O(n).\n",
-    "- Additionally, there are a few integer variables used for indices and lengths, which consume constant space.\n",
-    "- Overall, the space complexity of the function is O(min(n, m)) or simply O(n) in the worst case.\n",
-    "\n",
-    "**In summary,** the time complexity of the `length_of_longest_substring` function is O(n), and the space complexity is O(n) in the worst case due to the `char_index` dictionary. This algorithm provides an efficient way to find the length of the longest substring without repeating characters in linear time."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Longest Substring with K Distinct Characters:** Modify the problem to find the length of the longest substring with exactly K distinct characters. For example, given the input string \"abcabcbb\" and K = 2, the answer would be 4 because the longest substring with two distinct characters is \"abca.\"\n",
-    "2. **Longest Substring with Unique Characters:** Write a function to find the length of the longest substring in a given string where all characters are unique. For example, given the input string \"abcabcbb,\" the answer would be 4 because \"abcd\" is the longest substring with unique characters."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb b/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb
deleted file mode 100644
index df4b0d3..0000000
--- a/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb	
+++ /dev/null
@@ -1,189 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 424. Longest Repeating Character Replacement\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Longest Repeating Character Replacement problem on LeetCode, click here!](https://leetcode.com/problems/longest-repeating-character-replacement/)\n",
-    "\n",
-    "---\n",
-    "You are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.\n",
-    "\n",
-    "*Return the length of the longest substring containing the same letter you can get after performing the above operations.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `s.length` <= $10^5$\n",
-    "- `s` consists of only uppercase English letters.\n",
-    "- `0 <= k <= s.length`"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def characterReplacement(s, k):\n",
-    "    max_length = 0  # Initialize the maximum length\n",
-    "    max_count = 0   # Initialize the maximum count of repeating characters\n",
-    "    start = 0       # Initialize the start of the sliding window\n",
-    "    char_count = {} # Dictionary to store the count of each character\n",
-    "\n",
-    "    for end in range(len(s)):\n",
-    "        # Update the count of the current character in the dictionary\n",
-    "        char_count[s[end]] = char_count.get(s[end], 0) + 1\n",
-    "        \n",
-    "        # Update the maximum count of repeating characters\n",
-    "        max_count = max(max_count, char_count[s[end]])\n",
-    "        \n",
-    "        # Check if the current window size is greater than k\n",
-    "        if (end - start + 1) - max_count > k:\n",
-    "            # Move the start of the window to the right\n",
-    "            char_count[s[start]] -= 1\n",
-    "            start += 1\n",
-    "        \n",
-    "        # Update the maximum length\n",
-    "        max_length = max(max_length, end - start + 1)\n",
-    "\n",
-    "    return max_length"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We initialize some variables:\n",
-    "   - `max_length` to keep track of the maximum length of the substring containing the same letter.\n",
-    "   - `max_count` to keep track of the maximum count of repeating characters within the current window.\n",
-    "   - `start` to represent the start index of the sliding window.\n",
-    "   - `char_count` is a dictionary that will store the count of each character within the current window.\n",
-    "\n",
-    "2. We use a `for` loop to iterate through the characters of the string `s`.\n",
-    "\n",
-    "3. Inside the loop, we do the following for each character at position `end`:\n",
-    "   - Update the count of the current character in the `char_count` dictionary.\n",
-    "   - Update the `max_count` to be the maximum of the current `max_count` and the count of the current character. This keeps track of the maximum count of repeating characters within the current window.\n",
-    "\n",
-    "4. We check if the current window size (the difference between `end` and `start` plus one) minus the `max_count` is greater than `k`. This condition checks whether we have exceeded the allowed number of replacements (`k`) within the current window.\n",
-    "\n",
-    "   - If we have exceeded the allowed replacements, it means we need to shrink the window from the left side. We do this by moving the `start` of the window to the right and decrementing the count of the character at `s[start]` in the `char_count` dictionary. This effectively removes characters from the left side of the window until we have a valid window again.\n",
-    "\n",
-    "   - By doing this, we ensure that the difference between the current window size and the `max_count` is always less than or equal to `k`.\n",
-    "\n",
-    "5. After each character, we update the `max_length` to be the maximum of the current `max_length` and the size of the current window (`end - start + 1`).\n",
-    "\n",
-    "6. Finally, we return `max_length`, which holds the length of the longest substring containing the same letter with at most `k` replacements."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "s1 = \"ABAB\"\n",
-    "k1 = 2\n",
-    "print(characterReplacement(s1, k1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "s2 = \"AABABBA\"\n",
-    "k2 = 1\n",
-    "print(characterReplacement(s2, k2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The code uses a single `for` loop to iterate through the characters of the input string `s`. The loop runs from the beginning to the end of the string.\n",
-    "- Inside the loop, we perform constant-time operations such as updating the `char_count` dictionary and updating variables like `max_count` and `max_length`.\n",
-    "- The code's time complexity is primarily determined by the loop, which iterates through each character in the string once. Therefore, the time complexity is O(n), where n is the length of the input string `s`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The code uses several variables to store information, but the space they consume is constant and does not depend on the size of the input string. These variables include `max_length`, `max_count`, `start`, and `char_count`. Therefore, the space complexity is O(1) or constant space.\n",
-    "\n",
-    "**In summary**, the time complexity of the code is O(n), where n is the length of the input string, and the space complexity is O(1), as it uses a constant amount of additional space regardless of the input size. This algorithm efficiently solves the problem with a linear time complexity."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the code to find the longest substring containing the same letter with at most k replacements in O(n) time complexity and O(1) space complexity. Hint: You may need to update the approach to achieve this.\n",
-    "2. **Variation with Lowercase Letters:** Extend the problem to include both uppercase and lowercase English letters in the input string s. Write a function that can handle this extended input and still find the longest substring with at most k replacements efficiently."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb b/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb
deleted file mode 100644
index cf22c86..0000000
--- a/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb	
+++ /dev/null
@@ -1,260 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 76. Minimum Window Substring\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Minimum Window Substring problem on LeetCode, click here!](https://leetcode.com/problems/minimum-window-substring/)\n",
-    "\n",
-    "---\n",
-    "Given two strings `s` and `t` of lengths `m` and `n` respectively, return *the **minimum window substring** of `s` such that every character in `t` (**including duplicates**) is included in the window. If there is no such substring, return the empty string `\"\"`.*\n",
-    "\n",
-    "The testcases will be generated such that the answer is **unique**.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `m == s.length`\n",
-    "- `n == t.length`\n",
-    "- 1 <= `m, n` <= $10^5$\n",
-    "- `s` and `t` consist of uppercase and lowercase English letters.\n",
-    "\n",
-    "**Follow up**: Could you find an algorithm that runs in $O(m + n)$ time?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def min_window_substring(s, t):\n",
-    "    if not s or not t:\n",
-    "        return \"\"\n",
-    "\n",
-    "    # Initialize dictionaries to keep track of character counts for t and the current window in s.\n",
-    "    t_dict = {}\n",
-    "    current_window_dict = {}\n",
-    "    \n",
-    "    # Populate t_dict with character counts for string t.\n",
-    "    for char in t:\n",
-    "        t_dict[char] = t_dict.get(char, 0) + 1\n",
-    "\n",
-    "    # Initialize pointers for the sliding window.\n",
-    "    left = 0\n",
-    "    min_len = float('inf')\n",
-    "    min_window = \"\"\n",
-    "    required_chars = len(t_dict)\n",
-    "\n",
-    "    # Initialize a variable to keep track of how many required characters have been found in the current window.\n",
-    "    found_chars = 0\n",
-    "\n",
-    "    # Iterate over the string s using the right pointer.\n",
-    "    for right in range(len(s)):\n",
-    "        char = s[right]\n",
-    "\n",
-    "        # Update the current_window_dict.\n",
-    "        current_window_dict[char] = current_window_dict.get(char, 0) + 1\n",
-    "\n",
-    "        # Check if the current character is a required character and if its count matches the required count.\n",
-    "        if char in t_dict and current_window_dict[char] == t_dict[char]:\n",
-    "            found_chars += 1\n",
-    "\n",
-    "        # Try to minimize the window by moving the left pointer.\n",
-    "        while left <= right and found_chars == required_chars:\n",
-    "            # Calculate the current window size.\n",
-    "            window_size = right - left + 1\n",
-    "\n",
-    "            # If the current window is smaller than the minimum found so far, update min_len and min_window.\n",
-    "            if window_size < min_len:\n",
-    "                min_len = window_size\n",
-    "                min_window = s[left:right+1]\n",
-    "\n",
-    "            # Move the left pointer to the right to shrink the window.\n",
-    "            left_char = s[left]\n",
-    "            current_window_dict[left_char] -= 1\n",
-    "\n",
-    "            # Check if the character removed from the window was a required character.\n",
-    "            if left_char in t_dict and current_window_dict[left_char] < t_dict[left_char]:\n",
-    "                found_chars -= 1\n",
-    "\n",
-    "            # Move the left pointer further to continue shrinking the window.\n",
-    "            left += 1\n",
-    "\n",
-    "    return min_window"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by checking if either of the input strings `s` or `t` is empty. If either of them is empty, we return an empty string since there can't be any valid substring in this case.\n",
-    "\n",
-    "2. We initialize two dictionaries: `t_dict` and `current_window_dict`. These dictionaries will be used to keep track of character counts in the string `t` and the current window in string `s`, respectively.\n",
-    "\n",
-    "3. We populate the `t_dict` dictionary by iterating through string `t`. For each character, we increment its count in the dictionary using `t_dict.get(char, 0) + 1`. This allows us to count the occurrences of each character in `t`.\n",
-    "\n",
-    "4. We initialize two pointers: `left` and `right`. The `left` pointer will represent the start of the current window, and the `right` pointer will represent the end of the current window. We also initialize `min_len` to store the length of the minimum window found so far, and `min_window` to store the actual minimum window substring.\n",
-    "\n",
-    "5. We determine the number of required characters in `t` (i.e., the number of distinct characters in `t`) and store it in the variable `required_chars`.\n",
-    "\n",
-    "6. We initialize a variable `found_chars` to keep track of how many required characters have been found in the current window. Initially, it is set to 0.\n",
-    "\n",
-    "7. We iterate over the string `s` using the `right` pointer. In each iteration, we do the following:\n",
-    "\n",
-    "   - Update the `current_window_dict` by incrementing the count of the current character.\n",
-    "   \n",
-    "   - Check if the current character is a required character (present in `t_dict`) and if its count in the `current_window_dict` matches the required count from `t_dict`. If so, we increment `found_chars`.\n",
-    "\n",
-    "8. After updating the window, we attempt to minimize the window size by moving the `left` pointer to the right. In this step, we:\n",
-    "\n",
-    "   - Calculate the size of the current window.\n",
-    "   \n",
-    "   - If the current window size is smaller than the minimum found so far (`min_len`), we update `min_len` and `min_window` to store the current window substring.\n",
-    "   \n",
-    "   - Move the `left` pointer to the right to shrink the window.\n",
-    "   \n",
-    "   - Check if the character being removed from the window was a required character. If it was, we decrement `found_chars`.\n",
-    "   \n",
-    "   - Continue moving the `left` pointer further to continue shrinking the window if the window still contains all required characters.\n",
-    "\n",
-    "9. Finally, we return `min_window`, which will contain the minimum window substring that contains all characters from `t`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "BANC\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "print(min_window_substring(\"ADOBECODEBANC\", \"ABC\"))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "print(min_window_substring(\"a\", \"a\")) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "print(min_window_substring(\"a\", \"aa\")) "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `min_window_substring` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The main loop iterates through the string `s` from left to right using the `right` pointer. This loop runs in O(m) time, where 'm' is the length of string `s`.\n",
-    "\n",
-    "- Inside the loop, we have a while loop that moves the `left` pointer to the right to shrink the window as needed. In the worst case, this while loop can also run in O(m) time because in the worst case, we may have to move the `left` pointer all the way to the end of the string.\n",
-    "\n",
-    "- Within each iteration of the while loop, we perform constant time operations, such as updating dictionaries and comparing character counts.\n",
-    "\n",
-    "- The initialization of the `t_dict` dictionary takes O(n) time, where 'n' is the length of string `t`.\n",
-    "\n",
-    "Therefore, the overall time complexity of the function is O(m + n) because the dominant factor is the length of string `s`, and the length of string `t` has a smaller impact.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity is determined by the space used by the `t_dict` dictionary, the `current_window_dict` dictionary, and a few variables.\n",
-    "\n",
-    "- The `t_dict` dictionary stores character counts for string `t`. In the worst case, when all characters in `t` are distinct, this dictionary can have a maximum of 'n' key-value pairs. So, the space complexity for `t_dict` is O(n).\n",
-    "\n",
-    "- The `current_window_dict` dictionary stores character counts for the current window. In the worst case, it can have a maximum of 'm' key-value pairs. So, the space complexity for `current_window_dict` is also O(m).\n",
-    "\n",
-    "- Other variables used in the function, such as `left`, `right`, `min_len`, `min_window`, `required_chars`, and `found_chars`, are all of constant size and do not depend on the input sizes.\n",
-    "\n",
-    "**In summary**, the overall space complexity of the function is O(max(m, n)), which means it is determined by the larger of the two input strings' lengths. This is because the space used for `t_dict` and `current_window_dict` dominates the space complexity."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Smallest Distinct Substring:** Instead of finding the minimum window substring containing all characters, find the smallest distinct (unique) substring of s that contains all characters from t. This variation adds complexity because you must find a substring with distinct characters.\n",
-    "2. **No Extra Space:** Solve the problem without using any extra space, such as dictionaries or arrays, other than a constant amount of space. This is a significant optimization challenge."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb b/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb
deleted file mode 100644
index 04c0280..0000000
--- a/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb	
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 20. Valid Parentheses\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Valid Parentheses problem on LeetCode, click here!](https://leetcode.com/problems/valid-parentheses/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given a string s containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.\n",
-    "\n",
-    "An input string is valid if:\n",
-    "\n",
-    "1. Open brackets must be closed by the same type of brackets.\n",
-    "2. Open brackets must be closed in the correct order.\n",
-    "3. Every close bracket has a corresponding open bracket of the same type.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `s.length` <= $10^4$\n",
-    "- `s` consists of parentheses only `'()[]{}'`."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def isValid(s):\n",
-    "    # Create an empty stack to store opening brackets\n",
-    "    stack = []\n",
-    "    \n",
-    "    # Define a mapping for matching brackets\n",
-    "    bracket_mapping = {')': '(', '}': '{', ']': '['}\n",
-    "    \n",
-    "    # Iterate through the characters in the input string\n",
-    "    for char in s:\n",
-    "        # If the character is an opening bracket, push it onto the stack\n",
-    "        if char in bracket_mapping.values():\n",
-    "            stack.append(char)\n",
-    "        # If the character is a closing bracket\n",
-    "        elif char in bracket_mapping.keys():\n",
-    "            # Pop the top element from the stack if it exists, or use a dummy value '#'\n",
-    "            top_element = stack.pop() if stack else '#'\n",
-    "            # If the popped element does not match the corresponding opening bracket, return False\n",
-    "            if bracket_mapping[char] != top_element:\n",
-    "                return False\n",
-    "        # If the character is not a bracket, ignore it\n",
-    "        \n",
-    "    # After processing the entire string, if there are any unmatched opening brackets left in the stack, return False\n",
-    "    return len(stack) == 0"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining the `isValid` function that takes a single argument `s`, which is the input string containing parentheses and brackets.\n",
-    "\n",
-    "2. Inside the function, we create an empty stack, which is a list in Python, to store opening brackets as we encounter them in the input string. The stack will help us keep track of the brackets and their order.\n",
-    "\n",
-    "3. We define a `bracket_mapping` dictionary that maps each closing bracket to its corresponding opening bracket. This mapping will be used to check if a closing bracket matches the most recent opening bracket in the stack.\n",
-    "\n",
-    "4. We iterate through each character `char` in the input string `s` using a `for` loop.\n",
-    "\n",
-    "5. If the character `char` is an opening bracket (i.e., it exists in the `values()` of `bracket_mapping`), we push it onto the stack using the `append()` method.\n",
-    "\n",
-    "6. If the character `char` is a closing bracket (i.e., it exists in the `keys()` of `bracket_mapping`), we need to check if it matches the corresponding opening bracket. To do this, we pop the top element from the stack (if it exists) and store it in `top_element`. We use a dummy value `'#'` if the stack is empty to avoid errors.\n",
-    "\n",
-    "7. We then compare `top_element` with the corresponding opening bracket using `bracket_mapping[char]`. If they do not match, it means the string is not valid, and we return `False`.\n",
-    "\n",
-    "8. If the character `char` is not a bracket, we simply ignore it and continue the loop.\n",
-    "\n",
-    "9. After processing the entire string, we check if there are any unmatched opening brackets left in the stack. If the stack is empty, it means all opening brackets have been properly matched and closed, and we return `True`. Otherwise, we return `False`.\n",
-    "\n",
-    "10. Finally, we provide some test cases at the bottom of the code to demonstrate how the function works for different input strings."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "print(isValid(\"()\"))    "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "print(isValid(\"()[]{}\")) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "print(isValid(\"(]\"))     "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `isValid` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The function iterates through each character in the input string `s` once, performing constant-time operations for each character.\n",
-    "- Pushing and popping elements from the stack (list) also takes constant time in most cases.\n",
-    "- Therefore, the overall time complexity of the function is $O(n)$, where $n$ is the length of the input string `s`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity of the function is determined by the space used by the stack and the `bracket_mapping` dictionary.\n",
-    "- In the worst case, if the input string `s` consists entirely of opening brackets, the stack can potentially contain all of these opening brackets, resulting in a space complexity of $O(n)$ in terms of the stack.\n",
-    "- The `bracket_mapping` dictionary has a constant number of key-value pairs (3 pairs in this case).\n",
-    "- Therefore, the overall space complexity of the function is $O(n)$, where $n$ is the length of the input string `s`, mainly due to the stack space.\n",
-    "\n",
-    "**In summary:**\n",
-    "- The time complexity is $O(n)$ because we iterate through the string once.\n",
-    "- The space complexity is $O(n)$ due to the stack used to keep track of opening brackets."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Valid Expressions:** Modify the problem to validate not only brackets but also arithmetic expressions containing parentheses, such as \"$2 * (3 + 5) / (4 - 2)$\".\n",
-    "2. **Valid Parentheses Combinations:**\n",
-    "Write a function to generate all valid combinations of n pairs of parentheses, where $n$ is a positive integer. For example, for $n = 3$, the valid combinations are [\"((()))\", \"(()())\", \"(())()\", \"()(())\", \"()()()\"]."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb b/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb
deleted file mode 100644
index eb4858b..0000000
--- a/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 153. Find Minimum in Rotated Sorted Array\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Find Minimum in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:\n",
-    "\n",
-    "- `[4,5,6,7,0,1,2]` if it was rotated `4` times.\n",
-    "- `[0,1,2,4,5,6,7]` if it was rotated `7` times.\n",
-    "\n",
-    "Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.\n",
-    "\n",
-    "Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array.*\n",
-    "\n",
-    "You must write an algorithm that runs in $O(log\\ n)$ time.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `n == nums.length`\n",
-    "- `1 <= n <= 5000`\n",
-    "- `-5000 <= nums[i] <= 5000`\n",
-    "- All the integers of `nums` are **unique**.\n",
-    "- `nums` is sorted and rotated between `1` and `n` times."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def findMin(nums):\n",
-    "    left, right = 0, len(nums) - 1\n",
-    "    \n",
-    "    while left < right:\n",
-    "        mid = left + (right - left) // 2\n",
-    "        \n",
-    "        # If the mid element is greater than the rightmost element,\n",
-    "        # it means the minimum element is in the right half.\n",
-    "        if nums[mid] > nums[right]:\n",
-    "            left = mid + 1\n",
-    "        # If the mid element is less than or equal to the rightmost element,\n",
-    "        # it means the minimum element is in the left half or at the mid itself.\n",
-    "        else:\n",
-    "            right = mid\n",
-    "    \n",
-    "    # The loop will break when left and right converge to the minimum element.\n",
-    "    # At this point, left (or right) will be pointing to the minimum element.\n",
-    "    return nums[left]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The function `findMin(nums)` takes an input list `nums`, which represents the rotated sorted array.\n",
-    "\n",
-    "2. Initialize two pointers `left` and `right` to track the current search range. Initially, `left` is set to 0 (the beginning of the array), and `right` is set to `len(nums) - 1` (the end of the array).\n",
-    "\n",
-    "3. Enter a `while` loop with the condition `left < right`. This loop continues until `left` and `right` converge to a single element, which will be the minimum element in the rotated array.\n",
-    "\n",
-    "4. Inside the loop, calculate the middle index `mid` using integer division. This helps in finding the middle element of the current search range.\n",
-    "\n",
-    "5. Check if the element at the middle index (`nums[mid]`) is greater than the element at the rightmost index (`nums[right]`). If this condition is true, it means that the minimum element must be in the right half of the current search range. So, update `left` to `mid + 1`, effectively eliminating the left half of the search range.\n",
-    "\n",
-    "6. If the condition from step 5 is not met, it implies that the minimum element can be in the left half of the current search range or could be the element at the `mid` index itself. In this case, update `right` to `mid`, effectively eliminating the right half of the search range.\n",
-    "\n",
-    "7. Repeat steps 4-6 until `left` and `right` converge to the minimum element.\n",
-    "\n",
-    "8. When the loop exits, it means that `left` (or `right`) points to the minimum element in the array.\n",
-    "\n",
-    "9. Return `nums[left]` (or `nums[right]`) as the result, which is the minimum element in the rotated sorted array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1:\n",
-    "nums1 = [3, 4, 5, 1, 2]\n",
-    "print(findMin(nums1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "0\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-    "print(findMin(nums2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "11\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "nums3 = [11, 13, 15, 17]\n",
-    "print(findMin(nums3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of this code is $O(log\\ n)$, where '$n$' is the length of the input array `nums`. This is because the binary search algorithm is employed, which continually divides the search range in half with each iteration.\n",
-    "\n",
-    "In each iteration of the `while` loop, the search range is halved, and this process continues until the `left` and `right` pointers converge to the minimum element. In the worst case, it will take logarithmic time to reduce the search range to a single element.\n",
-    "\n",
-    "Therefore, the binary search used in this code runs in $O(log\\ n)$ time complexity.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity of this code is $O(1)$, which means it uses a constant amount of additional space that does not depend on the size of the input array.\n",
-    "\n",
-    "The algorithm uses a fixed number of variables (`left`, `right`, `mid`) to keep track of the search range and indices, but the number of these variables does not grow with the size of the input array. Hence, the space complexity is constant.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: $O(log\\ n)$\n",
-    "- Space Complexity: $O(1)$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. Solve the problem with a variation where the input array may contain duplicates.\n",
-    "\n",
-    "2. Modify the problem to return the index of the minimum element in the rotated sorted array, rather than the element itself. The algorithm should still run in $O(log\\ n)$ time.\n",
-    "\n",
-    "3. Implement a function that finds the maximum element in a rotated sorted array. How would you adapt the binary search algorithm to solve this problem efficiently in $O(log\\ n)$ time?\n",
-    "\n",
-    "4. Implement a function that can find the kth smallest element in a rotated sorted array. This is an extension of the original problem, and the algorithm should still run in $O(log\\ n)$ time."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb b/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb
deleted file mode 100644
index 7b3139d..0000000
--- a/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,237 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 33. Search in Rotated Sorted Array\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Search in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/search-in-rotated-sorted-array/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "There is an integer array `nums` sorted in ascending order (with **distinct** values).\n",
-    "\n",
-    "Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k (1 <= k < nums.length)` such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index 3 and become `[4,5,6,7,0,1,2]`.\n",
-    "\n",
-    "Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.*\n",
-    "\n",
-    "You must write an algorithm with $O(log\\ n)$ runtime complexity.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `1 <= nums.length <= 5000`\n",
-    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-    "- All values of `nums` are **unique**.\n",
-    "- `nums` is an ascending array that is possibly rotated.\n",
-    "- $-10^4$ <= `target` <= $10^4$"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def search(nums, target):\n",
-    "    left, right = 0, len(nums) - 1\n",
-    "\n",
-    "    # Find the pivot index using binary search\n",
-    "    while left < right:\n",
-    "        mid = left + (right - left) // 2\n",
-    "\n",
-    "        if nums[mid] > nums[right]:\n",
-    "            left = mid + 1\n",
-    "        else:\n",
-    "            right = mid\n",
-    "\n",
-    "    pivot = left\n",
-    "    left, right = 0, len(nums) - 1\n",
-    "\n",
-    "    # Determine which part of the array to search in\n",
-    "    if target >= nums[pivot] and target <= nums[right]:\n",
-    "        left = pivot\n",
-    "    else:\n",
-    "        right = pivot\n",
-    "\n",
-    "    # Perform binary search to find the target\n",
-    "    while left <= right:\n",
-    "        mid = left + (right - left) // 2\n",
-    "\n",
-    "        if nums[mid] == target:\n",
-    "            return mid\n",
-    "        elif nums[mid] < target:\n",
-    "            left = mid + 1\n",
-    "        else:\n",
-    "            right = mid - 1\n",
-    "\n",
-    "    return -1"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The `search` function takes two arguments: `nums`, which is the rotated sorted array, and `target`, which is the value we want to find in the array.\n",
-    "\n",
-    "2. We initialize two pointers, `left` and `right`, which represent the range in which we are going to perform binary search. Initially, `left` is set to 0, and `right` is set to the index of the last element in the array (`len(nums) - 1`).\n",
-    "\n",
-    "3. We start by finding the pivot index using binary search. The pivot index is the index at which the array is rotated. We use a `while` loop to continue the search until `left` is less than `right`.\n",
-    "\n",
-    "4. Inside the loop, we calculate the middle index `mid` using integer division. We then compare the value at `mid` with the value at `right`. If `nums[mid]` is greater than `nums[right]`, it means the pivot point lies to the right of `mid`, so we update `left` to `mid + 1`. Otherwise, the pivot point lies to the left of `mid`, so we update `right` to `mid`. This process continues until we find the pivot index.\n",
-    "\n",
-    "5. After finding the pivot index, we have divided the array into two parts: one part is sorted in ascending order, and the other part is also sorted in ascending order but rotated. We need to determine which part of the array contains the target value.\n",
-    "\n",
-    "6. We reset `left` and `right` pointers. If the target value is within the range `[nums[pivot], nums[right]]`, we set `left` to `pivot` (the start of the rotated part), indicating that we should search in the rotated part. Otherwise, we set `right` to `pivot` (the end of the sorted part), indicating that we should search in the sorted part.\n",
-    "\n",
-    "7. We then perform binary search again within the chosen range (`left` to `right`) to find the target element. The binary search continues until `left` is less than or equal to `right`.\n",
-    "\n",
-    "8. Inside the binary search loop, we calculate the middle index `mid` and compare `nums[mid]` with the target value. If they are equal, we return `mid` as the index of the target element. If `nums[mid]` is less than the target, we update `left` to `mid + 1` to search in the right half. If `nums[mid]` is greater than the target, we update `right` to `mid - 1` to search in the left half.\n",
-    "\n",
-    "9. If we exit the binary search loop without finding the target, we return -1 to indicate that the target is not present in the array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "nums1 = [4, 5, 6, 7, 0, 1, 2]\n",
-    "target1 = 0\n",
-    "result1 = search(nums1, target1)\n",
-    "print(result1) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-    "target2 = 3\n",
-    "result2 = search(nums2, target2)\n",
-    "print(result2) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "nums3 = [1]\n",
-    "target3 = 0\n",
-    "result3 = search(nums3, target3)\n",
-    "print(result3) "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. Finding the pivot index using binary search takes $O(log\\ n)$ time, where '$n$' is the number of elements in the array.\n",
-    "2. After finding the pivot index, performing binary search to find the target element also takes $O(log\\ n)$ time in the worst case.\n",
-    "\n",
-    "The dominant factor in the time complexity is the binary search, and since we perform two binary searches sequentially, the overall time complexity of the code is $O(log\\ n)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of the code is very minimal:\n",
-    "\n",
-    "1. We use a constant amount of additional space for variables such as `left`, `right`, `pivot`, and `mid`. These variables do not depend on the input size, so they contribute $O(1)$ space complexity.\n",
-    "2. The function does not use any additional data structures that scale with the input size.\n",
-    "\n",
-    "Therefore, the space complexity of the code is $O(1)$, which means it has a constant space complexity and does not depend on the size of the input array.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: $O(log\\ n)$\n",
-    "- Space Complexity: $O(1)$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Find Minimum Element in Rotated Sorted Array**: Write an algorithm to find the minimum element in a rotated sorted array. This is a variation of the problem where you don't need to search for a target value but instead find the smallest element.\n",
-    "\n",
-    "2. **Search in a Circularly Sorted Array**: Consider an array that is sorted in a circular manner (e.g., [4, 5, 6, 7, 0, 1, 2, 3]). Adapt the search algorithm to work for circularly sorted arrays while maintaining $O(log\\ n)$ complexity."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb b/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb
deleted file mode 100644
index c428c9f..0000000
--- a/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb	
+++ /dev/null
@@ -1,261 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 141. Linked List Cycle\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Linked List Cycle problem on LeetCode, click here!](https://leetcode.com/problems/linked-list-cycle/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given `head`, the head of a linked list, determine if the linked list has a cycle in it.\n",
-    "\n",
-    "There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is connected to. **Note that `pos` is not passed as a parameter.**\n",
-    "\n",
-    "Return `true` if there is a cycle in the linked list. Otherwise, return `false`.\n",
-    "\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of the nodes in the list is in the range $[0, 10^4]$.\n",
-    "- $-10^5$ <= Node.val <= $10^5$\n",
-    "- `pos` is `-1` or a **valid index** in the linked-list.\n",
-    "\n",
-    "**Follow up:** Can you solve it using $O(1)$ (i.e. constant) memory?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for singly-linked list.\n",
-    "class ListNode:\n",
-    "    def __init__(self, val=0):\n",
-    "        self.val = val\n",
-    "        self.next = None\n",
-    "\n",
-    "class Solution:\n",
-    "    def hasCycle(self, head: ListNode) -> bool:\n",
-    "        if not head:\n",
-    "            return False\n",
-    "        \n",
-    "        slow = head\n",
-    "        fast = head\n",
-    "        \n",
-    "        while fast and fast.next:\n",
-    "            slow = slow.next\n",
-    "            fast = fast.next.next\n",
-    "            \n",
-    "            if slow == fast:\n",
-    "                return True\n",
-    "        \n",
-    "        return False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The given code defines a class `Solution` with a method `hasCycle` for determining if a linked list has a cycle.\n",
-    "\n",
-    "The `ListNode` class represents nodes in the linked list, where each node has a `val` (a value associated with the node) and a `next` attribute (a reference to the next node in the list).\n",
-    "\n",
-    "The `hasCycle` method takes the `head` of a linked list as input and returns `True` if there is a cycle in the linked list, and `False` otherwise. It uses two pointers, `slow` and `fast`, to traverse the linked list. If there is a cycle, the `fast` pointer will eventually catch up to the `slow` pointer.\n",
-    "\n",
-    "The algorithm works as follows:\n",
-    "1. It checks if the input `head` is `None` (an empty list). If it is, the method returns `False` because an empty list can't have a cycle.\n",
-    "2. Two pointers, `slow` and `fast`, initially point to the `head` of the list.\n",
-    "3. The code enters a `while` loop where `fast` moves two steps at a time, and `slow` moves one step at a time.\n",
-    "4. If there is a cycle, `fast` will eventually catch up to `slow`, and the method returns `True`.\n",
-    "5. If the loop completes without finding a cycle, the method returns `False`."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to create a linked list with a cycle\n",
-    "def create_linked_list_with_cycle(values, pos):\n",
-    "    dummy = ListNode()\n",
-    "    current = dummy\n",
-    "    cycle_node = None\n",
-    "    \n",
-    "    for i, val in enumerate(values):\n",
-    "        current.next = ListNode(val)\n",
-    "        current = current.next\n",
-    "        \n",
-    "        if i == pos:\n",
-    "            cycle_node = current\n",
-    "    \n",
-    "    if cycle_node:\n",
-    "        current.next = cycle_node\n",
-    "    \n",
-    "    return dummy.next"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "This helper function is designed to create a linked list with a cycle for testing purposes. It takes two arguments:\n",
-    "\n",
-    "1. `values`: A list of values representing the nodes in the linked list.\n",
-    "2. `pos`: An integer representing the index of the node where the cycle begins.\n",
-    "\n",
-    "Here's how the function works:\n",
-    "\n",
-    "- It starts by creating an empty `ListNode` called `dummy`. This `dummy` node is used to simplify the creation of the linked list.\n",
-    "- It initializes a `current` pointer to the `dummy` node. This pointer will be used to traverse the linked list.\n",
-    "- It initializes a `cycle_node` variable to `None`. This variable will hold the reference to the node where the cycle begins, if any.\n",
-    "- It then iterates through the `values` list, creating a new `ListNode` for each value and appending it to the linked list.\n",
-    "- Inside the loop, it checks if the current index `i` is equal to the specified `pos`. If they match, it sets `cycle_node` to the current node. This simulates the creation of a cycle in the linked list.\n",
-    "- After the loop, if `cycle_node` is not `None` (indicating a cycle should be created), it connects the last node in the linked list to `cycle_node`, effectively creating a cycle.\n",
-    "- Finally, it returns the reference to the first node of the linked list (not the `dummy` node).\n",
-    "\n",
-    "This helper function allows you to easily create test cases where you can specify the values for the linked list and the position where the cycle starts. It's particularly useful for testing the code's ability to detect cycles in linked lists."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "head = [3,2,0,-4]\n",
-    "pos = 1\n",
-    "\n",
-    "head = create_linked_list_with_cycle(head, pos)\n",
-    "solution = Solution()\n",
-    "result = solution.hasCycle(head)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "head = [1,2]\n",
-    "pos = 0\n",
-    "\n",
-    "head = create_linked_list_with_cycle(head, pos)\n",
-    "solution = Solution()\n",
-    "result = solution.hasCycle(head)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "head = [1]\n",
-    "pos = -1\n",
-    "\n",
-    "head = create_linked_list_with_cycle(head, pos)\n",
-    "solution = Solution()\n",
-    "result = solution.hasCycle(head)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's discuss the time and space complexity of the code for detecting a cycle in a linked list:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this code is $O(n)$, where \"$n$\" is the number of nodes in the linked list.\n",
-    "\n",
-    "The reason for this is that both the `slow` and `fast` pointers traverse the linked list, and they move at different speeds. In the worst case, when there is no cycle, the `fast` pointer will reach the end of the list after going through approximately $n/2$ nodes. In the case of a cycle, it may take some extra iterations for the `fast` pointer to catch up to the `slow` pointer. However, the total number of iterations is still proportional to the number of nodes in the linked list, making it $O(n)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this code is $O(1)$, which means it uses constant space.\n",
-    "\n",
-    "The reason for this is that regardless of the size of the linked list, the code only uses two additional pointers (`slow` and `fast`) to traverse the list. These pointers do not depend on the size of the input linked list, so the space complexity remains constant.\n",
-    "\n",
-    "**In summary**, this code efficiently detects cycles in a linked list with a time complexity of $O(n)$ and a space complexity of $O(1)$. It is an example of an algorithm that solves a complex problem with minimal memory usage."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
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- "nbformat": 4,
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-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb b/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb
deleted file mode 100644
index 4ec3f3b..0000000
--- a/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb	
+++ /dev/null
@@ -1,277 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 143. Reorder List\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Reorder List problem on LeetCode, click here!](https://leetcode.com/problems/reorder-list/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given the head of a singly linked-list. The list can be represented as:\n",
-    "\n",
-    "$L_0 → L_1 → … → L_{n - 1} → L_n$\n",
-    "\n",
-    "Reorder the list to be on the following form:\n",
-    "\n",
-    "$L_0 → L_n → L_1 → L_{n - 1} → L_2 → L_{n - 2} → …$\n",
-    "\n",
-    "You may not modify the values in the list's nodes. Only nodes themselves may be changed.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the list is in the range $[1, 5 * 10^4]$.\n",
-    "- `1 <= Node.val <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "def reorderList(head):\n",
-    "    if not head or not head.next:\n",
-    "        return head\n",
-    "\n",
-    "    # Step 1: Find the middle of the linked list\n",
-    "    slow, fast = head, head\n",
-    "    while fast.next and fast.next.next:\n",
-    "        slow = slow.next\n",
-    "        fast = fast.next.next\n",
-    "\n",
-    "    # Split the list into two halves\n",
-    "    first_half = head\n",
-    "    second_half = slow.next\n",
-    "    slow.next = None\n",
-    "\n",
-    "    # Step 2: Reverse the second half of the linked list\n",
-    "    prev = None\n",
-    "    current = second_half\n",
-    "    while current:\n",
-    "        next_node = current.next\n",
-    "        current.next = prev\n",
-    "        prev = current\n",
-    "        current = next_node\n",
-    "    second_half = prev\n",
-    "\n",
-    "    # Step 3: Merge the first and reversed second halves alternately\n",
-    "    p1, p2 = first_half, second_half\n",
-    "    while p2:\n",
-    "        next_p1 = p1.next\n",
-    "        next_p2 = p2.next\n",
-    "        p1.next = p2\n",
-    "        p2.next = next_p1\n",
-    "        p1 = next_p1\n",
-    "        p2 = next_p2\n",
-    "\n",
-    "    return head"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The `reorderList` function is designed to reorder a singly linked list following the specific pattern described in the problem statement. Here's a detailed explanation of how the function works:\n",
-    "\n",
-    "1. **Base Cases Handling**:\n",
-    "   - The function first checks if the input linked list is empty (`not head`) or contains only one element (`not head.next`). If either of these conditions is met, the list cannot be reordered, so the function returns the original list as-is.\n",
-    "\n",
-    "2. **Finding the Middle of the Linked List**:\n",
-    "   - To reorder the list, we first need to find the middle point so that we can split the list into two halves. This is done using two pointers, `slow` and `fast`, initialized to the head of the list.\n",
-    "   - `slow` moves one step at a time while `fast` moves two steps at a time. When `fast` reaches the end of the list or the second-to-last node, `slow` will be at the middle node.\n",
-    "\n",
-    "3. **Splitting the List**:\n",
-    "   - After finding the middle node, we split the list into two halves:\n",
-    "     - The first half, `first_half`, contains nodes from the beginning up to the middle.\n",
-    "     - The second half, `second_half`, contains nodes from the middle to the end.\n",
-    "   - To split the list, we set the `next` pointer of the node before the middle node to `None`.\n",
-    "\n",
-    "4. **Reversing the Second Half**:\n",
-    "   - We reverse the second half of the linked list using the `prev`, `current`, and `next_node` pointers.\n",
-    "   - `prev` is initially `None`, and we iterate through the second half. For each node, we:\n",
-    "     - Set the `next` of the current node to `prev`, effectively reversing the next pointer direction.\n",
-    "     - Update `prev` to the current node.\n",
-    "     - Move to the next node using the `next_node`.\n",
-    "\n",
-    "5. **Merging the Two Halves Alternately**:\n",
-    "   - We now have two linked lists: `first_half` and the reversed `second_half`.\n",
-    "   - We merge these two lists alternately by adjusting the `next` pointers of the nodes.\n",
-    "   - `p1` and `p2` are pointers to the current nodes in `first_half` and `second_half`, respectively.\n",
-    "   - We iterate through both lists while reordering nodes:\n",
-    "     - Set the `next` of `p1` to `p2` to link a node from the first half to a node from the reversed second half.\n",
-    "     - Update `p1` and `p2` to their respective next nodes.\n",
-    "     - Repeat this process until we have processed all nodes in both halves.\n",
-    "\n",
-    "6. **Returning the Reordered List**:\n",
-    "   - After the merging process is complete, the linked list is reordered as specified.\n",
-    "   - The function returns the `head` of the reordered list.\n",
-    "\n",
-    "The overall result is a singly linked list that has been reordered according to the given pattern. The time complexity of this algorithm is O(N), where N is the number of nodes in the linked list, as we traverse the list once to find the middle, once to reverse the second half, and once to merge the two halves."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to print the linked list\n",
-    "def printLinkedList(head):\n",
-    "    result = []\n",
-    "    while head:\n",
-    "        result.append(head.val)\n",
-    "        head = head.next\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the helper functions:\n",
-    "**printLinkedList Function**:\n",
-    "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-    "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-    "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-    "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-    "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-    "\n",
-    "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 4, 2, 3]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "head1 = ListNode(1, ListNode(2, ListNode(3, ListNode(4))))\n",
-    "reorderList(head1)\n",
-    "print(printLinkedList(head1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 5, 2, 4, 3]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "head2 = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))\n",
-    "reorderList(head2)\n",
-    "print(printLinkedList(head2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `reorderList` function:\n",
-    "\n",
-    "**Time Complexity**:\n",
-    "1. **Finding the Middle of the Linked List**: In this step, we use two pointers, one moving one step at a time (`slow`) and the other moving two steps at a time (`fast`). This process takes $O(N/2)$ time, where $N$ is the number of nodes in the linked list.\n",
-    "\n",
-    "2. **Splitting the List**: After finding the middle, we split the list into two halves by setting the `next` pointer of the node before the middle node to `None`. This step takes $O(1)$ time.\n",
-    "\n",
-    "3. **Reversing the Second Half**: We reverse the second half of the linked list using a while loop that iterates through the second half of the list once. Therefore, this step also takes $O(N/2)$ time.\n",
-    "\n",
-    "4. **Merging the Two Halves Alternately**: In this step, we merge the two halves alternately by adjusting the `next` pointers of the nodes. We iterate through both halves once, so this step takes $O(N/2)$ time.\n",
-    "\n",
-    "Overall, the time complexity of the `reorderList` function is dominated by the steps involving reversing and merging the two halves, both of which take $O(N/2)$ time. Therefore, the total time complexity is $O(N)$.\n",
-    "\n",
-    "**Space Complexity**:\n",
-    "The space complexity of the `reorderList` function is primarily determined by the variables and data structures used within the function. Let's break down the space complexity components:\n",
-    "\n",
-    "1. **Constant Space Variables**: The variables `slow`, `fast`, `prev`, `current`, `next_node`, `p1`, and `p2` are used to traverse and manipulate the linked list. These variables occupy constant space, regardless of the input size. Therefore, they contribute $O(1)$ to the space complexity.\n",
-    "\n",
-    "2. **Split Linked Lists**: In the splitting step, we create two new linked list segments (`first_half` and `second_half`) that store references to nodes from the original linked list. These segments occupy space proportional to half of the input list, $O(N/2)$. \n",
-    "\n",
-    "3. **Reversed Second Half**: During the reversal step, we reverse the second half of the linked list in-place without creating any additional data structures. Therefore, it doesn't contribute to additional space complexity.\n",
-    "\n",
-    "4. **Overall**: Combining the above components, the total space complexity is $O(N/2)$, which simplifies to $O(N)$ in terms of space complexity.\n",
-    "\n",
-    "**In summary**, the `reorderList` function has a time complexity of $O(N)$ and a space complexity of $O(N)$ due to the creation of two new linked list segments while splitting the list. The constant space variables used for traversal do not significantly impact the overall space complexity."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Reorder by K Elements:** Generalize the reorderList function to reorder the list in a different pattern, where you reorder the list by $K$ elements at a time. For example, for $K=3$, you would reorder it as ($L_0 → L_1 → L_2 → L_n → L_{n-1} → L_{n-2} → ...$).\n",
-    "You may need to handle cases where the number of nodes is not a multiple of $K$.\n",
-    "\n",
-    "2. **Reorder by Odd and Even Nodes:** Modify the reorderList function to reorder the list in a pattern where odd-indexed nodes (1-based index) come before even-indexed nodes. For example, ($L_0 → L_n → L_1 → L_{n-1} → L_2 → L_{n-2} → ...$)."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
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-   "codemirror_mode": {
-    "name": "ipython",
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-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb b/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb
deleted file mode 100644
index 73dbf66..0000000
--- a/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb	
+++ /dev/null
@@ -1,268 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 19. Remove Nth Node From End of List\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Remove Nth Node From End of List problem on LeetCode, click here!](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the head of a linked list, remove the $n^{th}$ node from the end of the list and return its head.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the list is `sz`.\n",
-    "- `1 <= sz <= 30`\n",
-    "- `0 <= Node.val <= 100`\n",
-    "- `1 <= n <= sz`\n",
-    "\n",
-    "**Follow up:** Could you do this in one pass?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "def removeNthFromEnd(head, n):\n",
-    "    # Create a dummy node to handle the case of removing the head node.\n",
-    "    dummy = ListNode(0)\n",
-    "    dummy.next = head\n",
-    "    first = dummy\n",
-    "    second = dummy\n",
-    "    \n",
-    "    # Advance the first pointer by n+1 nodes.\n",
-    "    for _ in range(n + 1):\n",
-    "        first = first.next\n",
-    "    \n",
-    "    # Move both pointers simultaneously until first reaches the end.\n",
-    "    while first is not None:\n",
-    "        first = first.next\n",
-    "        second = second.next\n",
-    "    \n",
-    "    # Remove the nth node by updating the next pointer of the previous node.\n",
-    "    second.next = second.next.next\n",
-    "    \n",
-    "    return dummy.next  # Return the modified head of the linked list"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a `ListNode` class to represent individual nodes in a linked list. Each node has two attributes: `val` (the value of the node) and `next` (a reference to the next node in the list).\n",
-    "\n",
-    "2. The `removeNthFromEnd` function takes the head of a linked list (`head`) and the value `n`, which represents the position from the end of the list of the node to be removed. It returns the modified head of the linked list after the removal.\n",
-    "\n",
-    "3. We create a dummy node (`dummy`) at the beginning of the list. This dummy node simplifies the code by handling cases where the head of the list needs to be removed.\n",
-    "\n",
-    "4. We initialize two pointers, `first` and `second`, both initially pointing to the dummy node.\n",
-    "\n",
-    "5. To advance the `first` pointer by `n+1` nodes, we use a `for` loop. This positions the `first` pointer `n` nodes ahead of the `second` pointer.\n",
-    "\n",
-    "6. We then move both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This ensures that the `second` pointer ends up pointing to the node that needs to be removed.\n",
-    "\n",
-    "7. To remove the nth node from the end, we update the `next` pointer of the node pointed to by the `second` pointer to skip over the node to be removed.\n",
-    "\n",
-    "8. Finally, we return `dummy.next`, which is the modified head of the linked list without the removed node.\n",
-    "\n",
-    "9. Additionally, there are two helper functions:\n",
-    "   - `createLinkedList(values)` creates a linked list from a list of values.\n",
-    "   - `convertToList(head)` converts a linked list back into a list of values."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to create a linked list from a list of values.\n",
-    "def createLinkedList(values):\n",
-    "    if not values:\n",
-    "        return None\n",
-    "    head = ListNode(values[0])\n",
-    "    current = head\n",
-    "    for val in values[1:]:\n",
-    "        current.next = ListNode(val)\n",
-    "        current = current.next\n",
-    "    return head\n",
-    "\n",
-    "# Helper function to convert a linked list to a list of values.\n",
-    "def convertToList(head):\n",
-    "    result = []\n",
-    "    current = head\n",
-    "    while current:\n",
-    "        result.append(current.val)\n",
-    "        current = current.next\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the two helper functions:\n",
-    "\n",
-    "1. `createLinkedList(values)`: This function creates a linked list from a list of values.\n",
-    "\n",
-    "   - Input: `values` is a list containing values that you want to insert into the linked list.\n",
-    "   - Output: The function returns the head of the created linked list.\n",
-    "\n",
-    "   Explanation:\n",
-    "   - The function starts by checking if the `values` list is empty. If it is, it returns `None` to indicate an empty linked list.\n",
-    "   - If the `values` list is not empty, it creates the first node of the linked list with the value from the first element of `values`.\n",
-    "   - It then iterates through the remaining elements of `values` and creates new nodes for each value, connecting them together using the `next` attribute to form a linked list.\n",
-    "   - Finally, it returns the head of the linked list, which is the first node created.\n",
-    "   \n",
-    "\n",
-    "2. `convertToList(head)`: This function converts a linked list into a list of values.\n",
-    "\n",
-    "   - Input: `head` is the head node of the linked list that you want to convert.\n",
-    "   - Output: The function returns a list containing the values of the nodes in the linked list in order.\n",
-    "\n",
-    "   Explanation:\n",
-    "   - The function starts by initializing an empty list, `result`, to store the values of the linked list nodes.\n",
-    "   - It then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-    "   - The iteration continues until it reaches the end of the linked list (i.e., the `next` attribute of the current node becomes `None`).\n",
-    "   - After the iteration is complete, the function returns the `result` list, which now contains the values of all the nodes in the linked list in the same order as they appear in the linked list.\n",
-    "\n",
-    "These helper functions are useful for creating linked lists from lists of values and converting linked lists back into lists of values, making it easier to work with linked list examples and test cases in a more familiar list format."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 2, 3, 5]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "head1 = createLinkedList([1, 2, 3, 4, 5])\n",
-    "n1 = 2\n",
-    "new_head1 = removeNthFromEnd(head1, n1)\n",
-    "print(convertToList(new_head1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "head2 = createLinkedList([1])\n",
-    "n2 = 1\n",
-    "new_head2 = removeNthFromEnd(head2, n2)\n",
-    "print(convertToList(new_head2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "head3 = createLinkedList([1, 2])\n",
-    "n3 = 1\n",
-    "new_head3 = removeNthFromEnd(head3, n3)\n",
-    "print(convertToList(new_head3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Certainly! Let's analyze the time and space complexity of the `removeNthFromEnd` function in isolation:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The `removeNthFromEnd` function performs two passes through the linked list. In the worst case, it needs to traverse the entire linked list of length $N$.\n",
-    "- The first pass positions the `first` pointer $n$ nodes ahead of the `second` pointer. This pass takes $O(N)$ time because it involves iterating through all N nodes.\n",
-    "- The second pass moves both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This also takes $O(N)$ time in the worst case.\n",
-    "- Therefore, the overall time complexity of the `removeNthFromEnd` function is $O(N)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The `removeNthFromEnd` function uses a constant amount of extra space for its variables (e.g., `first`, `second`, and `dummy`). This space usage does not depend on the size of the linked list.\n",
-    "- The space complexity is $O(1)$, indicating that the space used by the function is constant and independent of the size of the input linked list.\n",
-    "\n",
-    "**In summary**, the `removeNthFromEnd` function has a time complexity of $O(N)$ and a space complexity of $O(1)$. It efficiently removes the nth node from the end of the linked list with a single pass through the list and constant extra space usage."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb b/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb
deleted file mode 100644
index 09de0a3..0000000
--- a/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb	
+++ /dev/null
@@ -1,297 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 206. Reverse Linked List\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Reverse Linked List problem on LeetCode, click here!](https://leetcode.com/problems/reverse-linked-list/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the `head` of a singly linked list, reverse the list, and return the reversed list.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the list is the range `[0, 5000]`.\n",
-    "- `-5000 <= Node.val <= 5000`\n",
-    "\n",
-    "**Follow up:** A linked list can be reversed either iteratively or recursively. Could you implement both?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for singly-linked list.\n",
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "# Function to reverse a linked list iteratively\n",
-    "def reverseListIterative(head):\n",
-    "    prev = None\n",
-    "    current = head\n",
-    "    \n",
-    "    while current:\n",
-    "        next_node = current.next\n",
-    "        current.next = prev\n",
-    "        prev = current\n",
-    "        current = next_node\n",
-    "    \n",
-    "    return prev\n",
-    "\n",
-    "# Function to reverse a linked list recursively\n",
-    "def reverseListRecursive(head):\n",
-    "    if not head or not head.next:\n",
-    "        return head\n",
-    "\n",
-    "    new_head = reverseListRecursive(head.next)\n",
-    "    head.next.next = head\n",
-    "    head.next = None\n",
-    "\n",
-    "    return new_head"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `reverseListIterative(head)`: This function reverses a linked list iteratively using a loop. Here's a step-by-step explanation:\n",
-    "\n",
-    "   - It takes the `head` of the input linked list as a parameter.\n",
-    "   - It initializes two pointers, `prev` and `current`, initially set to `None` and the `head` of the list, respectively.\n",
-    "   - It enters a `while` loop that continues until `current` reaches the end of the list (i.e., becomes `None`).\n",
-    "   - Inside the loop:\n",
-    "     - It stores the `next_node` which is the next node after `current`.\n",
-    "     - It updates the `next` pointer of `current` to point to the `prev` node. This effectively reverses the link direction.\n",
-    "     - It moves `prev` to `current` and `current` to `next_node`, advancing the pointers one step further in the list.\n",
-    "   - Once the loop completes, `prev` will be pointing to the new head of the reversed list (which was the last node of the original list), so it returns `prev`.\n",
-    "\n",
-    "2. `reverseListRecursive(head)`: This function reverses a linked list recursively. Here's how it works:\n",
-    "\n",
-    "   - It takes the `head` of the input linked list as a parameter.\n",
-    "   - The base case is checked: if `head` is `None` or `head.next` is `None`, meaning the list is empty or has only one node, it returns `head` as there's no need to reverse a list with zero or one element.\n",
-    "   - In the recursive case:\n",
-    "     - It calls itself with `head.next`, effectively moving down the list until it reaches the end.\n",
-    "     - Once it reaches the end, it starts reversing the links:\n",
-    "       - `head.next.next` is set to `head`, reversing the link between `head` and the next node.\n",
-    "       - `head.next` is set to `None` to avoid cycles.\n",
-    "   - Finally, it returns the new head of the reversed list, which will be the last node of the original list.\n",
-    "\n",
-    "**In summary,** `reverseListIterative` reverses the linked list by iterating through it and changing the next pointers, while `reverseListRecursive` reverses the linked list by recursively reaching the end and then reversing the links on the way back up the recursion stack. Both functions achieve the same result: reversing the linked list."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to create a linked list from a list of values\n",
-    "def createLinkedList(values):\n",
-    "    if not values:\n",
-    "        return None\n",
-    "    head = ListNode(values[0])\n",
-    "    current = head\n",
-    "    for val in values[1:]:\n",
-    "        current.next = ListNode(val)\n",
-    "        current = current.next\n",
-    "    return head\n",
-    "\n",
-    "# Helper function to convert a linked list to a list for testing\n",
-    "def linkedListToList(head):\n",
-    "    result = []\n",
-    "    current = head\n",
-    "    while current:\n",
-    "        result.append(current.val)\n",
-    "        current = current.next\n",
-    "    return result\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the two helper functions used in the provided code:\n",
-    "\n",
-    "1. `createLinkedList(values)`:\n",
-    "\n",
-    "   - Purpose: This function is used to create a linked list from a list of values. It takes a list of values as input and returns the head of the linked list.\n",
-    "\n",
-    "   - How it works:\n",
-    "     - It first checks if the input list `values` is empty. If it's empty, it returns `None` to indicate an empty linked list.\n",
-    "     - If `values` is not empty, it initializes a `head` node with the value of the first element in `values`.\n",
-    "     - It then iterates through the remaining values in `values`, creating new nodes for each value and linking them together to form a linked list.\n",
-    "     - Finally, it returns the `head` node of the newly created linked list.\n",
-    "\n",
-    "2. `linkedListToList(head)`:\n",
-    "\n",
-    "   - Purpose: This function is used to convert a linked list back into a Python list for testing and output purposes.\n",
-    "\n",
-    "   - How it works:\n",
-    "     - It takes the `head` of the linked list as input.\n",
-    "     - It initializes an empty list called `result`.\n",
-    "     - It then iterates through the linked list, starting from `head`, and appends each node's `val` (value) to the `result` list.\n",
-    "     - This process continues until the end of the linked list is reached.\n",
-    "     - Finally, it returns the `result` list, which contains the values from the linked list in the same order.\n",
-    "\n",
-    "These helper functions are used to simplify the process of creating linked lists from lists of values and converting linked lists back into lists, making it easier to work with linked lists in the provided examples and test cases."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[5, 4, 3, 2, 1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: Iterative\n",
-    "head1 = [1,2,3,4,5]\n",
-    "head1 = createLinkedList(head1)\n",
-    "reversed_head1 = reverseListIterative(head1)\n",
-    "print(linkedListToList(reversed_head1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[2, 1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2: Recursive\n",
-    "head2 = [1,2]\n",
-    "head2 = createLinkedList(head2)\n",
-    "reversed_head2 = reverseListRecursive(head2)\n",
-    "print(linkedListToList(reversed_head2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3: Empty list\n",
-    "head3 = []\n",
-    "head3 = createLinkedList(head3)\n",
-    "reversed_head3 = reverseListIterative(head3)\n",
-    "print(linkedListToList(reversed_head3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the two functions:\n",
-    "\n",
-    "1. `reverseListIterative(head)`:\n",
-    "\n",
-    "   - Time Complexity: $O(n)$\n",
-    "     - The iterative function visits each node in the linked list exactly once in a single pass, where \"$n$\" is the number of nodes in the list. Therefore, the time complexity is linear in the number of nodes.\n",
-    "\n",
-    "   - Space Complexity: $O(1)$\n",
-    "     - This function uses a constant amount of extra space for the `prev`, `current`, and `next_node` pointers. Regardless of the size of the input list, the amount of additional memory used remains the same, so the space complexity is constant.\n",
-    "\n",
-    "2. `reverseListRecursive(head)`:\n",
-    "\n",
-    "   - Time Complexity: $O(n)$\n",
-    "     - The recursive function also visits each node in the linked list exactly once. It recursively traverses the list from the head to the tail, reversing the links on the way back. Therefore, like the iterative approach, the time complexity is $O(n)$, where \"$n$\" is the number of nodes.\n",
-    "\n",
-    "   - Space Complexity: $O(n)$\n",
-    "     - The recursive function uses space on the call stack for each recursive call. In the worst case, when the list has \"$n$\" nodes, it will create \"$n$\" recursive function calls on the stack, resulting in a space complexity of $O(n)$. This is because each function call stores information about its state, including the `head` pointer, until it reaches the base case and starts returning.\n",
-    "\n",
-    "**In summary:**\n",
-    "\n",
-    "- The time complexity of both functions is $O(n)$ because they both visit each node once.\n",
-    "- The space complexity of `reverseListIterative` is $O(1)$ because it uses a constant amount of extra space.\n",
-    "- The space complexity of `reverseListRecursive` is $O(n)$ due to the recursive function calls and the associated call stack space.\n",
-    "\n",
-    "Both functions are efficient in terms of time complexity, but `reverseListIterative` is more memory-efficient because it uses a constant amount of additional space, while `reverseListRecursive` uses space on the call stack proportional to the number of nodes in the list."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **K-Group Reverse:** Modify the reverseListIterative function to reverse the linked list in groups of $K$ elements. For example, if the input linked list is [1, 2, 3, 4, 5, 6], and $K$ is 3, the output should be [3, 2, 1, 6, 5, 4].\n",
-    "\n",
-    "2. **Rotate Linked List:** Write a function to rotate a linked list to the right by $K$ places. For example, if the input is [1, 2, 3, 4, 5] and $K$ is 2, the output should be [4, 5, 1, 2, 3]."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb b/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb
deleted file mode 100644
index b433cf3..0000000
--- a/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb	
+++ /dev/null
@@ -1,277 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 21. Merge Two Sorted Lists\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Merge Two Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-two-sorted-lists/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given the heads of two sorted linked lists `list1` and `list2`.\n",
-    "\n",
-    "Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.\n",
-    "\n",
-    "*Return the head of the merged linked list.*\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in both lists is in the range `[0, 50]`.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "- Both `list1` and `list2` are sorted in non-decreasing order."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "def mergeTwoLists(list1, list2):\n",
-    "    # Create a dummy node to simplify the code.\n",
-    "    dummy = ListNode()\n",
-    "    current = dummy\n",
-    "\n",
-    "    while list1 and list2:\n",
-    "        if list1.val < list2.val:\n",
-    "            current.next = list1\n",
-    "            list1 = list1.next\n",
-    "        else:\n",
-    "            current.next = list2\n",
-    "            list2 = list2.next\n",
-    "        current = current.next\n",
-    "\n",
-    "    # Append the remaining elements from either list.\n",
-    "    if list1:\n",
-    "        current.next = list1\n",
-    "    else:\n",
-    "        current.next = list2\n",
-    "\n",
-    "    return dummy.next"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We define a class `ListNode` to represent the nodes of a singly-linked list. Each node has a `val` attribute to store the value of the node and a `next` attribute to point to the next node in the list.\n",
-    "\n",
-    "2. We define the `mergeTwoLists` function that takes two linked list heads, `list1` and `list2`, as input.\n",
-    "\n",
-    "3. We create a `dummy` node at the beginning of the merged list. The `dummy` node simplifies the code by serving as a placeholder, and its `next` attribute will point to the actual merged list.\n",
-    "\n",
-    "4. We initialize a `current` pointer to point to the `dummy` node initially. This `current` pointer helps us traverse and build the merged list.\n",
-    "\n",
-    "5. We enter a `while` loop that continues until either `list1` or `list2` becomes empty. Inside the loop, we compare the values of the nodes at the current positions of `list1` and `list2`.\n",
-    "\n",
-    "6. If the `val` of the node in `list1` is smaller than the `val` of the node in `list2`, we attach the node from `list1` to the merged list by updating the `next` attribute of the `current` node to point to the node in `list1`. We then move the `list1` pointer to the next node.\n",
-    "\n",
-    "7. If the `val` of the node in `list2` is smaller than or equal to the `val` of the node in `list1`, we attach the node from `list2` to the merged list in a similar manner. We move the `list2` pointer to the next node.\n",
-    "\n",
-    "8. After attaching a node to the merged list, we move the `current` pointer to the newly added node. This step is essential for keeping track of the end of the merged list.\n",
-    "\n",
-    "9. The loop continues until either `list1` or `list2` becomes empty.\n",
-    "\n",
-    "10. Once the loop exits, we check if there are any remaining elements in `list1` or `list2`. If `list1` is not empty, we attach the remaining elements of `list1` to the merged list. If `list2` is not empty, we attach the remaining elements of `list2`.\n",
-    "\n",
-    "11. Finally, we return the `next` attribute of the `dummy` node as the head of the merged list, which represents the sorted merged list."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to print the linked list\n",
-    "def printLinkedList(head):\n",
-    "    result = []\n",
-    "    while head:\n",
-    "        result.append(head.val)\n",
-    "        head = head.next\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the helper functions:\n",
-    "**printLinkedList Function**:\n",
-    "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-    "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-    "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-    "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-    "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-    "\n",
-    "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "[1, 1, 2, 3, 4, 4]"
-      ]
-     },
-     "execution_count": 3,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "# Input: list1 = [1,2,4], list2 = [1,3,4]\n",
-    "list1 = ListNode(1, ListNode(2, ListNode(4)))\n",
-    "list2 = ListNode(1, ListNode(3, ListNode(4)))\n",
-    "merged_list = mergeTwoLists(list1, list2)\n",
-    "printLinkedList(merged_list)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "[]"
-      ]
-     },
-     "execution_count": 4,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "# Input: list1 = [], list2 = []\n",
-    "list1 = None\n",
-    "list2 = None\n",
-    "merged_list = mergeTwoLists(list1, list2)\n",
-    "printLinkedList(merged_list)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "[0]"
-      ]
-     },
-     "execution_count": 5,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# list1 = [], list2 = [0]\n",
-    "list1 = None\n",
-    "list2 = ListNode(0)\n",
-    "merged_list = mergeTwoLists(list1, list2)\n",
-    "printLinkedList(merged_list)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `mergeTwoLists` function.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this function is $O(N + M$), where $N$ and $M$ are the lengths of `list1` and `list2`, respectively. Here's why:\n",
-    "\n",
-    "1. In the worst case, we need to traverse both `list1` and `list2` completely. This involves iterating through all the nodes in both lists once.\n",
-    "\n",
-    "2. The while loop runs until either `list1` or `list2` becomes empty. The number of iterations depends on the total number of nodes in both lists, which is N + M in the worst case.\n",
-    "\n",
-    "3. Inside the loop, we perform constant-time operations for each iteration, such as comparisons and updating pointers.\n",
-    "\n",
-    "Therefore, the dominant factor in the time complexity is the combined length of both input lists, resulting in a linear time complexity of $O(N + M)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this function is $O(1)$, which means it uses a constant amount of extra space regardless of the input sizes. Here's why:\n",
-    "\n",
-    "1. We create a few extra pointers (`dummy`, `current`) and temporary variables (`val`) to manage the merging process. These consume a fixed amount of memory, regardless of the input sizes. The number of these extra variables is independent of the lengths of `list1` and `list2`.\n",
-    "\n",
-    "2. We do not create a new data structure or allocate memory for the merged list. Instead, we rearrange the existing nodes from `list1` and `list2` to form the merged list. This operation does not consume additional memory proportional to the input sizes.\n",
-    "\n",
-    "**In summary**, the `mergeTwoLists` function has a time complexity of $O(N + M)$ and a space complexity of $O(1)$, making it an efficient solution for merging two sorted linked lists."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Merge K Sorted Lists:** Extend the problem to merge $K$ sorted linked lists instead of just two. You need to efficiently merge $K$ lists into one sorted list.\n",
-    "\n",
-    "2. **Merge Lists in a Zigzag Pattern:** Merge two sorted lists in a zigzag pattern. For example, given [1, 2, 4] and [1, 3, 5], the merged list should be [1, 1, 2, 3, 4, 5]."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb b/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb
deleted file mode 100644
index 0568867..0000000
--- a/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb	
+++ /dev/null
@@ -1,277 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 23. Merge k Sorted Lists\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Merge k Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-k-sorted-lists/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given an array of $k$ linked-lists lists, each linked-list is sorted in ascending order.\n",
-    "\n",
-    "Merge all the linked-lists into one sorted linked-list and return it.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `k == lists.length`\n",
-    "- 0 <= k <= $10^4$\n",
-    "- `0 <= lists[i].length <= 500`\n",
-    "- $-10^4$ <= lists[i][j] <= $10^4$\n",
-    "- `lists[i]` is sorted in ascending order.\n",
-    "- The sum of `lists[i].length` will not exceed $10^4$.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for singly-linked list.\n",
-    "class ListNode:\n",
-    "     def __init__(self, val=0, next=None):\n",
-    "         self.val = val\n",
-    "         self.next = next\n",
-    "\n",
-    "class Solution:\n",
-    "    def mergeKLists(self, lists: [[ListNode]]) -> ListNode:\n",
-    "        # Check if the input list of linked lists is empty\n",
-    "        if not lists or len(lists) == 0:\n",
-    "            return None\n",
-    "\n",
-    "        # Loop until there is only one merged list left\n",
-    "        while len(lists) > 1:\n",
-    "            mergedLists = []\n",
-    "\n",
-    "            # Merge pairs of lists\n",
-    "            for i in range(0, len(lists), 2):\n",
-    "                l1 = lists[i]\n",
-    "                l2 = lists[i + 1] if (i + 1) < len(lists) else None\n",
-    "                mergedLists.append(self.mergeList(l1, l2))\n",
-    "\n",
-    "            # Update the list of lists with merged results\n",
-    "            lists = mergedLists\n",
-    "        \n",
-    "        # Return the final merged list\n",
-    "        return lists[0]\n",
-    "\n",
-    "    def mergeList(self, l1, l2):\n",
-    "        # Create a dummy node to simplify merging\n",
-    "        dummy = ListNode()\n",
-    "        tail = dummy\n",
-    "\n",
-    "        # Merge the two sorted lists\n",
-    "        while l1 and l2:\n",
-    "            if l1.val < l2.val:\n",
-    "                tail.next = l1\n",
-    "                l1 = l1.next\n",
-    "            else:\n",
-    "                tail.next = l2\n",
-    "                l2 = l2.next\n",
-    "            tail = tail.next\n",
-    "\n",
-    "        # Append any remaining elements from l1 or l2 (if any)\n",
-    "        if l1:\n",
-    "            tail.next = l1\n",
-    "        if l2:\n",
-    "            tail.next = l2\n",
-    "\n",
-    "        # Return the merged result starting from the next of the dummy node\n",
-    "        return dummy.next\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code defines a Python class `Solution` with two methods for merging k sorted linked lists:\n",
-    "\n",
-    "1. `mergeKLists(self, lists: List[ListNode]) -> ListNode`: This method takes a list of k sorted linked lists as input and returns a single merged sorted linked list. It uses a divide-and-conquer approach to repeatedly merge pairs of lists until only one merged list remains.\n",
-    "\n",
-    "2. `mergeList(self, l1, l2)`: This method takes two sorted linked lists, `l1` and `l2`, as input and merges them into a single sorted linked list. It uses a dummy node to simplify the merging process.\n",
-    "\n",
-    "Here's a high-level overview of how the code works:\n",
-    "\n",
-    "- The `mergeKLists` method checks if the input list of linked lists is empty or contains no lists. If there are no lists, it returns `None`.\n",
-    "\n",
-    "- Inside a `while` loop, the code repeatedly merges pairs of linked lists until only one merged list remains in the `lists` array. It does this by iterating through the input lists in pairs and calling the `mergeList` method to merge each pair.\n",
-    "\n",
-    "- The `mergeList` method takes two sorted linked lists, `l1` and `l2`, and merges them into a single sorted linked list. It uses a dummy node (`dummy`) and a `tail` pointer to keep track of the merged list while comparing and merging elements from `l1` and `l2`.\n",
-    "\n",
-    "- After merging all pairs of lists and updating the `lists` array with the merged results, the loop continues until only one merged list remains in the `lists` array.\n",
-    "\n",
-    "- Finally, the `mergeKLists` method returns the merged list.\n",
-    "\n",
-    "Overall, this code efficiently merges k sorted linked lists using a divide-and-conquer strategy, resulting in a single merged sorted linked list as the output."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> "
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "# Input: lists = [[1,4,5],[1,3,4],[2,6]]\n",
-    "\n",
-    "lists1 = [\n",
-    "    ListNode(1, ListNode(4, ListNode(5))),\n",
-    "    ListNode(1, ListNode(3, ListNode(4))),\n",
-    "    ListNode(2, ListNode(6))\n",
-    "]\n",
-    "solution = Solution()\n",
-    "result1 = solution.mergeKLists(lists1)\n",
-    "\n",
-    "# Print the result\n",
-    "if result1:\n",
-    "    current = result1\n",
-    "    while current:\n",
-    "        print(current.val, end=\" -> \")\n",
-    "        current = current.next\n",
-    "else:\n",
-    "    print(\"None\")  # Print \"None\" for input with a single None element"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "None\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "# Input: lists = []\n",
-    "\n",
-    "lists2 = []\n",
-    "solution = Solution()\n",
-    "result2 = solution.mergeKLists(lists2)\n",
-    "\n",
-    "# Print the result\n",
-    "if result2:\n",
-    "    current = result2\n",
-    "    while current:\n",
-    "        print(current.val, end=\" -> \")\n",
-    "        current = current.next\n",
-    "else:\n",
-    "    print(\"None\")  # Print \"None\" for input with a single None element"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "None\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "# Input: lists = [[]]\n",
-    "\n",
-    "lists3 = [None]\n",
-    "solution = Solution()\n",
-    "result3 = solution.mergeKLists(lists3)\n",
-    "\n",
-    "# Print the result\n",
-    "if result3:\n",
-    "    current = result3\n",
-    "    while current:\n",
-    "        print(current.val, end=\" -> \")\n",
-    "        current = current.next\n",
-    "else:\n",
-    "    print(\"None\")  # Print \"None\" for input with a single None element"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. **Heap Initialization:** The code does not use a heap. Instead, it uses a divide-and-conquer approach. The initial check for empty input lists takes $O(1)$ time.\n",
-    "\n",
-    "2. **Merging:** The merging operation is performed in a divide-and-conquer fashion. In each iteration of the `while` loop, we merge pairs of linked lists, and the number of comparisons made is proportional to the total number of nodes across all the linked lists ($n$). In each merge step, we effectively process each node once. The number of iterations required to reduce k lists to 1 is $O(log\\ k)$.\n",
-    "\n",
-    "   Therefore, the overall time complexity of the code is $O(n * log\\ k)$, where n is the total number of nodes across all lists, and $k$ is the number of linked lists.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. **Heap Space:** The code doesn't use a heap data structure, so there's no additional space complexity due to a heap.\n",
-    "\n",
-    "2. **Merged Lists:** In the `mergeList` method, we create a new merged list. However, this list is not stored in memory for all lists; it's replaced with each merged pair. The space used for these merged lists is proportional to the size of the largest merged list, which is $O(n)$ in the worst case.\n",
-    "\n",
-    "3. **Additional Variables:** The code uses a few additional variables, such as `dummy` and `tail`, but these occupy a constant amount of space and don't depend on the input size.\n",
-    "\n",
-    "   Therefore, the overall space complexity of the code is $O(n)$, where n is the total number of nodes across all lists.\n",
-    "\n",
-    "In summary, the code's time complexity is $O(n * log(k))$, and its space complexity is $O(n)$. This code efficiently merges $k$ sorted linked lists using a divide-and-conquer approach with a relatively low space overhead."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb b/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb
deleted file mode 100644
index 2238b48..0000000
--- a/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 100: Same Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-    "\n",
-    "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-    "\n",
-    "**Constraints**\n",
-    "\n",
-    "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-    "2. $-10^4$ <= Node.val <= $10^4$\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def isSameTree(p, q):\n",
-    "    # Base case: If both p and q are None, the trees are the same.\n",
-    "    if not p and not q:\n",
-    "        return True\n",
-    "    \n",
-    "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-    "    if not p or not q:\n",
-    "        return False\n",
-    "    \n",
-    "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-    "    if p.val != q.val:\n",
-    "        return False\n",
-    "    \n",
-    "    # Recursively check the left and right subtrees of p and q.\n",
-    "    # If both subtrees are the same, the entire trees are the same.\n",
-    "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-    "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-    "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-    "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-    "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-    "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "\n",
-    "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-    "\n",
-    "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "print(isSameTree(p1, q1)) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "\n",
-    "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-    "\n",
-    "p2 = TreeNode(1, TreeNode(2), None)\n",
-    "q2 = TreeNode(1, None, TreeNode(2))\n",
-    "print(isSameTree(p2, q2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "\n",
-    "#Input: p = [1,2,1], q = [1,1,2]\n",
-    "\n",
-    "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-    "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-    "print(isSameTree(p3, q3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time Complexity\n",
-    "\n",
-    "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-    "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-    "\n",
-    "## Space Complexity\n",
-    "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-    "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-    "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-    "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-    "\n",
-    "## In summary:\n",
-    "\n",
-    "+ **Time Complexity:** $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-    "+ **Space Complexity:** $O(n)$ due to the recursive call stack."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb b/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb
deleted file mode 100644
index 9d2cc7f..0000000
--- a/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb	
+++ /dev/null
@@ -1,263 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 102. Binary Tree Level Order Traversal\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Binary Tree Level Order Traversal problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-level-order-traversal/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary tree, return *the level order traversal of its nodes' values*. (i.e., from left to right, level by level).\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-    "- `-1000 <= Node.val <= 1000`\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def levelOrder(root):\n",
-    "    # Check if the tree is empty, if so, return an empty list.\n",
-    "    if not root:\n",
-    "        return []\n",
-    "\n",
-    "    # Initialize an empty list to store the result.\n",
-    "    result = []\n",
-    "    \n",
-    "    # Initialize a queue with the root node to perform BFS.\n",
-    "    queue = [root]\n",
-    "\n",
-    "    while queue:\n",
-    "        # Initialize an empty list to store the values of nodes at the current level.\n",
-    "        level_values = []\n",
-    "        \n",
-    "        # Get the number of nodes at the current level.\n",
-    "        level_size = len(queue)\n",
-    "\n",
-    "        # Iterate through the nodes at the current level.\n",
-    "        for _ in range(level_size):\n",
-    "            # Dequeue the front node from the queue.\n",
-    "            node = queue.pop(0)\n",
-    "            \n",
-    "            # Append the value of the current node to the level_values list.\n",
-    "            level_values.append(node.val)\n",
-    "\n",
-    "            # Enqueue the left and right children of the current node if they exist.\n",
-    "            if node.left:\n",
-    "                queue.append(node.left)\n",
-    "            if node.right:\n",
-    "                queue.append(node.right)\n",
-    "\n",
-    "        # Append the level_values list (values at the current level) to the result list.\n",
-    "        result.append(level_values)\n",
-    "\n",
-    "    # Return the final result, which is a list of lists representing level order traversal.\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We define a `TreeNode` class that represents a node in the binary tree. Each `TreeNode` object has a value (`val`) and two child nodes: `left` and `right`.\n",
-    "\n",
-    "2. The `levelOrder` function takes the root of the binary tree as its input and returns the level order traversal of the tree as a list of lists.\n",
-    "\n",
-    "3. We start by checking if the input `root` is `None`, which indicates an empty tree. If the tree is empty, we return an empty list because there are no nodes to traverse.\n",
-    "\n",
-    "4. We initialize an empty list called `result` to store the final result, which will be a list of lists containing node values at each level.\n",
-    "\n",
-    "5. We initialize a queue called `queue` with the root node. This queue will be used for breadth-first traversal of the tree.\n",
-    "\n",
-    "6. We enter a while loop that continues until the `queue` is empty. Inside the loop, we perform the following steps:\n",
-    "\n",
-    "   - We initialize an empty list called `level_values` to store the values of nodes at the current level.\n",
-    "\n",
-    "   - We determine the number of nodes at the current level by getting the length of the `queue`. This is done to process nodes level by level.\n",
-    "\n",
-    "   - We iterate through the nodes at the current level using a for loop. For each node in the current level:\n",
-    "\n",
-    "     - We dequeue (remove) the front node from the `queue`.\n",
-    "\n",
-    "     - We append the value of the dequeued node to the `level_values` list, effectively collecting the values of nodes at the current level.\n",
-    "\n",
-    "     - If the dequeued node has a left child, we enqueue the left child to the `queue`.\n",
-    "\n",
-    "     - If the dequeued node has a right child, we enqueue the right child to the `queue`.\n",
-    "\n",
-    "   - After processing all nodes at the current level, we append the `level_values` list to the `result` list. This represents the values at the current level.\n",
-    "\n",
-    "7. The loop continues until all levels have been traversed, and the `queue` becomes empty.\n",
-    "\n",
-    "8. Finally, we return the `result` list, which contains lists of node values at each level, representing the level order traversal of the binary tree.\n",
-    "\n",
-    "The code effectively performs a breadth-first traversal of the binary tree, processing nodes level by level, and constructs the result list that represents the level order traversal."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[3], [9, 20], [15, 7]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Construct the tree\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(9)\n",
-    "root.right = TreeNode(20)\n",
-    "root.right.left = TreeNode(15)\n",
-    "root.right.right = TreeNode(7)\n",
-    "\n",
-    "result = levelOrder(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[1]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Using the same tree as before\n",
-    "root = TreeNode(1)\n",
-    "result = levelOrder(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# Creating a new tree for this example\n",
-    "root = None\n",
-    "result = levelOrder(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `levelOrder` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this function is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the breadth-first traversal.\n",
-    "\n",
-    "In the worst case, we have to enqueue and dequeue all nodes in the binary tree, which is proportional to the number of nodes (N). In each level, we process all nodes in that level, and since there are a total of log(N) levels in a balanced binary tree, the time complexity can also be approximated as O(N) for unbalanced trees.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this function is O(N), where N is the number of nodes in the binary tree. Here's how the space complexity breaks down:\n",
-    "\n",
-    "1. The `result` list stores the level order traversal, and in the worst case, it contains N/2 levels (for a completely unbalanced binary tree). So, the space used by `result` is O(N).\n",
-    "\n",
-    "2. The `queue` data structure is used for BFS traversal. In the worst case, it can store all nodes at the last level of the tree. In a balanced binary tree, the maximum number of nodes at any level is 2^(log(N)), which is still O(N). In the case of an unbalanced tree, it can be even worse. So, the space used by `queue` is O(N).\n",
-    "\n",
-    "Overall, the dominant factor in terms of space complexity is the `queue`, and the space complexity is O(N).\n",
-    "\n",
-    "In summary, the function's time complexity is O(N), and its space complexity is also O(N). It is an efficient and optimal solution for performing a level order traversal of a binary tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Reverse Level Order Traversal:** Modify the `levelOrder` function to return the level order traversal in reverse order (from bottom to top). For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[15, 7], [9, 20], [3]]`.\n",
-    "\n",
-    "2. **Zigzag Level Order Traversal:** Write a function that performs a level order traversal of a binary tree in a zigzag pattern. In a zigzag traversal, the nodes at even levels are traversed from left to right, and nodes at odd levels are traversed from right to left. For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[3], [20, 9], [15, 7]]`."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb b/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb
deleted file mode 100644
index 87c5eaa..0000000
--- a/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 104. Maximum Depth of Binary Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the `root` of a binary tree, return its maximum depth.\n",
-    "\n",
-    "A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def maxDepth(root):\n",
-    "    # Base case: If the root is None, the depth is 0.\n",
-    "    if root is None:\n",
-    "        return 0\n",
-    "    \n",
-    "    # Recursively calculate the maximum depth of the left and right subtrees.\n",
-    "    left_depth = maxDepth(root.left)\n",
-    "    right_depth = maxDepth(root.right)\n",
-    "    \n",
-    "    # Return the maximum depth of the tree by adding 1 to the maximum depth of the subtrees.\n",
-    "    return max(left_depth, right_depth) + 1"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Let's go through the code step by step to understand how it works:\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class to represent the nodes of the binary tree. Each node has three attributes:\n",
-    "   - `val`: the value stored in the node.\n",
-    "   - `left`: a reference to the left child node.\n",
-    "   - `right`: a reference to the right child node.\n",
-    "\n",
-    "2. The `maxDepth` function is the main function that calculates the maximum depth of the binary tree. It takes a single argument, `root`, which is the root node of the binary tree.\n",
-    "\n",
-    "3. In the `maxDepth` function, we have a base case:\n",
-    "   - If the `root` is `None`, it means we have reached the end of a branch of the tree (a leaf node or an empty subtree). In this case, the depth is 0 because there are no nodes to count.\n",
-    "\n",
-    "4. If the `root` is not `None`, we recursively calculate the maximum depth of the left and right subtrees:\n",
-    "   - We call `maxDepth` on the `root.left` to calculate the maximum depth of the left subtree and store it in the variable `left_depth`.\n",
-    "   - We call `maxDepth` on the `root.right` to calculate the maximum depth of the right subtree and store it in the variable `right_depth`.\n",
-    "\n",
-    "5. Finally, we return the maximum depth of the tree by taking the maximum of `left_depth` and `right_depth` and adding 1 to it. This is because we are counting the current level as well."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "# Input: root = [3,9,20,null,null,15,7]\n",
-    "\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(9)\n",
-    "root.right = TreeNode(20)\n",
-    "root.right.left = TreeNode(15)\n",
-    "root.right.right = TreeNode(7)\n",
-    "\n",
-    "print(maxDepth(root))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a230be4f-1f47-4919-87e3-62fc3f072d97",
-   "metadata": {},
-   "source": [
-    "In this example, we create a binary tree based on the given input `[3,9,20,null,null,15,7]` and calculate its maximum depth, which is `3`. The tree structure is as follows:\n",
-    "   ```\n",
-    "       3\n",
-    "      / \\\n",
-    "     9  20\n",
-    "       /  \\\n",
-    "      15   7\n",
-    "   ```\n",
-    "\n",
-    "So, the maximum depth is the length of the longest path from the root to a leaf node, which is 3 in this case."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "# Input: root = [1,null,2]\n",
-    "\n",
-    "root = TreeNode(1)\n",
-    "root.right = TreeNode(2)\n",
-    "\n",
-    "# Calculate the maximum depth of the tree and print the result\n",
-    "print(maxDepth(root))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The time complexity of the `maxDepth` function is $O(N)$, where $N$ is the number of nodes in the binary tree.\n",
-    "- This is because in the worst case, the function visits every node exactly once in a depth-first manner.\n",
-    "- The recursion explores all nodes of the tree, so the time complexity is linear with respect to the number of nodes.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity of the `maxDepth` function is $O(H)$, where H is the height of the binary tree.\n",
-    "- In the worst case, if the binary tree is completely unbalanced (skewed), the recursion stack can go as deep as the height of the tree.\n",
-    "- In the best case, if the binary tree is perfectly balanced, the height is $O(log\\ N)$, where $N$ is the number of nodes.\n",
-    "- Therefore, the space complexity can vary from $O(log\\ N)$ to $O(N)$ depending on the shape of the tree.\n",
-    "- In addition to the recursion stack, there is a small constant amount of space used for variables and function call overhead.\n",
-    "\n",
-    "**In summary:**\n",
-    "- The time complexity of the code is $O(N)$ as it visits each node once.\n",
-    "- The space complexity is $O(H)$, where H is the height of the tree, which can vary from $O(log\\ N)$ to $O(N)$ depending on the tree's shape."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb b/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb
deleted file mode 100644
index 0d5bbd6..0000000
--- a/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb	
+++ /dev/null
@@ -1,305 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 105. Construct Binary Tree from Preorder and Inorder Traversal\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Construct Binary Tree from Preorder and Inorder Traversal problem on LeetCode, click here!](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)\n",
-    "\n",
-    "---\n",
-    "Given two integer arrays `preorder` and `inorder` where `preorder` is the preorder traversal of a binary tree and `inorder` is the inorder traversal of the same tree, construct and return *the binary tree.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= preorder.length <= 3000`\n",
-    "- `inorder.length == preorder.length`\n",
-    "- `-3000 <= preorder[i], inorder[i] <= 3000`\n",
-    "- `preorder` and `inorder` consist of **unique** values.\n",
-    "- Each value of `inorder` also appears in `preorder`.\n",
-    "- `preorder` is **guaranteed** to be the preorder traversal of the tree.\n",
-    "- `inorder` is **guaranteed** to be the inorder traversal of the tree.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def buildTree(preorder, inorder):\n",
-    "    if not preorder:\n",
-    "        return None\n",
-    "\n",
-    "    # The first element in the preorder traversal is the root of the current subtree\n",
-    "    root_val = preorder[0]\n",
-    "    root = TreeNode(root_val)\n",
-    "\n",
-    "    # Find the index of the root value in the inorder traversal\n",
-    "    root_idx_inorder = inorder.index(root_val)\n",
-    "\n",
-    "    # Recursively build left and right subtrees\n",
-    "    root.left = buildTree(preorder[1:1 + root_idx_inorder], inorder[:root_idx_inorder])\n",
-    "    root.right = buildTree(preorder[1 + root_idx_inorder:], inorder[root_idx_inorder + 1:])\n",
-    "\n",
-    "    return root"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `class TreeNode:`: This is a class definition for a binary tree node. It has three attributes:\n",
-    "   - `val`: The value of the node.\n",
-    "   - `left`: A reference to the left child node.\n",
-    "   - `right`: A reference to the right child node.\n",
-    "\n",
-    "   This class is used to create instances of binary tree nodes, which will be used to build the binary tree.\n",
-    "\n",
-    "2. `def buildTree(preorder, inorder):`: This is the main function that constructs a binary tree from its preorder and inorder traversals.\n",
-    "\n",
-    "   - `preorder`: A list representing the preorder traversal of the binary tree.\n",
-    "   - `inorder`: A list representing the inorder traversal of the binary tree.\n",
-    "\n",
-    "3. `if not preorder: return None`: This line checks if the `preorder` list is empty. If it is, it means there are no nodes to construct, so the function returns `None`.\n",
-    "\n",
-    "4. `root_val = preorder[0]`: The value of the root node is extracted from the first element of the `preorder` list.\n",
-    "\n",
-    "5. `root = TreeNode(root_val)`: A new `TreeNode` object is created with the `root_val` as its value. This represents the root of the current subtree.\n",
-    "\n",
-    "6. `root_idx_inorder = inorder.index(root_val)`: The index of the `root_val` in the `inorder` list is found. This index indicates the position of the root node in the inorder traversal.\n",
-    "\n",
-    "7. `root.left`: The left subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The left subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists up to the `root_idx_inorder`.\n",
-    "\n",
-    "8. `root.right`: The right subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The right subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists starting from `root_idx_inorder + 1`.\n",
-    "\n",
-    "9. Finally, the function returns the `root` of the subtree it just constructed.\n",
-    "\n",
-    "The recursive approach used here divides the problem of constructing the entire binary tree into smaller subproblems, starting with the root node and then recursively building the left and right subtrees until the entire tree is constructed."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "2d82857b-4fbe-474f-bf33-17b55c20147b",
-   "metadata": {},
-   "source": [
-    "## Helper Function \n",
-    "\n",
-    "Here is a helper function to represent the binary tree elements."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "13e6f3e1-659d-479d-9521-75629b8c782f",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def treeToList(root):\n",
-    "    # Initialize an empty list to store the elements of the binary tree\n",
-    "    result = []\n",
-    "    \n",
-    "    # Initialize a queue for level-order traversal starting with the root\n",
-    "    queue = [root]\n",
-    "\n",
-    "    while queue:\n",
-    "        # Pop the front node from the queue\n",
-    "        node = queue.pop(0)\n",
-    "        \n",
-    "        # If the node is not None (i.e., a valid node), add its value to the result list\n",
-    "        if node:\n",
-    "            result.append(node.val)\n",
-    "            \n",
-    "            # Add the left and right children of the node to the queue\n",
-    "            queue.append(node.left)\n",
-    "            queue.append(node.right)\n",
-    "        else:\n",
-    "            # If the node is None, add None to the result to represent an empty node\n",
-    "            result.append(None)\n",
-    "\n",
-    "    # Remove any trailing None values from the result list\n",
-    "    while result and result[-1] is None:\n",
-    "        result.pop()\n",
-    "\n",
-    "    # Return the resulting list representing the binary tree elements\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "86cd4c9d-a120-4dd1-a1bd-530436012574",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Let's break down the `treeToList` helper function step by step:\n",
-    "\n",
-    "1. `def treeToList(root):`: This function takes the root of a binary tree as input and converts it into a list.\n",
-    "\n",
-    "2. `result = []`: Initialize an empty list called `result` to store the elements of the binary tree in list format.\n",
-    "\n",
-    "3. `queue = [root]`: Initialize a queue data structure for a level-order traversal, starting with the root node.\n",
-    "\n",
-    "4. `while queue:`: This starts a loop that continues until the queue is empty, indicating that all nodes have been processed.\n",
-    "\n",
-    "5. `node = queue.pop(0)`: Dequeue the first node from the queue, effectively processing it.\n",
-    "\n",
-    "6. `if node:`: Check if the `node` is not None, which means it's a valid node in the binary tree.\n",
-    "\n",
-    "    a. `result.append(node.val)`: If the node is valid, append its value (`node.val`) to the `result` list. This represents the value of the current node in the binary tree.\n",
-    "\n",
-    "    b. `queue.append(node.left)`: Enqueue the left child of the current node if it exists. This adds the left child to the queue for processing in the next iteration.\n",
-    "\n",
-    "    c. `queue.append(node.right)`: Enqueue the right child of the current node if it exists. This adds the right child to the queue for processing in the next iteration.\n",
-    "\n",
-    "7. `else:`: If the `node` is None, it represents an empty node in the binary tree.\n",
-    "\n",
-    "    a. `result.append(None)`: Append `None` to the `result` list to indicate an empty node.\n",
-    "\n",
-    "8. `while result and result[-1] is None:`: After the traversal is complete, there might be trailing `None` values in the `result` list. This loop removes any such trailing `None` values to ensure a clean representation of the tree's elements.\n",
-    "\n",
-    "9. `return result`: Return the `result` list, which now contains the elements of the binary tree in a format where `None` represents empty nodes and the order of elements reflects a level-order traversal of the tree.\n",
-    "\n",
-    "In summary, the `treeToList` function performs a level-order traversal of the binary tree using a queue, constructing a list where each element corresponds to a node's value or represents an empty node with `None`. This list represents the binary tree's elements in a structured format."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[3, 9, 20, None, None, 15, 7]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "preorder = [3, 9, 20, 15, 7]\n",
-    "inorder = [9, 3, 15, 20, 7]\n",
-    "result = buildTree(preorder, inorder)\n",
-    "\n",
-    "print(treeToList(result))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[-1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "preorder = [-1]\n",
-    "inorder = [-1]\n",
-    "result = buildTree(preorder, inorder)\n",
-    "\n",
-    "print(treeToList(result))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `buildTree` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of the `buildTree` function can be analyzed in terms of the number of nodes in the binary tree. Let's assume there are `n` nodes in the binary tree.\n",
-    "\n",
-    "1. Finding the root value in the `preorder` traversal takes O(1) time as it's just extracting the first element from the list.\n",
-    "2. Finding the index of the root value in the `inorder` traversal using `inorder.index(root_val)` takes O(n) time in the worst case because in the worst case, it might have to search through all `n` elements in the `inorder` list to find the index.\n",
-    "3. The recursive calls to `buildTree` for the left and right subtrees are made once for each node in the tree. Therefore, the recursive calls have a combined time complexity of O(n) as they process each node once.\n",
-    "\n",
-    "The total time complexity of the `buildTree` function is O(n) due to the recursive calls and finding the root's index in the `inorder` list.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity is determined by the space used by the function's call stack during recursion and any additional data structures used. \n",
-    "\n",
-    "1. The space used by the function call stack during recursion depends on the height of the binary tree. In the worst case, where the tree is highly unbalanced (e.g., a skewed tree), the space complexity for the call stack is O(n) as it can go as deep as the number of nodes in the tree.\n",
-    "\n",
-    "2. Additionally, the function creates TreeNode objects for each node in the binary tree. Therefore, the space complexity for these objects is also O(n).\n",
-    "\n",
-    "Overall, the space complexity of the `buildTree` function is O(n) due to the space used by the call stack and the TreeNode objects.\n",
-    "\n",
-    "In summary, the time complexity of `buildTree` is O(n), and the space complexity is O(n), where 'n' is the number of nodes in the binary tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Construct Binary Tree from Postorder and Inorder Traversal:**\n",
-    "   Modify the problem to construct a binary tree from its postorder and inorder traversals. Implement a function similar to `buildTree` but for postorder and inorder traversals.\n",
-    "\n",
-    "2. **Reconstruct Binary Tree with Duplicate Values:**\n",
-    "   Extend the problem to handle binary trees with duplicate values. Ensure that your solution correctly handles scenarios where there are duplicate values in the preorder and inorder traversals."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb b/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb
deleted file mode 100644
index b661fa1..0000000
--- a/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb	
+++ /dev/null
@@ -1,243 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 124. Binary Tree Maximum Path Sum\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Binary Tree Maximum Path Sum problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-maximum-path-sum/)\n",
-    "\n",
-    "---\n",
-    "A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.\n",
-    "\n",
-    "The **path sum** of a path is the sum of the node's values in the path.\n",
-    "\n",
-    "Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[1, 3*10^4]$.\n",
-    "- `-1000 <= Node.val <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to find the maximum path sum in a binary tree. A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. The path sum of a path is the sum of the node values in that path.\n",
-    "\n",
-    "In this problem, you are given the root of a binary tree, and you need to find the maximum path sum among all possible paths in the tree. Note that the path does not need to start at the root node or end at a leaf node; it can start and end at any nodes in the tree.\n",
-    "\n",
-    "To find the maximum path sum, you need to consider both left and right subtrees of each node while traversing the tree. At each node, you have two choices:\n",
-    "\n",
-    "1. Include the current node in the path: In this case, you add the value of the current node to the sum and continue exploring both the left and right subtrees for possible extensions of the path.\n",
-    "\n",
-    "2. Start a new path from the current node: In this case, you do not include the current node in the path sum, and you choose either the left subtree or the right subtree to start a new path.\n",
-    "\n",
-    "To solve this problem, you can use a recursive approach to traverse the binary tree. For each node, you calculate the maximum path sum that can pass through that node (option 1) and also return the maximum path sum that can be extended from that node (option 2). You keep track of the global maximum path sum encountered during the traversal.\n",
-    "\n",
-    "Ultimately, the maximum path sum among all paths in the tree will be the maximum of the global maximum path sum and the maximum path sum calculated for each node.\n",
-    "\n",
-    "The constraints for this problem include:\n",
-    "\n",
-    "- The number of nodes in the tree is in the range $[1, 3 * 10^4]$.\n",
-    "- Node values are in the range [-1000, 1000].\n",
-    "\n",
-    "By considering all possible paths through the tree, the algorithm aims to find the most significant sum of node values in any path in the binary tree.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "class Solution:\n",
-    "    def maxPathSum(self, root):\n",
-    "        def max_path_sum_helper(node):\n",
-    "            if not node:\n",
-    "                return 0\n",
-    "            \n",
-    "            # Recursively calculate the maximum path sum for left and right subtrees\n",
-    "            left_max = max(0, max_path_sum_helper(node.left))\n",
-    "            right_max = max(0, max_path_sum_helper(node.right))\n",
-    "            \n",
-    "            # Update the global maximum path sum\n",
-    "            self.max_sum = max(self.max_sum, left_max + right_max + node.val)\n",
-    "            \n",
-    "            # Return the maximum path sum starting from the current node\n",
-    "            return max(left_max, right_max) + node.val\n",
-    "        \n",
-    "        self.max_sum = float('-inf')  # Initialize the global maximum to negative infinity\n",
-    "        max_path_sum_helper(root)  # Start the recursive traversal from the root\n",
-    "        return self.max_sum"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The code starts by defining a `TreeNode` class to represent nodes in the binary tree. Each `TreeNode` has a `val` attribute representing its value, a `left` attribute pointing to the left child, and a `right` attribute pointing to the right child.\n",
-    "\n",
-    "2. Next, a `Solution` class is defined to contain the `maxPathSum` method, which will be used to find the maximum path sum in the binary tree.\n",
-    "\n",
-    "3. Within the `maxPathSum` method, a helper function called `max_path_sum_helper` is defined. This recursive function takes a single argument, `node`, which is the current node being considered during traversal.\n",
-    "\n",
-    "4. Inside the `max_path_sum_helper` function:\n",
-    "   - If `node` is `None` (i.e., the current node is a null node), it returns 0. This is the base case of the recursion.\n",
-    "   - It calculates the maximum path sum starting from the left subtree (`left_max`) and the right subtree (`right_max`). Importantly, it uses `max(0, ...)` to ensure that negative values (which would make the path sum smaller) are not considered. If a subtree's path sum is negative, it's better to not include it in the path.\n",
-    "   - The code then updates the global maximum path sum (`self.max_sum`) by checking if the sum of `left_max`, `right_max`, and the current node's value is greater than the current maximum.\n",
-    "   - Finally, it returns the maximum path sum starting from the current node, which is the maximum of `left_max` and `right_max` plus the current node's value.\n",
-    "\n",
-    "5. Before calling the `max_path_sum_helper` function, the `max_sum` attribute of the `Solution` class is initialized to negative infinity (`float('-inf')`). This is done to ensure that any valid path sum encountered during traversal will be greater than the initial value of `max_sum`.\n",
-    "\n",
-    "6. The `max_path_sum_helper` function is called with the root of the binary tree, effectively starting the traversal from the root node.\n",
-    "\n",
-    "7. Once the traversal is complete, the method returns the value of `self.max_sum`, which contains the maximum path sum found in the binary tree.\n",
-    "\n",
-    "8. Example usage at the bottom of the code demonstrates how to create binary trees and use the `maxPathSum` method to find the maximum path sum for two different tree structures.\n",
-    "\n",
-    "In summary, the code uses a recursive approach to explore all possible paths in the binary tree, ensuring that negative path sums are not considered. It keeps track of the maximum path sum encountered and returns that maximum value as the result."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "6\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "root1 = TreeNode(1)\n",
-    "root1.left = TreeNode(2)\n",
-    "root1.right = TreeNode(3)\n",
-    "solution = Solution()\n",
-    "print(solution.maxPathSum(root1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "42\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "root2 = TreeNode(-10)\n",
-    "root2.left = TreeNode(9)\n",
-    "root2.right = TreeNode(20)\n",
-    "root2.right.left = TreeNode(15)\n",
-    "root2.right.right = TreeNode(7)\n",
-    "print(solution.maxPathSum(root2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the given code:\n",
-    "\n",
-    "Time Complexity:\n",
-    "- The code uses a recursive depth-first traversal to explore each node in the binary tree exactly once.\n",
-    "- During the traversal, for each node, we perform constant time operations, such as updating the `max_sum` variable and calculating the maximum path sum starting from that node.\n",
-    "- Therefore, the time complexity of the code is O(N), where N is the number of nodes in the binary tree.\n",
-    "\n",
-    "Space Complexity:\n",
-    "- The space complexity of the code is determined by the space used in the call stack during the recursive traversal.\n",
-    "- In the worst case, the depth of the call stack can be equal to the height of the binary tree. In an unbalanced tree, this could be O(N), but in a balanced binary tree, the height is O(log N).\n",
-    "- Additionally, the code uses a constant amount of space for variables like `left_max`, `right_max`, and `max_sum`.\n",
-    "- Therefore, the overall space complexity is O(H), where H is the height of the binary tree. In the worst case, it's O(N), and in the best case (a balanced binary tree), it's O(log N).\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: O(N)\n",
-    "- Space Complexity: O(H), where H is the height of the binary tree, ranging from O(log N) to O(N) depending on the tree's balance."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Print the Maximum Path:**\n",
-    "   Modify the code to not only find the maximum path sum but also print the nodes involved in the maximum path. You can print the nodes in the correct order from the root to the leaf.\n",
-    "\n",
-    "2. **Path Sum Count:**\n",
-    "   Instead of finding the maximum path sum, find the count of unique paths that sum to a given target value. Each path can start and end anywhere in the tree."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb b/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb
deleted file mode 100644
index 5bbdf42..0000000
--- a/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 199. Binary Tree Right Side View\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Binary Tree Right Side View problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-right-side-view/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary tree, imagine yourself standing on the **right side** of it, return *the values of the nodes you can see ordered from top to bottom*.\n",
-    "\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def rightSideView(root):\n",
-    "    if not root:\n",
-    "        return []  # If the tree is empty, return an empty list.\n",
-    "\n",
-    "    result = []  # Initialize a list to store the right side view values.\n",
-    "    queue = [root]  # Initialize a queue for level-order traversal with the root node.\n",
-    "\n",
-    "    while queue:\n",
-    "        # Get the number of nodes at the current level.\n",
-    "        level_size = len(queue)\n",
-    "\n",
-    "        # Traverse all nodes at the current level and add the rightmost node to the result.\n",
-    "        for i in range(level_size):\n",
-    "            node = queue.pop(0)  # Dequeue the first node from the queue.\n",
-    "\n",
-    "            # If it's the rightmost node at this level, add its value to the result.\n",
-    "            if i == level_size - 1:\n",
-    "                result.append(node.val)\n",
-    "\n",
-    "            # Add the children of the current node to the queue.\n",
-    "            if node.left:\n",
-    "                queue.append(node.left)\n",
-    "            if node.right:\n",
-    "                queue.append(node.right)\n",
-    "\n",
-    "    return result  # Return the list of right side view values.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start with a binary tree represented using a TreeNode class. Each node has a value (`val`) and can have a left child (`left`) and a right child (`right`).\n",
-    "\n",
-    "2. The `rightSideView` function takes the root node of the binary tree as input and returns a list of values representing the nodes you can see when standing on the right side of the tree.\n",
-    "\n",
-    "3. We check if the `root` is `None`, which means the tree is empty. If it's empty, we return an empty list because there are no nodes to see.\n",
-    "\n",
-    "4. We initialize an empty list called `result` to store the right side view values.\n",
-    "\n",
-    "5. We initialize a queue called `queue` with the root node. This queue is used for a level-order traversal of the tree.\n",
-    "\n",
-    "6. We enter a while loop that continues until the `queue` is empty. Inside this loop, we perform the following steps for each level of the tree:\n",
-    "\n",
-    "   a. We determine the number of nodes at the current level by getting the length of the `queue`. This is important for processing nodes at the same level together.\n",
-    "\n",
-    "   b. We then iterate through the nodes at the current level using a for loop. For each node in the level, we dequeue it from the `queue` using `queue.pop(0)`.\n",
-    "\n",
-    "   c. If the node we dequeue is the rightmost node at the current level (determined by `i == level_size - 1`), we add its `val` to the `result` list. This is because, when standing on the right side of the tree, you can see the rightmost node at each level.\n",
-    "\n",
-    "   d. We enqueue the left and right children of the current node (if they exist) to the `queue`. This ensures that we process the next level in the subsequent iterations of the while loop.\n",
-    "\n",
-    "7. After processing all levels of the tree, the `result` list contains the values of the rightmost nodes in each level, ordered from top to bottom.\n",
-    "\n",
-    "8. Finally, we return the `result` list as the output of the function, which represents the right side view of the binary tree.\n",
-    "\n",
-    "The code effectively performs a level-order traversal of the binary tree while keeping track of the rightmost nodes at each level and adding them to the result list. This ensures that we obtain the correct order of nodes visible from the right side of the tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 3, 4]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Construct the tree\n",
-    "root1 = TreeNode(1)\n",
-    "root1.left = TreeNode(2)\n",
-    "root1.right = TreeNode(3)\n",
-    "root1.left.right = TreeNode(5)\n",
-    "root1.right.right = TreeNode(4)\n",
-    "print(rightSideView(root1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 3]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "root2 = TreeNode(1)\n",
-    "root2.right = TreeNode(3)\n",
-    "print(rightSideView(root2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# Creating a new tree for this example\n",
-    "root = None\n",
-    "result = rightSideView(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the given code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of the code is O(N), where N is the number of nodes in the binary tree. Here's why:\n",
-    "\n",
-    "1. We perform a level-order traversal of the tree using a queue. In the worst case, we visit all nodes once, which takes O(N) time since we visit each node exactly once.\n",
-    "\n",
-    "2. For each node, we perform constant-time operations like dequeuing it from the queue, checking if it's the rightmost node at the current level, and enqueuing its children (if they exist). These operations do not depend on the size of the tree, so they do not contribute to the overall time complexity.\n",
-    "\n",
-    "Therefore, the dominant factor in the time complexity is the level-order traversal, making it O(N).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of the code is also O(N). Here's why:\n",
-    "\n",
-    "1. We use a queue to perform level-order traversal. In the worst case, the queue can contain all nodes at the maximum width of the tree, which can be up to N/2 nodes for a completely unbalanced tree. Therefore, the space required for the queue is O(N).\n",
-    "\n",
-    "2. The `result` list stores the values of the rightmost nodes. In the worst case, when the binary tree is a complete binary tree, it can have roughly N/2 rightmost nodes. Therefore, the space required for the `result` list is also O(N).\n",
-    "\n",
-    "3. Other auxiliary variables like `level_size`, `i`, and `node` require only constant space and do not depend on the size of the tree.\n",
-    "\n",
-    "Combining the space used by the queue and the `result` list, we get a space complexity of O(N).\n",
-    "\n",
-    "In summary, the code has a time complexity of O(N) and a space complexity of O(N), where N is the number of nodes in the binary tree. These complexities are efficient and scale linearly with the size of the input tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Variant with Left Side View**: Modify the code to find and return the values of nodes visible from the left side of the binary tree instead of the right side.\n",
-    "\n",
-    "2. **Zigzag Right Side View**: Extend the code to return the right side view values in a zigzag order. In a zigzag order, you alternate between starting from the rightmost node at level 0, then the leftmost at level 1, rightmost at level 2, and so on."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb b/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb
deleted file mode 100644
index 3e542c9..0000000
--- a/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 226: Invert Binary Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the root of a binary tree, invert the tree, and return its root.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the tree is in the range `[0, 100]`.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def invertTree(root):\n",
-    "    # Base case: If the root is None or the tree is empty, return None.\n",
-    "    if not root:\n",
-    "        return None\n",
-    "\n",
-    "    # Swap the left and right subtrees of the current node.\n",
-    "    root.left, root.right = root.right, root.left\n",
-    "\n",
-    "    # Recursively invert the left and right subtrees.\n",
-    "    invertTree(root.left)\n",
-    "    invertTree(root.right)\n",
-    "\n",
-    "    return root"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a TreeNode class, which represents a node in a binary tree. Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-    "\n",
-    "2. The `invertTree` function is defined to invert the binary tree. It takes the root node of the tree as an argument.\n",
-    "\n",
-    "3. In the `invertTree` function, we have a base case to handle the scenario when the root is `None` (empty tree). In such cases, we return `None` because there's nothing to invert.\n",
-    "\n",
-    "4. For non-empty trees, we swap the left and right subtrees of the current node. This effectively inverts the tree at the current node.\n",
-    "\n",
-    "5. We then recursively call `invertTree` on the left and right subtrees to invert them.\n",
-    "\n",
-    "6. Finally, we return the root of the inverted tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Inverted Tree (Example 1): <__main__.TreeNode object at 0x7faa5c445ab0>\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "# Input: root = [4,2,7,1,3,6,9]\n",
-    "\n",
-    "root1 = TreeNode(4)\n",
-    "root1.left = TreeNode(2)\n",
-    "root1.right = TreeNode(7)\n",
-    "root1.left.left = TreeNode(1)\n",
-    "root1.left.right = TreeNode(3)\n",
-    "root1.right.left = TreeNode(6)\n",
-    "root1.right.right = TreeNode(9)\n",
-    "\n",
-    "inverted_root1 = invertTree(root1)\n",
-    "print(\"Inverted Tree (Example 1):\", inverted_root1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Inverted Tree (Example 2): <__main__.TreeNode object at 0x7faa5c444310>\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "# Input: root = [2,1,3]\n",
-    "\n",
-    "root2 = TreeNode(2)\n",
-    "root2.left = TreeNode(1)\n",
-    "root2.right = TreeNode(3)\n",
-    "\n",
-    "inverted_root2 = invertTree(root2)\n",
-    "print(\"Inverted Tree (Example 2):\", inverted_root2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Inverted Tree (Example 3): None\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "# Input: root = []\n",
-    "\n",
-    "root3 = None  # Empty tree\n",
-    "\n",
-    "inverted_root3 = invertTree(root3)\n",
-    "print(\"Inverted Tree (Example 3):\", inverted_root3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's discuss the time and space complexity of the provided Python code to invert a binary tree.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of the `invertTree` function is $O(n)$, where $n$ is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal of the tree. In the worst case, we have to visit all nodes in the tree.\n",
-    "\n",
-    "The reason for this time complexity is the depth-first traversal of the tree, where we recursively process the left and right subtrees of each node.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of the code is determined by the function call stack during the recursion. In the worst case, the space complexity is $O(h)$, where $h$ is the height of the binary tree. \n",
-    "\n",
-    "In a completely unbalanced binary tree (essentially a linked list), the height $h$ is equal to the number of nodes $n$, resulting in a space complexity of $O(n)$. This occurs when the tree is skewed to one side.\n",
-    "\n",
-    "In a balanced binary tree, such as a full binary tree, the height $h$ is $O(log\\ n)$, and the space complexity is $O(log\\ n)$.\n",
-    "\n",
-    "The space complexity depends on how balanced the tree is. In practical scenarios, binary trees are often approximately balanced, so the space complexity is typically closer to $O(log\\ n)$.\n",
-    "\n",
-    "**In summary:**\n",
-    "\n",
-    "- Time Complexity: $O(n)$ where $n$ is the number of nodes.\n",
-    "- Space Complexity: $O(h)$, where $h$ is the height of the binary tree. In the worst case, it can be $O(n)$, and in a balanced tree, it is $O(log\\ n)$.\n",
-    "\n",
-    "Keep in mind that these complexities are based on the recursive implementation provided. Iterative solutions can achieve the same task with $O(1)$ space complexity, using auxiliary data structures like stacks or queues to mimic the recursion stack."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb b/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb
deleted file mode 100644
index fdad9ba..0000000
--- a/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 230. Kth Smallest Element in a BST\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Kth Smallest Element in a BST problem on LeetCode, click here!](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary search tree, and an integer `k`, return *the $k^{th}$ smallest value (**1-indexed**) of all the values of the nodes in the tree.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is `n`.\n",
-    "- 1 <= `k <= n` <= $10^4$\n",
-    "- 0 <= `Node.val` <= $10^4$\n",
-    "\n",
-    "**Follow up**: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def kthSmallest(root, k):\n",
-    "    stack = []  # Initialize an empty stack to simulate the traversal\n",
-    "    while root or stack:\n",
-    "        while root:\n",
-    "            stack.append(root)  # Push the current node onto the stack\n",
-    "            root = root.left  # Move to the left child\n",
-    "        root = stack.pop()  # Pop a node from the stack\n",
-    "        k -= 1  # Decrement k since we've visited a node\n",
-    "        if k == 0:\n",
-    "            return root.val  # If k becomes 0, return the current node's value as the kth smallest\n",
-    "        root = root.right  # Move to the right child to continue the traversal"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class, which represents a node in the binary search tree (BST). Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-    "\n",
-    "2. The `kthSmallest` function takes two parameters: `root`, which is the root of the BST, and `k`, which represents the kth smallest element we want to find.\n",
-    "\n",
-    "3. We initialize an empty stack (`stack`) to simulate the traversal of the BST. The stack will be used to keep track of nodes as we traverse the tree in an iterative manner.\n",
-    "\n",
-    "4. We enter a while loop that continues until either `root` becomes `None` (indicating we have traversed the entire tree) or the `stack` is empty.\n",
-    "\n",
-    "5. Within the loop, we start another while loop to traverse as far left as possible in the BST. We repeatedly push nodes onto the `stack` and move to their left children until we reach the leftmost leaf node. This is the smallest node in the BST.\n",
-    "\n",
-    "6. Once we have reached the leftmost leaf node (the smallest node), we pop nodes from the `stack` one by one. As we pop each node, we decrement `k` by 1 to keep track of the number of nodes we have visited.\n",
-    "\n",
-    "7. If `k` becomes 0 after decrementing, it means we have found the kth smallest element. In this case, we return the `val` of the current node as the result.\n",
-    "\n",
-    "8. If `k` is still greater than 0, it means we haven't found the kth smallest element yet. In this case, we move to the right child of the current node to continue the traversal, as the kth smallest element, if it exists, will be in the right subtree of the current node.\n",
-    "\n",
-    "9. The process continues until we find the kth smallest element or traverse the entire tree.\n",
-    "\n",
-    "10. Finally, we return the kth smallest element found.\n",
-    "\n",
-    "The key idea in this code is to perform an in-order traversal of the BST while keeping track of the kth smallest element. By visiting nodes in ascending order, we can efficiently find the kth smallest element in O(h + k) time, where h is the height of the BST and k is the desired kth element. This approach is both concise and efficient for this problem."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "# Create the tree for the first example: root = [3,1,4,null,2], k = 1\n",
-    "root1 = TreeNode(3)\n",
-    "root1.left = TreeNode(1)\n",
-    "root1.right = TreeNode(4)\n",
-    "root1.left.right = TreeNode(2)\n",
-    "k1 = 1\n",
-    "print(kthSmallest(root1, k1)) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Create the tree for the second example: root = [5,3,6,2,4,null,null,1], k = 3\n",
-    "root2 = TreeNode(5)\n",
-    "root2.left = TreeNode(3)\n",
-    "root2.right = TreeNode(6)\n",
-    "root2.left.left = TreeNode(2)\n",
-    "root2.left.right = TreeNode(4)\n",
-    "root2.left.left.left = TreeNode(1)\n",
-    "k2 = 3\n",
-    "print(kthSmallest(root2, k2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the given code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this code is O(h + k), where:\n",
-    "- h is the height of the binary search tree (BST).\n",
-    "- k is the desired kth smallest element we want to find.\n",
-    "\n",
-    "1. In the worst case, where the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the time complexity is O(n + k).\n",
-    "\n",
-    "2. In the best case, where the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the time complexity is O(log(n) + k).\n",
-    "\n",
-    "So, the time complexity can vary from O(log(n) + k) in the best-case scenario to O(n + k) in the worst-case scenario. Typically, for balanced BSTs, the time complexity is closer to O(log(n) + k).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this code is O(h) due to the stack used for the iterative in-order traversal, where h is the height of the BST.\n",
-    "\n",
-    "1. In the worst case, when the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the space complexity is O(n) because the stack can potentially store all n nodes.\n",
-    "\n",
-    "2. In the best case, when the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the space complexity is O(log(n)) because the stack will have at most log(n) nodes.\n",
-    "\n",
-    "So, the space complexity depends on the height of the BST and can vary from O(log(n)) in the best-case scenario to O(n) in the worst-case scenario.\n",
-    "\n",
-    "In practical terms, for balanced BSTs or moderately unbalanced BSTs, the space complexity is usually close to O(log(n)), and the code is efficient in terms of space usage."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Kth Largest Element**: Modify the code to find the kth largest element in the BST instead of the kth smallest element.\n",
-    "\n",
-    "2. **Kth Smallest Element in Two BSTs**: Given two BSTs, find the kth smallest element when considering both BSTs as a single sorted list. This involves merging the two BSTs while finding the kth element efficiently."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb b/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb
deleted file mode 100644
index 0179ca4..0000000
--- a/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,237 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 235. Lowest Common Ancestor of a Binary Search Tree\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Lowest Common Ancestor of a Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)\n",
-    "\n",
-    "---\n",
-    "Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.\n",
-    "\n",
-    "According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[2, 10^5]$.\n",
-    "- $-10^9$ <= `Node.val` <= $10^9$\n",
-    "- All `Node.val` are **unique**.\n",
-    "- `p != q`\n",
-    "- `p` and `q` will exist in the BST.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def lowestCommonAncestor(root, p, q):\n",
-    "    # Ensure p is less than q\n",
-    "    if p.val > q.val:\n",
-    "        p, q = q, p\n",
-    "\n",
-    "    while root:\n",
-    "        # If the current node value is greater than both p and q, move left\n",
-    "        if root.val > q.val:\n",
-    "            root = root.left\n",
-    "        # If the current node value is less than both p and q, move right\n",
-    "        elif root.val < p.val:\n",
-    "            root = root.right\n",
-    "        # If the current node value is between p and q (inclusive), or it matches either p or q, it's the LCA\n",
-    "        else:\n",
-    "            return root"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `class TreeNode:` defines a simple class for representing nodes in a binary tree. Each `TreeNode` has three attributes:\n",
-    "   - `val`: The value of the node.\n",
-    "   - `left`: A reference to the left child node.\n",
-    "   - `right`: A reference to the right child node.\n",
-    "\n",
-    "2. `def lowestCommonAncestor(root, p, q):` is a function that takes three arguments:\n",
-    "   - `root`: The root node of the BST.\n",
-    "   - `p`: A TreeNode representing one of the nodes for which we want to find the LCA.\n",
-    "   - `q`: A TreeNode representing the other node for which we want to find the LCA.\n",
-    "\n",
-    "3. `if p.val > q.val:` checks if the value of `p` is greater than the value of `q`. In a BST, it's essential to ensure that `p` represents the smaller value, and `q` represents the larger value. If `p` is greater than `q`, the code swaps their values.\n",
-    "\n",
-    "4. The main logic is inside the `while` loop, which runs until `root` becomes `None`. It performs the following steps to find the LCA:\n",
-    "   - If the current `root` node's value is greater than the value of `q`, it means both `p` and `q` are on the left subtree of the current node. So, we move to the left child of the current node by setting `root = root.left`.\n",
-    "   - If the current `root` node's value is less than the value of `p`, it means both `p` and `q` are on the right subtree of the current node. So, we move to the right child of the current node by setting `root = root.right`.\n",
-    "   - If neither of the above conditions is met, it means the current `root` node's value is between `p` and `q`, or it matches either `p` or `q`. In this case, the current node is the lowest common ancestor (LCA), so we return `root`.\n",
-    "\n",
-    "The algorithm is efficient because it takes advantage of the properties of a BST. It eliminates subtrees that cannot contain the LCA by comparing the values of `p`, `q`, and the current `root` node. Eventually, it reaches the LCA node, and that node is returned as the result."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "6\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Construct the tree\n",
-    "root = TreeNode(6)\n",
-    "root.left = TreeNode(2)\n",
-    "root.right = TreeNode(8)\n",
-    "root.left.left = TreeNode(0)\n",
-    "root.left.right = TreeNode(4)\n",
-    "root.right.left = TreeNode(7)\n",
-    "root.right.right = TreeNode(9)\n",
-    "root.left.right.left = TreeNode(3)\n",
-    "root.left.right.right = TreeNode(5)\n",
-    "\n",
-    "p = root.left  # Node 2\n",
-    "q = root.right  # Node 8\n",
-    "\n",
-    "result = lowestCommonAncestor(root, p, q)\n",
-    "print(result.val)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Using the same tree as before\n",
-    "p = root.left  # Node 2\n",
-    "q = root.left.right  # Node 4\n",
-    "\n",
-    "result = lowestCommonAncestor(root, p, q)\n",
-    "print(result.val) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# Creating a new tree for this example\n",
-    "root2 = TreeNode(2)\n",
-    "root2.left = TreeNode(1)\n",
-    "\n",
-    "p = root2  # Node 2\n",
-    "q = root2.left  # Node 1\n",
-    "\n",
-    "result = lowestCommonAncestor(root2, p, q)\n",
-    "print(result.val)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided algorithm for finding the lowest common ancestor (LCA) in a Binary Search Tree (BST).\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of this algorithm is O(h), where \"h\" is the height of the BST. In the worst case, where the tree is completely unbalanced (essentially a linked list), the height of the tree is O(n), where \"n\" is the number of nodes in the tree. However, in a well-balanced BST, the height is logarithmic, which is O(log n).\n",
-    "\n",
-    "The reason for this time complexity is that the algorithm efficiently narrows down the search space by traversing either left or right subtrees based on the values of the target nodes `p` and `q`. It eliminates entire subtrees that cannot contain the LCA, leading to a relatively quick search.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity of the algorithm is O(1). This is because it uses a constant amount of extra space, regardless of the size of the input BST. The only variables used are `root`, `p`, and `q`, and there are no data structures like stacks or queues used for additional space. The algorithm performs a simple traversal without recursion, so it does not consume extra memory as the tree depth increases.\n",
-    "\n",
-    "**In summary**, the provided algorithm for finding the LCA in a BST is both time and space-efficient. Its time complexity is O(h), where \"h\" is the height of the tree, and its space complexity is O(1), making it suitable for practical use even in large BSTs."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "### Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple LCAs:** Extend the algorithm to find all the lowest common ancestors of two given nodes `p` and `q` in a BST. In some cases, there can be multiple LCAs.\n",
-    "\n",
-    "2. **LCA with k Nodes:** Given a BST and k nodes, find the lowest common ancestor of these k nodes. This is an extension of the problem where you must find the LCA of more than two nodes."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb b/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb
deleted file mode 100644
index 78a78b5..0000000
--- a/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb	
+++ /dev/null
@@ -1,291 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 297. Serialize and Deserialize Binary Tree\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Serialize and Deserialize Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)\n",
-    "\n",
-    "---\n",
-    "Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.\n",
-    "\n",
-    "Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.\n",
-    "\n",
-    "**Clarification:** The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-    "- `-1000 <= Node.val <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "TThe problem is to design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of characters or bytes so that it can be easily stored in a file, transmitted over a network, or otherwise persisted. Deserialization is the reverse process of converting the serialized data back into the original data structure.\n",
-    "\n",
-    "In the context of this problem:\n",
-    "\n",
-    "1. **Serialization**: You are given a binary tree, and your task is to convert it into a string representation such that you can later recreate the same binary tree from this string. The format of serialization is flexible, but it should allow you to reconstruct the original binary tree accurately.\n",
-    "\n",
-    "2. **Deserialization**: Given a serialized string, you need to reconstruct the binary tree it represents, ensuring that it is identical to the original tree.\n",
-    "\n",
-    "Here are some key points to consider in solving this problem:\n",
-    "\n",
-    "- The input can include any valid binary tree, including trees with nodes having integer values within the range [-1000, 1000].\n",
-    "- You don't necessarily need to follow a specific format for serialization, but it should be designed in a way that allows unambiguous deserialization.\n",
-    "- The goal is to serialize the tree structure and its values and then deserialize it back into the same structure and values.\n",
-    "\n",
-    "A common approach for serialization is to use a traversal method like preorder traversal, where you visit nodes in the order: root, left subtree, right subtree. This way, you can serialize the tree into a string, and during deserialization, you can reconstruct the tree by parsing the string in the same order.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode(object):\n",
-    "    def __init__(self, x):\n",
-    "        self.val = x\n",
-    "        self.left = None\n",
-    "        self.right = None\n",
-    "\n",
-    "class Codec:\n",
-    "    def serialize(self, root):\n",
-    "        serialized  = []\n",
-    "\n",
-    "        def dfs(node):\n",
-    "            if not node:\n",
-    "                # If the node is None, represent it as \"Null\" in the serialized string\n",
-    "                serialized .append(\"Null\")\n",
-    "                return\n",
-    "            # Convert the node's value to a string and add it to the serialized string\n",
-    "            serialized .append(str(node.val))\n",
-    "            dfs(node.left)\n",
-    "            dfs(node.right)\n",
-    "\n",
-    "        dfs(root)\n",
-    "        # Join the serialized values with \",\"\n",
-    "        return \",\".join(serialized )\n",
-    "\n",
-    "    def deserialize(self, data):\n",
-    "        vals = data.split(\",\")\n",
-    "        self.i = 0\n",
-    "\n",
-    "        def dfs():\n",
-    "            if vals[self.i] == \"Null\":\n",
-    "                self.i += 1\n",
-    "                return None\n",
-    "            # Convert the value to an integer to create a new node\n",
-    "            node = TreeNode(int(vals[self.i]))\n",
-    "            self.i += 1\n",
-    "            node.left = dfs()\n",
-    "            node.right = dfs()\n",
-    "            return node\n",
-    "\n",
-    "        return dfs()"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Here's a step-by-step explanation of the code:\n",
-    "\n",
-    "1. **Definition of TreeNode**:\n",
-    "   - The code defines a simple `TreeNode` class that represents a node in a binary tree. Each `TreeNode` has three attributes:\n",
-    "     - `val`: The integer value stored in the node.\n",
-    "     - `left`: A reference to the left child node (or None if there is no left child).\n",
-    "     - `right`: A reference to the right child node (or None if there is no right child).\n",
-    "   - This class allows you to create binary tree nodes with integer values and left and right child references.\n",
-    "\n",
-    "2. **Serialization (serialize method)**:\n",
-    "   - The `serialize` method takes a `root` node as input, which is the root of the binary tree that needs to be serialized.\n",
-    "   - It initializes an empty list `serialized` to store the serialized elements.\n",
-    "   - The `dfs` (depth-first search) function is defined within the `serialize` method to perform a preorder traversal of the binary tree.\n",
-    "   - In the `dfs` function:\n",
-    "     - If the current `node` is `None` (i.e., a leaf node or a child of a leaf node), it appends the string \"Null\" to the `serialized` list to represent the absence of a node.\n",
-    "     - If the current `node` is not `None`, it appends the string representation of the `node.val` to the `serialized` list and then recursively calls `dfs` on the left and right children.\n",
-    "   - After the `dfs` traversal, the `serialized` list contains the serialized binary tree elements.\n",
-    "   - The method returns a string obtained by joining the elements in the `serialized` list with commas.\n",
-    "\n",
-    "3. **Deserialization (deserialize method)**:\n",
-    "   - The `deserialize` method takes a serialized string `data` as input, which represents a binary tree in the custom format.\n",
-    "   - It splits the `data` string by commas to obtain a list of elements called `vals`.\n",
-    "   - The method initializes an index `self.i` to 0. This index keeps track of the current position in the `vals` list during deserialization.\n",
-    "   - The `dfs` function is defined within the `deserialize` method to perform the deserialization process using a recursive approach.\n",
-    "   - In the `dfs` function:\n",
-    "     - If the current element in `vals` is \"Null,\" it means there is no node at this position in the binary tree, so it returns `None`.\n",
-    "     - If the current element is not \"Null,\" it converts the element to an integer to create a new `TreeNode` with that value.\n",
-    "     - It then recursively calls `dfs` to set the left and right children of the current node.\n",
-    "   - The method returns the root node of the reconstructed binary tree.\n",
-    "\n",
-    "Overall, this code demonstrates a way to serialize a binary tree into a string format and then deserialize it back into the original tree structure, allowing for the representation of both the tree structure and the values stored in the nodes."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1,2,Null,Null,3,4,Null,Null,5,Null,Null\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "# Serialize a binary tree\n",
-    "root = TreeNode(1)\n",
-    "root.left = TreeNode(2)\n",
-    "root.right = TreeNode(3)\n",
-    "root.right.left = TreeNode(4)\n",
-    "root.right.right = TreeNode(5)\n",
-    "\n",
-    "codec = Codec()\n",
-    "serialized_tree = codec.serialize(root)\n",
-    "print(serialized_tree)\n",
-    "\n",
-    "# Deserialize the serialized tree\n",
-    "new_root = codec.deserialize(serialized_tree)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Null\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Example usage:\n",
-    "# Serialize a binary tree\n",
-    "root = None\n",
-    "\n",
-    "codec = Codec()\n",
-    "serialized_tree = codec.serialize(root)\n",
-    "print(serialized_tree)\n",
-    "\n",
-    "# Deserialize the serialized tree\n",
-    "new_root = codec.deserialize(serialized_tree)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for serializing and deserializing a binary tree:\n",
-    "\n",
-    "1. **Serialization (serialize method)**:\n",
-    "\n",
-    "   - Time Complexity:\n",
-    "     - The `serialize` method uses a depth-first traversal (preorder) of the binary tree.\n",
-    "     - It visits each node exactly once.\n",
-    "     - At each node, it performs constant-time operations (string conversion and list append).\n",
-    "     - Therefore, the time complexity for serialization is O(N), where N is the number of nodes in the binary tree.\n",
-    "\n",
-    "   - Space Complexity:\n",
-    "     - The space complexity for serialization includes the space used by the `serialized` list and the recursive call stack.\n",
-    "     - The `serialized` list stores the serialized elements, and its space is proportional to the number of nodes.\n",
-    "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-    "     - Therefore, the space complexity for serialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-    "\n",
-    "2. **Deserialization (deserialize method)**:\n",
-    "\n",
-    "   - Time Complexity:\n",
-    "     - The `deserialize` method splits the serialized string into a list of elements, which takes O(N) time.\n",
-    "     - The `dfs` function is a recursive function that visits each element in the list exactly once.\n",
-    "     - At each element, it performs constant-time operations (string comparison, integer conversion, and recursive function calls).\n",
-    "     - Therefore, the time complexity for deserialization is O(N).\n",
-    "\n",
-    "   - Space Complexity:\n",
-    "     - The space complexity for deserialization includes the space used by the `vals` list and the recursive call stack.\n",
-    "     - The `vals` list stores the elements from the serialized string, and its space is proportional to the number of nodes.\n",
-    "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-    "     - Therefore, the space complexity for deserialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-    "\n",
-    "Overall, the provided code has a time complexity of O(N) for both serialization and deserialization and a space complexity of O(N) in the majority of cases (unless the tree is severely unbalanced, in which case the height H dominates the space complexity). It efficiently serializes and deserializes a binary tree while using a linear amount of memory."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Efficient Serialization**:\n",
-    "   Optimize the serialization process to minimize the length of the serialized string. Consider using techniques like binary encoding to represent node values more efficiently.\n",
-    "\n",
-    "2. **Custom Serialization Format**:\n",
-    "   Modify the serialization and deserialization methods to use a custom format different from the one provided. Ensure that the new format allows for accurate reconstruction of the original binary tree."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/572. Subtree of Another Tree.ipynb b/_build/jupyter_execute/07. Trees/572. Subtree of Another Tree.ipynb
deleted file mode 100644
index 425f22d..0000000
--- a/_build/jupyter_execute/07. Trees/572. Subtree of Another Tree.ipynb	
+++ /dev/null
@@ -1,259 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 572: Subtree of Another Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Subtree of Another Tree problem on LeetCode, click here!](https://leetcode.com/problems/subtree-of-another-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "**Problem Description:**\n",
-    "\n",
-    "Given the roots of two binary trees `root` and `subRoot`, return `True` if there is a subtree of `root` with the same structure and node values as `subRoot`, and `False` otherwise.\n",
-    "\n",
-    "A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "1. The number of nodes in the `root` tree is in the range $[1, 2000]$.\n",
-    "2. The number of nodes in the `subRoot` tree is in the range $[1, 1000]$.\n",
-    "3. $-10^4$ <= root.val <= $10^4$\n",
-    "4. $-10^4$ <= subRoot.val <= $10^4$\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "class Solution:\n",
-    "    def isSubtree(self, root, subRoot):\n",
-    "        if not root:\n",
-    "            return False\n",
-    "        \n",
-    "        # Check if the current subtree is equal to the subRoot\n",
-    "        if self.isSameTree(root, subRoot):\n",
-    "            return True\n",
-    "        \n",
-    "        # Recursively check the left and right subtrees\n",
-    "        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)\n",
-    "    \n",
-    "    def isSameTree(self, tree1, tree2):\n",
-    "        if not tree1 and not tree2:\n",
-    "            return True\n",
-    "        if not tree1 or not tree2:\n",
-    "            return False\n",
-    "        \n",
-    "        return (\n",
-    "            tree1.val == tree2.val and\n",
-    "            self.isSameTree(tree1.left, tree2.left) and\n",
-    "            self.isSameTree(tree1.right, tree2.right)\n",
-    "        )"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "This code defines a TreeNode class for the binary tree nodes and a Solution class with two methods:\n",
-    "\n",
-    "1. `isSubtree`: This method checks if there is a subtree of the `root` tree with the same structure and node values as the `subRoot` tree. It uses a helper function `isSameTree` to compare two trees for equality.\n",
-    "\n",
-    "2. `isSameTree`: This helper method recursively compares two trees to check if they are the same in structure and node values.\n",
-    "\n",
-    "### Here's a detailed explanation of the code:\n",
-    "\n",
-    "1. **TreeNode Class:**\n",
-    "\n",
-    "The `TreeNode` class is defined to represent nodes in a binary tree. Each node has a `val` (the node's value) and may have a `left` and `right` child.\n",
-    "\n",
-    "2. **Solution Class:**\n",
-    "\n",
-    "    + The `Solution` class contains the solution for the problem and defines two important methods:\n",
-    "\n",
-    "    + `isSubtree(self, root, subRoot)`:\n",
-    "\n",
-    "        - This method checks whether `subRoot` is a subtree of `root`. It takes two tree nodes, `root` and `subRoot`, as input arguments.\n",
-    "        - If `root` is `None`, it returns `False` because there is no subtree to search.\n",
-    "        - It then checks if the current subtree with `root` as its root is equal to `subRoot` using the `isSameTree` method. If they are the same, it returns `True`.\n",
-    "        - If the current subtree is not the same as `subRoot`, it recursively checks the left and right subtrees of `root` to see if `subRoot` is a subtree of any of them.\n",
-    "        - It returns `True` if `subRoot` is found in either the left or right subtree; otherwise, it returns `False`.\n",
-    "\n",
-    "    + `isSameTree(self, tree1, tree2)`:\n",
-    "\n",
-    "        - This method checks whether two trees, `tree1` and `tree2`, are the same.\n",
-    "        - If both `tree1` and `tree2` are `None`, they are considered the same tree, so it returns `True`.\n",
-    "        - If only one of them is `None` (but not both), they are different trees, so it returns `False`.\n",
-    "        - If both `tree1` and `tree2` have values, it checks if their values are equal and recursively checks if their left and right subtrees are the same.\n",
-    "        - It returns `True` if the trees are the same; otherwise, it returns `False`.\n",
-    "\n",
-    "The code effectively uses recursion to traverse the binary trees and check for subtree equality. The `isSubtree` method starts the recursive search, and the `isSameTree` method is used to compare individual subtrees. The approach is efficient and avoids unnecessary checks when possible.\n",
-    "\n",
-    "This solution allows you to determine if there exists a subtree within the `root` tree that matches the structure and node values of `subRoot`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Is subRoot a subtree of root? True\n"
-     ]
-    }
-   ],
-   "source": [
-    "#root = [3,4,5,1,2]\n",
-    "#subRoot = [4,1,2]\n",
-    "\n",
-    "# Example input\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(4)\n",
-    "root.right = TreeNode(5)\n",
-    "root.left.left = TreeNode(1)\n",
-    "root.left.right = TreeNode(2)\n",
-    "\n",
-    "subRoot = TreeNode(4)\n",
-    "subRoot.left = TreeNode(1)\n",
-    "subRoot.right = TreeNode(2)\n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Test the isSubtree method\n",
-    "result = solution.isSubtree(root, subRoot)\n",
-    "print(\"Is subRoot a subtree of root?\", result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Is subRoot a subtree of root? False\n"
-     ]
-    }
-   ],
-   "source": [
-    "#root = [3,4,5,1,2,null,null,null,null,0] \n",
-    "#subRoot = [4,1,2]\n",
-    "\n",
-    "\n",
-    "# Example input\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(4)\n",
-    "root.right = TreeNode(5)\n",
-    "root.left.left = TreeNode(1)\n",
-    "root.left.right = TreeNode(2)\n",
-    "root.left.right.left = TreeNode(0)\n",
-    "\n",
-    "subRoot = TreeNode(4)\n",
-    "subRoot.left = TreeNode(1)\n",
-    "subRoot.right = TreeNode(2)\n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Test the isSubtree method\n",
-    "result = solution.isSubtree(root, subRoot)\n",
-    "print(\"Is subRoot a subtree of root?\", result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "86a34f0d-0789-498c-818c-2d30f84754d0",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code for the \"Subtree of Another Tree\" problem:\n",
-    "\n",
-    "**Time Complexity**\n",
-    "\n",
-    "The time complexity of the code primarily depends on the recursive traversal of the binary tree. In both the `isSubtree` and `isSameTree` functions, we visit each node in the binary trees once. Let's break down the time complexity:\n",
-    "\n",
-    "1. The `isSubtree` function:\n",
-    "   - In the worst case, we need to visit every node in the `root` tree.\n",
-    "   - For each node in the `root` tree, we call the `isSameTree` function, which has its own traversal.\n",
-    "   - So, the total time complexity is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree.\n",
-    "\n",
-    "2. The `isSameTree` function:\n",
-    "   - In the worst case, we visit every node in both `tree1` and `tree2`.\n",
-    "   - The number of recursive calls made is proportional to the number of nodes in the trees.\n",
-    "   - So, the time complexity of this function is $O(max(n, m))$, where $n$ and $m$ are the numbers of nodes in `tree1` and `tree2`, respectively.\n",
-    "\n",
-    "Overall, the time complexity of the entire code is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree. In practice, it may be less than $O(n * m)$ if a subtree mismatch is detected early during the traversal.\n",
-    "\n",
-    "**Space Complexity**\n",
-    "\n",
-    "The space complexity of the code is determined by the function call stack during recursion and the space used by the recursive functions. Let's analyze the space complexity:\n",
-    "\n",
-    "1. The `isSubtree` function:\n",
-    "   - It uses the call stack for recursion.\n",
-    "   - The maximum depth of the recursion is equal to the height of the `root` tree, which can be $O(n)$ in the worst case (unbalanced tree).\n",
-    "   - Additionally, the function doesn't use any significant extra space other than the recursion stack.\n",
-    "\n",
-    "2. The `isSameTree` function:\n",
-    "   - It also uses the call stack for recursion.\n",
-    "   - The maximum depth of the recursion is equal to the height of the `tree1` or `tree2`, whichever is greater.\n",
-    "   - So, the maximum space used for the call stack is $O(max(n, m))$.\n",
-    "\n",
-    "In summary, the space complexity of the code is $O(max(n, m))$ due to the function call stack. It scales with the maximum height of the trees being compared.\n",
-    "\n",
-    "Overall, the code is efficient and works well for trees with moderate sizes. However, it's important to keep in mind that the worst-case time complexity is $O(n * m)$, so for very large trees, the performance may degrade."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb b/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb
deleted file mode 100644
index 0cbe2fc..0000000
--- a/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 98. Validate Binary Search Tree\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Validate Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/validate-binary-search-tree/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary tree, *determine if it is a valid binary search tree (BST)*.\n",
-    "\n",
-    "A **valid BST** is defined as follows:\n",
-    "\n",
-    "- The left subtree of a node contains only nodes with keys **less than** the node's key.\n",
-    "- The right subtree of a node contains only nodes with keys **greater than** the node's key.\n",
-    "- Both the left and right subtrees must also be binary search trees. \n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[1, 10^4]$.\n",
-    "- $-2^{31}$ <= `Node.val` <= $2^{31} - 1$\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def is_valid_BST(root):\n",
-    "    def is_valid(node, min_val, max_val):\n",
-    "        # Base case: If the node is None, it's a valid BST.\n",
-    "        if node is None:\n",
-    "            return True\n",
-    "        \n",
-    "        # Check if the current node's value is within the valid range.\n",
-    "        if not (min_val < node.val < max_val):\n",
-    "            return False\n",
-    "        \n",
-    "        # Recursively check the left and right subtrees with updated ranges.\n",
-    "        return (is_valid(node.left, min_val, node.val) and\n",
-    "                is_valid(node.right, node.val, max_val))\n",
-    "    \n",
-    "    # Call the helper function starting with a wide range for root node.\n",
-    "    return is_valid(root, float('-inf'), float('inf'))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "- It defines a `TreeNode` class to represent nodes in the binary tree.\n",
-    "\n",
-    "- The `is_valid_BST` function takes the root node of the binary tree as input and returns `True` if the tree is a valid BST, and `False` otherwise.\n",
-    "\n",
-    "- Inside the `is_valid_BST` function, there is a helper function called `is_valid` that performs the actual validation using a recursive approach.\n",
-    "\n",
-    "- The `is_valid` function checks each node in the tree to ensure that it satisfies the properties of a BST:\n",
-    "  - The value of the current node must be within a valid range defined by `min_val` and `max_val`.\n",
-    "  - The left subtree of the current node should contain values less than the current node's value.\n",
-    "  - The right subtree of the current node should contain values greater than the current node's value.\n",
-    "\n",
-    "- If any of these conditions are violated, the function returns `False`, indicating that the tree is not a valid BST.\n",
-    "\n",
-    "- If all nodes satisfy these conditions, the function returns `True`, indicating that the tree is a valid BST.\n",
-    "\n",
-    "- The code calls the `is_valid` function with the root node and initial range values of negative infinity to positive infinity to start the validation.\n",
-    "\n",
-    "- The time complexity of the code is O(N), where N is the number of nodes in the tree, as it traverses each node once.\n",
-    "\n",
-    "- The space complexity depends on the height of the tree. In the average case for a balanced BST, the space complexity is O(log N), but in the worst case (skewed tree), it can be O(N) due to the recursive call stack.\n",
-    "\n",
-    "In summary, this code checks whether a given binary tree is a valid BST by recursively validating each node's value and its left and right subtrees while maintaining valid value ranges. If all nodes satisfy the BST properties, the tree is considered a valid BST."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "# Create the tree for the first example: [2, 1, 3]\n",
-    "root1 = TreeNode(2)\n",
-    "root1.left = TreeNode(1)\n",
-    "root1.right = TreeNode(3)\n",
-    "print(is_valid_BST(root1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Create the tree for the second example: [5, 1, 4, None, None, 3, 6]\n",
-    "root2 = TreeNode(5)\n",
-    "root2.left = TreeNode(1)\n",
-    "root2.right = TreeNode(4)\n",
-    "root2.right.left = TreeNode(3)\n",
-    "root2.right.right = TreeNode(6)\n",
-    "print(is_valid_BST(root2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `is_valid_BST` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The time complexity of the function primarily depends on the number of nodes in the binary tree.\n",
-    "- In the worst case, we may have to visit every node in the tree once to validate whether it's part of a valid BST.\n",
-    "- Since we are performing a depth-first search (DFS) traversal of the tree, the time complexity is O(N), where N is the number of nodes in the tree.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity is determined by the space used by the recursive call stack during the DFS traversal.\n",
-    "- In the worst case, if the tree is completely unbalanced (a skewed tree), the maximum depth of the call stack would be equal to the height of the tree.\n",
-    "- In a balanced BST, the height is approximately log2(N), where N is the number of nodes.\n",
-    "- Therefore, the space complexity of the call stack is O(log N) for a balanced BST.\n",
-    "- In the worst case (skewed tree), the space complexity can be O(N) as the height of the tree can be equal to N.\n",
-    "\n",
-    "**Overall:**\n",
-    "- Time Complexity: O(N)\n",
-    "- Space Complexity: O(log N) on average for a balanced BST, and O(N) in the worst case for a skewed tree.\n",
-    "\n",
-    "The space complexity is typically dominated by the recursive call stack, and it varies depending on the shape of the binary tree. In practice, for balanced binary trees, the space complexity is often close to O(log N), which is quite efficient."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Complex Constraints**: Modify the problem to have more complex constraints on the tree. For example, consider a binary tree with constraints like \"The left subtree of a node contains only nodes with keys less than or equal to the node's key,\" and \"The right subtree of a node contains only nodes with keys greater than or equal to the node's key.\" How would you adapt the code to handle these constraints?\n",
-    "\n",
-    "2. **Handling Duplicates**: Modify the code to handle binary search trees that allow duplicate values. In a standard BST, each value is unique, but in this case, multiple nodes can have the same value. The tree should still be considered valid if it follows the BST property."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb b/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb
deleted file mode 100644
index 5c54c25..0000000
--- a/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 208. Implement Trie (Prefix Tree)\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Implement Trie problem on LeetCode, click here!](https://leetcode.com/problems/implement-trie-prefix-tree/)\n",
-    "\n",
-    "---\n",
-    "A **trie** (pronounced as \"try\") or **prefix tree** is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.\n",
-    "\n",
-    "Implement the Trie class:\n",
-    "\n",
-    "- `Trie()` Initializes the trie object.\n",
-    "- `void insert(String word)` Inserts the string `word` into the trie.\n",
-    "- `boolean search(String word)` Returns `true` if the string `word` is in the trie (i.e., was inserted before), and `false` otherwise.\n",
-    "- `boolean startsWith(String prefix)` Returns `true` if there is a previously inserted string `word` that has the prefix `prefix`, and `false` otherwise.\n",
-    "\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= word.length, prefix.length <= 2000`\n",
-    "- `word` and `prefix` consist only of lowercase English letters.\n",
-    "- At most $3 * 10^4$ calls in total will be made to `insert`, `search`, and `startsWith`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand is to implement a data structure known as a Trie (pronounced \"try\") or Prefix Tree. A Trie is a tree-like data structure that is used to efficiently store and retrieve a set of strings or words. It's particularly useful for tasks involving string manipulation, such as autocomplete and spell-checking. The problem statement defines the following operations for implementing the Trie class:\n",
-    "\n",
-    "1. `Trie()`: This is the constructor that initializes the Trie object.\n",
-    "\n",
-    "2. `void insert(String word)`: This method inserts the given string `word` into the Trie. It effectively adds the characters of the word to the Trie's structure, creating nodes for each character if they don't already exist. At the end of the word, a special flag is set to indicate that this node represents the end of a valid word.\n",
-    "\n",
-    "3. `boolean search(String word)`: This method checks if the given string `word` is present in the Trie. It starts from the root of the Trie and traverses the Trie by following the characters in the word. If it successfully traverses the Trie and reaches the end of the word, it returns `true`, indicating that the word exists in the Trie; otherwise, it returns `false`.\n",
-    "\n",
-    "4. `boolean startsWith(String prefix)`: This method checks if there is any previously inserted word in the Trie that has the given `prefix`. It's similar to the `search` method but does not require that the prefix be a complete word. If there is any word in the Trie that starts with the given prefix, it returns `true`; otherwise, it returns `false`.\n",
-    "\n",
-    "The problem also provides an example scenario where these operations are called, demonstrating the expected output for each operation.\n",
-    "\n",
-    "In summary, the goal is to create a data structure (Trie) that efficiently stores a set of strings and provides methods to insert new strings, search for complete words, and check if a given prefix exists in the stored set of strings.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TrieNode:\n",
-    "    def __init__(self):\n",
-    "        self.children = {}           # A dictionary to store child nodes\n",
-    "        self.is_end_of_word = False  # Indicates if a word ends at this node\n",
-    "\n",
-    "class Trie:\n",
-    "    def __init__(self):\n",
-    "        self.root = TrieNode()       # Initialize the Trie with a root node\n",
-    "\n",
-    "    def insert(self, word):\n",
-    "        node = self.root             # Start from the root\n",
-    "        for char in word:\n",
-    "            if char not in node.children:\n",
-    "                node.children[char] = TrieNode()  # Create a new node if the character is not present\n",
-    "            node = node.children[char]           # Move to the child node\n",
-    "        node.is_end_of_word = True                # Mark the end of the inserted word\n",
-    "\n",
-    "    def search(self, word):\n",
-    "        node = self.root             # Start from the root\n",
-    "        for char in word:\n",
-    "            if char not in node.children:\n",
-    "                return False         # If the character is not found, the word is not in the Trie\n",
-    "            node = node.children[char]  # Move to the child node\n",
-    "        return node.is_end_of_word    # Check if the node represents the end of a valid word\n",
-    "\n",
-    "    def startsWith(self, prefix):\n",
-    "        node = self.root             # Start from the root\n",
-    "        for char in prefix:\n",
-    "            if char not in node.children:\n",
-    "                return False         # If the character is not found, no word starts with the prefix\n",
-    "            node = node.children[char]  # Move to the child node\n",
-    "        return True                  # Prefix found in the Trie"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Let's explain how the code works step by step:\n",
-    "\n",
-    "1. `TrieNode` Class:\n",
-    "   - `TrieNode` is a class that represents nodes in the Trie. Each node has two attributes:\n",
-    "     - `children`: This is a dictionary that stores child nodes. Each key in the dictionary represents a character, and the corresponding value is the child node for that character.\n",
-    "     - `is_end_of_word`: This boolean flag indicates whether the node represents the end of a complete word. Initially, it's set to `False` for all nodes.\n",
-    "\n",
-    "2. `Trie` Class:\n",
-    "   - `Trie` is the main class that implements the Trie data structure. It has the following methods:\n",
-    "\n",
-    "   - `__init__(self)`: The constructor initializes the Trie object by creating a root node, which serves as the starting point for all Trie operations.\n",
-    "\n",
-    "   - `insert(self, word)`: This method inserts a string `word` into the Trie. It starts from the root node and iterates through each character in the word.\n",
-    "     - If a character is not present as a child node, it creates a new node for that character.\n",
-    "     - It then moves to the child node and continues the process until the entire word is inserted.\n",
-    "     - Finally, it sets the `is_end_of_word` flag to `True` for the last node to mark the end of the inserted word.\n",
-    "\n",
-    "   - `search(self, word)`: This method checks if a complete word (string) exists in the Trie. It starts from the root node and iterates through each character in the word.\n",
-    "     - If a character is not found as a child node, it immediately returns `False` because the word is not in the Trie.\n",
-    "     - It continues to move to the child node for each character.\n",
-    "     - After reaching the end of the word, it checks if the `is_end_of_word` flag is `True` for the last node to confirm the presence of the word.\n",
-    "\n",
-    "   - `startsWith(self, prefix)`: This method checks if there is any previously inserted word in the Trie that starts with a given `prefix`. It follows the same logic as the `search` method but does not require the entire word to be present.\n",
-    "     - If the prefix is found in the Trie, it returns `True`; otherwise, it returns `False`.\n",
-    "\n",
-    "3. Example Usage:\n",
-    "   - The code demonstrates how to create a Trie object, insert words (\"apple\" and \"app\"), and perform operations on the Trie:\n",
-    "     - It inserts the word \"apple\" into the Trie.\n",
-    "     - It checks if \"apple\" is in the Trie, which returns `True`.\n",
-    "     - It checks if \"app\" is in the Trie, which returns `False`.\n",
-    "     - It checks if there is any word in the Trie that starts with \"app,\" which returns `True`.\n",
-    "     - It inserts the word \"app\" into the Trie.\n",
-    "     - It checks if \"app\" is in the Trie again, which now returns `True`.\n",
-    "\n",
-    "In summary, the code efficiently implements a Trie data structure with the ability to insert words, search for complete words, and check for the existence of words with a given prefix. The Trie is organized as a tree of nodes, with each node representing a character in the words being stored."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n",
-      "False\n",
-      "True\n",
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "\n",
-    "trie = Trie()\n",
-    "trie.insert(\"apple\")\n",
-    "print(trie.search(\"apple\"))   # Output: True\n",
-    "print(trie.search(\"app\"))     # Output: False\n",
-    "print(trie.startsWith(\"app\")) # Output: True\n",
-    "trie.insert(\"app\")\n",
-    "print(trie.search(\"app\"))     # Output: True"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the Trie implementation:\n",
-    "\n",
-    "1. Time Complexity:\n",
-    "   - Insertion (`insert` method): The time complexity of inserting a word into the Trie is O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to insert it into the Trie.\n",
-    "   - Search (`search` method): The time complexity of searching for a word in the Trie is also O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to determine if it exists in the Trie.\n",
-    "   - Starts with (`startsWith` method): The time complexity of checking if a prefix exists in the Trie is O(p), where p is the length of the prefix. In the worst case, you may need to traverse the entire prefix to determine its presence.\n",
-    "\n",
-    "   Therefore, for each operation (insertion, search, or starts with), the time complexity is O(m) or O(p), where m is the length of the word being operated on, and p is the length of the prefix.\n",
-    "\n",
-    "2. Space Complexity:\n",
-    "   - The space complexity of the Trie is determined by the number of nodes and characters stored in it. In the worst case, where none of the inserted words share common prefixes, the space complexity is O(N), where N is the total number of characters across all inserted words.\n",
-    "   - Each node in the Trie represents a character, and the number of nodes is directly related to the total number of characters.\n",
-    "\n",
-    "   In practice, the space complexity can be less than O(N) because common prefixes are shared among words in the Trie, leading to space optimization.\n",
-    "\n",
-    "In summary:\n",
-    "- Time complexity for each operation (insertion, search, starts with) is O(m) or O(p), where m is the length of the word and p is the length of the prefix.\n",
-    "- Space complexity is O(N) in the worst case, where N is the total number of characters across all inserted words. However, space optimization occurs due to common prefix sharing in practice, potentially reducing the actual space used."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Count Prefixes**:\n",
-    "   Implement a method to count the number of words in the Trie that have a specific prefix. This exercise requires you to maintain additional data in the Trie nodes to keep track of the count.\n",
-    "\n",
-    "2. **Auto-Complete Suggestions**:\n",
-    "   Implement an auto-complete feature using the Trie. Given a prefix, the program should return a list of suggested words based on the words previously inserted into the Trie."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
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-    "name": "ipython",
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\ No newline at end of file
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@@ -1,271 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 211. Design Add and Search Words Data Structure\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Design Add and Search Words Data Structure problem on LeetCode, click here!](https://leetcode.com/problems/design-add-and-search-words-data-structure/)\n",
-    "\n",
-    "---\n",
-    "Design a data structure that supports adding new words and finding if a string matches any previously added string.\n",
-    "\n",
-    "Implement the `WordDictionary` class:\n",
-    "\n",
-    "- `WordDictionary()` Initializes the object.\n",
-    "- `void addWord(word)` Adds `word` to the data structure, it can be matched later.\n",
-    "- `bool search(word)` Returns `true` if there is any string in the data structure that matches `word` or `false` otherwise. `word` may contain dots `'.'` where dots can be matched with any letter.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= word.length <= 25`\n",
-    "- `word` in `addWord` consists of lowercase English letters.\n",
-    "- `word` in `search` consist of `'.'` or lowercase English letters.\n",
-    "- There will be at most `2` dots in `word` for `search` queries.\n",
-    "- At most $10^4$ calls will be made to `addWord` and `search`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to design a data structure called `WordDictionary` that allows you to add words and search for words efficiently. The key feature of this data structure is that it should support wildcard search, where a dot ('.') can match any letter.\n",
-    "\n",
-    "Here are the specific requirements and explanations for the problem:\n",
-    "\n",
-    "1. Initialize the `WordDictionary` object using the constructor `WordDictionary()`.\n",
-    "\n",
-    "2. The `addWord(word)` method allows you to add a word to the data structure. Once added, you should be able to search for this word later. The words consist of lowercase English letters, and each word has a length between 1 and 25 characters.\n",
-    "\n",
-    "3. The `search(word)` method allows you to search for a word in the data structure. This method returns `True` if there is any string in the data structure that matches the given word or `False` otherwise. The word you are searching for may contain dots ('.'), where each dot can match any letter.\n",
-    "\n",
-    "4. The wildcard character ('.') in the `search` method allows for partial or fuzzy searching. For example, if the word dictionary contains the words \"bad,\" \"dad,\" and \"mad,\" then:\n",
-    "   - Searching for \"pad\" should return `False` because there is no exact match.\n",
-    "   - Searching for \"bad\" should return `True` because \"bad\" is in the dictionary.\n",
-    "   - Searching for \".ad\" should return `True` because it can match \"bad,\" \"dad,\" or \"mad.\"\n",
-    "   - Searching for \"b..\" should return `True` because it can match \"bad,\" \"bed,\" \"bet,\" etc.\n",
-    "\n",
-    "5. The problem specifies that there will be at most two dots ('.') in search queries, and there will be at most 10^4 calls to both `addWord` and `search`.\n",
-    "\n",
-    "In summary, you need to implement the `WordDictionary` class that efficiently supports adding words and searching for words, where the search can include wildcard characters ('.') that match any letter.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TrieNode:\n",
-    "    def __init__(self):\n",
-    "        # Initialize a TrieNode with children and a flag to indicate the end of a word.\n",
-    "        self.children = {}\n",
-    "        self.is_end_of_word = False\n",
-    "\n",
-    "class WordDictionary:\n",
-    "\n",
-    "    def __init__(self):\n",
-    "        # Initialize the WordDictionary with a root node.\n",
-    "        self.root = TrieNode()\n",
-    "\n",
-    "    def addWord(self, word: str) -> None:\n",
-    "        node = self.root\n",
-    "        # Iterate through the characters in the word.\n",
-    "        for char in word:\n",
-    "            if char not in node.children:\n",
-    "                # Create a new node for the character if it doesn't exist.\n",
-    "                node.children[char] = TrieNode()\n",
-    "            # Move to the next node.\n",
-    "            node = node.children[char]\n",
-    "        # Mark the end of the word by setting the flag to True.\n",
-    "        node.is_end_of_word = True\n",
-    "\n",
-    "    def search_helper(self, node: TrieNode, word: str) -> bool:\n",
-    "        if not word:\n",
-    "            # If there are no characters left in the word, check if we reached the end of a word in the trie.\n",
-    "            return node.is_end_of_word\n",
-    "        \n",
-    "        char = word[0]\n",
-    "        if char == '.':\n",
-    "            # If the character is a dot, explore all possible child nodes.\n",
-    "            for child in node.children.values():\n",
-    "                if self.search_helper(child, word[1:]):\n",
-    "                    return True\n",
-    "        elif char in node.children:\n",
-    "            # If the character is in the children of the current node, move to the next node.\n",
-    "            return self.search_helper(node.children[char], word[1:])\n",
-    "        \n",
-    "        # If the character is not found and is not a dot, the word is not in the trie.\n",
-    "        return False\n",
-    "\n",
-    "    def search(self, word: str) -> bool:\n",
-    "        # Start the search from the root of the trie.\n",
-    "        return self.search_helper(self.root, word)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "This code defines a `WordDictionary` class that uses a Trie data structure to efficiently store and search for words. Here's a step-by-step explanation:\n",
-    "\n",
-    "1. `TrieNode` class: This class represents nodes in the Trie data structure. Each node has two attributes: `children` (a dictionary to store child nodes) and `is_end_of_word` (a flag to indicate whether the node marks the end of a word).\n",
-    "\n",
-    "2. `WordDictionary` class: This class initializes with an empty Trie by creating a root node.\n",
-    "\n",
-    "3. `addWord` method: This method adds a word to the Trie. It iterates through each character in the word and creates Trie nodes as necessary. It sets the `is_end_of_word` flag to `True` for the final character of the word to mark the end of the word.\n",
-    "\n",
-    "4. `search_helper` method: This is a recursive helper function for searching words in the Trie. It takes a Trie node (`node`) and a word to search (`word`) as input. If there are no characters left in the word (`not word`), it checks if the current node marks the end of a word. If it does, it returns `True`.\n",
-    "\n",
-    "5. If the first character of the word is a dot ('.'), the function explores all possible child nodes by iterating through `node.children.values()` and recursively calls `search_helper` for each child with the remaining part of the word (`word[1:]`).\n",
-    "\n",
-    "6. If the first character of the word is not a dot and is found in the children of the current node, the function recursively moves to the next node by calling `search_helper` on the child node and the remaining part of the word.\n",
-    "\n",
-    "7. If the character is not found in the children and is not a dot, the word is not in the Trie, so the function returns `False`.\n",
-    "\n",
-    "8. `search` method: This is the public method for searching words. It initiates the search process by calling `search_helper` starting from the root node of the Trie.\n",
-    "\n",
-    "Overall, this code efficiently implements a Trie-based data structure that allows adding and searching for words, including support for wildcard characters represented by dots ('.')."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n",
-      "True\n",
-      "True\n",
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "\n",
-    "wordDictionary = WordDictionary()\n",
-    "wordDictionary.addWord(\"bad\")\n",
-    "wordDictionary.addWord(\"dad\")\n",
-    "wordDictionary.addWord(\"mad\")\n",
-    "print(wordDictionary.search(\"pad\"))  # Output: False\n",
-    "print(wordDictionary.search(\"bad\"))  # Output: True\n",
-    "print(wordDictionary.search(\".ad\"))  # Output: True\n",
-    "print(wordDictionary.search(\"b..\"))  # Output: True"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `WordDictionary` class methods:\n",
-    "\n",
-    "1. `addWord` Method:\n",
-    "\n",
-    "   - Time Complexity: O(L)\n",
-    "     - L is the length of the word being added.\n",
-    "     - In the worst case, we need to iterate through each character in the word and create nodes in the Trie. This takes O(L) time.\n",
-    "\n",
-    "   - Space Complexity: O(L)\n",
-    "     - The space complexity for storing the word in the Trie is also O(L) because we create nodes for each character in the word.\n",
-    "\n",
-    "2. `search_helper` Method (called by `search`):\n",
-    "\n",
-    "   - Time Complexity: O(2^M * L)\n",
-    "     - M is the maximum number of dots ('.') in the search word.\n",
-    "     - In the worst case, when there are M dots in the search word, we may explore all possible child nodes at each dot. This results in a branching factor of 26 (for lowercase letters) at each dot.\n",
-    "     - So, the time complexity is exponential in the number of dots and linear in the length of the word being searched (L).\n",
-    "\n",
-    "   - Space Complexity: O(L)\n",
-    "     - The space complexity for the recursive call stack is O(L) in the worst case because we may have to recurse to a depth equal to the length of the search word.\n",
-    "\n",
-    "3. `search` Method:\n",
-    "\n",
-    "   - Time Complexity: O(2^M * L)\n",
-    "     - The `search` method calls the `search_helper` method, which has the same time complexity explained above.\n",
-    "\n",
-    "   - Space Complexity: O(L)\n",
-    "     - The space complexity for the recursive call stack is O(L) in the worst case, as explained earlier.\n",
-    "\n",
-    "Overall, the time complexity of the `search` method is mainly affected by the number of dots ('.') in the search word. In the worst case, when there are multiple dots and many possibilities to explore at each dot, the time complexity can be exponential. However, for most practical cases, it performs reasonably well.\n",
-    "\n",
-    "The space complexity primarily depends on the space required to store the words in the Trie. It is proportional to the total number of characters in all added words and is linear with respect to the length of the words added.\n",
-    "\n",
-    "In summary:\n",
-    "\n",
-    "- Time Complexity for `addWord`: O(L)\n",
-    "- Time Complexity for `search`: O(2^M * L)\n",
-    "- Space Complexity: O(L) for storing words in the Trie, O(L) for the recursive call stack during search.\n",
-    "\n",
-    "Note: The space complexity analysis assumes that the dictionary contains a reasonable number of words and does not account for the overhead of Python objects or system-level memory allocation."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Dictionary Auto-Completion**: Extend the `WordDictionary` class to provide auto-completion suggestions based on the prefix of a word. Implement a method that returns a list of words that match the given prefix.\n",
-    "\n",
-    "2. **Support Word Deletion**: Extend the `WordDictionary` class to support word deletion. Add a method `deleteWord(word)` that removes a word from the data structure. Ensure that the search operation still works correctly after deletion."
-   ]
-  }
- ],
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-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb b/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb
deleted file mode 100644
index e582b9d..0000000
--- a/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb	
+++ /dev/null
@@ -1,310 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 212. Word Search II\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Word Search II problem on LeetCode, click here!](https://leetcode.com/problems/word-search-ii/)\n",
-    "\n",
-    "---\n",
-    "Given an `m x n` `board` of characters and a list of strings `words`, the task is to return *all words on the board*.\n",
-    "\n",
-    "Each word must be constructed from letters of sequentially **adjacent cells**, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.\n",
-    "\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == board.length`\n",
-    "- `n == board[i].length`\n",
-    "- `1 <= m, n <= 12`\n",
-    "- `board[i][j]` is a lowercase English letter.\n",
-    "- 1 <= `words.length` <= $3 * 10^4$\n",
-    "- `1 <= words[i].length <= 10`\n",
-    "- `words[i]` consists of lowercase English letters.\n",
-    "- All the strings of `words` are unique."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "To solve the \"Word Search II\" problem, you can use a Trie data structure and perform a depth-first search (DFS) on the board. Here's a step-by-step approach to solve the problem:\n",
-    "\n",
-    "1. Build a Trie from the given list of words:\n",
-    "   - Create a Trie data structure to store all the words from the input list.\n",
-    "   - For each word in the input list, insert it into the Trie, character by character. Make sure to mark the end of each word in the Trie.\n",
-    "\n",
-    "2. Perform DFS on the board:\n",
-    "   - Iterate through each cell (i, j) on the board.\n",
-    "   - For each cell, start a DFS traversal from that cell to search for words.\n",
-    "   - During the DFS traversal, maintain a current path string.\n",
-    "   - At each cell, check if the current path string, concatenated with the character in the cell, matches any word prefix in the Trie. If it does, continue the DFS.\n",
-    "   - If the current path string matches a word in the Trie, add it to the result.\n",
-    "\n",
-    "3. Implement DFS with backtracking:\n",
-    "   - During the DFS traversal, you will explore neighboring cells (horizontally and vertically) if they are within bounds and haven't been visited before.\n",
-    "   - To prevent revisiting the same cell within the same word, mark the cell as visited (e.g., change its character to a special character like '#') before exploring it and restore its original character after the DFS traversal.\n",
-    "\n",
-    "4. Return the result:\n",
-    "   - After completing the DFS traversal for all cells on the board, you will have collected all the valid words in the result set.\n",
-    "   - Convert the result set to a list and return it as the final output.\n",
-    "\n",
-    "This approach efficiently finds all words on the board by utilizing the Trie data structure to prune unnecessary DFS paths and avoiding duplicate visits to the same cell within the same word. It ensures that each word is constructed from letters of sequentially adjacent cells without using the same letter cell more than once in a word.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Define a TrieNode class to represent nodes in the Trie.\n",
-    "class TrieNode:\n",
-    "    def __init__(self):\n",
-    "        self.children = {}  # Dictionary to store child nodes.\n",
-    "        self.is_end_of_word = False  # Flag to mark the end of a word.\n",
-    "\n",
-    "# Main function to find words on the board.\n",
-    "def findWords(board, words):\n",
-    "    # DFS function to search for words starting from a given cell.\n",
-    "    def dfs(node, i, j, path):\n",
-    "        char = board[i][j]  # Get the character at the current cell.\n",
-    "        curr_node = node.children.get(char)  # Find the corresponding Trie node.\n",
-    "\n",
-    "        if not curr_node:\n",
-    "            return  # If no matching Trie node, stop the search.\n",
-    "\n",
-    "        path += char  # Add the character to the current path.\n",
-    "\n",
-    "        if curr_node.is_end_of_word:\n",
-    "            result.add(path)  # If the path matches a word, add it to the result.\n",
-    "\n",
-    "        temp, board[i][j] = board[i][j], \"#\"  # Mark the cell as visited.\n",
-    "\n",
-    "        # Explore neighboring cells (up, down, left, right).\n",
-    "        if i > 0:\n",
-    "            dfs(curr_node, i - 1, j, path)\n",
-    "        if i < m - 1:\n",
-    "            dfs(curr_node, i + 1, j, path)\n",
-    "        if j > 0:\n",
-    "            dfs(curr_node, i, j - 1, path)\n",
-    "        if j < n - 1:\n",
-    "            dfs(curr_node, i, j + 1, path)\n",
-    "\n",
-    "        board[i][j] = temp  # Restore the cell's original character.\n",
-    "\n",
-    "    # Function to build a Trie from the list of words.\n",
-    "    def buildTrie():\n",
-    "        root = TrieNode()  # Create a root node.\n",
-    "        for word in words:\n",
-    "            node = root\n",
-    "            for char in word:\n",
-    "                if char not in node.children:\n",
-    "                    node.children[char] = TrieNode()  # Create a new node if needed.\n",
-    "                node = node.children[char]\n",
-    "            node.is_end_of_word = True  # Mark the end of the word.\n",
-    "        return root\n",
-    "\n",
-    "    m, n = len(board), len(board[0])  # Get the dimensions of the board.\n",
-    "    root = buildTrie()  # Build the Trie from the list of words.\n",
-    "    result = set()  # Use a set to store unique words found on the board.\n",
-    "\n",
-    "    # Iterate through each cell on the board and start DFS from there.\n",
-    "    for i in range(m):\n",
-    "        for j in range(n):\n",
-    "            dfs(root, i, j, \"\")  # Start DFS from each cell on the board.\n",
-    "\n",
-    "    return list(result)  # Convert the set to a list and return the result."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code solves the \"Word Search II\" problem, where you have a grid of characters and a list of words. The goal is to find all words from the list that can be constructed by traversing adjacent cells (horizontally or vertically) in the grid, without using the same cell more than once for a single word.\n",
-    "\n",
-    "The code does the following:\n",
-    "\n",
-    "1. Defines a `TrieNode` class to represent nodes in a Trie data structure. Each node has children (other characters in the word) and a flag to mark the end of a word.\n",
-    "\n",
-    "2. Defines the `findWords` function, which takes the grid (`board`) and a list of words (`words`) as input.\n",
-    "\n",
-    "3. Inside the `findWords` function, there is a `dfs` (depth-first search) function. This function is used to explore the grid starting from a specific cell. It checks if the current path matches any word in the Trie and adds it to the result if found.\n",
-    "\n",
-    "4. The code also includes a `buildTrie` function that constructs a Trie data structure from the list of words. It iterates through each word, creating nodes for each character in the Trie and marking the end of words.\n",
-    "\n",
-    "5. The dimensions of the grid are obtained (m x n).\n",
-    "\n",
-    "6. A Trie is built using the `buildTrie` function.\n",
-    "\n",
-    "7. A set called `result` is used to store unique words found in the grid.\n",
-    "\n",
-    "8. The code iterates through each cell on the grid (using nested loops) and starts a DFS search from each cell.\n",
-    "\n",
-    "9. During the DFS, it explores neighboring cells (up, down, left, right) if they haven't been visited before and if the current path matches a prefix in the Trie.\n",
-    "\n",
-    "10. When a word is found during the DFS traversal, it is added to the `result` set.\n",
-    "\n",
-    "11. The grid cell is marked as visited during the DFS by changing its character to \"#\", and it is restored to its original character after the DFS traversal.\n",
-    "\n",
-    "12. Finally, the set of unique words found is converted to a list and returned as the output.\n",
-    "\n",
-    "The code effectively uses a Trie data structure to optimize the search and ensures that each word is constructed from adjacent cells without reusing the same cell for the same word."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "['eat', 'oath']\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "board1 = [[\"o\",\"a\",\"a\",\"n\"],[\"e\",\"t\",\"a\",\"e\"],[\"i\",\"h\",\"k\",\"r\"],[\"i\",\"f\",\"l\",\"v\"]]\n",
-    "words1 = [\"oath\",\"pea\",\"eat\",\"rain\"]\n",
-    "print(findWords(board1, words1))  # Output: [\"eat\", \"oath\"]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "c748b879-a2d8-4298-a42c-c7f61f9c7fc6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "\n",
-    "board2 = [[\"a\",\"b\"],[\"c\",\"d\"]]\n",
-    "words2 = [\"abcb\"]\n",
-    "print(findWords(board2, words2))  # Output: []"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. **Building the Trie (`buildTrie` function):** \n",
-    "   - Suppose there are `N` words in the input list, and the average length of a word is `L`.\n",
-    "   - Building the Trie takes O(N * L) time, as we iterate through each word and insert each character into the Trie.\n",
-    "\n",
-    "2. **DFS Traversal (`dfs` function):** \n",
-    "   - In the worst case, the DFS function explores all possible paths in the grid.\n",
-    "   - The maximum number of recursive calls for each cell is 4 (up, down, left, right).\n",
-    "   - The maximum depth of the DFS is limited by the length of the longest word in the Trie, which is at most `L`.\n",
-    "   - Therefore, the DFS traversal for each cell takes $O(4^L)$ time.\n",
-    "\n",
-    "3. **Iterating Through the Grid (`findWords` function):**\n",
-    "   - In the worst case, we iterate through all `m` rows and `n` columns in the grid.\n",
-    "   - For each cell, we start a DFS traversal.\n",
-    "   - Therefore, iterating through the entire grid takes $O(m * n * 4^L)$ time.\n",
-    "\n",
-    "Overall, the time complexity of the code is $O(N * L + m * n * 4^L)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. **Trie Data Structure:**\n",
-    "   - The space required to store the Trie depends on the number of unique characters in all the words.\n",
-    "   - In the worst case, where all words are unique and have no common prefixes, the space complexity of the Trie is O(N * L).\n",
-    "   - In practice, it may be less if there are common prefixes among words.\n",
-    "\n",
-    "2. **DFS Stack (Recursive Calls):**\n",
-    "   - The depth of the recursive call stack during DFS is at most `L`, where `L` is the length of the longest word in the Trie.\n",
-    "   - Therefore, the space complexity for the recursive call stack is O(L).\n",
-    "\n",
-    "3. **Result Set (`result` set):**\n",
-    "   - The space used to store the result set depends on the number of valid words found in the grid.\n",
-    "   - In the worst case, when all words are found, the space complexity of the result set is O(N * L).\n",
-    "\n",
-    "Overall, the space complexity of the code is O(N * L) for the Trie and O(L) for the recursive call stack.\n",
-    "\n",
-    "In summary, the time complexity is dominated by the DFS traversal and is influenced by the number of words, their lengths, and the size of the grid. The space complexity is mainly determined by the Trie structure and the recursive call stack depth during DFS."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Word Search Paths:**\n",
-    "   Extend the basic word search exercise to find all possible paths (sequences of cells) that spell out a given word on the board. Return a list of paths if the word can be formed, otherwise an empty list.\n",
-    "\n",
-    "\n",
-    "2. **Reverse Word Search:**\n",
-    "   Modify the code to search for words in reverse order on the board. Given a 2D grid of characters and a list of words, find all words from the list that can be formed by traversing the board in reverse (right to left, bottom to top, etc.)."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb b/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb
deleted file mode 100644
index f82ad98..0000000
--- a/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb	
+++ /dev/null
@@ -1,248 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 1046. Last Stone Weight\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Last Stone Weight problem on LeetCode, click here!](https://leetcode.com/problems/last-stone-weight/)\n",
-    "\n",
-    "---\n",
-    "You are given an array of integers `stones` where `stones[i]` is the weight of the `i`th stone.\n",
-    "\n",
-    "We are playing a game with the stones. On each turn, you choose the **heaviest two stones** and smash them together. Suppose the heaviest two stones have weights `x` and `y` with `x <= y`. The result of this smash is:\n",
-    "\n",
-    "1. If `x == y`, both stones are destroyed.\n",
-    "2. If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - x`.\n",
-    "\n",
-    "At the end of the game, there is **at most one** stone left.\n",
-    "\n",
-    "Return *the weight of the last remaining stone*. If there are no stones left, return `0`.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= stones.length <= 30`\n",
-    "- `1 <= stones[i] <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The \"Last Stone Weight\" problem involves a game played with a collection of stones, each stone having a certain weight. The goal is to find the weight of the last remaining stone after applying a specific set of rules.\n",
-    "\n",
-    "Here's how the game works:\n",
-    "\n",
-    "1. You start with an array of stones, where each stone is represented by its weight. The weights of the stones are given in the input as an array.\n",
-    "\n",
-    "2. In each turn of the game, you select the two heaviest stones from the remaining stones. If there is only one stone left, you've found the answer.\n",
-    "\n",
-    "3. If there are two stones with weights `x` and `y`, where `x` is less than or equal to `y`, the following happens:\n",
-    "   - If `x` is equal to `y`, both stones are completely destroyed, and they are removed from the array of stones.\n",
-    "   - If `x` is not equal to `y`, the stone with weight `x` is destroyed, and the stone with weight `y` now has a new weight of `y - x`. The stone with weight `x` is removed from the array, and the modified stone with weight `y - x` remains in the array.\n",
-    "\n",
-    "4. The game continues with the modified array of stones until there is at most one stone left.\n",
-    "\n",
-    "5. The goal is to find the weight of the last remaining stone, or if there are no stones left, return `0`.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import heapq\n",
-    "\n",
-    "def lastStoneWeight(stones):\n",
-    "    # Convert the input list to a max heap (negate the values to simulate a max heap)\n",
-    "    max_heap = [-stone for stone in stones]\n",
-    "    heapq.heapify(max_heap)\n",
-    "\n",
-    "    # Continue as long as there are more than 1 stone in the heap\n",
-    "    while len(max_heap) > 1:\n",
-    "        # Get the two heaviest stones\n",
-    "        y = -heapq.heappop(max_heap)  # Get the heaviest stone and negate it back to positive\n",
-    "        x = -heapq.heappop(max_heap)  # Get the second heaviest stone and negate it back to positive\n",
-    "\n",
-    "        # Calculate the new stone's weight after smashing\n",
-    "        if x != y:\n",
-    "            new_stone = y - x\n",
-    "            heapq.heappush(max_heap, -new_stone)  # Negate the value and push it back to the heap\n",
-    "\n",
-    "    # If there is one stone left, return its weight (negate it back to positive)\n",
-    "    if max_heap:\n",
-    "        return -max_heap[0]\n",
-    "    else:\n",
-    "        return 0"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a Python function called `lastStoneWeight` that takes a list of stone weights as input. It aims to find the weight of the last remaining stone after playing the stone smashing game described in the problem statement.\n",
-    "\n",
-    "Here's a step-by-step explanation of the code:\n",
-    "\n",
-    "1. We import the `heapq` library, which allows us to create and manipulate a heap (priority queue).\n",
-    "\n",
-    "2. Inside the `lastStoneWeight` function:\n",
-    "   - We create a max heap (negating stone weights to simulate a max heap) using the `heapq.heapify` function. This heap will keep track of the heaviest stones.\n",
-    "   \n",
-    "   - We enter a loop that continues as long as there are more than 1 stone in the heap.\n",
-    "   \n",
-    "   - Inside the loop, we:\n",
-    "     - Retrieve and negate the heaviest stone (`y`) from the heap.\n",
-    "     - Retrieve and negate the second heaviest stone (`x`) from the heap.\n",
-    "   \n",
-    "     - Calculate the new stone's weight after smashing (`y - x`), and negate it back to simulate a max heap.\n",
-    "     - Add the new stone back to the heap using `heapq.heappush`.\n",
-    "   \n",
-    "3. After the loop, if there is one stone left in the heap, we retrieve and negate it to get its actual weight, and return it as the result.\n",
-    "   \n",
-    "4. If there are no stones left (the heap is empty), we return `0`, indicating that no stones remain.\n",
-    "\n",
-    "5. Finally, we provide two test cases to demonstrate the function's usage and correctness.\n",
-    "\n",
-    "In summary, the code efficiently implements the stone smashing game by maintaining a max heap, selecting the heaviest stones in each turn, smashing them, and updating the heap until there is at most one stone left. It then returns the weight of the last remaining stone or 0 if there are no stones left."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "stones1 = [2, 7, 4, 1, 8, 1]\n",
-    "print(lastStoneWeight(stones1))  # Output: 1"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "b7e79514-4821-48d0-bb9d-0f2de8150945",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2: \n",
-    "\n",
-    "stones2 = [1]\n",
-    "print(lastStoneWeight(stones2))  # Output: 1"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided Python code for the \"Last Stone Weight\" problem:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. The `heapq.heapify` operation to create the max heap from the input stone weights has a time complexity of O(n), where n is the number of stones.\n",
-    "\n",
-    "2. The loop that continues until there are more than 1 stone in the heap performs the following operations in each iteration:\n",
-    "   - Retrieving and negating the heaviest stone (`y`) from the heap, which takes O(log n) time.\n",
-    "   - Retrieving and negating the second heaviest stone (`x`) from the heap, also taking O(log n) time.\n",
-    "   - Calculating the new stone's weight and pushing it back into the heap using `heapq.heappush`, which takes O(log n) time.\n",
-    "\n",
-    "3. In the worst case, the loop runs until there is only one stone left, which requires approximately (n-1) iterations.\n",
-    "\n",
-    "Overall, the dominant time complexity is O(n) for the initial heap creation, and within the loop, each iteration takes O(log n) time. Therefore, the overall time complexity of the algorithm is O(n log n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The space complexity is primarily determined by the space required for the max heap. In the worst case, this heap can contain all the stones, so the space complexity for the heap is O(n).\n",
-    "\n",
-    "2. The rest of the variables used in the function (e.g., `x`, `y`, `new_stone`) require only constant space, so they do not significantly contribute to the space complexity.\n",
-    "\n",
-    "Therefore, the overall space complexity of the algorithm is O(n) due to the space used for the max heap.\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: $O(n\\ log\\ n)$\n",
-    "- Space Complexity: $O(n)$\n",
-    "\n",
-    "The algorithm is efficient enough to handle the problem's constraints, as the worst-case time and space complexities are both linearithmic in terms of the number of stones."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Kth Last Stone Weight**: Modify the problem to find the weight of the Kth last remaining stone, where K is an integer input. Your function should work efficiently for large values of K.\n",
-    "\n",
-    "\n",
-    "2. **Multiple Stones Smash**: Modify the problem so that instead of smashing two stones at a time, you can smash up to K stones at each turn, where K is an integer input. Determine the weight of the last remaining stone in this variation."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb b/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb
deleted file mode 100644
index 6c43a8d..0000000
--- a/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb	
+++ /dev/null
@@ -1,238 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 295. Find Median from Data Stream\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Find Median from Data Stream problem on LeetCode, click here!](https://leetcode.com/problems/find-median-from-data-stream/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.\n",
-    "\n",
-    "Implement the MedianFinder class:\n",
-    "\n",
-    "- `MedianFinder()` initializes the `MedianFinder` object.\n",
-    "- `void addNum(int num)` adds the integer `num` from the data stream to the data structure.\n",
-    "- `double findMedian()` returns the median of all elements so far. Answers within $10^{-5}$ of the actual answer will be accepted.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- $-10^{5}$ <= `num` <= $10^{5}$\n",
-    "- There will be at least one element in the data structure before calling `findMedian`.\n",
-    "- At most $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-    "\n",
-    "**Follow-up:**\n",
-    "\n",
-    "1. If all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?\n",
-    "2. If `99%` of all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem, \"Find Median from Data Stream,\" deals with efficiently computing the median of a sequence of numbers as they are incrementally provided. The median is the middle value in an ordered list of numbers. When the list has an even number of elements, the median is the average of the two middle values.\n",
-    "\n",
-    "Here's a breakdown of the problem:\n",
-    "\n",
-    "1. **Initialization**: You are asked to implement a `MedianFinder` class that supports the following operations:\n",
-    "\n",
-    "    - `MedianFinder()`: Initializes the MedianFinder object.\n",
-    "    - `addNum(int num)`: Adds an integer `num` from the data stream to the data structure.\n",
-    "    - `findMedian()`: Returns the median of all elements added to the data structure.\n",
-    "\n",
-    "2. **Median Calculation**: The `findMedian` operation should efficiently compute the median, and the result should have a precision of at least $10^{-5}$ (i.e., answers within this range will be considered correct).\n",
-    "\n",
-    "3. **Examples**:\n",
-    "   \n",
-    "   - If you add numbers [1, 2], the median should be 1.5 because (1 + 2) / 2 = 1.5.\n",
-    "   - If you add numbers [1, 2, 3], the median should be 2 because it's the middle value.\n",
-    "   \n",
-    "4. **Constraints**:\n",
-    "\n",
-    "   - The integers provided in the data stream are in the range from $-10^5$ to $10^5$.\n",
-    "   - There will be at least one element in the data structure before calling `findMedian`.\n",
-    "   - At most, $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-    "\n",
-    "The problem can be solved by using two priority queues (heaps): one max-heap to store the smaller half of the numbers and one min-heap to store the larger half of the numbers. The max-heap ensures that the largest number in the smaller half is at the top, and the min-heap ensures that the smallest number in the larger half is at the top. These heaps are balanced to efficiently find the median when requested.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import heapq\n",
-    "\n",
-    "class MedianFinder:\n",
-    "\n",
-    "    def __init__(self):\n",
-    "        # Initialize two heaps: a max-heap (small_half) for the smaller half of numbers,\n",
-    "        # and a min-heap (large_half) for the larger half of numbers.\n",
-    "        self.small_half = []  # Max-heap for the smaller half of the numbers\n",
-    "        self.large_half = []  # Min-heap for the larger half of the numbers\n",
-    "\n",
-    "    def addNum(self, num: int) -> None:\n",
-    "        # Add the number to the appropriate heap based on its value\n",
-    "        if not self.small_half or num <= -self.small_half[0]:\n",
-    "            # If the number is less than or equal to the current maximum in the smaller half,\n",
-    "            # add it to the smaller half (max-heap)\n",
-    "            heapq.heappush(self.small_half, -num)\n",
-    "        else:\n",
-    "            # Otherwise, add it to the larger half (min-heap)\n",
-    "            heapq.heappush(self.large_half, num)\n",
-    "\n",
-    "        # Balance the heaps if necessary to ensure the size difference is at most 1\n",
-    "        if len(self.small_half) > len(self.large_half) + 1:\n",
-    "            # If the size of the smaller half is more than one greater than the size of the larger half,\n",
-    "            # move the maximum from the smaller half to the larger half to balance them.\n",
-    "            heapq.heappush(self.large_half, -heapq.heappop(self.small_half))\n",
-    "        elif len(self.large_half) > len(self.small_half):\n",
-    "            # If the size of the larger half is greater than the size of the smaller half,\n",
-    "            # move the minimum from the larger half to the smaller half to balance them.\n",
-    "            heapq.heappush(self.small_half, -heapq.heappop(self.large_half))\n",
-    "\n",
-    "    def findMedian(self) -> float:\n",
-    "        if len(self.small_half) == len(self.large_half):\n",
-    "            # If both halves have the same size, there's an even number of elements,\n",
-    "            # so the median is the average of the top elements in both heaps.\n",
-    "            return (-self.small_half[0] + self.large_half[0]) / 2.0\n",
-    "        else:\n",
-    "            # If the smaller half has more elements, it contains the median (odd number of elements).\n",
-    "            return -self.small_half[0]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a `MedianFinder` class for efficiently finding the median of a stream of numbers. It uses two heaps: a max-heap (`small_half`) to store the smaller half of the numbers and a min-heap (`large_half`) to store the larger half. Here's how the code works:\n",
-    "\n",
-    "1. The `__init__` method initializes the two heaps.\n",
-    "\n",
-    "2. The `addNum` method adds a number to the appropriate heap, ensuring that the heaps remain balanced. It does this by comparing the number to the current maximum in the smaller half and adding it to the appropriate heap. Then, it checks the sizes of the two heaps and rebalances them if necessary.\n",
-    "\n",
-    "3. The `findMedian` method calculates and returns the median. If both halves have the same size (even number of elements), it computes the average of the tops of both heaps. If the smaller half has more elements (odd number of elements), it returns the top of the max-heap, which is the median.\n",
-    "\n",
-    "In summary, this code efficiently maintains two heaps to divide the elements into smaller and larger halves, allowing for quick median retrieval. It offers a time complexity of O(log N) for adding elements and O(1) for finding the median, where N is the total number of elements processed. The space complexity is O(1) as the space used by the two heaps is independent of the input size."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1.5\n",
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "medianFinder = MedianFinder()\n",
-    "medianFinder.addNum(1)\n",
-    "medianFinder.addNum(2)\n",
-    "print(medianFinder.findMedian())  # Output: 1.5\n",
-    "medianFinder.addNum(3)\n",
-    "print(medianFinder.findMedian())  # Output: 2.0\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `MedianFinder` class:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. `MedianFinder()` Initialization: This operation has a time complexity of O(1). It simply initializes the two heaps.\n",
-    "\n",
-    "2. `addNum(int num)`: The time complexity for adding a number is O(log N), where N is the total number of elements processed so far. This is because, in the worst case, you might need to perform logarithmic operations on the heaps to maintain their balance.\n",
-    "\n",
-    "3. `findMedian()`: Finding the median has a time complexity of O(1). It simply involves accessing the tops of the two heaps, which takes constant time.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. `MedianFinder()` Initialization: The space complexity for initialization is O(1). It involves creating two empty heaps, which do not depend on the size of the data stream.\n",
-    "\n",
-    "2. `addNum(int num)`: The space complexity for adding a number is also O(1). It involves adding the number to one of the two heaps, which do not significantly affect the overall space complexity.\n",
-    "\n",
-    "3. `findMedian()`: The space complexity for finding the median is O(1). It doesn't require any additional data structures that grow with the input size.\n",
-    "\n",
-    "In summary, the time complexity of the `addNum` operation is O(log N), and the time complexity of the `findMedian` operation is O(1). The space complexity is O(1) as well, as the space used by the two heaps is independent of the size of the data stream. This solution efficiently maintains and retrieves the median, making it suitable for the given problem constraints."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Generalized kth Element**: Extend the `MedianFinder` class to support finding the kth element in the data stream efficiently. This means finding the kth smallest or kth largest element in the stream.\n",
-    "\n",
-    "2. **Sliding Window Median**: Given an array and a sliding window size, find the median within the window as it slides over the array. This is an extension of the problem to a moving window scenario."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb b/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb
deleted file mode 100644
index e6ab9e3..0000000
--- a/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb	
+++ /dev/null
@@ -1,235 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 703. Kth Largest Element in a Stream\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Kth Largest Element in a Stream problem on LeetCode, click here!](https://leetcode.com/problems/kth-largest-element-in-a-stream/)\n",
-    "\n",
-    "---\n",
-    "Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.\n",
-    "\n",
-    "Implement the `KthLargest` class with the following methods:\n",
-    "\n",
-    "- `KthLargest(int k, int[] nums)`: Initializes the object with the integer `k` and the stream of integers `nums`.\n",
-    "- `int add(int val)`: Appends the integer `val` to the stream and returns the element representing the `kth` largest element in the stream.\n",
-    "\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `k` <= $10^4$\n",
-    "- 0 <= `nums.length` <= $10^4$\n",
-    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-    "- $-10^4$ <= `val` <= $10^4$\n",
-    "- At most $10^4$ calls will be made to `add`.\n",
-    "- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem you're trying to solve is to design a class called `KthLargest` that can efficiently find the kth largest element in a stream of integers. You need to support two main operations:\n",
-    "\n",
-    "To solve this problem efficiently, you can use a min-heap data structure. Here's how the approach works:\n",
-    "\n",
-    "1. Initialize a min-heap to store the k largest elements. Initially, it's empty.\n",
-    "\n",
-    "2. In the `__init__` method, populate the min-heap with the first `k` elements from `nums`. This ensures that you have the k largest elements seen so far.\n",
-    "\n",
-    "3. Whenever you add a new element to the stream using the `add` method, follow these steps:\n",
-    "   - Add the new element to the min-heap.\n",
-    "   - If the size of the min-heap exceeds `k`, remove the smallest element from the min-heap. This ensures that you always have the k largest elements in the min-heap.\n",
-    "   - The smallest element in the min-heap (the root) will always represent the kth largest element in the stream.\n",
-    "\n",
-    "Here's why this approach works efficiently:\n",
-    "\n",
-    "- By using a min-heap, you can quickly maintain the k largest elements, and finding the smallest element in the heap (the root) takes constant time.\n",
-    "\n",
-    "- When you add a new element, the min-heap's size is kept at most `k`, which ensures that you only track the k largest elements and discard the smaller ones.\n",
-    "\n",
-    "- The time complexity for adding an element is O(log k), which is very efficient compared to sorting the entire stream, which would be O(n log n).\n",
-    "\n",
-    "- This approach meets the constraints of the problem, including handling large streams and having a low time complexity for both initialization and adding elements.\n",
-    "\n",
-    "In summary, the min-heap approach efficiently tracks the kth largest element in the stream by maintaining a heap of the k largest elements seen so far, updating it as new elements are added. This approach provides a fast and scalable solution to the problem.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import heapq\n",
-    "from typing import List  # Import the List type from the typing module\n",
-    "\n",
-    "class KthLargest:\n",
-    "\n",
-    "    def __init__(self, k: int, nums: List[int]):\n",
-    "        # Initialize the KthLargest object with k and nums.\n",
-    "        self.k = k\n",
-    "        # Create a min-heap to store the k largest elements.\n",
-    "        self.min_heap = []\n",
-    "        # Populate the min-heap with the first k elements from nums.\n",
-    "        for num in nums:\n",
-    "            self.add(num)\n",
-    "\n",
-    "    def add(self, val: int) -> int:\n",
-    "        # Add val to the min-heap.\n",
-    "        heapq.heappush(self.min_heap, val)\n",
-    "        # If the size of the min-heap exceeds k, remove the smallest element.\n",
-    "        if len(self.min_heap) > self.k:\n",
-    "            heapq.heappop(self.min_heap)\n",
-    "        # The root of the min-heap is the kth largest element.\n",
-    "        return self.min_heap[0]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "- We start by importing the necessary modules:\n",
-    "  - `heapq`: This module provides functions to create and manipulate heaps.\n",
-    "  - `List` from the `typing` module: This is used to specify the type of the `nums` parameter.\n",
-    "\n",
-    "- We define the `KthLargest` class:\n",
-    "  - The `__init__` method initializes the object with the integer `k` and the list of integers `nums`.\n",
-    "  - It also creates an empty min-heap called `self.min_heap` to store the k largest elements.\n",
-    "  - It populates the min-heap with the first `k` elements from `nums` by calling the `add` method.\n",
-    "\n",
-    "- The `add` method:\n",
-    "  - Adds the new integer `val` to the min-heap using `heapq.heappush`. This maintains the min-heap property.\n",
-    "  - Checks if the size of the min-heap exceeds `k`. If it does, it removes the smallest element (the k+1th largest) using `heapq.heappop`.\n",
-    "  - Finally, it returns the smallest element in the min-heap, which is always the kth largest element.\n",
-    "\n",
-    "Overall, this code implements a class that efficiently finds the kth largest element in a stream of integers by maintaining a min-heap of the k largest elements seen so far.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n",
-      "5\n",
-      "5\n",
-      "8\n",
-      "8\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "kthLargest = KthLargest(3, [4, 5, 8, 2])\n",
-    "print(kthLargest.add(3))  # Output: 4\n",
-    "print(kthLargest.add(5))  # Output: 5\n",
-    "print(kthLargest.add(10)) # Output: 5\n",
-    "print(kthLargest.add(9))  # Output: 8\n",
-    "print(kthLargest.add(4))  # Output: 8"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `KthLargest` class using the min-heap approach:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. `__init__` method:\n",
-    "   - In the `__init__` method, you iterate over the first `k` elements in `nums` and add them to the min-heap. Each `heapq.heappush` operation takes O(log k) time.\n",
-    "   - Therefore, the time complexity of the `__init__` method is O(k * log k).\n",
-    "\n",
-    "2. `add` method:\n",
-    "   - In the `add` method, you perform the following operations:\n",
-    "     - `heapq.heappush`: O(log k) to add the new element to the min-heap.\n",
-    "     - If the size of the min-heap exceeds `k`, you perform `heapq.heappop`, which is also O(log k).\n",
-    "     - Finally, you return the smallest element from the min-heap, which is O(1) because it's always the root.\n",
-    "   - Overall, the time complexity of the `add` method is O(log k).\n",
-    "\n",
-    "3. Overall, if you make `n` calls to the `add` method, the total time complexity is O(n * log k), where `n` is the total number of elements added to the stream.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. The space complexity is determined by the space used to store the min-heap and the instance variables.\n",
-    "2. The min-heap stores at most `k` elements at any given time, so its space complexity is O(k).\n",
-    "3. The instance variables (such as `self.k` and `self.min_heap`) have constant space requirements.\n",
-    "4. Therefore, the overall space complexity of the `KthLargest` class is O(k).\n",
-    "\n",
-    "In summary, the time complexity of the `KthLargest` class is O(n * log k) for `n` add operations, and the space complexity is O(k), where `k` is the parameter passed during initialization. This implementation efficiently maintains the kth largest element in the stream while meeting the problem's constraints."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Dynamic k:** Modify the `KthLargest` class to support dynamic changes in the value of `k`. Implement a method `update_k` that allows changing the value of `k` during the lifetime of the object. Ensure that the object can still correctly find the kth largest element based on the updated value of `k`.\n",
-    "\n",
-    "\n",
-    "2. **Implement kth Smallest:** Create a new class called `KthSmallest` that finds the kth smallest element in a stream instead of the kth largest. You may need to modify the data structure used in the implementation."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb b/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb
deleted file mode 100644
index 85d59ac..0000000
--- a/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb	
+++ /dev/null
@@ -1,274 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 39. Combination Sum\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Combination Sum problem on LeetCode, click here!](https://leetcode.com/problems/combination-sum/)\n",
-    "\n",
-    "---\n",
-    "Given an array of **distinct** integers `candidates` and a target integer `target`, return *a list of all **unique combinations** of `candidates` where the chosen numbers sum to the `target`*. You may return the combinations in **any order**. \n",
-    "\n",
-    "The **same** number may be chosen from `candidates` an **unlimited number** of times. Two combinations are unique if the **frequency** of at least one of the chosen numbers is different.\n",
-    "\n",
-    "The test cases are generated such that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= candidates.length <= 30`\n",
-    "- `2 <= candidates[i] <= 40`\n",
-    "- All elements of `candidates` are **distinct**.\n",
-    "- `1 <= target <= 40`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to find all unique combinations of numbers from a given array (candidates) such that their sum equals a target number. Here are the details of the problem:\n",
-    "\n",
-    "- You are given an array of distinct integers called \"candidates.\"\n",
-    "- You are also given a target integer called \"target.\"\n",
-    "\n",
-    "The goal is to find all unique combinations of numbers from the candidates array where the sum of the selected numbers is equal to the target. You can use the same number from the candidates array an unlimited number of times. A combination is considered unique if it has a different frequency (i.e., a different number of occurrences) of at least one chosen number compared to other combinations.\n",
-    "\n",
-    "For example:\n",
-    "\n",
-    "**Example 1:**\n",
-    "```python\n",
-    "Input: candidates = [2, 3, 6, 7], target = 7\n",
-    "Output: [[2, 2, 3], [7]]\n",
-    "```\n",
-    "In this example, there are two unique combinations that sum up to the target:\n",
-    "- 2 + 2 + 3 = 7\n",
-    "- 7 = 7\n",
-    "\n",
-    "**Example 2:**\n",
-    "```python\n",
-    "Input: candidates = [2, 3, 5], target = 8\n",
-    "Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n",
-    "```\n",
-    "Here, there are three unique combinations that sum up to the target.\n",
-    "\n",
-    "**Example 3:**\n",
-    "```python\n",
-    "Input: candidates = [2], target = 1\n",
-    "Output: []\n",
-    "```\n",
-    "In this case, there are no combinations that can be formed from the candidates to reach the target of 1, so the output is an empty list.\n",
-    "\n",
-    "The problem asks you to find and return these combinations in any order.\n",
-    "\n",
-    "The constraints for this problem include the length of the candidates array, the values of candidates, and the target value.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def combinationSum(candidates, target):\n",
-    "    # Define a recursive DFS function to find combinations\n",
-    "    def dfs(remaining, start, path):\n",
-    "        # If the remaining target is 0, we found a valid combination\n",
-    "        if remaining == 0:\n",
-    "            result.append(path)\n",
-    "            return\n",
-    "        # If the remaining target is negative, this path is invalid\n",
-    "        if remaining < 0:\n",
-    "            return\n",
-    "        # Iterate through candidates starting from 'start' to avoid duplicates\n",
-    "        for i in range(start, len(candidates)):\n",
-    "            # Explore the current candidate by subtracting it from the remaining target\n",
-    "            # Add the current candidate to the path\n",
-    "            dfs(remaining - candidates[i], i, path + [candidates[i]])\n",
-    "\n",
-    "    result = []  # Initialize an empty list to store the result\n",
-    "    candidates.sort()  # Sort the candidates for deduplication and early stopping\n",
-    "    dfs(target, 0, [])  # Start the DFS from the target value with an empty path\n",
-    "    return result  # Return the list of unique combinations"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The `combinationSum` function takes an array of distinct integers `candidates` and a target integer `target`. It returns a list of all unique combinations of `candidates` where the chosen numbers sum to the `target`. Each number in the `candidates` array can be used an unlimited number of times.\n",
-    "\n",
-    "Here's an overview of how the code works:\n",
-    "\n",
-    "1. The `combinationSum` function is defined, and it takes two arguments: `candidates` and `target`.\n",
-    "\n",
-    "2. Inside the function, there is a nested helper function called `backtrack`. This function is responsible for the actual combination generation using a recursive approach.\n",
-    "\n",
-    "3. The `backtrack` function is called with three arguments: `start`, `target`, and `path`. The `start` variable helps keep track of the current index in the `candidates` array, `target` keeps track of the remaining sum to be achieved, and `path` is a list that stores the current combination.\n",
-    "\n",
-    "4. Within the `backtrack` function, there are three main conditional statements:\n",
-    "   - If `target` becomes zero, it means we have found a combination that sums up to the target, so we add `path` to the `result` list.\n",
-    "   - If `target` becomes negative, it means the current combination doesn't work, so we return without doing anything.\n",
-    "   - Otherwise, we enter a loop that iterates over the `candidates` array, starting from the current index `start`.\n",
-    "\n",
-    "5. In the loop, the `backtrack` function is called recursively with the updated `target` and `path` after including the current candidate. This process explores different combinations by considering the current candidate or moving to the next candidate.\n",
-    "\n",
-    "6. After the loop completes, the `result` list contains all unique combinations that sum to the target.\n",
-    "\n",
-    "7. The `candidates` array is sorted to optimize the search. Sorting helps in avoiding unnecessary recursive branches and reduces the number of explored combinations.\n",
-    "\n",
-    "8. Finally, the `backtrack` function is initially called from the `combinationSum` function with `start` set to 0, `target` set to the original target value, and an empty `path`. The result is returned as a list of unique combinations.\n",
-    "\n",
-    "The code includes a few example cases to demonstrate how the function works with different inputs."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[2, 2, 3], [7]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example 1\n",
-    "candidates1 = [2, 3, 6, 7]\n",
-    "target1 = 7\n",
-    "print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "candidates2 = [2, 3, 5]\n",
-    "target2 = 8\n",
-    "print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "candidates3 = [2]\n",
-    "target3 = 1\n",
-    "print(combinationSum(candidates3, target3))  # Output: []"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `combinationSum` function.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of this function is influenced by the number of recursive calls made by the `backtrack` function and the amount of work done within each call. In the worst case, the function will explore all possible combinations.\n",
-    "\n",
-    "The worst-case scenario occurs when we have many combinations to reach the target. The time complexity can be expressed as $O(2^n)$, where 'n' is the maximum number of recursive calls. In this context, 'n' corresponds to the number of elements in the `candidates` list, and it may be up to 30 (as per the constraints). However, the actual number of combinations explored is generally much smaller because we eliminate branches when the target becomes negative, and we skip over candidates that cannot lead to a solution. The sorting step also helps to optimize the search.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity is determined by the auxiliary space used for storing the results and the stack space for the recursive calls.\n",
-    "\n",
-    "1. The space required for the `result` list can be up to $O(2^n)$ in the worst case because each combination is stored.\n",
-    "\n",
-    "2. The stack space for the recursive calls can also be up to O(n) in the worst case, where 'n' is the number of elements in the `candidates` list.\n",
-    "\n",
-    "Therefore, the overall space complexity of the function is $O(2^n + n)$. However, in practice, the space used for the results list is often the dominant factor, and the actual space used may be much smaller than 2^n because not all combinations are explored.\n",
-    "\n",
-    "In summary, the time complexity is exponential but typically smaller in practice due to optimization, and the space complexity is influenced by the number of results and the depth of the recursion."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Find the Number of Unique Combinations:** Instead of listing all unique combinations, modify the function to return the count of unique combinations that sum to the target.\n",
-    "\n",
-    "2. **Combinations with a Maximum Value:** Add a maximum value constraint for each combination, so they cannot exceed a certain value. Modify the code to find combinations respecting this constraint."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb b/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb
deleted file mode 100644
index 8d97d20..0000000
--- a/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb	
+++ /dev/null
@@ -1,290 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 79. Word Search\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Word Search problem on LeetCode, click here!](https://leetcode.com/problems/word-search/)\n",
-    "\n",
-    "---\n",
-    "Given an `m x n` grid of characters `board` and a string `word`, return *`true` if `word` exists in the grid.*\n",
-    "\n",
-    "The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == board.length`\n",
-    "- `n = board[i].length`\n",
-    "- `1 <= m, n <= 6`\n",
-    "- `1 <= word.length <= 15`\n",
-    "- `board` and `word` consists of only lowercase and uppercase English letters.\n",
-    "\n",
-    "**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to determine whether a given word can be constructed from characters in an `m x n` grid of characters. The word can be formed by moving sequentially through horizontally or vertically neighboring cells in the grid, and you cannot use the same cell more than once.\n",
-    "\n",
-    "Here are the specifics of the problem:\n",
-    "\n",
-    "- You are given an `m x n` grid, which is a 2D board of characters.\n",
-    "- You are also given a string `word` consisting of lowercase and uppercase English letters.\n",
-    "- Your task is to determine if the word can be found in the grid by moving from one cell to an adjacent cell (horizontally or vertically) and collecting the characters sequentially to form the word.\n",
-    "- You can't use the same cell more than once when forming the word.\n",
-    "- If it's possible to form the word in the grid, return `True`; otherwise, return `False`.\n",
-    "\n",
-    "For example, consider the following grid and word:\n",
-    "\n",
-    "```python\n",
-    "board = [[\"A\", \"B\", \"C\", \"E\"],\n",
-    "         [\"S\", \"F\", \"C\", \"S\"],\n",
-    "         [\"A\", \"D\", \"E\", \"E\"]]         \n",
-    "word = \"ABCCED\"\n",
-    "```\n",
-    "\n",
-    "In this case, the word \"ABCCED\" can be formed by moving through the cells (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) in the grid. Therefore, the function should return `True`.\n",
-    "\n",
-    "Conversely, if the word is impossible to construct, such as in the case of the word \"ABCB\" for the same grid, where there's no valid path that forms the word, the function should return `False`.\n",
-    "\n",
-    "The problem involves searching for a path through the grid that matches the characters in the word while respecting the constraints mentioned. It can be solved using depth-first search (DFS) or backtracking.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def exist(self, board: List[List[str]], word: str) -> bool:\n",
-    "        def dfs(i, j, k):\n",
-    "            # Check if the current position is out of bounds or doesn't match the character in the word.\n",
-    "            if not (0 <= i < len(board)) or not (0 <= j < len(board[0])) or board[i][j] != word[k]:\n",
-    "                return False\n",
-    "            \n",
-    "            # If we have matched all characters in the word, return True.\n",
-    "            if k == len(word) - 1:\n",
-    "                return True\n",
-    "\n",
-    "            # Temporarily mark the current cell to prevent revisiting.\n",
-    "            tmp, board[i][j] = board[i][j], \"/\"\n",
-    "\n",
-    "            # Explore adjacent cells by making recursive calls in all four directions.\n",
-    "            found = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)\n",
-    "\n",
-    "            # Restore the original character in the cell.\n",
-    "            board[i][j] = tmp\n",
-    "\n",
-    "            return found\n",
-    "\n",
-    "        # Iterate through all cells in the board.\n",
-    "        for i in range(len(board)):\n",
-    "            for j in range(len(board[0])):\n",
-    "                # Check if the 'dfs' function starting from this cell returns True.\n",
-    "                if dfs(i, j, 0):\n",
-    "                    return True\n",
-    "\n",
-    "        # If no match is found after exploring all cells, return False.\n",
-    "        return False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. Import the necessary module:\n",
-    "   - `from typing import List`: This line imports the `List` type from the `typing` module, which is used to specify the type of the `board` parameter in the method signature.\n",
-    "\n",
-    "2. Define the `Solution` class:\n",
-    "   - `class Solution:`: This line defines a class called `Solution` to encapsulate the solution for the word search problem.\n",
-    "\n",
-    "3. Define the `exist` method:\n",
-    "   - `def exist(self, board: List[List[str]], word: str) -> bool:`: This method is a member of the `Solution` class and takes two parameters: `board`, which is a 2D list of characters, and `word`, which is a string. It returns a boolean value, indicating whether the word exists in the board.\n",
-    "\n",
-    "4. Implement the inner `dfs` function:\n",
-    "   - `def dfs(i, j, k):`: This is a helper function used for depth-first search (DFS). It is called recursively to explore the board and search for the word. It takes three parameters: `i` and `j` representing the current position on the board, and `k` representing the current index in the `word` string.\n",
-    "\n",
-    "   - Inside `dfs`, it checks if the current position is out of bounds or if the character at the current position on the board does not match the character at the current index in the `word`. If either of these conditions is met, it returns `False`.\n",
-    "\n",
-    "   - If `k` is equal to the length of the `word` minus one, it means we have successfully matched all characters in the `word`, so it returns `True`.\n",
-    "\n",
-    "   - The function then temporarily replaces the character at the current position with a marker, \"/\", to prevent revisiting the same cell. It then explores adjacent cells by making recursive calls in all four directions: up, down, left, and right. If any of these recursive calls return `True`, it propagates the `True` result up the call stack. After the exploration is complete, it restores the original character in the cell.\n",
-    "\n",
-    "5. Iterate through the board:\n",
-    "   - The outer loop iterates through all cells in the `board` using two nested loops. For each cell, it checks if the `dfs` function starting from that cell (with `k` initially set to 0) returns `True`. If it does, it means the word exists, so the method returns `True`.\n",
-    "\n",
-    "6. Return `False`:\n",
-    "   - If the loop completes without finding the word in the board, the method returns `False`.\n",
-    "\n",
-    "To use this code, you can create an instance of the `Solution` class and call the `exist` method on that instance, passing the `board` and the `word` as arguments to check if the word exists in the board."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Test case 1: Word \"ABCCED\" can be found in the grid\n",
-    "board1 = [[\"A\",\"B\",\"C\",\"E\"],\n",
-    "          [\"S\",\"F\",\"C\",\"S\"],\n",
-    "          [\"A\",\"D\",\"E\",\"E\"]]\n",
-    "word1 = \"ABCCED\"\n",
-    "print(solution.exist(board1, word1))  # Expected output: True"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "\n",
-    "# Test case 2: Word \"SEE\" can be found in the grid\n",
-    "word2 = \"SEE\"\n",
-    "print(solution.exist(board1, word2))  # Expected output: True"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "\n",
-    "# Test case 3: Word \"ABCB\" cannot be found in the grid\n",
-    "word3 = \"ABCB\"\n",
-    "print(solution.exist(board1, word3))  # Expected output: False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `exist` method in the given code.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The primary work in this code is done by the `dfs` function, which performs a depth-first search through the grid. The time complexity of the `dfs` function is $O(4^(n))$, where \"n\" is the length of the `word`. This is because, in the worst case, the function explores all possible directions (up, down, left, right) for each character in the `word`.\n",
-    "\n",
-    "The outer loops iterate through all cells in the `board`, which has dimensions `m x n`. So, the total number of iterations of `dfs` is O(m * n). However, since the `dfs` function has a higher time complexity, the overall time complexity is still dominated by the `dfs` function.\n",
-    "\n",
-    "Therefore, the overall time complexity of the `exist` method is $O(m * n * 4^n)$, where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. The primary space usage comes from the recursive calls in the `dfs` function. The depth of the recursion can be at most equal to the length of the `word`, which is O(n).\n",
-    "\n",
-    "2. Additionally, the `tmp` variable and the recursive calls on the call stack consume a small amount of additional memory.\n",
-    "\n",
-    "3. The marking of visited cells with the \"/\" character temporarily modifies the `board` in-place but doesn't significantly contribute to the space complexity.\n",
-    "\n",
-    "Therefore, the overall space complexity is O(n) due to the recursive call stack.\n",
-    "\n",
-    "In summary, the time complexity is $O(m * n * 4^n)$, and the space complexity is O(n), where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word. The time complexity is exponential in the worst case due to the depth-first search through all possible paths in the grid."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the code to find all unique words that can be constructed from the grid. Instead of returning `True` or `False`, return a list of words found. Ensure that the same cell is not used more than once in constructing each word.\n",
-    "\n",
-    "2. **Board Generation:** Create a function to generate random grids of characters and then use the `exist` method to solve them. Test the efficiency and correctness of your code by generating larger boards and words."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb b/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb
deleted file mode 100644
index 56b7647..0000000
--- a/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb	
+++ /dev/null
@@ -1,311 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 133. Clone Graph\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Clone Graph problem on LeetCode, click here!](https://leetcode.com/problems/clone-graph/)\n",
-    "\n",
-    "---\n",
-    "Given a reference of a node in a **connected** undirected graph.\n",
-    "\n",
-    "Return a **deep copy** (clone) of the graph.\n",
-    "\n",
-    "Each node in the graph contains a value (`int`) and a list (`List[Node]`) of its neighbors.\n",
-    "\n",
-    "```\n",
-    "class Node {\n",
-    "    public int val;\n",
-    "    public List<Node> neighbors;\n",
-    "}\n",
-    "```\n",
-    "\n",
-    "**Test case format:**\n",
-    "\n",
-    "For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with `val == 1`, the second node with `val == 2`, and so on. The graph is represented in the test case using an adjacency list.\n",
-    "\n",
-    "**An adjacency list** is a collection of unordered **lists** used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.\n",
-    "\n",
-    "The given node will always be the first node with `val = 1`. You must return the **copy of the given node** as a reference to the cloned graph.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the graph is in the range `[0, 100]`.\n",
-    "- `1 <= Node.val <= 100`\n",
-    "- `Node.val` is unique for each node.\n",
-    "- There are no repeated edges and no self-loops in the graph.\n",
-    "- The Graph is connected and all nodes can be visited starting from the given node."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand is to clone a connected undirected graph, which means we need to create a deep copy of the original graph while preserving its structure and relationships between nodes. The graph is represented using adjacency lists, where each node has a value (an integer) and a list of its neighbors.\n",
-    "\n",
-    "Here are the key aspects of the problem:\n",
-    "\n",
-    "1. **Input**: You are given a reference to one of the nodes in the original graph. This reference node serves as the entry point to the graph. The entire graph can be explored starting from this node.\n",
-    "\n",
-    "2. **Output**: The goal is to create a deep copy of the graph and return a reference to the cloned graph. The cloned graph should have the same structure and relationships as the original one but should be a separate instance in memory.\n",
-    "\n",
-    "3. **Graph Structure**: The graph is composed of nodes (vertices) and edges. Nodes have values (integers) and are connected to other nodes through edges (undirected connections). Each node has a list of neighbors, which are other nodes it is connected to.\n",
-    "\n",
-    "4. **Connected Graph**: The problem assumes that the graph is connected, meaning that you can start from the reference node provided and traverse the entire graph by following the edges. This ensures that there are no isolated components in the graph.\n",
-    "\n",
-    "To solve this problem, you typically need to perform a graph traversal (e.g., depth-first search or breadth-first search) to visit all the nodes in the original graph, create corresponding nodes in the clone, and establish the same connections between nodes in the cloned graph as in the original one. To avoid duplication of nodes, you also need to keep track of visited nodes to ensure that the cloning process is efficient and that nodes are not duplicated in the clone.\n",
-    "\n",
-    "The problem can be challenging because you need to create a deep copy of the graph while handling cyclic dependencies between nodes and ensuring that the structure and relationships of the original graph are preserved in the clone. The solution requires recursion or a data structure to maintain the mapping between original nodes and their clones to achieve an efficient and accurate cloning process.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a Node.\n",
-    "class Node:\n",
-    "    def __init__(self, val=0, neighbors=None):\n",
-    "        self.val = val\n",
-    "        self.neighbors = neighbors if neighbors is not None else []\n",
-    "\n",
-    "class Solution:\n",
-    "    def cloneGraph(self, node: 'Node') -> 'Node':\n",
-    "        if not node:\n",
-    "            return None\n",
-    "\n",
-    "        visited = {}  # Dictionary to keep track of visited nodes\n",
-    "\n",
-    "        def dfs(original_node):\n",
-    "            if original_node in visited:\n",
-    "                return visited[original_node]\n",
-    "\n",
-    "            # Create a copy of the original node\n",
-    "            copy_node = Node(original_node.val)\n",
-    "            visited[original_node] = copy_node\n",
-    "\n",
-    "            # Recursively clone the neighbors of the original node\n",
-    "            for neighbor in original_node.neighbors:\n",
-    "                copy_node.neighbors.append(dfs(neighbor))\n",
-    "\n",
-    "            return copy_node\n",
-    "\n",
-    "        return dfs(node)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The given code defines a Python class, `Node`, and a solution class, `Solution`, to clone a connected undirected graph represented by nodes and their neighbors. Here's an explanation of the code:\n",
-    "\n",
-    "1. The `Node` class defines a node in the graph. Each node has a value (`val`) which is an integer and a list of neighbors (`neighbors`) that are other nodes in the graph.\n",
-    "\n",
-    "2. The `Solution` class contains the method `cloneGraph`, which takes a reference to the first node of the graph (`node`) as input and returns a deep copy of the entire graph, preserving the structure and relationships between nodes.\n",
-    "\n",
-    "3. Inside the `cloneGraph` method:\n",
-    "   - It first checks if the input `node` is `None`. If it's `None`, it means there is no graph to clone, and the method returns `None`.\n",
-    "\n",
-    "   - It initializes a `visited` dictionary to keep track of visited nodes. This dictionary is used to ensure that each node is cloned only once to avoid duplication.\n",
-    "\n",
-    "   - The `dfs` (depth-first search) function is defined within the `cloneGraph` method. It takes an original node as input and returns its corresponding clone. This function performs the actual cloning of the graph.\n",
-    "\n",
-    "   - Inside the `dfs` function:\n",
-    "     - It checks if the original node has been visited before by looking it up in the `visited` dictionary. If it has been visited, it returns the corresponding clone from the `visited` dictionary.\n",
-    "\n",
-    "     - If the original node is not visited, it creates a new node, `copy_node`, with the same value as the original node. This is the first step in creating the clone of the original node.\n",
-    "\n",
-    "     - It then recursively clones the neighbors of the original node by iterating through the `neighbors` list of the original node and appending the corresponding cloned neighbors to the `neighbors` list of the `copy_node`. This step ensures that the relationships between nodes are preserved in the clone.\n",
-    "\n",
-    "     - The `copy_node` is added to the `visited` dictionary with the original node as the key and the clone as the value.\n",
-    "\n",
-    "     - Finally, the `copy_node` is returned as the result of the DFS for the original node.\n",
-    "\n",
-    "4. The `cloneGraph` method returns the result of the DFS on the input `node`, which is the clone of the entire graph with its structure and relationships intact.\n",
-    "\n",
-    "In summary, the code uses depth-first search (DFS) to traverse the original graph, creating a clone for each node and its neighbors. It ensures that each node is cloned only once to avoid duplication and returns a reference to the cloned graph."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[1, 2, 4]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "def get_adjacency_list(node):\n",
-    "    visited = {}  # Dictionary to keep track of visited nodes\n",
-    "\n",
-    "    def dfs(node):\n",
-    "        if not node:\n",
-    "            return []\n",
-    "\n",
-    "        if node in visited:\n",
-    "            return visited[node]\n",
-    "\n",
-    "        neighbors = [n.val for n in node.neighbors]\n",
-    "        visited[node] = neighbors\n",
-    "\n",
-    "        for neighbor in node.neighbors:\n",
-    "            dfs(neighbor)\n",
-    "\n",
-    "    dfs(node)\n",
-    "    result = []\n",
-    "\n",
-    "    def get_node_and_neighbors(node):\n",
-    "        result.append([node.val] + visited[node])\n",
-    "\n",
-    "    get_node_and_neighbors(node)\n",
-    "    return result\n",
-    "\n",
-    "\n",
-    "# Testing the code with the provided examples\n",
-    "adjList1 = [[2, 4], [1, 3], [2, 4], [1, 3]]\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Create the graph from adjList\n",
-    "nodes = [Node(i) for i in range(1, len(adjList1) + 1)]\n",
-    "for i, adj in enumerate(adjList1):\n",
-    "    nodes[i].neighbors = [nodes[j - 1] for j in adj]\n",
-    "\n",
-    "# Clone the graph and get its adjacency list\n",
-    "cloned_node1 = solution.cloneGraph(nodes[0])\n",
-    "cloned_adjList1 = get_adjacency_list(cloned_node1)\n",
-    "\n",
-    "# Print the cloned graph as an adjacency list\n",
-    "print(cloned_adjList1)\n",
-    "\n",
-    "# Output should match adjList1\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[1]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "# Input: adjList = [[]]\n",
-    "\n",
-    "adjList2 = [[]]\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Create the graph from adjList\n",
-    "node1 = Node(1)\n",
-    "\n",
-    "# Clone the graph and get its adjacency list\n",
-    "cloned_node1 = solution.cloneGraph(node1)\n",
-    "cloned_adjList2 = get_adjacency_list(cloned_node1)\n",
-    "\n",
-    "# Print the cloned graph as an adjacency list\n",
-    "print(cloned_adjList2)\n",
-    "\n",
-    "# Output should be [[]], which represents an empty graph"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for cloning a graph using depth-first search (DFS).\n",
-    "\n",
-    "Time Complexity:\n",
-    "1. Creating the graph: The time complexity to create the original graph from `adjList` is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. We iterate through each node and its neighbors in `adjList`.\n",
-    "\n",
-    "2. Cloning the graph (DFS): The depth-first search (DFS) is used to traverse the original graph and create a clone. The time complexity of DFS is O(V + E), as we visit each node and each edge once.\n",
-    "\n",
-    "Overall, the time complexity is O(V + E) for both creating the original graph and cloning it.\n",
-    "\n",
-    "Space Complexity:\n",
-    "1. Storage for the original graph: We need to store the original graph, which includes all nodes and their adjacency lists. In the worst case, this requires O(V + E) space, where V is the number of nodes, and E is the number of edges.\n",
-    "\n",
-    "2. Storage for the visited dictionary: The `visited` dictionary stores mappings from original nodes to their corresponding clones. In the worst case, this requires O(V) space because each node is visited once.\n",
-    "\n",
-    "3. Recursive stack for DFS: The space used for the recursive call stack during DFS can be up to O(V) in the worst case, where V is the number of nodes.\n",
-    "\n",
-    "Overall, the space complexity is O(V + E) for storing the original graph, and O(V) for auxiliary data structures, including the `visited` dictionary and the DFS call stack. Therefore, the overall space complexity is O(V + E).\n",
-    "\n",
-    "In summary, the time and space complexities of the provided code for cloning a graph are both O(V + E), where V is the number of nodes, and E is the number of edges in the graph."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Cyclic Graph**: Design a graph with cycles (nodes connected in a cycle). Test if the code can correctly clone graphs with cycles and doesn't fall into an infinite loop during traversal.\n",
-    "\n",
-    "\n",
-    "2. **Partial Graph Cloning**: Modify the code to clone only a subset of the graph starting from a given reference node, rather than the entire graph. Implement this as an extension to the problem."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb b/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb
deleted file mode 100644
index 0dd53e9..0000000
--- a/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 178. Graph Valid Tree\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Graph Valid Tree problem on LintCode, click here!](https://www.lintcode.com/problem/178/)\n",
-    "\n",
-    "---\n",
-    "Given `n` nodes labeled from `0` to `n - 1` and a list of `undirected` edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- You can assume that no duplicate edges will appear in edges. Since all edges are `undirected`, `[0, 1]` is the same as `[1, 0]` and thus will not appear together in edges."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to determine whether a given set of undirected edges can form a valid tree. In this context, a \"tree\" is a specific type of graph with the following properties:\n",
-    "\n",
-    "1. It is a connected graph, meaning there is a path between any pair of nodes in the graph.\n",
-    "2. It has no cycles, which means there are no closed loops or circuits in the graph.\n",
-    "3. It has exactly `n - 1` edges, where `n` is the number of nodes in the tree.\n",
-    "\n",
-    "The problem is typically given with two pieces of information:\n",
-    "\n",
-    "1. The number of nodes (`n`), which are usually labeled from 0 to `n-1`.\n",
-    "2. A list of undirected edges, where each edge is represented as a pair of nodes.\n",
-    "\n",
-    "The task is to write a function or algorithm to determine whether the provided set of edges can be arranged to form a valid tree based on the properties mentioned above.\n",
-    "\n",
-    "To solve this problem, you need to check the following conditions:\n",
-    "\n",
-    "1. There should be exactly `n - 1` edges to ensure that the graph is connected and tree-like.\n",
-    "2. There should be no cycles in the graph. If there are any cycles, the set of edges cannot form a tree.\n",
-    "3. The graph should be connected, which means you can reach any node from any other node through a series of edges.\n",
-    "\n",
-    "The solution typically involves using graph traversal algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the graph and check these conditions. If all conditions are met, the given edges form a valid tree; otherwise, they do not.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def validTree(n, edges):\n",
-    "    if len(edges) != n - 1:\n",
-    "        return False  # A tree with n nodes must have n-1 edges.\n",
-    "\n",
-    "    # Create an adjacency list to represent the graph.\n",
-    "    adj_list = {i: [] for i in range(n)}\n",
-    "    for u, v in edges:\n",
-    "        adj_list[u].append(v)\n",
-    "        adj_list[v].append(u)\n",
-    "\n",
-    "    visited = set()\n",
-    "\n",
-    "    # Define a recursive DFS function.\n",
-    "    def dfs(node, parent):\n",
-    "        if node in visited:\n",
-    "            return False  # If we encounter a visited node, there's a cycle.\n",
-    "        visited.add(node)\n",
-    "\n",
-    "        for neighbor in adj_list[node]:\n",
-    "            if neighbor != parent:\n",
-    "                if not dfs(neighbor, node):\n",
-    "                    return False\n",
-    "\n",
-    "        return True\n",
-    "\n",
-    "    # Start DFS from any node (e.g., node 0).\n",
-    "    if not dfs(0, -1):\n",
-    "        return False  # If there's a cycle, it's not a valid tree.\n",
-    "\n",
-    "    return len(visited) == n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The `validTree` function checks if a given set of undirected edges forms a valid tree. A valid tree must satisfy these conditions:\n",
-    "\n",
-    "1. It has exactly `n - 1` edges, where `n` is the number of nodes.\n",
-    "2. There are no cycles in the graph.\n",
-    "3. The graph is connected, meaning all nodes are reachable from any starting node.\n",
-    "\n",
-    "The code accomplishes this as follows:\n",
-    "\n",
-    "- It first checks if the number of edges is equal to `n - 1`. If not, it immediately returns `False` since a tree with `n` nodes must have `n - 1` edges to be connected.\n",
-    "\n",
-    "- It then constructs an adjacency list (`adj_list`) to represent the graph efficiently. This data structure stores the neighbors of each node.\n",
-    "\n",
-    "- The code uses a `visited` set to keep track of nodes visited during depth-first search (DFS).\n",
-    "\n",
-    "- The `dfs` function is a recursive function that checks for cycles in the graph. It does this by visiting nodes and marking them as visited. If it encounters a node that has already been visited (indicating a cycle), it returns `False`.\n",
-    "\n",
-    "- The DFS function explores the neighbors of each node, avoiding revisiting the parent node to prevent cycles.\n",
-    "\n",
-    "- After the DFS is complete, the code checks if the number of visited nodes is equal to `n`, ensuring the graph is connected.\n",
-    "\n",
-    "- Finally, if all conditions are met, the code returns `True`, indicating that the given edges form a valid tree. Otherwise, it returns `False`.\n",
-    "\n",
-    "The code provides examples to demonstrate how to use the `validTree` function to check whether a set of edges forms a valid tree or not."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "n1 = 5\n",
-    "edges1 = [[0, 1], [0, 2], [0, 3], [1, 4]]\n",
-    "print(validTree(n1, edges1))  # Output: true"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "n2 = 5\n",
-    "edges2 = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]\n",
-    "print(validTree(n2, edges2))  # Output: false"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexities of the `validTree` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. Constructing the adjacency list (`adj_list`) takes O(E) time, where E is the number of edges in the input graph (|edges|).\n",
-    "\n",
-    "2. The DFS function, when considering all nodes, has a time complexity of O(V + E), where V is the number of nodes in the graph. This is because, in the worst case, the DFS function visits each node and each edge once.\n",
-    "\n",
-    "3. Checking if the number of visited nodes is equal to `n` takes O(1) time.\n",
-    "\n",
-    "Overall, the time complexity of the `validTree` function is O(E + V) in the worst case.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The adjacency list (`adj_list`) consumes additional space to store the graph structure, taking O(V + E) space. This is because it stores information about each node and its adjacent nodes.\n",
-    "\n",
-    "2. The `visited` set keeps track of visited nodes, and its space complexity is O(V) in the worst case, as there can be at most V unique nodes in the set.\n",
-    "\n",
-    "3. The depth of the function call stack during DFS can be at most V, which contributes O(V) space to the space complexity.\n",
-    "\n",
-    "Overall, the space complexity of the `validTree` function is O(V + E) due to the adjacency list and O(V) due to the `visited` set and DFS call stack.\n",
-    "\n",
-    "In summary, the time complexity is O(E + V), and the space complexity is O(V + E). These complexities are based on the worst-case scenario, where all nodes and edges are considered. In practical cases, the actual complexities may be smaller, depending on the specific structure of the graph."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Minimum Spanning Tree (MST)**: Implement an algorithm to find the minimum spanning tree of a connected graph using either Kruskal's or Prim's algorithm. Compare the MST to the original graph to determine if the original graph is a tree. This exercise reinforces the concept of tree properties.\n",
-    "\n",
-    "2. **Detecting Cycles**: Modify the code to not only check if the input forms a tree but also to identify and print out any cycles present in the graph if it's not a tree. This exercise enhances your ability to detect and visualize cycles in a graph."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb b/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb
deleted file mode 100644
index 5a878a4..0000000
--- a/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb	
+++ /dev/null
@@ -1,236 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 200. Number of Islands\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Number of Islands problem on LeetCode, click here!](https://leetcode.com/problems/number-of-islands/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given an `m x n` 2D binary grid `grid` which represents a map of `'1'`s (land) and `'0'`s (water), return *the number of islands*. \n",
-    "\n",
-    "An **island** is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == grid.length`\n",
-    "- `n == grid[i].length`\n",
-    "- `1 <= m, n <= 300`\n",
-    "- `grid[i][j]` is `'0'` or `'1'`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is about counting the number of islands in a 2D binary grid. In this grid, '1' represents land, and '0' represents water. An island is defined as a group of '1' cells that are adjacent to each other either horizontally or vertically. It's important to note that diagonal connections do not count. Additionally, it's assumed that all four edges of the grid are surrounded by water, meaning the grid is finite and doesn't \"wrap around.\"\n",
-    "\n",
-    "The objective is to determine how many distinct islands exist in the grid. To clarify, an island consists of connected '1' cells, and these connections can only occur horizontally or vertically. If there are no '1' cells in the grid, the number of islands would be zero.\n",
-    "\n",
-    "The problem often involves using graph traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to identify and count these islands. You start at one '1' cell and explore its adjacent cells to determine the extent of the island. Once you've visited all the '1' cells in an island, you move on to the next unvisited '1' cell and repeat the process until you've counted all the islands in the grid.\n",
-    "\n",
-    "In essence, the problem asks you to analyze the grid to find groups of connected '1' cells and count them as individual islands. The goal is to return the total count of islands in the given grid.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def numIslands(self, grid: List[List[str]]) -> int:\n",
-    "        if not grid:\n",
-    "            return 0\n",
-    "\n",
-    "        m, n = len(grid), len(grid[0])  # Get the dimensions of the grid\n",
-    "        num_islands = 0  # Initialize the count of islands\n",
-    "\n",
-    "        def dfs(row, col):\n",
-    "            if row < 0 or row >= m or col < 0 or col >= n or grid[row][col] == '0':\n",
-    "                return\n",
-    "\n",
-    "            grid[row][col] = '0'  # Mark the current cell as visited by changing it to '0'\n",
-    "            directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Define four possible directions to move\n",
-    "\n",
-    "            for dr, dc in directions:\n",
-    "                # Explore adjacent cells\n",
-    "                dfs(row + dr, col + dc)\n",
-    "\n",
-    "        for i in range(m):\n",
-    "            for j in range(n):\n",
-    "                if grid[i][j] == '1':\n",
-    "                    num_islands += 1  # Increment the island count\n",
-    "                    dfs(i, j)  # Start a DFS from the current '1' cell to explore the entire island\n",
-    "\n",
-    "        return num_islands\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code defines a Python class called `Solution` with a method called `numIslands`. This class is designed to count the number of islands in a 2D binary grid.\n",
-    "\n",
-    "The `numIslands` method takes a 2D grid as input, where '1' represents land, and '0' represents water. It uses a depth-first search (DFS) approach to identify and count islands. The key steps in the code include:\n",
-    "\n",
-    "1. Checking if the input grid is empty, and if so, returning 0 (indicating there are no islands).\n",
-    "\n",
-    "2. Determining the dimensions of the grid (number of rows and columns).\n",
-    "\n",
-    "3. Initializing a variable `num_islands` to keep track of the count of islands.\n",
-    "\n",
-    "4. Defining a nested function `dfs` to perform depth-first search. This function recursively explores adjacent land cells connected to the current cell and marks them as visited by changing '1' to '0'.\n",
-    "\n",
-    "5. In the `dfs` function, it checks for boundary conditions and whether the current cell is '0'. If these conditions are met, it returns without further exploration.\n",
-    "\n",
-    "6. The `dfs` function explores neighboring cells in all four directions (up, down, left, right) using a loop.\n",
-    "\n",
-    "7. The main loop iterates through each cell of the grid. When a '1' cell is encountered, it increments the `num_islands` count and starts a DFS from that cell to explore the entire island.\n",
-    "\n",
-    "8. Finally, the method returns the total count of islands found in the grid.\n",
-    "\n",
-    "This code encapsulates the solution in a class, allowing you to create an instance of the `Solution` class and call the `numIslands` method on it to count the number of islands in a given grid."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 1 Output: 1\n"
-     ]
-    }
-   ],
-   "source": [
-    "grid1 = [\n",
-    "  [\"1\", \"1\", \"1\", \"1\", \"0\"],\n",
-    "  [\"1\", \"1\", \"0\", \"1\", \"0\"],\n",
-    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-    "  [\"0\", \"0\", \"0\", \"0\", \"0\"]\n",
-    "]\n",
-    "\n",
-    "solution = Solution()\n",
-    "result1 = solution.numIslands(grid1)\n",
-    "print(\"Example 1 Output:\", result1)  # Expected output: 1\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 2 Output: 3\n"
-     ]
-    }
-   ],
-   "source": [
-    "grid2 = [\n",
-    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-    "  [\"0\", \"0\", \"1\", \"0\", \"0\"],\n",
-    "  [\"0\", \"0\", \"0\", \"1\", \"1\"]\n",
-    "]\n",
-    "\n",
-    "result2 = solution.numIslands(grid2)\n",
-    "print(\"Example 2 Output:\", result2)  # Expected output: 3\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "Time Complexity:\n",
-    "- The code uses a depth-first search (DFS) to explore the grid and count the number of islands.\n",
-    "- In the worst case, it visits every cell in the grid exactly once.\n",
-    "- The DFS function explores neighboring cells in four directions (up, down, left, right), so it has a time complexity of $O(4^{max(m, n)})$, where m is the number of rows and n is the number of columns.\n",
-    "- Therefore, the overall time complexity of the code is O(m * n), where m is the number of rows, and n is the number of columns.\n",
-    "\n",
-    "Space Complexity:\n",
-    "- The space complexity is determined by the recursive calls and the depth of the DFS stack.\n",
-    "- In the worst case, when the entire grid consists of '1's, the depth of the DFS recursion can be as deep as max(m, n).\n",
-    "- Therefore, the space complexity of the code is O(max(m, n)), representing the maximum depth of the DFS stack.\n",
-    "\n",
-    "In practical terms, for reasonably sized grids, the time and space complexity is linear, and the code should perform well. However, it's important to keep in mind that the worst-case time complexity is O(m * n), and the space complexity can be O(max(m, n))."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Identify Largest Island:** In addition to counting the number of islands, modify the code to identify the largest island in the grid and return its size.\n",
-    "\n",
-    "2. **Optimize for Space:** Modify the solution to reduce its space complexity to O(1) without modifying the input grid. This is a more memory-efficient version of the problem."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb b/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb
deleted file mode 100644
index 4176cab..0000000
--- a/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb	
+++ /dev/null
@@ -1,243 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 207. Course Schedule\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Course Schedule problem on LeetCode, click here!](https://leetcode.com/problems/course-schedule/)\n",
-    "\n",
-    "---\n",
-    "There are a total of `numCourses` courses to take, labeled from 0 to `numCourses - 1`. You are given an array `prerequisites` where `prerequisites[i] = [ai, bi]` indicates that you **must** take course `bi` first if you want to take course `ai`. \n",
-    "\n",
-    "- For example, the pair `[0, 1]`, indicates that to take course `0` you have to first take course `1`.\n",
-    "\n",
-    "Return `true` if you can finish all courses. Otherwise, return `false`.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `numCourses` <= 2000\n",
-    "- 0 <= `prerequisites.length` <= 5000\n",
-    "- `prerequisites[i].length` == 2\n",
-    "- 0 <= `ai`, `bi` < `numCourses`\n",
-    "- All the pairs `prerequisites[i]` are **unique**."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to determine whether it's possible to complete a set of courses given a list of prerequisites. You are given the following:\n",
-    "\n",
-    "- `numCourses`: The total number of courses to be taken, labeled from 0 to `numCourses - 1`.\n",
-    "- `prerequisites`: A list of prerequisite courses, where each element `prerequisites[i]` is a pair `[ai, bi]`, indicating that you must take course `bi` before you can take course `ai`.\n",
-    "\n",
-    "The goal is to check whether it's possible to complete all the courses while respecting the prerequisite requirements. In other words, you need to determine if there are any circular dependencies in the course prerequisites that would prevent you from taking all the courses.\n",
-    "\n",
-    "For example:\n",
-    "\n",
-    "- If `numCourses` is 2, and `prerequisites` is `[[1, 0]]`, it means you have two courses, and you must finish course 0 before you can take course 1. In this case, it's possible to complete all courses, so the function should return `True`.\n",
-    "\n",
-    "- If `numCourses` is 2, and `prerequisites` is `[[1, 0], [0, 1]]`, it means you have two courses, but there's a circular dependency where course 1 requires course 0 and course 0 requires course 1. In this case, it's impossible to complete all courses, so the function should return `False`.\n",
-    "\n",
-    "The problem is essentially about checking for the presence of cycles in a directed graph, where the courses are nodes, and the prerequisites represent directed edges between them. If there are no cycles, it's possible to complete all courses; otherwise, it's not possible.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from collections import defaultdict, deque\n",
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:\n",
-    "        # Create a graph using an adjacency list to represent the courses and their prerequisites.\n",
-    "        graph = defaultdict(list)  # Initialize an empty adjacency list.\n",
-    "        in_degree = [0] * numCourses  # Initialize an array to store in-degrees of courses.\n",
-    "\n",
-    "        # Populate the graph and calculate in-degrees.\n",
-    "        for course, prerequisite in prerequisites:\n",
-    "            graph[prerequisite].append(course)  # Add the course as a neighbor to its prerequisite.\n",
-    "            in_degree[course] += 1  # Increment the in-degree of the course.\n",
-    "\n",
-    "        # Initialize a queue with courses that have no prerequisites.\n",
-    "        queue = deque()\n",
-    "        for course in range(numCourses):\n",
-    "            if in_degree[course] == 0:\n",
-    "                queue.append(course)\n",
-    "\n",
-    "        # Perform topological sorting.\n",
-    "        while queue:\n",
-    "            course = queue.popleft()  # Take a course with no prerequisites.\n",
-    "            numCourses -= 1  # Decrement the count of remaining courses.\n",
-    "\n",
-    "            for neighbor in graph[course]:\n",
-    "                in_degree[neighbor] -= 1  # Remove the prerequisite relationship.\n",
-    "                if in_degree[neighbor] == 0:\n",
-    "                    queue.append(neighbor)  # If no more prerequisites, add to the queue.\n",
-    "\n",
-    "        # If all courses were successfully taken (numCourses becomes 0), return True.\n",
-    "        return numCourses == 0"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "This Python code is designed to determine whether it is possible to complete all the required courses given a set of prerequisites. It uses a topological sorting algorithm to accomplish this. Here's a brief explanation of the code:\n",
-    "\n",
-    "1. The code defines a `Solution` class with a `canFinish` method that takes two parameters: `numCourses` (the total number of courses) and `prerequisites` (a list of pairs indicating which course must be taken before another).\n",
-    "\n",
-    "2. It initializes an empty graph as an adjacency list using a `defaultdict` to represent the courses and their prerequisites. It also initializes a list called `in_degree` to keep track of the in-degrees of courses, which is used in the topological sorting process.\n",
-    "\n",
-    "3. The code then populates the graph by iterating through the `prerequisites` list. For each pair `(ai, bi)`, it adds course `ai` as a neighbor to its prerequisite course `bi` in the graph and increments the in-degree of course `ai`.\n",
-    "\n",
-    "4. Next, it initializes a queue with courses that have no prerequisites. It does this by iterating through all the courses and adding those with an in-degree of 0 to the queue.\n",
-    "\n",
-    "5. The code performs a topological sorting of the courses using a while loop. In each iteration of the loop, it dequeues a course from the front of the queue (a course with no prerequisites).\n",
-    "\n",
-    "6. For each neighbor (course that depends on the dequeued course), it decrements the in-degree of the neighbor. If the neighbor's in-degree becomes 0, it means all of its prerequisites have been taken, so the neighbor is added to the queue.\n",
-    "\n",
-    "7. The loop continues until there are no more courses with no prerequisites to dequeue, and the in-degrees are updated accordingly.\n",
-    "\n",
-    "8. After the loop, if all courses were successfully taken (i.e., `numCourses` becomes 0), it means that there were no circular dependencies, and it's possible to finish all courses. In this case, the method returns `True`. Otherwise, if there are remaining courses (indicating a circular dependency), it returns `False`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "numCourses1 = 2\n",
-    "prerequisites1 = [[1, 0]]\n",
-    "sol = Solution()\n",
-    "print(sol.canFinish(numCourses1, prerequisites1))  # Output: True"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "numCourses2 = 2\n",
-    "prerequisites2 = [[1, 0], [0, 1]]\n",
-    "print(sol.canFinish(numCourses2, prerequisites2))  # Output: False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `canFinish` method:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. Constructing the graph: In the first loop where we iterate through the `prerequisites` list, we populate the graph with prerequisites. This loop has a time complexity of O(E), where E is the number of prerequisites (edges).\n",
-    "\n",
-    "2. Initializing in-degrees and finding courses with no prerequisites: In the worst case, we iterate through all `numCourses` courses to initialize in-degrees and find the courses with no prerequisites. This has a time complexity of O(V), where V is the number of courses (vertices).\n",
-    "\n",
-    "3. Topological Sorting: In the while loop, we perform topological sorting, processing each course once and its outgoing edges. In the worst case, each course is processed once. This has a time complexity of O(V + E).\n",
-    "\n",
-    "The overall time complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. Graph Representation: The space complexity for storing the graph as an adjacency list is O(E), where E is the number of prerequisites.\n",
-    "\n",
-    "2. In-Degree Array: The space complexity for storing the in-degrees of courses is O(V), where V is the number of courses.\n",
-    "\n",
-    "3. Queue: The space complexity for the queue is O(V), as it may contain all the courses.\n",
-    "\n",
-    "The overall space complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-    "\n",
-    "In most practical cases, the number of prerequisites (E) is much smaller than the number of courses (V), so the time and space complexity can often be considered as O(V)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimizing Course Scheduling**: Given a list of courses with their durations and prerequisites, find the most efficient way to schedule these courses to minimize the time it takes to complete all of them.\n",
-    "\n",
-    "2. **Detecting Cycles in Course Dependencies**: Modify the original problem to not just determine if you can finish all courses but also to identify the specific courses that form a cycle in the prerequisite dependencies."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb b/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb
deleted file mode 100644
index a2789c4..0000000
--- a/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb	
+++ /dev/null
@@ -1,230 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 3651. Number of Connected Components in an Undirected Graph\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Number of Connected Components in an Undirected Graph problem on LintCode, click here!](https://www.lintcode.com/problem/3651/)\n",
-    "\n",
-    "---\n",
-    "In this problem, there is an undirected graph with `n` nodes. There is also an `edges` array. Where `edges[i] = [a, b]` means that there is an edge between *node a* and *node b* in the graph.\n",
-    "\n",
-    "You need to return *the number of connected components in that graph*."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand is to determine the number of connected components in an undirected graph. Here's a more detailed explanation of the problem:\n",
-    "\n",
-    "- You are given an undirected graph, which consists of nodes and edges. The graph may have multiple distinct connected components.\n",
-    "\n",
-    "- A connected component is a group of nodes within the graph where each node can be reached from every other node in the same component. In other words, there is a path between any two nodes in the same connected component, but there are no paths connecting nodes from different components.\n",
-    "\n",
-    "- Your task is to write a function that takes as input the number of nodes `n` and a list of edges `edges`, where each edge is represented as a pair of nodes `[a, b]` indicating an edge between node `a` and node `b`.\n",
-    "\n",
-    "- The goal is to determine the number of distinct connected components in the graph.\n",
-    "\n",
-    "- You need to implement an algorithm that can identify these connected components and count how many there are in the given graph.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def countConnectedComponents(n, edges):\n",
-    "    # Define a helper function for depth-first search (DFS).\n",
-    "    def dfs(node):\n",
-    "        if not visited[node]:  # If the node has not been visited yet:\n",
-    "            visited[node] = True  # Mark it as visited.\n",
-    "            for neighbor in graph[node]:  # Explore all neighbors of the current node.\n",
-    "                dfs(neighbor)  # Recursively call DFS on the neighbor.\n",
-    "\n",
-    "    # Create an empty adjacency list to represent the graph.\n",
-    "    graph = [[] for _ in range(n)]\n",
-    "\n",
-    "    # Initialize a boolean list to keep track of visited nodes.\n",
-    "    visited = [False] * n\n",
-    "\n",
-    "    # Populate the adjacency list with edges from the input.\n",
-    "    for a, b in edges:\n",
-    "        graph[a].append(b)\n",
-    "        graph[b].append(a)  # Since the graph is undirected, we add edges in both directions.\n",
-    "\n",
-    "    # Initialize a counter to keep track of the number of connected components.\n",
-    "    count = 0\n",
-    "\n",
-    "    # Iterate through all nodes in the graph.\n",
-    "    for i in range(n):\n",
-    "        if not visited[i]:  # If the node has not been visited:\n",
-    "            count += 1  # Increment the count since we've found a new connected component.\n",
-    "            dfs(i)  # Start a depth-first search from the unvisited node to explore the component.\n",
-    "\n",
-    "    return count"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided Python code calculates the number of connected components in an undirected graph. Here's a high-level explanation of how the code works:\n",
-    "\n",
-    "1. The `countConnectedComponents` function takes two arguments: `n` (the number of nodes in the graph) and `edges` (a list of edge pairs representing connections between nodes).\n",
-    "\n",
-    "2. Inside the function, there is a helper function called `dfs` (depth-first search), which is used to explore and mark nodes as visited.\n",
-    "\n",
-    "3. An empty adjacency list called `graph` is created to represent the graph. The `graph` is a list of lists where each element represents a node, and the list associated with each node contains its neighboring nodes.\n",
-    "\n",
-    "4. Another list called `visited` is created to keep track of visited nodes. Initially, all nodes are marked as not visited (`False`).\n",
-    "\n",
-    "5. The code populates the `graph` by iterating through the `edges` list and adding each edge to the adjacency list. Since the graph is undirected, both directions of each edge are added.\n",
-    "\n",
-    "6. A counter variable called `count` is initialized to keep track of the number of connected components.\n",
-    "\n",
-    "7. The code iterates through all nodes from `0` to `n-1`. For each node, it checks whether it has been visited yet. If it hasn't been visited, it means a new connected component is found. The counter `count` is incremented, and a depth-first search (DFS) is initiated from this unvisited node to explore the connected component.\n",
-    "\n",
-    "8. The `dfs` function recursively visits all nodes in the connected component, marking them as visited. It does this by checking each neighbor of the current node and recursively calling `dfs` on unvisited neighbors.\n",
-    "\n",
-    "9. After the loop through all nodes, the `count` variable contains the total number of connected components in the graph.\n",
-    "\n",
-    "10. The function returns the value of `count`, which represents the number of connected components.\n",
-    "\n",
-    "11. Two examples are provided at the end to demonstrate how to use the function with different inputs."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "n1 = 3\n",
-    "edges1 = [[0, 1], [0, 2]]\n",
-    "print(countConnectedComponents(n1, edges1))  # Output: 1"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "n2 = 6\n",
-    "edges2 = [[0, 1], [1, 2], [2, 3], [4, 5]]\n",
-    "print(countConnectedComponents(n2, edges2))  # Output: 2"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the code for counting the number of connected components in an undirected graph using depth-first search (DFS).\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The code iterates through all nodes in the graph (from 0 to `n-1`).\n",
-    "- For each unvisited node, it initiates a depth-first search (DFS) from that node to explore the connected component.\n",
-    "- In the worst case, every node may belong to a separate connected component, and you perform a DFS for each node.\n",
-    "- The time complexity of the DFS for a single connected component is O(V + E), where V is the number of nodes in the component, and E is the number of edges.\n",
-    "- In the worst case, you perform O(n) DFS calls, each with its own O(V + E) complexity.\n",
-    "\n",
-    "So, the overall time complexity of the code is O(n * (V + E)). In the worst case, if the graph is connected (a single connected component), this simplifies to O(n * (n + m)), where m is the number of edges.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The main data structures used are the `graph` (adjacency list), `visited` (boolean array), and the recursive call stack for DFS.\n",
-    "- The `graph` has a space complexity of O(n + m), where n is the number of nodes, and m is the number of edges.\n",
-    "- The `visited` boolean array also has a space complexity of O(n) to keep track of whether nodes have been visited.\n",
-    "- The depth of the recursive call stack for DFS can go up to O(n) in the worst case, when all nodes are part of a single connected component.\n",
-    "\n",
-    "So, the overall space complexity of the code is O(n + m) for data structures and O(n) for the call stack. In the worst case, this simplifies to O(n + m).\n",
-    "\n",
-    "In summary, the time complexity is O(n * (V + E)), and the space complexity is O(n + m)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Disconnected Components with Multiple Queries:** You are given an undirected graph with `n` nodes and `m` edges. Initially, the graph is disconnected, and you need to process `q` queries. Each query consists of adding an edge to connect two nodes. After each query, you need to output the number of connected components in the updated graph.\n",
-    "\n",
-    "2. **Maximum Size of Connected Components:** Given an undirected graph with `n` nodes and `m` edges, find the size of the largest connected component in the graph. Additionally, identify the nodes that belong to this largest component. You can assume there is at least one connected component in the graph.\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb b/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb
deleted file mode 100644
index c685b23..0000000
--- a/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb	
+++ /dev/null
@@ -1,255 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 417. Pacific Atlantic Water Flow\n",
-    "\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Pacific Atlantic Water Flow problem on LeetCode, click here!](https://leetcode.com/problems/pacific-atlantic-water-flow/)\n",
-    "\n",
-    "---\n",
-    "There is an `m x n` rectangular island that borders both the **Pacific Ocean** and **Atlantic Ocean**. The **Pacific Ocea**n touches the island's left and top edges, and the **Atlantic Ocean** touches the island's right and bottom edges.\n",
-    "\n",
-    "The island is partitioned into a grid of square cells. You are given an `m x n` integer matrix `heights` where `heights[r][c]` represents the **height above sea level** of the cell at coordinate `(r, c)`.\n",
-    "\n",
-    "The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is **less than or equal to** the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.\n",
-    "\n",
-    "Return *a **2D list** of grid coordinates `result` where $result[i] = [r_i, c_i]$ denotes that rain water can flow from cell $(r_i, c_i)$ to **both** the Pacific and Atlantic oceans*.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == heights.length`\n",
-    "- `n == heights[r].length`\n",
-    "- `1 <= m, n <= 200`\n",
-    "- 0 <= `heights[r][c]` <= $10^5$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand involves a scenario with a rectangular island surrounded by both the Pacific Ocean and the Atlantic Ocean. The Pacific Ocean is in contact with the island's left and top edges, while the Atlantic Ocean is in contact with the island's right and bottom edges. The goal is to determine which cells on the island can allow rainwater to flow to both the Pacific and Atlantic Oceans.\n",
-    "\n",
-    "Here are the key details and constraints of the problem:\n",
-    "\n",
-    "1. The island is divided into a grid of square cells.\n",
-    "\n",
-    "2. Each cell is represented by an integer value in a matrix `heights`. The value at `heights[r][c]` represents the height above sea level of the cell at coordinates `(r, c)` on the island.\n",
-    "\n",
-    "3. Rainwater can flow from a cell to neighboring cells directly north, south, east, or west only if the neighboring cell's height is less than or equal to the current cell's height.\n",
-    "\n",
-    "4. Water can flow from any cell adjacent to the ocean into the ocean. This means water can flow from cells on the edge of the island to the Pacific Ocean and the Atlantic Ocean.\n",
-    "\n",
-    "The task is to find and return a 2D list of grid coordinates where rainwater can flow from a cell to both the Pacific and Atlantic Oceans. In other words, you need to identify the cells that can send water to both the left and top edges (Pacific) and the right and bottom edges (Atlantic) of the island.\n",
-    "\n",
-    "The problem is solved by performing a depth-first search (DFS) starting from the ocean edges and marking the cells that can flow to each ocean. Then, the algorithm identifies cells that are reachable from both oceans and returns their coordinates as the result.\n",
-    "\n",
-    "The problem involves both traversal of the island and backtracking to find the solution, which makes it a classic graph traversal problem.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:\n",
-    "        # Check if the input matrix is empty\n",
-    "        if not heights:\n",
-    "            return []\n",
-    "        \n",
-    "        m, n = len(heights), len(heights[0])\n",
-    "        \n",
-    "        # Initialize boolean matrices to keep track of cells reachable from each ocean\n",
-    "        pacific_reachable = [[False] * n for _ in range(m)]\n",
-    "        atlantic_reachable = [[False] * n for _ in range(m)]\n",
-    "        \n",
-    "        # Depth-First Search (DFS) function to mark cells that can be reached from an ocean\n",
-    "        def dfs(x, y, reachable):\n",
-    "            if reachable[x][y]:\n",
-    "                return\n",
-    "            reachable[x][y] = True\n",
-    "            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:\n",
-    "                new_x, new_y = x + dx, y + dy\n",
-    "                if 0 <= new_x < m and 0 <= new_y < n and heights[new_x][new_y] >= heights[x][y]:\n",
-    "                    dfs(new_x, new_y, reachable)\n",
-    "        \n",
-    "        # Start DFS from the Pacific Ocean (left and top edges)\n",
-    "        for i in range(m):\n",
-    "            dfs(i, 0, pacific_reachable)\n",
-    "            dfs(i, n - 1, atlantic_reachable)\n",
-    "        \n",
-    "        for j in range(n):\n",
-    "            dfs(0, j, pacific_reachable)\n",
-    "            dfs(m - 1, j, atlantic_reachable)\n",
-    "        \n",
-    "        result = []\n",
-    "        \n",
-    "        # Find cells that can flow to both the Pacific and Atlantic oceans\n",
-    "        for i in range(m):\n",
-    "            for j in range(n):\n",
-    "                if pacific_reachable[i][j] and atlantic_reachable[i][j]:\n",
-    "                    result.append([i, j])\n",
-    "        \n",
-    "        return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a class called `Solution` with a method named `pacificAtlantic`. This method takes a 2D matrix `heights`, where each element represents the height above sea level of a cell on a rectangular island. The goal of the method is to find and return a list of grid coordinates (cells) from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-    "\n",
-    "The code uses a Depth-First Search (DFS) algorithm to traverse the island, starting from the ocean edges, and marks cells that are reachable from either the Pacific or Atlantic Ocean. It does this by creating two boolean matrices, `pacific_reachable` and `atlantic_reachable`, to keep track of cells that can be reached from each ocean.\n",
-    "\n",
-    "The code iterates through the entire island, initiating DFS searches from the edges of the island. If a cell is reachable from an ocean (either Pacific or Atlantic), it is marked as reachable in the corresponding boolean matrix. The DFS function ensures that the traversal follows the rule that water can flow to a neighboring cell if the neighboring cell's height is less than or equal to the current cell's height.\n",
-    "\n",
-    "After marking all reachable cells from both oceans, the code then looks for cells that are reachable from both the Pacific and Atlantic Oceans. If a cell satisfies this condition, it is added to the `result` list.\n",
-    "\n",
-    "Finally, the code returns the `result` list, which contains the grid coordinates of cells from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-    "\n",
-    "The code provides a solution to the problem of finding cells on the island where water can reach both oceans, demonstrating a traversal algorithm that explores the connectivity of cells on the island."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "heights1 = [\n",
-    "    [1,2,2,3,5],\n",
-    "    [3,2,3,4,4],\n",
-    "    [2,4,5,3,1],\n",
-    "    [6,7,1,4,5],\n",
-    "    [5,1,1,2,4]\n",
-    "]\n",
-    "\n",
-    "solution = Solution()\n",
-    "print(solution.pacificAtlantic(heights1)) # Expected Output: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[0, 0]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "heights2 = [[1]]\n",
-    "print(solution.pacificAtlantic(heights2)) # Expected Output: [[0, 0]]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Time Complexity:\n",
-    "The time complexity of this solution is O(m * n), where 'm' is the number of rows in the input matrix and 'n' is the number of columns. Here's a breakdown of the time complexity:\n",
-    "\n",
-    "1. Creating the `pacific_reachable` and `atlantic_reachable` matrices with dimensions m x n takes O(m * n) time.\n",
-    "\n",
-    "2. Initiating Depth-First Search (DFS) from the edges of the island takes O(m * n) time because it processes each cell once.\n",
-    "\n",
-    "3. The DFS function visits each cell at most once, and its time complexity is O(1) for each cell. The number of cells visited by the DFS in the worst case is m * n.\n",
-    "\n",
-    "4. The final step of identifying and appending cells to the result list takes O(m * n) time in the worst case.\n",
-    "\n",
-    "Overall, the time complexity is dominated by the DFS and is O(m * n).\n",
-    "\n",
-    "Space Complexity:\n",
-    "The space complexity of this solution is also O(m * n). Here's how the space is used:\n",
-    "\n",
-    "1. Two boolean matrices, `pacific_reachable` and `atlantic_reachable`, are created with dimensions m x n. Each matrix takes O(m * n) space, resulting in O(2 * m * n) space usage.\n",
-    "\n",
-    "2. The depth-first search (DFS) stack space during the recursive calls can go as deep as the diagonal of the grid, which is at most min(m, n). This additional space for the call stack is relatively small compared to the boolean matrices and can be considered O(min(m, n)).\n",
-    "\n",
-    "3. The `result` list stores the grid coordinates, and its size is determined by the number of cells that can flow to both oceans. In the worst case, this list can contain all m * n cells, so it takes O(m * n) space.\n",
-    "\n",
-    "The overall space complexity is the sum of these components, which is O(m * n) + O(min(m, n)) + O(m * n). In big O notation, we can simplify this to O(m * n)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the algorithm to find the maximum flow of water from any cell to both the Pacific and Atlantic Oceans. In other words, find the cell(s) that can send the most water to both oceans.\n",
-    "\n",
-    "2. **Efficiency Challenge:** Optimize the time and space complexity of the algorithm while maintaining correctness. Aim to reduce both time and space complexity as much as possible."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
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-   "version": "3.10.8"
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\ No newline at end of file
diff --git a/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb b/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb
deleted file mode 100644
index 19db846..0000000
--- a/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb	
+++ /dev/null
@@ -1,250 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 892. Alien Dictionary\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Alien Dictionary problem on LintCode, click here!](https://www.lintcode.com/problem/892/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of **non-empty** words from the dictionary, where words are **sorted lexicographically by the rules of this new language**. Derive the order of letters in this language.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- You may assume all letters are in lowercase.\n",
-    "- The dictionary is invalid, if string a is prefix of string b and b is appear before a.\n",
-    "- If the order is invalid, return an empty string.\n",
-    "- There may be multiple valid order of letters, return the smallest in normal lexicographical order.\n",
-    "- The letters in **one** string are of the same rank by default and are sorted in Human dictionary order."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand involves determining the order of letters in a new alien language that uses the Latin alphabet. However, the order of letters in this alien language is unknown. To solve this problem, we are given a list of non-empty words from the dictionary, where the words are sorted lexicographically based on the rules of the alien language. The goal is to derive the correct order of letters in this new language.\n",
-    "\n",
-    "Here's a more detailed explanation of the problem:\n",
-    "\n",
-    "1. The alphabet: The new alien language uses the Latin alphabet, which consists of lowercase letters.\n",
-    "\n",
-    "2. Dictionary: We are provided with a list of non-empty words. These words are sorted in lexicographical order based on the rules of the alien language. The order of letters in the words reflects the correct order in the new language.\n",
-    "\n",
-    "3. Invalid order: The problem specifies that the dictionary is considered invalid if there exists a situation where one string 'a' is a prefix of another string 'b,' and 'b' appears before 'a' in the dictionary. This condition ensures that there are no inconsistencies in the order of letters.\n",
-    "\n",
-    "4. Multiple valid orders: It's possible that there are multiple valid orders of letters that satisfy the provided dictionary. However, we are instructed to return the smallest valid order in normal lexicographical order.\n",
-    "\n",
-    "To solve this problem, you need to analyze the given list of words and determine the correct order of letters in the alien language while adhering to the specified constraints. You should return the order as a string in lexicographical order if it's valid, and an empty string if the order is invalid.\n",
-    "\n",
-    "The key to solving this problem is to construct a directed graph of letter relationships based on the order information provided by the words in the dictionary. Then, perform a topological sort on this graph to derive the order of letters, ensuring that the order is consistent and valid according to the given constraints.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from collections import defaultdict, deque\n",
-    "\n",
-    "def alienOrder(words):\n",
-    "    # Create an adjacency list to represent the graph\n",
-    "    graph = defaultdict(list)\n",
-    "    \n",
-    "    # Create a dictionary to store the in-degrees of each letter\n",
-    "    in_degree = {}\n",
-    "    \n",
-    "    # Initialize in-degrees to 0 for all letters\n",
-    "    for word in words:\n",
-    "        for char in word:\n",
-    "            if char not in in_degree:\n",
-    "                in_degree[char] = 0\n",
-    "    \n",
-    "    # Build the graph and update in-degrees\n",
-    "    for i in range(1, len(words)):\n",
-    "        word1, word2 = words[i-1], words[i]\n",
-    "        min_length = min(len(word1), len(word2))\n",
-    "        \n",
-    "        # Compare characters in the two words\n",
-    "        j = 0\n",
-    "        while j < min_length and word1[j] == word2[j]:\n",
-    "            j += 1\n",
-    "        \n",
-    "        if j < min_length:\n",
-    "            # If word1[j] is lexicographically before word2[j], add an edge\n",
-    "            graph[word1[j]].append(word2[j])\n",
-    "            in_degree[word2[j]] += 1\n",
-    "            \n",
-    "            # Break the loop to avoid further comparisons\n",
-    "            break\n",
-    "    \n",
-    "    # Perform topological sorting using Kahn's algorithm\n",
-    "    result = []\n",
-    "    queue = deque([char for char in in_degree if in_degree[char] == 0])\n",
-    "    \n",
-    "    while queue:\n",
-    "        char = queue.popleft()\n",
-    "        result.append(char)\n",
-    "        \n",
-    "        for neighbor in graph[char]:\n",
-    "            in_degree[neighbor] -= 1\n",
-    "            if in_degree[neighbor] == 0:\n",
-    "                queue.append(neighbor)\n",
-    "    \n",
-    "    # Check if there is a valid order\n",
-    "    if len(result) < len(in_degree):\n",
-    "        return \"\"\n",
-    "    \n",
-    "    return \"\".join(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code is designed to determine the order of letters in an unknown alien language that uses the Latin alphabet. It takes a list of non-empty words from a dictionary as input, where the words are sorted lexicographically based on the rules of this alien language.\n",
-    "\n",
-    "The code follows these main steps:\n",
-    "\n",
-    "1. Create a data structure to represent the graph of letters and their relationships in the alien language. It uses a defaultdict to store a list of letters that come after each letter. Additionally, it uses a dictionary called `in_degree` to keep track of the in-degrees (the number of letters that precede a letter) for each letter.\n",
-    "\n",
-    "2. Initialize the in-degrees of all letters to 0. This is done by iterating through all the words and characters in the words, adding each character to the `in_degree` dictionary with an initial in-degree of 0.\n",
-    "\n",
-    "3. Build the graph by comparing adjacent words in the dictionary. For each pair of adjacent words, the code finds the first differing character between the two words. If such a character exists, it means there's an order relationship between the letters represented by these characters. The code updates the graph and in-degrees accordingly.\n",
-    "\n",
-    "4. After constructing the graph, the code performs a topological sorting of the letters using Kahn's algorithm. It starts by initializing an empty result list and a queue containing letters with in-degrees of 0. The algorithm repeatedly removes a letter with an in-degree of 0 from the queue, adds it to the result list, and updates the in-degrees of its neighbors. This process continues until the queue is empty.\n",
-    "\n",
-    "5. Finally, the code checks if the topological sorting was successful by comparing the length of the result list with the number of unique letters in the input. If the lengths differ, it means there's a cycle in the graph, and the order is invalid. In such a case, an empty string is returned. Otherwise, the result list is joined together to form the smallest order of letters in normal lexicographical order, and it is returned as the output.\n",
-    "\n",
-    "The code provides a way to determine the order of letters in the alien language while handling various edge cases, including checking for invalid order conditions."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "wrtef\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "words1 = [\"wrt\", \"wrf\", \"er\", \"ett\", \"rftt\"]\n",
-    "print(alienOrder(words1))  # Output: \"wertf\""
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "zx\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "words2 = [\"z\", \"x\"]\n",
-    "print(alienOrder(words2))  # Output: \"zx\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for determining the order of letters in an alien language.\n",
-    "\n",
-    "Time Complexity:\n",
-    "1. Building the graph: The code iterates through the list of words and compares adjacent words. In the worst case, where all words have the same prefix, it takes O(N) time to build the graph, where N is the total number of characters in all the words.\n",
-    "2. Topological sorting: Performing a topological sort on the graph takes O(V + E) time, where V is the number of unique letters (vertices) and E is the number of relationships between letters (edges) in the graph.\n",
-    "\n",
-    "Overall, the time complexity is O(N + V + E), where N is the total number of characters in the words, V is the number of unique letters, and E is the number of order relationships between letters.\n",
-    "\n",
-    "Space Complexity:\n",
-    "1. Graph and In-degrees: The code uses data structures to represent the graph and in-degrees. The space complexity for these data structures is O(V + E), where V is the number of unique letters, and E is the number of order relationships between letters in the words.\n",
-    "2. Result List: The result list stores the order of letters, which can have a maximum size of V, where V is the number of unique letters.\n",
-    "3. Queue: The space used by the queue during topological sorting is also O(V).\n",
-    "\n",
-    "Overall, the space complexity is O(V + E) due to the data structures used for graph representation and the space needed for storing the result and the queue.\n",
-    "\n",
-    "In the worst case, where there are many unique letters and many order relationships, the space complexity is dominated by the number of unique letters and their relationships.\n",
-    "\n",
-    "In summary, the time complexity of the code is O(N + V + E), and the space complexity is O(V + E). The actual performance will depend on the specific input data, such as the number of unique letters and the relationships between them in the alien language."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple Valid Orders**: Extend the problem to handle cases where there are multiple valid orders of letters in the alien language. Modify the code to return all valid orders instead of just one. Be mindful of performance.\n",
-    "\n",
-    "2. **Detect Invalid Dictionary**: Given a dictionary of words, write a function to detect whether the dictionary is valid or not based on the constraints mentioned in the problem statement."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
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- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb b/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb
deleted file mode 100644
index 9539363..0000000
--- a/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb	
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# House Robber\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "---\n",
-    "You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and **it will automatically contact the police if two adjacent houses were broken into on the same night.**\n",
-    "\n",
-    "Given an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight **without alerting the police**.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= nums.length <= 100`\n",
-    "- `0 <= nums[i] <= 400`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is a classic dynamic programming challenge known as the \"House Robber\" problem. The scenario is that you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, but the constraint is that adjacent houses have security systems connected. If two adjacent houses are broken into on the same night, the security systems will automatically contact the police.\n",
-    "\n",
-    "The goal is to determine the maximum amount of money you can rob tonight without alerting the police. You need to find a strategy to maximize your earnings while avoiding robbing two adjacent houses.\n",
-    "\n",
-    "The input to the problem is an array `nums`, where `nums[i]` represents the amount of money in the `i`-th house. The task is to return the maximum amount of money that can be robbed without triggering the security system.\n",
-    "\n",
-    "For example:\n",
-    "```\n",
-    "Input: nums = [1,2,3,1]\n",
-    "Output: 4\n",
-    "Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).\n",
-    "Total amount you can rob = 1 + 3 = 4.\n",
-    "\n",
-    "Input: nums = [2,7,9,3,1]\n",
-    "Output: 12\n",
-    "Explanation: Rob house 1 (money = 2), rob house 3 (money = 9), and rob house 5 (money = 1).\n",
-    "Total amount you can rob = 2 + 9 + 1 = 12.\n",
-    "```\n",
-    "\n",
-    "The challenge involves designing an algorithm to efficiently solve this problem and find the optimal strategy for robbing houses to maximize the stolen money while adhering to the constraint of not robbing adjacent houses. The provided solution utilizes dynamic programming to achieve this.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def rob(self, nums: List[int]) -> int:\n",
-    "        # Initialize variables to keep track of the maximum amount of money robbed at the previous and current houses\n",
-    "        prev, curr = 0, 0\n",
-    "\n",
-    "        # Iterate through the houses\n",
-    "        for n in nums:\n",
-    "            # Calculate the maximum amount of money that can be robbed at the current house\n",
-    "            # It is the maximum of either robbing the current house and the money robbed at the house before the previous one (prev),\n",
-    "            # or skipping the current house and taking the maximum amount from the previous house (curr)\n",
-    "            temp = max(n + prev, curr)\n",
-    "\n",
-    "            # Update the variables for the next iteration\n",
-    "            prev, curr = curr, temp\n",
-    "\n",
-    "        # The final result is the maximum amount of money that can be robbed at the last house\n",
-    "        return curr"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code is a Python implementation of the \"House Robber\" problem using a dynamic programming approach. It's part of a class `Solution` and includes a method `rob` that takes a list of integers `nums` as input and returns the maximum amount of money that can be robbed without alerting the police.\n",
-    "\n",
-    "The key idea of the algorithm is to maintain two variables, `prev` and `curr`, which represent the maximum amount of money that can be robbed up to the previous house and the current house, respectively. The algorithm iterates through the houses, updating these variables based on the maximum amount that can be robbed at the current house, considering the constraint of not robbing adjacent houses.\n",
-    "\n",
-    "The loop calculates a temporary variable `temp` that represents the maximum amount of money that can be robbed at the current house. It considers two scenarios: either robbing the current house and adding the money robbed at the house before the previous one (`n + prev`), or skipping the current house and taking the maximum amount from the previous house (`curr`). The maximum of these two scenarios is assigned to `temp`.\n",
-    "\n",
-    "The `prev` and `curr` variables are then updated for the next iteration. `prev` is set to the previous value of `curr`, and `curr` is set to the calculated `temp`.\n",
-    "\n",
-    "The final result is the value stored in the `curr` variable, representing the maximum amount of money that can be robbed at the last house. This value is returned as the result of the `rob` method.\n",
-    "\n",
-    "The provided test cases at the end demonstrate the usage of the `rob` method with different input arrays."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "sol = Solution()\n",
-    "\n",
-    "nums1 = [1, 2, 3, 1]\n",
-    "print(sol.rob(nums1))  # Output: 4"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "12\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [2, 7, 9, 3, 1]\n",
-    "print(sol.rob(nums2))  # Output: 12"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code.\n",
-    "\n",
-    "### Time Complexity:\n",
-    "\n",
-    "The time complexity of the code is O(n), where n is the number of houses. The algorithm iterates through the array once, and in each iteration, it performs constant time operations. The loop runs for each house, and the number of operations inside the loop is independent of the input size. Therefore, the overall time complexity is linear, O(n).\n",
-    "\n",
-    "### Space Complexity:\n",
-    "\n",
-    "The space complexity is O(1), constant space. The algorithm uses only a constant amount of extra space regardless of the input size. The variables `prev`, `curr`, and `temp` are the only additional variables used, and they do not depend on the size of the input array. Thus, the space complexity is constant, O(1).\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: O(n)\n",
-    "- Space Complexity: O(1)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Variation with Maximum Number of Houses to Rob:** Modify the problem to include an additional constraint: you are allowed to rob at most 'k' houses without triggering the alarm. Find the maximum amount you can rob under this new constraint.\n",
-    "\n",
-    "2. **Robbing in a Circular Street:** Extend the problem to a circular street where the first and last houses are adjacent. You cannot rob both the first and last houses simultaneously. Find the maximum amount you can rob in this circular setting.\n",
-    "\n",
-    "\n",
-    "**Link to Problem:** [To see the House Robber problem on LeetCode, click here!](https://leetcode.com/problems/house-robber/description/)\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb b/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb
deleted file mode 100644
index d210017..0000000
--- a/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb	
+++ /dev/null
@@ -1,243 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# House Robber II\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "---\n",
-    "You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle**. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.\n",
-    "\n",
-    "Given an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight without alerting the police*.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= nums.length <= 100`\n",
-    "- `0 <= nums[i] <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "Certainly! The problem statement describes a scenario where you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. However, there is a security system in place: adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.\n",
-    "\n",
-    "The twist in this problem is that the houses are arranged in a circle. That means the first house is the neighbor of the last one. So, robbing the first house has implications for the last house, and vice versa.\n",
-    "\n",
-    "The goal is to find the maximum amount of money you can rob tonight without alerting the police. You need to implement a function that takes an integer array `nums` representing the amount of money in each house and returns the maximum amount of money that can be robbed without triggering the security system.\n",
-    "\n",
-    "Here are a few examples to illustrate the problem:\n",
-    "\n",
-    "1. **Example 1:**\n",
-    "   ```python\n",
-    "   Input: nums = [2, 3, 2]\n",
-    "   Output: 3\n",
-    "   Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2) because they are adjacent houses. The optimal strategy is to rob house 2 (money = 3).\n",
-    "   ```\n",
-    "\n",
-    "2. **Example 2:**\n",
-    "   ```python\n",
-    "   Input: nums = [1, 2, 3, 1]\n",
-    "   Output: 4\n",
-    "   Explanation: Robbing house 1 (money = 1) and then robbing house 3 (money = 3) results in a total amount of 4. Robbing house 2 is skipped to avoid alerting the police.\n",
-    "   ```\n",
-    "\n",
-    "3. **Example 3:**\n",
-    "   ```python\n",
-    "   Input: nums = [1, 2, 3]\n",
-    "   Output: 3\n",
-    "   Explanation: Since the houses are in a circle, you can choose either house 1, 2, or 3. The optimal strategy is to rob house 3 for a total amount of 3.\n",
-    "   ```\n",
-    "\n",
-    "The constraints for the problem are that the length of the `nums` array is between 1 and 100, and the amount of money in each house (`nums[i]`) is between 0 and 1000.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def rob(self, nums: List[int]) -> int:\n",
-    "        def simple_rob(nums):\n",
-    "            # Helper function for basic house robbery problem\n",
-    "            prev, curr = 0, 0\n",
-    "            for num in nums:\n",
-    "                # Calculate the maximum amount of money that can be robbed\n",
-    "                # without alerting the police for each house\n",
-    "                prev, curr = curr, max(prev + num, curr)\n",
-    "            return curr\n",
-    "\n",
-    "        # Check the base cases where the length of nums is 1 or 2\n",
-    "        if len(nums) == 1:\n",
-    "            return nums[0]\n",
-    "        elif len(nums) == 2:\n",
-    "            return max(nums[0], nums[1])\n",
-    "\n",
-    "        # Case 1: First house is robbed, last house is not\n",
-    "        case1 = simple_rob(nums[:-1])\n",
-    "\n",
-    "        # Case 2: First house is not robbed, last house is\n",
-    "        case2 = simple_rob(nums[1:])\n",
-    "\n",
-    "        # Maximum of the two cases\n",
-    "        return max(case1, case2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code defines a class `Solution` with a method `rob`, which aims to determine the maximum amount of money a professional robber can steal from houses arranged in a circle without alerting the police. Each house contains a certain amount of money, and adjacent houses have a security system that triggers if both are robbed on the same night.\n",
-    "\n",
-    "The code uses dynamic programming to solve the problem efficiently. It contains a helper function `simple_rob` that computes the maximum amount of money that can be robbed from a linear arrangement of houses. The main method `rob` handles the circular arrangement by considering two cases: robbing the first house and not robbing the last, and not robbing the first house and robbing the last. The result is the maximum of these two cases.\n",
-    "\n",
-    "The class is instantiated as `solution`, and the `rob` method is applied to three different scenarios (test cases) to demonstrate its functionality. The final output represents the maximum amount of money the robber can steal without triggering the security system."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "nums1 = [2, 3, 2]\n",
-    "print(solution.rob(nums1))  # Output: 3"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [1, 2, 3, 1]\n",
-    "print(solution.rob(nums2))  # Output: 4"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e6e1554f-ae77-410c-8af2-e59e5428166c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums3 = [1, 2, 3]\n",
-    "print(solution.rob(nums3))  # Output: 3"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "### Time Complexity:\n",
-    "The time complexity is primarily determined by the `simple_rob` helper function, which has a time complexity of O(N), where N is the number of houses in the input array. The main `rob` function calls `simple_rob` twice, once for the case of robbing the first house and not robbing the last, and once for the case of not robbing the first house and robbing the last. Therefore, the overall time complexity is O(N).\n",
-    "\n",
-    "### Space Complexity:\n",
-    "The space complexity is O(1) because the space used by the algorithm is constant, regardless of the size of the input array. The only variables used are `prev` and `curr` in the `simple_rob` function, which represent the previous and current maximum amounts of money that can be robbed without alerting the police. These variables are updated iteratively, and no additional data structures are used that scale with the input size. Therefore, the space complexity is constant, or O(1).\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: O(N)\n",
-    "- Space Complexity: O(1)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Optimize the space complexity of the solution. Can you solve the problem with O(1) space complexity without sacrificing the time complexity?\n",
-    "\n",
-    "2. **Circular Robbery Variations:** Consider variations of the circular house robbery problem, such as allowing the robber to skip a certain number of houses or restricting the number of consecutive houses that can be robbed. Modify the solution to accommodate these variations.\n",
-    "\n",
-    "**Link to Problem:** [To see the House Robber II problem on LeetCode, click here!](https://leetcode.com/problems/house-robber-ii/)"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb b/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb
deleted file mode 100644
index 32c5da3..0000000
--- a/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb	
+++ /dev/null
@@ -1,204 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 70. Climbing Stairs\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Climbing Stairs problem on LeetCode, click here!](https://leetcode.com/problems/climbing-stairs/)\n",
-    "\n",
-    "---\n",
-    "You are climbing a staircase. It takes `n` steps to reach the top.\n",
-    "\n",
-    "Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= n <= 45`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The \"Climbing Stairs\" problem is a classic algorithmic problem that asks you to find the number of distinct ways to climb a staircase with a given number of steps. Here's the problem statement:\n",
-    "\n",
-    "You are given a staircase with 'n' steps. Each time you can either climb 1 step or 2 steps. You need to find out how many distinct ways there are to reach the top of the staircase.\n",
-    "\n",
-    "For example, if 'n' is 2, there are two ways to climb to the top:\n",
-    "\n",
-    "1. Climbing two steps at once.\n",
-    "2. Climbing one step, and then another step.\n",
-    "\n",
-    "If 'n' is 3, there are three ways:\n",
-    "\n",
-    "1. Climbing one step, then one step, and finally one step.\n",
-    "2. Climbing one step, then two steps.\n",
-    "3. Climbing two steps, then one step.\n",
-    "\n",
-    "The problem essentially asks you to find a solution that counts the possible combinations of 1-step and 2-step climbs that can be taken to reach the top of the staircase. It's a classic example of a dynamic programming problem where you can break it down into smaller subproblems and build a solution incrementally, as shown in the code I provided earlier.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class Solution:\n",
-    "    def climbStairs(self, n: int) -> int:\n",
-    "        # If there are 1 or 2 steps, there are exactly n ways to reach the top.\n",
-    "        if n <= 2:\n",
-    "            return n\n",
-    "\n",
-    "        # Create an array to store the number of ways to reach each step.\n",
-    "        dp = [0] * (n + 1)\n",
-    "\n",
-    "        # There is one way to reach the first step (base case).\n",
-    "        dp[1] = 1\n",
-    "\n",
-    "        # There are two ways to reach the second step (base case).\n",
-    "        dp[2] = 2\n",
-    "\n",
-    "        # Calculate the number of ways to reach each step from the previous steps.\n",
-    "        for i in range(3, n + 1):\n",
-    "            # The number of ways to reach step 'i' is the sum of the ways to reach\n",
-    "            # the previous two steps, as you can take 1 or 2 steps at a time.\n",
-    "            dp[i] = dp[i - 1] + dp[i - 2]\n",
-    "\n",
-    "        # The result is the number of ways to reach the top, which is stored in dp[n].\n",
-    "        return dp[n]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a class `Solution` with a method `climbStairs` that takes an integer `n` as input and returns the number of distinct ways to climb a staircase with `n` steps.\n",
-    "\n",
-    "1. If `n` is 1 or 2, the function returns `n` directly because there are only 1 or 2 distinct ways to reach the top in those cases.\n",
-    "\n",
-    "2. For `n` greater than 2, the code uses dynamic programming to calculate the number of ways to reach each step. It creates an array `dp` of length `n + 1` to store these values.\n",
-    "\n",
-    "3. The `dp[1]` and `dp[2]` are initialized to 1 and 2, respectively. This is because there is only one way to reach the first step (by taking 1 step), and there are two ways to reach the second step (by taking 2 steps or two 1-step climbs).\n",
-    "\n",
-    "4. The code then uses a loop to fill in the `dp` array for steps 3 and above. For each step `i` from 3 to `n`, it calculates the number of ways to reach that step by adding the number of ways to reach the previous two steps, as you can either take 1 step or 2 steps at a time.\n",
-    "\n",
-    "5. Finally, the function returns the value stored in `dp[n]`, which represents the total number of distinct ways to climb the staircase with `n` steps."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 1 - Input: n = 2, Output: 2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "n = 2\n",
-    "solution = Solution()\n",
-    "print(f\"Example 1 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 2"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 2 - Input: n = 3, Output: 3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "n = 3\n",
-    "print(f\"Example 2 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 3\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Time Complexity:\n",
-    "The time complexity of the solution is O(n) because we use a single for loop that iterates from 3 to n to fill the `dp` array. In each iteration, we perform constant time operations (addition and assignment). Therefore, the time complexity is linear with respect to the input value `n`.\n",
-    "\n",
-    "Space Complexity:\n",
-    "The space complexity of the solution is O(n) as well. We use an array `dp` of length `n + 1` to store the number of ways to reach each step. The size of this array is directly proportional to the input value `n`, which makes the space complexity linear in terms of `n`. The rest of the variables used in the function (e.g., `i`) occupy constant space and do not contribute significantly to the overall space complexity.\n",
-    "\n",
-    "In summary, the time complexity is O(n), and the space complexity is O(n) for this dynamic programming solution to the \"Climbing Stairs\" problem."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Generalized Step Sizes**: Instead of being limited to 1 step or 2 steps, consider a scenario where you can take steps of sizes `[1, 2, 3]`. Implement a function that calculates the number of ways to climb the staircase with these different step sizes.\n",
-    "\n",
-    "2. **Minimize Steps**: Given an integer `n`, find the minimum number of steps required to reach the top of the staircase. Return both the minimum number of steps and the specific sequence of steps taken."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/Book.ipynb b/_build/jupyter_execute/Book.ipynb
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index 6c891d7..0000000
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+++ /dev/null
@@ -1,914 +0,0 @@
-{
- "cells": [
-  {
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-   "metadata": {},
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-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: backcall in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.2.0)\r\n",
-      "Requirement already satisfied: decorator in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (5.1.1)\r\n",
-      "Requirement already satisfied: jedi>=0.16 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.19.1)\r\n",
-      "Requirement already satisfied: pickleshare in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.7.5)\r\n",
-      "Requirement already satisfied: prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (3.0.39)\r\n",
-      "Requirement already satisfied: stack-data in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.6.3)\r\n",
-      "Requirement already satisfied: exceptiongroup in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (1.1.3)\r\n",
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-     ]
-    },
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-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: parso<0.9.0,>=0.8.3 in /home/codespace/.local/lib/python3.10/site-packages (from jedi>=0.16->ipython->myst-nb~=0.17.1->jupyter-book) (0.8.3)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: python-dateutil>=2.8.2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-client>=6.1.12->nbclient->myst-nb~=0.17.1->jupyter-book) (2.8.2)\r\n",
-      "Requirement already satisfied: platformdirs>=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-core!=5.0.*,>=4.12->nbclient->myst-nb~=0.17.1->jupyter-book) (3.11.0)\r\n"
-     ]
-    },
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-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py<3.0.0,>=1.0.0->myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.1.2)\r\n",
-      "Requirement already satisfied: ptyprocess>=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect>4.3->ipython->myst-nb~=0.17.1->jupyter-book) (0.7.0)\r\n",
-      "Requirement already satisfied: wcwidth in /home/codespace/.local/lib/python3.10/site-packages (from prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.8)\r\n"
-     ]
-    },
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-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: greenlet!=0.4.17 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sqlalchemy<3,>=1.3.12->jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (3.0.1)\r\n"
-     ]
-    },
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-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: soupsieve>1.2 in /home/codespace/.local/lib/python3.10/site-packages (from beautifulsoup4->pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (2.5)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: executing>=1.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.0.0)\r\n",
-      "Requirement already satisfied: asttokens>=2.1.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.4.0)\r\n",
-      "Requirement already satisfied: pure-eval in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.2)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Note: you may need to restart the kernel to use updated packages.\n"
-     ]
-    }
-   ],
-   "source": [
-    "!python -m pip install --upgrade pip\n",
-    "%pip install jupyter-book"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions\n"
-     ]
-    }
-   ],
-   "source": [
-    "%cd /workspaces/LeetCode-Solutions"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Traceback (most recent call last):\r\n",
-      "  File \"/home/codespace/.python/current/bin/jupyter-book\", line 8, in <module>\r\n",
-      "    sys.exit(main())\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1157, in __call__\r\n",
-      "    return self.main(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1078, in main\r\n",
-      "    rv = self.invoke(ctx)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
-      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
-      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1434, in invoke\r\n",
-      "    return ctx.invoke(self.callback, **ctx.params)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 783, in invoke\r\n",
-      "    return __callback(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx_external_toc/cli.py\", line 139, in migrate_toc\r\n",
-      "    toc = migrate_jupyter_book(Path(toc_file))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx_external_toc/tools.py\", line 327, in migrate_jupyter_book\r\n",
-      "    raise MalformedError(\"no top-level 'file' key found\")\r\n",
-      "sphinx_external_toc.parsing.MalformedError: no top-level 'file' key found\r\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book toc migrate /workspaces/LeetCode-Solutions/_toc.yml -o /workspaces/LeetCode-Solutions/_toc.yml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions\r\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\r\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\r\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[etoc] Changing master_doc to 'README'\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\r\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\r\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\r\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 77 source files that are out of date\r\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 77 added, 0 changed, 0 removed\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[  1%] \u001b[35m01. Array Hashing/1. Two Sum\u001b[39;49;00m                          \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.07 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  2%] \u001b[35m01. Array Hashing/128. Longest Consecutive Sequence\u001b[39;49;00m   \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.17 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[  3%] \u001b[35m01. Array Hashing/217. Contains Duplicate\u001b[39;49;00m             \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.19 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[  5%] \u001b[35m01. Array Hashing/238. Product of Array Except Self\u001b[39;49;00m   \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 1.26 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  6%] \u001b[35m01. Array Hashing/242. Valid Anagram\u001b[39;49;00m                  \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 1.18 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  7%] \u001b[35m01. Array Hashing/347. Top K Frequent Elements\u001b[39;49;00m        \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.13 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  9%] \u001b[35m01. Array Hashing/49. Group Anagrams\u001b[39;49;00m                  \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 4.02 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 10%] \u001b[35m01. Array Hashing/659. Encode and Decode Strings\u001b[39;49;00m      \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 11%] \u001b[35m01. Array Hashing/README\u001b[39;49;00m                              \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 12%] \u001b[35m02. Two Pointers/11. Container With Most Water\u001b[39;49;00m        \r",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.26 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 14%] \u001b[35m02. Two Pointers/125. Valid Palindrome\u001b[39;49;00m                \r",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 1.07 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 15%] \u001b[35m02. Two Pointers/15. 3Sum\u001b[39;49;00m                             \r",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.00 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 16%] \u001b[35m02. Two Pointers/README\u001b[39;49;00m                               \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 18%] \u001b[35m03. Sliding Window/121. Best Time to Buy and Sell Stock\u001b[39;49;00m\r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.89 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 19%] \u001b[35m03. Sliding Window/3. Longest Substring Without Repeating Characters\u001b[39;49;00m\r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 1.16 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 20%] \u001b[35m03. Sliding Window/424. Longest Repeating Character Replacement\u001b[39;49;00m\r",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 22%] \u001b[35m03. Sliding Window/76. Minimum Window Substring\u001b[39;49;00m       \r",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 0.87 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 23%] \u001b[35m03. Sliding Window/README\u001b[39;49;00m                             \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 24%] \u001b[35m04. Stack/20. Valid Parentheses\u001b[39;49;00m                       \r",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 4.02 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 25%] \u001b[35m04. Stack/README\u001b[39;49;00m                                      \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 27%] \u001b[35m05. Binary Search/153. Find Minimum in Rotated Sorted Array\u001b[39;49;00m\r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 0.98 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 28%] \u001b[35m05. Binary Search/33. Search in Rotated Sorted Array\u001b[39;49;00m  \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 29%] \u001b[35m05. Binary Search/README\u001b[39;49;00m                              \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 31%] \u001b[35m06. Linked List/141. Linked List Cycle\u001b[39;49;00m                \r",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 0.96 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 32%] \u001b[35m06. Linked List/143. Reorder List\u001b[39;49;00m                     \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "^C\r\n",
-      "\r\n",
-      "\u001b[91mException occurred:\u001b[39;49;00m\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 512, in _async_cleanup_kernel\r\n",
-      "    await ensure_async(self.km.cleanup_resources())\r\n",
-      "AttributeError: 'NoneType' object has no attribute 'cleanup_resources'\r\n",
-      "\u001b[91mThe full traceback has been saved in /tmp/sphinx-err-tl19fjln.log, if you want to report the issue to the developers.\u001b[39;49;00m\r\n",
-      "Please also report this if it was a user error, so that a better error message can be provided next time.\r\n",
-      "A bug report can be filed in the tracker at <https://github.com/sphinx-doc/sphinx/issues>. Thanks!\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py\", line 89, in wrapper\r\n",
-      "    self._ready.set_result(None)\r\n",
-      "asyncio.exceptions.InvalidStateError: invalid state\r\n",
-      "\r\n",
-      "During handling of the above exception, another exception occurred:\r\n",
-      "\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 505, in _async_cleanup_kernel\r\n",
-      "    await ensure_async(self.km.shutdown_kernel(now=now))\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py\", line 189, in ensure_async\r\n",
-      "    result = await obj\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py\", line 92, in wrapper\r\n",
-      "    self._ready.set_exception(e)\r\n",
-      "asyncio.exceptions.InvalidStateError: invalid state\r\n",
-      "\r\n",
-      "During handling of the above exception, another exception occurred:\r\n",
-      "\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/sphinx.py\", line 171, in build_sphinx\r\n",
-      "    app.build(force_all, filenames)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/application.py\", line 329, in build\r\n",
-      "    self.builder.build_update()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 288, in build_update\r\n",
-      "    self.build(to_build,\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 302, in build\r\n",
-      "    updated_docnames = set(self.read())\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 409, in read\r\n",
-      "    self._read_serial(docnames)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 430, in _read_serial\r\n",
-      "    self.read_doc(docname)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 483, in read_doc\r\n",
-      "    publisher.publish()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/core.py\", line 217, in publish\r\n",
-      "    self.document = self.reader.read(self.source, self.parser,\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/io.py\", line 103, in read\r\n",
-      "    self.parse()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/readers/__init__.py\", line 78, in parse\r\n",
-      "    self.parser.parse(self.input, document)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/sphinx_.py\", line 150, in parse\r\n",
-      "    with create_client(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/base.py\", line 83, in __enter__\r\n",
-      "    self.start_client()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/direct.py\", line 40, in start_client\r\n",
-      "    result = single_nb_execution(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_cache/executors/utils.py\", line 58, in single_nb_execution\r\n",
-      "    executenb(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 1305, in execute\r\n",
-      "    return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute()\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py\", line 173, in wrapped\r\n",
-      "    return loop.run_until_complete(inner)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/asyncio/base_events.py\", line 649, in run_until_complete\r\n",
-      "    return future.result()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 689, in async_execute\r\n",
-      "    async with self.async_setup_kernel(**kwargs):\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/contextlib.py\", line 206, in __aexit__\r\n",
-      "    await anext(self.gen)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 656, in async_setup_kernel\r\n",
-      "    await self._async_cleanup_kernel()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 512, in _async_cleanup_kernel\r\n",
-      "    await ensure_async(self.km.cleanup_resources())\r\n",
-      "AttributeError: 'NoneType' object has no attribute 'cleanup_resources'\r\n",
-      "\r\n",
-      "The above exception was the direct cause of the following exception:\r\n",
-      "\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/home/codespace/.python/current/bin/jupyter-book\", line 8, in <module>\r\n",
-      "    sys.exit(main())\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1157, in __call__\r\n",
-      "    return self.main(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1078, in main\r\n",
-      "    rv = self.invoke(ctx)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
-      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1434, in invoke\r\n",
-      "    return ctx.invoke(self.callback, **ctx.params)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 783, in invoke\r\n",
-      "    return __callback(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py\", line 317, in build\r\n",
-      "    builder_specific_actions(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py\", line 525, in builder_specific_actions\r\n",
-      "    raise RuntimeError(_message_box(msg, color=\"red\", doprint=False)) from result\r\n",
-      "RuntimeError: \r\n",
-      "\u001b[91m===============================================================================\u001b[0m\r\n",
-      "\r\n",
-      "There was an error in building your book. Look above for the cause.\r\n",
-      "\r\n",
-      "\u001b[91m===============================================================================\u001b[0m\r\n",
-      "\r\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build ."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book build /workspaces/LeetCode-Solutions/ --builder pdfhtml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/html": [
-       "<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>"
-      ],
-      "text/plain": [
-       "<IPython.core.display.HTML object>"
-      ]
-     },
-     "execution_count": 5,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "from IPython.display import HTML\n",
-    "HTML('<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>')\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Your book template can be found at\n",
-      "\n",
-      "    mybook/\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book create mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions/mybook\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_build/html\n",
-      "[sphinxcontrib-bibtex] Beware that docutils versions 0.18 and 0.19 (you are running 0.18.1) are known to generate invalid html for citations. If this issue affects you, please use docutils<0.18 (or >=0.20 once released) instead. For more details, see https://sourceforge.net/p/docutils/patches/195/\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "[etoc] Changing master_doc to 'intro'\n",
-      "checking bibtex cache... out of date\n",
-      "parsing bibtex file /workspaces/LeetCode-Solutions/mybook/references.bib... parsed 5 entries\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='force', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/mybook/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 14 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 14 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.35 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.97 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 0.83 seconds [mystnb]\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mintro\u001b[39;49;00m                                                 \n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mintro\u001b[39;49;00m                                                  \n",
-      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
-      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
-      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
-      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
-      "[etoc] missing index.html written as redirect to 'intro.html'\n",
-      "\u001b[01mbuild succeeded, 2 warnings.\u001b[39;49;00m\n",
-      "\n",
-      "The HTML pages are in mybook/_build/html.\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Finished generating HTML for book.\n",
-      "Your book's HTML pages are here:\n",
-      "    mybook/_build/html/\n",
-      "You can look at your book by opening this file in a browser:\n",
-      "    mybook/_build/html/index.html\n",
-      "Or paste this line directly into your browser bar:\n",
-      "    file:///workspaces/LeetCode-Solutions/mybook/_build/html/index.html            \n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "format: jb-book\n",
-      "root: intro\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book toc from-project /workspaces/LeetCode-Solutions/mybook/ -f jb-book"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 2
-}
\ No newline at end of file
diff --git a/_config.yml b/_config.yml
deleted file mode 100644
index 3def3c7..0000000
--- a/_config.yml
+++ /dev/null
@@ -1,31 +0,0 @@
-# Jupyter Book configuration
-
-# Project information
-title: "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges" # Python Problem-Solving Pro: Conquering the Blind 75 LeetCode Challenges
-author: Mohsen Tabibian and Saeed Samadidana
-email: mohsentabibian@gmail.com
-description: Explore 75 essential problems frequently asked in tech interviews. Each solution includes detailed explanations and time/space complexity analysis. Boost your coding skills and interview preparedness!
-
-repository: 
-   url: https://github.com/mohsentabibian/LeetCode-Solutions
-   path: docs #LeetCode-Solutions/
-   branch: main
-
-# Execute notebooks when building (set to false if you don't want to run the code)
-execute:
-  execute_notebooks: auto
-
-# HTML settings
-html:
-  theme: jekyll-theme-Architect #jupyter
-  logo: logo.png
-  toc:
-    display: true
-  use_issues_button: true
-  use_repository_button: true
-
-#Build settings
-#repository_url: https://github.com/mohsentabibian/LeetCode-Solutions  # URL to the GitHub repository
-
-# Add GitHub buttons to the book
-# See https://jupyterbook.org/customize/config.html #add-a-link-to-your-repository
\ No newline at end of file
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+++ /dev/null
@@ -1,88 +0,0 @@
-format: jb-book
-root: README
-chapters:
-  - file: 01. Array Hashing/README
-    sections:
-      - file: 01. Array Hashing/1. Two Sum
-      - file: 01. Array Hashing/128. Longest Consecutive Sequence
-      - file: 01. Array Hashing/217. Contains Duplicate
-      - file: 01. Array Hashing/238. Product of Array Except Self
-      - file: 01. Array Hashing/242. Valid Anagram
-      - file: 01. Array Hashing/347. Top K Frequent Elements
-      - file: 01. Array Hashing/49. Group Anagrams
-      - file: 01. Array Hashing/659. Encode and Decode Strings
-  - file: 02. Two Pointers/README
-    sections:
-      - file: 02. Two Pointers/11. Container With Most Water
-      - file: 02. Two Pointers/125. Valid Palindrome
-      - file: 02. Two Pointers/15. 3Sum
-  - file: 03. Sliding Window/README
-    sections:
-      - file: 03. Sliding Window/121. Best Time to Buy and Sell Stock
-      - file: 03. Sliding Window/3. Longest Substring Without Repeating Characters
-      - file: 03. Sliding Window/424. Longest Repeating Character Replacement
-      - file: 03. Sliding Window/76. Minimum Window Substring
-  - file: 04. Stack/README
-    sections:
-      - file: 04. Stack/20. Valid Parentheses
-  - file: 05. Binary Search/README
-    sections:
-      - file: 05. Binary Search/153. Find Minimum in Rotated Sorted Array
-      - file: 05. Binary Search/33. Search in Rotated Sorted Array
-  - file: 06. Linked List/README
-    sections:
-      - file: 06. Linked List/141. Linked List Cycle
-      - file: 06. Linked List/143. Reorder List
-      - file: 06. Linked List/19. Remove Nth Node From End of List
-      - file: 06. Linked List/206. Reverse Linked List
-      - file: 06. Linked List/21. Merge Two Sorted Lists
-      - file: 06. Linked List/23. Merge k Sorted Lists
-  - file: 07. Trees/README
-    sections:
-      - file: 07. Trees/100. Same Tree
-      - file: 07. Trees/102. Binary Tree Level Order Traversal
-      - file: 07. Trees/104. Maximum Depth of Binary Tree
-      - file: 07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal
-      - file: 07. Trees/124. Binary Tree Maximum Path Sum
-      - file: 07. Trees/199. Binary Tree Right Side View
-      - file: 07. Trees/226. Invert Binary Tree
-      - file: 07. Trees/230. Kth Smallest Element in a BST
-      - file: 07. Trees/235. Lowest Common Ancestor of a Binary Search Tree
-      - file: 07. Trees/297. Serialize and Deserialize Binary Tree
-      - file: 07. Trees/572. Subtree of Another Tree
-      - file: 07. Trees/98. Validate Binary Search Tree
-  - file: 08. Tries/README
-    sections:
-      - file: 08. Tries/208. Implement Trie (Prefix Tree)
-      - file: 08. Tries/211. Design Add and Search Words Data Structure
-      - file: 08. Tries/212. Word Search II
-  - file: 09. Heap - Priority Queue/README
-    sections:
-      - file: 09. Heap - Priority Queue/1046. Last Stone Weight
-      - file: 09. Heap - Priority Queue/295. Find Median from Data Stream
-      - file: 09. Heap - Priority Queue/703. Kth Largest Element in a Stream
-  - file: 10. Backtracking/README
-    sections:
-      - file: 10. Backtracking/39. Combination Sum
-      - file: 10. Backtracking/79. Word Search
-  - file: 11. Graphs/README
-    sections:
-      - file: 11. Graphs/133. Clone Graph
-      - file: 11. Graphs/178. Graph Valid Tree
-      - file: 11. Graphs/200. Number of Islands
-      - file: 11. Graphs/207. Course Schedule
-      - file: 11. Graphs/3651. Number of Connected Components in an Undirected Graph
-      - file: 11. Graphs/417. Pacific Atlantic Water Flow
-  - file: 12. Advanced Graphs/README
-    sections:
-      - file: 12. Advanced Graphs/892. Alien Dictionary
-  - file: 13. One-D Dynamic Programming/README
-    sections:
-      - file: 13. One-D Dynamic Programming/70. Climbing Stairs
-      - file: 13. One-D Dynamic Programming/198. House Robber
-      - file: 13. One-D Dynamic Programming/213. House Robber II
-  - file: 14. Two-D Dynamic Programming/README
-  - file: 15. Greedy/README
-  - file: 16. Intervals/README
-  - file: 17. Math Geometry/README
-  - file: 18. Bit Manipulation/README
\ No newline at end of file
diff --git a/docs/01. Array Hashing/1. Two Sum.ipynb b/docs/01. Array Hashing/1. Two Sum.ipynb
deleted file mode 100644
index b3b517f..0000000
--- a/docs/01. Array Hashing/1. Two Sum.ipynb	
+++ /dev/null
@@ -1,211 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 100: Same Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-        "\n",
-        "---\n",
-        "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-        "\n",
-        "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-        "\n",
-        "**Constraints**\n",
-        "\n",
-        "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-        "2. $-10^4$ <= `Node.val` <= $10^4$"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def isSameTree(p, q):\n",
-        "    # Base case: If both p and q are None, the trees are the same.\n",
-        "    if not p and not q:\n",
-        "        return True\n",
-        "    \n",
-        "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-        "    if not p or not q:\n",
-        "        return False\n",
-        "    \n",
-        "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-        "    if p.val != q.val:\n",
-        "        return False\n",
-        "    \n",
-        "    # Recursively check the left and right subtrees of p and q.\n",
-        "    # If both subtrees are the same, the entire trees are the same.\n",
-        "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-        "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-        "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-        "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-        "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-        "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "\n",
-        "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-        "\n",
-        "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "print(isSameTree(p1, q1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "\n",
-        "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-        "\n",
-        "p2 = TreeNode(1, TreeNode(2), None)\n",
-        "q2 = TreeNode(1, None, TreeNode(2))\n",
-        "print(isSameTree(p2, q2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "\n",
-        "#Input: p = [1,2,1], q = [1,1,2]\n",
-        "\n",
-        "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-        "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-        "print(isSameTree(p3, q3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "**Time Complexity**\n",
-        "\n",
-        "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-        "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-        "\n",
-        "**Space Complexity**\n",
-        "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-        "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-        "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-        "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "+ Time Complexity: $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-        "+ Space Complexity: $O(n)$ due to the recursive call stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/01. Array Hashing/128. Longest Consecutive Sequence.ipynb b/docs/01. Array Hashing/128. Longest Consecutive Sequence.ipynb
deleted file mode 100644
index a77a03d..0000000
--- a/docs/01. Array Hashing/128. Longest Consecutive Sequence.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 128. Longest Consecutive Sequence\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Consecutive Sequence problem on LeetCode, click here!](https://leetcode.com/problems/longest-consecutive-sequence/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an unsorted array of integers `nums`, return *the length of the longest consecutive elements sequence.*\n",
-        "\n",
-        "You must write an algorithm that runs in `O(n)` time.\n",
-        "\n",
-        "*Constraints:**\n",
-        "- 0 <= `nums.length` <= $10^5$\n",
-        "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 58,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def longestConsecutive(nums):\n",
-        "    if not nums:\n",
-        "        return 0\n",
-        "\n",
-        "    num_set = set(nums)\n",
-        "    max_length = 0\n",
-        "\n",
-        "    for num in num_set:\n",
-        "        if num - 1 not in num_set:\n",
-        "            current_num = num\n",
-        "            current_length = 1\n",
-        "\n",
-        "            while current_num + 1 in num_set:\n",
-        "                current_num += 1\n",
-        "                current_length += 1\n",
-        "\n",
-        "            max_length = max(max_length, current_length)\n",
-        "\n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining the `longestConsecutive` function that takes the `nums` array as input.\n",
-        "\n",
-        "2. We check if the `nums` array is empty. If it's empty, there are no consecutive elements, so we return 0.\n",
-        "\n",
-        "3. We create a Python set called `num_set` and insert all elements from the `nums` array into it. Using a set allows us to efficiently check for the existence of elements in O(1) time.\n",
-        "\n",
-        "4. We initialize a variable `max_length` to 0. This variable will keep track of the maximum length of consecutive elements sequence found.\n",
-        "\n",
-        "5. We iterate through the elements of the `num_set`. For each element `num`, we check if `num - 1` exists in the `num_set`. If it doesn't exist, it means `num` is the starting element of a potential consecutive sequence.\n",
-        "\n",
-        "6. Inside the loop, we initialize two variables: `current_num` to the current element `num` and `current_length` to 1. We start with a length of 1 because `num` itself is part of the sequence.\n",
-        "\n",
-        "7. We then enter a while loop that continues as long as `current_num + 1` exists in the `num_set`. This means we are incrementing the consecutive sequence.\n",
-        "\n",
-        "8. Inside the while loop, we increment `current_num` by 1 and also increment `current_length` by 1 to account for the next consecutive element.\n",
-        "\n",
-        "9. We compare `current_length` with the `max_length` and update `max_length` if the current sequence is longer.\n",
-        "\n",
-        "10. After the while loop, we move to the next element in the outer loop and repeat the process.\n",
-        "\n",
-        "11. Finally, we return the `max_length` as the result, which represents the length of the longest consecutive elements sequence in the `nums` array.\n",
-        "\n",
-        "The key idea here is to use a set to efficiently check for the existence of elements and to iterate through the elements, considering each element as the potential start of a consecutive sequence. By doing this, we can find the longest consecutive sequence in O(n) time complexity, where n is the number of elements in the array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 59,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "nums1 = [100, 4, 200, 1, 3, 2]\n",
-        "print(longestConsecutive(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 60,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "9\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "nums2 = [0, 3, 7, 2, 5, 8, 4, 6, 0, 1]\n",
-        "print(longestConsecutive(nums2)) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The code is designed to run in O(n) time complexity, where n is the number of elements in the `nums` array. Here's the breakdown:\n",
-        "\n",
-        "1. Constructing the `num_set` by inserting all elements from `nums` into it takes O(n) time because we perform an insertion operation for each element in `nums`.\n",
-        "\n",
-        "2. The main loop iterates through the elements in `num_set`. In the worst case, each element is visited only once, so the loop itself takes O(n) time.\n",
-        "\n",
-        "3. Within the loop, we have a while loop that may also take O(n) time in the worst case. However, this while loop is nested inside the main loop, so its overall time complexity remains O(n).\n",
-        "\n",
-        "Therefore, the overall time complexity of the code is O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of the code is determined by the space used by the `num_set` and a few additional variables. Here's the breakdown:\n",
-        "\n",
-        "1. `num_set` is a set that stores the unique elements from `nums`. In the worst case, it can store all n elements from `nums`, so the space complexity is O(n).\n",
-        "\n",
-        "2. The additional variables used, such as `max_length`, `current_num`, and `current_length`, have constant space requirements and do not depend on the size of the input array.\n",
-        "\n",
-        "Therefore, the overall space complexity of the code is O(n) due to the space used by the `num_set`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. Write an algorithm that not only returns the length of the longest consecutive elements sequence but also returns the actual consecutive sequence itself.\n",
-        "\n",
-        "2. Extend the problem to allow elements to be considered consecutive if they are within a certain absolute difference (e.g., less than or equal to k) instead of exactly 1. Write an algorithm that finds the longest such sequence and runs in O(n) time."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/01. Array Hashing/217. Contains Duplicate.ipynb b/docs/01. Array Hashing/217. Contains Duplicate.ipynb
deleted file mode 100644
index 57bb55d..0000000
--- a/docs/01. Array Hashing/217. Contains Duplicate.ipynb	
+++ /dev/null
@@ -1,218 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 217. Contains Duplicate\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Contains Duplicate problem on LeetCode, click here!](https://leetcode.com/problems/contains-duplicate/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `nums.length` <= $10^5$\n",
-        "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 19,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def containsDuplicate(nums):\n",
-        "    # Create an empty set to store unique elements\n",
-        "    unique_elements = set()\n",
-        "    \n",
-        "    # Iterate through the array\n",
-        "    for num in nums:\n",
-        "        # If the element is already in the set, return True\n",
-        "        if num in unique_elements:\n",
-        "            return True\n",
-        "        # Otherwise, add it to the set\n",
-        "        else:\n",
-        "            unique_elements.add(num)\n",
-        "    \n",
-        "    # If the loop completes without finding duplicates, return False\n",
-        "    return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `containsDuplicate` function takes a single argument, `nums`, which is the input integer array.\n",
-        "\n",
-        "2. Inside the function, an empty set called `unique_elements` is created. This set will be used to keep track of unique elements in the input array.\n",
-        "\n",
-        "3. The function then iterates through the input array `nums` using a `for` loop.\n",
-        "\n",
-        "4. For each element `num` in the array, it checks whether `num` is already in the `unique_elements` set using the `if num in unique_elements:` condition.\n",
-        "\n",
-        "5. If `num` is already in the set, it means there is a duplicate element in the array, and the function immediately returns `True`.\n",
-        "\n",
-        "6. If `num` is not in the set, it is added to the `unique_elements` set using `unique_elements.add(num)`.\n",
-        "\n",
-        "7. The loop continues to the next element, and the process repeats.\n",
-        "\n",
-        "8. If the loop completes without finding any duplicates, it means that all elements in the array are distinct, and the function returns `False`.\n",
-        "\n",
-        "The code efficiently utilizes a set data structure to keep track of unique elements while iterating through the array, allowing it to quickly detect duplicate elements. This code meets the problem's requirements and constraints, as explained earlier."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "nums1 = [1, 2, 3, 1]\n",
-        "print(containsDuplicate(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "nums2 = [1, 2, 3, 4]\n",
-        "print(containsDuplicate(nums2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 22,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "nums3 = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]\n",
-        "print(containsDuplicate(nums3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for the \"Contains Duplicate\" problem:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The primary operation that affects the time complexity is the `for` loop that iterates through the input array `nums`. In the worst case, the loop will iterate through all `n` elements of the array, where `n` is the length of the input array.\n",
-        "\n",
-        "- Iterating through the array: O(n)\n",
-        "\n",
-        "Inside the loop, we perform two operations:\n",
-        "\n",
-        "1. Checking whether an element exists in the `unique_elements` set (`if num in unique_elements`). This operation has an average time complexity of O(1) for a set.\n",
-        "2. Adding an element to the `unique_elements` set (`unique_elements.add(num)`), which also has an average time complexity of O(1) for a set.\n",
-        "\n",
-        "Since these operations are performed for each element in the array, the overall time complexity remains O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity is determined by the additional data structures used in the code, which are the `unique_elements` set.\n",
-        "\n",
-        "- `unique_elements` set: This set stores unique elements from the input array. In the worst case, if all elements in the input array are distinct, the set will store all `n` elements.\n",
-        "\n",
-        "Therefore, the space complexity is O(n) because, in the worst case, the set's size will grow linearly with the input array's size.\n",
-        "\n",
-        "**In summary**, the time complexity of the code is O(n), where `n` is the length of the input array, and the space complexity is also O(n) in the worst case. This code is efficient and meets the constraints of the problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K Duplicates:** Modify the problem to return `true` if there are exactly K duplicate elements in the array. Write a function that takes an additional integer parameter K and returns `true` if there are exactly K duplicates.\n",
-        "2. **Frequency Count:** Write a function that returns a list of all the elements that appear more than once in the array, along with their frequencies."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/01. Array Hashing/238. Product of Array Except Self.ipynb b/docs/01. Array Hashing/238. Product of Array Except Self.ipynb
deleted file mode 100644
index 1ff9a9c..0000000
--- a/docs/01. Array Hashing/238. Product of Array Except Self.ipynb	
+++ /dev/null
@@ -1,197 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 238. Product of Array Except Self\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Product of Array Except Self problem on LeetCode, click here!](https://leetcode.com/problems/product-of-array-except-self/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array `nums`, return *an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.*\n",
-        "\n",
-        "The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-        "\n",
-        "You must write an algorithm that runs in O(n) time and without using the division operation.\n",
-        "\n",
-        "Constraints:\n",
-        "- 2 <= `nums.length` <= $10^5$\n",
-        "- `-30 <= nums[i] <= 30`\n",
-        "- The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-        "\n",
-        "**Follow-up:** Can you solve the problem in O(1) extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 50,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def productExceptSelf(nums):\n",
-        "    n = len(nums)\n",
-        "    \n",
-        "    # Initialize two arrays to store products to the left and right of each element\n",
-        "    left_products = [1] * n\n",
-        "    right_products = [1] * n\n",
-        "    \n",
-        "    # Calculate products to the left of each element\n",
-        "    left_product = 1\n",
-        "    for i in range(n):\n",
-        "        left_products[i] = left_product\n",
-        "        left_product *= nums[i]\n",
-        "    \n",
-        "    # Calculate products to the right of each element\n",
-        "    right_product = 1\n",
-        "    for i in range(n - 1, -1, -1):\n",
-        "        right_products[i] = right_product\n",
-        "        right_product *= nums[i]\n",
-        "    \n",
-        "    # Calculate the final answer using left and right products\n",
-        "    answer = [left_products[i] * right_products[i] for i in range(n)]\n",
-        "    \n",
-        "    return answer"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `n` is initialized as the length of the `nums` array.\n",
-        "2. Two arrays, `left_products` and `right_products`, are created, each with a length of `n` and filled with ones initially.\n",
-        "3. `left_product` is initialized to 1.\n",
-        "4. A loop iterates through the `nums` array, from left to right. For each element at index `i`, it stores the product of all elements to its left (including itself) in the `left_products` array and updates `left_product` accordingly.\n",
-        "5. `right_product` is initialized to 1.\n",
-        "6. Another loop iterates through the `nums` array in reverse order, from right to left. For each element at index `i`, it stores the product of all elements to its right (including itself) in the `right_products` array and updates `right_product`.\n",
-        "7. The final `answer` list is constructed by multiplying corresponding elements from the `left_products` and `right_products` arrays for each index `i`.\n",
-        "8. The function returns the `answer` list, which contains the desired results.\n",
-        "\n",
-        "Overall, this algorithm efficiently computes the product of all elements except the current element, as required. It runs in O(n) time complexity and uses O(n) extra space, which is within the problem's constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 52,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[24, 12, 8, 6]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "nums = [1,2,3,4]\n",
-        "result1 = productExceptSelf(nums)\n",
-        "print(result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 53,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[0, 0, 9, 0, 0]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums = [-1,1,0,-3,3]\n",
-        "result2 = productExceptSelf(nums)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The first loop iterates through the `nums` array to calculate the products to the left of each element. This loop runs in O(n) time, where n is the length of the `nums` array.\n",
-        "2. The second loop also iterates through the `nums` array in reverse order to calculate the products to the right of each element. This loop also runs in O(n) time.\n",
-        "3. The final loop constructs the `answer` list by multiplying elements from the `left_products` and `right_products` arrays. This loop runs in O(n) time.\n",
-        "\n",
-        "Since all the loops are independent and sequential, the overall time complexity is O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. Two additional arrays, `left_products` and `right_products`, are created with a length of `n`, where `n` is the length of the `nums` array. Therefore, these arrays consume O(n) extra space.\n",
-        "2. The `left_product` and `right_product` variables are used to keep track of the product to the left and right of each element, respectively. These variables occupy O(1) extra space.\n",
-        "3. The `answer` list, which contains the final results, also consumes O(n) space, but it's not counted toward the extra space complexity analysis as per the problem's constraints.\n",
-        "\n",
-        "Overall, the space complexity is O(n) for this solution."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Handle Zeros in the Input**: Modify the solution to handle cases where the input array contains zeros. For example, if `nums = [0, 1, 2, 3, 0]`, the output should be `[0, 0, 0, 0, 0]` because any element multiplied by zero results in zero.\n",
-        "\n",
-        "2. **In-Place Modification**: Attempt to solve the problem with in-place modification of the `nums` array, where the output is stored directly in the input array without using any additional space."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/01. Array Hashing/242. Valid Anagram.ipynb b/docs/01. Array Hashing/242. Valid Anagram.ipynb
deleted file mode 100644
index d165899..0000000
--- a/docs/01. Array Hashing/242. Valid Anagram.ipynb	
+++ /dev/null
@@ -1,198 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 242. Valid Anagram\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Anagram problem on LeetCode, click here!](https://leetcode.com/problems/valid-anagram/)\n",
-        "\n",
-        "---\n",
-        "Given two strings `s` and `t`, return `true` if `t` is an anagram of `s`, and `false` otherwise.\n",
-        "\n",
-        "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `s.length`, `t.length` <= $5 * 10^4$\n",
-        "- `s` and `t` consist of lowercase English letters.\n",
-        "\n",
-        "**Follow-up:** What if the inputs contain Unicode characters? How would you adapt your solution to such a case?\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 29,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isAnagram(s: str, t: str) -> bool:\n",
-        "    # Check if the lengths of s and t are not equal\n",
-        "    if len(s) != len(t):\n",
-        "        return False\n",
-        "\n",
-        "    # Create dictionaries to store character frequencies for s and t\n",
-        "    countS, countT = {}, {}\n",
-        "\n",
-        "    # Count character frequencies in s\n",
-        "    for i in range(len(s)):\n",
-        "        countS[s[i]] = 1 + countS.get(s[i], 0)\n",
-        "\n",
-        "    # Count character frequencies in t\n",
-        "    for i in range(len(t)):\n",
-        "        countT[t[i]] = 1 + countT.get(t[i], 0)\n",
-        "\n",
-        "    # Check if the character frequencies in s and t are the same\n",
-        "    return countS == countT"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The code defines a class called `Solution` with a method `isAnagram` that takes two input strings, `s` and `t`, and returns a boolean (`True` or `False`) based on whether `t` is an anagram of `s`.\n",
-        "\n",
-        "2. The first check performed in the method is whether the lengths of `s` and `t` are equal. If they are not equal, it immediately returns `False` because strings of different lengths cannot be anagrams of each other.\n",
-        "\n",
-        "3. Two dictionaries, `countS` and `countT`, are created to store the frequency of characters in `s` and `t`, respectively. These dictionaries will be used to count the occurrences of each character in the strings.\n",
-        "\n",
-        "4. The code then enters a loop to iterate through each character in string `s` using a `for` loop. Inside the loop, it updates the `countS` dictionary. The line `countS[s[i]] = 1 + countS.get(s[i], 0)` increments the count of character `s[i]` in `countS` by 1. If the character is not already in the dictionary, it initializes the count to 1.\n",
-        "\n",
-        "5. Similarly, the code iterates through each character in string `t` and updates the `countT` dictionary in the same way.\n",
-        "\n",
-        "6. After counting the character frequencies in both strings, the code compares the two dictionaries using `countS == countT`. If the dictionaries are equal, it means that both strings have the same character frequencies, and therefore, `t` is an anagram of `s`. In this case, the method returns `True`.\n",
-        "\n",
-        "7. If the dictionaries are not equal, the method returns `False`, indicating that `t` is not an anagram of `s`.\n",
-        "\n",
-        "In summary, this code efficiently determines whether two input strings are anagrams by counting the character frequencies in each string using dictionaries and then comparing these counts. If the character frequencies match, the strings are considered anagrams, and the method returns `True`; otherwise, it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 30,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "s = \"anagram\"\n",
-        "t = \"nagaram\"\n",
-        "print(isAnagram(s, t))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "s = \"rat\"\n",
-        "t = \"car\"\n",
-        "print(isAnagram(s, t))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. The code first checks whether the lengths of the input strings `s` and `t` are equal, which takes constant time. This check has a time complexity of O(1).\n",
-        "\n",
-        "2. The code then iterates through both strings, `s` and `t`, once to count the character frequencies. Since both strings have a maximum length of 5 * 10^4, the worst-case scenario is that the code iterates through 5 * 10^4 characters in each string. This results in a linear time complexity of O(n), where n is the length of the longer of the two input strings (`s` or `t`).\n",
-        "\n",
-        "3. Finally, the code compares the two dictionaries `countS` and `countT` to check if the character frequencies match. This comparison takes O(n) time in the worst case, where n is the length of the longer string.\n",
-        "\n",
-        "Overall, the time complexity of the code is O(n), where n is the length of the longer input string.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The code uses two dictionaries, `countS` and `countT`, to store the character frequencies of the input strings `s` and `t`. In the worst case, both dictionaries can contain all unique characters from the input strings. Since the input strings can have a maximum length of 5 * 10^4, the space complexity for these dictionaries is O(5 * 10^4) or simply O(n), where n is the length of the longer input string.\n",
-        "\n",
-        "2. The code uses a few additional variables for bookkeeping, but these variables have constant space requirements and do not depend on the input size. Therefore, they do not significantly impact the overall space complexity.\n",
-        "\n",
-        "**In summary**, the space complexity of the code is O(n), where n is the length of the longer input string."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Anagram Chains**: Given a list of words, find the longest chain of anagrams, where each word in the chain is an anagram of the previous word. For example, given [\"bat\", \"tab\", \"cat\", \"tac\", \"dog\"], the longest anagram chain is [\"bat\", \"tab\", \"cat\", \"tac\"].\n",
-        "\n",
-        "2. **Minimum Deletions to Make Anagrams**: Given two strings `s` and `t`, find the minimum number of character deletions required in both strings to make them anagrams of each other. For example, if `s = \"hello\"` and `t = \"billion\"`, you can remove the characters \"heo\" from `s` and \"billi\" from `t` to make them anagrams (\"lo\" and \"on\" are left in the two strings)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/01. Array Hashing/347. Top K Frequent Elements.ipynb b/docs/01. Array Hashing/347. Top K Frequent Elements.ipynb
deleted file mode 100644
index ecfa494..0000000
--- a/docs/01. Array Hashing/347. Top K Frequent Elements.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 347. Top K Frequent Elements\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Top K Frequent Elements problem on LeetCode, click here!](https://leetcode.com/problems/top-k-frequent-elements/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array `nums` and an integer `k`, return the `k` most frequent elements. You may return the answer in **any order**.\n",
-        "\n",
-        "Constraints:\n",
-        "- 1 <= `nums.length` <= $10^5$\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- `k` is in the range `[1, the number of unique elements in the array]`.\n",
-        "- It is **guaranteed** that the answer is **unique**.\n",
-        "\n",
-        "**Follow-up:** Your algorithm's time complexity must be better than $O(n\\ log\\ n)$, where n is the array's size.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 46,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def topKFrequent(nums, k):\n",
-        "    # Create a dictionary to store the frequency of each element\n",
-        "    count = {}\n",
-        "    # Create a list of lists to store elements with the same frequency\n",
-        "    frequency = [[] for i in range(len(nums) + 1)]\n",
-        "\n",
-        "    # Count the frequency of each element in nums\n",
-        "    for n in nums:\n",
-        "        count[n] = 1 + count.get(n, 0)\n",
-        "\n",
-        "    # Place elements in the freq list according to their frequency\n",
-        "    for n, c in count.items():\n",
-        "        frequency[c].append(n)\n",
-        "\n",
-        "    res = []\n",
-        "    # Traverse freq list from the end (higher frequencies)\n",
-        "    for i in range(len(frequency) - 1, 0, -1):\n",
-        "        for n in frequency[i]:\n",
-        "            res.append(n)\n",
-        "            if len(res) == k:\n",
-        "                return res"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `def topKFrequent(nums, k):`: This is a function that takes two arguments: `nums`, which is the input array of integers, and `k`, which is the number of most frequent elements to return.\n",
-        "\n",
-        "2. `count = {}`: This dictionary `count` will be used to store the frequency of each unique element in the `nums` array. The keys are elements from the input array, and the values are their corresponding frequencies.\n",
-        "\n",
-        "3. `frequency = [[] for i in range(len(nums) + 1)]`: This creates a list of empty lists called `frequency`. It's used to store elements based on their frequencies, similar to the `freq` list in the previous code. The size of this list is set to be one greater than the length of the input `nums`.\n",
-        "\n",
-        "4. `for n in nums:`: This loop iterates through each element `n` in the input `nums` array.\n",
-        "\n",
-        "5. `count[n] = 1 + count.get(n, 0)`: This line counts the frequency of each element `n` in the `nums` array. It uses the `count.get(n, 0)` method to retrieve the current count of `n` from the `count` dictionary. If `n` is not in the dictionary, it defaults to 0. It then increments the count by 1.\n",
-        "\n",
-        "6. After the above loop, the `count` dictionary will contain counts of each unique element in the `nums` array.\n",
-        "\n",
-        "7. `for n, c in count.items():`: This loop iterates through the items (key-value pairs) of the `count` dictionary. `n` represents the element, and `c` represents its frequency.\n",
-        "\n",
-        "8. `frequency[c].append(n)`: This line places the element `n` into the bucket corresponding to its frequency `c`. Buckets are represented by the `frequency` list. For example, if an element `n` has a frequency of 3, it will be added to `frequency[3]`.\n",
-        "\n",
-        "9. After this loop, the `frequency` list will contain buckets of elements grouped by their frequencies.\n",
-        "\n",
-        "10. `res = []`: This list `res` will be used to store the k most frequent elements.\n",
-        "\n",
-        "11. `for i in range(len(freq) - 1, 0, -1):`: This loop iterates in reverse order through the `frequency` list, starting from the highest frequency and going down to 1. Note that there is a typo here; it should be `frequency` instead of `freq`.\n",
-        "\n",
-        "12. `for n in frequency[i]:`: For each element `n` in the current bucket (i.e., elements with frequency `i`), it appends `n` to the result list `res`.\n",
-        "\n",
-        "13. `if len(res) == k:`: Once `res` contains k elements, the code exits and returns the result.\n",
-        "\n",
-        "The code efficiently finds the k most frequent elements in the input array without using sorting, similar to the previous explanation, with the only difference being the variable names (e.g., `frequency` instead of `freq`)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 48,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 2]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "nums = [1,1,1,2,2,3]\n",
-        "k = 2\n",
-        "result1 = topKFrequent(nums, k)\n",
-        "print(result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 49,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums = [1]\n",
-        "k = 1\n",
-        "result2 = topKFrequent(nums, k)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The first loop that counts the frequency of elements by iterating through `nums` has a time complexity of O(n), where n is the number of elements in the input array `nums`.\n",
-        "\n",
-        "2. The second loop iterates through the keys in the `count` dictionary, which has at most `k` unique elements (as per the constraints). Therefore, the second loop also has a time complexity of O(k).\n",
-        "\n",
-        "3. The third loop iterates through the `frequency` list, which has a length equal to the maximum frequency of elements in `nums`. In the worst case, this loop can have a time complexity of O(n), where n is the number of elements in `nums`.\n",
-        "\n",
-        "Overall, the dominant factor in terms of time complexity is the loop that iterates through the `count` dictionary. So, the total time complexity of the code is O(n + k).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The `count` dictionary stores the frequency of each unique element in `nums`. In the worst case, it can have at most `k` unique elements, so the space complexity for `count` is O(k).\n",
-        "\n",
-        "2. The `frequency` list is used to store elements grouped by their frequencies. It has a length of `len(nums) + 1`, which can be at most 105 based on the constraints. Therefore, the space complexity for `frequency` is O(105).\n",
-        "\n",
-        "3. The `res` list stores the k most frequent elements, which can be at most `k` elements, so the space complexity for `res` is O(k).\n",
-        "\n",
-        "**In summary**, the space complexity is dominated by the `count` dictionary and the `frequency` list, both of which have a space complexity of O(k + 105)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimized k Most Frequent Elements:** Modify the code to find the k most frequent elements in an array while ensuring that the time complexity is O(n + klogk). You can use a priority queue (heap) to achieve this.\n",
-        "\n",
-        "2. **Handling Duplicate Frequencies:** Extend the code to handle cases where multiple elements have the same frequency and are among the k most frequent elements. Ensure that the output contains exactly k elements."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/01. Array Hashing/49. Group Anagrams.ipynb b/docs/01. Array Hashing/49. Group Anagrams.ipynb
deleted file mode 100644
index afe89ec..0000000
--- a/docs/01. Array Hashing/49. Group Anagrams.ipynb	
+++ /dev/null
@@ -1,229 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 49. Group Anagrams\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Group Anagrams problem on LeetCode, click here!](https://leetcode.com/problems/group-anagrams/)\n",
-        "\n",
-        "---\n",
-        "Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.\n",
-        "\n",
-        "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `strs.length <= $10^4$\n",
-        "- `0 <= strs[i].length <= 100`\n",
-        "- `strs[i]` consists of lowercase English letters.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 36,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict\n",
-        "\n",
-        "\n",
-        "def groupAnagrams(strs):\n",
-        "    ans = defaultdict(list)\n",
-        "\n",
-        "    # Iterate through the list of input strings\n",
-        "    for s in strs:\n",
-        "        # Initialize a list to represent character counts for each character (a-z)\n",
-        "        count = [0] * 26\n",
-        "\n",
-        "        # Count the occurrences of each character in the current word\n",
-        "        for c in s:\n",
-        "            count[ord(c) - ord(\"a\")] += 1\n",
-        "\n",
-        "        # Use a tuple of character counts as the key and append the word to the anagram group\n",
-        "        ans[tuple(count)].append(s)\n",
-        "\n",
-        "    # Convert the values (lists of anagrams) to a list of lists\n",
-        "    return list(ans.values())"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. Inside the method, a `defaultdict` called `ans` is created to store anagram groups. This dictionary will have a list as its default value, meaning that each key in the dictionary will be associated with an empty list by default.\n",
-        "\n",
-        "2. The code then iterates through the list of input strings, `strs`, using a `for` loop. For each word `s` in `strs`, it performs the following steps:\n",
-        "\n",
-        "   - It initializes a list `count` of length 26, where each element represents the count of a specific character (a-z) in the word `s`.\n",
-        "\n",
-        "   - The code then iterates through the characters of `s`. For each character `c`, it increments the corresponding count in the `count` list based on its ASCII value.\n",
-        "\n",
-        "   - After counting the characters in `s`, it converts the `count` list into a tuple `tuple(count)` to use as a key for the `ans` dictionary.\n",
-        "\n",
-        "   - It appends the word `s` to the list associated with the key (tuple) in the `ans` dictionary. This groups all anagrams of the same word together under the same key.\n",
-        "\n",
-        "3. After processing all words in `strs`, the code converts the values of the `ans` dictionary (which are lists of anagrams) to a list of lists using the `list()` constructor.\n",
-        "\n",
-        "4. Finally, the code returns the list of anagram groups, which is the result of grouping the anagrams in the input list.\n",
-        "\n",
-        "In summary, the code efficiently groups anagrams together by counting the characters in each word and using a dictionary to store them under the same key. The result is a list of lists, where each inner list represents a group of anagrams."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 37,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: Group anagrams from a list of words\n",
-        "strs1 = [\"eat\", \"tea\", \"tan\", \"ate\", \"nat\", \"bat\"]\n",
-        "result1 = groupAnagrams(strs1)\n",
-        "print(result1)\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 38,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2: Group anagrams from a list with an empty string\n",
-        "strs2 = [\"\"]\n",
-        "result2 = groupAnagrams(strs2)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 40,
-      "id": "c8367e2a-e7d5-4e18-8235-2870e341a04b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['a']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3: Group anagrams from a list with a single word\n",
-        "strs3 = [\"a\"]\n",
-        "result3 = groupAnagrams(strs3)\n",
-        "print(result3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `groupAnagrams` method:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. First, we create an instance of the `Solution` class, which is an O(1) operation.\n",
-        "\n",
-        "2. Inside the `groupAnagrams` method, we iterate through the list of input strings `strs` once in a loop:\n",
-        "   - For each word in `strs`, we perform character counting, which is done in O(K) time, where K is the maximum length of a word in the list.\n",
-        "\n",
-        "3. The overall time complexity of the method is O(N * K), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. We use a `defaultdict` called `ans` to store the anagram groups. In the worst case, each word belongs to a distinct anagram group, so the space complexity for `ans` is O(N).\n",
-        "\n",
-        "2. Within the loop, we create a `count` list of length 26 to store character counts for each word. This is a fixed-size list and does not depend on the input size, so it has a constant space complexity of O(26), which is effectively O(1).\n",
-        "\n",
-        "3. The space used for other variables is also constant and does not depend on the input size.\n",
-        "\n",
-        "4. The final result, `result`, is a list of lists that contains the grouped anagrams. In the worst case, each word is a distinct anagram group, so the space complexity for `result` is O(N).\n",
-        "\n",
-        "5. The overall space complexity of the method is O(N) due to the space used by `ans` and `result`. The space used by the `count` list and other variables is constant and does not significantly affect the overall space complexity.\n",
-        "\n",
-        "**In summary,** the time complexity of the `groupAnagrams` method is O(N * K), and the space complexity is O(N), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Anagramic Palindromes**: Given a list of strings, find all the groups of anagrams where each group contains words that can be rearranged into a palindrome. For example, in the list [\"bat\", \"tab\", \"act\", \"tac\", \"dog\"], there are two groups: [\"bat\", \"tab\", \"act\", \"tac\"] and [\"dog\"].\n",
-        "\n",
-        "2. **Largest Anagram Group**: Given a list of strings, find the largest group of anagrams. Return the list of anagrams in that group. If there are multiple largest groups, return all of them."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/01. Array Hashing/659. Encode and Decode Strings.ipynb b/docs/01. Array Hashing/659. Encode and Decode Strings.ipynb
deleted file mode 100644
index 037c476..0000000
--- a/docs/01. Array Hashing/659. Encode and Decode Strings.ipynb	
+++ /dev/null
@@ -1,204 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 659· Encode and Decode Strings\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Encode and Decode Strings problem on LintCode, click here!](https://www.lintcode.com/problem/659/)\n",
-        "\n",
-        "---\n",
-        "Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.\n",
-        "\n",
-        "Please implement `encode` and `decode`.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 54,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class Codec:\n",
-        "\n",
-        "    def encode(self, strs):\n",
-        "        \"\"\"Encodes a list of strings to a single string.\n",
-        "        \n",
-        "        :type strs: List[str]\n",
-        "        :rtype: str\n",
-        "        \"\"\"\n",
-        "        encoded = \"\"\n",
-        "        for s in strs:\n",
-        "            encoded += s.replace(':', '::') + ':;'\n",
-        "        return encoded\n",
-        "\n",
-        "    def decode(self, s):\n",
-        "        \"\"\"Decodes a single string to a list of strings.\n",
-        "        \n",
-        "        :type s: str\n",
-        "        :rtype: List[str]\n",
-        "        \"\"\"\n",
-        "        decoded = []\n",
-        "        i = 0\n",
-        "        while i < len(s):\n",
-        "            end = s.find(':;', i)\n",
-        "            if end == -1:\n",
-        "                end = len(s)\n",
-        "            decoded.append(s[i:end].replace('::', ':'))\n",
-        "            i = end + 2\n",
-        "        return decoded"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a class called `Codec` to encapsulate the encoding and decoding operations for a list of strings.\n",
-        "\n",
-        "2. In the `encode` method, we take a list of strings (`strs`) as input and return a single encoded string. The encoding process involves concatenating the strings together with a delimiter `:;`. We iterate through each string in the input list, replace any occurrence of `:` with `::` (to avoid conflicts with the delimiter), and then append `:;` to indicate the end of that string.\n",
-        "\n",
-        "3. In the `decode` method, we take an encoded string (`s`) as input and return a list of strings. The decoding process involves splitting the encoded string using the `:;` delimiter. We start from the beginning of the encoded string (`i = 0`) and repeatedly find the next occurrence of `:;`. We extract the substring between the current position (`i`) and the next delimiter (`end`). Before adding it to the result list, we replace any `::` with `:` to revert the encoding. We then update the current position `i` to be after the delimiter, so we can find the next substring in the next iteration.\n",
-        "\n",
-        "4. Finally, we create an instance of the `Codec` class and test it with two examples:\n",
-        "\n",
-        "By using the `:;` delimiter and handling the `::` escaping, this code can encode and decode lists of strings, preserving the original content even if it contains colons or the delimiter itself."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 55,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['lint', 'code', 'love', 'you']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "codec = Codec()\n",
-        "\n",
-        "input1 = [\"lint\", \"code\", \"love\", \"you\"]\n",
-        "encoded1 = codec.encode(input1)\n",
-        "decoded1 = codec.decode(encoded1)\n",
-        "print(decoded1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 57,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['we', 'say', ':', 'yes']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "codec = Codec() \n",
-        "\n",
-        "input2 = [\"we\", \"say\", \":\", \"yes\"]\n",
-        "encoded2 = codec.encode(input2)\n",
-        "decoded2 = codec.decode(encoded2)\n",
-        "print(decoded2)  "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `encode` and `decode` methods in the provided code.\n",
-        "\n",
-        "**Encode Method (`encode`):**\n",
-        "\n",
-        "- Time Complexity: O(n * m)\n",
-        "  - Here, `n` is the number of strings in the input list `strs`, and `m` is the average length of these strings.\n",
-        "  - In the worst case, for each string in the list, we iterate over its characters once to replace any colons (`:`) with double colons (`::`) and then append `:;`. This is done for each string in the list.\n",
-        "\n",
-        "- Space Complexity: O(n * m)\n",
-        "  - The space complexity is also O(n * m) because we build the encoded string by concatenating the modified strings. In the worst case, the encoded string can be of the same length as the original strings.\n",
-        "\n",
-        "**Decode Method (`decode`):**\n",
-        "\n",
-        "- Time Complexity: O(n * m)\n",
-        "  - Similar to the encode method, we iterate through the encoded string in the decode method. In the worst case, we may have to scan each character in the encoded string to find the `:;` delimiter.\n",
-        "\n",
-        "- Space Complexity: O(n * m)\n",
-        "  - The space complexity of the decode method is also O(n * m) because we build the list of decoded strings, which can be of the same length as the encoded string.\n",
-        "\n",
-        "Overall, both the `encode` and `decode` methods have time and space complexities of O(n * m), where `n` is the number of strings in the input list, and `m` is the average length of these strings. The space complexity arises from storing the encoded or decoded strings, and the time complexity arises from iterating through the characters in these strings."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Delimiters**: Modify the code to allow for multiple delimiters, not just one. Each string could have its own delimiter, and the code should be able to handle this complexity.\n",
-        "\n",
-        "2. **Optimize Encoding**: Modify the `encode` method to achieve a more efficient encoding algorithm. Try to minimize the length of the encoded string while ensuring that it can be correctly decoded."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/01. Array Hashing/README.md b/docs/01. Array Hashing/README.md
deleted file mode 100644
index a331dab..0000000
--- a/docs/01. Array Hashing/README.md	
+++ /dev/null
@@ -1,27 +0,0 @@
-# Array & Hashing Solutions - Blind 75 LeetCode Problems
-![Array & Hashing Solutions](https://img.shields.io/badge/Array_&_Hashing_Solutions-blue)![Blind 75 LeetCode](https://img.shields.io/badge/Blind_75_LeetCode-Problems-blue?labelColor=red)
-
-This repository contains my solutions to Array & Hashing Problems from Blind 75 LeetCode problems. I've organized the solutions by categories for easier navigation and reference. Each problem solution is presented in Jupyter Notebook format (`.ipynb`).
-
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [217. Contains Duplicate](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/217.%20Contains%20Duplicate.ipynb)                            | Easy       |
-| [242. Valid Anagram](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/242.%20Valid%20Anagram.ipynb)                                      | Easy       |
-| [1. Two Sum](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/1.%20Two%20Sum.ipynb)                                                      | Easy       |
-| [49. Group Anagrams](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/49.%20Group%20Anagrams.ipynb)                                      | Medium     |
-| [347. Top K Frequent Elements](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.ipynb)              | Medium     |
-| [238. Product of Array Except Self](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.ipynb)  | Medium     |
-| [659. Encode and Decode Strings](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.ipynb)          | Medium     |
-| [128. Longest Consecutive Sequence](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.ipynb)      | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/02. Two Pointers/11. Container With Most Water.ipynb b/docs/02. Two Pointers/11. Container With Most Water.ipynb
deleted file mode 100644
index 284c832..0000000
--- a/docs/02. Two Pointers/11. Container With Most Water.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 11. Container With Most Water\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Container With Most Water problem on LeetCode, click here!](https://leetcode.com/problems/container-with-most-water/)\n",
-        "\n",
-        "---\n",
-        "You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the $i^{th}$ line are `(i, 0)` and `(i, height[i])`.\n",
-        "\n",
-        "Find two lines that together with the x-axis form a container, such that the container contains the most water.\n",
-        "\n",
-        "Return *the maximum amount of water a container can store.*\n",
-        "\n",
-        "**Notice** that you may not slant the container.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `n == height.length`\n",
-        "- 2 <= `n` <= $10^5$\n",
-        "- 0 <= `height[i]` <= $10^4$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def maxArea(height):\n",
-        "    # Initialize the maximum area to 0.\n",
-        "    max_area = 0\n",
-        "    # Initialize two pointers, one at the beginning (left) and one at the end (right) of the array.\n",
-        "    left = 0\n",
-        "    right = len(height) - 1\n",
-        "\n",
-        "    # Iterate until the left pointer is less than the right pointer.\n",
-        "    while left < right:\n",
-        "        # Calculate the height of the container, which is the minimum height of the two lines.\n",
-        "        h = min(height[left], height[right])\n",
-        "        # Calculate the width of the container, which is the distance between the two pointers.\n",
-        "        width = right - left\n",
-        "        # Calculate the area of the container using height and width and update max_area if it's greater.\n",
-        "        max_area = max(max_area, h * width)\n",
-        "\n",
-        "        # Move the pointer with the shorter line inward.\n",
-        "        if height[left] < height[right]:\n",
-        "            left += 1\n",
-        "        else:\n",
-        "            right -= 1\n",
-        "\n",
-        "    # Continue this process until left < right, and then return the maximum area.\n",
-        "    return max_area"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start with initializing `max_area` to 0, which will store the maximum area of water that can be contained by the lines.\n",
-        "\n",
-        "2. Two pointers, `left` and `right`, are initialized to the beginning and end of the input `height` array.\n",
-        "\n",
-        "3. The main loop runs while `left` is less than `right`. This ensures that we are checking all possible pairs of lines in a systematic way.\n",
-        "\n",
-        "4. Inside the loop:\n",
-        "   - We calculate the height of the container `h` as the minimum of the heights at the `left` and `right` pointers. This represents the height of the water the container can hold.\n",
-        "   - We calculate the width of the container as the difference between `right` and `left`. This represents the distance between the two lines.\n",
-        "   - We calculate the area of the container using `h * width` and update `max_area` if the calculated area is greater than the current `max_area`.\n",
-        "\n",
-        "5. After calculating the area and updating `max_area`, we move one of the pointers. We move the pointer pointing to the shorter line inward because moving the pointer pointing to the taller line won't increase the height and will only decrease the width, which will result in a smaller area.\n",
-        "\n",
-        "6. We continue this process until `left` is no longer less than `right`, indicating that we have checked all possible pairs of lines.\n",
-        "\n",
-        "7. Finally, we return the `max_area` which contains the maximum area of water that can be contained by two lines from the input array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 49\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "height1 = [1, 8, 6, 2, 5, 4, 8, 3, 7]\n",
-        "result1 = maxArea(height1)\n",
-        "print(\"Output:\", result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "height2 = [1, 1]\n",
-        "result2 = maxArea(height2)\n",
-        "print(\"Output:\", result2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 45\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Additional Example\n",
-        "height3 = [3, 9, 3, 4, 7, 2, 12, 6]\n",
-        "result3 = maxArea(height3)\n",
-        "print(\"Output:\", result3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `maxArea` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The function uses a two-pointer approach to find the maximum area of water. In the worst case, both the left and right pointers traverse the entire input array once. Therefore, the time complexity is O(n), where n is the number of elements in the `height` array.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The function uses a constant amount of extra space for variables like `max_area`, `left`, `right`, `h`, and `width`. It does not use any data structures whose space consumption depends on the input size. As a result, the space complexity is O(1), which means it has constant space complexity.\n",
-        "\n",
-        "**In summary,** the `maxArea` function has a time complexity of O(n) and a space complexity of O(1), making it an efficient algorithm for finding the maximum area of water that can be contained by two lines in the input array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the maxArea function to not only return the maximum area but also the indices of the two lines that form the container with the maximum area.\n",
-        "2. **Variant Problem:** Instead of finding the maximum area of water, find the minimum amount of water required to fill all the gaps between the lines. You'll need to return the total water volume needed."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/02. Two Pointers/125. Valid Palindrome.ipynb b/docs/02. Two Pointers/125. Valid Palindrome.ipynb
deleted file mode 100644
index 8520b3b..0000000
--- a/docs/02. Two Pointers/125. Valid Palindrome.ipynb	
+++ /dev/null
@@ -1,218 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 125. Valid Palindrome\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Palindrome problem on LeetCode, click here!](https://leetcode.com/problems/valid-palindrome/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.\n",
-        "\n",
-        "Given a string `s`, *return `true` if it is a **palindrome**, or `false` otherwise.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $2*10^5$\n",
-        "- `s` consists only of printable ASCII characters.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isPalindrome(s):\n",
-        "    # Step 1: Convert the input string to lowercase\n",
-        "    s = s.lower()\n",
-        "    \n",
-        "    # Step 2: Remove all non-alphanumeric characters from the string\n",
-        "    cleaned_s = ''.join(c for c in s if c.isalnum())\n",
-        "    \n",
-        "    # Step 3: Initialize two pointers, one at the beginning and one at the end of the cleaned string\n",
-        "    left, right = 0, len(cleaned_s) - 1\n",
-        "    \n",
-        "    # Step 4: Compare characters using the two pointers while moving them towards each other\n",
-        "    while left < right:\n",
-        "        if cleaned_s[left] != cleaned_s[right]:\n",
-        "            return False  # If characters don't match, it's not a palindrome\n",
-        "        left += 1\n",
-        "        right -= 1\n",
-        "    \n",
-        "    # Step 5: If the loop completes without returning False, it's a palindrome\n",
-        "    return True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `isPalindrome` function takes an input string `s`.\n",
-        "\n",
-        "2. It starts by converting the input string to lowercase using `s.lower()`. This step ensures that the function is case-insensitive when checking for palindromes.\n",
-        "\n",
-        "3. Next, it removes all non-alphanumeric characters from the string and stores the result in the `cleaned_s` variable. It does this by iterating through each character in the input string and only keeping characters that are alphanumeric (letters and digits). This step removes spaces, punctuation, and other non-alphanumeric characters.\n",
-        "\n",
-        "4. The function then initializes two pointers, `left` and `right`, which will be used to compare characters from the beginning and end of the cleaned string.\n",
-        "\n",
-        "5. In a while loop, it compares characters using the two pointers while moving them towards each other. The loop continues until the `left` pointer is less than the `right` pointer. This is done to check all characters in the cleaned string.\n",
-        "\n",
-        "6. Inside the loop, it checks if the characters at the `left` and `right` pointers don't match. If they don't match, it means the input is not a palindrome, so the function returns `False`.\n",
-        "\n",
-        "7. If the characters at the `left` and `right` pointers match, the pointers are moved closer to each other by incrementing `left` and decrementing `right`.\n",
-        "\n",
-        "8. The loop continues to compare characters until it either finds a mismatch (returning `False`) or until the `left` pointer is greater than or equal to the `right` pointer. If the loop completes without finding a mismatch, it means the input is a palindrome, so the function returns `True`.\n",
-        "\n",
-        "In summary, the code converts the input string to lowercase, removes non-alphanumeric characters, and then uses two pointers to compare characters from the cleaned string's beginning and end. If all characters match during the comparison, the function returns `True`, indicating that the input is a palindrome. If any characters do not match, it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "input1 = \"A man, a plan, a canal: Panama\"\n",
-        "output1 = isPalindrome(input1)\n",
-        "print(output1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "input2 = \"race a car\"\n",
-        "output2 = isPalindrome(input2)\n",
-        "print(output2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "input3 = \"\"\n",
-        "output3 = isPalindrome(input3)\n",
-        "print(output3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `isPalindrome` function that uses two pointers:\n",
-        "\n",
-        "1. **Time Complexity:**\n",
-        "   - Converting the input string to lowercase using `s.lower()` takes O(n) time, where 'n' is the length of the input string.\n",
-        "   - Removing non-alphanumeric characters using the list comprehension `''.join(c for c in s if c.isalnum())` also takes O(n) time because it iterates through each character in the string once.\n",
-        "   - The comparison of characters using the two pointers (`left` and `right`) is done in a loop that runs until `left` is less than `right`. In the worst case, the loop iterates through half of the string, so it takes O(n/2) time, which is still considered O(n).\n",
-        "\n",
-        "   Therefore, the overall time complexity of the `isPalindrome` function is O(n), where 'n' is the length of the input string.\n",
-        "\n",
-        "2. **Space Complexity:**\n",
-        "   - The space complexity is primarily determined by the additional string `cleaned_s` created to store the cleaned version of the input string. In the worst case, if all characters in the input string are alphanumeric, this cleaned string can be as large as the original input string.\n",
-        "   - Additionally, the function uses two pointers, `left` and `right`, which consume constant space and do not depend on the size of the input.\n",
-        "\n",
-        "**In summary,** the overall space complexity of the `isPalindrome` function is O(n) in the worst case, where 'n' is the length of the input string. The space complexity is dominated by the space required for `cleaned_s`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Longest Palindromic Substring:** Given a string `s`, find and return the longest palindromic substring within `s`. For example, if `s = \"babad\"`, the function should return `\"bab\"` or `\"aba\"`.\n",
-        "2. **Palindrome with Limited Characters:** Given a string `s` and a list of characters `allowed_chars`, check if `s` is a palindrome considering only the characters from `allowed_chars`. Ignore all other characters. For example, if `s = \"A man, a plan, a canal: Panama\"` and `allowed_chars = ['a', 'm', 'n', 'p']`, the function should return `True`."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/02. Two Pointers/15. 3Sum.ipynb b/docs/02. Two Pointers/15. 3Sum.ipynb
deleted file mode 100644
index 06de549..0000000
--- a/docs/02. Two Pointers/15. 3Sum.ipynb	
+++ /dev/null
@@ -1,250 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 15. 3Sum\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the 3Sum problem on LeetCode, click here!](https://leetcode.com/problems/3sum/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array nums, *return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.*\n",
-        "\n",
-        "**Notice** that the solution set must not contain duplicate triplets.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `3 <= nums.length <= 3000`\n",
-        "- $-10^5$ <= `nums[i]` <= $10^5$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def threeSum(nums):\n",
-        "    # Sort the input array in ascending order\n",
-        "    nums.sort()\n",
-        "    triplets = []\n",
-        "    \n",
-        "    for i in range(len(nums) - 2):\n",
-        "        # Skip duplicates to avoid duplicate triplets\n",
-        "        if i > 0 and nums[i] == nums[i - 1]:\n",
-        "            continue\n",
-        "        \n",
-        "        left, right = i + 1, len(nums) - 1\n",
-        "        \n",
-        "        while left < right:\n",
-        "            # Calculate the total sum of the current triplet\n",
-        "            total = nums[i] + nums[left] + nums[right]\n",
-        "            \n",
-        "            if total == 0:\n",
-        "                # If the total is zero, we found a valid triplet\n",
-        "                triplets.append([nums[i], nums[left], nums[right]])\n",
-        "                \n",
-        "                # Skip duplicates of left and right pointers\n",
-        "                while left < right and nums[left] == nums[left + 1]:\n",
-        "                    left += 1\n",
-        "                while left < right and nums[right] == nums[right - 1]:\n",
-        "                    right -= 1\n",
-        "                \n",
-        "                # Move the pointers to the next unique elements\n",
-        "                left += 1\n",
-        "                right -= 1\n",
-        "            elif total < 0:\n",
-        "                # If the total is negative, we need to increase the sum by moving the left pointer to the right\n",
-        "                left += 1\n",
-        "            else:\n",
-        "                # If the total is positive, we need to decrease the sum by moving the right pointer to the left\n",
-        "                right -= 1\n",
-        "    \n",
-        "    return triplets"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. It starts by sorting the input array `nums` in ascending order. Sorting helps in efficiently finding triplets that sum up to zero.\n",
-        "\n",
-        "2. It initializes an empty list `triplets` to store the unique triplets that meet the given conditions.\n",
-        "\n",
-        "3. The code then iterates through the `nums` array using a loop with index `i`. This loop will consider each element of the array as a potential starting point for a triplet.\n",
-        "\n",
-        "4. Inside the loop, it checks for duplicates and skips them. This is done to ensure that the solution set does not contain duplicate triplets. If `nums[i]` is the same as the previous element `nums[i-1]`, it continues to the next iteration of the loop.\n",
-        "\n",
-        "5. Two pointers, `left` and `right`, are initialized. `left` starts just after the current element `nums[i]`, and `right` starts at the end of the array.\n",
-        "\n",
-        "6. The code enters another loop with `left` and `right` pointers, trying to find a triplet that sums up to zero.\n",
-        "\n",
-        "7. It calculates the `total` sum of the current triplet as `nums[i] + nums[left] + nums[right]`.\n",
-        "\n",
-        "8. If `total` is zero, it means a valid triplet is found, so it appends `[nums[i], nums[left], nums[right]]` to the `triplets` list. Then, it moves both the `left` and `right` pointers to their next unique elements while skipping duplicates.\n",
-        "\n",
-        "9. If `total` is less than zero, it means the sum is negative, so it increments the `left` pointer to move towards a larger sum.\n",
-        "\n",
-        "10. If `total` is greater than zero, it means the sum is positive, so it decrements the `right` pointer to move towards a smaller sum.\n",
-        "\n",
-        "11. This process continues until all possible triplets have been considered.\n",
-        "\n",
-        "12. Finally, the function returns the `triplets` list, which contains all the unique triplets that sum up to zero in the sorted `nums` array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[-1, -1, 2], [-1, 0, 1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "nums = [-1, 0, 1, 2, -1, -4]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "nums = [0, 1, 1]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 0, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "nums = [0, 0, 0]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `threeSum` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. Sorting the input array `nums` takes O(n log n) time, where n is the length of the array.\n",
-        "\n",
-        "2. The outer loop runs for each element of the array, so it iterates O(n) times.\n",
-        "\n",
-        "3. Inside the outer loop, we have a while loop with two pointers (`left` and `right`). In the worst case, the while loop can iterate O(n) times (when all elements are unique).\n",
-        "\n",
-        "4. Inside the while loop, we have constant time operations (additions, comparisons, and list appends).\n",
-        "\n",
-        "Overall, the time complexity of the `threeSum` function is dominated by the sorting step, so it is O(n log n) due to the sorting. The other operations inside the loops contribute linearly but are dominated by the sorting step.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The space used by the `triplets` list to store the output triplets is O(k), where k is the number of unique triplets that sum up to zero. In the worst case, this can be O(n^2/3) because there can be roughly O(n^2) possible triplets, and many of them may sum up to zero.\n",
-        "\n",
-        "2. Other than the `triplets` list, the function uses only a constant amount of additional space for variables like `i`, `left`, `right`, and `total`.\n",
-        "\n",
-        "Overall, the space complexity of the `threeSum` function is O(k), where k is the number of unique triplets that meet the criteria. It's important to note that the space complexity does not depend on the size of the input array but rather on the number of valid triplets found.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: O(n log n) due to sorting (dominant factor) and O(n^2) in the worst case for the nested loops.\n",
-        "- Space Complexity: O(k) for storing the output triplets, where k is the number of unique triplets that meet the criteria."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K-Sum Problem:** Generalize the problem to the K-Sum problem, where instead of finding triplets that sum to zero, you need to find K elements that sum to a target value. Implement a function for this generalization.\n",
-        "2. **Count the Number of Unique Triplets:** Modify the algorithm to count the number of unique triplets that satisfy the conditions instead of returning the triplets themselves. This will require some modifications to handle duplicates efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/02. Two Pointers/README.md b/docs/02. Two Pointers/README.md
deleted file mode 100644
index 0d4c7cf..0000000
--- a/docs/02. Two Pointers/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Two Pointers Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)                            | Easy       |
-| [15. 3Sum](https://leetcode.com/problems/3sum/)                                                       | Medium     |
-| [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)           | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb b/docs/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb
deleted file mode 100644
index 554c940..0000000
--- a/docs/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb	
+++ /dev/null
@@ -1,191 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 121. Best Time to Buy and Sell Stock\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Best Time to Buy and Sell Stock problem on LeetCode, click here!](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given an array `prices` where `prices[i]` is the price of a given stock on the $i^{th}$ day.\n",
-        "\n",
-        "You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.\n",
-        "\n",
-        "*Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `prices.length` <= $10^5$\n",
-        "- 0 <= `prices[i]` <= $10^4$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 65,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def maxProfit(prices):\n",
-        "    if not prices:\n",
-        "        return 0\n",
-        "    \n",
-        "    min_price = prices[0]\n",
-        "    max_profit = 0\n",
-        "    \n",
-        "    for price in prices:\n",
-        "        # Update the minimum price if a lower price is encountered\n",
-        "        min_price = min(min_price, price)\n",
-        "        # Calculate the profit if selling at the current price\n",
-        "        max_profit = max(max_profit, price - min_price)\n",
-        "    \n",
-        "    return max_profit"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- The `maxProfit` function takes a single argument, `prices`, which is a list of stock prices.\n",
-        "- The first line of the function checks if the `prices` list is empty. If it's empty, there are no prices to analyze, so the function returns 0 (no profit can be made).\n",
-        "\n",
-        "- Then, we initialize two variables:\n",
-        "  - `min_price`: This variable will keep track of the minimum price seen so far. We initialize it with the price of the first day in the list (`prices[0]`).\n",
-        "  - `max_profit`: This variable will keep track of the maximum profit we can achieve. We initialize it to 0 because initially, we haven't made any profit.\n",
-        "\n",
-        "- We start iterating through the `prices` list, where `price` represents the stock price for the current day.\n",
-        "\n",
-        "- In each iteration, we check if the current `price` is lower than the `min_price` we've seen so far. If it is, we update `min_price` to the current `price`. This is because we want to buy the stock at the lowest possible price.\n",
-        "\n",
-        "- Next, we calculate the profit we would make if we sold the stock at the current `price` (assuming we bought it at `min_price`). We do this by subtracting `min_price` from the current `price`. This represents the profit for the current day.\n",
-        "\n",
-        "- We also use the `max` function to keep track of the maximum profit seen so far. If the profit calculated for the current day is greater than the `max_profit` we've seen previously, we update `max_profit` with this higher value.\n",
-        "\n",
-        "- Finally, after iterating through all the days in the `prices` list, we return `max_profit`, which represents the maximum profit that can be achieved by buying and selling a single stock."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 66,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "5\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "prices1 = [7, 1, 5, 3, 6, 4]\n",
-        "print(maxProfit(prices1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 67,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "0\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "prices2 = [7, 6, 4, 3, 1]\n",
-        "print(maxProfit(prices2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code iterates through the `prices` array once, examining each price exactly once.\n",
-        "- Within the loop, there are constant-time operations such as comparisons, additions, and subtractions.\n",
-        "- Therefore, the time complexity of this code is O(n), where n is the length of the `prices` array. It performs a linear number of operations relative to the size of the input.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the code is constant, O(1).\n",
-        "- Regardless of the size of the input `prices` array, the code only uses a fixed amount of additional memory to store two variables (`min_price` and `max_profit`) and a few loop variables.\n",
-        "- The memory usage does not depend on the size of the input, so it is considered constant space complexity.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: O(n) where n is the length of the `prices` array.\n",
-        "- Space Complexity: O(1) (constant space usage)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Transactions Allowed:** Modify the problem to allow multiple transactions (buy and sell multiple times), but you must sell before buying again. Find the maximum profit in this scenario.\n",
-        "2. **K Transactions Allowed:** Extend the problem to allow at most k transactions. Find the maximum profit considering this constraint.\n",
-        "3. **With Transaction Fee:** Introduce a transaction fee for each buy/sell operation. Modify the code to maximize profit while considering the transaction fee."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb b/docs/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb
deleted file mode 100644
index 4a516de..0000000
--- a/docs/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb	
+++ /dev/null
@@ -1,214 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 3. Longest Substring Without Repeating Characters\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Substring Without Repeating Characters problem on LeetCode, click here!](https://leetcode.com/problems/longest-substring-without-repeating-characters/)\n",
-        "\n",
-        "---\n",
-        "Given a string `s`, find the length of the **longest substring** without repeating characters.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 0 <= `s.length` <= $5 * 10^4$\n",
-        "- `s` consists of English letters, digits, symbols and spaces.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 61,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def length_of_longest_substring(s):\n",
-        "    # Create a dictionary to store the index of each character's last occurrence.\n",
-        "    char_index = {}\n",
-        "    \n",
-        "    # Initialize variables to keep track of the start and end of the current substring.\n",
-        "    start = 0\n",
-        "    max_length = 0\n",
-        "    \n",
-        "    for end in range(len(s)):\n",
-        "        # If the character is in the dictionary and its last occurrence is after the start of the current substring,\n",
-        "        # update the start of the substring to the next character after its last occurrence.\n",
-        "        if s[end] in char_index and char_index[s[end]] >= start:\n",
-        "            start = char_index[s[end]] + 1\n",
-        "        \n",
-        "        # Update the last occurrence index of the current character.\n",
-        "        char_index[s[end]] = end\n",
-        "        \n",
-        "        # Calculate the length of the current substring and update the maximum length if necessary.\n",
-        "        current_length = end - start + 1\n",
-        "        max_length = max(max_length, current_length)\n",
-        "    \n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a function called `length_of_longest_substring` that takes a single argument `s`, which is the input string for which we want to find the length of the longest substring without repeating characters.\n",
-        "\n",
-        "2. We create a dictionary called `char_index` to store the index of the last occurrence of each character in the input string. This dictionary will help us keep track of where each character was last seen.\n",
-        "\n",
-        "3. We initialize two variables, `start` and `max_length`. \n",
-        "   - `start` represents the start index of the current substring without repeating characters. It starts at 0.\n",
-        "   - `max_length` is used to keep track of the maximum length found so far and is initially set to 0.\n",
-        "\n",
-        "4. We then iterate through the input string `s` using a for loop, where the loop variable `end` represents the current end index of the substring we are considering.\n",
-        "\n",
-        "5. Inside the loop:\n",
-        "   - We check if the character `s[end]` is already in the `char_index` dictionary and if its last occurrence is within or after the current substring. If so, it means that we've encountered a repeating character, and we need to update the `start` of the substring to the next character after the last occurrence of `s[end]`. This ensures that we have a new valid substring without repeating characters.\n",
-        "   - We then update the `char_index` dictionary by storing the current index `end` as the last occurrence index of the character `s[end]`.\n",
-        "\n",
-        "6. Next, we calculate the length of the current substring without repeating characters, which is `current_length = end - start + 1`. We add 1 to account for the fact that the indices are zero-based.\n",
-        "\n",
-        "7. We update the `max_length` with the maximum of its current value and the `current_length`. This step ensures that we keep track of the longest valid substring we have encountered so far.\n",
-        "\n",
-        "8. The loop continues to iterate through the input string, and at the end, we return the `max_length`, which represents the length of the longest substring without repeating characters."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 62,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(length_of_longest_substring(\"abcabcbb\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 63,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(length_of_longest_substring(\"bbbbb\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 64,
-      "id": "cf04eba0-0ead-4e8b-ba33-4316ea64a9d3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(length_of_longest_substring(\"pwwkew\"))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "Let's analyze the time and space complexity of the `length_of_longest_substring` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The function iterates through the input string `s` using a single for loop. The loop runs from the beginning of the string to the end once.\n",
-        "- Inside the loop, we perform constant-time operations such as dictionary lookups and updates, comparisons, and arithmetic operations.\n",
-        "- Therefore, the time complexity of the function is O(n), where n is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The primary data structure that consumes space in this function is the `char_index` dictionary.\n",
-        "- In the worst case, if there are no repeating characters in the input string, the dictionary can store all unique characters in the string.\n",
-        "- Therefore, the space complexity is O(min(n, m)), where n is the length of the input string `s`, and m is the number of unique characters in the string. In the worst case, when all characters are unique, m is equal to n, so the space complexity is O(n).\n",
-        "- Additionally, there are a few integer variables used for indices and lengths, which consume constant space.\n",
-        "- Overall, the space complexity of the function is O(min(n, m)) or simply O(n) in the worst case.\n",
-        "\n",
-        "**In summary,** the time complexity of the `length_of_longest_substring` function is O(n), and the space complexity is O(n) in the worst case due to the `char_index` dictionary. This algorithm provides an efficient way to find the length of the longest substring without repeating characters in linear time."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Longest Substring with K Distinct Characters:** Modify the problem to find the length of the longest substring with exactly K distinct characters. For example, given the input string \"abcabcbb\" and K = 2, the answer would be 4 because the longest substring with two distinct characters is \"abca.\"\n",
-        "2. **Longest Substring with Unique Characters:** Write a function to find the length of the longest substring in a given string where all characters are unique. For example, given the input string \"abcabcbb,\" the answer would be 4 because \"abcd\" is the longest substring with unique characters."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb b/docs/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb
deleted file mode 100644
index 07e96a2..0000000
--- a/docs/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb	
+++ /dev/null
@@ -1,195 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 424. Longest Repeating Character Replacement\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Repeating Character Replacement problem on LeetCode, click here!](https://leetcode.com/problems/longest-repeating-character-replacement/)\n",
-        "\n",
-        "---\n",
-        "You are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.\n",
-        "\n",
-        "*Return the length of the longest substring containing the same letter you can get after performing the above operations.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $10^5$\n",
-        "- `s` consists of only uppercase English letters.\n",
-        "- `0 <= k <= s.length`"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 58,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def characterReplacement(s, k):\n",
-        "    max_length = 0  # Initialize the maximum length\n",
-        "    max_count = 0   # Initialize the maximum count of repeating characters\n",
-        "    start = 0       # Initialize the start of the sliding window\n",
-        "    char_count = {} # Dictionary to store the count of each character\n",
-        "\n",
-        "    for end in range(len(s)):\n",
-        "        # Update the count of the current character in the dictionary\n",
-        "        char_count[s[end]] = char_count.get(s[end], 0) + 1\n",
-        "        \n",
-        "        # Update the maximum count of repeating characters\n",
-        "        max_count = max(max_count, char_count[s[end]])\n",
-        "        \n",
-        "        # Check if the current window size is greater than k\n",
-        "        if (end - start + 1) - max_count > k:\n",
-        "            # Move the start of the window to the right\n",
-        "            char_count[s[start]] -= 1\n",
-        "            start += 1\n",
-        "        \n",
-        "        # Update the maximum length\n",
-        "        max_length = max(max_length, end - start + 1)\n",
-        "\n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We initialize some variables:\n",
-        "   - `max_length` to keep track of the maximum length of the substring containing the same letter.\n",
-        "   - `max_count` to keep track of the maximum count of repeating characters within the current window.\n",
-        "   - `start` to represent the start index of the sliding window.\n",
-        "   - `char_count` is a dictionary that will store the count of each character within the current window.\n",
-        "\n",
-        "2. We use a `for` loop to iterate through the characters of the string `s`.\n",
-        "\n",
-        "3. Inside the loop, we do the following for each character at position `end`:\n",
-        "   - Update the count of the current character in the `char_count` dictionary.\n",
-        "   - Update the `max_count` to be the maximum of the current `max_count` and the count of the current character. This keeps track of the maximum count of repeating characters within the current window.\n",
-        "\n",
-        "4. We check if the current window size (the difference between `end` and `start` plus one) minus the `max_count` is greater than `k`. This condition checks whether we have exceeded the allowed number of replacements (`k`) within the current window.\n",
-        "\n",
-        "   - If we have exceeded the allowed replacements, it means we need to shrink the window from the left side. We do this by moving the `start` of the window to the right and decrementing the count of the character at `s[start]` in the `char_count` dictionary. This effectively removes characters from the left side of the window until we have a valid window again.\n",
-        "\n",
-        "   - By doing this, we ensure that the difference between the current window size and the `max_count` is always less than or equal to `k`.\n",
-        "\n",
-        "5. After each character, we update the `max_length` to be the maximum of the current `max_length` and the size of the current window (`end - start + 1`).\n",
-        "\n",
-        "6. Finally, we return `max_length`, which holds the length of the longest substring containing the same letter with at most `k` replacements."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 59,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "s1 = \"ABAB\"\n",
-        "k1 = 2\n",
-        "print(characterReplacement(s1, k1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 60,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "s2 = \"AABABBA\"\n",
-        "k2 = 1\n",
-        "print(characterReplacement(s2, k2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code uses a single `for` loop to iterate through the characters of the input string `s`. The loop runs from the beginning to the end of the string.\n",
-        "- Inside the loop, we perform constant-time operations such as updating the `char_count` dictionary and updating variables like `max_count` and `max_length`.\n",
-        "- The code's time complexity is primarily determined by the loop, which iterates through each character in the string once. Therefore, the time complexity is O(n), where n is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The code uses several variables to store information, but the space they consume is constant and does not depend on the size of the input string. These variables include `max_length`, `max_count`, `start`, and `char_count`. Therefore, the space complexity is O(1) or constant space.\n",
-        "\n",
-        "**In summary**, the time complexity of the code is O(n), where n is the length of the input string, and the space complexity is O(1), as it uses a constant amount of additional space regardless of the input size. This algorithm efficiently solves the problem with a linear time complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the code to find the longest substring containing the same letter with at most k replacements in O(n) time complexity and O(1) space complexity. Hint: You may need to update the approach to achieve this.\n",
-        "2. **Variation with Lowercase Letters:** Extend the problem to include both uppercase and lowercase English letters in the input string s. Write a function that can handle this extended input and still find the longest substring with at most k replacements efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/03. Sliding Window/76. Minimum Window Substring.ipynb b/docs/03. Sliding Window/76. Minimum Window Substring.ipynb
deleted file mode 100644
index 33c2a44..0000000
--- a/docs/03. Sliding Window/76. Minimum Window Substring.ipynb	
+++ /dev/null
@@ -1,273 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 76. Minimum Window Substring\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Minimum Window Substring problem on LeetCode, click here!](https://leetcode.com/problems/minimum-window-substring/)\n",
-        "\n",
-        "---\n",
-        "Given two strings `s` and `t` of lengths `m` and `n` respectively, return *the **minimum window substring** of `s` such that every character in `t` (**including duplicates**) is included in the window. If there is no such substring, return the empty string `\"\"`.*\n",
-        "\n",
-        "The testcases will be generated such that the answer is **unique**.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `m == s.length`\n",
-        "- `n == t.length`\n",
-        "- 1 <= `m, n` <= $10^5$\n",
-        "- `s` and `t` consist of uppercase and lowercase English letters.\n",
-        "\n",
-        "**Follow up**: Could you find an algorithm that runs in $O(m + n)$ time?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 54,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def min_window_substring(s, t):\n",
-        "    if not s or not t:\n",
-        "        return \"\"\n",
-        "\n",
-        "    # Initialize dictionaries to keep track of character counts for t and the current window in s.\n",
-        "    t_dict = {}\n",
-        "    current_window_dict = {}\n",
-        "    \n",
-        "    # Populate t_dict with character counts for string t.\n",
-        "    for char in t:\n",
-        "        t_dict[char] = t_dict.get(char, 0) + 1\n",
-        "\n",
-        "    # Initialize pointers for the sliding window.\n",
-        "    left = 0\n",
-        "    min_len = float('inf')\n",
-        "    min_window = \"\"\n",
-        "    required_chars = len(t_dict)\n",
-        "\n",
-        "    # Initialize a variable to keep track of how many required characters have been found in the current window.\n",
-        "    found_chars = 0\n",
-        "\n",
-        "    # Iterate over the string s using the right pointer.\n",
-        "    for right in range(len(s)):\n",
-        "        char = s[right]\n",
-        "\n",
-        "        # Update the current_window_dict.\n",
-        "        current_window_dict[char] = current_window_dict.get(char, 0) + 1\n",
-        "\n",
-        "        # Check if the current character is a required character and if its count matches the required count.\n",
-        "        if char in t_dict and current_window_dict[char] == t_dict[char]:\n",
-        "            found_chars += 1\n",
-        "\n",
-        "        # Try to minimize the window by moving the left pointer.\n",
-        "        while left <= right and found_chars == required_chars:\n",
-        "            # Calculate the current window size.\n",
-        "            window_size = right - left + 1\n",
-        "\n",
-        "            # If the current window is smaller than the minimum found so far, update min_len and min_window.\n",
-        "            if window_size < min_len:\n",
-        "                min_len = window_size\n",
-        "                min_window = s[left:right+1]\n",
-        "\n",
-        "            # Move the left pointer to the right to shrink the window.\n",
-        "            left_char = s[left]\n",
-        "            current_window_dict[left_char] -= 1\n",
-        "\n",
-        "            # Check if the character removed from the window was a required character.\n",
-        "            if left_char in t_dict and current_window_dict[left_char] < t_dict[left_char]:\n",
-        "                found_chars -= 1\n",
-        "\n",
-        "            # Move the left pointer further to continue shrinking the window.\n",
-        "            left += 1\n",
-        "\n",
-        "    return min_window"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by checking if either of the input strings `s` or `t` is empty. If either of them is empty, we return an empty string since there can't be any valid substring in this case.\n",
-        "\n",
-        "2. We initialize two dictionaries: `t_dict` and `current_window_dict`. These dictionaries will be used to keep track of character counts in the string `t` and the current window in string `s`, respectively.\n",
-        "\n",
-        "3. We populate the `t_dict` dictionary by iterating through string `t`. For each character, we increment its count in the dictionary using `t_dict.get(char, 0) + 1`. This allows us to count the occurrences of each character in `t`.\n",
-        "\n",
-        "4. We initialize two pointers: `left` and `right`. The `left` pointer will represent the start of the current window, and the `right` pointer will represent the end of the current window. We also initialize `min_len` to store the length of the minimum window found so far, and `min_window` to store the actual minimum window substring.\n",
-        "\n",
-        "5. We determine the number of required characters in `t` (i.e., the number of distinct characters in `t`) and store it in the variable `required_chars`.\n",
-        "\n",
-        "6. We initialize a variable `found_chars` to keep track of how many required characters have been found in the current window. Initially, it is set to 0.\n",
-        "\n",
-        "7. We iterate over the string `s` using the `right` pointer. In each iteration, we do the following:\n",
-        "\n",
-        "   - Update the `current_window_dict` by incrementing the count of the current character.\n",
-        "   \n",
-        "   - Check if the current character is a required character (present in `t_dict`) and if its count in the `current_window_dict` matches the required count from `t_dict`. If so, we increment `found_chars`.\n",
-        "\n",
-        "8. After updating the window, we attempt to minimize the window size by moving the `left` pointer to the right. In this step, we:\n",
-        "\n",
-        "   - Calculate the size of the current window.\n",
-        "   \n",
-        "   - If the current window size is smaller than the minimum found so far (`min_len`), we update `min_len` and `min_window` to store the current window substring.\n",
-        "   \n",
-        "   - Move the `left` pointer to the right to shrink the window.\n",
-        "   \n",
-        "   - Check if the character being removed from the window was a required character. If it was, we decrement `found_chars`.\n",
-        "   \n",
-        "   - Continue moving the `left` pointer further to continue shrinking the window if the window still contains all required characters.\n",
-        "\n",
-        "9. Finally, we return `min_window`, which will contain the minimum window substring that contains all characters from `t`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 55,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "BANC\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(min_window_substring(\"ADOBECODEBANC\", \"ABC\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 56,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "a\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(min_window_substring(\"a\", \"a\")) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 57,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": []
-        },
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(min_window_substring(\"a\", \"aa\")) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `min_window_substring` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The main loop iterates through the string `s` from left to right using the `right` pointer. This loop runs in O(m) time, where 'm' is the length of string `s`.\n",
-        "\n",
-        "- Inside the loop, we have a while loop that moves the `left` pointer to the right to shrink the window as needed. In the worst case, this while loop can also run in O(m) time because in the worst case, we may have to move the `left` pointer all the way to the end of the string.\n",
-        "\n",
-        "- Within each iteration of the while loop, we perform constant time operations, such as updating dictionaries and comparing character counts.\n",
-        "\n",
-        "- The initialization of the `t_dict` dictionary takes O(n) time, where 'n' is the length of string `t`.\n",
-        "\n",
-        "Therefore, the overall time complexity of the function is O(m + n) because the dominant factor is the length of string `s`, and the length of string `t` has a smaller impact.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity is determined by the space used by the `t_dict` dictionary, the `current_window_dict` dictionary, and a few variables.\n",
-        "\n",
-        "- The `t_dict` dictionary stores character counts for string `t`. In the worst case, when all characters in `t` are distinct, this dictionary can have a maximum of 'n' key-value pairs. So, the space complexity for `t_dict` is O(n).\n",
-        "\n",
-        "- The `current_window_dict` dictionary stores character counts for the current window. In the worst case, it can have a maximum of 'm' key-value pairs. So, the space complexity for `current_window_dict` is also O(m).\n",
-        "\n",
-        "- Other variables used in the function, such as `left`, `right`, `min_len`, `min_window`, `required_chars`, and `found_chars`, are all of constant size and do not depend on the input sizes.\n",
-        "\n",
-        "**In summary**, the overall space complexity of the function is O(max(m, n)), which means it is determined by the larger of the two input strings' lengths. This is because the space used for `t_dict` and `current_window_dict` dominates the space complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Smallest Distinct Substring:** Instead of finding the minimum window substring containing all characters, find the smallest distinct (unique) substring of s that contains all characters from t. This variation adds complexity because you must find a substring with distinct characters.\n",
-        "2. **No Extra Space:** Solve the problem without using any extra space, such as dictionaries or arrays, other than a constant amount of space. This is a significant optimization challenge."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/03. Sliding Window/README.md b/docs/03. Sliding Window/README.md
deleted file mode 100644
index e1294ff..0000000
--- a/docs/03. Sliding Window/README.md	
+++ /dev/null
@@ -1,19 +0,0 @@
-# Sliding Window Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [121. Best Time to Buy And Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | Easy       |
-| [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) | Medium |
-| [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) | Medium |
-| [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)             | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/04. Stack/20. Valid Parentheses.ipynb b/docs/04. Stack/20. Valid Parentheses.ipynb
deleted file mode 100644
index c47292f..0000000
--- a/docs/04. Stack/20. Valid Parentheses.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 20. Valid Parentheses\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Parentheses problem on LeetCode, click here!](https://leetcode.com/problems/valid-parentheses/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given a string s containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.\n",
-        "\n",
-        "An input string is valid if:\n",
-        "\n",
-        "1. Open brackets must be closed by the same type of brackets.\n",
-        "2. Open brackets must be closed in the correct order.\n",
-        "3. Every close bracket has a corresponding open bracket of the same type.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $10^4$\n",
-        "- `s` consists of parentheses only `'()[]{}'`."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 49,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isValid(s):\n",
-        "    # Create an empty stack to store opening brackets\n",
-        "    stack = []\n",
-        "    \n",
-        "    # Define a mapping for matching brackets\n",
-        "    bracket_mapping = {')': '(', '}': '{', ']': '['}\n",
-        "    \n",
-        "    # Iterate through the characters in the input string\n",
-        "    for char in s:\n",
-        "        # If the character is an opening bracket, push it onto the stack\n",
-        "        if char in bracket_mapping.values():\n",
-        "            stack.append(char)\n",
-        "        # If the character is a closing bracket\n",
-        "        elif char in bracket_mapping.keys():\n",
-        "            # Pop the top element from the stack if it exists, or use a dummy value '#'\n",
-        "            top_element = stack.pop() if stack else '#'\n",
-        "            # If the popped element does not match the corresponding opening bracket, return False\n",
-        "            if bracket_mapping[char] != top_element:\n",
-        "                return False\n",
-        "        # If the character is not a bracket, ignore it\n",
-        "        \n",
-        "    # After processing the entire string, if there are any unmatched opening brackets left in the stack, return False\n",
-        "    return len(stack) == 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining the `isValid` function that takes a single argument `s`, which is the input string containing parentheses and brackets.\n",
-        "\n",
-        "2. Inside the function, we create an empty stack, which is a list in Python, to store opening brackets as we encounter them in the input string. The stack will help us keep track of the brackets and their order.\n",
-        "\n",
-        "3. We define a `bracket_mapping` dictionary that maps each closing bracket to its corresponding opening bracket. This mapping will be used to check if a closing bracket matches the most recent opening bracket in the stack.\n",
-        "\n",
-        "4. We iterate through each character `char` in the input string `s` using a `for` loop.\n",
-        "\n",
-        "5. If the character `char` is an opening bracket (i.e., it exists in the `values()` of `bracket_mapping`), we push it onto the stack using the `append()` method.\n",
-        "\n",
-        "6. If the character `char` is a closing bracket (i.e., it exists in the `keys()` of `bracket_mapping`), we need to check if it matches the corresponding opening bracket. To do this, we pop the top element from the stack (if it exists) and store it in `top_element`. We use a dummy value `'#'` if the stack is empty to avoid errors.\n",
-        "\n",
-        "7. We then compare `top_element` with the corresponding opening bracket using `bracket_mapping[char]`. If they do not match, it means the string is not valid, and we return `False`.\n",
-        "\n",
-        "8. If the character `char` is not a bracket, we simply ignore it and continue the loop.\n",
-        "\n",
-        "9. After processing the entire string, we check if there are any unmatched opening brackets left in the stack. If the stack is empty, it means all opening brackets have been properly matched and closed, and we return `True`. Otherwise, we return `False`.\n",
-        "\n",
-        "10. Finally, we provide some test cases at the bottom of the code to demonstrate how the function works for different input strings."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 50,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(isValid(\"()\"))    "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 51,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(isValid(\"()[]{}\")) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 52,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(isValid(\"(]\"))     "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `isValid` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The function iterates through each character in the input string `s` once, performing constant-time operations for each character.\n",
-        "- Pushing and popping elements from the stack (list) also takes constant time in most cases.\n",
-        "- Therefore, the overall time complexity of the function is $O(n)$, where $n$ is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the function is determined by the space used by the stack and the `bracket_mapping` dictionary.\n",
-        "- In the worst case, if the input string `s` consists entirely of opening brackets, the stack can potentially contain all of these opening brackets, resulting in a space complexity of $O(n)$ in terms of the stack.\n",
-        "- The `bracket_mapping` dictionary has a constant number of key-value pairs (3 pairs in this case).\n",
-        "- Therefore, the overall space complexity of the function is $O(n)$, where $n$ is the length of the input string `s`, mainly due to the stack space.\n",
-        "\n",
-        "**In summary:**\n",
-        "- The time complexity is $O(n)$ because we iterate through the string once.\n",
-        "- The space complexity is $O(n)$ due to the stack used to keep track of opening brackets."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Valid Expressions:** Modify the problem to validate not only brackets but also arithmetic expressions containing parentheses, such as \"$2 * (3 + 5) / (4 - 2)$\".\n",
-        "2. **Valid Parentheses Combinations:**\n",
-        "Write a function to generate all valid combinations of n pairs of parentheses, where $n$ is a positive integer. For example, for $n = 3$, the valid combinations are [\"((()))\", \"(()())\", \"(())()\", \"()(())\", \"()()()\"]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/04. Stack/README.md b/docs/04. Stack/README.md
deleted file mode 100644
index 2ed115b..0000000
--- a/docs/04. Stack/README.md	
+++ /dev/null
@@ -1,16 +0,0 @@
-# Stack Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)                            | Easy       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb b/docs/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb
deleted file mode 100644
index 98f2331..0000000
--- a/docs/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,228 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 153. Find Minimum in Rotated Sorted Array\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Find Minimum in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:\n",
-        "\n",
-        "- `[4,5,6,7,0,1,2]` if it was rotated `4` times.\n",
-        "- `[0,1,2,4,5,6,7]` if it was rotated `7` times.\n",
-        "\n",
-        "Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.\n",
-        "\n",
-        "Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array.*\n",
-        "\n",
-        "You must write an algorithm that runs in $O(log\\ n)$ time.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `n == nums.length`\n",
-        "- `1 <= n <= 5000`\n",
-        "- `-5000 <= nums[i] <= 5000`\n",
-        "- All the integers of `nums` are **unique**.\n",
-        "- `nums` is sorted and rotated between `1` and `n` times."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 39,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def findMin(nums):\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "    \n",
-        "    while left < right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "        \n",
-        "        # If the mid element is greater than the rightmost element,\n",
-        "        # it means the minimum element is in the right half.\n",
-        "        if nums[mid] > nums[right]:\n",
-        "            left = mid + 1\n",
-        "        # If the mid element is less than or equal to the rightmost element,\n",
-        "        # it means the minimum element is in the left half or at the mid itself.\n",
-        "        else:\n",
-        "            right = mid\n",
-        "    \n",
-        "    # The loop will break when left and right converge to the minimum element.\n",
-        "    # At this point, left (or right) will be pointing to the minimum element.\n",
-        "    return nums[left]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The function `findMin(nums)` takes an input list `nums`, which represents the rotated sorted array.\n",
-        "\n",
-        "2. Initialize two pointers `left` and `right` to track the current search range. Initially, `left` is set to 0 (the beginning of the array), and `right` is set to `len(nums) - 1` (the end of the array).\n",
-        "\n",
-        "3. Enter a `while` loop with the condition `left < right`. This loop continues until `left` and `right` converge to a single element, which will be the minimum element in the rotated array.\n",
-        "\n",
-        "4. Inside the loop, calculate the middle index `mid` using integer division. This helps in finding the middle element of the current search range.\n",
-        "\n",
-        "5. Check if the element at the middle index (`nums[mid]`) is greater than the element at the rightmost index (`nums[right]`). If this condition is true, it means that the minimum element must be in the right half of the current search range. So, update `left` to `mid + 1`, effectively eliminating the left half of the search range.\n",
-        "\n",
-        "6. If the condition from step 5 is not met, it implies that the minimum element can be in the left half of the current search range or could be the element at the `mid` index itself. In this case, update `right` to `mid`, effectively eliminating the right half of the search range.\n",
-        "\n",
-        "7. Repeat steps 4-6 until `left` and `right` converge to the minimum element.\n",
-        "\n",
-        "8. When the loop exits, it means that `left` (or `right`) points to the minimum element in the array.\n",
-        "\n",
-        "9. Return `nums[left]` (or `nums[right]`) as the result, which is the minimum element in the rotated sorted array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 43,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1:\n",
-        "nums1 = [3, 4, 5, 1, 2]\n",
-        "print(findMin(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 41,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "0\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "print(findMin(nums2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 42,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "11\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "nums3 = [11, 13, 15, 17]\n",
-        "print(findMin(nums3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this code is $O(log\\ n)$, where '$n$' is the length of the input array `nums`. This is because the binary search algorithm is employed, which continually divides the search range in half with each iteration.\n",
-        "\n",
-        "In each iteration of the `while` loop, the search range is halved, and this process continues until the `left` and `right` pointers converge to the minimum element. In the worst case, it will take logarithmic time to reduce the search range to a single element.\n",
-        "\n",
-        "Therefore, the binary search used in this code runs in $O(log\\ n)$ time complexity.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of this code is $O(1)$, which means it uses a constant amount of additional space that does not depend on the size of the input array.\n",
-        "\n",
-        "The algorithm uses a fixed number of variables (`left`, `right`, `mid`) to keep track of the search range and indices, but the number of these variables does not grow with the size of the input array. Hence, the space complexity is constant.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: $O(log\\ n)$\n",
-        "- Space Complexity: $O(1)$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. Solve the problem with a variation where the input array may contain duplicates.\n",
-        "\n",
-        "2. Modify the problem to return the index of the minimum element in the rotated sorted array, rather than the element itself. The algorithm should still run in $O(log\\ n)$ time.\n",
-        "\n",
-        "3. Implement a function that finds the maximum element in a rotated sorted array. How would you adapt the binary search algorithm to solve this problem efficiently in $O(log\\ n)$ time?\n",
-        "\n",
-        "4. Implement a function that can find the kth smallest element in a rotated sorted array. This is an extension of the original problem, and the algorithm should still run in $O(log\\ n)$ time."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/05. Binary Search/33. Search in Rotated Sorted Array.ipynb b/docs/05. Binary Search/33. Search in Rotated Sorted Array.ipynb
deleted file mode 100644
index d8d4202..0000000
--- a/docs/05. Binary Search/33. Search in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 33. Search in Rotated Sorted Array\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Search in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/search-in-rotated-sorted-array/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "There is an integer array `nums` sorted in ascending order (with **distinct** values).\n",
-        "\n",
-        "Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k (1 <= k < nums.length)` such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index 3 and become `[4,5,6,7,0,1,2]`.\n",
-        "\n",
-        "Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.*\n",
-        "\n",
-        "You must write an algorithm with $O(log\\ n)$ runtime complexity.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `1 <= nums.length <= 5000`\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- All values of `nums` are **unique**.\n",
-        "- `nums` is an ascending array that is possibly rotated.\n",
-        "- $-10^4$ <= `target` <= $10^4$"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 44,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def search(nums, target):\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "\n",
-        "    # Find the pivot index using binary search\n",
-        "    while left < right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "\n",
-        "        if nums[mid] > nums[right]:\n",
-        "            left = mid + 1\n",
-        "        else:\n",
-        "            right = mid\n",
-        "\n",
-        "    pivot = left\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "\n",
-        "    # Determine which part of the array to search in\n",
-        "    if target >= nums[pivot] and target <= nums[right]:\n",
-        "        left = pivot\n",
-        "    else:\n",
-        "        right = pivot\n",
-        "\n",
-        "    # Perform binary search to find the target\n",
-        "    while left <= right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "\n",
-        "        if nums[mid] == target:\n",
-        "            return mid\n",
-        "        elif nums[mid] < target:\n",
-        "            left = mid + 1\n",
-        "        else:\n",
-        "            right = mid - 1\n",
-        "\n",
-        "    return -1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `search` function takes two arguments: `nums`, which is the rotated sorted array, and `target`, which is the value we want to find in the array.\n",
-        "\n",
-        "2. We initialize two pointers, `left` and `right`, which represent the range in which we are going to perform binary search. Initially, `left` is set to 0, and `right` is set to the index of the last element in the array (`len(nums) - 1`).\n",
-        "\n",
-        "3. We start by finding the pivot index using binary search. The pivot index is the index at which the array is rotated. We use a `while` loop to continue the search until `left` is less than `right`.\n",
-        "\n",
-        "4. Inside the loop, we calculate the middle index `mid` using integer division. We then compare the value at `mid` with the value at `right`. If `nums[mid]` is greater than `nums[right]`, it means the pivot point lies to the right of `mid`, so we update `left` to `mid + 1`. Otherwise, the pivot point lies to the left of `mid`, so we update `right` to `mid`. This process continues until we find the pivot index.\n",
-        "\n",
-        "5. After finding the pivot index, we have divided the array into two parts: one part is sorted in ascending order, and the other part is also sorted in ascending order but rotated. We need to determine which part of the array contains the target value.\n",
-        "\n",
-        "6. We reset `left` and `right` pointers. If the target value is within the range `[nums[pivot], nums[right]]`, we set `left` to `pivot` (the start of the rotated part), indicating that we should search in the rotated part. Otherwise, we set `right` to `pivot` (the end of the sorted part), indicating that we should search in the sorted part.\n",
-        "\n",
-        "7. We then perform binary search again within the chosen range (`left` to `right`) to find the target element. The binary search continues until `left` is less than or equal to `right`.\n",
-        "\n",
-        "8. Inside the binary search loop, we calculate the middle index `mid` and compare `nums[mid]` with the target value. If they are equal, we return `mid` as the index of the target element. If `nums[mid]` is less than the target, we update `left` to `mid + 1` to search in the right half. If `nums[mid]` is greater than the target, we update `right` to `mid - 1` to search in the left half.\n",
-        "\n",
-        "9. If we exit the binary search loop without finding the target, we return -1 to indicate that the target is not present in the array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 45,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "nums1 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "target1 = 0\n",
-        "result1 = search(nums1, target1)\n",
-        "print(result1) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 46,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "-1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "target2 = 3\n",
-        "result2 = search(nums2, target2)\n",
-        "print(result2) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 47,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "-1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "nums3 = [1]\n",
-        "target3 = 0\n",
-        "result3 = search(nums3, target3)\n",
-        "print(result3) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. Finding the pivot index using binary search takes $O(log\\ n)$ time, where '$n$' is the number of elements in the array.\n",
-        "2. After finding the pivot index, performing binary search to find the target element also takes $O(log\\ n)$ time in the worst case.\n",
-        "\n",
-        "The dominant factor in the time complexity is the binary search, and since we perform two binary searches sequentially, the overall time complexity of the code is $O(log\\ n)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is very minimal:\n",
-        "\n",
-        "1. We use a constant amount of additional space for variables such as `left`, `right`, `pivot`, and `mid`. These variables do not depend on the input size, so they contribute $O(1)$ space complexity.\n",
-        "2. The function does not use any additional data structures that scale with the input size.\n",
-        "\n",
-        "Therefore, the space complexity of the code is $O(1)$, which means it has a constant space complexity and does not depend on the size of the input array.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: $O(log\\ n)$\n",
-        "- Space Complexity: $O(1)$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Find Minimum Element in Rotated Sorted Array**: Write an algorithm to find the minimum element in a rotated sorted array. This is a variation of the problem where you don't need to search for a target value but instead find the smallest element.\n",
-        "\n",
-        "2. **Search in a Circularly Sorted Array**: Consider an array that is sorted in a circular manner (e.g., [4, 5, 6, 7, 0, 1, 2, 3]). Adapt the search algorithm to work for circularly sorted arrays while maintaining $O(log\\ n)$ complexity."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/05. Binary Search/README.md b/docs/05. Binary Search/README.md
deleted file mode 100644
index 6086b9b..0000000
--- a/docs/05. Binary Search/README.md	
+++ /dev/null
@@ -1,17 +0,0 @@
-# Binary search Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [153. Find Minimum In Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) | Medium |
-| [33. Search In Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)   | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/06. Linked List/141. Linked List Cycle.ipynb b/docs/06. Linked List/141. Linked List Cycle.ipynb
deleted file mode 100644
index 0336bc8..0000000
--- a/docs/06. Linked List/141. Linked List Cycle.ipynb	
+++ /dev/null
@@ -1,271 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 141. Linked List Cycle\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Linked List Cycle problem on LeetCode, click here!](https://leetcode.com/problems/linked-list-cycle/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given `head`, the head of a linked list, determine if the linked list has a cycle in it.\n",
-        "\n",
-        "There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is connected to. **Note that `pos` is not passed as a parameter.**\n",
-        "\n",
-        "Return `true` if there is a cycle in the linked list. Otherwise, return `false`.\n",
-        "\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of the nodes in the list is in the range $[0, 10^4]$.\n",
-        "- $-10^5$ <= Node.val <= $10^5$\n",
-        "- `pos` is `-1` or a **valid index** in the linked-list.\n",
-        "\n",
-        "**Follow up:** Can you solve it using $O(1)$ (i.e. constant) memory?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "    def __init__(self, val=0):\n",
-        "        self.val = val\n",
-        "        self.next = None\n",
-        "\n",
-        "class Solution:\n",
-        "    def hasCycle(self, head: ListNode) -> bool:\n",
-        "        if not head:\n",
-        "            return False\n",
-        "        \n",
-        "        slow = head\n",
-        "        fast = head\n",
-        "        \n",
-        "        while fast and fast.next:\n",
-        "            slow = slow.next\n",
-        "            fast = fast.next.next\n",
-        "            \n",
-        "            if slow == fast:\n",
-        "                return True\n",
-        "        \n",
-        "        return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The given code defines a class `Solution` with a method `hasCycle` for determining if a linked list has a cycle.\n",
-        "\n",
-        "The `ListNode` class represents nodes in the linked list, where each node has a `val` (a value associated with the node) and a `next` attribute (a reference to the next node in the list).\n",
-        "\n",
-        "The `hasCycle` method takes the `head` of a linked list as input and returns `True` if there is a cycle in the linked list, and `False` otherwise. It uses two pointers, `slow` and `fast`, to traverse the linked list. If there is a cycle, the `fast` pointer will eventually catch up to the `slow` pointer.\n",
-        "\n",
-        "The algorithm works as follows:\n",
-        "1. It checks if the input `head` is `None` (an empty list). If it is, the method returns `False` because an empty list can't have a cycle.\n",
-        "2. Two pointers, `slow` and `fast`, initially point to the `head` of the list.\n",
-        "3. The code enters a `while` loop where `fast` moves two steps at a time, and `slow` moves one step at a time.\n",
-        "4. If there is a cycle, `fast` will eventually catch up to `slow`, and the method returns `True`.\n",
-        "5. If the loop completes without finding a cycle, the method returns `False`."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list with a cycle\n",
-        "def create_linked_list_with_cycle(values, pos):\n",
-        "    dummy = ListNode()\n",
-        "    current = dummy\n",
-        "    cycle_node = None\n",
-        "    \n",
-        "    for i, val in enumerate(values):\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "        \n",
-        "        if i == pos:\n",
-        "            cycle_node = current\n",
-        "    \n",
-        "    if cycle_node:\n",
-        "        current.next = cycle_node\n",
-        "    \n",
-        "    return dummy.next"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "This helper function is designed to create a linked list with a cycle for testing purposes. It takes two arguments:\n",
-        "\n",
-        "1. `values`: A list of values representing the nodes in the linked list.\n",
-        "2. `pos`: An integer representing the index of the node where the cycle begins.\n",
-        "\n",
-        "Here's how the function works:\n",
-        "\n",
-        "- It starts by creating an empty `ListNode` called `dummy`. This `dummy` node is used to simplify the creation of the linked list.\n",
-        "- It initializes a `current` pointer to the `dummy` node. This pointer will be used to traverse the linked list.\n",
-        "- It initializes a `cycle_node` variable to `None`. This variable will hold the reference to the node where the cycle begins, if any.\n",
-        "- It then iterates through the `values` list, creating a new `ListNode` for each value and appending it to the linked list.\n",
-        "- Inside the loop, it checks if the current index `i` is equal to the specified `pos`. If they match, it sets `cycle_node` to the current node. This simulates the creation of a cycle in the linked list.\n",
-        "- After the loop, if `cycle_node` is not `None` (indicating a cycle should be created), it connects the last node in the linked list to `cycle_node`, effectively creating a cycle.\n",
-        "- Finally, it returns the reference to the first node of the linked list (not the `dummy` node).\n",
-        "\n",
-        "This helper function allows you to easily create test cases where you can specify the values for the linked list and the position where the cycle starts. It's particularly useful for testing the code's ability to detect cycles in linked lists."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "head = [3,2,0,-4]\n",
-        "pos = 1\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "head = [1,2]\n",
-        "pos = 0\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 22,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "head = [1]\n",
-        "pos = -1\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's discuss the time and space complexity of the code for detecting a cycle in a linked list:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this code is $O(n)$, where \"$n$\" is the number of nodes in the linked list.\n",
-        "\n",
-        "The reason for this is that both the `slow` and `fast` pointers traverse the linked list, and they move at different speeds. In the worst case, when there is no cycle, the `fast` pointer will reach the end of the list after going through approximately $n/2$ nodes. In the case of a cycle, it may take some extra iterations for the `fast` pointer to catch up to the `slow` pointer. However, the total number of iterations is still proportional to the number of nodes in the linked list, making it $O(n)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this code is $O(1)$, which means it uses constant space.\n",
-        "\n",
-        "The reason for this is that regardless of the size of the linked list, the code only uses two additional pointers (`slow` and `fast`) to traverse the list. These pointers do not depend on the size of the input linked list, so the space complexity remains constant.\n",
-        "\n",
-        "**In summary**, this code efficiently detects cycles in a linked list with a time complexity of $O(n)$ and a space complexity of $O(1)$. It is an example of an algorithm that solves a complex problem with minimal memory usage."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/06. Linked List/143. Reorder List.ipynb b/docs/06. Linked List/143. Reorder List.ipynb
deleted file mode 100644
index a83545b..0000000
--- a/docs/06. Linked List/143. Reorder List.ipynb	
+++ /dev/null
@@ -1,285 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 143. Reorder List\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Reorder List problem on LeetCode, click here!](https://leetcode.com/problems/reorder-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given the head of a singly linked-list. The list can be represented as:\n",
-        "\n",
-        "$L_0 → L_1 → … → L_{n - 1} → L_n$\n",
-        "\n",
-        "Reorder the list to be on the following form:\n",
-        "\n",
-        "$L_0 → L_n → L_1 → L_{n - 1} → L_2 → L_{n - 2} → …$\n",
-        "\n",
-        "You may not modify the values in the list's nodes. Only nodes themselves may be changed.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is in the range $[1, 5 * 10^4]$.\n",
-        "- `1 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 28,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def reorderList(head):\n",
-        "    if not head or not head.next:\n",
-        "        return head\n",
-        "\n",
-        "    # Step 1: Find the middle of the linked list\n",
-        "    slow, fast = head, head\n",
-        "    while fast.next and fast.next.next:\n",
-        "        slow = slow.next\n",
-        "        fast = fast.next.next\n",
-        "\n",
-        "    # Split the list into two halves\n",
-        "    first_half = head\n",
-        "    second_half = slow.next\n",
-        "    slow.next = None\n",
-        "\n",
-        "    # Step 2: Reverse the second half of the linked list\n",
-        "    prev = None\n",
-        "    current = second_half\n",
-        "    while current:\n",
-        "        next_node = current.next\n",
-        "        current.next = prev\n",
-        "        prev = current\n",
-        "        current = next_node\n",
-        "    second_half = prev\n",
-        "\n",
-        "    # Step 3: Merge the first and reversed second halves alternately\n",
-        "    p1, p2 = first_half, second_half\n",
-        "    while p2:\n",
-        "        next_p1 = p1.next\n",
-        "        next_p2 = p2.next\n",
-        "        p1.next = p2\n",
-        "        p2.next = next_p1\n",
-        "        p1 = next_p1\n",
-        "        p2 = next_p2\n",
-        "\n",
-        "    return head"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `reorderList` function is designed to reorder a singly linked list following the specific pattern described in the problem statement. Here's a detailed explanation of how the function works:\n",
-        "\n",
-        "1. **Base Cases Handling**:\n",
-        "   - The function first checks if the input linked list is empty (`not head`) or contains only one element (`not head.next`). If either of these conditions is met, the list cannot be reordered, so the function returns the original list as-is.\n",
-        "\n",
-        "2. **Finding the Middle of the Linked List**:\n",
-        "   - To reorder the list, we first need to find the middle point so that we can split the list into two halves. This is done using two pointers, `slow` and `fast`, initialized to the head of the list.\n",
-        "   - `slow` moves one step at a time while `fast` moves two steps at a time. When `fast` reaches the end of the list or the second-to-last node, `slow` will be at the middle node.\n",
-        "\n",
-        "3. **Splitting the List**:\n",
-        "   - After finding the middle node, we split the list into two halves:\n",
-        "     - The first half, `first_half`, contains nodes from the beginning up to the middle.\n",
-        "     - The second half, `second_half`, contains nodes from the middle to the end.\n",
-        "   - To split the list, we set the `next` pointer of the node before the middle node to `None`.\n",
-        "\n",
-        "4. **Reversing the Second Half**:\n",
-        "   - We reverse the second half of the linked list using the `prev`, `current`, and `next_node` pointers.\n",
-        "   - `prev` is initially `None`, and we iterate through the second half. For each node, we:\n",
-        "     - Set the `next` of the current node to `prev`, effectively reversing the next pointer direction.\n",
-        "     - Update `prev` to the current node.\n",
-        "     - Move to the next node using the `next_node`.\n",
-        "\n",
-        "5. **Merging the Two Halves Alternately**:\n",
-        "   - We now have two linked lists: `first_half` and the reversed `second_half`.\n",
-        "   - We merge these two lists alternately by adjusting the `next` pointers of the nodes.\n",
-        "   - `p1` and `p2` are pointers to the current nodes in `first_half` and `second_half`, respectively.\n",
-        "   - We iterate through both lists while reordering nodes:\n",
-        "     - Set the `next` of `p1` to `p2` to link a node from the first half to a node from the reversed second half.\n",
-        "     - Update `p1` and `p2` to their respective next nodes.\n",
-        "     - Repeat this process until we have processed all nodes in both halves.\n",
-        "\n",
-        "6. **Returning the Reordered List**:\n",
-        "   - After the merging process is complete, the linked list is reordered as specified.\n",
-        "   - The function returns the `head` of the reordered list.\n",
-        "\n",
-        "The overall result is a singly linked list that has been reordered according to the given pattern. The time complexity of this algorithm is O(N), where N is the number of nodes in the linked list, as we traverse the list once to find the middle, once to reverse the second half, and once to merge the two halves."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 29,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to print the linked list\n",
-        "def printLinkedList(head):\n",
-        "    result = []\n",
-        "    while head:\n",
-        "        result.append(head.val)\n",
-        "        head = head.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the helper functions:\n",
-        "**printLinkedList Function**:\n",
-        "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-        "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-        "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-        "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-        "\n",
-        "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 30,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 4, 2, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "head1 = ListNode(1, ListNode(2, ListNode(3, ListNode(4))))\n",
-        "reorderList(head1)\n",
-        "print(printLinkedList(head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 5, 2, 4, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "head2 = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))\n",
-        "reorderList(head2)\n",
-        "print(printLinkedList(head2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `reorderList` function:\n",
-        "\n",
-        "**Time Complexity**:\n",
-        "1. **Finding the Middle of the Linked List**: In this step, we use two pointers, one moving one step at a time (`slow`) and the other moving two steps at a time (`fast`). This process takes $O(N/2)$ time, where $N$ is the number of nodes in the linked list.\n",
-        "\n",
-        "2. **Splitting the List**: After finding the middle, we split the list into two halves by setting the `next` pointer of the node before the middle node to `None`. This step takes $O(1)$ time.\n",
-        "\n",
-        "3. **Reversing the Second Half**: We reverse the second half of the linked list using a while loop that iterates through the second half of the list once. Therefore, this step also takes $O(N/2)$ time.\n",
-        "\n",
-        "4. **Merging the Two Halves Alternately**: In this step, we merge the two halves alternately by adjusting the `next` pointers of the nodes. We iterate through both halves once, so this step takes $O(N/2)$ time.\n",
-        "\n",
-        "Overall, the time complexity of the `reorderList` function is dominated by the steps involving reversing and merging the two halves, both of which take $O(N/2)$ time. Therefore, the total time complexity is $O(N)$.\n",
-        "\n",
-        "**Space Complexity**:\n",
-        "The space complexity of the `reorderList` function is primarily determined by the variables and data structures used within the function. Let's break down the space complexity components:\n",
-        "\n",
-        "1. **Constant Space Variables**: The variables `slow`, `fast`, `prev`, `current`, `next_node`, `p1`, and `p2` are used to traverse and manipulate the linked list. These variables occupy constant space, regardless of the input size. Therefore, they contribute $O(1)$ to the space complexity.\n",
-        "\n",
-        "2. **Split Linked Lists**: In the splitting step, we create two new linked list segments (`first_half` and `second_half`) that store references to nodes from the original linked list. These segments occupy space proportional to half of the input list, $O(N/2)$. \n",
-        "\n",
-        "3. **Reversed Second Half**: During the reversal step, we reverse the second half of the linked list in-place without creating any additional data structures. Therefore, it doesn't contribute to additional space complexity.\n",
-        "\n",
-        "4. **Overall**: Combining the above components, the total space complexity is $O(N/2)$, which simplifies to $O(N)$ in terms of space complexity.\n",
-        "\n",
-        "**In summary**, the `reorderList` function has a time complexity of $O(N)$ and a space complexity of $O(N)$ due to the creation of two new linked list segments while splitting the list. The constant space variables used for traversal do not significantly impact the overall space complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Reorder by K Elements:** Generalize the reorderList function to reorder the list in a different pattern, where you reorder the list by $K$ elements at a time. For example, for $K=3$, you would reorder it as ($L_0 → L_1 → L_2 → L_n → L_{n-1} → L_{n-2} → ...$).\n",
-        "You may need to handle cases where the number of nodes is not a multiple of $K$.\n",
-        "\n",
-        "2. **Reorder by Odd and Even Nodes:** Modify the reorderList function to reorder the list in a pattern where odd-indexed nodes (1-based index) come before even-indexed nodes. For example, ($L_0 → L_n → L_1 → L_{n-1} → L_2 → L_{n-2} → ...$)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/06. Linked List/19. Remove Nth Node From End of List.ipynb b/docs/06. Linked List/19. Remove Nth Node From End of List.ipynb
deleted file mode 100644
index 263124e..0000000
--- a/docs/06. Linked List/19. Remove Nth Node From End of List.ipynb	
+++ /dev/null
@@ -1,278 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 19. Remove Nth Node From End of List\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Remove Nth Node From End of List problem on LeetCode, click here!](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the head of a linked list, remove the $n^{th}$ node from the end of the list and return its head.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is `sz`.\n",
-        "- `1 <= sz <= 30`\n",
-        "- `0 <= Node.val <= 100`\n",
-        "- `1 <= n <= sz`\n",
-        "\n",
-        "**Follow up:** Could you do this in one pass?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 23,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def removeNthFromEnd(head, n):\n",
-        "    # Create a dummy node to handle the case of removing the head node.\n",
-        "    dummy = ListNode(0)\n",
-        "    dummy.next = head\n",
-        "    first = dummy\n",
-        "    second = dummy\n",
-        "    \n",
-        "    # Advance the first pointer by n+1 nodes.\n",
-        "    for _ in range(n + 1):\n",
-        "        first = first.next\n",
-        "    \n",
-        "    # Move both pointers simultaneously until first reaches the end.\n",
-        "    while first is not None:\n",
-        "        first = first.next\n",
-        "        second = second.next\n",
-        "    \n",
-        "    # Remove the nth node by updating the next pointer of the previous node.\n",
-        "    second.next = second.next.next\n",
-        "    \n",
-        "    return dummy.next  # Return the modified head of the linked list"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a `ListNode` class to represent individual nodes in a linked list. Each node has two attributes: `val` (the value of the node) and `next` (a reference to the next node in the list).\n",
-        "\n",
-        "2. The `removeNthFromEnd` function takes the head of a linked list (`head`) and the value `n`, which represents the position from the end of the list of the node to be removed. It returns the modified head of the linked list after the removal.\n",
-        "\n",
-        "3. We create a dummy node (`dummy`) at the beginning of the list. This dummy node simplifies the code by handling cases where the head of the list needs to be removed.\n",
-        "\n",
-        "4. We initialize two pointers, `first` and `second`, both initially pointing to the dummy node.\n",
-        "\n",
-        "5. To advance the `first` pointer by `n+1` nodes, we use a `for` loop. This positions the `first` pointer `n` nodes ahead of the `second` pointer.\n",
-        "\n",
-        "6. We then move both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This ensures that the `second` pointer ends up pointing to the node that needs to be removed.\n",
-        "\n",
-        "7. To remove the nth node from the end, we update the `next` pointer of the node pointed to by the `second` pointer to skip over the node to be removed.\n",
-        "\n",
-        "8. Finally, we return `dummy.next`, which is the modified head of the linked list without the removed node.\n",
-        "\n",
-        "9. Additionally, there are two helper functions:\n",
-        "   - `createLinkedList(values)` creates a linked list from a list of values.\n",
-        "   - `convertToList(head)` converts a linked list back into a list of values."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list from a list of values.\n",
-        "def createLinkedList(values):\n",
-        "    if not values:\n",
-        "        return None\n",
-        "    head = ListNode(values[0])\n",
-        "    current = head\n",
-        "    for val in values[1:]:\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "    return head\n",
-        "\n",
-        "# Helper function to convert a linked list to a list of values.\n",
-        "def convertToList(head):\n",
-        "    result = []\n",
-        "    current = head\n",
-        "    while current:\n",
-        "        result.append(current.val)\n",
-        "        current = current.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the two helper functions:\n",
-        "\n",
-        "1. `createLinkedList(values)`: This function creates a linked list from a list of values.\n",
-        "\n",
-        "   - Input: `values` is a list containing values that you want to insert into the linked list.\n",
-        "   - Output: The function returns the head of the created linked list.\n",
-        "\n",
-        "   Explanation:\n",
-        "   - The function starts by checking if the `values` list is empty. If it is, it returns `None` to indicate an empty linked list.\n",
-        "   - If the `values` list is not empty, it creates the first node of the linked list with the value from the first element of `values`.\n",
-        "   - It then iterates through the remaining elements of `values` and creates new nodes for each value, connecting them together using the `next` attribute to form a linked list.\n",
-        "   - Finally, it returns the head of the linked list, which is the first node created.\n",
-        "   \n",
-        "\n",
-        "2. `convertToList(head)`: This function converts a linked list into a list of values.\n",
-        "\n",
-        "   - Input: `head` is the head node of the linked list that you want to convert.\n",
-        "   - Output: The function returns a list containing the values of the nodes in the linked list in order.\n",
-        "\n",
-        "   Explanation:\n",
-        "   - The function starts by initializing an empty list, `result`, to store the values of the linked list nodes.\n",
-        "   - It then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - The iteration continues until it reaches the end of the linked list (i.e., the `next` attribute of the current node becomes `None`).\n",
-        "   - After the iteration is complete, the function returns the `result` list, which now contains the values of all the nodes in the linked list in the same order as they appear in the linked list.\n",
-        "\n",
-        "These helper functions are useful for creating linked lists from lists of values and converting linked lists back into lists of values, making it easier to work with linked list examples and test cases in a more familiar list format."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 2, 3, 5]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "head1 = createLinkedList([1, 2, 3, 4, 5])\n",
-        "n1 = 2\n",
-        "new_head1 = removeNthFromEnd(head1, n1)\n",
-        "print(convertToList(new_head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 26,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "head2 = createLinkedList([1])\n",
-        "n2 = 1\n",
-        "new_head2 = removeNthFromEnd(head2, n2)\n",
-        "print(convertToList(new_head2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 27,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "head3 = createLinkedList([1, 2])\n",
-        "n3 = 1\n",
-        "new_head3 = removeNthFromEnd(head3, n3)\n",
-        "print(convertToList(new_head3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Certainly! Let's analyze the time and space complexity of the `removeNthFromEnd` function in isolation:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The `removeNthFromEnd` function performs two passes through the linked list. In the worst case, it needs to traverse the entire linked list of length $N$.\n",
-        "- The first pass positions the `first` pointer $n$ nodes ahead of the `second` pointer. This pass takes $O(N)$ time because it involves iterating through all N nodes.\n",
-        "- The second pass moves both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This also takes $O(N)$ time in the worst case.\n",
-        "- Therefore, the overall time complexity of the `removeNthFromEnd` function is $O(N)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The `removeNthFromEnd` function uses a constant amount of extra space for its variables (e.g., `first`, `second`, and `dummy`). This space usage does not depend on the size of the linked list.\n",
-        "- The space complexity is $O(1)$, indicating that the space used by the function is constant and independent of the size of the input linked list.\n",
-        "\n",
-        "**In summary**, the `removeNthFromEnd` function has a time complexity of $O(N)$ and a space complexity of $O(1)$. It efficiently removes the nth node from the end of the linked list with a single pass through the list and constant extra space usage."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/06. Linked List/206. Reverse Linked List.ipynb b/docs/06. Linked List/206. Reverse Linked List.ipynb
deleted file mode 100644
index c97de2d..0000000
--- a/docs/06. Linked List/206. Reverse Linked List.ipynb	
+++ /dev/null
@@ -1,316 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 206. Reverse Linked List\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Reverse Linked List problem on LeetCode, click here!](https://leetcode.com/problems/reverse-linked-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the `head` of a singly linked list, reverse the list, and return the reversed list.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is the range `[0, 5000]`.\n",
-        "- `-5000 <= Node.val <= 5000`\n",
-        "\n",
-        "**Follow up:** A linked list can be reversed either iteratively or recursively. Could you implement both?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 33,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "# Function to reverse a linked list iteratively\n",
-        "def reverseListIterative(head):\n",
-        "    prev = None\n",
-        "    current = head\n",
-        "    \n",
-        "    while current:\n",
-        "        next_node = current.next\n",
-        "        current.next = prev\n",
-        "        prev = current\n",
-        "        current = next_node\n",
-        "    \n",
-        "    return prev\n",
-        "\n",
-        "# Function to reverse a linked list recursively\n",
-        "def reverseListRecursive(head):\n",
-        "    if not head or not head.next:\n",
-        "        return head\n",
-        "\n",
-        "    new_head = reverseListRecursive(head.next)\n",
-        "    head.next.next = head\n",
-        "    head.next = None\n",
-        "\n",
-        "    return new_head"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `reverseListIterative(head)`: This function reverses a linked list iteratively using a loop. Here's a step-by-step explanation:\n",
-        "\n",
-        "   - It takes the `head` of the input linked list as a parameter.\n",
-        "   - It initializes two pointers, `prev` and `current`, initially set to `None` and the `head` of the list, respectively.\n",
-        "   - It enters a `while` loop that continues until `current` reaches the end of the list (i.e., becomes `None`).\n",
-        "   - Inside the loop:\n",
-        "     - It stores the `next_node` which is the next node after `current`.\n",
-        "     - It updates the `next` pointer of `current` to point to the `prev` node. This effectively reverses the link direction.\n",
-        "     - It moves `prev` to `current` and `current` to `next_node`, advancing the pointers one step further in the list.\n",
-        "   - Once the loop completes, `prev` will be pointing to the new head of the reversed list (which was the last node of the original list), so it returns `prev`.\n",
-        "\n",
-        "2. `reverseListRecursive(head)`: This function reverses a linked list recursively. Here's how it works:\n",
-        "\n",
-        "   - It takes the `head` of the input linked list as a parameter.\n",
-        "   - The base case is checked: if `head` is `None` or `head.next` is `None`, meaning the list is empty or has only one node, it returns `head` as there's no need to reverse a list with zero or one element.\n",
-        "   - In the recursive case:\n",
-        "     - It calls itself with `head.next`, effectively moving down the list until it reaches the end.\n",
-        "     - Once it reaches the end, it starts reversing the links:\n",
-        "       - `head.next.next` is set to `head`, reversing the link between `head` and the next node.\n",
-        "       - `head.next` is set to `None` to avoid cycles.\n",
-        "   - Finally, it returns the new head of the reversed list, which will be the last node of the original list.\n",
-        "\n",
-        "**In summary,** `reverseListIterative` reverses the linked list by iterating through it and changing the next pointers, while `reverseListRecursive` reverses the linked list by recursively reaching the end and then reversing the links on the way back up the recursion stack. Both functions achieve the same result: reversing the linked list."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 36,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list from a list of values\n",
-        "def createLinkedList(values):\n",
-        "    if not values:\n",
-        "        return None\n",
-        "    head = ListNode(values[0])\n",
-        "    current = head\n",
-        "    for val in values[1:]:\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "    return head\n",
-        "\n",
-        "# Helper function to convert a linked list to a list for testing\n",
-        "def linkedListToList(head):\n",
-        "    result = []\n",
-        "    current = head\n",
-        "    while current:\n",
-        "        result.append(current.val)\n",
-        "        current = current.next\n",
-        "    return result\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the two helper functions used in the provided code:\n",
-        "\n",
-        "1. `createLinkedList(values)`:\n",
-        "\n",
-        "   - Purpose: This function is used to create a linked list from a list of values. It takes a list of values as input and returns the head of the linked list.\n",
-        "\n",
-        "   - How it works:\n",
-        "     - It first checks if the input list `values` is empty. If it's empty, it returns `None` to indicate an empty linked list.\n",
-        "     - If `values` is not empty, it initializes a `head` node with the value of the first element in `values`.\n",
-        "     - It then iterates through the remaining values in `values`, creating new nodes for each value and linking them together to form a linked list.\n",
-        "     - Finally, it returns the `head` node of the newly created linked list.\n",
-        "\n",
-        "2. `linkedListToList(head)`:\n",
-        "\n",
-        "   - Purpose: This function is used to convert a linked list back into a Python list for testing and output purposes.\n",
-        "\n",
-        "   - How it works:\n",
-        "     - It takes the `head` of the linked list as input.\n",
-        "     - It initializes an empty list called `result`.\n",
-        "     - It then iterates through the linked list, starting from `head`, and appends each node's `val` (value) to the `result` list.\n",
-        "     - This process continues until the end of the linked list is reached.\n",
-        "     - Finally, it returns the `result` list, which contains the values from the linked list in the same order.\n",
-        "\n",
-        "These helper functions are used to simplify the process of creating linked lists from lists of values and converting linked lists back into lists, making it easier to work with linked lists in the provided examples and test cases."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 35,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[1, 1, 2, 3, 4, 4]"
-            ]
-          },
-          "execution_count": 35,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 1: Iterative\n",
-        "head1 = [1,2,3,4,5]\n",
-        "head1 = createLinkedList(head1)\n",
-        "reversed_head1 = reverseListIterative(head1)\n",
-        "print(linkedListToList(reversed_head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[]"
-            ]
-          },
-          "execution_count": 37,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 2: Recursive\n",
-        "head2 = [1,2]\n",
-        "head2 = createLinkedList(head2)\n",
-        "reversed_head2 = reverseListRecursive(head2)\n",
-        "print(linkedListToList(reversed_head2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 38,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[0]"
-            ]
-          },
-          "execution_count": 38,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 3: Empty list\n",
-        "head3 = []\n",
-        "head3 = createLinkedList(head3)\n",
-        "reversed_head3 = reverseListIterative(head3)\n",
-        "print(linkedListToList(reversed_head3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the two functions:\n",
-        "\n",
-        "1. `reverseListIterative(head)`:\n",
-        "\n",
-        "   - Time Complexity: $O(n)$\n",
-        "     - The iterative function visits each node in the linked list exactly once in a single pass, where \"$n$\" is the number of nodes in the list. Therefore, the time complexity is linear in the number of nodes.\n",
-        "\n",
-        "   - Space Complexity: $O(1)$\n",
-        "     - This function uses a constant amount of extra space for the `prev`, `current`, and `next_node` pointers. Regardless of the size of the input list, the amount of additional memory used remains the same, so the space complexity is constant.\n",
-        "\n",
-        "2. `reverseListRecursive(head)`:\n",
-        "\n",
-        "   - Time Complexity: $O(n)$\n",
-        "     - The recursive function also visits each node in the linked list exactly once. It recursively traverses the list from the head to the tail, reversing the links on the way back. Therefore, like the iterative approach, the time complexity is $O(n)$, where \"$n$\" is the number of nodes.\n",
-        "\n",
-        "   - Space Complexity: $O(n)$\n",
-        "     - The recursive function uses space on the call stack for each recursive call. In the worst case, when the list has \"$n$\" nodes, it will create \"$n$\" recursive function calls on the stack, resulting in a space complexity of $O(n)$. This is because each function call stores information about its state, including the `head` pointer, until it reaches the base case and starts returning.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "- The time complexity of both functions is $O(n)$ because they both visit each node once.\n",
-        "- The space complexity of `reverseListIterative` is $O(1)$ because it uses a constant amount of extra space.\n",
-        "- The space complexity of `reverseListRecursive` is $O(n)$ due to the recursive function calls and the associated call stack space.\n",
-        "\n",
-        "Both functions are efficient in terms of time complexity, but `reverseListIterative` is more memory-efficient because it uses a constant amount of additional space, while `reverseListRecursive` uses space on the call stack proportional to the number of nodes in the list."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K-Group Reverse:** Modify the reverseListIterative function to reverse the linked list in groups of $K$ elements. For example, if the input linked list is [1, 2, 3, 4, 5, 6], and $K$ is 3, the output should be [3, 2, 1, 6, 5, 4].\n",
-        "\n",
-        "2. **Rotate Linked List:** Write a function to rotate a linked list to the right by $K$ places. For example, if the input is [1, 2, 3, 4, 5] and $K$ is 2, the output should be [4, 5, 1, 2, 3]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/06. Linked List/21. Merge Two Sorted Lists.ipynb b/docs/06. Linked List/21. Merge Two Sorted Lists.ipynb
deleted file mode 100644
index 876d633..0000000
--- a/docs/06. Linked List/21. Merge Two Sorted Lists.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 21. Merge Two Sorted Lists\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Merge Two Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-two-sorted-lists/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given the heads of two sorted linked lists `list1` and `list2`.\n",
-        "\n",
-        "Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.\n",
-        "\n",
-        "*Return the head of the merged linked list.*\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in both lists is in the range `[0, 50]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "- Both `list1` and `list2` are sorted in non-decreasing order."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def mergeTwoLists(list1, list2):\n",
-        "    # Create a dummy node to simplify the code.\n",
-        "    dummy = ListNode()\n",
-        "    current = dummy\n",
-        "\n",
-        "    while list1 and list2:\n",
-        "        if list1.val < list2.val:\n",
-        "            current.next = list1\n",
-        "            list1 = list1.next\n",
-        "        else:\n",
-        "            current.next = list2\n",
-        "            list2 = list2.next\n",
-        "        current = current.next\n",
-        "\n",
-        "    # Append the remaining elements from either list.\n",
-        "    if list1:\n",
-        "        current.next = list1\n",
-        "    else:\n",
-        "        current.next = list2\n",
-        "\n",
-        "    return dummy.next"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a class `ListNode` to represent the nodes of a singly-linked list. Each node has a `val` attribute to store the value of the node and a `next` attribute to point to the next node in the list.\n",
-        "\n",
-        "2. We define the `mergeTwoLists` function that takes two linked list heads, `list1` and `list2`, as input.\n",
-        "\n",
-        "3. We create a `dummy` node at the beginning of the merged list. The `dummy` node simplifies the code by serving as a placeholder, and its `next` attribute will point to the actual merged list.\n",
-        "\n",
-        "4. We initialize a `current` pointer to point to the `dummy` node initially. This `current` pointer helps us traverse and build the merged list.\n",
-        "\n",
-        "5. We enter a `while` loop that continues until either `list1` or `list2` becomes empty. Inside the loop, we compare the values of the nodes at the current positions of `list1` and `list2`.\n",
-        "\n",
-        "6. If the `val` of the node in `list1` is smaller than the `val` of the node in `list2`, we attach the node from `list1` to the merged list by updating the `next` attribute of the `current` node to point to the node in `list1`. We then move the `list1` pointer to the next node.\n",
-        "\n",
-        "7. If the `val` of the node in `list2` is smaller than or equal to the `val` of the node in `list1`, we attach the node from `list2` to the merged list in a similar manner. We move the `list2` pointer to the next node.\n",
-        "\n",
-        "8. After attaching a node to the merged list, we move the `current` pointer to the newly added node. This step is essential for keeping track of the end of the merged list.\n",
-        "\n",
-        "9. The loop continues until either `list1` or `list2` becomes empty.\n",
-        "\n",
-        "10. Once the loop exits, we check if there are any remaining elements in `list1` or `list2`. If `list1` is not empty, we attach the remaining elements of `list1` to the merged list. If `list2` is not empty, we attach the remaining elements of `list2`.\n",
-        "\n",
-        "11. Finally, we return the `next` attribute of the `dummy` node as the head of the merged list, which represents the sorted merged list."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to print the linked list\n",
-        "def printLinkedList(head):\n",
-        "    result = []\n",
-        "    while head:\n",
-        "        result.append(head.val)\n",
-        "        head = head.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the helper functions:\n",
-        "**printLinkedList Function**:\n",
-        "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-        "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-        "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-        "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-        "\n",
-        "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 1:\n",
-        "# Input: list1 = [1,2,4], list2 = [1,3,4]\n",
-        "list1 = ListNode(1, ListNode(2, ListNode(4)))\n",
-        "list2 = ListNode(1, ListNode(3, ListNode(4)))\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 2:\n",
-        "# Input: list1 = [], list2 = []\n",
-        "list1 = None\n",
-        "list2 = None\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 3:\n",
-        "# list1 = [], list2 = [0]\n",
-        "list1 = None\n",
-        "list2 = ListNode(0)\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `mergeTwoLists` function.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this function is $O(N + M$), where $N$ and $M$ are the lengths of `list1` and `list2`, respectively. Here's why:\n",
-        "\n",
-        "1. In the worst case, we need to traverse both `list1` and `list2` completely. This involves iterating through all the nodes in both lists once.\n",
-        "\n",
-        "2. The while loop runs until either `list1` or `list2` becomes empty. The number of iterations depends on the total number of nodes in both lists, which is N + M in the worst case.\n",
-        "\n",
-        "3. Inside the loop, we perform constant-time operations for each iteration, such as comparisons and updating pointers.\n",
-        "\n",
-        "Therefore, the dominant factor in the time complexity is the combined length of both input lists, resulting in a linear time complexity of $O(N + M)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this function is $O(1)$, which means it uses a constant amount of extra space regardless of the input sizes. Here's why:\n",
-        "\n",
-        "1. We create a few extra pointers (`dummy`, `current`) and temporary variables (`val`) to manage the merging process. These consume a fixed amount of memory, regardless of the input sizes. The number of these extra variables is independent of the lengths of `list1` and `list2`.\n",
-        "\n",
-        "2. We do not create a new data structure or allocate memory for the merged list. Instead, we rearrange the existing nodes from `list1` and `list2` to form the merged list. This operation does not consume additional memory proportional to the input sizes.\n",
-        "\n",
-        "**In summary**, the `mergeTwoLists` function has a time complexity of $O(N + M)$ and a space complexity of $O(1)$, making it an efficient solution for merging two sorted linked lists."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Merge K Sorted Lists:** Extend the problem to merge $K$ sorted linked lists instead of just two. You need to efficiently merge $K$ lists into one sorted list.\n",
-        "\n",
-        "2. **Merge Lists in a Zigzag Pattern:** Merge two sorted lists in a zigzag pattern. For example, given [1, 2, 4] and [1, 3, 5], the merged list should be [1, 1, 2, 3, 4, 5]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": "sql",
-      "file_extension": "",
-      "mimetype": "",
-      "name": "sql",
-      "version": "3.32.3"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/06. Linked List/23. Merge k Sorted Lists.ipynb b/docs/06. Linked List/23. Merge k Sorted Lists.ipynb
deleted file mode 100644
index a850734..0000000
--- a/docs/06. Linked List/23. Merge k Sorted Lists.ipynb	
+++ /dev/null
@@ -1,285 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 23. Merge k Sorted Lists\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Merge k Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-k-sorted-lists/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given an array of $k$ linked-lists lists, each linked-list is sorted in ascending order.\n",
-        "\n",
-        "Merge all the linked-lists into one sorted linked-list and return it.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `k == lists.length`\n",
-        "- 0 <= k <= $10^4$\n",
-        "- `0 <= lists[i].length <= 500`\n",
-        "- $-10^4$ <= lists[i][j] <= $10^4$\n",
-        "- `lists[i]` is sorted in ascending order.\n",
-        "- The sum of `lists[i].length` will not exceed $10^4$.\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "     def __init__(self, val=0, next=None):\n",
-        "         self.val = val\n",
-        "         self.next = next\n",
-        "\n",
-        "class Solution:\n",
-        "    def mergeKLists(self, lists: [[ListNode]]) -> ListNode:\n",
-        "        # Check if the input list of linked lists is empty\n",
-        "        if not lists or len(lists) == 0:\n",
-        "            return None\n",
-        "\n",
-        "        # Loop until there is only one merged list left\n",
-        "        while len(lists) > 1:\n",
-        "            mergedLists = []\n",
-        "\n",
-        "            # Merge pairs of lists\n",
-        "            for i in range(0, len(lists), 2):\n",
-        "                l1 = lists[i]\n",
-        "                l2 = lists[i + 1] if (i + 1) < len(lists) else None\n",
-        "                mergedLists.append(self.mergeList(l1, l2))\n",
-        "\n",
-        "            # Update the list of lists with merged results\n",
-        "            lists = mergedLists\n",
-        "        \n",
-        "        # Return the final merged list\n",
-        "        return lists[0]\n",
-        "\n",
-        "    def mergeList(self, l1, l2):\n",
-        "        # Create a dummy node to simplify merging\n",
-        "        dummy = ListNode()\n",
-        "        tail = dummy\n",
-        "\n",
-        "        # Merge the two sorted lists\n",
-        "        while l1 and l2:\n",
-        "            if l1.val < l2.val:\n",
-        "                tail.next = l1\n",
-        "                l1 = l1.next\n",
-        "            else:\n",
-        "                tail.next = l2\n",
-        "                l2 = l2.next\n",
-        "            tail = tail.next\n",
-        "\n",
-        "        # Append any remaining elements from l1 or l2 (if any)\n",
-        "        if l1:\n",
-        "            tail.next = l1\n",
-        "        if l2:\n",
-        "            tail.next = l2\n",
-        "\n",
-        "        # Return the merged result starting from the next of the dummy node\n",
-        "        return dummy.next\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided code defines a Python class `Solution` with two methods for merging k sorted linked lists:\n",
-        "\n",
-        "1. `mergeKLists(self, lists: List[ListNode]) -> ListNode`: This method takes a list of k sorted linked lists as input and returns a single merged sorted linked list. It uses a divide-and-conquer approach to repeatedly merge pairs of lists until only one merged list remains.\n",
-        "\n",
-        "2. `mergeList(self, l1, l2)`: This method takes two sorted linked lists, `l1` and `l2`, as input and merges them into a single sorted linked list. It uses a dummy node to simplify the merging process.\n",
-        "\n",
-        "Here's a high-level overview of how the code works:\n",
-        "\n",
-        "- The `mergeKLists` method checks if the input list of linked lists is empty or contains no lists. If there are no lists, it returns `None`.\n",
-        "\n",
-        "- Inside a `while` loop, the code repeatedly merges pairs of linked lists until only one merged list remains in the `lists` array. It does this by iterating through the input lists in pairs and calling the `mergeList` method to merge each pair.\n",
-        "\n",
-        "- The `mergeList` method takes two sorted linked lists, `l1` and `l2`, and merges them into a single sorted linked list. It uses a dummy node (`dummy`) and a `tail` pointer to keep track of the merged list while comparing and merging elements from `l1` and `l2`.\n",
-        "\n",
-        "- After merging all pairs of lists and updating the `lists` array with the merged results, the loop continues until only one merged list remains in the `lists` array.\n",
-        "\n",
-        "- Finally, the `mergeKLists` method returns the merged list.\n",
-        "\n",
-        "Overall, this code efficiently merges k sorted linked lists using a divide-and-conquer strategy, resulting in a single merged sorted linked list as the output."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> "
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: lists = [[1,4,5],[1,3,4],[2,6]]\n",
-        "\n",
-        "lists1 = [\n",
-        "    ListNode(1, ListNode(4, ListNode(5))),\n",
-        "    ListNode(1, ListNode(3, ListNode(4))),\n",
-        "    ListNode(2, ListNode(6))\n",
-        "]\n",
-        "solution = Solution()\n",
-        "result1 = solution.mergeKLists(lists1)\n",
-        "\n",
-        "# Print the result\n",
-        "if result1:\n",
-        "    current = result1\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: lists = []\n",
-        "\n",
-        "lists2 = []\n",
-        "solution = Solution()\n",
-        "result2 = solution.mergeKLists(lists2)\n",
-        "\n",
-        "# Print the result\n",
-        "if result2:\n",
-        "    current = result2\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "# Input: lists = [[]]\n",
-        "\n",
-        "lists3 = [None]\n",
-        "solution = Solution()\n",
-        "result3 = solution.mergeKLists(lists3)\n",
-        "\n",
-        "# Print the result\n",
-        "if result3:\n",
-        "    current = result3\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. **Heap Initialization:** The code does not use a heap. Instead, it uses a divide-and-conquer approach. The initial check for empty input lists takes $O(1)$ time.\n",
-        "\n",
-        "2. **Merging:** The merging operation is performed in a divide-and-conquer fashion. In each iteration of the `while` loop, we merge pairs of linked lists, and the number of comparisons made is proportional to the total number of nodes across all the linked lists ($n$). In each merge step, we effectively process each node once. The number of iterations required to reduce k lists to 1 is $O(log\\ k)$.\n",
-        "\n",
-        "   Therefore, the overall time complexity of the code is $O(n * log\\ k)$, where n is the total number of nodes across all lists, and $k$ is the number of linked lists.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. **Heap Space:** The code doesn't use a heap data structure, so there's no additional space complexity due to a heap.\n",
-        "\n",
-        "2. **Merged Lists:** In the `mergeList` method, we create a new merged list. However, this list is not stored in memory for all lists; it's replaced with each merged pair. The space used for these merged lists is proportional to the size of the largest merged list, which is $O(n)$ in the worst case.\n",
-        "\n",
-        "3. **Additional Variables:** The code uses a few additional variables, such as `dummy` and `tail`, but these occupy a constant amount of space and don't depend on the input size.\n",
-        "\n",
-        "   Therefore, the overall space complexity of the code is $O(n)$, where n is the total number of nodes across all lists.\n",
-        "\n",
-        "In summary, the code's time complexity is $O(n * log(k))$, and its space complexity is $O(n)$. This code efficiently merges $k$ sorted linked lists using a divide-and-conquer approach with a relatively low space overhead."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/06. Linked List/README.md b/docs/06. Linked List/README.md
deleted file mode 100644
index fb7b3ee..0000000
--- a/docs/06. Linked List/README.md	
+++ /dev/null
@@ -1,21 +0,0 @@
-# Linked List Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)                       | Easy       |
-| [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)                   | Easy       |
-| [143. Reorder List](https://leetcode.com/problems/reorder-list/)                                     | Medium     |
-| [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | Medium |
-| [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)                           | Medium     |
-| [23. Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)                       | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/07. Trees/100. Same Tree.ipynb b/docs/07. Trees/100. Same Tree.ipynb
deleted file mode 100644
index a270fab..0000000
--- a/docs/07. Trees/100. Same Tree.ipynb	
+++ /dev/null
@@ -1,211 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 100: Same Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-        "\n",
-        "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-        "\n",
-        "**Constraints**\n",
-        "\n",
-        "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-        "2. $-10^4$ <= Node.val <= $10^4$\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def isSameTree(p, q):\n",
-        "    # Base case: If both p and q are None, the trees are the same.\n",
-        "    if not p and not q:\n",
-        "        return True\n",
-        "    \n",
-        "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-        "    if not p or not q:\n",
-        "        return False\n",
-        "    \n",
-        "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-        "    if p.val != q.val:\n",
-        "        return False\n",
-        "    \n",
-        "    # Recursively check the left and right subtrees of p and q.\n",
-        "    # If both subtrees are the same, the entire trees are the same.\n",
-        "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-        "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-        "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-        "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-        "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-        "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "\n",
-        "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-        "\n",
-        "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "print(isSameTree(p1, q1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "\n",
-        "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-        "\n",
-        "p2 = TreeNode(1, TreeNode(2), None)\n",
-        "q2 = TreeNode(1, None, TreeNode(2))\n",
-        "print(isSameTree(p2, q2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "\n",
-        "#Input: p = [1,2,1], q = [1,1,2]\n",
-        "\n",
-        "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-        "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-        "print(isSameTree(p3, q3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time Complexity\n",
-        "\n",
-        "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-        "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-        "\n",
-        "## Space Complexity\n",
-        "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-        "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-        "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-        "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-        "\n",
-        "## In summary:\n",
-        "\n",
-        "+ **Time Complexity:** $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-        "+ **Space Complexity:** $O(n)$ due to the recursive call stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/102. Binary Tree Level Order Traversal.ipynb b/docs/07. Trees/102. Binary Tree Level Order Traversal.ipynb
deleted file mode 100644
index 8cb08b7..0000000
--- a/docs/07. Trees/102. Binary Tree Level Order Traversal.ipynb	
+++ /dev/null
@@ -1,271 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 102. Binary Tree Level Order Traversal\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Level Order Traversal problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-level-order-traversal/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, return *the level order traversal of its nodes' values*. (i.e., from left to right, level by level).\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-        "- `-1000 <= Node.val <= 1000`\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def levelOrder(root):\n",
-        "    # Check if the tree is empty, if so, return an empty list.\n",
-        "    if not root:\n",
-        "        return []\n",
-        "\n",
-        "    # Initialize an empty list to store the result.\n",
-        "    result = []\n",
-        "    \n",
-        "    # Initialize a queue with the root node to perform BFS.\n",
-        "    queue = [root]\n",
-        "\n",
-        "    while queue:\n",
-        "        # Initialize an empty list to store the values of nodes at the current level.\n",
-        "        level_values = []\n",
-        "        \n",
-        "        # Get the number of nodes at the current level.\n",
-        "        level_size = len(queue)\n",
-        "\n",
-        "        # Iterate through the nodes at the current level.\n",
-        "        for _ in range(level_size):\n",
-        "            # Dequeue the front node from the queue.\n",
-        "            node = queue.pop(0)\n",
-        "            \n",
-        "            # Append the value of the current node to the level_values list.\n",
-        "            level_values.append(node.val)\n",
-        "\n",
-        "            # Enqueue the left and right children of the current node if they exist.\n",
-        "            if node.left:\n",
-        "                queue.append(node.left)\n",
-        "            if node.right:\n",
-        "                queue.append(node.right)\n",
-        "\n",
-        "        # Append the level_values list (values at the current level) to the result list.\n",
-        "        result.append(level_values)\n",
-        "\n",
-        "    # Return the final result, which is a list of lists representing level order traversal.\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a `TreeNode` class that represents a node in the binary tree. Each `TreeNode` object has a value (`val`) and two child nodes: `left` and `right`.\n",
-        "\n",
-        "2. The `levelOrder` function takes the root of the binary tree as its input and returns the level order traversal of the tree as a list of lists.\n",
-        "\n",
-        "3. We start by checking if the input `root` is `None`, which indicates an empty tree. If the tree is empty, we return an empty list because there are no nodes to traverse.\n",
-        "\n",
-        "4. We initialize an empty list called `result` to store the final result, which will be a list of lists containing node values at each level.\n",
-        "\n",
-        "5. We initialize a queue called `queue` with the root node. This queue will be used for breadth-first traversal of the tree.\n",
-        "\n",
-        "6. We enter a while loop that continues until the `queue` is empty. Inside the loop, we perform the following steps:\n",
-        "\n",
-        "   - We initialize an empty list called `level_values` to store the values of nodes at the current level.\n",
-        "\n",
-        "   - We determine the number of nodes at the current level by getting the length of the `queue`. This is done to process nodes level by level.\n",
-        "\n",
-        "   - We iterate through the nodes at the current level using a for loop. For each node in the current level:\n",
-        "\n",
-        "     - We dequeue (remove) the front node from the `queue`.\n",
-        "\n",
-        "     - We append the value of the dequeued node to the `level_values` list, effectively collecting the values of nodes at the current level.\n",
-        "\n",
-        "     - If the dequeued node has a left child, we enqueue the left child to the `queue`.\n",
-        "\n",
-        "     - If the dequeued node has a right child, we enqueue the right child to the `queue`.\n",
-        "\n",
-        "   - After processing all nodes at the current level, we append the `level_values` list to the `result` list. This represents the values at the current level.\n",
-        "\n",
-        "7. The loop continues until all levels have been traversed, and the `queue` becomes empty.\n",
-        "\n",
-        "8. Finally, we return the `result` list, which contains lists of node values at each level, representing the level order traversal of the binary tree.\n",
-        "\n",
-        "The code effectively performs a breadth-first traversal of the binary tree, processing nodes level by level, and constructs the result list that represents the level order traversal."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[3], [9, 20], [15, 7]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(9)\n",
-        "root.right = TreeNode(20)\n",
-        "root.right.left = TreeNode(15)\n",
-        "root.right.right = TreeNode(7)\n",
-        "\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Using the same tree as before\n",
-        "root = TreeNode(1)\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root = None\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `levelOrder` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this function is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the breadth-first traversal.\n",
-        "\n",
-        "In the worst case, we have to enqueue and dequeue all nodes in the binary tree, which is proportional to the number of nodes (N). In each level, we process all nodes in that level, and since there are a total of log(N) levels in a balanced binary tree, the time complexity can also be approximated as O(N) for unbalanced trees.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this function is O(N), where N is the number of nodes in the binary tree. Here's how the space complexity breaks down:\n",
-        "\n",
-        "1. The `result` list stores the level order traversal, and in the worst case, it contains N/2 levels (for a completely unbalanced binary tree). So, the space used by `result` is O(N).\n",
-        "\n",
-        "2. The `queue` data structure is used for BFS traversal. In the worst case, it can store all nodes at the last level of the tree. In a balanced binary tree, the maximum number of nodes at any level is 2^(log(N)), which is still O(N). In the case of an unbalanced tree, it can be even worse. So, the space used by `queue` is O(N).\n",
-        "\n",
-        "Overall, the dominant factor in terms of space complexity is the `queue`, and the space complexity is O(N).\n",
-        "\n",
-        "In summary, the function's time complexity is O(N), and its space complexity is also O(N). It is an efficient and optimal solution for performing a level order traversal of a binary tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Reverse Level Order Traversal:** Modify the `levelOrder` function to return the level order traversal in reverse order (from bottom to top). For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[15, 7], [9, 20], [3]]`.\n",
-        "\n",
-        "2. **Zigzag Level Order Traversal:** Write a function that performs a level order traversal of a binary tree in a zigzag pattern. In a zigzag traversal, the nodes at even levels are traversed from left to right, and nodes at odd levels are traversed from right to left. For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[3], [20, 9], [15, 7]]`."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/104. Maximum Depth of Binary Tree.ipynb b/docs/07. Trees/104. Maximum Depth of Binary Tree.ipynb
deleted file mode 100644
index 213989d..0000000
--- a/docs/07. Trees/104. Maximum Depth of Binary Tree.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 104. Maximum Depth of Binary Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the `root` of a binary tree, return its maximum depth.\n",
-        "\n",
-        "A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 19,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def maxDepth(root):\n",
-        "    # Base case: If the root is None, the depth is 0.\n",
-        "    if root is None:\n",
-        "        return 0\n",
-        "    \n",
-        "    # Recursively calculate the maximum depth of the left and right subtrees.\n",
-        "    left_depth = maxDepth(root.left)\n",
-        "    right_depth = maxDepth(root.right)\n",
-        "    \n",
-        "    # Return the maximum depth of the tree by adding 1 to the maximum depth of the subtrees.\n",
-        "    return max(left_depth, right_depth) + 1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's go through the code step by step to understand how it works:\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class to represent the nodes of the binary tree. Each node has three attributes:\n",
-        "   - `val`: the value stored in the node.\n",
-        "   - `left`: a reference to the left child node.\n",
-        "   - `right`: a reference to the right child node.\n",
-        "\n",
-        "2. The `maxDepth` function is the main function that calculates the maximum depth of the binary tree. It takes a single argument, `root`, which is the root node of the binary tree.\n",
-        "\n",
-        "3. In the `maxDepth` function, we have a base case:\n",
-        "   - If the `root` is `None`, it means we have reached the end of a branch of the tree (a leaf node or an empty subtree). In this case, the depth is 0 because there are no nodes to count.\n",
-        "\n",
-        "4. If the `root` is not `None`, we recursively calculate the maximum depth of the left and right subtrees:\n",
-        "   - We call `maxDepth` on the `root.left` to calculate the maximum depth of the left subtree and store it in the variable `left_depth`.\n",
-        "   - We call `maxDepth` on the `root.right` to calculate the maximum depth of the right subtree and store it in the variable `right_depth`.\n",
-        "\n",
-        "5. Finally, we return the maximum depth of the tree by taking the maximum of `left_depth` and `right_depth` and adding 1 to it. This is because we are counting the current level as well."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: root = [3,9,20,null,null,15,7]\n",
-        "\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(9)\n",
-        "root.right = TreeNode(20)\n",
-        "root.right.left = TreeNode(15)\n",
-        "root.right.right = TreeNode(7)\n",
-        "\n",
-        "print(maxDepth(root))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a230be4f-1f47-4919-87e3-62fc3f072d97",
-      "metadata": {},
-      "source": [
-        "In this example, we create a binary tree based on the given input `[3,9,20,null,null,15,7]` and calculate its maximum depth, which is `3`. The tree structure is as follows:\n",
-        "   ```\n",
-        "       3\n",
-        "      / \\\n",
-        "     9  20\n",
-        "       /  \\\n",
-        "      15   7\n",
-        "   ```\n",
-        "\n",
-        "So, the maximum depth is the length of the longest path from the root to a leaf node, which is 3 in this case."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: root = [1,null,2]\n",
-        "\n",
-        "root = TreeNode(1)\n",
-        "root.right = TreeNode(2)\n",
-        "\n",
-        "# Calculate the maximum depth of the tree and print the result\n",
-        "print(maxDepth(root))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The time complexity of the `maxDepth` function is $O(N)$, where $N$ is the number of nodes in the binary tree.\n",
-        "- This is because in the worst case, the function visits every node exactly once in a depth-first manner.\n",
-        "- The recursion explores all nodes of the tree, so the time complexity is linear with respect to the number of nodes.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the `maxDepth` function is $O(H)$, where H is the height of the binary tree.\n",
-        "- In the worst case, if the binary tree is completely unbalanced (skewed), the recursion stack can go as deep as the height of the tree.\n",
-        "- In the best case, if the binary tree is perfectly balanced, the height is $O(log\\ N)$, where $N$ is the number of nodes.\n",
-        "- Therefore, the space complexity can vary from $O(log\\ N)$ to $O(N)$ depending on the shape of the tree.\n",
-        "- In addition to the recursion stack, there is a small constant amount of space used for variables and function call overhead.\n",
-        "\n",
-        "**In summary:**\n",
-        "- The time complexity of the code is $O(N)$ as it visits each node once.\n",
-        "- The space complexity is $O(H)$, where H is the height of the tree, which can vary from $O(log\\ N)$ to $O(N)$ depending on the tree's shape."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb b/docs/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb
deleted file mode 100644
index 31c9e69..0000000
--- a/docs/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb	
+++ /dev/null
@@ -1,313 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 105. Construct Binary Tree from Preorder and Inorder Traversal\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Construct Binary Tree from Preorder and Inorder Traversal problem on LeetCode, click here!](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)\n",
-        "\n",
-        "---\n",
-        "Given two integer arrays `preorder` and `inorder` where `preorder` is the preorder traversal of a binary tree and `inorder` is the inorder traversal of the same tree, construct and return *the binary tree.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= preorder.length <= 3000`\n",
-        "- `inorder.length == preorder.length`\n",
-        "- `-3000 <= preorder[i], inorder[i] <= 3000`\n",
-        "- `preorder` and `inorder` consist of **unique** values.\n",
-        "- Each value of `inorder` also appears in `preorder`.\n",
-        "- `preorder` is **guaranteed** to be the preorder traversal of the tree.\n",
-        "- `inorder` is **guaranteed** to be the inorder traversal of the tree.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def buildTree(preorder, inorder):\n",
-        "    if not preorder:\n",
-        "        return None\n",
-        "\n",
-        "    # The first element in the preorder traversal is the root of the current subtree\n",
-        "    root_val = preorder[0]\n",
-        "    root = TreeNode(root_val)\n",
-        "\n",
-        "    # Find the index of the root value in the inorder traversal\n",
-        "    root_idx_inorder = inorder.index(root_val)\n",
-        "\n",
-        "    # Recursively build left and right subtrees\n",
-        "    root.left = buildTree(preorder[1:1 + root_idx_inorder], inorder[:root_idx_inorder])\n",
-        "    root.right = buildTree(preorder[1 + root_idx_inorder:], inorder[root_idx_inorder + 1:])\n",
-        "\n",
-        "    return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `class TreeNode:`: This is a class definition for a binary tree node. It has three attributes:\n",
-        "   - `val`: The value of the node.\n",
-        "   - `left`: A reference to the left child node.\n",
-        "   - `right`: A reference to the right child node.\n",
-        "\n",
-        "   This class is used to create instances of binary tree nodes, which will be used to build the binary tree.\n",
-        "\n",
-        "2. `def buildTree(preorder, inorder):`: This is the main function that constructs a binary tree from its preorder and inorder traversals.\n",
-        "\n",
-        "   - `preorder`: A list representing the preorder traversal of the binary tree.\n",
-        "   - `inorder`: A list representing the inorder traversal of the binary tree.\n",
-        "\n",
-        "3. `if not preorder: return None`: This line checks if the `preorder` list is empty. If it is, it means there are no nodes to construct, so the function returns `None`.\n",
-        "\n",
-        "4. `root_val = preorder[0]`: The value of the root node is extracted from the first element of the `preorder` list.\n",
-        "\n",
-        "5. `root = TreeNode(root_val)`: A new `TreeNode` object is created with the `root_val` as its value. This represents the root of the current subtree.\n",
-        "\n",
-        "6. `root_idx_inorder = inorder.index(root_val)`: The index of the `root_val` in the `inorder` list is found. This index indicates the position of the root node in the inorder traversal.\n",
-        "\n",
-        "7. `root.left`: The left subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The left subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists up to the `root_idx_inorder`.\n",
-        "\n",
-        "8. `root.right`: The right subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The right subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists starting from `root_idx_inorder + 1`.\n",
-        "\n",
-        "9. Finally, the function returns the `root` of the subtree it just constructed.\n",
-        "\n",
-        "The recursive approach used here divides the problem of constructing the entire binary tree into smaller subproblems, starting with the root node and then recursively building the left and right subtrees until the entire tree is constructed."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "2d82857b-4fbe-474f-bf33-17b55c20147b",
-      "metadata": {},
-      "source": [
-        "## Helper Function \n",
-        "\n",
-        "Here is a helper function to represent the binary tree elements."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "13e6f3e1-659d-479d-9521-75629b8c782f",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def treeToList(root):\n",
-        "    # Initialize an empty list to store the elements of the binary tree\n",
-        "    result = []\n",
-        "    \n",
-        "    # Initialize a queue for level-order traversal starting with the root\n",
-        "    queue = [root]\n",
-        "\n",
-        "    while queue:\n",
-        "        # Pop the front node from the queue\n",
-        "        node = queue.pop(0)\n",
-        "        \n",
-        "        # If the node is not None (i.e., a valid node), add its value to the result list\n",
-        "        if node:\n",
-        "            result.append(node.val)\n",
-        "            \n",
-        "            # Add the left and right children of the node to the queue\n",
-        "            queue.append(node.left)\n",
-        "            queue.append(node.right)\n",
-        "        else:\n",
-        "            # If the node is None, add None to the result to represent an empty node\n",
-        "            result.append(None)\n",
-        "\n",
-        "    # Remove any trailing None values from the result list\n",
-        "    while result and result[-1] is None:\n",
-        "        result.pop()\n",
-        "\n",
-        "    # Return the resulting list representing the binary tree elements\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "86cd4c9d-a120-4dd1-a1bd-530436012574",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's break down the `treeToList` helper function step by step:\n",
-        "\n",
-        "1. `def treeToList(root):`: This function takes the root of a binary tree as input and converts it into a list.\n",
-        "\n",
-        "2. `result = []`: Initialize an empty list called `result` to store the elements of the binary tree in list format.\n",
-        "\n",
-        "3. `queue = [root]`: Initialize a queue data structure for a level-order traversal, starting with the root node.\n",
-        "\n",
-        "4. `while queue:`: This starts a loop that continues until the queue is empty, indicating that all nodes have been processed.\n",
-        "\n",
-        "5. `node = queue.pop(0)`: Dequeue the first node from the queue, effectively processing it.\n",
-        "\n",
-        "6. `if node:`: Check if the `node` is not None, which means it's a valid node in the binary tree.\n",
-        "\n",
-        "    a. `result.append(node.val)`: If the node is valid, append its value (`node.val`) to the `result` list. This represents the value of the current node in the binary tree.\n",
-        "\n",
-        "    b. `queue.append(node.left)`: Enqueue the left child of the current node if it exists. This adds the left child to the queue for processing in the next iteration.\n",
-        "\n",
-        "    c. `queue.append(node.right)`: Enqueue the right child of the current node if it exists. This adds the right child to the queue for processing in the next iteration.\n",
-        "\n",
-        "7. `else:`: If the `node` is None, it represents an empty node in the binary tree.\n",
-        "\n",
-        "    a. `result.append(None)`: Append `None` to the `result` list to indicate an empty node.\n",
-        "\n",
-        "8. `while result and result[-1] is None:`: After the traversal is complete, there might be trailing `None` values in the `result` list. This loop removes any such trailing `None` values to ensure a clean representation of the tree's elements.\n",
-        "\n",
-        "9. `return result`: Return the `result` list, which now contains the elements of the binary tree in a format where `None` represents empty nodes and the order of elements reflects a level-order traversal of the tree.\n",
-        "\n",
-        "In summary, the `treeToList` function performs a level-order traversal of the binary tree using a queue, constructing a list where each element corresponds to a node's value or represents an empty node with `None`. This list represents the binary tree's elements in a structured format."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[3, 9, 20, None, None, 15, 7]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "preorder = [3, 9, 20, 15, 7]\n",
-        "inorder = [9, 3, 15, 20, 7]\n",
-        "result = buildTree(preorder, inorder)\n",
-        "\n",
-        "print(treeToList(result))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[-1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "preorder = [-1]\n",
-        "inorder = [-1]\n",
-        "result = buildTree(preorder, inorder)\n",
-        "\n",
-        "print(treeToList(result))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `buildTree` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of the `buildTree` function can be analyzed in terms of the number of nodes in the binary tree. Let's assume there are `n` nodes in the binary tree.\n",
-        "\n",
-        "1. Finding the root value in the `preorder` traversal takes O(1) time as it's just extracting the first element from the list.\n",
-        "2. Finding the index of the root value in the `inorder` traversal using `inorder.index(root_val)` takes O(n) time in the worst case because in the worst case, it might have to search through all `n` elements in the `inorder` list to find the index.\n",
-        "3. The recursive calls to `buildTree` for the left and right subtrees are made once for each node in the tree. Therefore, the recursive calls have a combined time complexity of O(n) as they process each node once.\n",
-        "\n",
-        "The total time complexity of the `buildTree` function is O(n) due to the recursive calls and finding the root's index in the `inorder` list.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity is determined by the space used by the function's call stack during recursion and any additional data structures used. \n",
-        "\n",
-        "1. The space used by the function call stack during recursion depends on the height of the binary tree. In the worst case, where the tree is highly unbalanced (e.g., a skewed tree), the space complexity for the call stack is O(n) as it can go as deep as the number of nodes in the tree.\n",
-        "\n",
-        "2. Additionally, the function creates TreeNode objects for each node in the binary tree. Therefore, the space complexity for these objects is also O(n).\n",
-        "\n",
-        "Overall, the space complexity of the `buildTree` function is O(n) due to the space used by the call stack and the TreeNode objects.\n",
-        "\n",
-        "In summary, the time complexity of `buildTree` is O(n), and the space complexity is O(n), where 'n' is the number of nodes in the binary tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Construct Binary Tree from Postorder and Inorder Traversal:**\n",
-        "   Modify the problem to construct a binary tree from its postorder and inorder traversals. Implement a function similar to `buildTree` but for postorder and inorder traversals.\n",
-        "\n",
-        "2. **Reconstruct Binary Tree with Duplicate Values:**\n",
-        "   Extend the problem to handle binary trees with duplicate values. Ensure that your solution correctly handles scenarios where there are duplicate values in the preorder and inorder traversals."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/124. Binary Tree Maximum Path Sum.ipynb b/docs/07. Trees/124. Binary Tree Maximum Path Sum.ipynb
deleted file mode 100644
index 59d8e7d..0000000
--- a/docs/07. Trees/124. Binary Tree Maximum Path Sum.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 124. Binary Tree Maximum Path Sum\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Maximum Path Sum problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-maximum-path-sum/)\n",
-        "\n",
-        "---\n",
-        "A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.\n",
-        "\n",
-        "The **path sum** of a path is the sum of the node's values in the path.\n",
-        "\n",
-        "Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[1, 3*10^4]$.\n",
-        "- `-1000 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to find the maximum path sum in a binary tree. A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. The path sum of a path is the sum of the node values in that path.\n",
-        "\n",
-        "In this problem, you are given the root of a binary tree, and you need to find the maximum path sum among all possible paths in the tree. Note that the path does not need to start at the root node or end at a leaf node; it can start and end at any nodes in the tree.\n",
-        "\n",
-        "To find the maximum path sum, you need to consider both left and right subtrees of each node while traversing the tree. At each node, you have two choices:\n",
-        "\n",
-        "1. Include the current node in the path: In this case, you add the value of the current node to the sum and continue exploring both the left and right subtrees for possible extensions of the path.\n",
-        "\n",
-        "2. Start a new path from the current node: In this case, you do not include the current node in the path sum, and you choose either the left subtree or the right subtree to start a new path.\n",
-        "\n",
-        "To solve this problem, you can use a recursive approach to traverse the binary tree. For each node, you calculate the maximum path sum that can pass through that node (option 1) and also return the maximum path sum that can be extended from that node (option 2). You keep track of the global maximum path sum encountered during the traversal.\n",
-        "\n",
-        "Ultimately, the maximum path sum among all paths in the tree will be the maximum of the global maximum path sum and the maximum path sum calculated for each node.\n",
-        "\n",
-        "The constraints for this problem include:\n",
-        "\n",
-        "- The number of nodes in the tree is in the range $[1, 3 * 10^4]$.\n",
-        "- Node values are in the range [-1000, 1000].\n",
-        "\n",
-        "By considering all possible paths through the tree, the algorithm aims to find the most significant sum of node values in any path in the binary tree.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "class Solution:\n",
-        "    def maxPathSum(self, root):\n",
-        "        def max_path_sum_helper(node):\n",
-        "            if not node:\n",
-        "                return 0\n",
-        "            \n",
-        "            # Recursively calculate the maximum path sum for left and right subtrees\n",
-        "            left_max = max(0, max_path_sum_helper(node.left))\n",
-        "            right_max = max(0, max_path_sum_helper(node.right))\n",
-        "            \n",
-        "            # Update the global maximum path sum\n",
-        "            self.max_sum = max(self.max_sum, left_max + right_max + node.val)\n",
-        "            \n",
-        "            # Return the maximum path sum starting from the current node\n",
-        "            return max(left_max, right_max) + node.val\n",
-        "        \n",
-        "        self.max_sum = float('-inf')  # Initialize the global maximum to negative infinity\n",
-        "        max_path_sum_helper(root)  # Start the recursive traversal from the root\n",
-        "        return self.max_sum"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The code starts by defining a `TreeNode` class to represent nodes in the binary tree. Each `TreeNode` has a `val` attribute representing its value, a `left` attribute pointing to the left child, and a `right` attribute pointing to the right child.\n",
-        "\n",
-        "2. Next, a `Solution` class is defined to contain the `maxPathSum` method, which will be used to find the maximum path sum in the binary tree.\n",
-        "\n",
-        "3. Within the `maxPathSum` method, a helper function called `max_path_sum_helper` is defined. This recursive function takes a single argument, `node`, which is the current node being considered during traversal.\n",
-        "\n",
-        "4. Inside the `max_path_sum_helper` function:\n",
-        "   - If `node` is `None` (i.e., the current node is a null node), it returns 0. This is the base case of the recursion.\n",
-        "   - It calculates the maximum path sum starting from the left subtree (`left_max`) and the right subtree (`right_max`). Importantly, it uses `max(0, ...)` to ensure that negative values (which would make the path sum smaller) are not considered. If a subtree's path sum is negative, it's better to not include it in the path.\n",
-        "   - The code then updates the global maximum path sum (`self.max_sum`) by checking if the sum of `left_max`, `right_max`, and the current node's value is greater than the current maximum.\n",
-        "   - Finally, it returns the maximum path sum starting from the current node, which is the maximum of `left_max` and `right_max` plus the current node's value.\n",
-        "\n",
-        "5. Before calling the `max_path_sum_helper` function, the `max_sum` attribute of the `Solution` class is initialized to negative infinity (`float('-inf')`). This is done to ensure that any valid path sum encountered during traversal will be greater than the initial value of `max_sum`.\n",
-        "\n",
-        "6. The `max_path_sum_helper` function is called with the root of the binary tree, effectively starting the traversal from the root node.\n",
-        "\n",
-        "7. Once the traversal is complete, the method returns the value of `self.max_sum`, which contains the maximum path sum found in the binary tree.\n",
-        "\n",
-        "8. Example usage at the bottom of the code demonstrates how to create binary trees and use the `maxPathSum` method to find the maximum path sum for two different tree structures.\n",
-        "\n",
-        "In summary, the code uses a recursive approach to explore all possible paths in the binary tree, ensuring that negative path sums are not considered. It keeps track of the maximum path sum encountered and returns that maximum value as the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "6\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "root1 = TreeNode(1)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(3)\n",
-        "solution = Solution()\n",
-        "print(solution.maxPathSum(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "42\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "root2 = TreeNode(-10)\n",
-        "root2.left = TreeNode(9)\n",
-        "root2.right = TreeNode(20)\n",
-        "root2.right.left = TreeNode(15)\n",
-        "root2.right.right = TreeNode(7)\n",
-        "print(solution.maxPathSum(root2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "Time Complexity:\n",
-        "- The code uses a recursive depth-first traversal to explore each node in the binary tree exactly once.\n",
-        "- During the traversal, for each node, we perform constant time operations, such as updating the `max_sum` variable and calculating the maximum path sum starting from that node.\n",
-        "- Therefore, the time complexity of the code is O(N), where N is the number of nodes in the binary tree.\n",
-        "\n",
-        "Space Complexity:\n",
-        "- The space complexity of the code is determined by the space used in the call stack during the recursive traversal.\n",
-        "- In the worst case, the depth of the call stack can be equal to the height of the binary tree. In an unbalanced tree, this could be O(N), but in a balanced binary tree, the height is O(log N).\n",
-        "- Additionally, the code uses a constant amount of space for variables like `left_max`, `right_max`, and `max_sum`.\n",
-        "- Therefore, the overall space complexity is O(H), where H is the height of the binary tree. In the worst case, it's O(N), and in the best case (a balanced binary tree), it's O(log N).\n",
-        "\n",
-        "In summary:\n",
-        "- Time Complexity: O(N)\n",
-        "- Space Complexity: O(H), where H is the height of the binary tree, ranging from O(log N) to O(N) depending on the tree's balance."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Print the Maximum Path:**\n",
-        "   Modify the code to not only find the maximum path sum but also print the nodes involved in the maximum path. You can print the nodes in the correct order from the root to the leaf.\n",
-        "\n",
-        "2. **Path Sum Count:**\n",
-        "   Instead of finding the maximum path sum, find the count of unique paths that sum to a given target value. Each path can start and end anywhere in the tree."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/199. Binary Tree Right Side View.ipynb b/docs/07. Trees/199. Binary Tree Right Side View.ipynb
deleted file mode 100644
index 55dc944..0000000
--- a/docs/07. Trees/199. Binary Tree Right Side View.ipynb	
+++ /dev/null
@@ -1,257 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 199. Binary Tree Right Side View\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Right Side View problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-right-side-view/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, imagine yourself standing on the **right side** of it, return *the values of the nodes you can see ordered from top to bottom*.\n",
-        "\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def rightSideView(root):\n",
-        "    if not root:\n",
-        "        return []  # If the tree is empty, return an empty list.\n",
-        "\n",
-        "    result = []  # Initialize a list to store the right side view values.\n",
-        "    queue = [root]  # Initialize a queue for level-order traversal with the root node.\n",
-        "\n",
-        "    while queue:\n",
-        "        # Get the number of nodes at the current level.\n",
-        "        level_size = len(queue)\n",
-        "\n",
-        "        # Traverse all nodes at the current level and add the rightmost node to the result.\n",
-        "        for i in range(level_size):\n",
-        "            node = queue.pop(0)  # Dequeue the first node from the queue.\n",
-        "\n",
-        "            # If it's the rightmost node at this level, add its value to the result.\n",
-        "            if i == level_size - 1:\n",
-        "                result.append(node.val)\n",
-        "\n",
-        "            # Add the children of the current node to the queue.\n",
-        "            if node.left:\n",
-        "                queue.append(node.left)\n",
-        "            if node.right:\n",
-        "                queue.append(node.right)\n",
-        "\n",
-        "    return result  # Return the list of right side view values.\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start with a binary tree represented using a TreeNode class. Each node has a value (`val`) and can have a left child (`left`) and a right child (`right`).\n",
-        "\n",
-        "2. The `rightSideView` function takes the root node of the binary tree as input and returns a list of values representing the nodes you can see when standing on the right side of the tree.\n",
-        "\n",
-        "3. We check if the `root` is `None`, which means the tree is empty. If it's empty, we return an empty list because there are no nodes to see.\n",
-        "\n",
-        "4. We initialize an empty list called `result` to store the right side view values.\n",
-        "\n",
-        "5. We initialize a queue called `queue` with the root node. This queue is used for a level-order traversal of the tree.\n",
-        "\n",
-        "6. We enter a while loop that continues until the `queue` is empty. Inside this loop, we perform the following steps for each level of the tree:\n",
-        "\n",
-        "   a. We determine the number of nodes at the current level by getting the length of the `queue`. This is important for processing nodes at the same level together.\n",
-        "\n",
-        "   b. We then iterate through the nodes at the current level using a for loop. For each node in the level, we dequeue it from the `queue` using `queue.pop(0)`.\n",
-        "\n",
-        "   c. If the node we dequeue is the rightmost node at the current level (determined by `i == level_size - 1`), we add its `val` to the `result` list. This is because, when standing on the right side of the tree, you can see the rightmost node at each level.\n",
-        "\n",
-        "   d. We enqueue the left and right children of the current node (if they exist) to the `queue`. This ensures that we process the next level in the subsequent iterations of the while loop.\n",
-        "\n",
-        "7. After processing all levels of the tree, the `result` list contains the values of the rightmost nodes in each level, ordered from top to bottom.\n",
-        "\n",
-        "8. Finally, we return the `result` list as the output of the function, which represents the right side view of the binary tree.\n",
-        "\n",
-        "The code effectively performs a level-order traversal of the binary tree while keeping track of the rightmost nodes at each level and adding them to the result list. This ensures that we obtain the correct order of nodes visible from the right side of the tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 3, 4]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root1 = TreeNode(1)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(3)\n",
-        "root1.left.right = TreeNode(5)\n",
-        "root1.right.right = TreeNode(4)\n",
-        "print(rightSideView(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "root2 = TreeNode(1)\n",
-        "root2.right = TreeNode(3)\n",
-        "print(rightSideView(root2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root = None\n",
-        "result = rightSideView(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of the code is O(N), where N is the number of nodes in the binary tree. Here's why:\n",
-        "\n",
-        "1. We perform a level-order traversal of the tree using a queue. In the worst case, we visit all nodes once, which takes O(N) time since we visit each node exactly once.\n",
-        "\n",
-        "2. For each node, we perform constant-time operations like dequeuing it from the queue, checking if it's the rightmost node at the current level, and enqueuing its children (if they exist). These operations do not depend on the size of the tree, so they do not contribute to the overall time complexity.\n",
-        "\n",
-        "Therefore, the dominant factor in the time complexity is the level-order traversal, making it O(N).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is also O(N). Here's why:\n",
-        "\n",
-        "1. We use a queue to perform level-order traversal. In the worst case, the queue can contain all nodes at the maximum width of the tree, which can be up to N/2 nodes for a completely unbalanced tree. Therefore, the space required for the queue is O(N).\n",
-        "\n",
-        "2. The `result` list stores the values of the rightmost nodes. In the worst case, when the binary tree is a complete binary tree, it can have roughly N/2 rightmost nodes. Therefore, the space required for the `result` list is also O(N).\n",
-        "\n",
-        "3. Other auxiliary variables like `level_size`, `i`, and `node` require only constant space and do not depend on the size of the tree.\n",
-        "\n",
-        "Combining the space used by the queue and the `result` list, we get a space complexity of O(N).\n",
-        "\n",
-        "In summary, the code has a time complexity of O(N) and a space complexity of O(N), where N is the number of nodes in the binary tree. These complexities are efficient and scale linearly with the size of the input tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Variant with Left Side View**: Modify the code to find and return the values of nodes visible from the left side of the binary tree instead of the right side.\n",
-        "\n",
-        "2. **Zigzag Right Side View**: Extend the code to return the right side view values in a zigzag order. In a zigzag order, you alternate between starting from the rightmost node at level 0, then the leftmost at level 1, rightmost at level 2, and so on."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/226. Invert Binary Tree.ipynb b/docs/07. Trees/226. Invert Binary Tree.ipynb
deleted file mode 100644
index 8ea4609..0000000
--- a/docs/07. Trees/226. Invert Binary Tree.ipynb	
+++ /dev/null
@@ -1,223 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 226: Invert Binary Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the root of a binary tree, invert the tree, and return its root.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the tree is in the range `[0, 100]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def invertTree(root):\n",
-        "    # Base case: If the root is None or the tree is empty, return None.\n",
-        "    if not root:\n",
-        "        return None\n",
-        "\n",
-        "    # Swap the left and right subtrees of the current node.\n",
-        "    root.left, root.right = root.right, root.left\n",
-        "\n",
-        "    # Recursively invert the left and right subtrees.\n",
-        "    invertTree(root.left)\n",
-        "    invertTree(root.right)\n",
-        "\n",
-        "    return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a TreeNode class, which represents a node in a binary tree. Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-        "\n",
-        "2. The `invertTree` function is defined to invert the binary tree. It takes the root node of the tree as an argument.\n",
-        "\n",
-        "3. In the `invertTree` function, we have a base case to handle the scenario when the root is `None` (empty tree). In such cases, we return `None` because there's nothing to invert.\n",
-        "\n",
-        "4. For non-empty trees, we swap the left and right subtrees of the current node. This effectively inverts the tree at the current node.\n",
-        "\n",
-        "5. We then recursively call `invertTree` on the left and right subtrees to invert them.\n",
-        "\n",
-        "6. Finally, we return the root of the inverted tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inverted Tree (Example 1): <__main__.TreeNode object at 0x270c9c0>\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: root = [4,2,7,1,3,6,9]\n",
-        "\n",
-        "root1 = TreeNode(4)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(7)\n",
-        "root1.left.left = TreeNode(1)\n",
-        "root1.left.right = TreeNode(3)\n",
-        "root1.right.left = TreeNode(6)\n",
-        "root1.right.right = TreeNode(9)\n",
-        "\n",
-        "inverted_root1 = invertTree(root1)\n",
-        "print(\"Inverted Tree (Example 1):\", inverted_root1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inorder Traversal of Inverted Tree (Example 2): <__main__.TreeNode object at 0x2255fc0>\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: root = [2,1,3]\n",
-        "\n",
-        "root2 = TreeNode(2)\n",
-        "root2.left = TreeNode(1)\n",
-        "root2.right = TreeNode(3)\n",
-        "\n",
-        "inverted_root2 = invertTree(root2)\n",
-        "print(\"Inverted Tree (Example 2):\", inverted_root2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inverted Tree (Example 3): None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "# Input: root = []\n",
-        "\n",
-        "root3 = None  # Empty tree\n",
-        "\n",
-        "inverted_root3 = invertTree(root3)\n",
-        "print(\"Inverted Tree (Example 3):\", inverted_root3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's discuss the time and space complexity of the provided Python code to invert a binary tree.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of the `invertTree` function is $O(n)$, where $n$ is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal of the tree. In the worst case, we have to visit all nodes in the tree.\n",
-        "\n",
-        "The reason for this time complexity is the depth-first traversal of the tree, where we recursively process the left and right subtrees of each node.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is determined by the function call stack during the recursion. In the worst case, the space complexity is $O(h)$, where $h$ is the height of the binary tree. \n",
-        "\n",
-        "In a completely unbalanced binary tree (essentially a linked list), the height $h$ is equal to the number of nodes $n$, resulting in a space complexity of $O(n)$. This occurs when the tree is skewed to one side.\n",
-        "\n",
-        "In a balanced binary tree, such as a full binary tree, the height $h$ is $O(log\\ n)$, and the space complexity is $O(log\\ n)$.\n",
-        "\n",
-        "The space complexity depends on how balanced the tree is. In practical scenarios, binary trees are often approximately balanced, so the space complexity is typically closer to $O(log\\ n)$.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "- Time Complexity: $O(n)$ where $n$ is the number of nodes.\n",
-        "- Space Complexity: $O(h)$, where $h$ is the height of the binary tree. In the worst case, it can be $O(n)$, and in a balanced tree, it is $O(log\\ n)$.\n",
-        "\n",
-        "Keep in mind that these complexities are based on the recursive implementation provided. Iterative solutions can achieve the same task with $O(1)$ space complexity, using auxiliary data structures like stacks or queues to mimic the recursion stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/230. Kth Smallest Element in a BST.ipynb b/docs/07. Trees/230. Kth Smallest Element in a BST.ipynb
deleted file mode 100644
index 673ced2..0000000
--- a/docs/07. Trees/230. Kth Smallest Element in a BST.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 230. Kth Smallest Element in a BST\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Kth Smallest Element in a BST problem on LeetCode, click here!](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary search tree, and an integer `k`, return *the $k^{th}$ smallest value (**1-indexed**) of all the values of the nodes in the tree.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is `n`.\n",
-        "- 1 <= `k <= n` <= $10^4$\n",
-        "- 0 <= `Node.val` <= $10^4$\n",
-        "\n",
-        "**Follow up**: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def kthSmallest(root, k):\n",
-        "    stack = []  # Initialize an empty stack to simulate the traversal\n",
-        "    while root or stack:\n",
-        "        while root:\n",
-        "            stack.append(root)  # Push the current node onto the stack\n",
-        "            root = root.left  # Move to the left child\n",
-        "        root = stack.pop()  # Pop a node from the stack\n",
-        "        k -= 1  # Decrement k since we've visited a node\n",
-        "        if k == 0:\n",
-        "            return root.val  # If k becomes 0, return the current node's value as the kth smallest\n",
-        "        root = root.right  # Move to the right child to continue the traversal"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in the binary search tree (BST). Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-        "\n",
-        "2. The `kthSmallest` function takes two parameters: `root`, which is the root of the BST, and `k`, which represents the kth smallest element we want to find.\n",
-        "\n",
-        "3. We initialize an empty stack (`stack`) to simulate the traversal of the BST. The stack will be used to keep track of nodes as we traverse the tree in an iterative manner.\n",
-        "\n",
-        "4. We enter a while loop that continues until either `root` becomes `None` (indicating we have traversed the entire tree) or the `stack` is empty.\n",
-        "\n",
-        "5. Within the loop, we start another while loop to traverse as far left as possible in the BST. We repeatedly push nodes onto the `stack` and move to their left children until we reach the leftmost leaf node. This is the smallest node in the BST.\n",
-        "\n",
-        "6. Once we have reached the leftmost leaf node (the smallest node), we pop nodes from the `stack` one by one. As we pop each node, we decrement `k` by 1 to keep track of the number of nodes we have visited.\n",
-        "\n",
-        "7. If `k` becomes 0 after decrementing, it means we have found the kth smallest element. In this case, we return the `val` of the current node as the result.\n",
-        "\n",
-        "8. If `k` is still greater than 0, it means we haven't found the kth smallest element yet. In this case, we move to the right child of the current node to continue the traversal, as the kth smallest element, if it exists, will be in the right subtree of the current node.\n",
-        "\n",
-        "9. The process continues until we find the kth smallest element or traverse the entire tree.\n",
-        "\n",
-        "10. Finally, we return the kth smallest element found.\n",
-        "\n",
-        "The key idea in this code is to perform an in-order traversal of the BST while keeping track of the kth smallest element. By visiting nodes in ascending order, we can efficiently find the kth smallest element in O(h + k) time, where h is the height of the BST and k is the desired kth element. This approach is both concise and efficient for this problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Create the tree for the first example: root = [3,1,4,null,2], k = 1\n",
-        "root1 = TreeNode(3)\n",
-        "root1.left = TreeNode(1)\n",
-        "root1.right = TreeNode(4)\n",
-        "root1.left.right = TreeNode(2)\n",
-        "k1 = 1\n",
-        "print(kthSmallest(root1, k1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Create the tree for the second example: root = [5,3,6,2,4,null,null,1], k = 3\n",
-        "root2 = TreeNode(5)\n",
-        "root2.left = TreeNode(3)\n",
-        "root2.right = TreeNode(6)\n",
-        "root2.left.left = TreeNode(2)\n",
-        "root2.left.right = TreeNode(4)\n",
-        "root2.left.left.left = TreeNode(1)\n",
-        "k2 = 3\n",
-        "print(kthSmallest(root2, k2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this code is O(h + k), where:\n",
-        "- h is the height of the binary search tree (BST).\n",
-        "- k is the desired kth smallest element we want to find.\n",
-        "\n",
-        "1. In the worst case, where the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the time complexity is O(n + k).\n",
-        "\n",
-        "2. In the best case, where the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the time complexity is O(log(n) + k).\n",
-        "\n",
-        "So, the time complexity can vary from O(log(n) + k) in the best-case scenario to O(n + k) in the worst-case scenario. Typically, for balanced BSTs, the time complexity is closer to O(log(n) + k).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this code is O(h) due to the stack used for the iterative in-order traversal, where h is the height of the BST.\n",
-        "\n",
-        "1. In the worst case, when the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the space complexity is O(n) because the stack can potentially store all n nodes.\n",
-        "\n",
-        "2. In the best case, when the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the space complexity is O(log(n)) because the stack will have at most log(n) nodes.\n",
-        "\n",
-        "So, the space complexity depends on the height of the BST and can vary from O(log(n)) in the best-case scenario to O(n) in the worst-case scenario.\n",
-        "\n",
-        "In practical terms, for balanced BSTs or moderately unbalanced BSTs, the space complexity is usually close to O(log(n)), and the code is efficient in terms of space usage."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Kth Largest Element**: Modify the code to find the kth largest element in the BST instead of the kth smallest element.\n",
-        "\n",
-        "2. **Kth Smallest Element in Two BSTs**: Given two BSTs, find the kth smallest element when considering both BSTs as a single sorted list. This involves merging the two BSTs while finding the kth element efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb b/docs/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb
deleted file mode 100644
index 666bcd9..0000000
--- a/docs/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 235. Lowest Common Ancestor of a Binary Search Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Lowest Common Ancestor of a Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)\n",
-        "\n",
-        "---\n",
-        "Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.\n",
-        "\n",
-        "According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[2, 10^5]$.\n",
-        "- $-10^9$ <= `Node.val` <= $10^9$\n",
-        "- All `Node.val` are **unique**.\n",
-        "- `p != q`\n",
-        "- `p` and `q` will exist in the BST.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def lowestCommonAncestor(root, p, q):\n",
-        "    # Ensure p is less than q\n",
-        "    if p.val > q.val:\n",
-        "        p, q = q, p\n",
-        "\n",
-        "    while root:\n",
-        "        # If the current node value is greater than both p and q, move left\n",
-        "        if root.val > q.val:\n",
-        "            root = root.left\n",
-        "        # If the current node value is less than both p and q, move right\n",
-        "        elif root.val < p.val:\n",
-        "            root = root.right\n",
-        "        # If the current node value is between p and q (inclusive), or it matches either p or q, it's the LCA\n",
-        "        else:\n",
-        "            return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `class TreeNode:` defines a simple class for representing nodes in a binary tree. Each `TreeNode` has three attributes:\n",
-        "   - `val`: The value of the node.\n",
-        "   - `left`: A reference to the left child node.\n",
-        "   - `right`: A reference to the right child node.\n",
-        "\n",
-        "2. `def lowestCommonAncestor(root, p, q):` is a function that takes three arguments:\n",
-        "   - `root`: The root node of the BST.\n",
-        "   - `p`: A TreeNode representing one of the nodes for which we want to find the LCA.\n",
-        "   - `q`: A TreeNode representing the other node for which we want to find the LCA.\n",
-        "\n",
-        "3. `if p.val > q.val:` checks if the value of `p` is greater than the value of `q`. In a BST, it's essential to ensure that `p` represents the smaller value, and `q` represents the larger value. If `p` is greater than `q`, the code swaps their values.\n",
-        "\n",
-        "4. The main logic is inside the `while` loop, which runs until `root` becomes `None`. It performs the following steps to find the LCA:\n",
-        "   - If the current `root` node's value is greater than the value of `q`, it means both `p` and `q` are on the left subtree of the current node. So, we move to the left child of the current node by setting `root = root.left`.\n",
-        "   - If the current `root` node's value is less than the value of `p`, it means both `p` and `q` are on the right subtree of the current node. So, we move to the right child of the current node by setting `root = root.right`.\n",
-        "   - If neither of the above conditions is met, it means the current `root` node's value is between `p` and `q`, or it matches either `p` or `q`. In this case, the current node is the lowest common ancestor (LCA), so we return `root`.\n",
-        "\n",
-        "The algorithm is efficient because it takes advantage of the properties of a BST. It eliminates subtrees that cannot contain the LCA by comparing the values of `p`, `q`, and the current `root` node. Eventually, it reaches the LCA node, and that node is returned as the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "6\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root = TreeNode(6)\n",
-        "root.left = TreeNode(2)\n",
-        "root.right = TreeNode(8)\n",
-        "root.left.left = TreeNode(0)\n",
-        "root.left.right = TreeNode(4)\n",
-        "root.right.left = TreeNode(7)\n",
-        "root.right.right = TreeNode(9)\n",
-        "root.left.right.left = TreeNode(3)\n",
-        "root.left.right.right = TreeNode(5)\n",
-        "\n",
-        "p = root.left  # Node 2\n",
-        "q = root.right  # Node 8\n",
-        "\n",
-        "result = lowestCommonAncestor(root, p, q)\n",
-        "print(result.val)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Using the same tree as before\n",
-        "p = root.left  # Node 2\n",
-        "q = root.left.right  # Node 4\n",
-        "\n",
-        "result = lowestCommonAncestor(root, p, q)\n",
-        "print(result.val) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root2 = TreeNode(2)\n",
-        "root2.left = TreeNode(1)\n",
-        "\n",
-        "p = root2  # Node 2\n",
-        "q = root2.left  # Node 1\n",
-        "\n",
-        "result = lowestCommonAncestor(root2, p, q)\n",
-        "print(result.val)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided algorithm for finding the lowest common ancestor (LCA) in a Binary Search Tree (BST).\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this algorithm is O(h), where \"h\" is the height of the BST. In the worst case, where the tree is completely unbalanced (essentially a linked list), the height of the tree is O(n), where \"n\" is the number of nodes in the tree. However, in a well-balanced BST, the height is logarithmic, which is O(log n).\n",
-        "\n",
-        "The reason for this time complexity is that the algorithm efficiently narrows down the search space by traversing either left or right subtrees based on the values of the target nodes `p` and `q`. It eliminates entire subtrees that cannot contain the LCA, leading to a relatively quick search.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of the algorithm is O(1). This is because it uses a constant amount of extra space, regardless of the size of the input BST. The only variables used are `root`, `p`, and `q`, and there are no data structures like stacks or queues used for additional space. The algorithm performs a simple traversal without recursion, so it does not consume extra memory as the tree depth increases.\n",
-        "\n",
-        "**In summary**, the provided algorithm for finding the LCA in a BST is both time and space-efficient. Its time complexity is O(h), where \"h\" is the height of the tree, and its space complexity is O(1), making it suitable for practical use even in large BSTs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "### Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple LCAs:** Extend the algorithm to find all the lowest common ancestors of two given nodes `p` and `q` in a BST. In some cases, there can be multiple LCAs.\n",
-        "\n",
-        "2. **LCA with k Nodes:** Given a BST and k nodes, find the lowest common ancestor of these k nodes. This is an extension of the problem where you must find the LCA of more than two nodes."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb b/docs/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb
deleted file mode 100644
index 75c48ea..0000000
--- a/docs/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb	
+++ /dev/null
@@ -1,297 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 297. Serialize and Deserialize Binary Tree\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Serialize and Deserialize Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.\n",
-        "\n",
-        "Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.\n",
-        "\n",
-        "**Clarification:** The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-        "- `-1000 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "TThe problem is to design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of characters or bytes so that it can be easily stored in a file, transmitted over a network, or otherwise persisted. Deserialization is the reverse process of converting the serialized data back into the original data structure.\n",
-        "\n",
-        "In the context of this problem:\n",
-        "\n",
-        "1. **Serialization**: You are given a binary tree, and your task is to convert it into a string representation such that you can later recreate the same binary tree from this string. The format of serialization is flexible, but it should allow you to reconstruct the original binary tree accurately.\n",
-        "\n",
-        "2. **Deserialization**: Given a serialized string, you need to reconstruct the binary tree it represents, ensuring that it is identical to the original tree.\n",
-        "\n",
-        "Here are some key points to consider in solving this problem:\n",
-        "\n",
-        "- The input can include any valid binary tree, including trees with nodes having integer values within the range [-1000, 1000].\n",
-        "- You don't necessarily need to follow a specific format for serialization, but it should be designed in a way that allows unambiguous deserialization.\n",
-        "- The goal is to serialize the tree structure and its values and then deserialize it back into the same structure and values.\n",
-        "\n",
-        "A common approach for serialization is to use a traversal method like preorder traversal, where you visit nodes in the order: root, left subtree, right subtree. This way, you can serialize the tree into a string, and during deserialization, you can reconstruct the tree by parsing the string in the same order.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode(object):\n",
-        "    def __init__(self, x):\n",
-        "        self.val = x\n",
-        "        self.left = None\n",
-        "        self.right = None\n",
-        "\n",
-        "class Codec:\n",
-        "    def serialize(self, root):\n",
-        "        serialized  = []\n",
-        "\n",
-        "        def dfs(node):\n",
-        "            if not node:\n",
-        "                # If the node is None, represent it as \"Null\" in the serialized string\n",
-        "                serialized .append(\"Null\")\n",
-        "                return\n",
-        "            # Convert the node's value to a string and add it to the serialized string\n",
-        "            serialized .append(str(node.val))\n",
-        "            dfs(node.left)\n",
-        "            dfs(node.right)\n",
-        "\n",
-        "        dfs(root)\n",
-        "        # Join the serialized values with \",\"\n",
-        "        return \",\".join(serialized )\n",
-        "\n",
-        "    def deserialize(self, data):\n",
-        "        vals = data.split(\",\")\n",
-        "        self.i = 0\n",
-        "\n",
-        "        def dfs():\n",
-        "            if vals[self.i] == \"Null\":\n",
-        "                self.i += 1\n",
-        "                return None\n",
-        "            # Convert the value to an integer to create a new node\n",
-        "            node = TreeNode(int(vals[self.i]))\n",
-        "            self.i += 1\n",
-        "            node.left = dfs()\n",
-        "            node.right = dfs()\n",
-        "            return node\n",
-        "\n",
-        "        return dfs()"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Here's a step-by-step explanation of the code:\n",
-        "\n",
-        "1. **Definition of TreeNode**:\n",
-        "   - The code defines a simple `TreeNode` class that represents a node in a binary tree. Each `TreeNode` has three attributes:\n",
-        "     - `val`: The integer value stored in the node.\n",
-        "     - `left`: A reference to the left child node (or None if there is no left child).\n",
-        "     - `right`: A reference to the right child node (or None if there is no right child).\n",
-        "   - This class allows you to create binary tree nodes with integer values and left and right child references.\n",
-        "\n",
-        "2. **Serialization (serialize method)**:\n",
-        "   - The `serialize` method takes a `root` node as input, which is the root of the binary tree that needs to be serialized.\n",
-        "   - It initializes an empty list `serialized` to store the serialized elements.\n",
-        "   - The `dfs` (depth-first search) function is defined within the `serialize` method to perform a preorder traversal of the binary tree.\n",
-        "   - In the `dfs` function:\n",
-        "     - If the current `node` is `None` (i.e., a leaf node or a child of a leaf node), it appends the string \"Null\" to the `serialized` list to represent the absence of a node.\n",
-        "     - If the current `node` is not `None`, it appends the string representation of the `node.val` to the `serialized` list and then recursively calls `dfs` on the left and right children.\n",
-        "   - After the `dfs` traversal, the `serialized` list contains the serialized binary tree elements.\n",
-        "   - The method returns a string obtained by joining the elements in the `serialized` list with commas.\n",
-        "\n",
-        "3. **Deserialization (deserialize method)**:\n",
-        "   - The `deserialize` method takes a serialized string `data` as input, which represents a binary tree in the custom format.\n",
-        "   - It splits the `data` string by commas to obtain a list of elements called `vals`.\n",
-        "   - The method initializes an index `self.i` to 0. This index keeps track of the current position in the `vals` list during deserialization.\n",
-        "   - The `dfs` function is defined within the `deserialize` method to perform the deserialization process using a recursive approach.\n",
-        "   - In the `dfs` function:\n",
-        "     - If the current element in `vals` is \"Null,\" it means there is no node at this position in the binary tree, so it returns `None`.\n",
-        "     - If the current element is not \"Null,\" it converts the element to an integer to create a new `TreeNode` with that value.\n",
-        "     - It then recursively calls `dfs` to set the left and right children of the current node.\n",
-        "   - The method returns the root node of the reconstructed binary tree.\n",
-        "\n",
-        "Overall, this code demonstrates a way to serialize a binary tree into a string format and then deserialize it back into the original tree structure, allowing for the representation of both the tree structure and the values stored in the nodes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1,2,Null,Null,3,4,Null,Null,5,Null,Null\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Serialize a binary tree\n",
-        "root = TreeNode(1)\n",
-        "root.left = TreeNode(2)\n",
-        "root.right = TreeNode(3)\n",
-        "root.right.left = TreeNode(4)\n",
-        "root.right.right = TreeNode(5)\n",
-        "\n",
-        "codec = Codec()\n",
-        "serialized_tree = codec.serialize(root)\n",
-        "print(serialized_tree)\n",
-        "\n",
-        "# Deserialize the serialized tree\n",
-        "new_root = codec.deserialize(serialized_tree)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Null\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Example usage:\n",
-        "# Serialize a binary tree\n",
-        "root = None\n",
-        "\n",
-        "codec = Codec()\n",
-        "serialized_tree = codec.serialize(root)\n",
-        "print(serialized_tree)\n",
-        "\n",
-        "# Deserialize the serialized tree\n",
-        "new_root = codec.deserialize(serialized_tree)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for serializing and deserializing a binary tree:\n",
-        "\n",
-        "1. **Serialization (serialize method)**:\n",
-        "\n",
-        "   - Time Complexity:\n",
-        "     - The `serialize` method uses a depth-first traversal (preorder) of the binary tree.\n",
-        "     - It visits each node exactly once.\n",
-        "     - At each node, it performs constant-time operations (string conversion and list append).\n",
-        "     - Therefore, the time complexity for serialization is O(N), where N is the number of nodes in the binary tree.\n",
-        "\n",
-        "   - Space Complexity:\n",
-        "     - The space complexity for serialization includes the space used by the `serialized` list and the recursive call stack.\n",
-        "     - The `serialized` list stores the serialized elements, and its space is proportional to the number of nodes.\n",
-        "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-        "     - Therefore, the space complexity for serialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-        "\n",
-        "2. **Deserialization (deserialize method)**:\n",
-        "\n",
-        "   - Time Complexity:\n",
-        "     - The `deserialize` method splits the serialized string into a list of elements, which takes O(N) time.\n",
-        "     - The `dfs` function is a recursive function that visits each element in the list exactly once.\n",
-        "     - At each element, it performs constant-time operations (string comparison, integer conversion, and recursive function calls).\n",
-        "     - Therefore, the time complexity for deserialization is O(N).\n",
-        "\n",
-        "   - Space Complexity:\n",
-        "     - The space complexity for deserialization includes the space used by the `vals` list and the recursive call stack.\n",
-        "     - The `vals` list stores the elements from the serialized string, and its space is proportional to the number of nodes.\n",
-        "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-        "     - Therefore, the space complexity for deserialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-        "\n",
-        "Overall, the provided code has a time complexity of O(N) for both serialization and deserialization and a space complexity of O(N) in the majority of cases (unless the tree is severely unbalanced, in which case the height H dominates the space complexity). It efficiently serializes and deserializes a binary tree while using a linear amount of memory."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Efficient Serialization**:\n",
-        "   Optimize the serialization process to minimize the length of the serialized string. Consider using techniques like binary encoding to represent node values more efficiently.\n",
-        "\n",
-        "2. **Custom Serialization Format**:\n",
-        "   Modify the serialization and deserialization methods to use a custom format different from the one provided. Ensure that the new format allows for accurate reconstruction of the original binary tree."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/572. Subtree of Another Tree.ipynb b/docs/07. Trees/572. Subtree of Another Tree.ipynb
deleted file mode 100644
index 8a41c3b..0000000
--- a/docs/07. Trees/572. Subtree of Another Tree.ipynb	
+++ /dev/null
@@ -1,265 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 572: Subtree of Another Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Subtree of Another Tree problem on LeetCode, click here!](https://leetcode.com/problems/subtree-of-another-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "**Problem Description:**\n",
-        "\n",
-        "Given the roots of two binary trees `root` and `subRoot`, return `True` if there is a subtree of `root` with the same structure and node values as `subRoot`, and `False` otherwise.\n",
-        "\n",
-        "A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "1. The number of nodes in the `root` tree is in the range $[1, 2000]$.\n",
-        "2. The number of nodes in the `subRoot` tree is in the range $[1, 1000]$.\n",
-        "3. $-10^4$ <= root.val <= $10^4$\n",
-        "4. $-10^4$ <= subRoot.val <= $10^4$\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "class Solution:\n",
-        "    def isSubtree(self, root, subRoot):\n",
-        "        if not root:\n",
-        "            return False\n",
-        "        \n",
-        "        # Check if the current subtree is equal to the subRoot\n",
-        "        if self.isSameTree(root, subRoot):\n",
-        "            return True\n",
-        "        \n",
-        "        # Recursively check the left and right subtrees\n",
-        "        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)\n",
-        "    \n",
-        "    def isSameTree(self, tree1, tree2):\n",
-        "        if not tree1 and not tree2:\n",
-        "            return True\n",
-        "        if not tree1 or not tree2:\n",
-        "            return False\n",
-        "        \n",
-        "        return (\n",
-        "            tree1.val == tree2.val and\n",
-        "            self.isSameTree(tree1.left, tree2.left) and\n",
-        "            self.isSameTree(tree1.right, tree2.right)\n",
-        "        )"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "This code defines a TreeNode class for the binary tree nodes and a Solution class with two methods:\n",
-        "\n",
-        "1. `isSubtree`: This method checks if there is a subtree of the `root` tree with the same structure and node values as the `subRoot` tree. It uses a helper function `isSameTree` to compare two trees for equality.\n",
-        "\n",
-        "2. `isSameTree`: This helper method recursively compares two trees to check if they are the same in structure and node values.\n",
-        "\n",
-        "### Here's a detailed explanation of the code:\n",
-        "\n",
-        "1. **TreeNode Class:**\n",
-        "\n",
-        "The `TreeNode` class is defined to represent nodes in a binary tree. Each node has a `val` (the node's value) and may have a `left` and `right` child.\n",
-        "\n",
-        "2. **Solution Class:**\n",
-        "\n",
-        "    + The `Solution` class contains the solution for the problem and defines two important methods:\n",
-        "\n",
-        "    + `isSubtree(self, root, subRoot)`:\n",
-        "\n",
-        "        - This method checks whether `subRoot` is a subtree of `root`. It takes two tree nodes, `root` and `subRoot`, as input arguments.\n",
-        "        - If `root` is `None`, it returns `False` because there is no subtree to search.\n",
-        "        - It then checks if the current subtree with `root` as its root is equal to `subRoot` using the `isSameTree` method. If they are the same, it returns `True`.\n",
-        "        - If the current subtree is not the same as `subRoot`, it recursively checks the left and right subtrees of `root` to see if `subRoot` is a subtree of any of them.\n",
-        "        - It returns `True` if `subRoot` is found in either the left or right subtree; otherwise, it returns `False`.\n",
-        "\n",
-        "    + `isSameTree(self, tree1, tree2)`:\n",
-        "\n",
-        "        - This method checks whether two trees, `tree1` and `tree2`, are the same.\n",
-        "        - If both `tree1` and `tree2` are `None`, they are considered the same tree, so it returns `True`.\n",
-        "        - If only one of them is `None` (but not both), they are different trees, so it returns `False`.\n",
-        "        - If both `tree1` and `tree2` have values, it checks if their values are equal and recursively checks if their left and right subtrees are the same.\n",
-        "        - It returns `True` if the trees are the same; otherwise, it returns `False`.\n",
-        "\n",
-        "The code effectively uses recursion to traverse the binary trees and check for subtree equality. The `isSubtree` method starts the recursive search, and the `isSameTree` method is used to compare individual subtrees. The approach is efficient and avoids unnecessary checks when possible.\n",
-        "\n",
-        "This solution allows you to determine if there exists a subtree within the `root` tree that matches the structure and node values of `subRoot`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Is subRoot a subtree of root? True\n"
-          ]
-        }
-      ],
-      "source": [
-        "#root = [3,4,5,1,2]\n",
-        "#subRoot = [4,1,2]\n",
-        "\n",
-        "# Example input\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(4)\n",
-        "root.right = TreeNode(5)\n",
-        "root.left.left = TreeNode(1)\n",
-        "root.left.right = TreeNode(2)\n",
-        "\n",
-        "subRoot = TreeNode(4)\n",
-        "subRoot.left = TreeNode(1)\n",
-        "subRoot.right = TreeNode(2)\n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test the isSubtree method\n",
-        "result = solution.isSubtree(root, subRoot)\n",
-        "print(\"Is subRoot a subtree of root?\", result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Is subRoot a subtree of root? False\n"
-          ]
-        }
-      ],
-      "source": [
-        "#root = [3,4,5,1,2,null,null,null,null,0] \n",
-        "#subRoot = [4,1,2]\n",
-        "\n",
-        "\n",
-        "# Example input\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(4)\n",
-        "root.right = TreeNode(5)\n",
-        "root.left.left = TreeNode(1)\n",
-        "root.left.right = TreeNode(2)\n",
-        "root.left.right.left = TreeNode(0)\n",
-        "\n",
-        "subRoot = TreeNode(4)\n",
-        "subRoot.left = TreeNode(1)\n",
-        "subRoot.right = TreeNode(2)\n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test the isSubtree method\n",
-        "result = solution.isSubtree(root, subRoot)\n",
-        "print(\"Is subRoot a subtree of root?\", result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "86a34f0d-0789-498c-818c-2d30f84754d0",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code for the \"Subtree of Another Tree\" problem:\n",
-        "\n",
-        "**Time Complexity**\n",
-        "\n",
-        "The time complexity of the code primarily depends on the recursive traversal of the binary tree. In both the `isSubtree` and `isSameTree` functions, we visit each node in the binary trees once. Let's break down the time complexity:\n",
-        "\n",
-        "1. The `isSubtree` function:\n",
-        "   - In the worst case, we need to visit every node in the `root` tree.\n",
-        "   - For each node in the `root` tree, we call the `isSameTree` function, which has its own traversal.\n",
-        "   - So, the total time complexity is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree.\n",
-        "\n",
-        "2. The `isSameTree` function:\n",
-        "   - In the worst case, we visit every node in both `tree1` and `tree2`.\n",
-        "   - The number of recursive calls made is proportional to the number of nodes in the trees.\n",
-        "   - So, the time complexity of this function is $O(max(n, m))$, where $n$ and $m$ are the numbers of nodes in `tree1` and `tree2`, respectively.\n",
-        "\n",
-        "Overall, the time complexity of the entire code is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree. In practice, it may be less than $O(n * m)$ if a subtree mismatch is detected early during the traversal.\n",
-        "\n",
-        "**Space Complexity**\n",
-        "\n",
-        "The space complexity of the code is determined by the function call stack during recursion and the space used by the recursive functions. Let's analyze the space complexity:\n",
-        "\n",
-        "1. The `isSubtree` function:\n",
-        "   - It uses the call stack for recursion.\n",
-        "   - The maximum depth of the recursion is equal to the height of the `root` tree, which can be $O(n)$ in the worst case (unbalanced tree).\n",
-        "   - Additionally, the function doesn't use any significant extra space other than the recursion stack.\n",
-        "\n",
-        "2. The `isSameTree` function:\n",
-        "   - It also uses the call stack for recursion.\n",
-        "   - The maximum depth of the recursion is equal to the height of the `tree1` or `tree2`, whichever is greater.\n",
-        "   - So, the maximum space used for the call stack is $O(max(n, m))$.\n",
-        "\n",
-        "In summary, the space complexity of the code is $O(max(n, m))$ due to the function call stack. It scales with the maximum height of the trees being compared.\n",
-        "\n",
-        "Overall, the code is efficient and works well for trees with moderate sizes. However, it's important to keep in mind that the worst-case time complexity is $O(n * m)$, so for very large trees, the performance may degrade."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/98. Validate Binary Search Tree.ipynb b/docs/07. Trees/98. Validate Binary Search Tree.ipynb
deleted file mode 100644
index ad86882..0000000
--- a/docs/07. Trees/98. Validate Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 98. Validate Binary Search Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Validate Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/validate-binary-search-tree/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, *determine if it is a valid binary search tree (BST)*.\n",
-        "\n",
-        "A **valid BST** is defined as follows:\n",
-        "\n",
-        "- The left subtree of a node contains only nodes with keys **less than** the node's key.\n",
-        "- The right subtree of a node contains only nodes with keys **greater than** the node's key.\n",
-        "- Both the left and right subtrees must also be binary search trees. \n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[1, 10^4]$.\n",
-        "- $-2^{31}$ <= `Node.val` <= $2^{31} - 1$\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def is_valid_BST(root):\n",
-        "    def is_valid(node, min_val, max_val):\n",
-        "        # Base case: If the node is None, it's a valid BST.\n",
-        "        if node is None:\n",
-        "            return True\n",
-        "        \n",
-        "        # Check if the current node's value is within the valid range.\n",
-        "        if not (min_val < node.val < max_val):\n",
-        "            return False\n",
-        "        \n",
-        "        # Recursively check the left and right subtrees with updated ranges.\n",
-        "        return (is_valid(node.left, min_val, node.val) and\n",
-        "                is_valid(node.right, node.val, max_val))\n",
-        "    \n",
-        "    # Call the helper function starting with a wide range for root node.\n",
-        "    return is_valid(root, float('-inf'), float('inf'))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- It defines a `TreeNode` class to represent nodes in the binary tree.\n",
-        "\n",
-        "- The `is_valid_BST` function takes the root node of the binary tree as input and returns `True` if the tree is a valid BST, and `False` otherwise.\n",
-        "\n",
-        "- Inside the `is_valid_BST` function, there is a helper function called `is_valid` that performs the actual validation using a recursive approach.\n",
-        "\n",
-        "- The `is_valid` function checks each node in the tree to ensure that it satisfies the properties of a BST:\n",
-        "  - The value of the current node must be within a valid range defined by `min_val` and `max_val`.\n",
-        "  - The left subtree of the current node should contain values less than the current node's value.\n",
-        "  - The right subtree of the current node should contain values greater than the current node's value.\n",
-        "\n",
-        "- If any of these conditions are violated, the function returns `False`, indicating that the tree is not a valid BST.\n",
-        "\n",
-        "- If all nodes satisfy these conditions, the function returns `True`, indicating that the tree is a valid BST.\n",
-        "\n",
-        "- The code calls the `is_valid` function with the root node and initial range values of negative infinity to positive infinity to start the validation.\n",
-        "\n",
-        "- The time complexity of the code is O(N), where N is the number of nodes in the tree, as it traverses each node once.\n",
-        "\n",
-        "- The space complexity depends on the height of the tree. In the average case for a balanced BST, the space complexity is O(log N), but in the worst case (skewed tree), it can be O(N) due to the recursive call stack.\n",
-        "\n",
-        "In summary, this code checks whether a given binary tree is a valid BST by recursively validating each node's value and its left and right subtrees while maintaining valid value ranges. If all nodes satisfy the BST properties, the tree is considered a valid BST."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Create the tree for the first example: [2, 1, 3]\n",
-        "root1 = TreeNode(2)\n",
-        "root1.left = TreeNode(1)\n",
-        "root1.right = TreeNode(3)\n",
-        "print(is_valid_BST(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Create the tree for the second example: [5, 1, 4, None, None, 3, 6]\n",
-        "root2 = TreeNode(5)\n",
-        "root2.left = TreeNode(1)\n",
-        "root2.right = TreeNode(4)\n",
-        "root2.right.left = TreeNode(3)\n",
-        "root2.right.right = TreeNode(6)\n",
-        "print(is_valid_BST(root2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `is_valid_BST` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The time complexity of the function primarily depends on the number of nodes in the binary tree.\n",
-        "- In the worst case, we may have to visit every node in the tree once to validate whether it's part of a valid BST.\n",
-        "- Since we are performing a depth-first search (DFS) traversal of the tree, the time complexity is O(N), where N is the number of nodes in the tree.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity is determined by the space used by the recursive call stack during the DFS traversal.\n",
-        "- In the worst case, if the tree is completely unbalanced (a skewed tree), the maximum depth of the call stack would be equal to the height of the tree.\n",
-        "- In a balanced BST, the height is approximately log2(N), where N is the number of nodes.\n",
-        "- Therefore, the space complexity of the call stack is O(log N) for a balanced BST.\n",
-        "- In the worst case (skewed tree), the space complexity can be O(N) as the height of the tree can be equal to N.\n",
-        "\n",
-        "**Overall:**\n",
-        "- Time Complexity: O(N)\n",
-        "- Space Complexity: O(log N) on average for a balanced BST, and O(N) in the worst case for a skewed tree.\n",
-        "\n",
-        "The space complexity is typically dominated by the recursive call stack, and it varies depending on the shape of the binary tree. In practice, for balanced binary trees, the space complexity is often close to O(log N), which is quite efficient."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Complex Constraints**: Modify the problem to have more complex constraints on the tree. For example, consider a binary tree with constraints like \"The left subtree of a node contains only nodes with keys less than or equal to the node's key,\" and \"The right subtree of a node contains only nodes with keys greater than or equal to the node's key.\" How would you adapt the code to handle these constraints?\n",
-        "\n",
-        "2. **Handling Duplicates**: Modify the code to handle binary search trees that allow duplicate values. In a standard BST, each value is unique, but in this case, multiple nodes can have the same value. The tree should still be considered valid if it follows the BST property."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/07. Trees/README.md b/docs/07. Trees/README.md
deleted file mode 100644
index 1eba5d4..0000000
--- a/docs/07. Trees/README.md	
+++ /dev/null
@@ -1,26 +0,0 @@
-# Trees Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)                          | Easy       |
-| [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)     | Easy       |
-| [100. Same Tree](https://leetcode.com/problems/same-tree/)                                           | Easy       |
-| [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)               | Easy       |
-| [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/) | Easy |
-| [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/) | Medium |
-| [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)       | Medium     |
-| [230. Kth Smallest Element In a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)   | Medium     |
-| [105. Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) | Medium |
-| [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)     | Hard       |
-| [297. Serialize And Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/) | Hard |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/08. Tries/208. Implement Trie (Prefix Tree).ipynb b/docs/08. Tries/208. Implement Trie (Prefix Tree).ipynb
deleted file mode 100644
index 9da2a17..0000000
--- a/docs/08. Tries/208. Implement Trie (Prefix Tree).ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 208. Implement Trie (Prefix Tree)\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Implement Trie problem on LeetCode, click here!](https://leetcode.com/problems/implement-trie-prefix-tree/)\n",
-        "\n",
-        "---\n",
-        "A **trie** (pronounced as \"try\") or **prefix tree** is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.\n",
-        "\n",
-        "Implement the Trie class:\n",
-        "\n",
-        "- `Trie()` Initializes the trie object.\n",
-        "- `void insert(String word)` Inserts the string `word` into the trie.\n",
-        "- `boolean search(String word)` Returns `true` if the string `word` is in the trie (i.e., was inserted before), and `false` otherwise.\n",
-        "- `boolean startsWith(String prefix)` Returns `true` if there is a previously inserted string `word` that has the prefix `prefix`, and `false` otherwise.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= word.length, prefix.length <= 2000`\n",
-        "- `word` and `prefix` consist only of lowercase English letters.\n",
-        "- At most $3 * 10^4$ calls in total will be made to `insert`, `search`, and `startsWith`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to implement a data structure known as a Trie (pronounced \"try\") or Prefix Tree. A Trie is a tree-like data structure that is used to efficiently store and retrieve a set of strings or words. It's particularly useful for tasks involving string manipulation, such as autocomplete and spell-checking. The problem statement defines the following operations for implementing the Trie class:\n",
-        "\n",
-        "1. `Trie()`: This is the constructor that initializes the Trie object.\n",
-        "\n",
-        "2. `void insert(String word)`: This method inserts the given string `word` into the Trie. It effectively adds the characters of the word to the Trie's structure, creating nodes for each character if they don't already exist. At the end of the word, a special flag is set to indicate that this node represents the end of a valid word.\n",
-        "\n",
-        "3. `boolean search(String word)`: This method checks if the given string `word` is present in the Trie. It starts from the root of the Trie and traverses the Trie by following the characters in the word. If it successfully traverses the Trie and reaches the end of the word, it returns `true`, indicating that the word exists in the Trie; otherwise, it returns `false`.\n",
-        "\n",
-        "4. `boolean startsWith(String prefix)`: This method checks if there is any previously inserted word in the Trie that has the given `prefix`. It's similar to the `search` method but does not require that the prefix be a complete word. If there is any word in the Trie that starts with the given prefix, it returns `true`; otherwise, it returns `false`.\n",
-        "\n",
-        "The problem also provides an example scenario where these operations are called, demonstrating the expected output for each operation.\n",
-        "\n",
-        "In summary, the goal is to create a data structure (Trie) that efficiently stores a set of strings and provides methods to insert new strings, search for complete words, and check if a given prefix exists in the stored set of strings.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        self.children = {}           # A dictionary to store child nodes\n",
-        "        self.is_end_of_word = False  # Indicates if a word ends at this node\n",
-        "\n",
-        "class Trie:\n",
-        "    def __init__(self):\n",
-        "        self.root = TrieNode()       # Initialize the Trie with a root node\n",
-        "\n",
-        "    def insert(self, word):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                node.children[char] = TrieNode()  # Create a new node if the character is not present\n",
-        "            node = node.children[char]           # Move to the child node\n",
-        "        node.is_end_of_word = True                # Mark the end of the inserted word\n",
-        "\n",
-        "    def search(self, word):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                return False         # If the character is not found, the word is not in the Trie\n",
-        "            node = node.children[char]  # Move to the child node\n",
-        "        return node.is_end_of_word    # Check if the node represents the end of a valid word\n",
-        "\n",
-        "    def startsWith(self, prefix):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in prefix:\n",
-        "            if char not in node.children:\n",
-        "                return False         # If the character is not found, no word starts with the prefix\n",
-        "            node = node.children[char]  # Move to the child node\n",
-        "        return True                  # Prefix found in the Trie"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's explain how the code works step by step:\n",
-        "\n",
-        "1. `TrieNode` Class:\n",
-        "   - `TrieNode` is a class that represents nodes in the Trie. Each node has two attributes:\n",
-        "     - `children`: This is a dictionary that stores child nodes. Each key in the dictionary represents a character, and the corresponding value is the child node for that character.\n",
-        "     - `is_end_of_word`: This boolean flag indicates whether the node represents the end of a complete word. Initially, it's set to `False` for all nodes.\n",
-        "\n",
-        "2. `Trie` Class:\n",
-        "   - `Trie` is the main class that implements the Trie data structure. It has the following methods:\n",
-        "\n",
-        "   - `__init__(self)`: The constructor initializes the Trie object by creating a root node, which serves as the starting point for all Trie operations.\n",
-        "\n",
-        "   - `insert(self, word)`: This method inserts a string `word` into the Trie. It starts from the root node and iterates through each character in the word.\n",
-        "     - If a character is not present as a child node, it creates a new node for that character.\n",
-        "     - It then moves to the child node and continues the process until the entire word is inserted.\n",
-        "     - Finally, it sets the `is_end_of_word` flag to `True` for the last node to mark the end of the inserted word.\n",
-        "\n",
-        "   - `search(self, word)`: This method checks if a complete word (string) exists in the Trie. It starts from the root node and iterates through each character in the word.\n",
-        "     - If a character is not found as a child node, it immediately returns `False` because the word is not in the Trie.\n",
-        "     - It continues to move to the child node for each character.\n",
-        "     - After reaching the end of the word, it checks if the `is_end_of_word` flag is `True` for the last node to confirm the presence of the word.\n",
-        "\n",
-        "   - `startsWith(self, prefix)`: This method checks if there is any previously inserted word in the Trie that starts with a given `prefix`. It follows the same logic as the `search` method but does not require the entire word to be present.\n",
-        "     - If the prefix is found in the Trie, it returns `True`; otherwise, it returns `False`.\n",
-        "\n",
-        "3. Example Usage:\n",
-        "   - The code demonstrates how to create a Trie object, insert words (\"apple\" and \"app\"), and perform operations on the Trie:\n",
-        "     - It inserts the word \"apple\" into the Trie.\n",
-        "     - It checks if \"apple\" is in the Trie, which returns `True`.\n",
-        "     - It checks if \"app\" is in the Trie, which returns `False`.\n",
-        "     - It checks if there is any word in the Trie that starts with \"app,\" which returns `True`.\n",
-        "     - It inserts the word \"app\" into the Trie.\n",
-        "     - It checks if \"app\" is in the Trie again, which now returns `True`.\n",
-        "\n",
-        "In summary, the code efficiently implements a Trie data structure with the ability to insert words, search for complete words, and check for the existence of words with a given prefix. The Trie is organized as a tree of nodes, with each node representing a character in the words being stored."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n",
-            "False\n",
-            "True\n",
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "\n",
-        "trie = Trie()\n",
-        "trie.insert(\"apple\")\n",
-        "print(trie.search(\"apple\"))   # Output: True\n",
-        "print(trie.search(\"app\"))     # Output: False\n",
-        "print(trie.startsWith(\"app\")) # Output: True\n",
-        "trie.insert(\"app\")\n",
-        "print(trie.search(\"app\"))     # Output: True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the Trie implementation:\n",
-        "\n",
-        "1. Time Complexity:\n",
-        "   - Insertion (`insert` method): The time complexity of inserting a word into the Trie is O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to insert it into the Trie.\n",
-        "   - Search (`search` method): The time complexity of searching for a word in the Trie is also O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to determine if it exists in the Trie.\n",
-        "   - Starts with (`startsWith` method): The time complexity of checking if a prefix exists in the Trie is O(p), where p is the length of the prefix. In the worst case, you may need to traverse the entire prefix to determine its presence.\n",
-        "\n",
-        "   Therefore, for each operation (insertion, search, or starts with), the time complexity is O(m) or O(p), where m is the length of the word being operated on, and p is the length of the prefix.\n",
-        "\n",
-        "2. Space Complexity:\n",
-        "   - The space complexity of the Trie is determined by the number of nodes and characters stored in it. In the worst case, where none of the inserted words share common prefixes, the space complexity is O(N), where N is the total number of characters across all inserted words.\n",
-        "   - Each node in the Trie represents a character, and the number of nodes is directly related to the total number of characters.\n",
-        "\n",
-        "   In practice, the space complexity can be less than O(N) because common prefixes are shared among words in the Trie, leading to space optimization.\n",
-        "\n",
-        "In summary:\n",
-        "- Time complexity for each operation (insertion, search, starts with) is O(m) or O(p), where m is the length of the word and p is the length of the prefix.\n",
-        "- Space complexity is O(N) in the worst case, where N is the total number of characters across all inserted words. However, space optimization occurs due to common prefix sharing in practice, potentially reducing the actual space used."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Count Prefixes**:\n",
-        "   Implement a method to count the number of words in the Trie that have a specific prefix. This exercise requires you to maintain additional data in the Trie nodes to keep track of the count.\n",
-        "\n",
-        "2. **Auto-Complete Suggestions**:\n",
-        "   Implement an auto-complete feature using the Trie. Given a prefix, the program should return a list of suggested words based on the words previously inserted into the Trie."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/08. Tries/211. Design Add and Search Words Data Structure.ipynb b/docs/08. Tries/211. Design Add and Search Words Data Structure.ipynb
deleted file mode 100644
index 2de472c..0000000
--- a/docs/08. Tries/211. Design Add and Search Words Data Structure.ipynb	
+++ /dev/null
@@ -1,275 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 211. Design Add and Search Words Data Structure\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Design Add and Search Words Data Structure problem on LeetCode, click here!](https://leetcode.com/problems/design-add-and-search-words-data-structure/)\n",
-        "\n",
-        "---\n",
-        "Design a data structure that supports adding new words and finding if a string matches any previously added string.\n",
-        "\n",
-        "Implement the `WordDictionary` class:\n",
-        "\n",
-        "- `WordDictionary()` Initializes the object.\n",
-        "- `void addWord(word)` Adds `word` to the data structure, it can be matched later.\n",
-        "- `bool search(word)` Returns `true` if there is any string in the data structure that matches `word` or `false` otherwise. `word` may contain dots `'.'` where dots can be matched with any letter.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= word.length <= 25`\n",
-        "- `word` in `addWord` consists of lowercase English letters.\n",
-        "- `word` in `search` consist of `'.'` or lowercase English letters.\n",
-        "- There will be at most `2` dots in `word` for `search` queries.\n",
-        "- At most $10^4$ calls will be made to `addWord` and `search`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to design a data structure called `WordDictionary` that allows you to add words and search for words efficiently. The key feature of this data structure is that it should support wildcard search, where a dot ('.') can match any letter.\n",
-        "\n",
-        "Here are the specific requirements and explanations for the problem:\n",
-        "\n",
-        "1. Initialize the `WordDictionary` object using the constructor `WordDictionary()`.\n",
-        "\n",
-        "2. The `addWord(word)` method allows you to add a word to the data structure. Once added, you should be able to search for this word later. The words consist of lowercase English letters, and each word has a length between 1 and 25 characters.\n",
-        "\n",
-        "3. The `search(word)` method allows you to search for a word in the data structure. This method returns `True` if there is any string in the data structure that matches the given word or `False` otherwise. The word you are searching for may contain dots ('.'), where each dot can match any letter.\n",
-        "\n",
-        "4. The wildcard character ('.') in the `search` method allows for partial or fuzzy searching. For example, if the word dictionary contains the words \"bad,\" \"dad,\" and \"mad,\" then:\n",
-        "   - Searching for \"pad\" should return `False` because there is no exact match.\n",
-        "   - Searching for \"bad\" should return `True` because \"bad\" is in the dictionary.\n",
-        "   - Searching for \".ad\" should return `True` because it can match \"bad,\" \"dad,\" or \"mad.\"\n",
-        "   - Searching for \"b..\" should return `True` because it can match \"bad,\" \"bed,\" \"bet,\" etc.\n",
-        "\n",
-        "5. The problem specifies that there will be at most two dots ('.') in search queries, and there will be at most 10^4 calls to both `addWord` and `search`.\n",
-        "\n",
-        "In summary, you need to implement the `WordDictionary` class that efficiently supports adding words and searching for words, where the search can include wildcard characters ('.') that match any letter.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        # Initialize a TrieNode with children and a flag to indicate the end of a word.\n",
-        "        self.children = {}\n",
-        "        self.is_end_of_word = False\n",
-        "\n",
-        "class WordDictionary:\n",
-        "\n",
-        "    def __init__(self):\n",
-        "        # Initialize the WordDictionary with a root node.\n",
-        "        self.root = TrieNode()\n",
-        "\n",
-        "    def addWord(self, word: str) -> None:\n",
-        "        node = self.root\n",
-        "        # Iterate through the characters in the word.\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                # Create a new node for the character if it doesn't exist.\n",
-        "                node.children[char] = TrieNode()\n",
-        "            # Move to the next node.\n",
-        "            node = node.children[char]\n",
-        "        # Mark the end of the word by setting the flag to True.\n",
-        "        node.is_end_of_word = True\n",
-        "\n",
-        "    def search_helper(self, node: TrieNode, word: str) -> bool:\n",
-        "        if not word:\n",
-        "            # If there are no characters left in the word, check if we reached the end of a word in the trie.\n",
-        "            return node.is_end_of_word\n",
-        "        \n",
-        "        char = word[0]\n",
-        "        if char == '.':\n",
-        "            # If the character is a dot, explore all possible child nodes.\n",
-        "            for child in node.children.values():\n",
-        "                if self.search_helper(child, word[1:]):\n",
-        "                    return True\n",
-        "        elif char in node.children:\n",
-        "            # If the character is in the children of the current node, move to the next node.\n",
-        "            return self.search_helper(node.children[char], word[1:])\n",
-        "        \n",
-        "        # If the character is not found and is not a dot, the word is not in the trie.\n",
-        "        return False\n",
-        "\n",
-        "    def search(self, word: str) -> bool:\n",
-        "        # Start the search from the root of the trie.\n",
-        "        return self.search_helper(self.root, word)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "This code defines a `WordDictionary` class that uses a Trie data structure to efficiently store and search for words. Here's a step-by-step explanation:\n",
-        "\n",
-        "1. `TrieNode` class: This class represents nodes in the Trie data structure. Each node has two attributes: `children` (a dictionary to store child nodes) and `is_end_of_word` (a flag to indicate whether the node marks the end of a word).\n",
-        "\n",
-        "2. `WordDictionary` class: This class initializes with an empty Trie by creating a root node.\n",
-        "\n",
-        "3. `addWord` method: This method adds a word to the Trie. It iterates through each character in the word and creates Trie nodes as necessary. It sets the `is_end_of_word` flag to `True` for the final character of the word to mark the end of the word.\n",
-        "\n",
-        "4. `search_helper` method: This is a recursive helper function for searching words in the Trie. It takes a Trie node (`node`) and a word to search (`word`) as input. If there are no characters left in the word (`not word`), it checks if the current node marks the end of a word. If it does, it returns `True`.\n",
-        "\n",
-        "5. If the first character of the word is a dot ('.'), the function explores all possible child nodes by iterating through `node.children.values()` and recursively calls `search_helper` for each child with the remaining part of the word (`word[1:]`).\n",
-        "\n",
-        "6. If the first character of the word is not a dot and is found in the children of the current node, the function recursively moves to the next node by calling `search_helper` on the child node and the remaining part of the word.\n",
-        "\n",
-        "7. If the character is not found in the children and is not a dot, the word is not in the Trie, so the function returns `False`.\n",
-        "\n",
-        "8. `search` method: This is the public method for searching words. It initiates the search process by calling `search_helper` starting from the root node of the Trie.\n",
-        "\n",
-        "Overall, this code efficiently implements a Trie-based data structure that allows adding and searching for words, including support for wildcard characters represented by dots ('.')."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n",
-            "True\n",
-            "True\n",
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "\n",
-        "wordDictionary = WordDictionary()\n",
-        "wordDictionary.addWord(\"bad\")\n",
-        "wordDictionary.addWord(\"dad\")\n",
-        "wordDictionary.addWord(\"mad\")\n",
-        "print(wordDictionary.search(\"pad\"))  # Output: False\n",
-        "print(wordDictionary.search(\"bad\"))  # Output: True\n",
-        "print(wordDictionary.search(\".ad\"))  # Output: True\n",
-        "print(wordDictionary.search(\"b..\"))  # Output: True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `WordDictionary` class methods:\n",
-        "\n",
-        "1. `addWord` Method:\n",
-        "\n",
-        "   - Time Complexity: O(L)\n",
-        "     - L is the length of the word being added.\n",
-        "     - In the worst case, we need to iterate through each character in the word and create nodes in the Trie. This takes O(L) time.\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for storing the word in the Trie is also O(L) because we create nodes for each character in the word.\n",
-        "\n",
-        "2. `search_helper` Method (called by `search`):\n",
-        "\n",
-        "   - Time Complexity: O(2^M * L)\n",
-        "     - M is the maximum number of dots ('.') in the search word.\n",
-        "     - In the worst case, when there are M dots in the search word, we may explore all possible child nodes at each dot. This results in a branching factor of 26 (for lowercase letters) at each dot.\n",
-        "     - So, the time complexity is exponential in the number of dots and linear in the length of the word being searched (L).\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for the recursive call stack is O(L) in the worst case because we may have to recurse to a depth equal to the length of the search word.\n",
-        "\n",
-        "3. `search` Method:\n",
-        "\n",
-        "   - Time Complexity: O(2^M * L)\n",
-        "     - The `search` method calls the `search_helper` method, which has the same time complexity explained above.\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for the recursive call stack is O(L) in the worst case, as explained earlier.\n",
-        "\n",
-        "Overall, the time complexity of the `search` method is mainly affected by the number of dots ('.') in the search word. In the worst case, when there are multiple dots and many possibilities to explore at each dot, the time complexity can be exponential. However, for most practical cases, it performs reasonably well.\n",
-        "\n",
-        "The space complexity primarily depends on the space required to store the words in the Trie. It is proportional to the total number of characters in all added words and is linear with respect to the length of the words added.\n",
-        "\n",
-        "In summary:\n",
-        "\n",
-        "- Time Complexity for `addWord`: O(L)\n",
-        "- Time Complexity for `search`: O(2^M * L)\n",
-        "- Space Complexity: O(L) for storing words in the Trie, O(L) for the recursive call stack during search.\n",
-        "\n",
-        "Note: The space complexity analysis assumes that the dictionary contains a reasonable number of words and does not account for the overhead of Python objects or system-level memory allocation."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Dictionary Auto-Completion**: Extend the `WordDictionary` class to provide auto-completion suggestions based on the prefix of a word. Implement a method that returns a list of words that match the given prefix.\n",
-        "\n",
-        "2. **Support Word Deletion**: Extend the `WordDictionary` class to support word deletion. Add a method `deleteWord(word)` that removes a word from the data structure. Ensure that the search operation still works correctly after deletion."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/08. Tries/212. Word Search II.ipynb b/docs/08. Tries/212. Word Search II.ipynb
deleted file mode 100644
index 71597b8..0000000
--- a/docs/08. Tries/212. Word Search II.ipynb	
+++ /dev/null
@@ -1,316 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 212. Word Search II\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Word Search II problem on LeetCode, click here!](https://leetcode.com/problems/word-search-ii/)\n",
-        "\n",
-        "---\n",
-        "Given an `m x n` `board` of characters and a list of strings `words`, the task is to return *all words on the board*.\n",
-        "\n",
-        "Each word must be constructed from letters of sequentially **adjacent cells**, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == board.length`\n",
-        "- `n == board[i].length`\n",
-        "- `1 <= m, n <= 12`\n",
-        "- `board[i][j]` is a lowercase English letter.\n",
-        "- 1 <= `words.length` <= $3 * 10^4$\n",
-        "- `1 <= words[i].length <= 10`\n",
-        "- `words[i]` consists of lowercase English letters.\n",
-        "- All the strings of `words` are unique."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "To solve the \"Word Search II\" problem, you can use a Trie data structure and perform a depth-first search (DFS) on the board. Here's a step-by-step approach to solve the problem:\n",
-        "\n",
-        "1. Build a Trie from the given list of words:\n",
-        "   - Create a Trie data structure to store all the words from the input list.\n",
-        "   - For each word in the input list, insert it into the Trie, character by character. Make sure to mark the end of each word in the Trie.\n",
-        "\n",
-        "2. Perform DFS on the board:\n",
-        "   - Iterate through each cell (i, j) on the board.\n",
-        "   - For each cell, start a DFS traversal from that cell to search for words.\n",
-        "   - During the DFS traversal, maintain a current path string.\n",
-        "   - At each cell, check if the current path string, concatenated with the character in the cell, matches any word prefix in the Trie. If it does, continue the DFS.\n",
-        "   - If the current path string matches a word in the Trie, add it to the result.\n",
-        "\n",
-        "3. Implement DFS with backtracking:\n",
-        "   - During the DFS traversal, you will explore neighboring cells (horizontally and vertically) if they are within bounds and haven't been visited before.\n",
-        "   - To prevent revisiting the same cell within the same word, mark the cell as visited (e.g., change its character to a special character like '#') before exploring it and restore its original character after the DFS traversal.\n",
-        "\n",
-        "4. Return the result:\n",
-        "   - After completing the DFS traversal for all cells on the board, you will have collected all the valid words in the result set.\n",
-        "   - Convert the result set to a list and return it as the final output.\n",
-        "\n",
-        "This approach efficiently finds all words on the board by utilizing the Trie data structure to prune unnecessary DFS paths and avoiding duplicate visits to the same cell within the same word. It ensures that each word is constructed from letters of sequentially adjacent cells without using the same letter cell more than once in a word.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Define a TrieNode class to represent nodes in the Trie.\n",
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        self.children = {}  # Dictionary to store child nodes.\n",
-        "        self.is_end_of_word = False  # Flag to mark the end of a word.\n",
-        "\n",
-        "# Main function to find words on the board.\n",
-        "def findWords(board, words):\n",
-        "    # DFS function to search for words starting from a given cell.\n",
-        "    def dfs(node, i, j, path):\n",
-        "        char = board[i][j]  # Get the character at the current cell.\n",
-        "        curr_node = node.children.get(char)  # Find the corresponding Trie node.\n",
-        "\n",
-        "        if not curr_node:\n",
-        "            return  # If no matching Trie node, stop the search.\n",
-        "\n",
-        "        path += char  # Add the character to the current path.\n",
-        "\n",
-        "        if curr_node.is_end_of_word:\n",
-        "            result.add(path)  # If the path matches a word, add it to the result.\n",
-        "\n",
-        "        temp, board[i][j] = board[i][j], \"#\"  # Mark the cell as visited.\n",
-        "\n",
-        "        # Explore neighboring cells (up, down, left, right).\n",
-        "        if i > 0:\n",
-        "            dfs(curr_node, i - 1, j, path)\n",
-        "        if i < m - 1:\n",
-        "            dfs(curr_node, i + 1, j, path)\n",
-        "        if j > 0:\n",
-        "            dfs(curr_node, i, j - 1, path)\n",
-        "        if j < n - 1:\n",
-        "            dfs(curr_node, i, j + 1, path)\n",
-        "\n",
-        "        board[i][j] = temp  # Restore the cell's original character.\n",
-        "\n",
-        "    # Function to build a Trie from the list of words.\n",
-        "    def buildTrie():\n",
-        "        root = TrieNode()  # Create a root node.\n",
-        "        for word in words:\n",
-        "            node = root\n",
-        "            for char in word:\n",
-        "                if char not in node.children:\n",
-        "                    node.children[char] = TrieNode()  # Create a new node if needed.\n",
-        "                node = node.children[char]\n",
-        "            node.is_end_of_word = True  # Mark the end of the word.\n",
-        "        return root\n",
-        "\n",
-        "    m, n = len(board), len(board[0])  # Get the dimensions of the board.\n",
-        "    root = buildTrie()  # Build the Trie from the list of words.\n",
-        "    result = set()  # Use a set to store unique words found on the board.\n",
-        "\n",
-        "    # Iterate through each cell on the board and start DFS from there.\n",
-        "    for i in range(m):\n",
-        "        for j in range(n):\n",
-        "            dfs(root, i, j, \"\")  # Start DFS from each cell on the board.\n",
-        "\n",
-        "    return list(result)  # Convert the set to a list and return the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code solves the \"Word Search II\" problem, where you have a grid of characters and a list of words. The goal is to find all words from the list that can be constructed by traversing adjacent cells (horizontally or vertically) in the grid, without using the same cell more than once for a single word.\n",
-        "\n",
-        "The code does the following:\n",
-        "\n",
-        "1. Defines a `TrieNode` class to represent nodes in a Trie data structure. Each node has children (other characters in the word) and a flag to mark the end of a word.\n",
-        "\n",
-        "2. Defines the `findWords` function, which takes the grid (`board`) and a list of words (`words`) as input.\n",
-        "\n",
-        "3. Inside the `findWords` function, there is a `dfs` (depth-first search) function. This function is used to explore the grid starting from a specific cell. It checks if the current path matches any word in the Trie and adds it to the result if found.\n",
-        "\n",
-        "4. The code also includes a `buildTrie` function that constructs a Trie data structure from the list of words. It iterates through each word, creating nodes for each character in the Trie and marking the end of words.\n",
-        "\n",
-        "5. The dimensions of the grid are obtained (m x n).\n",
-        "\n",
-        "6. A Trie is built using the `buildTrie` function.\n",
-        "\n",
-        "7. A set called `result` is used to store unique words found in the grid.\n",
-        "\n",
-        "8. The code iterates through each cell on the grid (using nested loops) and starts a DFS search from each cell.\n",
-        "\n",
-        "9. During the DFS, it explores neighboring cells (up, down, left, right) if they haven't been visited before and if the current path matches a prefix in the Trie.\n",
-        "\n",
-        "10. When a word is found during the DFS traversal, it is added to the `result` set.\n",
-        "\n",
-        "11. The grid cell is marked as visited during the DFS by changing its character to \"#\", and it is restored to its original character after the DFS traversal.\n",
-        "\n",
-        "12. Finally, the set of unique words found is converted to a list and returned as the output.\n",
-        "\n",
-        "The code effectively uses a Trie data structure to optimize the search and ensures that each word is constructed from adjacent cells without reusing the same cell for the same word."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['eat', 'oath']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "board1 = [[\"o\",\"a\",\"a\",\"n\"],[\"e\",\"t\",\"a\",\"e\"],[\"i\",\"h\",\"k\",\"r\"],[\"i\",\"f\",\"l\",\"v\"]]\n",
-        "words1 = [\"oath\",\"pea\",\"eat\",\"rain\"]\n",
-        "print(findWords(board1, words1))  # Output: [\"eat\", \"oath\"]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "c748b879-a2d8-4298-a42c-c7f61f9c7fc6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "\n",
-        "board2 = [[\"a\",\"b\"],[\"c\",\"d\"]]\n",
-        "words2 = [\"abcb\"]\n",
-        "print(findWords(board2, words2))  # Output: []"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. **Building the Trie (`buildTrie` function):** \n",
-        "   - Suppose there are `N` words in the input list, and the average length of a word is `L`.\n",
-        "   - Building the Trie takes O(N * L) time, as we iterate through each word and insert each character into the Trie.\n",
-        "\n",
-        "2. **DFS Traversal (`dfs` function):** \n",
-        "   - In the worst case, the DFS function explores all possible paths in the grid.\n",
-        "   - The maximum number of recursive calls for each cell is 4 (up, down, left, right).\n",
-        "   - The maximum depth of the DFS is limited by the length of the longest word in the Trie, which is at most `L`.\n",
-        "   - Therefore, the DFS traversal for each cell takes $O(4^L)$ time.\n",
-        "\n",
-        "3. **Iterating Through the Grid (`findWords` function):**\n",
-        "   - In the worst case, we iterate through all `m` rows and `n` columns in the grid.\n",
-        "   - For each cell, we start a DFS traversal.\n",
-        "   - Therefore, iterating through the entire grid takes $O(m * n * 4^L)$ time.\n",
-        "\n",
-        "Overall, the time complexity of the code is $O(N * L + m * n * 4^L)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. **Trie Data Structure:**\n",
-        "   - The space required to store the Trie depends on the number of unique characters in all the words.\n",
-        "   - In the worst case, where all words are unique and have no common prefixes, the space complexity of the Trie is O(N * L).\n",
-        "   - In practice, it may be less if there are common prefixes among words.\n",
-        "\n",
-        "2. **DFS Stack (Recursive Calls):**\n",
-        "   - The depth of the recursive call stack during DFS is at most `L`, where `L` is the length of the longest word in the Trie.\n",
-        "   - Therefore, the space complexity for the recursive call stack is O(L).\n",
-        "\n",
-        "3. **Result Set (`result` set):**\n",
-        "   - The space used to store the result set depends on the number of valid words found in the grid.\n",
-        "   - In the worst case, when all words are found, the space complexity of the result set is O(N * L).\n",
-        "\n",
-        "Overall, the space complexity of the code is O(N * L) for the Trie and O(L) for the recursive call stack.\n",
-        "\n",
-        "In summary, the time complexity is dominated by the DFS traversal and is influenced by the number of words, their lengths, and the size of the grid. The space complexity is mainly determined by the Trie structure and the recursive call stack depth during DFS."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Word Search Paths:**\n",
-        "   Extend the basic word search exercise to find all possible paths (sequences of cells) that spell out a given word on the board. Return a list of paths if the word can be formed, otherwise an empty list.\n",
-        "\n",
-        "\n",
-        "2. **Reverse Word Search:**\n",
-        "   Modify the code to search for words in reverse order on the board. Given a 2D grid of characters and a list of words, find all words from the list that can be formed by traversing the board in reverse (right to left, bottom to top, etc.)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/08. Tries/README.md b/docs/08. Tries/README.md
deleted file mode 100644
index 22c3bcd..0000000
--- a/docs/08. Tries/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Trie Problems - Blind 75 LeetCode
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [208. Implement Trie Prefix Tree](https://leetcode.com/problems/implement-trie-prefix-tree/)         | Medium     |
-| [211. Design Add And Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/) | Medium |
-| [212. Word Search II](https://leetcode.com/problems/word-search-ii/)                                 | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb b/docs/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb
deleted file mode 100644
index cb66b6a..0000000
--- a/docs/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb	
+++ /dev/null
@@ -1,254 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 1046. Last Stone Weight\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Last Stone Weight problem on LeetCode, click here!](https://leetcode.com/problems/last-stone-weight/)\n",
-        "\n",
-        "---\n",
-        "You are given an array of integers `stones` where `stones[i]` is the weight of the `i`th stone.\n",
-        "\n",
-        "We are playing a game with the stones. On each turn, you choose the **heaviest two stones** and smash them together. Suppose the heaviest two stones have weights `x` and `y` with `x <= y`. The result of this smash is:\n",
-        "\n",
-        "1. If `x == y`, both stones are destroyed.\n",
-        "2. If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - x`.\n",
-        "\n",
-        "At the end of the game, there is **at most one** stone left.\n",
-        "\n",
-        "Return *the weight of the last remaining stone*. If there are no stones left, return `0`.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= stones.length <= 30`\n",
-        "- `1 <= stones[i] <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The \"Last Stone Weight\" problem involves a game played with a collection of stones, each stone having a certain weight. The goal is to find the weight of the last remaining stone after applying a specific set of rules.\n",
-        "\n",
-        "Here's how the game works:\n",
-        "\n",
-        "1. You start with an array of stones, where each stone is represented by its weight. The weights of the stones are given in the input as an array.\n",
-        "\n",
-        "2. In each turn of the game, you select the two heaviest stones from the remaining stones. If there is only one stone left, you've found the answer.\n",
-        "\n",
-        "3. If there are two stones with weights `x` and `y`, where `x` is less than or equal to `y`, the following happens:\n",
-        "   - If `x` is equal to `y`, both stones are completely destroyed, and they are removed from the array of stones.\n",
-        "   - If `x` is not equal to `y`, the stone with weight `x` is destroyed, and the stone with weight `y` now has a new weight of `y - x`. The stone with weight `x` is removed from the array, and the modified stone with weight `y - x` remains in the array.\n",
-        "\n",
-        "4. The game continues with the modified array of stones until there is at most one stone left.\n",
-        "\n",
-        "5. The goal is to find the weight of the last remaining stone, or if there are no stones left, return `0`.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "\n",
-        "def lastStoneWeight(stones):\n",
-        "    # Convert the input list to a max heap (negate the values to simulate a max heap)\n",
-        "    max_heap = [-stone for stone in stones]\n",
-        "    heapq.heapify(max_heap)\n",
-        "\n",
-        "    # Continue as long as there are more than 1 stone in the heap\n",
-        "    while len(max_heap) > 1:\n",
-        "        # Get the two heaviest stones\n",
-        "        y = -heapq.heappop(max_heap)  # Get the heaviest stone and negate it back to positive\n",
-        "        x = -heapq.heappop(max_heap)  # Get the second heaviest stone and negate it back to positive\n",
-        "\n",
-        "        # Calculate the new stone's weight after smashing\n",
-        "        if x != y:\n",
-        "            new_stone = y - x\n",
-        "            heapq.heappush(max_heap, -new_stone)  # Negate the value and push it back to the heap\n",
-        "\n",
-        "    # If there is one stone left, return its weight (negate it back to positive)\n",
-        "    if max_heap:\n",
-        "        return -max_heap[0]\n",
-        "    else:\n",
-        "        return 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a Python function called `lastStoneWeight` that takes a list of stone weights as input. It aims to find the weight of the last remaining stone after playing the stone smashing game described in the problem statement.\n",
-        "\n",
-        "Here's a step-by-step explanation of the code:\n",
-        "\n",
-        "1. We import the `heapq` library, which allows us to create and manipulate a heap (priority queue).\n",
-        "\n",
-        "2. Inside the `lastStoneWeight` function:\n",
-        "   - We create a max heap (negating stone weights to simulate a max heap) using the `heapq.heapify` function. This heap will keep track of the heaviest stones.\n",
-        "   \n",
-        "   - We enter a loop that continues as long as there are more than 1 stone in the heap.\n",
-        "   \n",
-        "   - Inside the loop, we:\n",
-        "     - Retrieve and negate the heaviest stone (`y`) from the heap.\n",
-        "     - Retrieve and negate the second heaviest stone (`x`) from the heap.\n",
-        "   \n",
-        "     - Calculate the new stone's weight after smashing (`y - x`), and negate it back to simulate a max heap.\n",
-        "     - Add the new stone back to the heap using `heapq.heappush`.\n",
-        "   \n",
-        "3. After the loop, if there is one stone left in the heap, we retrieve and negate it to get its actual weight, and return it as the result.\n",
-        "   \n",
-        "4. If there are no stones left (the heap is empty), we return `0`, indicating that no stones remain.\n",
-        "\n",
-        "5. Finally, we provide two test cases to demonstrate the function's usage and correctness.\n",
-        "\n",
-        "In summary, the code efficiently implements the stone smashing game by maintaining a max heap, selecting the heaviest stones in each turn, smashing them, and updating the heap until there is at most one stone left. It then returns the weight of the last remaining stone or 0 if there are no stones left."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "stones1 = [2, 7, 4, 1, 8, 1]\n",
-        "print(lastStoneWeight(stones1))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "b7e79514-4821-48d0-bb9d-0f2de8150945",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2: \n",
-        "\n",
-        "stones2 = [1]\n",
-        "print(lastStoneWeight(stones2))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided Python code for the \"Last Stone Weight\" problem:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The `heapq.heapify` operation to create the max heap from the input stone weights has a time complexity of O(n), where n is the number of stones.\n",
-        "\n",
-        "2. The loop that continues until there are more than 1 stone in the heap performs the following operations in each iteration:\n",
-        "   - Retrieving and negating the heaviest stone (`y`) from the heap, which takes O(log n) time.\n",
-        "   - Retrieving and negating the second heaviest stone (`x`) from the heap, also taking O(log n) time.\n",
-        "   - Calculating the new stone's weight and pushing it back into the heap using `heapq.heappush`, which takes O(log n) time.\n",
-        "\n",
-        "3. In the worst case, the loop runs until there is only one stone left, which requires approximately (n-1) iterations.\n",
-        "\n",
-        "Overall, the dominant time complexity is O(n) for the initial heap creation, and within the loop, each iteration takes O(log n) time. Therefore, the overall time complexity of the algorithm is O(n log n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The space complexity is primarily determined by the space required for the max heap. In the worst case, this heap can contain all the stones, so the space complexity for the heap is O(n).\n",
-        "\n",
-        "2. The rest of the variables used in the function (e.g., `x`, `y`, `new_stone`) require only constant space, so they do not significantly contribute to the space complexity.\n",
-        "\n",
-        "Therefore, the overall space complexity of the algorithm is O(n) due to the space used for the max heap.\n",
-        "\n",
-        "In summary:\n",
-        "- Time Complexity: $O(n\\ log\\ n)$\n",
-        "- Space Complexity: $O(n)$\n",
-        "\n",
-        "The algorithm is efficient enough to handle the problem's constraints, as the worst-case time and space complexities are both linearithmic in terms of the number of stones."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Kth Last Stone Weight**: Modify the problem to find the weight of the Kth last remaining stone, where K is an integer input. Your function should work efficiently for large values of K.\n",
-        "\n",
-        "\n",
-        "2. **Multiple Stones Smash**: Modify the problem so that instead of smashing two stones at a time, you can smash up to K stones at each turn, where K is an integer input. Determine the weight of the last remaining stone in this variation."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb b/docs/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb
deleted file mode 100644
index 847dd60..0000000
--- a/docs/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 295. Find Median from Data Stream\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Find Median from Data Stream problem on LeetCode, click here!](https://leetcode.com/problems/find-median-from-data-stream/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.\n",
-        "\n",
-        "Implement the MedianFinder class:\n",
-        "\n",
-        "- `MedianFinder()` initializes the `MedianFinder` object.\n",
-        "- `void addNum(int num)` adds the integer `num` from the data stream to the data structure.\n",
-        "- `double findMedian()` returns the median of all elements so far. Answers within $10^{-5}$ of the actual answer will be accepted.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- $-10^{5}$ <= `num` <= $10^{5}$\n",
-        "- There will be at least one element in the data structure before calling `findMedian`.\n",
-        "- At most $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-        "\n",
-        "**Follow-up:**\n",
-        "\n",
-        "1. If all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?\n",
-        "2. If `99%` of all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem, \"Find Median from Data Stream,\" deals with efficiently computing the median of a sequence of numbers as they are incrementally provided. The median is the middle value in an ordered list of numbers. When the list has an even number of elements, the median is the average of the two middle values.\n",
-        "\n",
-        "Here's a breakdown of the problem:\n",
-        "\n",
-        "1. **Initialization**: You are asked to implement a `MedianFinder` class that supports the following operations:\n",
-        "\n",
-        "    - `MedianFinder()`: Initializes the MedianFinder object.\n",
-        "    - `addNum(int num)`: Adds an integer `num` from the data stream to the data structure.\n",
-        "    - `findMedian()`: Returns the median of all elements added to the data structure.\n",
-        "\n",
-        "2. **Median Calculation**: The `findMedian` operation should efficiently compute the median, and the result should have a precision of at least $10^{-5}$ (i.e., answers within this range will be considered correct).\n",
-        "\n",
-        "3. **Examples**:\n",
-        "   \n",
-        "   - If you add numbers [1, 2], the median should be 1.5 because (1 + 2) / 2 = 1.5.\n",
-        "   - If you add numbers [1, 2, 3], the median should be 2 because it's the middle value.\n",
-        "   \n",
-        "4. **Constraints**:\n",
-        "\n",
-        "   - The integers provided in the data stream are in the range from $-10^5$ to $10^5$.\n",
-        "   - There will be at least one element in the data structure before calling `findMedian`.\n",
-        "   - At most, $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-        "\n",
-        "The problem can be solved by using two priority queues (heaps): one max-heap to store the smaller half of the numbers and one min-heap to store the larger half of the numbers. The max-heap ensures that the largest number in the smaller half is at the top, and the min-heap ensures that the smallest number in the larger half is at the top. These heaps are balanced to efficiently find the median when requested.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "\n",
-        "class MedianFinder:\n",
-        "\n",
-        "    def __init__(self):\n",
-        "        # Initialize two heaps: a max-heap (small_half) for the smaller half of numbers,\n",
-        "        # and a min-heap (large_half) for the larger half of numbers.\n",
-        "        self.small_half = []  # Max-heap for the smaller half of the numbers\n",
-        "        self.large_half = []  # Min-heap for the larger half of the numbers\n",
-        "\n",
-        "    def addNum(self, num: int) -> None:\n",
-        "        # Add the number to the appropriate heap based on its value\n",
-        "        if not self.small_half or num <= -self.small_half[0]:\n",
-        "            # If the number is less than or equal to the current maximum in the smaller half,\n",
-        "            # add it to the smaller half (max-heap)\n",
-        "            heapq.heappush(self.small_half, -num)\n",
-        "        else:\n",
-        "            # Otherwise, add it to the larger half (min-heap)\n",
-        "            heapq.heappush(self.large_half, num)\n",
-        "\n",
-        "        # Balance the heaps if necessary to ensure the size difference is at most 1\n",
-        "        if len(self.small_half) > len(self.large_half) + 1:\n",
-        "            # If the size of the smaller half is more than one greater than the size of the larger half,\n",
-        "            # move the maximum from the smaller half to the larger half to balance them.\n",
-        "            heapq.heappush(self.large_half, -heapq.heappop(self.small_half))\n",
-        "        elif len(self.large_half) > len(self.small_half):\n",
-        "            # If the size of the larger half is greater than the size of the smaller half,\n",
-        "            # move the minimum from the larger half to the smaller half to balance them.\n",
-        "            heapq.heappush(self.small_half, -heapq.heappop(self.large_half))\n",
-        "\n",
-        "    def findMedian(self) -> float:\n",
-        "        if len(self.small_half) == len(self.large_half):\n",
-        "            # If both halves have the same size, there's an even number of elements,\n",
-        "            # so the median is the average of the top elements in both heaps.\n",
-        "            return (-self.small_half[0] + self.large_half[0]) / 2.0\n",
-        "        else:\n",
-        "            # If the smaller half has more elements, it contains the median (odd number of elements).\n",
-        "            return -self.small_half[0]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a `MedianFinder` class for efficiently finding the median of a stream of numbers. It uses two heaps: a max-heap (`small_half`) to store the smaller half of the numbers and a min-heap (`large_half`) to store the larger half. Here's how the code works:\n",
-        "\n",
-        "1. The `__init__` method initializes the two heaps.\n",
-        "\n",
-        "2. The `addNum` method adds a number to the appropriate heap, ensuring that the heaps remain balanced. It does this by comparing the number to the current maximum in the smaller half and adding it to the appropriate heap. Then, it checks the sizes of the two heaps and rebalances them if necessary.\n",
-        "\n",
-        "3. The `findMedian` method calculates and returns the median. If both halves have the same size (even number of elements), it computes the average of the tops of both heaps. If the smaller half has more elements (odd number of elements), it returns the top of the max-heap, which is the median.\n",
-        "\n",
-        "In summary, this code efficiently maintains two heaps to divide the elements into smaller and larger halves, allowing for quick median retrieval. It offers a time complexity of O(log N) for adding elements and O(1) for finding the median, where N is the total number of elements processed. The space complexity is O(1) as the space used by the two heaps is independent of the input size."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1.5\n",
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "medianFinder = MedianFinder()\n",
-        "medianFinder.addNum(1)\n",
-        "medianFinder.addNum(2)\n",
-        "print(medianFinder.findMedian())  # Output: 1.5\n",
-        "medianFinder.addNum(3)\n",
-        "print(medianFinder.findMedian())  # Output: 2.0\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `MedianFinder` class:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. `MedianFinder()` Initialization: This operation has a time complexity of O(1). It simply initializes the two heaps.\n",
-        "\n",
-        "2. `addNum(int num)`: The time complexity for adding a number is O(log N), where N is the total number of elements processed so far. This is because, in the worst case, you might need to perform logarithmic operations on the heaps to maintain their balance.\n",
-        "\n",
-        "3. `findMedian()`: Finding the median has a time complexity of O(1). It simply involves accessing the tops of the two heaps, which takes constant time.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. `MedianFinder()` Initialization: The space complexity for initialization is O(1). It involves creating two empty heaps, which do not depend on the size of the data stream.\n",
-        "\n",
-        "2. `addNum(int num)`: The space complexity for adding a number is also O(1). It involves adding the number to one of the two heaps, which do not significantly affect the overall space complexity.\n",
-        "\n",
-        "3. `findMedian()`: The space complexity for finding the median is O(1). It doesn't require any additional data structures that grow with the input size.\n",
-        "\n",
-        "In summary, the time complexity of the `addNum` operation is O(log N), and the time complexity of the `findMedian` operation is O(1). The space complexity is O(1) as well, as the space used by the two heaps is independent of the size of the data stream. This solution efficiently maintains and retrieves the median, making it suitable for the given problem constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Generalized kth Element**: Extend the `MedianFinder` class to support finding the kth element in the data stream efficiently. This means finding the kth smallest or kth largest element in the stream.\n",
-        "\n",
-        "2. **Sliding Window Median**: Given an array and a sliding window size, find the median within the window as it slides over the array. This is an extension of the problem to a moving window scenario."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb b/docs/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb
deleted file mode 100644
index 2e1522d..0000000
--- a/docs/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb	
+++ /dev/null
@@ -1,239 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 703. Kth Largest Element in a Stream\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Kth Largest Element in a Stream problem on LeetCode, click here!](https://leetcode.com/problems/kth-largest-element-in-a-stream/)\n",
-        "\n",
-        "---\n",
-        "Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.\n",
-        "\n",
-        "Implement the `KthLargest` class with the following methods:\n",
-        "\n",
-        "- `KthLargest(int k, int[] nums)`: Initializes the object with the integer `k` and the stream of integers `nums`.\n",
-        "- `int add(int val)`: Appends the integer `val` to the stream and returns the element representing the `kth` largest element in the stream.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `k` <= $10^4$\n",
-        "- 0 <= `nums.length` <= $10^4$\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- $-10^4$ <= `val` <= $10^4$\n",
-        "- At most $10^4$ calls will be made to `add`.\n",
-        "- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem you're trying to solve is to design a class called `KthLargest` that can efficiently find the kth largest element in a stream of integers. You need to support two main operations:\n",
-        "\n",
-        "To solve this problem efficiently, you can use a min-heap data structure. Here's how the approach works:\n",
-        "\n",
-        "1. Initialize a min-heap to store the k largest elements. Initially, it's empty.\n",
-        "\n",
-        "2. In the `__init__` method, populate the min-heap with the first `k` elements from `nums`. This ensures that you have the k largest elements seen so far.\n",
-        "\n",
-        "3. Whenever you add a new element to the stream using the `add` method, follow these steps:\n",
-        "   - Add the new element to the min-heap.\n",
-        "   - If the size of the min-heap exceeds `k`, remove the smallest element from the min-heap. This ensures that you always have the k largest elements in the min-heap.\n",
-        "   - The smallest element in the min-heap (the root) will always represent the kth largest element in the stream.\n",
-        "\n",
-        "Here's why this approach works efficiently:\n",
-        "\n",
-        "- By using a min-heap, you can quickly maintain the k largest elements, and finding the smallest element in the heap (the root) takes constant time.\n",
-        "\n",
-        "- When you add a new element, the min-heap's size is kept at most `k`, which ensures that you only track the k largest elements and discard the smaller ones.\n",
-        "\n",
-        "- The time complexity for adding an element is O(log k), which is very efficient compared to sorting the entire stream, which would be O(n log n).\n",
-        "\n",
-        "- This approach meets the constraints of the problem, including handling large streams and having a low time complexity for both initialization and adding elements.\n",
-        "\n",
-        "In summary, the min-heap approach efficiently tracks the kth largest element in the stream by maintaining a heap of the k largest elements seen so far, updating it as new elements are added. This approach provides a fast and scalable solution to the problem.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "from typing import List  # Import the List type from the typing module\n",
-        "\n",
-        "class KthLargest:\n",
-        "\n",
-        "    def __init__(self, k: int, nums: List[int]):\n",
-        "        # Initialize the KthLargest object with k and nums.\n",
-        "        self.k = k\n",
-        "        # Create a min-heap to store the k largest elements.\n",
-        "        self.min_heap = []\n",
-        "        # Populate the min-heap with the first k elements from nums.\n",
-        "        for num in nums:\n",
-        "            self.add(num)\n",
-        "\n",
-        "    def add(self, val: int) -> int:\n",
-        "        # Add val to the min-heap.\n",
-        "        heapq.heappush(self.min_heap, val)\n",
-        "        # If the size of the min-heap exceeds k, remove the smallest element.\n",
-        "        if len(self.min_heap) > self.k:\n",
-        "            heapq.heappop(self.min_heap)\n",
-        "        # The root of the min-heap is the kth largest element.\n",
-        "        return self.min_heap[0]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- We start by importing the necessary modules:\n",
-        "  - `heapq`: This module provides functions to create and manipulate heaps.\n",
-        "  - `List` from the `typing` module: This is used to specify the type of the `nums` parameter.\n",
-        "\n",
-        "- We define the `KthLargest` class:\n",
-        "  - The `__init__` method initializes the object with the integer `k` and the list of integers `nums`.\n",
-        "  - It also creates an empty min-heap called `self.min_heap` to store the k largest elements.\n",
-        "  - It populates the min-heap with the first `k` elements from `nums` by calling the `add` method.\n",
-        "\n",
-        "- The `add` method:\n",
-        "  - Adds the new integer `val` to the min-heap using `heapq.heappush`. This maintains the min-heap property.\n",
-        "  - Checks if the size of the min-heap exceeds `k`. If it does, it removes the smallest element (the k+1th largest) using `heapq.heappop`.\n",
-        "  - Finally, it returns the smallest element in the min-heap, which is always the kth largest element.\n",
-        "\n",
-        "Overall, this code implements a class that efficiently finds the kth largest element in a stream of integers by maintaining a min-heap of the k largest elements seen so far.\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n",
-            "5\n",
-            "5\n",
-            "8\n",
-            "8\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "kthLargest = KthLargest(3, [4, 5, 8, 2])\n",
-        "print(kthLargest.add(3))  # Output: 4\n",
-        "print(kthLargest.add(5))  # Output: 5\n",
-        "print(kthLargest.add(10)) # Output: 5\n",
-        "print(kthLargest.add(9))  # Output: 8\n",
-        "print(kthLargest.add(4))  # Output: 8"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `KthLargest` class using the min-heap approach:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. `__init__` method:\n",
-        "   - In the `__init__` method, you iterate over the first `k` elements in `nums` and add them to the min-heap. Each `heapq.heappush` operation takes O(log k) time.\n",
-        "   - Therefore, the time complexity of the `__init__` method is O(k * log k).\n",
-        "\n",
-        "2. `add` method:\n",
-        "   - In the `add` method, you perform the following operations:\n",
-        "     - `heapq.heappush`: O(log k) to add the new element to the min-heap.\n",
-        "     - If the size of the min-heap exceeds `k`, you perform `heapq.heappop`, which is also O(log k).\n",
-        "     - Finally, you return the smallest element from the min-heap, which is O(1) because it's always the root.\n",
-        "   - Overall, the time complexity of the `add` method is O(log k).\n",
-        "\n",
-        "3. Overall, if you make `n` calls to the `add` method, the total time complexity is O(n * log k), where `n` is the total number of elements added to the stream.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The space complexity is determined by the space used to store the min-heap and the instance variables.\n",
-        "2. The min-heap stores at most `k` elements at any given time, so its space complexity is O(k).\n",
-        "3. The instance variables (such as `self.k` and `self.min_heap`) have constant space requirements.\n",
-        "4. Therefore, the overall space complexity of the `KthLargest` class is O(k).\n",
-        "\n",
-        "In summary, the time complexity of the `KthLargest` class is O(n * log k) for `n` add operations, and the space complexity is O(k), where `k` is the parameter passed during initialization. This implementation efficiently maintains the kth largest element in the stream while meeting the problem's constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Dynamic k:** Modify the `KthLargest` class to support dynamic changes in the value of `k`. Implement a method `update_k` that allows changing the value of `k` during the lifetime of the object. Ensure that the object can still correctly find the kth largest element based on the updated value of `k`.\n",
-        "\n",
-        "\n",
-        "2. **Implement kth Smallest:** Create a new class called `KthSmallest` that finds the kth smallest element in a stream instead of the kth largest. You may need to modify the data structure used in the implementation."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/09. Heap - Priority Queue/README.md b/docs/09. Heap - Priority Queue/README.md
deleted file mode 100644
index 2eac70c..0000000
--- a/docs/09. Heap - Priority Queue/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Heap / Priority Queue Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [703. Kth Largest Element in a Stream](https://leetcode.com/problems/kth-largest-element-in-a-stream/)     | Easy       |
-| [1046. Last Stone Weight](https://leetcode.com/problems/last-stone-weight/)     | Easy       |
-| [295. Find Median From Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)     | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/10. Backtracking/39. Combination Sum.ipynb b/docs/10. Backtracking/39. Combination Sum.ipynb
deleted file mode 100644
index 228a62e..0000000
--- a/docs/10. Backtracking/39. Combination Sum.ipynb	
+++ /dev/null
@@ -1,282 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 39. Combination Sum\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Combination Sum problem on LeetCode, click here!](https://leetcode.com/problems/combination-sum/)\n",
-        "\n",
-        "---\n",
-        "Given an array of **distinct** integers `candidates` and a target integer `target`, return *a list of all **unique combinations** of `candidates` where the chosen numbers sum to the `target`*. You may return the combinations in **any order**. \n",
-        "\n",
-        "The **same** number may be chosen from `candidates` an **unlimited number** of times. Two combinations are unique if the **frequency** of at least one of the chosen numbers is different.\n",
-        "\n",
-        "The test cases are generated such that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= candidates.length <= 30`\n",
-        "- `2 <= candidates[i] <= 40`\n",
-        "- All elements of `candidates` are **distinct**.\n",
-        "- `1 <= target <= 40`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to find all unique combinations of numbers from a given array (candidates) such that their sum equals a target number. Here are the details of the problem:\n",
-        "\n",
-        "- You are given an array of distinct integers called \"candidates.\"\n",
-        "- You are also given a target integer called \"target.\"\n",
-        "\n",
-        "The goal is to find all unique combinations of numbers from the candidates array where the sum of the selected numbers is equal to the target. You can use the same number from the candidates array an unlimited number of times. A combination is considered unique if it has a different frequency (i.e., a different number of occurrences) of at least one chosen number compared to other combinations.\n",
-        "\n",
-        "For example:\n",
-        "\n",
-        "**Example 1:**\n",
-        "```python\n",
-        "Input: candidates = [2, 3, 6, 7], target = 7\n",
-        "Output: [[2, 2, 3], [7]]\n",
-        "```\n",
-        "In this example, there are two unique combinations that sum up to the target:\n",
-        "- 2 + 2 + 3 = 7\n",
-        "- 7 = 7\n",
-        "\n",
-        "**Example 2:**\n",
-        "```python\n",
-        "Input: candidates = [2, 3, 5], target = 8\n",
-        "Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n",
-        "```\n",
-        "Here, there are three unique combinations that sum up to the target.\n",
-        "\n",
-        "**Example 3:**\n",
-        "```python\n",
-        "Input: candidates = [2], target = 1\n",
-        "Output: []\n",
-        "```\n",
-        "In this case, there are no combinations that can be formed from the candidates to reach the target of 1, so the output is an empty list.\n",
-        "\n",
-        "The problem asks you to find and return these combinations in any order.\n",
-        "\n",
-        "The constraints for this problem include the length of the candidates array, the values of candidates, and the target value.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def combinationSum(candidates, target):\n",
-        "    # Define a recursive DFS function to find combinations\n",
-        "    def dfs(remaining, start, path):\n",
-        "        # If the remaining target is 0, we found a valid combination\n",
-        "        if remaining == 0:\n",
-        "            result.append(path)\n",
-        "            return\n",
-        "        # If the remaining target is negative, this path is invalid\n",
-        "        if remaining < 0:\n",
-        "            return\n",
-        "        # Iterate through candidates starting from 'start' to avoid duplicates\n",
-        "        for i in range(start, len(candidates)):\n",
-        "            # Explore the current candidate by subtracting it from the remaining target\n",
-        "            # Add the current candidate to the path\n",
-        "            dfs(remaining - candidates[i], i, path + [candidates[i]])\n",
-        "\n",
-        "    result = []  # Initialize an empty list to store the result\n",
-        "    candidates.sort()  # Sort the candidates for deduplication and early stopping\n",
-        "    dfs(target, 0, [])  # Start the DFS from the target value with an empty path\n",
-        "    return result  # Return the list of unique combinations"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `combinationSum` function takes an array of distinct integers `candidates` and a target integer `target`. It returns a list of all unique combinations of `candidates` where the chosen numbers sum to the `target`. Each number in the `candidates` array can be used an unlimited number of times.\n",
-        "\n",
-        "Here's an overview of how the code works:\n",
-        "\n",
-        "1. The `combinationSum` function is defined, and it takes two arguments: `candidates` and `target`.\n",
-        "\n",
-        "2. Inside the function, there is a nested helper function called `backtrack`. This function is responsible for the actual combination generation using a recursive approach.\n",
-        "\n",
-        "3. The `backtrack` function is called with three arguments: `start`, `target`, and `path`. The `start` variable helps keep track of the current index in the `candidates` array, `target` keeps track of the remaining sum to be achieved, and `path` is a list that stores the current combination.\n",
-        "\n",
-        "4. Within the `backtrack` function, there are three main conditional statements:\n",
-        "   - If `target` becomes zero, it means we have found a combination that sums up to the target, so we add `path` to the `result` list.\n",
-        "   - If `target` becomes negative, it means the current combination doesn't work, so we return without doing anything.\n",
-        "   - Otherwise, we enter a loop that iterates over the `candidates` array, starting from the current index `start`.\n",
-        "\n",
-        "5. In the loop, the `backtrack` function is called recursively with the updated `target` and `path` after including the current candidate. This process explores different combinations by considering the current candidate or moving to the next candidate.\n",
-        "\n",
-        "6. After the loop completes, the `result` list contains all unique combinations that sum to the target.\n",
-        "\n",
-        "7. The `candidates` array is sorted to optimize the search. Sorting helps in avoiding unnecessary recursive branches and reduces the number of explored combinations.\n",
-        "\n",
-        "8. Finally, the `backtrack` function is initially called from the `combinationSum` function with `start` set to 0, `target` set to the original target value, and an empty `path`. The result is returned as a list of unique combinations.\n",
-        "\n",
-        "The code includes a few example cases to demonstrate how the function works with different inputs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[2, 2, 3], [7]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example 1\n",
-        "candidates1 = [2, 3, 6, 7]\n",
-        "target1 = 7\n",
-        "print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "candidates2 = [2, 3, 5]\n",
-        "target2 = 8\n",
-        "print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "candidates3 = [2]\n",
-        "target3 = 1\n",
-        "print(combinationSum(candidates3, target3))  # Output: []"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `combinationSum` function.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this function is influenced by the number of recursive calls made by the `backtrack` function and the amount of work done within each call. In the worst case, the function will explore all possible combinations.\n",
-        "\n",
-        "The worst-case scenario occurs when we have many combinations to reach the target. The time complexity can be expressed as $O(2^n)$, where 'n' is the maximum number of recursive calls. In this context, 'n' corresponds to the number of elements in the `candidates` list, and it may be up to 30 (as per the constraints). However, the actual number of combinations explored is generally much smaller because we eliminate branches when the target becomes negative, and we skip over candidates that cannot lead to a solution. The sorting step also helps to optimize the search.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity is determined by the auxiliary space used for storing the results and the stack space for the recursive calls.\n",
-        "\n",
-        "1. The space required for the `result` list can be up to $O(2^n)$ in the worst case because each combination is stored.\n",
-        "\n",
-        "2. The stack space for the recursive calls can also be up to O(n) in the worst case, where 'n' is the number of elements in the `candidates` list.\n",
-        "\n",
-        "Therefore, the overall space complexity of the function is $O(2^n + n)$. However, in practice, the space used for the results list is often the dominant factor, and the actual space used may be much smaller than 2^n because not all combinations are explored.\n",
-        "\n",
-        "In summary, the time complexity is exponential but typically smaller in practice due to optimization, and the space complexity is influenced by the number of results and the depth of the recursion."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Find the Number of Unique Combinations:** Instead of listing all unique combinations, modify the function to return the count of unique combinations that sum to the target.\n",
-        "\n",
-        "2. **Combinations with a Maximum Value:** Add a maximum value constraint for each combination, so they cannot exceed a certain value. Modify the code to find combinations respecting this constraint."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/10. Backtracking/79. Word Search.ipynb b/docs/10. Backtracking/79. Word Search.ipynb
deleted file mode 100644
index cee509b..0000000
--- a/docs/10. Backtracking/79. Word Search.ipynb	
+++ /dev/null
@@ -1,298 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 79. Word Search\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Word Search problem on LeetCode, click here!](https://leetcode.com/problems/word-search/)\n",
-        "\n",
-        "---\n",
-        "Given an `m x n` grid of characters `board` and a string `word`, return *`true` if `word` exists in the grid.*\n",
-        "\n",
-        "The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == board.length`\n",
-        "- `n = board[i].length`\n",
-        "- `1 <= m, n <= 6`\n",
-        "- `1 <= word.length <= 15`\n",
-        "- `board` and `word` consists of only lowercase and uppercase English letters.\n",
-        "\n",
-        "**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether a given word can be constructed from characters in an `m x n` grid of characters. The word can be formed by moving sequentially through horizontally or vertically neighboring cells in the grid, and you cannot use the same cell more than once.\n",
-        "\n",
-        "Here are the specifics of the problem:\n",
-        "\n",
-        "- You are given an `m x n` grid, which is a 2D board of characters.\n",
-        "- You are also given a string `word` consisting of lowercase and uppercase English letters.\n",
-        "- Your task is to determine if the word can be found in the grid by moving from one cell to an adjacent cell (horizontally or vertically) and collecting the characters sequentially to form the word.\n",
-        "- You can't use the same cell more than once when forming the word.\n",
-        "- If it's possible to form the word in the grid, return `True`; otherwise, return `False`.\n",
-        "\n",
-        "For example, consider the following grid and word:\n",
-        "\n",
-        "```python\n",
-        "board = [[\"A\", \"B\", \"C\", \"E\"],\n",
-        "         [\"S\", \"F\", \"C\", \"S\"],\n",
-        "         [\"A\", \"D\", \"E\", \"E\"]]         \n",
-        "word = \"ABCCED\"\n",
-        "```\n",
-        "\n",
-        "In this case, the word \"ABCCED\" can be formed by moving through the cells (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) in the grid. Therefore, the function should return `True`.\n",
-        "\n",
-        "Conversely, if the word is impossible to construct, such as in the case of the word \"ABCB\" for the same grid, where there's no valid path that forms the word, the function should return `False`.\n",
-        "\n",
-        "The problem involves searching for a path through the grid that matches the characters in the word while respecting the constraints mentioned. It can be solved using depth-first search (DFS) or backtracking.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def exist(self, board: List[List[str]], word: str) -> bool:\n",
-        "        def dfs(i, j, k):\n",
-        "            # Check if the current position is out of bounds or doesn't match the character in the word.\n",
-        "            if not (0 <= i < len(board)) or not (0 <= j < len(board[0])) or board[i][j] != word[k]:\n",
-        "                return False\n",
-        "            \n",
-        "            # If we have matched all characters in the word, return True.\n",
-        "            if k == len(word) - 1:\n",
-        "                return True\n",
-        "\n",
-        "            # Temporarily mark the current cell to prevent revisiting.\n",
-        "            tmp, board[i][j] = board[i][j], \"/\"\n",
-        "\n",
-        "            # Explore adjacent cells by making recursive calls in all four directions.\n",
-        "            found = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)\n",
-        "\n",
-        "            # Restore the original character in the cell.\n",
-        "            board[i][j] = tmp\n",
-        "\n",
-        "            return found\n",
-        "\n",
-        "        # Iterate through all cells in the board.\n",
-        "        for i in range(len(board)):\n",
-        "            for j in range(len(board[0])):\n",
-        "                # Check if the 'dfs' function starting from this cell returns True.\n",
-        "                if dfs(i, j, 0):\n",
-        "                    return True\n",
-        "\n",
-        "        # If no match is found after exploring all cells, return False.\n",
-        "        return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. Import the necessary module:\n",
-        "   - `from typing import List`: This line imports the `List` type from the `typing` module, which is used to specify the type of the `board` parameter in the method signature.\n",
-        "\n",
-        "2. Define the `Solution` class:\n",
-        "   - `class Solution:`: This line defines a class called `Solution` to encapsulate the solution for the word search problem.\n",
-        "\n",
-        "3. Define the `exist` method:\n",
-        "   - `def exist(self, board: List[List[str]], word: str) -> bool:`: This method is a member of the `Solution` class and takes two parameters: `board`, which is a 2D list of characters, and `word`, which is a string. It returns a boolean value, indicating whether the word exists in the board.\n",
-        "\n",
-        "4. Implement the inner `dfs` function:\n",
-        "   - `def dfs(i, j, k):`: This is a helper function used for depth-first search (DFS). It is called recursively to explore the board and search for the word. It takes three parameters: `i` and `j` representing the current position on the board, and `k` representing the current index in the `word` string.\n",
-        "\n",
-        "   - Inside `dfs`, it checks if the current position is out of bounds or if the character at the current position on the board does not match the character at the current index in the `word`. If either of these conditions is met, it returns `False`.\n",
-        "\n",
-        "   - If `k` is equal to the length of the `word` minus one, it means we have successfully matched all characters in the `word`, so it returns `True`.\n",
-        "\n",
-        "   - The function then temporarily replaces the character at the current position with a marker, \"/\", to prevent revisiting the same cell. It then explores adjacent cells by making recursive calls in all four directions: up, down, left, and right. If any of these recursive calls return `True`, it propagates the `True` result up the call stack. After the exploration is complete, it restores the original character in the cell.\n",
-        "\n",
-        "5. Iterate through the board:\n",
-        "   - The outer loop iterates through all cells in the `board` using two nested loops. For each cell, it checks if the `dfs` function starting from that cell (with `k` initially set to 0) returns `True`. If it does, it means the word exists, so the method returns `True`.\n",
-        "\n",
-        "6. Return `False`:\n",
-        "   - If the loop completes without finding the word in the board, the method returns `False`.\n",
-        "\n",
-        "To use this code, you can create an instance of the `Solution` class and call the `exist` method on that instance, passing the `board` and the `word` as arguments to check if the word exists in the board."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test case 1: Word \"ABCCED\" can be found in the grid\n",
-        "board1 = [[\"A\",\"B\",\"C\",\"E\"],\n",
-        "          [\"S\",\"F\",\"C\",\"S\"],\n",
-        "          [\"A\",\"D\",\"E\",\"E\"]]\n",
-        "word1 = \"ABCCED\"\n",
-        "print(solution.exist(board1, word1))  # Expected output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "\n",
-        "# Test case 2: Word \"SEE\" can be found in the grid\n",
-        "word2 = \"SEE\"\n",
-        "print(solution.exist(board1, word2))  # Expected output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "\n",
-        "# Test case 3: Word \"ABCB\" cannot be found in the grid\n",
-        "word3 = \"ABCB\"\n",
-        "print(solution.exist(board1, word3))  # Expected output: False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `exist` method in the given code.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The primary work in this code is done by the `dfs` function, which performs a depth-first search through the grid. The time complexity of the `dfs` function is $O(4^(n))$, where \"n\" is the length of the `word`. This is because, in the worst case, the function explores all possible directions (up, down, left, right) for each character in the `word`.\n",
-        "\n",
-        "The outer loops iterate through all cells in the `board`, which has dimensions `m x n`. So, the total number of iterations of `dfs` is O(m * n). However, since the `dfs` function has a higher time complexity, the overall time complexity is still dominated by the `dfs` function.\n",
-        "\n",
-        "Therefore, the overall time complexity of the `exist` method is $O(m * n * 4^n)$, where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The primary space usage comes from the recursive calls in the `dfs` function. The depth of the recursion can be at most equal to the length of the `word`, which is O(n).\n",
-        "\n",
-        "2. Additionally, the `tmp` variable and the recursive calls on the call stack consume a small amount of additional memory.\n",
-        "\n",
-        "3. The marking of visited cells with the \"/\" character temporarily modifies the `board` in-place but doesn't significantly contribute to the space complexity.\n",
-        "\n",
-        "Therefore, the overall space complexity is O(n) due to the recursive call stack.\n",
-        "\n",
-        "In summary, the time complexity is $O(m * n * 4^n)$, and the space complexity is O(n), where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word. The time complexity is exponential in the worst case due to the depth-first search through all possible paths in the grid."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the code to find all unique words that can be constructed from the grid. Instead of returning `True` or `False`, return a list of words found. Ensure that the same cell is not used more than once in constructing each word.\n",
-        "\n",
-        "2. **Board Generation:** Create a function to generate random grids of characters and then use the `exist` method to solve them. Test the efficiency and correctness of your code by generating larger boards and words."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/10. Backtracking/README.md b/docs/10. Backtracking/README.md
deleted file mode 100644
index 0d1ea13..0000000
--- a/docs/10. Backtracking/README.md	
+++ /dev/null
@@ -1,17 +0,0 @@
-# Backtracking Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [39. Combination Sum](https://leetcode.com/problems/combination-sum/)        | Medium     |
-| [79. Word Search](https://leetcode.com/problems/word-search/)                | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/11. Graphs/133. Clone Graph.ipynb b/docs/11. Graphs/133. Clone Graph.ipynb
deleted file mode 100644
index 07351b6..0000000
--- a/docs/11. Graphs/133. Clone Graph.ipynb	
+++ /dev/null
@@ -1,317 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 133. Clone Graph\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Clone Graph problem on LeetCode, click here!](https://leetcode.com/problems/clone-graph/)\n",
-        "\n",
-        "---\n",
-        "Given a reference of a node in a **connected** undirected graph.\n",
-        "\n",
-        "Return a **deep copy** (clone) of the graph.\n",
-        "\n",
-        "Each node in the graph contains a value (`int`) and a list (`List[Node]`) of its neighbors.\n",
-        "\n",
-        "```\n",
-        "class Node {\n",
-        "    public int val;\n",
-        "    public List<Node> neighbors;\n",
-        "}\n",
-        "```\n",
-        "\n",
-        "**Test case format:**\n",
-        "\n",
-        "For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with `val == 1`, the second node with `val == 2`, and so on. The graph is represented in the test case using an adjacency list.\n",
-        "\n",
-        "**An adjacency list** is a collection of unordered **lists** used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.\n",
-        "\n",
-        "The given node will always be the first node with `val = 1`. You must return the **copy of the given node** as a reference to the cloned graph.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the graph is in the range `[0, 100]`.\n",
-        "- `1 <= Node.val <= 100`\n",
-        "- `Node.val` is unique for each node.\n",
-        "- There are no repeated edges and no self-loops in the graph.\n",
-        "- The Graph is connected and all nodes can be visited starting from the given node."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to clone a connected undirected graph, which means we need to create a deep copy of the original graph while preserving its structure and relationships between nodes. The graph is represented using adjacency lists, where each node has a value (an integer) and a list of its neighbors.\n",
-        "\n",
-        "Here are the key aspects of the problem:\n",
-        "\n",
-        "1. **Input**: You are given a reference to one of the nodes in the original graph. This reference node serves as the entry point to the graph. The entire graph can be explored starting from this node.\n",
-        "\n",
-        "2. **Output**: The goal is to create a deep copy of the graph and return a reference to the cloned graph. The cloned graph should have the same structure and relationships as the original one but should be a separate instance in memory.\n",
-        "\n",
-        "3. **Graph Structure**: The graph is composed of nodes (vertices) and edges. Nodes have values (integers) and are connected to other nodes through edges (undirected connections). Each node has a list of neighbors, which are other nodes it is connected to.\n",
-        "\n",
-        "4. **Connected Graph**: The problem assumes that the graph is connected, meaning that you can start from the reference node provided and traverse the entire graph by following the edges. This ensures that there are no isolated components in the graph.\n",
-        "\n",
-        "To solve this problem, you typically need to perform a graph traversal (e.g., depth-first search or breadth-first search) to visit all the nodes in the original graph, create corresponding nodes in the clone, and establish the same connections between nodes in the cloned graph as in the original one. To avoid duplication of nodes, you also need to keep track of visited nodes to ensure that the cloning process is efficient and that nodes are not duplicated in the clone.\n",
-        "\n",
-        "The problem can be challenging because you need to create a deep copy of the graph while handling cyclic dependencies between nodes and ensuring that the structure and relationships of the original graph are preserved in the clone. The solution requires recursion or a data structure to maintain the mapping between original nodes and their clones to achieve an efficient and accurate cloning process.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a Node.\n",
-        "class Node:\n",
-        "    def __init__(self, val=0, neighbors=None):\n",
-        "        self.val = val\n",
-        "        self.neighbors = neighbors if neighbors is not None else []\n",
-        "\n",
-        "class Solution:\n",
-        "    def cloneGraph(self, node: 'Node') -> 'Node':\n",
-        "        if not node:\n",
-        "            return None\n",
-        "\n",
-        "        visited = {}  # Dictionary to keep track of visited nodes\n",
-        "\n",
-        "        def dfs(original_node):\n",
-        "            if original_node in visited:\n",
-        "                return visited[original_node]\n",
-        "\n",
-        "            # Create a copy of the original node\n",
-        "            copy_node = Node(original_node.val)\n",
-        "            visited[original_node] = copy_node\n",
-        "\n",
-        "            # Recursively clone the neighbors of the original node\n",
-        "            for neighbor in original_node.neighbors:\n",
-        "                copy_node.neighbors.append(dfs(neighbor))\n",
-        "\n",
-        "            return copy_node\n",
-        "\n",
-        "        return dfs(node)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The given code defines a Python class, `Node`, and a solution class, `Solution`, to clone a connected undirected graph represented by nodes and their neighbors. Here's an explanation of the code:\n",
-        "\n",
-        "1. The `Node` class defines a node in the graph. Each node has a value (`val`) which is an integer and a list of neighbors (`neighbors`) that are other nodes in the graph.\n",
-        "\n",
-        "2. The `Solution` class contains the method `cloneGraph`, which takes a reference to the first node of the graph (`node`) as input and returns a deep copy of the entire graph, preserving the structure and relationships between nodes.\n",
-        "\n",
-        "3. Inside the `cloneGraph` method:\n",
-        "   - It first checks if the input `node` is `None`. If it's `None`, it means there is no graph to clone, and the method returns `None`.\n",
-        "\n",
-        "   - It initializes a `visited` dictionary to keep track of visited nodes. This dictionary is used to ensure that each node is cloned only once to avoid duplication.\n",
-        "\n",
-        "   - The `dfs` (depth-first search) function is defined within the `cloneGraph` method. It takes an original node as input and returns its corresponding clone. This function performs the actual cloning of the graph.\n",
-        "\n",
-        "   - Inside the `dfs` function:\n",
-        "     - It checks if the original node has been visited before by looking it up in the `visited` dictionary. If it has been visited, it returns the corresponding clone from the `visited` dictionary.\n",
-        "\n",
-        "     - If the original node is not visited, it creates a new node, `copy_node`, with the same value as the original node. This is the first step in creating the clone of the original node.\n",
-        "\n",
-        "     - It then recursively clones the neighbors of the original node by iterating through the `neighbors` list of the original node and appending the corresponding cloned neighbors to the `neighbors` list of the `copy_node`. This step ensures that the relationships between nodes are preserved in the clone.\n",
-        "\n",
-        "     - The `copy_node` is added to the `visited` dictionary with the original node as the key and the clone as the value.\n",
-        "\n",
-        "     - Finally, the `copy_node` is returned as the result of the DFS for the original node.\n",
-        "\n",
-        "4. The `cloneGraph` method returns the result of the DFS on the input `node`, which is the clone of the entire graph with its structure and relationships intact.\n",
-        "\n",
-        "In summary, the code uses depth-first search (DFS) to traverse the original graph, creating a clone for each node and its neighbors. It ensures that each node is cloned only once to avoid duplication and returns a reference to the cloned graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1, 2, 4]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "def get_adjacency_list(node):\n",
-        "    visited = {}  # Dictionary to keep track of visited nodes\n",
-        "\n",
-        "    def dfs(node):\n",
-        "        if not node:\n",
-        "            return []\n",
-        "\n",
-        "        if node in visited:\n",
-        "            return visited[node]\n",
-        "\n",
-        "        neighbors = [n.val for n in node.neighbors]\n",
-        "        visited[node] = neighbors\n",
-        "\n",
-        "        for neighbor in node.neighbors:\n",
-        "            dfs(neighbor)\n",
-        "\n",
-        "    dfs(node)\n",
-        "    result = []\n",
-        "\n",
-        "    def get_node_and_neighbors(node):\n",
-        "        result.append([node.val] + visited[node])\n",
-        "\n",
-        "    get_node_and_neighbors(node)\n",
-        "    return result\n",
-        "\n",
-        "\n",
-        "# Testing the code with the provided examples\n",
-        "adjList1 = [[2, 4], [1, 3], [2, 4], [1, 3]]\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Create the graph from adjList\n",
-        "nodes = [Node(i) for i in range(1, len(adjList1) + 1)]\n",
-        "for i, adj in enumerate(adjList1):\n",
-        "    nodes[i].neighbors = [nodes[j - 1] for j in adj]\n",
-        "\n",
-        "# Clone the graph and get its adjacency list\n",
-        "cloned_node1 = solution.cloneGraph(nodes[0])\n",
-        "cloned_adjList1 = get_adjacency_list(cloned_node1)\n",
-        "\n",
-        "# Print the cloned graph as an adjacency list\n",
-        "print(cloned_adjList1)\n",
-        "\n",
-        "# Output should match adjList1\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "# Input: adjList = [[]]\n",
-        "\n",
-        "adjList2 = [[]]\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Create the graph from adjList\n",
-        "node1 = Node(1)\n",
-        "\n",
-        "# Clone the graph and get its adjacency list\n",
-        "cloned_node1 = solution.cloneGraph(node1)\n",
-        "cloned_adjList2 = get_adjacency_list(cloned_node1)\n",
-        "\n",
-        "# Print the cloned graph as an adjacency list\n",
-        "print(cloned_adjList2)\n",
-        "\n",
-        "# Output should be [[]], which represents an empty graph"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for cloning a graph using depth-first search (DFS).\n",
-        "\n",
-        "Time Complexity:\n",
-        "1. Creating the graph: The time complexity to create the original graph from `adjList` is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. We iterate through each node and its neighbors in `adjList`.\n",
-        "\n",
-        "2. Cloning the graph (DFS): The depth-first search (DFS) is used to traverse the original graph and create a clone. The time complexity of DFS is O(V + E), as we visit each node and each edge once.\n",
-        "\n",
-        "Overall, the time complexity is O(V + E) for both creating the original graph and cloning it.\n",
-        "\n",
-        "Space Complexity:\n",
-        "1. Storage for the original graph: We need to store the original graph, which includes all nodes and their adjacency lists. In the worst case, this requires O(V + E) space, where V is the number of nodes, and E is the number of edges.\n",
-        "\n",
-        "2. Storage for the visited dictionary: The `visited` dictionary stores mappings from original nodes to their corresponding clones. In the worst case, this requires O(V) space because each node is visited once.\n",
-        "\n",
-        "3. Recursive stack for DFS: The space used for the recursive call stack during DFS can be up to O(V) in the worst case, where V is the number of nodes.\n",
-        "\n",
-        "Overall, the space complexity is O(V + E) for storing the original graph, and O(V) for auxiliary data structures, including the `visited` dictionary and the DFS call stack. Therefore, the overall space complexity is O(V + E).\n",
-        "\n",
-        "In summary, the time and space complexities of the provided code for cloning a graph are both O(V + E), where V is the number of nodes, and E is the number of edges in the graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Cyclic Graph**: Design a graph with cycles (nodes connected in a cycle). Test if the code can correctly clone graphs with cycles and doesn't fall into an infinite loop during traversal.\n",
-        "\n",
-        "\n",
-        "2. **Partial Graph Cloning**: Modify the code to clone only a subset of the graph starting from a given reference node, rather than the entire graph. Implement this as an extension to the problem."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/11. Graphs/178. Graph Valid Tree.ipynb b/docs/11. Graphs/178. Graph Valid Tree.ipynb
deleted file mode 100644
index e0235d6..0000000
--- a/docs/11. Graphs/178. Graph Valid Tree.ipynb	
+++ /dev/null
@@ -1,248 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 178. Graph Valid Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Graph Valid Tree problem on LintCode, click here!](https://www.lintcode.com/problem/178/)\n",
-        "\n",
-        "---\n",
-        "Given `n` nodes labeled from `0` to `n - 1` and a list of `undirected` edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- You can assume that no duplicate edges will appear in edges. Since all edges are `undirected`, `[0, 1]` is the same as `[1, 0]` and thus will not appear together in edges."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether a given set of undirected edges can form a valid tree. In this context, a \"tree\" is a specific type of graph with the following properties:\n",
-        "\n",
-        "1. It is a connected graph, meaning there is a path between any pair of nodes in the graph.\n",
-        "2. It has no cycles, which means there are no closed loops or circuits in the graph.\n",
-        "3. It has exactly `n - 1` edges, where `n` is the number of nodes in the tree.\n",
-        "\n",
-        "The problem is typically given with two pieces of information:\n",
-        "\n",
-        "1. The number of nodes (`n`), which are usually labeled from 0 to `n-1`.\n",
-        "2. A list of undirected edges, where each edge is represented as a pair of nodes.\n",
-        "\n",
-        "The task is to write a function or algorithm to determine whether the provided set of edges can be arranged to form a valid tree based on the properties mentioned above.\n",
-        "\n",
-        "To solve this problem, you need to check the following conditions:\n",
-        "\n",
-        "1. There should be exactly `n - 1` edges to ensure that the graph is connected and tree-like.\n",
-        "2. There should be no cycles in the graph. If there are any cycles, the set of edges cannot form a tree.\n",
-        "3. The graph should be connected, which means you can reach any node from any other node through a series of edges.\n",
-        "\n",
-        "The solution typically involves using graph traversal algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the graph and check these conditions. If all conditions are met, the given edges form a valid tree; otherwise, they do not.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def validTree(n, edges):\n",
-        "    if len(edges) != n - 1:\n",
-        "        return False  # A tree with n nodes must have n-1 edges.\n",
-        "\n",
-        "    # Create an adjacency list to represent the graph.\n",
-        "    adj_list = {i: [] for i in range(n)}\n",
-        "    for u, v in edges:\n",
-        "        adj_list[u].append(v)\n",
-        "        adj_list[v].append(u)\n",
-        "\n",
-        "    visited = set()\n",
-        "\n",
-        "    # Define a recursive DFS function.\n",
-        "    def dfs(node, parent):\n",
-        "        if node in visited:\n",
-        "            return False  # If we encounter a visited node, there's a cycle.\n",
-        "        visited.add(node)\n",
-        "\n",
-        "        for neighbor in adj_list[node]:\n",
-        "            if neighbor != parent:\n",
-        "                if not dfs(neighbor, node):\n",
-        "                    return False\n",
-        "\n",
-        "        return True\n",
-        "\n",
-        "    # Start DFS from any node (e.g., node 0).\n",
-        "    if not dfs(0, -1):\n",
-        "        return False  # If there's a cycle, it's not a valid tree.\n",
-        "\n",
-        "    return len(visited) == n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `validTree` function checks if a given set of undirected edges forms a valid tree. A valid tree must satisfy these conditions:\n",
-        "\n",
-        "1. It has exactly `n - 1` edges, where `n` is the number of nodes.\n",
-        "2. There are no cycles in the graph.\n",
-        "3. The graph is connected, meaning all nodes are reachable from any starting node.\n",
-        "\n",
-        "The code accomplishes this as follows:\n",
-        "\n",
-        "- It first checks if the number of edges is equal to `n - 1`. If not, it immediately returns `False` since a tree with `n` nodes must have `n - 1` edges to be connected.\n",
-        "\n",
-        "- It then constructs an adjacency list (`adj_list`) to represent the graph efficiently. This data structure stores the neighbors of each node.\n",
-        "\n",
-        "- The code uses a `visited` set to keep track of nodes visited during depth-first search (DFS).\n",
-        "\n",
-        "- The `dfs` function is a recursive function that checks for cycles in the graph. It does this by visiting nodes and marking them as visited. If it encounters a node that has already been visited (indicating a cycle), it returns `False`.\n",
-        "\n",
-        "- The DFS function explores the neighbors of each node, avoiding revisiting the parent node to prevent cycles.\n",
-        "\n",
-        "- After the DFS is complete, the code checks if the number of visited nodes is equal to `n`, ensuring the graph is connected.\n",
-        "\n",
-        "- Finally, if all conditions are met, the code returns `True`, indicating that the given edges form a valid tree. Otherwise, it returns `False`.\n",
-        "\n",
-        "The code provides examples to demonstrate how to use the `validTree` function to check whether a set of edges forms a valid tree or not."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "n1 = 5\n",
-        "edges1 = [[0, 1], [0, 2], [0, 3], [1, 4]]\n",
-        "print(validTree(n1, edges1))  # Output: true"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "n2 = 5\n",
-        "edges2 = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]\n",
-        "print(validTree(n2, edges2))  # Output: false"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexities of the `validTree` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. Constructing the adjacency list (`adj_list`) takes O(E) time, where E is the number of edges in the input graph (|edges|).\n",
-        "\n",
-        "2. The DFS function, when considering all nodes, has a time complexity of O(V + E), where V is the number of nodes in the graph. This is because, in the worst case, the DFS function visits each node and each edge once.\n",
-        "\n",
-        "3. Checking if the number of visited nodes is equal to `n` takes O(1) time.\n",
-        "\n",
-        "Overall, the time complexity of the `validTree` function is O(E + V) in the worst case.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The adjacency list (`adj_list`) consumes additional space to store the graph structure, taking O(V + E) space. This is because it stores information about each node and its adjacent nodes.\n",
-        "\n",
-        "2. The `visited` set keeps track of visited nodes, and its space complexity is O(V) in the worst case, as there can be at most V unique nodes in the set.\n",
-        "\n",
-        "3. The depth of the function call stack during DFS can be at most V, which contributes O(V) space to the space complexity.\n",
-        "\n",
-        "Overall, the space complexity of the `validTree` function is O(V + E) due to the adjacency list and O(V) due to the `visited` set and DFS call stack.\n",
-        "\n",
-        "In summary, the time complexity is O(E + V), and the space complexity is O(V + E). These complexities are based on the worst-case scenario, where all nodes and edges are considered. In practical cases, the actual complexities may be smaller, depending on the specific structure of the graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Minimum Spanning Tree (MST)**: Implement an algorithm to find the minimum spanning tree of a connected graph using either Kruskal's or Prim's algorithm. Compare the MST to the original graph to determine if the original graph is a tree. This exercise reinforces the concept of tree properties.\n",
-        "\n",
-        "2. **Detecting Cycles**: Modify the code to not only check if the input forms a tree but also to identify and print out any cycles present in the graph if it's not a tree. This exercise enhances your ability to detect and visualize cycles in a graph."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/11. Graphs/200. Number of Islands.ipynb b/docs/11. Graphs/200. Number of Islands.ipynb
deleted file mode 100644
index a8a7727..0000000
--- a/docs/11. Graphs/200. Number of Islands.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 200. Number of Islands\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Number of Islands problem on LeetCode, click here!](https://leetcode.com/problems/number-of-islands/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an `m x n` 2D binary grid `grid` which represents a map of `'1'`s (land) and `'0'`s (water), return *the number of islands*. \n",
-        "\n",
-        "An **island** is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == grid.length`\n",
-        "- `n == grid[i].length`\n",
-        "- `1 <= m, n <= 300`\n",
-        "- `grid[i][j]` is `'0'` or `'1'`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is about counting the number of islands in a 2D binary grid. In this grid, '1' represents land, and '0' represents water. An island is defined as a group of '1' cells that are adjacent to each other either horizontally or vertically. It's important to note that diagonal connections do not count. Additionally, it's assumed that all four edges of the grid are surrounded by water, meaning the grid is finite and doesn't \"wrap around.\"\n",
-        "\n",
-        "The objective is to determine how many distinct islands exist in the grid. To clarify, an island consists of connected '1' cells, and these connections can only occur horizontally or vertically. If there are no '1' cells in the grid, the number of islands would be zero.\n",
-        "\n",
-        "The problem often involves using graph traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to identify and count these islands. You start at one '1' cell and explore its adjacent cells to determine the extent of the island. Once you've visited all the '1' cells in an island, you move on to the next unvisited '1' cell and repeat the process until you've counted all the islands in the grid.\n",
-        "\n",
-        "In essence, the problem asks you to analyze the grid to find groups of connected '1' cells and count them as individual islands. The goal is to return the total count of islands in the given grid.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def numIslands(self, grid: List[List[str]]) -> int:\n",
-        "        if not grid:\n",
-        "            return 0\n",
-        "\n",
-        "        m, n = len(grid), len(grid[0])  # Get the dimensions of the grid\n",
-        "        num_islands = 0  # Initialize the count of islands\n",
-        "\n",
-        "        def dfs(row, col):\n",
-        "            if row < 0 or row >= m or col < 0 or col >= n or grid[row][col] == '0':\n",
-        "                return\n",
-        "\n",
-        "            grid[row][col] = '0'  # Mark the current cell as visited by changing it to '0'\n",
-        "            directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Define four possible directions to move\n",
-        "\n",
-        "            for dr, dc in directions:\n",
-        "                # Explore adjacent cells\n",
-        "                dfs(row + dr, col + dc)\n",
-        "\n",
-        "        for i in range(m):\n",
-        "            for j in range(n):\n",
-        "                if grid[i][j] == '1':\n",
-        "                    num_islands += 1  # Increment the island count\n",
-        "                    dfs(i, j)  # Start a DFS from the current '1' cell to explore the entire island\n",
-        "\n",
-        "        return num_islands\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided code defines a Python class called `Solution` with a method called `numIslands`. This class is designed to count the number of islands in a 2D binary grid.\n",
-        "\n",
-        "The `numIslands` method takes a 2D grid as input, where '1' represents land, and '0' represents water. It uses a depth-first search (DFS) approach to identify and count islands. The key steps in the code include:\n",
-        "\n",
-        "1. Checking if the input grid is empty, and if so, returning 0 (indicating there are no islands).\n",
-        "\n",
-        "2. Determining the dimensions of the grid (number of rows and columns).\n",
-        "\n",
-        "3. Initializing a variable `num_islands` to keep track of the count of islands.\n",
-        "\n",
-        "4. Defining a nested function `dfs` to perform depth-first search. This function recursively explores adjacent land cells connected to the current cell and marks them as visited by changing '1' to '0'.\n",
-        "\n",
-        "5. In the `dfs` function, it checks for boundary conditions and whether the current cell is '0'. If these conditions are met, it returns without further exploration.\n",
-        "\n",
-        "6. The `dfs` function explores neighboring cells in all four directions (up, down, left, right) using a loop.\n",
-        "\n",
-        "7. The main loop iterates through each cell of the grid. When a '1' cell is encountered, it increments the `num_islands` count and starts a DFS from that cell to explore the entire island.\n",
-        "\n",
-        "8. Finally, the method returns the total count of islands found in the grid.\n",
-        "\n",
-        "This code encapsulates the solution in a class, allowing you to create an instance of the `Solution` class and call the `numIslands` method on it to count the number of islands in a given grid."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 1 Output: 1\n"
-          ]
-        }
-      ],
-      "source": [
-        "grid1 = [\n",
-        "  [\"1\", \"1\", \"1\", \"1\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"1\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"0\", \"0\", \"0\"]\n",
-        "]\n",
-        "\n",
-        "solution = Solution()\n",
-        "result1 = solution.numIslands(grid1)\n",
-        "print(\"Example 1 Output:\", result1)  # Expected output: 1\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 2 Output: 3\n"
-          ]
-        }
-      ],
-      "source": [
-        "grid2 = [\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"1\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"0\", \"1\", \"1\"]\n",
-        "]\n",
-        "\n",
-        "result2 = solution.numIslands(grid2)\n",
-        "print(\"Example 2 Output:\", result2)  # Expected output: 3\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "Time Complexity:\n",
-        "- The code uses a depth-first search (DFS) to explore the grid and count the number of islands.\n",
-        "- In the worst case, it visits every cell in the grid exactly once.\n",
-        "- The DFS function explores neighboring cells in four directions (up, down, left, right), so it has a time complexity of $O(4^{max(m, n)})$, where m is the number of rows and n is the number of columns.\n",
-        "- Therefore, the overall time complexity of the code is O(m * n), where m is the number of rows, and n is the number of columns.\n",
-        "\n",
-        "Space Complexity:\n",
-        "- The space complexity is determined by the recursive calls and the depth of the DFS stack.\n",
-        "- In the worst case, when the entire grid consists of '1's, the depth of the DFS recursion can be as deep as max(m, n).\n",
-        "- Therefore, the space complexity of the code is O(max(m, n)), representing the maximum depth of the DFS stack.\n",
-        "\n",
-        "In practical terms, for reasonably sized grids, the time and space complexity is linear, and the code should perform well. However, it's important to keep in mind that the worst-case time complexity is O(m * n), and the space complexity can be O(max(m, n))."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Identify Largest Island:** In addition to counting the number of islands, modify the code to identify the largest island in the grid and return its size.\n",
-        "\n",
-        "2. **Optimize for Space:** Modify the solution to reduce its space complexity to O(1) without modifying the input grid. This is a more memory-efficient version of the problem."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/11. Graphs/207. Course Schedule.ipynb b/docs/11. Graphs/207. Course Schedule.ipynb
deleted file mode 100644
index 37624f6..0000000
--- a/docs/11. Graphs/207. Course Schedule.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 207. Course Schedule\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Course Schedule problem on LeetCode, click here!](https://leetcode.com/problems/course-schedule/)\n",
-        "\n",
-        "---\n",
-        "There are a total of `numCourses` courses to take, labeled from 0 to `numCourses - 1`. You are given an array `prerequisites` where `prerequisites[i] = [ai, bi]` indicates that you **must** take course `bi` first if you want to take course `ai`. \n",
-        "\n",
-        "- For example, the pair `[0, 1]`, indicates that to take course `0` you have to first take course `1`.\n",
-        "\n",
-        "Return `true` if you can finish all courses. Otherwise, return `false`.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `numCourses` <= 2000\n",
-        "- 0 <= `prerequisites.length` <= 5000\n",
-        "- `prerequisites[i].length` == 2\n",
-        "- 0 <= `ai`, `bi` < `numCourses`\n",
-        "- All the pairs `prerequisites[i]` are **unique**."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether it's possible to complete a set of courses given a list of prerequisites. You are given the following:\n",
-        "\n",
-        "- `numCourses`: The total number of courses to be taken, labeled from 0 to `numCourses - 1`.\n",
-        "- `prerequisites`: A list of prerequisite courses, where each element `prerequisites[i]` is a pair `[ai, bi]`, indicating that you must take course `bi` before you can take course `ai`.\n",
-        "\n",
-        "The goal is to check whether it's possible to complete all the courses while respecting the prerequisite requirements. In other words, you need to determine if there are any circular dependencies in the course prerequisites that would prevent you from taking all the courses.\n",
-        "\n",
-        "For example:\n",
-        "\n",
-        "- If `numCourses` is 2, and `prerequisites` is `[[1, 0]]`, it means you have two courses, and you must finish course 0 before you can take course 1. In this case, it's possible to complete all courses, so the function should return `True`.\n",
-        "\n",
-        "- If `numCourses` is 2, and `prerequisites` is `[[1, 0], [0, 1]]`, it means you have two courses, but there's a circular dependency where course 1 requires course 0 and course 0 requires course 1. In this case, it's impossible to complete all courses, so the function should return `False`.\n",
-        "\n",
-        "The problem is essentially about checking for the presence of cycles in a directed graph, where the courses are nodes, and the prerequisites represent directed edges between them. If there are no cycles, it's possible to complete all courses; otherwise, it's not possible.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict, deque\n",
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:\n",
-        "        # Create a graph using an adjacency list to represent the courses and their prerequisites.\n",
-        "        graph = defaultdict(list)  # Initialize an empty adjacency list.\n",
-        "        in_degree = [0] * numCourses  # Initialize an array to store in-degrees of courses.\n",
-        "\n",
-        "        # Populate the graph and calculate in-degrees.\n",
-        "        for course, prerequisite in prerequisites:\n",
-        "            graph[prerequisite].append(course)  # Add the course as a neighbor to its prerequisite.\n",
-        "            in_degree[course] += 1  # Increment the in-degree of the course.\n",
-        "\n",
-        "        # Initialize a queue with courses that have no prerequisites.\n",
-        "        queue = deque()\n",
-        "        for course in range(numCourses):\n",
-        "            if in_degree[course] == 0:\n",
-        "                queue.append(course)\n",
-        "\n",
-        "        # Perform topological sorting.\n",
-        "        while queue:\n",
-        "            course = queue.popleft()  # Take a course with no prerequisites.\n",
-        "            numCourses -= 1  # Decrement the count of remaining courses.\n",
-        "\n",
-        "            for neighbor in graph[course]:\n",
-        "                in_degree[neighbor] -= 1  # Remove the prerequisite relationship.\n",
-        "                if in_degree[neighbor] == 0:\n",
-        "                    queue.append(neighbor)  # If no more prerequisites, add to the queue.\n",
-        "\n",
-        "        # If all courses were successfully taken (numCourses becomes 0), return True.\n",
-        "        return numCourses == 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "This Python code is designed to determine whether it is possible to complete all the required courses given a set of prerequisites. It uses a topological sorting algorithm to accomplish this. Here's a brief explanation of the code:\n",
-        "\n",
-        "1. The code defines a `Solution` class with a `canFinish` method that takes two parameters: `numCourses` (the total number of courses) and `prerequisites` (a list of pairs indicating which course must be taken before another).\n",
-        "\n",
-        "2. It initializes an empty graph as an adjacency list using a `defaultdict` to represent the courses and their prerequisites. It also initializes a list called `in_degree` to keep track of the in-degrees of courses, which is used in the topological sorting process.\n",
-        "\n",
-        "3. The code then populates the graph by iterating through the `prerequisites` list. For each pair `(ai, bi)`, it adds course `ai` as a neighbor to its prerequisite course `bi` in the graph and increments the in-degree of course `ai`.\n",
-        "\n",
-        "4. Next, it initializes a queue with courses that have no prerequisites. It does this by iterating through all the courses and adding those with an in-degree of 0 to the queue.\n",
-        "\n",
-        "5. The code performs a topological sorting of the courses using a while loop. In each iteration of the loop, it dequeues a course from the front of the queue (a course with no prerequisites).\n",
-        "\n",
-        "6. For each neighbor (course that depends on the dequeued course), it decrements the in-degree of the neighbor. If the neighbor's in-degree becomes 0, it means all of its prerequisites have been taken, so the neighbor is added to the queue.\n",
-        "\n",
-        "7. The loop continues until there are no more courses with no prerequisites to dequeue, and the in-degrees are updated accordingly.\n",
-        "\n",
-        "8. After the loop, if all courses were successfully taken (i.e., `numCourses` becomes 0), it means that there were no circular dependencies, and it's possible to finish all courses. In this case, the method returns `True`. Otherwise, if there are remaining courses (indicating a circular dependency), it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "numCourses1 = 2\n",
-        "prerequisites1 = [[1, 0]]\n",
-        "sol = Solution()\n",
-        "print(sol.canFinish(numCourses1, prerequisites1))  # Output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "numCourses2 = 2\n",
-        "prerequisites2 = [[1, 0], [0, 1]]\n",
-        "print(sol.canFinish(numCourses2, prerequisites2))  # Output: False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `canFinish` method:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. Constructing the graph: In the first loop where we iterate through the `prerequisites` list, we populate the graph with prerequisites. This loop has a time complexity of O(E), where E is the number of prerequisites (edges).\n",
-        "\n",
-        "2. Initializing in-degrees and finding courses with no prerequisites: In the worst case, we iterate through all `numCourses` courses to initialize in-degrees and find the courses with no prerequisites. This has a time complexity of O(V), where V is the number of courses (vertices).\n",
-        "\n",
-        "3. Topological Sorting: In the while loop, we perform topological sorting, processing each course once and its outgoing edges. In the worst case, each course is processed once. This has a time complexity of O(V + E).\n",
-        "\n",
-        "The overall time complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. Graph Representation: The space complexity for storing the graph as an adjacency list is O(E), where E is the number of prerequisites.\n",
-        "\n",
-        "2. In-Degree Array: The space complexity for storing the in-degrees of courses is O(V), where V is the number of courses.\n",
-        "\n",
-        "3. Queue: The space complexity for the queue is O(V), as it may contain all the courses.\n",
-        "\n",
-        "The overall space complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-        "\n",
-        "In most practical cases, the number of prerequisites (E) is much smaller than the number of courses (V), so the time and space complexity can often be considered as O(V)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimizing Course Scheduling**: Given a list of courses with their durations and prerequisites, find the most efficient way to schedule these courses to minimize the time it takes to complete all of them.\n",
-        "\n",
-        "2. **Detecting Cycles in Course Dependencies**: Modify the original problem to not just determine if you can finish all courses but also to identify the specific courses that form a cycle in the prerequisite dependencies."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb b/docs/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb
deleted file mode 100644
index fb186f2..0000000
--- a/docs/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb	
+++ /dev/null
@@ -1,236 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 3651. Number of Connected Components in an Undirected Graph\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Number of Connected Components in an Undirected Graph problem on LintCode, click here!](https://www.lintcode.com/problem/3651/)\n",
-        "\n",
-        "---\n",
-        "In this problem, there is an undirected graph with `n` nodes. There is also an `edges` array. Where `edges[i] = [a, b]` means that there is an edge between *node a* and *node b* in the graph.\n",
-        "\n",
-        "You need to return *the number of connected components in that graph*."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to determine the number of connected components in an undirected graph. Here's a more detailed explanation of the problem:\n",
-        "\n",
-        "- You are given an undirected graph, which consists of nodes and edges. The graph may have multiple distinct connected components.\n",
-        "\n",
-        "- A connected component is a group of nodes within the graph where each node can be reached from every other node in the same component. In other words, there is a path between any two nodes in the same connected component, but there are no paths connecting nodes from different components.\n",
-        "\n",
-        "- Your task is to write a function that takes as input the number of nodes `n` and a list of edges `edges`, where each edge is represented as a pair of nodes `[a, b]` indicating an edge between node `a` and node `b`.\n",
-        "\n",
-        "- The goal is to determine the number of distinct connected components in the graph.\n",
-        "\n",
-        "- You need to implement an algorithm that can identify these connected components and count how many there are in the given graph.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def countConnectedComponents(n, edges):\n",
-        "    # Define a helper function for depth-first search (DFS).\n",
-        "    def dfs(node):\n",
-        "        if not visited[node]:  # If the node has not been visited yet:\n",
-        "            visited[node] = True  # Mark it as visited.\n",
-        "            for neighbor in graph[node]:  # Explore all neighbors of the current node.\n",
-        "                dfs(neighbor)  # Recursively call DFS on the neighbor.\n",
-        "\n",
-        "    # Create an empty adjacency list to represent the graph.\n",
-        "    graph = [[] for _ in range(n)]\n",
-        "\n",
-        "    # Initialize a boolean list to keep track of visited nodes.\n",
-        "    visited = [False] * n\n",
-        "\n",
-        "    # Populate the adjacency list with edges from the input.\n",
-        "    for a, b in edges:\n",
-        "        graph[a].append(b)\n",
-        "        graph[b].append(a)  # Since the graph is undirected, we add edges in both directions.\n",
-        "\n",
-        "    # Initialize a counter to keep track of the number of connected components.\n",
-        "    count = 0\n",
-        "\n",
-        "    # Iterate through all nodes in the graph.\n",
-        "    for i in range(n):\n",
-        "        if not visited[i]:  # If the node has not been visited:\n",
-        "            count += 1  # Increment the count since we've found a new connected component.\n",
-        "            dfs(i)  # Start a depth-first search from the unvisited node to explore the component.\n",
-        "\n",
-        "    return count"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided Python code calculates the number of connected components in an undirected graph. Here's a high-level explanation of how the code works:\n",
-        "\n",
-        "1. The `countConnectedComponents` function takes two arguments: `n` (the number of nodes in the graph) and `edges` (a list of edge pairs representing connections between nodes).\n",
-        "\n",
-        "2. Inside the function, there is a helper function called `dfs` (depth-first search), which is used to explore and mark nodes as visited.\n",
-        "\n",
-        "3. An empty adjacency list called `graph` is created to represent the graph. The `graph` is a list of lists where each element represents a node, and the list associated with each node contains its neighboring nodes.\n",
-        "\n",
-        "4. Another list called `visited` is created to keep track of visited nodes. Initially, all nodes are marked as not visited (`False`).\n",
-        "\n",
-        "5. The code populates the `graph` by iterating through the `edges` list and adding each edge to the adjacency list. Since the graph is undirected, both directions of each edge are added.\n",
-        "\n",
-        "6. A counter variable called `count` is initialized to keep track of the number of connected components.\n",
-        "\n",
-        "7. The code iterates through all nodes from `0` to `n-1`. For each node, it checks whether it has been visited yet. If it hasn't been visited, it means a new connected component is found. The counter `count` is incremented, and a depth-first search (DFS) is initiated from this unvisited node to explore the connected component.\n",
-        "\n",
-        "8. The `dfs` function recursively visits all nodes in the connected component, marking them as visited. It does this by checking each neighbor of the current node and recursively calling `dfs` on unvisited neighbors.\n",
-        "\n",
-        "9. After the loop through all nodes, the `count` variable contains the total number of connected components in the graph.\n",
-        "\n",
-        "10. The function returns the value of `count`, which represents the number of connected components.\n",
-        "\n",
-        "11. Two examples are provided at the end to demonstrate how to use the function with different inputs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "n1 = 3\n",
-        "edges1 = [[0, 1], [0, 2]]\n",
-        "print(countConnectedComponents(n1, edges1))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "n2 = 6\n",
-        "edges2 = [[0, 1], [1, 2], [2, 3], [4, 5]]\n",
-        "print(countConnectedComponents(n2, edges2))  # Output: 2"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the code for counting the number of connected components in an undirected graph using depth-first search (DFS).\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code iterates through all nodes in the graph (from 0 to `n-1`).\n",
-        "- For each unvisited node, it initiates a depth-first search (DFS) from that node to explore the connected component.\n",
-        "- In the worst case, every node may belong to a separate connected component, and you perform a DFS for each node.\n",
-        "- The time complexity of the DFS for a single connected component is O(V + E), where V is the number of nodes in the component, and E is the number of edges.\n",
-        "- In the worst case, you perform O(n) DFS calls, each with its own O(V + E) complexity.\n",
-        "\n",
-        "So, the overall time complexity of the code is O(n * (V + E)). In the worst case, if the graph is connected (a single connected component), this simplifies to O(n * (n + m)), where m is the number of edges.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The main data structures used are the `graph` (adjacency list), `visited` (boolean array), and the recursive call stack for DFS.\n",
-        "- The `graph` has a space complexity of O(n + m), where n is the number of nodes, and m is the number of edges.\n",
-        "- The `visited` boolean array also has a space complexity of O(n) to keep track of whether nodes have been visited.\n",
-        "- The depth of the recursive call stack for DFS can go up to O(n) in the worst case, when all nodes are part of a single connected component.\n",
-        "\n",
-        "So, the overall space complexity of the code is O(n + m) for data structures and O(n) for the call stack. In the worst case, this simplifies to O(n + m).\n",
-        "\n",
-        "In summary, the time complexity is O(n * (V + E)), and the space complexity is O(n + m)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Disconnected Components with Multiple Queries:** You are given an undirected graph with `n` nodes and `m` edges. Initially, the graph is disconnected, and you need to process `q` queries. Each query consists of adding an edge to connect two nodes. After each query, you need to output the number of connected components in the updated graph.\n",
-        "\n",
-        "2. **Maximum Size of Connected Components:** Given an undirected graph with `n` nodes and `m` edges, find the size of the largest connected component in the graph. Additionally, identify the nodes that belong to this largest component. You can assume there is at least one connected component in the graph.\n"
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/11. Graphs/417. Pacific Atlantic Water Flow.ipynb b/docs/11. Graphs/417. Pacific Atlantic Water Flow.ipynb
deleted file mode 100644
index eb34497..0000000
--- a/docs/11. Graphs/417. Pacific Atlantic Water Flow.ipynb	
+++ /dev/null
@@ -1,261 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 417. Pacific Atlantic Water Flow\n",
-        "\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Pacific Atlantic Water Flow problem on LeetCode, click here!](https://leetcode.com/problems/pacific-atlantic-water-flow/)\n",
-        "\n",
-        "---\n",
-        "There is an `m x n` rectangular island that borders both the **Pacific Ocean** and **Atlantic Ocean**. The **Pacific Ocea**n touches the island's left and top edges, and the **Atlantic Ocean** touches the island's right and bottom edges.\n",
-        "\n",
-        "The island is partitioned into a grid of square cells. You are given an `m x n` integer matrix `heights` where `heights[r][c]` represents the **height above sea level** of the cell at coordinate `(r, c)`.\n",
-        "\n",
-        "The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is **less than or equal to** the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.\n",
-        "\n",
-        "Return *a **2D list** of grid coordinates `result` where $result[i] = [r_i, c_i]$ denotes that rain water can flow from cell $(r_i, c_i)$ to **both** the Pacific and Atlantic oceans*.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == heights.length`\n",
-        "- `n == heights[r].length`\n",
-        "- `1 <= m, n <= 200`\n",
-        "- 0 <= `heights[r][c]` <= $10^5$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand involves a scenario with a rectangular island surrounded by both the Pacific Ocean and the Atlantic Ocean. The Pacific Ocean is in contact with the island's left and top edges, while the Atlantic Ocean is in contact with the island's right and bottom edges. The goal is to determine which cells on the island can allow rainwater to flow to both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "Here are the key details and constraints of the problem:\n",
-        "\n",
-        "1. The island is divided into a grid of square cells.\n",
-        "\n",
-        "2. Each cell is represented by an integer value in a matrix `heights`. The value at `heights[r][c]` represents the height above sea level of the cell at coordinates `(r, c)` on the island.\n",
-        "\n",
-        "3. Rainwater can flow from a cell to neighboring cells directly north, south, east, or west only if the neighboring cell's height is less than or equal to the current cell's height.\n",
-        "\n",
-        "4. Water can flow from any cell adjacent to the ocean into the ocean. This means water can flow from cells on the edge of the island to the Pacific Ocean and the Atlantic Ocean.\n",
-        "\n",
-        "The task is to find and return a 2D list of grid coordinates where rainwater can flow from a cell to both the Pacific and Atlantic Oceans. In other words, you need to identify the cells that can send water to both the left and top edges (Pacific) and the right and bottom edges (Atlantic) of the island.\n",
-        "\n",
-        "The problem is solved by performing a depth-first search (DFS) starting from the ocean edges and marking the cells that can flow to each ocean. Then, the algorithm identifies cells that are reachable from both oceans and returns their coordinates as the result.\n",
-        "\n",
-        "The problem involves both traversal of the island and backtracking to find the solution, which makes it a classic graph traversal problem.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:\n",
-        "        # Check if the input matrix is empty\n",
-        "        if not heights:\n",
-        "            return []\n",
-        "        \n",
-        "        m, n = len(heights), len(heights[0])\n",
-        "        \n",
-        "        # Initialize boolean matrices to keep track of cells reachable from each ocean\n",
-        "        pacific_reachable = [[False] * n for _ in range(m)]\n",
-        "        atlantic_reachable = [[False] * n for _ in range(m)]\n",
-        "        \n",
-        "        # Depth-First Search (DFS) function to mark cells that can be reached from an ocean\n",
-        "        def dfs(x, y, reachable):\n",
-        "            if reachable[x][y]:\n",
-        "                return\n",
-        "            reachable[x][y] = True\n",
-        "            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:\n",
-        "                new_x, new_y = x + dx, y + dy\n",
-        "                if 0 <= new_x < m and 0 <= new_y < n and heights[new_x][new_y] >= heights[x][y]:\n",
-        "                    dfs(new_x, new_y, reachable)\n",
-        "        \n",
-        "        # Start DFS from the Pacific Ocean (left and top edges)\n",
-        "        for i in range(m):\n",
-        "            dfs(i, 0, pacific_reachable)\n",
-        "            dfs(i, n - 1, atlantic_reachable)\n",
-        "        \n",
-        "        for j in range(n):\n",
-        "            dfs(0, j, pacific_reachable)\n",
-        "            dfs(m - 1, j, atlantic_reachable)\n",
-        "        \n",
-        "        result = []\n",
-        "        \n",
-        "        # Find cells that can flow to both the Pacific and Atlantic oceans\n",
-        "        for i in range(m):\n",
-        "            for j in range(n):\n",
-        "                if pacific_reachable[i][j] and atlantic_reachable[i][j]:\n",
-        "                    result.append([i, j])\n",
-        "        \n",
-        "        return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a class called `Solution` with a method named `pacificAtlantic`. This method takes a 2D matrix `heights`, where each element represents the height above sea level of a cell on a rectangular island. The goal of the method is to find and return a list of grid coordinates (cells) from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "The code uses a Depth-First Search (DFS) algorithm to traverse the island, starting from the ocean edges, and marks cells that are reachable from either the Pacific or Atlantic Ocean. It does this by creating two boolean matrices, `pacific_reachable` and `atlantic_reachable`, to keep track of cells that can be reached from each ocean.\n",
-        "\n",
-        "The code iterates through the entire island, initiating DFS searches from the edges of the island. If a cell is reachable from an ocean (either Pacific or Atlantic), it is marked as reachable in the corresponding boolean matrix. The DFS function ensures that the traversal follows the rule that water can flow to a neighboring cell if the neighboring cell's height is less than or equal to the current cell's height.\n",
-        "\n",
-        "After marking all reachable cells from both oceans, the code then looks for cells that are reachable from both the Pacific and Atlantic Oceans. If a cell satisfies this condition, it is added to the `result` list.\n",
-        "\n",
-        "Finally, the code returns the `result` list, which contains the grid coordinates of cells from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "The code provides a solution to the problem of finding cells on the island where water can reach both oceans, demonstrating a traversal algorithm that explores the connectivity of cells on the island."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "heights1 = [\n",
-        "    [1,2,2,3,5],\n",
-        "    [3,2,3,4,4],\n",
-        "    [2,4,5,3,1],\n",
-        "    [6,7,1,4,5],\n",
-        "    [5,1,1,2,4]\n",
-        "]\n",
-        "\n",
-        "solution = Solution()\n",
-        "print(solution.pacificAtlantic(heights1)) # Expected Output: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 26,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "heights2 = [[1]]\n",
-        "print(solution.pacificAtlantic(heights2)) # Expected Output: [[0, 0]]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Time Complexity:\n",
-        "The time complexity of this solution is O(m * n), where 'm' is the number of rows in the input matrix and 'n' is the number of columns. Here's a breakdown of the time complexity:\n",
-        "\n",
-        "1. Creating the `pacific_reachable` and `atlantic_reachable` matrices with dimensions m x n takes O(m * n) time.\n",
-        "\n",
-        "2. Initiating Depth-First Search (DFS) from the edges of the island takes O(m * n) time because it processes each cell once.\n",
-        "\n",
-        "3. The DFS function visits each cell at most once, and its time complexity is O(1) for each cell. The number of cells visited by the DFS in the worst case is m * n.\n",
-        "\n",
-        "4. The final step of identifying and appending cells to the result list takes O(m * n) time in the worst case.\n",
-        "\n",
-        "Overall, the time complexity is dominated by the DFS and is O(m * n).\n",
-        "\n",
-        "Space Complexity:\n",
-        "The space complexity of this solution is also O(m * n). Here's how the space is used:\n",
-        "\n",
-        "1. Two boolean matrices, `pacific_reachable` and `atlantic_reachable`, are created with dimensions m x n. Each matrix takes O(m * n) space, resulting in O(2 * m * n) space usage.\n",
-        "\n",
-        "2. The depth-first search (DFS) stack space during the recursive calls can go as deep as the diagonal of the grid, which is at most min(m, n). This additional space for the call stack is relatively small compared to the boolean matrices and can be considered O(min(m, n)).\n",
-        "\n",
-        "3. The `result` list stores the grid coordinates, and its size is determined by the number of cells that can flow to both oceans. In the worst case, this list can contain all m * n cells, so it takes O(m * n) space.\n",
-        "\n",
-        "The overall space complexity is the sum of these components, which is O(m * n) + O(min(m, n)) + O(m * n). In big O notation, we can simplify this to O(m * n)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the algorithm to find the maximum flow of water from any cell to both the Pacific and Atlantic Oceans. In other words, find the cell(s) that can send the most water to both oceans.\n",
-        "\n",
-        "2. **Efficiency Challenge:** Optimize the time and space complexity of the algorithm while maintaining correctness. Aim to reduce both time and space complexity as much as possible."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/11. Graphs/README.md b/docs/11. Graphs/README.md
deleted file mode 100644
index aad262c..0000000
--- a/docs/11. Graphs/README.md	
+++ /dev/null
@@ -1,23 +0,0 @@
-# Graphs Problems - Blind 75 LeetCode
-
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)                           | Medium     |
-| [133. Clone Graph](https://leetcode.com/problems/clone-graph/)                                       | Medium     |
-| [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)       | Medium     |
-| [207. Course Schedule](https://leetcode.com/problems/course-schedule/)                                 | Medium     |
-| [323. Number of Connected Components In An Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/) | Medium |
-| [261. Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)                               | Medium     |
-
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/12. Advanced Graphs/892. Alien Dictionary.ipynb b/docs/12. Advanced Graphs/892. Alien Dictionary.ipynb
deleted file mode 100644
index 81f0e5b..0000000
--- a/docs/12. Advanced Graphs/892. Alien Dictionary.ipynb	
+++ /dev/null
@@ -1,256 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 892. Alien Dictionary\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Alien Dictionary problem on LintCode, click here!](https://www.lintcode.com/problem/892/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of **non-empty** words from the dictionary, where words are **sorted lexicographically by the rules of this new language**. Derive the order of letters in this language.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- You may assume all letters are in lowercase.\n",
-        "- The dictionary is invalid, if string a is prefix of string b and b is appear before a.\n",
-        "- If the order is invalid, return an empty string.\n",
-        "- There may be multiple valid order of letters, return the smallest in normal lexicographical order.\n",
-        "- The letters in **one** string are of the same rank by default and are sorted in Human dictionary order."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand involves determining the order of letters in a new alien language that uses the Latin alphabet. However, the order of letters in this alien language is unknown. To solve this problem, we are given a list of non-empty words from the dictionary, where the words are sorted lexicographically based on the rules of the alien language. The goal is to derive the correct order of letters in this new language.\n",
-        "\n",
-        "Here's a more detailed explanation of the problem:\n",
-        "\n",
-        "1. The alphabet: The new alien language uses the Latin alphabet, which consists of lowercase letters.\n",
-        "\n",
-        "2. Dictionary: We are provided with a list of non-empty words. These words are sorted in lexicographical order based on the rules of the alien language. The order of letters in the words reflects the correct order in the new language.\n",
-        "\n",
-        "3. Invalid order: The problem specifies that the dictionary is considered invalid if there exists a situation where one string 'a' is a prefix of another string 'b,' and 'b' appears before 'a' in the dictionary. This condition ensures that there are no inconsistencies in the order of letters.\n",
-        "\n",
-        "4. Multiple valid orders: It's possible that there are multiple valid orders of letters that satisfy the provided dictionary. However, we are instructed to return the smallest valid order in normal lexicographical order.\n",
-        "\n",
-        "To solve this problem, you need to analyze the given list of words and determine the correct order of letters in the alien language while adhering to the specified constraints. You should return the order as a string in lexicographical order if it's valid, and an empty string if the order is invalid.\n",
-        "\n",
-        "The key to solving this problem is to construct a directed graph of letter relationships based on the order information provided by the words in the dictionary. Then, perform a topological sort on this graph to derive the order of letters, ensuring that the order is consistent and valid according to the given constraints.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict, deque\n",
-        "\n",
-        "def alienOrder(words):\n",
-        "    # Create an adjacency list to represent the graph\n",
-        "    graph = defaultdict(list)\n",
-        "    \n",
-        "    # Create a dictionary to store the in-degrees of each letter\n",
-        "    in_degree = {}\n",
-        "    \n",
-        "    # Initialize in-degrees to 0 for all letters\n",
-        "    for word in words:\n",
-        "        for char in word:\n",
-        "            if char not in in_degree:\n",
-        "                in_degree[char] = 0\n",
-        "    \n",
-        "    # Build the graph and update in-degrees\n",
-        "    for i in range(1, len(words)):\n",
-        "        word1, word2 = words[i-1], words[i]\n",
-        "        min_length = min(len(word1), len(word2))\n",
-        "        \n",
-        "        # Compare characters in the two words\n",
-        "        j = 0\n",
-        "        while j < min_length and word1[j] == word2[j]:\n",
-        "            j += 1\n",
-        "        \n",
-        "        if j < min_length:\n",
-        "            # If word1[j] is lexicographically before word2[j], add an edge\n",
-        "            graph[word1[j]].append(word2[j])\n",
-        "            in_degree[word2[j]] += 1\n",
-        "            \n",
-        "            # Break the loop to avoid further comparisons\n",
-        "            break\n",
-        "    \n",
-        "    # Perform topological sorting using Kahn's algorithm\n",
-        "    result = []\n",
-        "    queue = deque([char for char in in_degree if in_degree[char] == 0])\n",
-        "    \n",
-        "    while queue:\n",
-        "        char = queue.popleft()\n",
-        "        result.append(char)\n",
-        "        \n",
-        "        for neighbor in graph[char]:\n",
-        "            in_degree[neighbor] -= 1\n",
-        "            if in_degree[neighbor] == 0:\n",
-        "                queue.append(neighbor)\n",
-        "    \n",
-        "    # Check if there is a valid order\n",
-        "    if len(result) < len(in_degree):\n",
-        "        return \"\"\n",
-        "    \n",
-        "    return \"\".join(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code is designed to determine the order of letters in an unknown alien language that uses the Latin alphabet. It takes a list of non-empty words from a dictionary as input, where the words are sorted lexicographically based on the rules of this alien language.\n",
-        "\n",
-        "The code follows these main steps:\n",
-        "\n",
-        "1. Create a data structure to represent the graph of letters and their relationships in the alien language. It uses a defaultdict to store a list of letters that come after each letter. Additionally, it uses a dictionary called `in_degree` to keep track of the in-degrees (the number of letters that precede a letter) for each letter.\n",
-        "\n",
-        "2. Initialize the in-degrees of all letters to 0. This is done by iterating through all the words and characters in the words, adding each character to the `in_degree` dictionary with an initial in-degree of 0.\n",
-        "\n",
-        "3. Build the graph by comparing adjacent words in the dictionary. For each pair of adjacent words, the code finds the first differing character between the two words. If such a character exists, it means there's an order relationship between the letters represented by these characters. The code updates the graph and in-degrees accordingly.\n",
-        "\n",
-        "4. After constructing the graph, the code performs a topological sorting of the letters using Kahn's algorithm. It starts by initializing an empty result list and a queue containing letters with in-degrees of 0. The algorithm repeatedly removes a letter with an in-degree of 0 from the queue, adds it to the result list, and updates the in-degrees of its neighbors. This process continues until the queue is empty.\n",
-        "\n",
-        "5. Finally, the code checks if the topological sorting was successful by comparing the length of the result list with the number of unique letters in the input. If the lengths differ, it means there's a cycle in the graph, and the order is invalid. In such a case, an empty string is returned. Otherwise, the result list is joined together to form the smallest order of letters in normal lexicographical order, and it is returned as the output.\n",
-        "\n",
-        "The code provides a way to determine the order of letters in the alien language while handling various edge cases, including checking for invalid order conditions."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "wrtef\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "words1 = [\"wrt\", \"wrf\", \"er\", \"ett\", \"rftt\"]\n",
-        "print(alienOrder(words1))  # Output: \"wertf\""
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "zx\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "words2 = [\"z\", \"x\"]\n",
-        "print(alienOrder(words2))  # Output: \"zx\""
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for determining the order of letters in an alien language.\n",
-        "\n",
-        "Time Complexity:\n",
-        "1. Building the graph: The code iterates through the list of words and compares adjacent words. In the worst case, where all words have the same prefix, it takes O(N) time to build the graph, where N is the total number of characters in all the words.\n",
-        "2. Topological sorting: Performing a topological sort on the graph takes O(V + E) time, where V is the number of unique letters (vertices) and E is the number of relationships between letters (edges) in the graph.\n",
-        "\n",
-        "Overall, the time complexity is O(N + V + E), where N is the total number of characters in the words, V is the number of unique letters, and E is the number of order relationships between letters.\n",
-        "\n",
-        "Space Complexity:\n",
-        "1. Graph and In-degrees: The code uses data structures to represent the graph and in-degrees. The space complexity for these data structures is O(V + E), where V is the number of unique letters, and E is the number of order relationships between letters in the words.\n",
-        "2. Result List: The result list stores the order of letters, which can have a maximum size of V, where V is the number of unique letters.\n",
-        "3. Queue: The space used by the queue during topological sorting is also O(V).\n",
-        "\n",
-        "Overall, the space complexity is O(V + E) due to the data structures used for graph representation and the space needed for storing the result and the queue.\n",
-        "\n",
-        "In the worst case, where there are many unique letters and many order relationships, the space complexity is dominated by the number of unique letters and their relationships.\n",
-        "\n",
-        "In summary, the time complexity of the code is O(N + V + E), and the space complexity is O(V + E). The actual performance will depend on the specific input data, such as the number of unique letters and the relationships between them in the alien language."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Valid Orders**: Extend the problem to handle cases where there are multiple valid orders of letters in the alien language. Modify the code to return all valid orders instead of just one. Be mindful of performance.\n",
-        "\n",
-        "2. **Detect Invalid Dictionary**: Given a dictionary of words, write a function to detect whether the dictionary is valid or not based on the constraints mentioned in the problem statement."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/12. Advanced Graphs/README.md b/docs/12. Advanced Graphs/README.md
deleted file mode 100644
index f362de7..0000000
--- a/docs/12. Advanced Graphs/README.md	
+++ /dev/null
@@ -1,16 +0,0 @@
-# Advanced Graphs Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [269. Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)                               | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/13. One-D Dynamic Programming/198. House Robber.ipynb b/docs/13. One-D Dynamic Programming/198. House Robber.ipynb
deleted file mode 100644
index 8d32e21..0000000
--- a/docs/13. One-D Dynamic Programming/198. House Robber.ipynb	
+++ /dev/null
@@ -1,103 +0,0 @@
-{
-  "metadata": {
-    "kernelspec": {
-      "name": "python",
-      "display_name": "Python (Pyodide)",
-      "language": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat_minor": 5,
-  "nbformat": 4,
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "source": "# House Robber\n**Difficulty:** Medium\n\n---\nYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and **it will automatically contact the police if two adjacent houses were broken into on the same night.**\n\nGiven an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight **without alerting the police**.*\n\n**Constraints:**\n- `1 <= nums.length <= 100`\n- `0 <= nums[i] <= 400`",
-      "metadata": {},
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Probelm Explanation:\n\nThe problem is a classic dynamic programming challenge known as the \"House Robber\" problem. The scenario is that you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, but the constraint is that adjacent houses have security systems connected. If two adjacent houses are broken into on the same night, the security systems will automatically contact the police.\n\nThe goal is to determine the maximum amount of money you can rob tonight without alerting the police. You need to find a strategy to maximize your earnings while avoiding robbing two adjacent houses.\n\nThe input to the problem is an array `nums`, where `nums[i]` represents the amount of money in the `i`-th house. The task is to return the maximum amount of money that can be robbed without triggering the security system.\n\nFor example:\n```\nInput: nums = [1,2,3,1]\nOutput: 4\nExplanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).\nTotal amount you can rob = 1 + 3 = 4.\n\nInput: nums = [2,7,9,3,1]\nOutput: 12\nExplanation: Rob house 1 (money = 2), rob house 3 (money = 9), and rob house 5 (money = 1).\nTotal amount you can rob = 2 + 9 + 1 = 12.\n```\n\nThe challenge involves designing an algorithm to efficiently solve this problem and find the optimal strategy for robbing houses to maximize the stolen money while adhering to the constraint of not robbing adjacent houses. The provided solution utilizes dynamic programming to achieve this.\n\n## Solution:\nHere's a Python function to implement this algorithm:",
-      "metadata": {},
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76"
-    },
-    {
-      "cell_type": "code",
-      "source": "from typing import List\n\nclass Solution:\n    def rob(self, nums: List[int]) -> int:\n        # Initialize variables to keep track of the maximum amount of money robbed at the previous and current houses\n        prev, curr = 0, 0\n\n        # Iterate through the houses\n        for n in nums:\n            # Calculate the maximum amount of money that can be robbed at the current house\n            # It is the maximum of either robbing the current house and the money robbed at the house before the previous one (prev),\n            # or skipping the current house and taking the maximum amount from the previous house (curr)\n            temp = max(n + prev, curr)\n\n            # Update the variables for the next iteration\n            prev, curr = curr, temp\n\n        # The final result is the maximum amount of money that can be robbed at the last house\n        return curr",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 4,
-      "outputs": [],
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Explanation:\n\nThe provided code is a Python implementation of the \"House Robber\" problem using a dynamic programming approach. It's part of a class `Solution` and includes a method `rob` that takes a list of integers `nums` as input and returns the maximum amount of money that can be robbed without alerting the police.\n\nThe key idea of the algorithm is to maintain two variables, `prev` and `curr`, which represent the maximum amount of money that can be robbed up to the previous house and the current house, respectively. The algorithm iterates through the houses, updating these variables based on the maximum amount that can be robbed at the current house, considering the constraint of not robbing adjacent houses.\n\nThe loop calculates a temporary variable `temp` that represents the maximum amount of money that can be robbed at the current house. It considers two scenarios: either robbing the current house and adding the money robbed at the house before the previous one (`n + prev`), or skipping the current house and taking the maximum amount from the previous house (`curr`). The maximum of these two scenarios is assigned to `temp`.\n\nThe `prev` and `curr` variables are then updated for the next iteration. `prev` is set to the previous value of `curr`, and `curr` is set to the calculated `temp`.\n\nThe final result is the value stored in the `curr` variable, representing the maximum amount of money that can be robbed at the last house. This value is returned as the result of the `rob` method.\n\nThe provided test cases at the end demonstrate the usage of the `rob` method with different input arrays.",
-      "metadata": {},
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Test cases:\n\nHere's how you can use this solution:",
-      "metadata": {},
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 1:\nsol = Solution()\n\nnums1 = [1, 2, 3, 1]\nprint(sol.rob(nums1))  # Output: 4",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 5,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "4\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums2 = [2, 7, 9, 3, 1]\nprint(sol.rob(nums2))  # Output: 12",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 6,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "12\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Time and Space Complexity Analysis\n\nLet's analyze the time and space complexity of the provided code.\n\n### Time Complexity:\n\nThe time complexity of the code is O(n), where n is the number of houses. The algorithm iterates through the array once, and in each iteration, it performs constant time operations. The loop runs for each house, and the number of operations inside the loop is independent of the input size. Therefore, the overall time complexity is linear, O(n).\n\n### Space Complexity:\n\nThe space complexity is O(1), constant space. The algorithm uses only a constant amount of extra space regardless of the input size. The variables `prev`, `curr`, and `temp` are the only additional variables used, and they do not depend on the size of the input array. Thus, the space complexity is constant, O(1).\n\nIn summary:\n- Time Complexity: O(n)\n- Space Complexity: O(1)",
-      "metadata": {},
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Challenging Exercises:\n\n1. **Variation with Maximum Number of Houses to Rob:** Modify the problem to include an additional constraint: you are allowed to rob at most 'k' houses without triggering the alarm. Find the maximum amount you can rob under this new constraint.\n\n2. **Robbing in a Circular Street:** Extend the problem to a circular street where the first and last houses are adjacent. You cannot rob both the first and last houses simultaneously. Find the maximum amount you can rob in this circular setting.\n\n\n**Link to Problem:** [To see the House Robber problem on LeetCode, click here!](https://leetcode.com/problems/house-robber/description/)\n",
-      "metadata": {},
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6"
-    }
-  ]
-}
\ No newline at end of file
diff --git a/docs/13. One-D Dynamic Programming/213. House Robber II.ipynb b/docs/13. One-D Dynamic Programming/213. House Robber II.ipynb
deleted file mode 100644
index b1080e8..0000000
--- a/docs/13. One-D Dynamic Programming/213. House Robber II.ipynb	
+++ /dev/null
@@ -1,119 +0,0 @@
-{
-  "metadata": {
-    "kernelspec": {
-      "name": "python",
-      "display_name": "Python (Pyodide)",
-      "language": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat_minor": 5,
-  "nbformat": 4,
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "source": "# House Robber II\n**Difficulty:** Medium\n\n---\nYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle**. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.\n\nGiven an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight without alerting the police*.\n\n**Constraints:**\n- `1 <= nums.length <= 100`\n- `0 <= nums[i] <= 1000`",
-      "metadata": {},
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Probelm Explanation:\n\nCertainly! The problem statement describes a scenario where you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. However, there is a security system in place: adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.\n\nThe twist in this problem is that the houses are arranged in a circle. That means the first house is the neighbor of the last one. So, robbing the first house has implications for the last house, and vice versa.\n\nThe goal is to find the maximum amount of money you can rob tonight without alerting the police. You need to implement a function that takes an integer array `nums` representing the amount of money in each house and returns the maximum amount of money that can be robbed without triggering the security system.\n\nHere are a few examples to illustrate the problem:\n\n1. **Example 1:**\n   ```python\n   Input: nums = [2, 3, 2]\n   Output: 3\n   Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2) because they are adjacent houses. The optimal strategy is to rob house 2 (money = 3).\n   ```\n\n2. **Example 2:**\n   ```python\n   Input: nums = [1, 2, 3, 1]\n   Output: 4\n   Explanation: Robbing house 1 (money = 1) and then robbing house 3 (money = 3) results in a total amount of 4. Robbing house 2 is skipped to avoid alerting the police.\n   ```\n\n3. **Example 3:**\n   ```python\n   Input: nums = [1, 2, 3]\n   Output: 3\n   Explanation: Since the houses are in a circle, you can choose either house 1, 2, or 3. The optimal strategy is to rob house 3 for a total amount of 3.\n   ```\n\nThe constraints for the problem are that the length of the `nums` array is between 1 and 100, and the amount of money in each house (`nums[i]`) is between 0 and 1000.\n\n## Solution:\nHere's a Python function to implement this algorithm:",
-      "metadata": {},
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76"
-    },
-    {
-      "cell_type": "code",
-      "source": "from typing import List\n\nclass Solution:\n    def rob(self, nums: List[int]) -> int:\n        def simple_rob(nums):\n            # Helper function for basic house robbery problem\n            prev, curr = 0, 0\n            for num in nums:\n                # Calculate the maximum amount of money that can be robbed\n                # without alerting the police for each house\n                prev, curr = curr, max(prev + num, curr)\n            return curr\n\n        # Check the base cases where the length of nums is 1 or 2\n        if len(nums) == 1:\n            return nums[0]\n        elif len(nums) == 2:\n            return max(nums[0], nums[1])\n\n        # Case 1: First house is robbed, last house is not\n        case1 = simple_rob(nums[:-1])\n\n        # Case 2: First house is not robbed, last house is\n        case2 = simple_rob(nums[1:])\n\n        # Maximum of the two cases\n        return max(case1, case2)",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 7,
-      "outputs": [],
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Explanation:\n\nThe provided code defines a class `Solution` with a method `rob`, which aims to determine the maximum amount of money a professional robber can steal from houses arranged in a circle without alerting the police. Each house contains a certain amount of money, and adjacent houses have a security system that triggers if both are robbed on the same night.\n\nThe code uses dynamic programming to solve the problem efficiently. It contains a helper function `simple_rob` that computes the maximum amount of money that can be robbed from a linear arrangement of houses. The main method `rob` handles the circular arrangement by considering two cases: robbing the first house and not robbing the last, and not robbing the first house and robbing the last. The result is the maximum of these two cases.\n\nThe class is instantiated as `solution`, and the `rob` method is applied to three different scenarios (test cases) to demonstrate its functionality. The final output represents the maximum amount of money the robber can steal without triggering the security system.",
-      "metadata": {},
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Test cases:\n\nHere's how you can use this solution:",
-      "metadata": {},
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 1:\n\n# Create an instance of the Solution class\nsolution = Solution()\n\nnums1 = [2, 3, 2]\nprint(solution.rob(nums1))  # Output: 3",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 8,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "3\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums2 = [1, 2, 3, 1]\nprint(solution.rob(nums2))  # Output: 4",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 9,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "4\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums3 = [1, 2, 3]\nprint(solution.rob(nums3))  # Output: 3",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 10,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "3\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "e6e1554f-ae77-410c-8af2-e59e5428166c"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Time and Space Complexity Analysis\n\nLet's analyze the time and space complexity of the provided code:\n\n### Time Complexity:\nThe time complexity is primarily determined by the `simple_rob` helper function, which has a time complexity of O(N), where N is the number of houses in the input array. The main `rob` function calls `simple_rob` twice, once for the case of robbing the first house and not robbing the last, and once for the case of not robbing the first house and robbing the last. Therefore, the overall time complexity is O(N).\n\n### Space Complexity:\nThe space complexity is O(1) because the space used by the algorithm is constant, regardless of the size of the input array. The only variables used are `prev` and `curr` in the `simple_rob` function, which represent the previous and current maximum amounts of money that can be robbed without alerting the police. These variables are updated iteratively, and no additional data structures are used that scale with the input size. Therefore, the space complexity is constant, or O(1).\n\nIn summary:\n- Time Complexity: O(N)\n- Space Complexity: O(1)",
-      "metadata": {},
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Challenging Exercises:\n\n1. **Optimization Challenge:** Optimize the space complexity of the solution. Can you solve the problem with O(1) space complexity without sacrificing the time complexity?\n\n2. **Circular Robbery Variations:** Consider variations of the circular house robbery problem, such as allowing the robber to skip a certain number of houses or restricting the number of consecutive houses that can be robbed. Modify the solution to accommodate these variations.\n\n**Link to Problem:** [To see the House Robber II problem on LeetCode, click here!](https://leetcode.com/problems/house-robber-ii/)",
-      "metadata": {},
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6"
-    }
-  ]
-}
\ No newline at end of file
diff --git a/docs/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb b/docs/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb
deleted file mode 100644
index b727489..0000000
--- a/docs/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb	
+++ /dev/null
@@ -1,210 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 70. Climbing Stairs\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Climbing Stairs problem on LeetCode, click here!](https://leetcode.com/problems/climbing-stairs/)\n",
-        "\n",
-        "---\n",
-        "You are climbing a staircase. It takes `n` steps to reach the top.\n",
-        "\n",
-        "Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= n <= 45`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The \"Climbing Stairs\" problem is a classic algorithmic problem that asks you to find the number of distinct ways to climb a staircase with a given number of steps. Here's the problem statement:\n",
-        "\n",
-        "You are given a staircase with 'n' steps. Each time you can either climb 1 step or 2 steps. You need to find out how many distinct ways there are to reach the top of the staircase.\n",
-        "\n",
-        "For example, if 'n' is 2, there are two ways to climb to the top:\n",
-        "\n",
-        "1. Climbing two steps at once.\n",
-        "2. Climbing one step, and then another step.\n",
-        "\n",
-        "If 'n' is 3, there are three ways:\n",
-        "\n",
-        "1. Climbing one step, then one step, and finally one step.\n",
-        "2. Climbing one step, then two steps.\n",
-        "3. Climbing two steps, then one step.\n",
-        "\n",
-        "The problem essentially asks you to find a solution that counts the possible combinations of 1-step and 2-step climbs that can be taken to reach the top of the staircase. It's a classic example of a dynamic programming problem where you can break it down into smaller subproblems and build a solution incrementally, as shown in the code I provided earlier.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class Solution:\n",
-        "    def climbStairs(self, n: int) -> int:\n",
-        "        # If there are 1 or 2 steps, there are exactly n ways to reach the top.\n",
-        "        if n <= 2:\n",
-        "            return n\n",
-        "\n",
-        "        # Create an array to store the number of ways to reach each step.\n",
-        "        dp = [0] * (n + 1)\n",
-        "\n",
-        "        # There is one way to reach the first step (base case).\n",
-        "        dp[1] = 1\n",
-        "\n",
-        "        # There are two ways to reach the second step (base case).\n",
-        "        dp[2] = 2\n",
-        "\n",
-        "        # Calculate the number of ways to reach each step from the previous steps.\n",
-        "        for i in range(3, n + 1):\n",
-        "            # The number of ways to reach step 'i' is the sum of the ways to reach\n",
-        "            # the previous two steps, as you can take 1 or 2 steps at a time.\n",
-        "            dp[i] = dp[i - 1] + dp[i - 2]\n",
-        "\n",
-        "        # The result is the number of ways to reach the top, which is stored in dp[n].\n",
-        "        return dp[n]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a class `Solution` with a method `climbStairs` that takes an integer `n` as input and returns the number of distinct ways to climb a staircase with `n` steps.\n",
-        "\n",
-        "1. If `n` is 1 or 2, the function returns `n` directly because there are only 1 or 2 distinct ways to reach the top in those cases.\n",
-        "\n",
-        "2. For `n` greater than 2, the code uses dynamic programming to calculate the number of ways to reach each step. It creates an array `dp` of length `n + 1` to store these values.\n",
-        "\n",
-        "3. The `dp[1]` and `dp[2]` are initialized to 1 and 2, respectively. This is because there is only one way to reach the first step (by taking 1 step), and there are two ways to reach the second step (by taking 2 steps or two 1-step climbs).\n",
-        "\n",
-        "4. The code then uses a loop to fill in the `dp` array for steps 3 and above. For each step `i` from 3 to `n`, it calculates the number of ways to reach that step by adding the number of ways to reach the previous two steps, as you can either take 1 step or 2 steps at a time.\n",
-        "\n",
-        "5. Finally, the function returns the value stored in `dp[n]`, which represents the total number of distinct ways to climb the staircase with `n` steps."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 1 - Input: n = 2, Output: 2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "n = 2\n",
-        "solution = Solution()\n",
-        "print(f\"Example 1 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 2"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 2 - Input: n = 3, Output: 3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "n = 3\n",
-        "print(f\"Example 2 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 3\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Time Complexity:\n",
-        "The time complexity of the solution is O(n) because we use a single for loop that iterates from 3 to n to fill the `dp` array. In each iteration, we perform constant time operations (addition and assignment). Therefore, the time complexity is linear with respect to the input value `n`.\n",
-        "\n",
-        "Space Complexity:\n",
-        "The space complexity of the solution is O(n) as well. We use an array `dp` of length `n + 1` to store the number of ways to reach each step. The size of this array is directly proportional to the input value `n`, which makes the space complexity linear in terms of `n`. The rest of the variables used in the function (e.g., `i`) occupy constant space and do not contribute significantly to the overall space complexity.\n",
-        "\n",
-        "In summary, the time complexity is O(n), and the space complexity is O(n) for this dynamic programming solution to the \"Climbing Stairs\" problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Generalized Step Sizes**: Instead of being limited to 1 step or 2 steps, consider a scenario where you can take steps of sizes `[1, 2, 3]`. Implement a function that calculates the number of ways to climb the staircase with these different step sizes.\n",
-        "\n",
-        "2. **Minimize Steps**: Given an integer `n`, find the minimum number of steps required to reach the top of the staircase. Return both the minimum number of steps and the specific sequence of steps taken."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/13. One-D Dynamic Programming/README.md b/docs/13. One-D Dynamic Programming/README.md
deleted file mode 100644
index 73f4de3..0000000
--- a/docs/13. One-D Dynamic Programming/README.md	
+++ /dev/null
@@ -1,25 +0,0 @@
-# 1-D Dynamic Programming Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)                                   | Easy       |
-| [198. House Robber](https://leetcode.com/problems/house-robber/)                                       | Medium     |
-| [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)                                 | Medium     |
-| [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)       | Medium     |
-| [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)                   | Medium     |
-| [91. Decode Ways](https://leetcode.com/problems/decode-ways/)                                         | Medium     |
-| [322. Coin Change](https://leetcode.com/problems/coin-change/)                                         | Medium     |
-| [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)             | Medium     |
-| [139. Word Break](https://leetcode.com/problems/word-break/)                                           | Medium     |
-| [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/14. Two-D Dynamic Programming/README.md b/docs/14. Two-D Dynamic Programming/README.md
deleted file mode 100644
index 8bab876..0000000
--- a/docs/14. Two-D Dynamic Programming/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### 2-D Dynamic Programming
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [62. Unique Paths](https://leetcode.com/problems/unique-paths/)                                         | Medium     |
-| [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)       | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/15. Greedy/README.md b/docs/15. Greedy/README.md
deleted file mode 100644
index 82b6543..0000000
--- a/docs/15. Greedy/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Greedy
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)                                   | Easy       |
-| [55. Jump Game](https://leetcode.com/problems/jump-game/)                                                 | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/16. Intervals/README.md b/docs/16. Intervals/README.md
deleted file mode 100644
index a57713e..0000000
--- a/docs/16. Intervals/README.md	
+++ /dev/null
@@ -1,50 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Intervals
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [57. Insert Interval](https://leetcode.com/problems/insert-interval/)                                     | Hard       |
-| [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)                                   | Medium     |
-| [435. Non Overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)             | Medium     |
-| [252. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)                                       | Easy       |
-| [253. Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)                                 | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/17. Math Geometry/README.md b/docs/17. Math Geometry/README.md
deleted file mode 100644
index 8b2f180..0000000
--- a/docs/17. Math Geometry/README.md	
+++ /dev/null
@@ -1,48 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Math & Geometry
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [48. Rotate Image](https://leetcode.com/problems/rotate-image/)                                           | Medium     |
-| [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)                                         | Medium     |
-| [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)                               | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/18. Bit Manipulation/README.md b/docs/18. Bit Manipulation/README.md
deleted file mode 100644
index 05b2fef..0000000
--- a/docs/18. Bit Manipulation/README.md	
+++ /dev/null
@@ -1,51 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Bit Manupulation
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)                                   | Easy       |
-| [338. Counting Bits](https://leetcode.com/problems/counting-bits/)                                         | Medium     |
-| [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)                                           | Easy       |
-| [268. Missing Number](https://leetcode.com/problems/missing-number/)                                       | Easy       |
-| [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)                             | Medium     |
-
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/Book.ipynb b/docs/Book.ipynb
deleted file mode 100644
index e7b4f46..0000000
--- a/docs/Book.ipynb
+++ /dev/null
@@ -1,400 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!python -m pip install --upgrade pip\n",
-    "%pip install jupyter-book"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions\n"
-     ]
-    }
-   ],
-   "source": [
-    "%cd /workspaces/LeetCode-Solutions"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book toc migrate /workspaces/LeetCode-Solutions/_toc.yml -o /workspaces/LeetCode-Solutions/_toc.yml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "\u001b[01mmaking output directory... \u001b[39;49;00mdone\n",
-      "[etoc] Changing master_doc to 'README'\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 77 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 77 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]  \n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.26 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.00 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]     \n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.92 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 1.28 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executed notebook in 0.91 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executing notebook using local CWD [mystnb]      \n",
-      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executed notebook in 0.91 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executed notebook in 1.17 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executed notebook in 1.16 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executing notebook using local CWD [mystnb] \n",
-      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executed notebook in 1.27 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executed notebook in 4.03 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executing notebook using local CWD [mystnb]   \n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/Book.ipynb: Executing notebook using local CWD [mystnb]                          \n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mSECURITY\u001b[39;49;00m                                              \n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb:30008: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Executing notebook failed: CellTimeoutError [mystnb.exec]\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Notebook exception traceback saved in: /workspaces/LeetCode-Solutions/_build/html/reports/Book.err.log [mystnb.exec]\u001b[39;49;00m\n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CODE_OF_CONDUCT.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/SECURITY.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mSECURITY\u001b[39;49;00m                                               raph\u001b[39;49;00m0m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:8: WARNING: 'myst' reference target not found: #ways-to-contribute\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:9: WARNING: 'myst' reference target not found: #getting-started\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:10: WARNING: 'myst' reference target not found: #contributing-process\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:11: WARNING: 'myst' reference target not found: #code-of-conduct\u001b[39;49;00m\n",
-      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
-      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
-      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
-      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
-      "[etoc] missing index.html written as redirect to 'README.html'\n",
-      "\u001b[01mbuild succeeded, 11 warnings.\u001b[39;49;00m\n",
-      "\n",
-      "The HTML pages are in _build/html.\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Finished generating HTML for book.\n",
-      "Your book's HTML pages are here:\n",
-      "    _build/html/\n",
-      "You can look at your book by opening this file in a browser:\n",
-      "    _build/html/index.html\n",
-      "Or paste this line directly into your browser bar:\n",
-      "    file:///workspaces/LeetCode-Solutions/_build/html/index.html            \n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build ."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book build /workspaces/LeetCode-Solutions/ --builder pdfhtml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/html": [
-       "<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>"
-      ],
-      "text/plain": [
-       "<IPython.core.display.HTML object>"
-      ]
-     },
-     "execution_count": 5,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "from IPython.display import HTML\n",
-    "HTML('<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>')\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Your book template can be found at\n",
-      "\n",
-      "    mybook/\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book create mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions/mybook\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_build/html\n",
-      "[sphinxcontrib-bibtex] Beware that docutils versions 0.18 and 0.19 (you are running 0.18.1) are known to generate invalid html for citations. If this issue affects you, please use docutils<0.18 (or >=0.20 once released) instead. For more details, see https://sourceforge.net/p/docutils/patches/195/\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "[etoc] Changing master_doc to 'intro'\n",
-      "checking bibtex cache... out of date\n",
-      "parsing bibtex file /workspaces/LeetCode-Solutions/mybook/references.bib... parsed 5 entries\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='force', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/mybook/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 14 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 14 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.35 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.97 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 0.83 seconds [mystnb]\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mintro\u001b[39;49;00m                                                 \n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mintro\u001b[39;49;00m                                                  \n",
-      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
-      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
-      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
-      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
-      "[etoc] missing index.html written as redirect to 'intro.html'\n",
-      "\u001b[01mbuild succeeded, 2 warnings.\u001b[39;49;00m\n",
-      "\n",
-      "The HTML pages are in mybook/_build/html.\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Finished generating HTML for book.\n",
-      "Your book's HTML pages are here:\n",
-      "    mybook/_build/html/\n",
-      "You can look at your book by opening this file in a browser:\n",
-      "    mybook/_build/html/index.html\n",
-      "Or paste this line directly into your browser bar:\n",
-      "    file:///workspaces/LeetCode-Solutions/mybook/_build/html/index.html            \n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "format: jb-book\n",
-      "root: intro\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book toc from-project /workspaces/LeetCode-Solutions/mybook/ -f jb-book"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 2
-}
diff --git a/docs/CODE_OF_CONDUCT.md b/docs/CODE_OF_CONDUCT.md
deleted file mode 100644
index 5fe5041..0000000
--- a/docs/CODE_OF_CONDUCT.md
+++ /dev/null
@@ -1,128 +0,0 @@
-# Contributor Covenant Code of Conduct
-
-## Our Pledge
-
-We as members, contributors, and leaders pledge to make participation in our
-community a harassment-free experience for everyone, regardless of age, body
-size, visible or invisible disability, ethnicity, sex characteristics, gender
-identity and expression, level of experience, education, socio-economic status,
-nationality, personal appearance, race, religion, or sexual identity
-and orientation.
-
-We pledge to act and interact in ways that contribute to an open, welcoming,
-diverse, inclusive, and healthy community.
-
-## Our Standards
-
-Examples of behavior that contributes to a positive environment for our
-community include:
-
-* Demonstrating empathy and kindness toward other people
-* Being respectful of differing opinions, viewpoints, and experiences
-* Giving and gracefully accepting constructive feedback
-* Accepting responsibility and apologizing to those affected by our mistakes,
-  and learning from the experience
-* Focusing on what is best not just for us as individuals, but for the
-  overall community
-
-Examples of unacceptable behavior include:
-
-* The use of sexualized language or imagery, and sexual attention or
-  advances of any kind
-* Trolling, insulting or derogatory comments, and personal or political attacks
-* Public or private harassment
-* Publishing others' private information, such as a physical or email
-  address, without their explicit permission
-* Other conduct which could reasonably be considered inappropriate in a
-  professional setting
-
-## Enforcement Responsibilities
-
-Community leaders are responsible for clarifying and enforcing our standards of
-acceptable behavior and will take appropriate and fair corrective action in
-response to any behavior that they deem inappropriate, threatening, offensive,
-or harmful.
-
-Community leaders have the right and responsibility to remove, edit, or reject
-comments, commits, code, wiki edits, issues, and other contributions that are
-not aligned to this Code of Conduct, and will communicate reasons for moderation
-decisions when appropriate.
-
-## Scope
-
-This Code of Conduct applies within all community spaces, and also applies when
-an individual is officially representing the community in public spaces.
-Examples of representing our community include using an official e-mail address,
-posting via an official social media account, or acting as an appointed
-representative at an online or offline event.
-
-## Enforcement
-
-Instances of abusive, harassing, or otherwise unacceptable behavior may be
-reported to the community leaders responsible for enforcement at
-Email.
-All complaints will be reviewed and investigated promptly and fairly.
-
-All community leaders are obligated to respect the privacy and security of the
-reporter of any incident.
-
-## Enforcement Guidelines
-
-Community leaders will follow these Community Impact Guidelines in determining
-the consequences for any action they deem in violation of this Code of Conduct:
-
-### 1. Correction
-
-**Community Impact**: Use of inappropriate language or other behavior deemed
-unprofessional or unwelcome in the community.
-
-**Consequence**: A private, written warning from community leaders, providing
-clarity around the nature of the violation and an explanation of why the
-behavior was inappropriate. A public apology may be requested.
-
-### 2. Warning
-
-**Community Impact**: A violation through a single incident or series
-of actions.
-
-**Consequence**: A warning with consequences for continued behavior. No
-interaction with the people involved, including unsolicited interaction with
-those enforcing the Code of Conduct, for a specified period of time. This
-includes avoiding interactions in community spaces as well as external channels
-like social media. Violating these terms may lead to a temporary or
-permanent ban.
-
-### 3. Temporary Ban
-
-**Community Impact**: A serious violation of community standards, including
-sustained inappropriate behavior.
-
-**Consequence**: A temporary ban from any sort of interaction or public
-communication with the community for a specified period of time. No public or
-private interaction with the people involved, including unsolicited interaction
-with those enforcing the Code of Conduct, is allowed during this period.
-Violating these terms may lead to a permanent ban.
-
-### 4. Permanent Ban
-
-**Community Impact**: Demonstrating a pattern of violation of community
-standards, including sustained inappropriate behavior,  harassment of an
-individual, or aggression toward or disparagement of classes of individuals.
-
-**Consequence**: A permanent ban from any sort of public interaction within
-the community.
-
-## Attribution
-
-This Code of Conduct is adapted from the [Contributor Covenant][homepage],
-version 2.0, available at
-https://www.contributor-covenant.org/version/2/0/code_of_conduct.html.
-
-Community Impact Guidelines were inspired by [Mozilla's code of conduct
-enforcement ladder](https://github.com/mozilla/diversity).
-
-[homepage]: https://www.contributor-covenant.org
-
-For answers to common questions about this code of conduct, see the FAQ at
-https://www.contributor-covenant.org/faq. Translations are available at
-https://www.contributor-covenant.org/translations.
diff --git a/docs/CONTRIBUTING.md b/docs/CONTRIBUTING.md
deleted file mode 100644
index ca794ef..0000000
--- a/docs/CONTRIBUTING.md
+++ /dev/null
@@ -1,79 +0,0 @@
-# Contributing Guidelines
-
-Thank you for considering contributing to the Blind 75 LeetCode repository! We welcome contributions from the community to help improve this resource and make it even more valuable.
-
-Please take a moment to review this document to understand how you can contribute.
-
-## Table of Contents
-- [Ways to Contribute](#ways-to-contribute)
-- [Getting Started](#getting-started)
-- [Contributing Process](#contributing-process)
-- [Code of Conduct](#code-of-conduct)
-
-## Ways to Contribute
-
-There are several ways you can contribute to this repository:
-
-1. **Suggest Improvements**: If you have ideas for making the Blind 75 LeetCode repository better, please open an issue to discuss your suggestions.
-
-2. **Report Issues**: If you encounter any issues with the existing content or code, please open an issue and provide as much detail as possible.
-
-3. **Contribute Solutions**: You can contribute solutions to the problems listed in the repository. Please follow the guidelines mentioned below when submitting your solutions.
-
-4. **Enhance Documentation**: Improve documentation, including explanations and time/space complexity analysis, for existing solutions.
-
-5. **Fix Bugs**: If you find and fix any bugs in the existing code or documentation, you are welcome to submit a pull request.
-
-## Getting Started
-
-Before you start contributing, please ensure you have the following prerequisites:
-
-- A GitHub account.
-- Familiarity with Git and GitHub.
-- Code editor or IDE for making changes.
-
-## Contributing Process
-
-Here's the general process for contributing:
-
-1. **Fork the Repository**: Click the "Fork" button at the top of this repository to create your copy.
-
-2. **Clone Your Fork**: Clone your fork to your local machine.
-
-    ```
-    git clone https://github.com/your-username/Blind-75-LeetCode.git
-    ```
-
-3. **Create a Branch**: Create a new branch for your contribution.
-
-    ```
-    git checkout -b feature/your-feature-name
-    ```
-
-4. **Make Changes**: Implement your changes, following the guidelines provided.
-
-5. **Test Your Changes**: Ensure that your changes do not introduce new issues and are consistent with the repository's style and structure.
-
-6. **Commit Changes**: Commit your changes with a meaningful commit message.
-
-    ```
-    git commit -m "Add solution to problem X"
-    ```
-
-7. **Push Changes**: Push your changes to your fork on GitHub.
-
-    ```
-    git push origin feature/your-feature-name
-    ```
-
-8. **Create a Pull Request**: Go to the original repository and create a pull request. Please provide a detailed description of your changes.
-
-9. **Review and Collaboration**: Your pull request will be reviewed, and you may need to make further changes. Be responsive to comments and feedback.
-
-10. **Merge**: Once your pull request is approved, it will be merged into the main branch.
-
-## Code of Conduct
-
-Please note that we have a [Code of Conduct](CODE_OF_CONDUCT.md) that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.
-
-We appreciate your contributions to the Blind 75 LeetCode repository and look forward to working together to make it an even better resource for the community!
diff --git a/docs/LICENSE b/docs/LICENSE
deleted file mode 100644
index 8818d1f..0000000
--- a/docs/LICENSE
+++ /dev/null
@@ -1,21 +0,0 @@
-MIT License
-
-Copyright (c) 2023 Mohsen Tabibian
-
-Permission is hereby granted, free of charge, to any person obtaining a copy
-of this software and associated documentation files (the "Software"), to deal
-in the Software without restriction, including without limitation the rights
-to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
-copies of the Software, and to permit persons to whom the Software is
-furnished to do so, subject to the following conditions:
-
-The above copyright notice and this permission notice shall be included in all
-copies or substantial portions of the Software.
-
-THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
-IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
-FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
-AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
-LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
-OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
-SOFTWARE.
diff --git a/docs/README.md b/docs/README.md
deleted file mode 100644
index a9b8b2e..0000000
--- a/docs/README.md
+++ /dev/null
@@ -1,216 +0,0 @@
-# Blind 75 LeetCode Solutions
-
-![Blind 75 LeetCode](https://img.shields.io/badge/Blind_75_LeetCode-Solutions-blue?labelColor=red)
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Array & Hashing
-![Array & Hashing](https://img.shields.io/badge/Array_&_Hashing-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [217. Contains Duplicate](https://leetcode.com/problems/contains-duplicate/)                        | Easy       |
-| [242. Valid Anagram](https://leetcode.com/problems/valid-anagram/)                                  | Easy       |
-| [1. Two Sum](https://leetcode.com/problems/two-sum/)                                              | Easy       |
-| [49. Group Anagrams](https://leetcode.com/problems/group-anagrams/)                                | Medium     |
-| [347. Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/)              | Medium     |
-| [238. Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/)    | Medium     |
-| [659. Encode and Decode Strings](https://leetcode.com/problems/encode-and-decode-strings/)          | Medium     |
-| [128. Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/)    | Hard       |
-
-### Two Pointers
-![Array & Hashing](https://img.shields.io/badge/Two_Pointers-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)                            | Easy       |
-| [15. 3Sum](https://leetcode.com/problems/3sum/)                                                       | Medium     |
-| [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)           | Medium     |
-
-### Sliding Window
-![Array & Hashing](https://img.shields.io/badge/Sliding_Window-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [121. Best Time to Buy And Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | Easy       |
-| [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) | Medium |
-| [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) | Medium |
-| [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)             | Hard       |
-
-### Stack
-![Array & Hashing](https://img.shields.io/badge/Stack-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)                            | Easy       |
-
-### Binary Search
-![Array & Hashing](https://img.shields.io/badge/Binary_Search-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [153. Find Minimum In Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) | Medium |
-| [33. Search In Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)   | Medium     |
-
-### Linked List
-![Array & Hashing](https://img.shields.io/badge/Linked_List-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)                       | Easy       |
-| [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)                   | Easy       |
-| [143. Reorder List](https://leetcode.com/problems/reorder-list/)                                     | Medium     |
-| [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | Medium |
-| [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)                           | Medium     |
-| [23. Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)                       | Hard       |
-
-### Trees
-![Array & Hashing](https://img.shields.io/badge/Trees-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)                          | Easy       |
-| [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)     | Easy       |
-| [100. Same Tree](https://leetcode.com/problems/same-tree/)                                           | Easy       |
-| [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)               | Easy       |
-| [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/) | Easy |
-| [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/) | Medium |
-| [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)       | Medium     |
-| [230. Kth Smallest Element In a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)   | Medium     |
-| [105. Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) | Medium |
-| [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)     | Hard       |
-| [297. Serialize And Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/) | Hard |
-
-### Tries
-![Array & Hashing](https://img.shields.io/badge/Tries-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [208. Implement Trie Prefix Tree](https://leetcode.com/problems/implement-trie-prefix-tree/)         | Medium     |
-| [211. Design Add And Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/) | Medium |
-| [212. Word Search II](https://leetcode.com/problems/word-search-ii/)                                 | Hard       |
-
-### Heap / Priority Queue
-![Array & Hashing](https://img.shields.io/badge/Heap_/_Priority_Queue-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [295. Find Median From Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)     | Hard       |
-
-### Backtracking
-![Array & Hashing](https://img.shields.io/badge/Backtracking-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [39. Combination Sum](https://leetcode.com/problems/combination-sum/)                                 | Medium     |
-| [79. Word Search](https://leetcode.com/problems/word-search/)                                         | Medium     |
-
-### Graphs
-![Array & Hashing](https://img.shields.io/badge/Graphs-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)                           | Medium     |
-| [133. Clone Graph](https://leetcode.com/problems/clone-graph/)                                       | Medium     |
-| [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)       | Medium     |
-| [207. Course Schedule](https://leetcode.com/problems/course-schedule/)                                 | Medium     |
-| [323. Number of Connected Components In An Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/) | Medium |
-| [261. Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)                               | Medium     |
-
-### Advanced Graphs
-![Array & Hashing](https://img.shields.io/badge/Advanced_Graphs-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [269. Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)                               | Hard       |
-
-### 1-D Dynamic Programming
-![Array & Hashing](https://img.shields.io/badge/1_D_Dynamic_Programming-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)                                   | Easy       |
-| [198. House Robber](https://leetcode.com/problems/house-robber/)                                       | Easy       |
-| [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)                                 | Medium     |
-| [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)       | Medium     |
-| [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)                   | Medium     |
-| [91. Decode Ways](https://leetcode.com/problems/decode-ways/)                                         | Medium     |
-| [322. Coin Change](https://leetcode.com/problems/coin-change/)                                         | Medium     |
-| [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)             | Medium     |
-| [139. Word Break](https://leetcode.com/problems/word-break/)                                           | Medium     |
-| [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | Medium     |
-
-### 2-D Dynamic Programming
-![Array & Hashing](https://img.shields.io/badge/2_D_Dynamic_Programming-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [62. Unique Paths](https://leetcode.com/problems/unique-paths/)                                         | Medium     |
-| [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)       | Medium     |
-
-### Greedy
-![Array & Hashing](https://img.shields.io/badge/Greedy-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)                                   | Easy       |
-| [55. Jump Game](https://leetcode.com/problems/jump-game/)                                                 | Medium     |
-
-### Intervals
-![Array & Hashing](https://img.shields.io/badge/Intervals-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [57. Insert Interval](https://leetcode.com/problems/insert-interval/)                                     | Hard       |
-| [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)                                   | Medium     |
-| [435. Non Overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)             | Medium     |
-| [252. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)                                       | Easy       |
-| [253. Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)                                 | Medium     |
-
-### Math & Geometry
-![Array & Hashing](https://img.shields.io/badge/Math_&_Geometry-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [48. Rotate Image](https://leetcode.com/problems/rotate-image/)                                           | Medium     |
-| [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)                                         | Medium     |
-| [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)                               | Medium     |
-
-### Bit Manipulation
-![Array & Hashing](https://img.shields.io/badge/Bit_Manipulation-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)                                   | Easy       |
-| [338. Counting Bits](https://leetcode.com/problems/counting-bits/)                                         | Medium     |
-| [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)                                           | Easy       |
-| [268. Missing Number](https://leetcode.com/problems/missing-number/)                                       | Easy       |
-| [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)                             | Medium     |
-
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-## Code of Conduct
-
-We have a [Code of Conduct](CODE_OF_CONDUCT.md) that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.
-
-## Security Policy
-
-We take security seriously. Please review our [Security Policy](SECURITY.md) for information on reporting security vulnerabilities.
-
-Happy coding!
diff --git a/docs/SECURITY.md b/docs/SECURITY.md
deleted file mode 100644
index 7c19993..0000000
--- a/docs/SECURITY.md
+++ /dev/null
@@ -1,42 +0,0 @@
-# Security Policy
-
-## Reporting a Vulnerability
-
-We take the security of the Blind 75 LeetCode repository seriously. If you believe you have found a security vulnerability, please help us by following these steps:
-
-1. **Do Not Disclose Publicly**: Please do not disclose the issue publicly until it has been addressed by our team.
-
-2. **Report Privately**: Create a GitHub issue with the title "Security Vulnerability Report" and provide a detailed description of the vulnerability. You can use a sample report template as follows:
-
-    ```
-    **Description**
-    A clear and concise description of the security vulnerability.
-
-    **Steps to Reproduce**
-    Provide detailed steps on how to reproduce the vulnerability.
-
-    **Expected Behavior**
-    Describe what you expected to happen.
-
-    **Actual Behavior**
-    Describe what actually happened.
-
-    **Additional Information**
-    Any additional information that can help us understand and reproduce the issue, such as environment details, configuration, or screenshots.
-    ```
-
-3. **Secure Communication**: If the vulnerability involves sensitive data, please encrypt your communication with our PGP key. Contact us for our PGP key information.
-
-4. **Response**: We will acknowledge your report within 5 business days and provide an estimated timeline for the resolution.
-
-5. **Fix and Disclosure**: We will work to fix the issue promptly and privately. Once resolved, we will provide a security advisory and give credit to the reporter if desired.
-
-6. **Responsible Disclosure**: We encourage responsible disclosure. Please do not take advantage of the vulnerability or attempt to exploit it for malicious purposes.
-
-## Security Updates
-
-We are committed to promptly addressing and resolving any security issues. We recommend that you stay up-to-date with the latest releases of this repository, as they may include security updates and fixes.
-
-## Contact
-
-If you have any questions or need further assistance regarding security, you can contact us at mohsentabibian@gmail.com.
diff --git a/docs/_build/.doctrees/01. Array Hashing/1. Two Sum.doctree b/docs/_build/.doctrees/01. Array Hashing/1. Two Sum.doctree
deleted file mode 100644
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diff --git a/docs/_build/.doctrees/06. Linked List/19. Remove Nth Node From End of List.doctree b/docs/_build/.doctrees/06. Linked List/19. Remove Nth Node From End of List.doctree
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diff --git a/docs/_build/.doctrees/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.doctree b/docs/_build/.doctrees/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.doctree
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diff --git a/docs/_build/.doctrees/07. Trees/297. Serialize and Deserialize Binary Tree.doctree b/docs/_build/.doctrees/07. Trees/297. Serialize and Deserialize Binary Tree.doctree
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diff --git a/docs/_build/.doctrees/08. Tries/208. Implement Trie (Prefix Tree).doctree b/docs/_build/.doctrees/08. Tries/208. Implement Trie (Prefix Tree).doctree
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diff --git a/docs/_build/.doctrees/08. Tries/211. Design Add and Search Words Data Structure.doctree b/docs/_build/.doctrees/08. Tries/211. Design Add and Search Words Data Structure.doctree
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diff --git a/docs/_build/.doctrees/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.doctree b/docs/_build/.doctrees/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.doctree
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diff --git a/docs/_build/.doctrees/11. Graphs/3651. Number of Connected Components in an Undirected Graph.doctree b/docs/_build/.doctrees/11. Graphs/3651. Number of Connected Components in an Undirected Graph.doctree
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="same-tree">
-<h1>100: Same Tree<a class="headerlink" href="#same-tree" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/same-tree/">To see the Same Tree problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given the roots of two binary trees <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code>, write a function to check if they are the same or not.</p>
-<p>Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.</p>
-<p><strong>Constraints</strong></p>
-<ol class="arabic simple">
-<li><p>The number of nodes in both trees is in the range <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">100]</span></code>.</p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^4\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">Node.val</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-</ol>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Definition for a binary tree node.</span>
-<span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="c1"># Initialize a TreeNode with a value (val), left child, and right child.</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">def</span> <span class="nf">isSameTree</span><span class="p">(</span><span class="n">p</span><span class="p">,</span> <span class="n">q</span><span class="p">):</span>
-    <span class="c1"># Base case: If both p and q are None, the trees are the same.</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">p</span> <span class="ow">and</span> <span class="ow">not</span> <span class="n">q</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">True</span>
-    
-    <span class="c1"># Base case: If either p or q is None (but not both), the trees are different.</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">p</span> <span class="ow">or</span> <span class="ow">not</span> <span class="n">q</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">False</span>
-    
-    <span class="c1"># Check if the values of the current nodes (p.val and q.val) are equal.</span>
-    <span class="k">if</span> <span class="n">p</span><span class="o">.</span><span class="n">val</span> <span class="o">!=</span> <span class="n">q</span><span class="o">.</span><span class="n">val</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">False</span>
-    
-    <span class="c1"># Recursively check the left and right subtrees of p and q.</span>
-    <span class="c1"># If both subtrees are the same, the entire trees are the same.</span>
-    <span class="k">return</span> <span class="n">isSameTree</span><span class="p">(</span><span class="n">p</span><span class="o">.</span><span class="n">left</span><span class="p">,</span> <span class="n">q</span><span class="o">.</span><span class="n">left</span><span class="p">)</span> <span class="ow">and</span> <span class="n">isSameTree</span><span class="p">(</span><span class="n">p</span><span class="o">.</span><span class="n">right</span><span class="p">,</span> <span class="n">q</span><span class="o">.</span><span class="n">right</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-</div>
-<p>In this code, we define a <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class to represent binary tree nodes and a <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function to check if two binary trees are the same. The function uses recursive traversal to compare the trees’ structures and values.</p>
-<ol class="arabic simple">
-<li><p>We start by defining a <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class, which represents a node in a binary tree. Each node has a <code class="docutils literal notranslate"><span class="pre">val</span></code> (the node’s value), a <code class="docutils literal notranslate"><span class="pre">left</span></code> child, and a right child. This class will help us create and work with binary trees.</p></li>
-<li><p>Next, we define the <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function, which checks if two binary trees (<code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code>) are the same.</p>
-<ul class="simple">
-<li><p>The base case for the recursion is when both <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code> are <code class="docutils literal notranslate"><span class="pre">None</span></code>. In this case, they are considered the same, so we return <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>If either <code class="docutils literal notranslate"><span class="pre">p</span></code> or <code class="docutils literal notranslate"><span class="pre">q</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code> (but not both), they cannot be the same, so we return <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-<li><p>If the values of the current nodes <code class="docutils literal notranslate"><span class="pre">p.val</span></code> and <code class="docutils literal notranslate"><span class="pre">q.val</span></code> are not equal, we return <code class="docutils literal notranslate"><span class="pre">False</span></code> because the trees cannot be the same.</p></li>
-<li><p>Finally, we recursively check the left and right subtrees of <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code> to see if they are the same.</p></li>
-</ul>
-</li>
-</ol>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 1</span>
-
-<span class="c1">#Input: `p = [1,2,3]`, `q = [1,2,3]`</span>
-
-<span class="n">p1</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">),</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">))</span>
-<span class="n">q1</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">),</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">))</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isSameTree</span><span class="p">(</span><span class="n">p1</span><span class="p">,</span> <span class="n">q1</span><span class="p">))</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 2</span>
-
-<span class="c1">#Input: `p = [1,2]`, `q = [1,null,2]`</span>
-
-<span class="n">p2</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">),</span> <span class="kc">None</span><span class="p">)</span>
-<span class="n">q2</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="kc">None</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">))</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isSameTree</span><span class="p">(</span><span class="n">p2</span><span class="p">,</span> <span class="n">q2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 3</span>
-
-<span class="c1">#Input: p = [1,2,1], q = [1,1,2]</span>
-
-<span class="n">p3</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">),</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">))</span>
-<span class="n">q3</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">),</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">))</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isSameTree</span><span class="p">(</span><span class="n">p3</span><span class="p">,</span> <span class="n">q3</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p><strong>Time Complexity</strong></p>
-<p>The time complexity of the <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function can be analyzed as follows:</p>
-<p>In the worst case, the function needs to visit every node in both trees once to determine if they are the same.
-Since each node is visited exactly once, the time complexity is <span class="math notranslate nohighlight">\(O(n)\)</span>, where <span class="math notranslate nohighlight">\(n\)</span> is the total number of nodes in the input trees.</p>
-<p><strong>Space Complexity</strong>
-The space complexity of the <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function can be analyzed as follows:</p>
-<p>The space used by the function’s call stack during recursion is proportional to the maximum depth of the binary trees.
-In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be <span class="math notranslate nohighlight">\(n\)</span>, where <span class="math notranslate nohighlight">\(n\)</span> is the total number of nodes in the input trees.
-Therefore, the space complexity is <span class="math notranslate nohighlight">\(O(n)\)</span> due to the recursive call stack.
-In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.</p>
-<p><strong>In summary:</strong></p>
-<ul class="simple">
-<li><p>Time Complexity: <span class="math notranslate nohighlight">\(O(n)\)</span> where <span class="math notranslate nohighlight">\(n\)</span> is the total number of nodes in the input trees.</p></li>
-<li><p>Space Complexity: <span class="math notranslate nohighlight">\(O(n)\)</span> due to the recursive call stack.</p></li>
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-    <h1>128. Longest Consecutive Sequence</h1>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="longest-consecutive-sequence">
-<h1>128. Longest Consecutive Sequence<a class="headerlink" href="#longest-consecutive-sequence" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/longest-consecutive-sequence/">To see the Longest Consecutive Sequence problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given an unsorted array of integers <code class="docutils literal notranslate"><span class="pre">nums</span></code>, return <em>the length of the longest consecutive elements sequence.</em></p>
-<p>You must write an algorithm that runs in <code class="docutils literal notranslate"><span class="pre">O(n)</span></code> time.</p>
-<p><em>Constraints:</em>*</p>
-<ul class="simple">
-<li><p>0 &lt;= <code class="docutils literal notranslate"><span class="pre">nums.length</span></code> &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^9\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">nums[i]</span></code> &lt;= <span class="math notranslate nohighlight">\(10^9\)</span></p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">longestConsecutive</span><span class="p">(</span><span class="n">nums</span><span class="p">):</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">nums</span><span class="p">:</span>
-        <span class="k">return</span> <span class="mi">0</span>
-
-    <span class="n">num_set</span> <span class="o">=</span> <span class="nb">set</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span>
-    <span class="n">max_length</span> <span class="o">=</span> <span class="mi">0</span>
-
-    <span class="k">for</span> <span class="n">num</span> <span class="ow">in</span> <span class="n">num_set</span><span class="p">:</span>
-        <span class="k">if</span> <span class="n">num</span> <span class="o">-</span> <span class="mi">1</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">num_set</span><span class="p">:</span>
-            <span class="n">current_num</span> <span class="o">=</span> <span class="n">num</span>
-            <span class="n">current_length</span> <span class="o">=</span> <span class="mi">1</span>
-
-            <span class="k">while</span> <span class="n">current_num</span> <span class="o">+</span> <span class="mi">1</span> <span class="ow">in</span> <span class="n">num_set</span><span class="p">:</span>
-                <span class="n">current_num</span> <span class="o">+=</span> <span class="mi">1</span>
-                <span class="n">current_length</span> <span class="o">+=</span> <span class="mi">1</span>
-
-            <span class="n">max_length</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">max_length</span><span class="p">,</span> <span class="n">current_length</span><span class="p">)</span>
-
-    <span class="k">return</span> <span class="n">max_length</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We start by defining the <code class="docutils literal notranslate"><span class="pre">longestConsecutive</span></code> function that takes the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array as input.</p></li>
-<li><p>We check if the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array is empty. If it’s empty, there are no consecutive elements, so we return 0.</p></li>
-<li><p>We create a Python set called <code class="docutils literal notranslate"><span class="pre">num_set</span></code> and insert all elements from the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array into it. Using a set allows us to efficiently check for the existence of elements in O(1) time.</p></li>
-<li><p>We initialize a variable <code class="docutils literal notranslate"><span class="pre">max_length</span></code> to 0. This variable will keep track of the maximum length of consecutive elements sequence found.</p></li>
-<li><p>We iterate through the elements of the <code class="docutils literal notranslate"><span class="pre">num_set</span></code>. For each element <code class="docutils literal notranslate"><span class="pre">num</span></code>, we check if <code class="docutils literal notranslate"><span class="pre">num</span> <span class="pre">-</span> <span class="pre">1</span></code> exists in the <code class="docutils literal notranslate"><span class="pre">num_set</span></code>. If it doesn’t exist, it means <code class="docutils literal notranslate"><span class="pre">num</span></code> is the starting element of a potential consecutive sequence.</p></li>
-<li><p>Inside the loop, we initialize two variables: <code class="docutils literal notranslate"><span class="pre">current_num</span></code> to the current element <code class="docutils literal notranslate"><span class="pre">num</span></code> and <code class="docutils literal notranslate"><span class="pre">current_length</span></code> to 1. We start with a length of 1 because <code class="docutils literal notranslate"><span class="pre">num</span></code> itself is part of the sequence.</p></li>
-<li><p>We then enter a while loop that continues as long as <code class="docutils literal notranslate"><span class="pre">current_num</span> <span class="pre">+</span> <span class="pre">1</span></code> exists in the <code class="docutils literal notranslate"><span class="pre">num_set</span></code>. This means we are incrementing the consecutive sequence.</p></li>
-<li><p>Inside the while loop, we increment <code class="docutils literal notranslate"><span class="pre">current_num</span></code> by 1 and also increment <code class="docutils literal notranslate"><span class="pre">current_length</span></code> by 1 to account for the next consecutive element.</p></li>
-<li><p>We compare <code class="docutils literal notranslate"><span class="pre">current_length</span></code> with the <code class="docutils literal notranslate"><span class="pre">max_length</span></code> and update <code class="docutils literal notranslate"><span class="pre">max_length</span></code> if the current sequence is longer.</p></li>
-<li><p>After the while loop, we move to the next element in the outer loop and repeat the process.</p></li>
-<li><p>Finally, we return the <code class="docutils literal notranslate"><span class="pre">max_length</span></code> as the result, which represents the length of the longest consecutive elements sequence in the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array.</p></li>
-</ol>
-<p>The key idea here is to use a set to efficiently check for the existence of elements and to iterate through the elements, considering each element as the potential start of a consecutive sequence. By doing this, we can find the longest consecutive sequence in O(n) time complexity, where n is the number of elements in the array.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="n">nums1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">100</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">200</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">longestConsecutive</span><span class="p">(</span><span class="n">nums1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>4
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-
-<span class="n">nums2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">7</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">5</span><span class="p">,</span> <span class="mi">8</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">longestConsecutive</span><span class="p">(</span><span class="n">nums2</span><span class="p">))</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>9
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong>
-The code is designed to run in O(n) time complexity, where n is the number of elements in the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array. Here’s the breakdown:</p>
-<ol class="arabic simple">
-<li><p>Constructing the <code class="docutils literal notranslate"><span class="pre">num_set</span></code> by inserting all elements from <code class="docutils literal notranslate"><span class="pre">nums</span></code> into it takes O(n) time because we perform an insertion operation for each element in <code class="docutils literal notranslate"><span class="pre">nums</span></code>.</p></li>
-<li><p>The main loop iterates through the elements in <code class="docutils literal notranslate"><span class="pre">num_set</span></code>. In the worst case, each element is visited only once, so the loop itself takes O(n) time.</p></li>
-<li><p>Within the loop, we have a while loop that may also take O(n) time in the worst case. However, this while loop is nested inside the main loop, so its overall time complexity remains O(n).</p></li>
-</ol>
-<p>Therefore, the overall time complexity of the code is O(n).</p>
-<p><strong>Space Complexity:</strong>
-The space complexity of the code is determined by the space used by the <code class="docutils literal notranslate"><span class="pre">num_set</span></code> and a few additional variables. Here’s the breakdown:</p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">num_set</span></code> is a set that stores the unique elements from <code class="docutils literal notranslate"><span class="pre">nums</span></code>. In the worst case, it can store all n elements from <code class="docutils literal notranslate"><span class="pre">nums</span></code>, so the space complexity is O(n).</p></li>
-<li><p>The additional variables used, such as <code class="docutils literal notranslate"><span class="pre">max_length</span></code>, <code class="docutils literal notranslate"><span class="pre">current_num</span></code>, and <code class="docutils literal notranslate"><span class="pre">current_length</span></code>, have constant space requirements and do not depend on the size of the input array.</p></li>
-</ol>
-<p>Therefore, the overall space complexity of the code is O(n) due to the space used by the <code class="docutils literal notranslate"><span class="pre">num_set</span></code>.</p>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="contains-duplicate">
-<h1>217. Contains Duplicate<a class="headerlink" href="#contains-duplicate" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/contains-duplicate/">To see the Contains Duplicate problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given an integer array <code class="docutils literal notranslate"><span class="pre">nums</span></code>, return <code class="docutils literal notranslate"><span class="pre">true</span></code> if any value appears <strong>at least twice</strong> in the array, and return <code class="docutils literal notranslate"><span class="pre">false</span></code> if every element is distinct.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">nums.length</span></code> &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^9\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">nums[i]</span></code> &lt;= <span class="math notranslate nohighlight">\(10^9\)</span></p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">containsDuplicate</span><span class="p">(</span><span class="n">nums</span><span class="p">):</span>
-    <span class="c1"># Create an empty set to store unique elements</span>
-    <span class="n">unique_elements</span> <span class="o">=</span> <span class="nb">set</span><span class="p">()</span>
-    
-    <span class="c1"># Iterate through the array</span>
-    <span class="k">for</span> <span class="n">num</span> <span class="ow">in</span> <span class="n">nums</span><span class="p">:</span>
-        <span class="c1"># If the element is already in the set, return True</span>
-        <span class="k">if</span> <span class="n">num</span> <span class="ow">in</span> <span class="n">unique_elements</span><span class="p">:</span>
-            <span class="k">return</span> <span class="kc">True</span>
-        <span class="c1"># Otherwise, add it to the set</span>
-        <span class="k">else</span><span class="p">:</span>
-            <span class="n">unique_elements</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="n">num</span><span class="p">)</span>
-    
-    <span class="c1"># If the loop completes without finding duplicates, return False</span>
-    <span class="k">return</span> <span class="kc">False</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">containsDuplicate</span></code> function takes a single argument, <code class="docutils literal notranslate"><span class="pre">nums</span></code>, which is the input integer array.</p></li>
-<li><p>Inside the function, an empty set called <code class="docutils literal notranslate"><span class="pre">unique_elements</span></code> is created. This set will be used to keep track of unique elements in the input array.</p></li>
-<li><p>The function then iterates through the input array <code class="docutils literal notranslate"><span class="pre">nums</span></code> using a <code class="docutils literal notranslate"><span class="pre">for</span></code> loop.</p></li>
-<li><p>For each element <code class="docutils literal notranslate"><span class="pre">num</span></code> in the array, it checks whether <code class="docutils literal notranslate"><span class="pre">num</span></code> is already in the <code class="docutils literal notranslate"><span class="pre">unique_elements</span></code> set using the <code class="docutils literal notranslate"><span class="pre">if</span> <span class="pre">num</span> <span class="pre">in</span> <span class="pre">unique_elements:</span></code> condition.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">num</span></code> is already in the set, it means there is a duplicate element in the array, and the function immediately returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">num</span></code> is not in the set, it is added to the <code class="docutils literal notranslate"><span class="pre">unique_elements</span></code> set using <code class="docutils literal notranslate"><span class="pre">unique_elements.add(num)</span></code>.</p></li>
-<li><p>The loop continues to the next element, and the process repeats.</p></li>
-<li><p>If the loop completes without finding any duplicates, it means that all elements in the array are distinct, and the function returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-</ol>
-<p>The code efficiently utilizes a set data structure to keep track of unique elements while iterating through the array, allowing it to quickly detect duplicate elements. This code meets the problem’s requirements and constraints, as explained earlier.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="n">nums1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">containsDuplicate</span><span class="p">(</span><span class="n">nums1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">nums2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">4</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">containsDuplicate</span><span class="p">(</span><span class="n">nums2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3</span>
-<span class="n">nums3</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">containsDuplicate</span><span class="p">(</span><span class="n">nums3</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code for the “Contains Duplicate” problem:</p>
-<p><strong>Time Complexity:</strong></p>
-<p>The primary operation that affects the time complexity is the <code class="docutils literal notranslate"><span class="pre">for</span></code> loop that iterates through the input array <code class="docutils literal notranslate"><span class="pre">nums</span></code>. In the worst case, the loop will iterate through all <code class="docutils literal notranslate"><span class="pre">n</span></code> elements of the array, where <code class="docutils literal notranslate"><span class="pre">n</span></code> is the length of the input array.</p>
-<ul class="simple">
-<li><p>Iterating through the array: O(n)</p></li>
-</ul>
-<p>Inside the loop, we perform two operations:</p>
-<ol class="arabic simple">
-<li><p>Checking whether an element exists in the <code class="docutils literal notranslate"><span class="pre">unique_elements</span></code> set (<code class="docutils literal notranslate"><span class="pre">if</span> <span class="pre">num</span> <span class="pre">in</span> <span class="pre">unique_elements</span></code>). This operation has an average time complexity of O(1) for a set.</p></li>
-<li><p>Adding an element to the <code class="docutils literal notranslate"><span class="pre">unique_elements</span></code> set (<code class="docutils literal notranslate"><span class="pre">unique_elements.add(num)</span></code>), which also has an average time complexity of O(1) for a set.</p></li>
-</ol>
-<p>Since these operations are performed for each element in the array, the overall time complexity remains O(n).</p>
-<p><strong>Space Complexity:</strong></p>
-<p>The space complexity is determined by the additional data structures used in the code, which are the <code class="docutils literal notranslate"><span class="pre">unique_elements</span></code> set.</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">unique_elements</span></code> set: This set stores unique elements from the input array. In the worst case, if all elements in the input array are distinct, the set will store all <code class="docutils literal notranslate"><span class="pre">n</span></code> elements.</p></li>
-</ul>
-<p>Therefore, the space complexity is O(n) because, in the worst case, the set’s size will grow linearly with the input array’s size.</p>
-<p><strong>In summary</strong>, the time complexity of the code is O(n), where <code class="docutils literal notranslate"><span class="pre">n</span></code> is the length of the input array, and the space complexity is also O(n) in the worst case. This code is efficient and meets the constraints of the problem.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>K Duplicates:</strong> Modify the problem to return <code class="docutils literal notranslate"><span class="pre">true</span></code> if there are exactly K duplicate elements in the array. Write a function that takes an additional integer parameter K and returns <code class="docutils literal notranslate"><span class="pre">true</span></code> if there are exactly K duplicates.</p></li>
-<li><p><strong>Frequency Count:</strong> Write a function that returns a list of all the elements that appear more than once in the array, along with their frequencies.</p></li>
-</ol>
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-    <h1>238. Product of Array Except Self</h1>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="product-of-array-except-self">
-<h1>238. Product of Array Except Self<a class="headerlink" href="#product-of-array-except-self" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/product-of-array-except-self/">To see the Product of Array Except Self problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given an integer array <code class="docutils literal notranslate"><span class="pre">nums</span></code>, return <em>an array <code class="docutils literal notranslate"><span class="pre">answer</span></code> such that <code class="docutils literal notranslate"><span class="pre">answer[i]</span></code> is equal to the product of all the elements of <code class="docutils literal notranslate"><span class="pre">nums</span></code> except <code class="docutils literal notranslate"><span class="pre">nums[i]</span></code>.</em></p>
-<p>The product of any prefix or suffix of <code class="docutils literal notranslate"><span class="pre">nums</span></code> is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> integer.</p>
-<p>You must write an algorithm that runs in O(n) time and without using the division operation.</p>
-<p>Constraints:</p>
-<ul class="simple">
-<li><p>2 &lt;= <code class="docutils literal notranslate"><span class="pre">nums.length</span></code> &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-30</span> <span class="pre">&lt;=</span> <span class="pre">nums[i]</span> <span class="pre">&lt;=</span> <span class="pre">30</span></code></p></li>
-<li><p>The product of any prefix or suffix of <code class="docutils literal notranslate"><span class="pre">nums</span></code> is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> integer.</p></li>
-</ul>
-<p><strong>Follow-up:</strong> Can you solve the problem in O(1) extra space complexity? (The output array <strong>does not</strong> count as extra space for space complexity analysis.)</p>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">productExceptSelf</span><span class="p">(</span><span class="n">nums</span><span class="p">):</span>
-    <span class="n">n</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span>
-    
-    <span class="c1"># Initialize two arrays to store products to the left and right of each element</span>
-    <span class="n">left_products</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">]</span> <span class="o">*</span> <span class="n">n</span>
-    <span class="n">right_products</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">]</span> <span class="o">*</span> <span class="n">n</span>
-    
-    <span class="c1"># Calculate products to the left of each element</span>
-    <span class="n">left_product</span> <span class="o">=</span> <span class="mi">1</span>
-    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
-        <span class="n">left_products</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">=</span> <span class="n">left_product</span>
-        <span class="n">left_product</span> <span class="o">*=</span> <span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">]</span>
-    
-    <span class="c1"># Calculate products to the right of each element</span>
-    <span class="n">right_product</span> <span class="o">=</span> <span class="mi">1</span>
-    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="mi">1</span><span class="p">,</span> <span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="o">-</span><span class="mi">1</span><span class="p">):</span>
-        <span class="n">right_products</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">=</span> <span class="n">right_product</span>
-        <span class="n">right_product</span> <span class="o">*=</span> <span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">]</span>
-    
-    <span class="c1"># Calculate the final answer using left and right products</span>
-    <span class="n">answer</span> <span class="o">=</span> <span class="p">[</span><span class="n">left_products</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">*</span> <span class="n">right_products</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">)]</span>
-    
-    <span class="k">return</span> <span class="n">answer</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">n</span></code> is initialized as the length of the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array.</p></li>
-<li><p>Two arrays, <code class="docutils literal notranslate"><span class="pre">left_products</span></code> and <code class="docutils literal notranslate"><span class="pre">right_products</span></code>, are created, each with a length of <code class="docutils literal notranslate"><span class="pre">n</span></code> and filled with ones initially.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">left_product</span></code> is initialized to 1.</p></li>
-<li><p>A loop iterates through the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array, from left to right. For each element at index <code class="docutils literal notranslate"><span class="pre">i</span></code>, it stores the product of all elements to its left (including itself) in the <code class="docutils literal notranslate"><span class="pre">left_products</span></code> array and updates <code class="docutils literal notranslate"><span class="pre">left_product</span></code> accordingly.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">right_product</span></code> is initialized to 1.</p></li>
-<li><p>Another loop iterates through the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array in reverse order, from right to left. For each element at index <code class="docutils literal notranslate"><span class="pre">i</span></code>, it stores the product of all elements to its right (including itself) in the <code class="docutils literal notranslate"><span class="pre">right_products</span></code> array and updates <code class="docutils literal notranslate"><span class="pre">right_product</span></code>.</p></li>
-<li><p>The final <code class="docutils literal notranslate"><span class="pre">answer</span></code> list is constructed by multiplying corresponding elements from the <code class="docutils literal notranslate"><span class="pre">left_products</span></code> and <code class="docutils literal notranslate"><span class="pre">right_products</span></code> arrays for each index <code class="docutils literal notranslate"><span class="pre">i</span></code>.</p></li>
-<li><p>The function returns the <code class="docutils literal notranslate"><span class="pre">answer</span></code> list, which contains the desired results.</p></li>
-</ol>
-<p>Overall, this algorithm efficiently computes the product of all elements except the current element, as required. It runs in O(n) time complexity and uses O(n) extra space, which is within the problem’s constraints.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-<span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">4</span><span class="p">]</span>
-<span class="n">result1</span> <span class="o">=</span> <span class="n">productExceptSelf</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result1</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[24, 12, 8, 6]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="o">-</span><span class="mi">3</span><span class="p">,</span><span class="mi">3</span><span class="p">]</span>
-<span class="n">result2</span> <span class="o">=</span> <span class="n">productExceptSelf</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result2</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[0, 0, 9, 0, 0]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The first loop iterates through the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array to calculate the products to the left of each element. This loop runs in O(n) time, where n is the length of the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array.</p></li>
-<li><p>The second loop also iterates through the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array in reverse order to calculate the products to the right of each element. This loop also runs in O(n) time.</p></li>
-<li><p>The final loop constructs the <code class="docutils literal notranslate"><span class="pre">answer</span></code> list by multiplying elements from the <code class="docutils literal notranslate"><span class="pre">left_products</span></code> and <code class="docutils literal notranslate"><span class="pre">right_products</span></code> arrays. This loop runs in O(n) time.</p></li>
-</ol>
-<p>Since all the loops are independent and sequential, the overall time complexity is O(n).</p>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>Two additional arrays, <code class="docutils literal notranslate"><span class="pre">left_products</span></code> and <code class="docutils literal notranslate"><span class="pre">right_products</span></code>, are created with a length of <code class="docutils literal notranslate"><span class="pre">n</span></code>, where <code class="docutils literal notranslate"><span class="pre">n</span></code> is the length of the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array. Therefore, these arrays consume O(n) extra space.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">left_product</span></code> and <code class="docutils literal notranslate"><span class="pre">right_product</span></code> variables are used to keep track of the product to the left and right of each element, respectively. These variables occupy O(1) extra space.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">answer</span></code> list, which contains the final results, also consumes O(n) space, but it’s not counted toward the extra space complexity analysis as per the problem’s constraints.</p></li>
-</ol>
-<p>Overall, the space complexity is O(n) for this solution.</p>
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-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Handle Zeros in the Input</strong>: Modify the solution to handle cases where the input array contains zeros. For example, if <code class="docutils literal notranslate"><span class="pre">nums</span> <span class="pre">=</span> <span class="pre">[0,</span> <span class="pre">1,</span> <span class="pre">2,</span> <span class="pre">3,</span> <span class="pre">0]</span></code>, the output should be <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">0,</span> <span class="pre">0,</span> <span class="pre">0,</span> <span class="pre">0]</span></code> because any element multiplied by zero results in zero.</p></li>
-<li><p><strong>In-Place Modification</strong>: Attempt to solve the problem with in-place modification of the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array, where the output is stored directly in the input array without using any additional space.</p></li>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="valid-anagram">
-<h1>242. Valid Anagram<a class="headerlink" href="#valid-anagram" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/valid-anagram/">To see the Valid Anagram problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given two strings <code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code>, return <code class="docutils literal notranslate"><span class="pre">true</span></code> if <code class="docutils literal notranslate"><span class="pre">t</span></code> is an anagram of <code class="docutils literal notranslate"><span class="pre">s</span></code>, and <code class="docutils literal notranslate"><span class="pre">false</span></code> otherwise.</p>
-<p>An <strong>Anagram</strong> is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">s.length</span></code>, <code class="docutils literal notranslate"><span class="pre">t.length</span></code> &lt;= <span class="math notranslate nohighlight">\(5 * 10^4\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code> consist of lowercase English letters.</p></li>
-</ul>
-<p><strong>Follow-up:</strong> What if the inputs contain Unicode characters? How would you adapt your solution to such a case?</p>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">isAnagram</span><span class="p">(</span><span class="n">s</span><span class="p">:</span> <span class="nb">str</span><span class="p">,</span> <span class="n">t</span><span class="p">:</span> <span class="nb">str</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="nb">bool</span><span class="p">:</span>
-    <span class="c1"># Check if the lengths of s and t are not equal</span>
-    <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="n">s</span><span class="p">)</span> <span class="o">!=</span> <span class="nb">len</span><span class="p">(</span><span class="n">t</span><span class="p">):</span>
-        <span class="k">return</span> <span class="kc">False</span>
-
-    <span class="c1"># Create dictionaries to store character frequencies for s and t</span>
-    <span class="n">countS</span><span class="p">,</span> <span class="n">countT</span> <span class="o">=</span> <span class="p">{},</span> <span class="p">{}</span>
-
-    <span class="c1"># Count character frequencies in s</span>
-    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">s</span><span class="p">)):</span>
-        <span class="n">countS</span><span class="p">[</span><span class="n">s</span><span class="p">[</span><span class="n">i</span><span class="p">]]</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">+</span> <span class="n">countS</span><span class="o">.</span><span class="n">get</span><span class="p">(</span><span class="n">s</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="mi">0</span><span class="p">)</span>
-
-    <span class="c1"># Count character frequencies in t</span>
-    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">t</span><span class="p">)):</span>
-        <span class="n">countT</span><span class="p">[</span><span class="n">t</span><span class="p">[</span><span class="n">i</span><span class="p">]]</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">+</span> <span class="n">countT</span><span class="o">.</span><span class="n">get</span><span class="p">(</span><span class="n">t</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="mi">0</span><span class="p">)</span>
-
-    <span class="c1"># Check if the character frequencies in s and t are the same</span>
-    <span class="k">return</span> <span class="n">countS</span> <span class="o">==</span> <span class="n">countT</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>The code defines a class called <code class="docutils literal notranslate"><span class="pre">Solution</span></code> with a method <code class="docutils literal notranslate"><span class="pre">isAnagram</span></code> that takes two input strings, <code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code>, and returns a boolean (<code class="docutils literal notranslate"><span class="pre">True</span></code> or <code class="docutils literal notranslate"><span class="pre">False</span></code>) based on whether <code class="docutils literal notranslate"><span class="pre">t</span></code> is an anagram of <code class="docutils literal notranslate"><span class="pre">s</span></code>.</p></li>
-<li><p>The first check performed in the method is whether the lengths of <code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code> are equal. If they are not equal, it immediately returns <code class="docutils literal notranslate"><span class="pre">False</span></code> because strings of different lengths cannot be anagrams of each other.</p></li>
-<li><p>Two dictionaries, <code class="docutils literal notranslate"><span class="pre">countS</span></code> and <code class="docutils literal notranslate"><span class="pre">countT</span></code>, are created to store the frequency of characters in <code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code>, respectively. These dictionaries will be used to count the occurrences of each character in the strings.</p></li>
-<li><p>The code then enters a loop to iterate through each character in string <code class="docutils literal notranslate"><span class="pre">s</span></code> using a <code class="docutils literal notranslate"><span class="pre">for</span></code> loop. Inside the loop, it updates the <code class="docutils literal notranslate"><span class="pre">countS</span></code> dictionary. The line <code class="docutils literal notranslate"><span class="pre">countS[s[i]]</span> <span class="pre">=</span> <span class="pre">1</span> <span class="pre">+</span> <span class="pre">countS.get(s[i],</span> <span class="pre">0)</span></code> increments the count of character <code class="docutils literal notranslate"><span class="pre">s[i]</span></code> in <code class="docutils literal notranslate"><span class="pre">countS</span></code> by 1. If the character is not already in the dictionary, it initializes the count to 1.</p></li>
-<li><p>Similarly, the code iterates through each character in string <code class="docutils literal notranslate"><span class="pre">t</span></code> and updates the <code class="docutils literal notranslate"><span class="pre">countT</span></code> dictionary in the same way.</p></li>
-<li><p>After counting the character frequencies in both strings, the code compares the two dictionaries using <code class="docutils literal notranslate"><span class="pre">countS</span> <span class="pre">==</span> <span class="pre">countT</span></code>. If the dictionaries are equal, it means that both strings have the same character frequencies, and therefore, <code class="docutils literal notranslate"><span class="pre">t</span></code> is an anagram of <code class="docutils literal notranslate"><span class="pre">s</span></code>. In this case, the method returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>If the dictionaries are not equal, the method returns <code class="docutils literal notranslate"><span class="pre">False</span></code>, indicating that <code class="docutils literal notranslate"><span class="pre">t</span></code> is not an anagram of <code class="docutils literal notranslate"><span class="pre">s</span></code>.</p></li>
-</ol>
-<p>In summary, this code efficiently determines whether two input strings are anagrams by counting the character frequencies in each string using dictionaries and then comparing these counts. If the character frequencies match, the strings are considered anagrams, and the method returns <code class="docutils literal notranslate"><span class="pre">True</span></code>; otherwise, it returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="n">s</span> <span class="o">=</span> <span class="s2">&quot;anagram&quot;</span>
-<span class="n">t</span> <span class="o">=</span> <span class="s2">&quot;nagaram&quot;</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isAnagram</span><span class="p">(</span><span class="n">s</span><span class="p">,</span> <span class="n">t</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">s</span> <span class="o">=</span> <span class="s2">&quot;rat&quot;</span>
-<span class="n">t</span> <span class="o">=</span> <span class="s2">&quot;car&quot;</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isAnagram</span><span class="p">(</span><span class="n">s</span><span class="p">,</span> <span class="n">t</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The code first checks whether the lengths of the input strings <code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code> are equal, which takes constant time. This check has a time complexity of O(1).</p></li>
-<li><p>The code then iterates through both strings, <code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code>, once to count the character frequencies. Since both strings have a maximum length of 5 * 10^4, the worst-case scenario is that the code iterates through 5 * 10^4 characters in each string. This results in a linear time complexity of O(n), where n is the length of the longer of the two input strings (<code class="docutils literal notranslate"><span class="pre">s</span></code> or <code class="docutils literal notranslate"><span class="pre">t</span></code>).</p></li>
-<li><p>Finally, the code compares the two dictionaries <code class="docutils literal notranslate"><span class="pre">countS</span></code> and <code class="docutils literal notranslate"><span class="pre">countT</span></code> to check if the character frequencies match. This comparison takes O(n) time in the worst case, where n is the length of the longer string.</p></li>
-</ol>
-<p>Overall, the time complexity of the code is O(n), where n is the length of the longer input string.</p>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The code uses two dictionaries, <code class="docutils literal notranslate"><span class="pre">countS</span></code> and <code class="docutils literal notranslate"><span class="pre">countT</span></code>, to store the character frequencies of the input strings <code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code>. In the worst case, both dictionaries can contain all unique characters from the input strings. Since the input strings can have a maximum length of 5 * 10^4, the space complexity for these dictionaries is O(5 * 10^4) or simply O(n), where n is the length of the longer input string.</p></li>
-<li><p>The code uses a few additional variables for bookkeeping, but these variables have constant space requirements and do not depend on the input size. Therefore, they do not significantly impact the overall space complexity.</p></li>
-</ol>
-<p><strong>In summary</strong>, the space complexity of the code is O(n), where n is the length of the longer input string.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Anagram Chains</strong>: Given a list of words, find the longest chain of anagrams, where each word in the chain is an anagram of the previous word. For example, given [“bat”, “tab”, “cat”, “tac”, “dog”], the longest anagram chain is [“bat”, “tab”, “cat”, “tac”].</p></li>
-<li><p><strong>Minimum Deletions to Make Anagrams</strong>: Given two strings <code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code>, find the minimum number of character deletions required in both strings to make them anagrams of each other. For example, if <code class="docutils literal notranslate"><span class="pre">s</span> <span class="pre">=</span> <span class="pre">&quot;hello&quot;</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span> <span class="pre">=</span> <span class="pre">&quot;billion&quot;</span></code>, you can remove the characters “heo” from <code class="docutils literal notranslate"><span class="pre">s</span></code> and “billi” from <code class="docutils literal notranslate"><span class="pre">t</span></code> to make them anagrams (“lo” and “on” are left in the two strings).</p></li>
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-    <h1>347. Top K Frequent Elements</h1>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="top-k-frequent-elements">
-<h1>347. Top K Frequent Elements<a class="headerlink" href="#top-k-frequent-elements" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/top-k-frequent-elements/">To see the Top K Frequent Elements problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given an integer array <code class="docutils literal notranslate"><span class="pre">nums</span></code> and an integer <code class="docutils literal notranslate"><span class="pre">k</span></code>, return the <code class="docutils literal notranslate"><span class="pre">k</span></code> most frequent elements. You may return the answer in <strong>any order</strong>.</p>
-<p>Constraints:</p>
-<ul class="simple">
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">nums.length</span></code> &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^4\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">nums[i]</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">k</span></code> is in the range <code class="docutils literal notranslate"><span class="pre">[1,</span> <span class="pre">the</span> <span class="pre">number</span> <span class="pre">of</span> <span class="pre">unique</span> <span class="pre">elements</span> <span class="pre">in</span> <span class="pre">the</span> <span class="pre">array]</span></code>.</p></li>
-<li><p>It is <strong>guaranteed</strong> that the answer is <strong>unique</strong>.</p></li>
-</ul>
-<p><strong>Follow-up:</strong> Your algorithm’s time complexity must be better than <span class="math notranslate nohighlight">\(O(n\ log\ n)\)</span>, where n is the array’s size.</p>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">topKFrequent</span><span class="p">(</span><span class="n">nums</span><span class="p">,</span> <span class="n">k</span><span class="p">):</span>
-    <span class="c1"># Create a dictionary to store the frequency of each element</span>
-    <span class="n">count</span> <span class="o">=</span> <span class="p">{}</span>
-    <span class="c1"># Create a list of lists to store elements with the same frequency</span>
-    <span class="n">frequency</span> <span class="o">=</span> <span class="p">[[]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)]</span>
-
-    <span class="c1"># Count the frequency of each element in nums</span>
-    <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="n">nums</span><span class="p">:</span>
-        <span class="n">count</span><span class="p">[</span><span class="n">n</span><span class="p">]</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">+</span> <span class="n">count</span><span class="o">.</span><span class="n">get</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="mi">0</span><span class="p">)</span>
-
-    <span class="c1"># Place elements in the freq list according to their frequency</span>
-    <span class="k">for</span> <span class="n">n</span><span class="p">,</span> <span class="n">c</span> <span class="ow">in</span> <span class="n">count</span><span class="o">.</span><span class="n">items</span><span class="p">():</span>
-        <span class="n">frequency</span><span class="p">[</span><span class="n">c</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">n</span><span class="p">)</span>
-
-    <span class="n">res</span> <span class="o">=</span> <span class="p">[]</span>
-    <span class="c1"># Traverse freq list from the end (higher frequencies)</span>
-    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">frequency</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="o">-</span><span class="mi">1</span><span class="p">):</span>
-        <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="n">frequency</span><span class="p">[</span><span class="n">i</span><span class="p">]:</span>
-            <span class="n">res</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">n</span><span class="p">)</span>
-            <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="n">res</span><span class="p">)</span> <span class="o">==</span> <span class="n">k</span><span class="p">:</span>
-                <span class="k">return</span> <span class="n">res</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">def</span> <span class="pre">topKFrequent(nums,</span> <span class="pre">k):</span></code>: This is a function that takes two arguments: <code class="docutils literal notranslate"><span class="pre">nums</span></code>, which is the input array of integers, and <code class="docutils literal notranslate"><span class="pre">k</span></code>, which is the number of most frequent elements to return.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">count</span> <span class="pre">=</span> <span class="pre">{}</span></code>: This dictionary <code class="docutils literal notranslate"><span class="pre">count</span></code> will be used to store the frequency of each unique element in the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array. The keys are elements from the input array, and the values are their corresponding frequencies.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">frequency</span> <span class="pre">=</span> <span class="pre">[[]</span> <span class="pre">for</span> <span class="pre">i</span> <span class="pre">in</span> <span class="pre">range(len(nums)</span> <span class="pre">+</span> <span class="pre">1)]</span></code>: This creates a list of empty lists called <code class="docutils literal notranslate"><span class="pre">frequency</span></code>. It’s used to store elements based on their frequencies, similar to the <code class="docutils literal notranslate"><span class="pre">freq</span></code> list in the previous code. The size of this list is set to be one greater than the length of the input <code class="docutils literal notranslate"><span class="pre">nums</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">for</span> <span class="pre">n</span> <span class="pre">in</span> <span class="pre">nums:</span></code>: This loop iterates through each element <code class="docutils literal notranslate"><span class="pre">n</span></code> in the input <code class="docutils literal notranslate"><span class="pre">nums</span></code> array.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">count[n]</span> <span class="pre">=</span> <span class="pre">1</span> <span class="pre">+</span> <span class="pre">count.get(n,</span> <span class="pre">0)</span></code>: This line counts the frequency of each element <code class="docutils literal notranslate"><span class="pre">n</span></code> in the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array. It uses the <code class="docutils literal notranslate"><span class="pre">count.get(n,</span> <span class="pre">0)</span></code> method to retrieve the current count of <code class="docutils literal notranslate"><span class="pre">n</span></code> from the <code class="docutils literal notranslate"><span class="pre">count</span></code> dictionary. If <code class="docutils literal notranslate"><span class="pre">n</span></code> is not in the dictionary, it defaults to 0. It then increments the count by 1.</p></li>
-<li><p>After the above loop, the <code class="docutils literal notranslate"><span class="pre">count</span></code> dictionary will contain counts of each unique element in the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">for</span> <span class="pre">n,</span> <span class="pre">c</span> <span class="pre">in</span> <span class="pre">count.items():</span></code>: This loop iterates through the items (key-value pairs) of the <code class="docutils literal notranslate"><span class="pre">count</span></code> dictionary. <code class="docutils literal notranslate"><span class="pre">n</span></code> represents the element, and <code class="docutils literal notranslate"><span class="pre">c</span></code> represents its frequency.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">frequency[c].append(n)</span></code>: This line places the element <code class="docutils literal notranslate"><span class="pre">n</span></code> into the bucket corresponding to its frequency <code class="docutils literal notranslate"><span class="pre">c</span></code>. Buckets are represented by the <code class="docutils literal notranslate"><span class="pre">frequency</span></code> list. For example, if an element <code class="docutils literal notranslate"><span class="pre">n</span></code> has a frequency of 3, it will be added to <code class="docutils literal notranslate"><span class="pre">frequency[3]</span></code>.</p></li>
-<li><p>After this loop, the <code class="docutils literal notranslate"><span class="pre">frequency</span></code> list will contain buckets of elements grouped by their frequencies.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">res</span> <span class="pre">=</span> <span class="pre">[]</span></code>: This list <code class="docutils literal notranslate"><span class="pre">res</span></code> will be used to store the k most frequent elements.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">for</span> <span class="pre">i</span> <span class="pre">in</span> <span class="pre">range(len(freq)</span> <span class="pre">-</span> <span class="pre">1,</span> <span class="pre">0,</span> <span class="pre">-1):</span></code>: This loop iterates in reverse order through the <code class="docutils literal notranslate"><span class="pre">frequency</span></code> list, starting from the highest frequency and going down to 1. Note that there is a typo here; it should be <code class="docutils literal notranslate"><span class="pre">frequency</span></code> instead of <code class="docutils literal notranslate"><span class="pre">freq</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">for</span> <span class="pre">n</span> <span class="pre">in</span> <span class="pre">frequency[i]:</span></code>: For each element <code class="docutils literal notranslate"><span class="pre">n</span></code> in the current bucket (i.e., elements with frequency <code class="docutils literal notranslate"><span class="pre">i</span></code>), it appends <code class="docutils literal notranslate"><span class="pre">n</span></code> to the result list <code class="docutils literal notranslate"><span class="pre">res</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">if</span> <span class="pre">len(res)</span> <span class="pre">==</span> <span class="pre">k:</span></code>: Once <code class="docutils literal notranslate"><span class="pre">res</span></code> contains k elements, the code exits and returns the result.</p></li>
-</ol>
-<p>The code efficiently finds the k most frequent elements in the input array without using sorting, similar to the previous explanation, with the only difference being the variable names (e.g., <code class="docutils literal notranslate"><span class="pre">frequency</span></code> instead of <code class="docutils literal notranslate"><span class="pre">freq</span></code>).</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-<span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">]</span>
-<span class="n">k</span> <span class="o">=</span> <span class="mi">2</span>
-<span class="n">result1</span> <span class="o">=</span> <span class="n">topKFrequent</span><span class="p">(</span><span class="n">nums</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result1</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[1, 2]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">]</span>
-<span class="n">k</span> <span class="o">=</span> <span class="mi">1</span>
-<span class="n">result2</span> <span class="o">=</span> <span class="n">topKFrequent</span><span class="p">(</span><span class="n">nums</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result2</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[1]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The first loop that counts the frequency of elements by iterating through <code class="docutils literal notranslate"><span class="pre">nums</span></code> has a time complexity of O(n), where n is the number of elements in the input array <code class="docutils literal notranslate"><span class="pre">nums</span></code>.</p></li>
-<li><p>The second loop iterates through the keys in the <code class="docutils literal notranslate"><span class="pre">count</span></code> dictionary, which has at most <code class="docutils literal notranslate"><span class="pre">k</span></code> unique elements (as per the constraints). Therefore, the second loop also has a time complexity of O(k).</p></li>
-<li><p>The third loop iterates through the <code class="docutils literal notranslate"><span class="pre">frequency</span></code> list, which has a length equal to the maximum frequency of elements in <code class="docutils literal notranslate"><span class="pre">nums</span></code>. In the worst case, this loop can have a time complexity of O(n), where n is the number of elements in <code class="docutils literal notranslate"><span class="pre">nums</span></code>.</p></li>
-</ol>
-<p>Overall, the dominant factor in terms of time complexity is the loop that iterates through the <code class="docutils literal notranslate"><span class="pre">count</span></code> dictionary. So, the total time complexity of the code is O(n + k).</p>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">count</span></code> dictionary stores the frequency of each unique element in <code class="docutils literal notranslate"><span class="pre">nums</span></code>. In the worst case, it can have at most <code class="docutils literal notranslate"><span class="pre">k</span></code> unique elements, so the space complexity for <code class="docutils literal notranslate"><span class="pre">count</span></code> is O(k).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">frequency</span></code> list is used to store elements grouped by their frequencies. It has a length of <code class="docutils literal notranslate"><span class="pre">len(nums)</span> <span class="pre">+</span> <span class="pre">1</span></code>, which can be at most 105 based on the constraints. Therefore, the space complexity for <code class="docutils literal notranslate"><span class="pre">frequency</span></code> is O(105).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">res</span></code> list stores the k most frequent elements, which can be at most <code class="docutils literal notranslate"><span class="pre">k</span></code> elements, so the space complexity for <code class="docutils literal notranslate"><span class="pre">res</span></code> is O(k).</p></li>
-</ol>
-<p><strong>In summary</strong>, the space complexity is dominated by the <code class="docutils literal notranslate"><span class="pre">count</span></code> dictionary and the <code class="docutils literal notranslate"><span class="pre">frequency</span></code> list, both of which have a space complexity of O(k + 105).</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Optimized k Most Frequent Elements:</strong> Modify the code to find the k most frequent elements in an array while ensuring that the time complexity is O(n + klogk). You can use a priority queue (heap) to achieve this.</p></li>
-<li><p><strong>Handling Duplicate Frequencies:</strong> Extend the code to handle cases where multiple elements have the same frequency and are among the k most frequent elements. Ensure that the output contains exactly k elements.</p></li>
-</ol>
-</section>
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-<h1>49. Group Anagrams<a class="headerlink" href="#group-anagrams" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/group-anagrams/">To see the Group Anagrams problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given an array of strings <code class="docutils literal notranslate"><span class="pre">strs</span></code>, group <strong>the anagrams</strong> together. You can return the answer in <strong>any order</strong>.</p>
-<p>An <strong>Anagram</strong> is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>1 &lt;= `strs.length &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">0</span> <span class="pre">&lt;=</span> <span class="pre">strs[i].length</span> <span class="pre">&lt;=</span> <span class="pre">100</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">strs[i]</span></code> consists of lowercase English letters.</p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">from</span> <span class="nn">collections</span> <span class="kn">import</span> <span class="n">defaultdict</span>
-
-
-<span class="k">def</span> <span class="nf">groupAnagrams</span><span class="p">(</span><span class="n">strs</span><span class="p">):</span>
-    <span class="n">ans</span> <span class="o">=</span> <span class="n">defaultdict</span><span class="p">(</span><span class="nb">list</span><span class="p">)</span>
-
-    <span class="c1"># Iterate through the list of input strings</span>
-    <span class="k">for</span> <span class="n">s</span> <span class="ow">in</span> <span class="n">strs</span><span class="p">:</span>
-        <span class="c1"># Initialize a list to represent character counts for each character (a-z)</span>
-        <span class="n">count</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="mi">26</span>
-
-        <span class="c1"># Count the occurrences of each character in the current word</span>
-        <span class="k">for</span> <span class="n">c</span> <span class="ow">in</span> <span class="n">s</span><span class="p">:</span>
-            <span class="n">count</span><span class="p">[</span><span class="nb">ord</span><span class="p">(</span><span class="n">c</span><span class="p">)</span> <span class="o">-</span> <span class="nb">ord</span><span class="p">(</span><span class="s2">&quot;a&quot;</span><span class="p">)]</span> <span class="o">+=</span> <span class="mi">1</span>
-
-        <span class="c1"># Use a tuple of character counts as the key and append the word to the anagram group</span>
-        <span class="n">ans</span><span class="p">[</span><span class="nb">tuple</span><span class="p">(</span><span class="n">count</span><span class="p">)]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">s</span><span class="p">)</span>
-
-    <span class="c1"># Convert the values (lists of anagrams) to a list of lists</span>
-    <span class="k">return</span> <span class="nb">list</span><span class="p">(</span><span class="n">ans</span><span class="o">.</span><span class="n">values</span><span class="p">())</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>Inside the method, a <code class="docutils literal notranslate"><span class="pre">defaultdict</span></code> called <code class="docutils literal notranslate"><span class="pre">ans</span></code> is created to store anagram groups. This dictionary will have a list as its default value, meaning that each key in the dictionary will be associated with an empty list by default.</p></li>
-<li><p>The code then iterates through the list of input strings, <code class="docutils literal notranslate"><span class="pre">strs</span></code>, using a <code class="docutils literal notranslate"><span class="pre">for</span></code> loop. For each word <code class="docutils literal notranslate"><span class="pre">s</span></code> in <code class="docutils literal notranslate"><span class="pre">strs</span></code>, it performs the following steps:</p>
-<ul class="simple">
-<li><p>It initializes a list <code class="docutils literal notranslate"><span class="pre">count</span></code> of length 26, where each element represents the count of a specific character (a-z) in the word <code class="docutils literal notranslate"><span class="pre">s</span></code>.</p></li>
-<li><p>The code then iterates through the characters of <code class="docutils literal notranslate"><span class="pre">s</span></code>. For each character <code class="docutils literal notranslate"><span class="pre">c</span></code>, it increments the corresponding count in the <code class="docutils literal notranslate"><span class="pre">count</span></code> list based on its ASCII value.</p></li>
-<li><p>After counting the characters in <code class="docutils literal notranslate"><span class="pre">s</span></code>, it converts the <code class="docutils literal notranslate"><span class="pre">count</span></code> list into a tuple <code class="docutils literal notranslate"><span class="pre">tuple(count)</span></code> to use as a key for the <code class="docutils literal notranslate"><span class="pre">ans</span></code> dictionary.</p></li>
-<li><p>It appends the word <code class="docutils literal notranslate"><span class="pre">s</span></code> to the list associated with the key (tuple) in the <code class="docutils literal notranslate"><span class="pre">ans</span></code> dictionary. This groups all anagrams of the same word together under the same key.</p></li>
-</ul>
-</li>
-<li><p>After processing all words in <code class="docutils literal notranslate"><span class="pre">strs</span></code>, the code converts the values of the <code class="docutils literal notranslate"><span class="pre">ans</span></code> dictionary (which are lists of anagrams) to a list of lists using the <code class="docutils literal notranslate"><span class="pre">list()</span></code> constructor.</p></li>
-<li><p>Finally, the code returns the list of anagram groups, which is the result of grouping the anagrams in the input list.</p></li>
-</ol>
-<p>In summary, the code efficiently groups anagrams together by counting the characters in each word and using a dictionary to store them under the same key. The result is a list of lists, where each inner list represents a group of anagrams.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: Group anagrams from a list of words</span>
-<span class="n">strs1</span> <span class="o">=</span> <span class="p">[</span><span class="s2">&quot;eat&quot;</span><span class="p">,</span> <span class="s2">&quot;tea&quot;</span><span class="p">,</span> <span class="s2">&quot;tan&quot;</span><span class="p">,</span> <span class="s2">&quot;ate&quot;</span><span class="p">,</span> <span class="s2">&quot;nat&quot;</span><span class="p">,</span> <span class="s2">&quot;bat&quot;</span><span class="p">]</span>
-<span class="n">result1</span> <span class="o">=</span> <span class="n">groupAnagrams</span><span class="p">(</span><span class="n">strs1</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result1</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[&#39;eat&#39;, &#39;tea&#39;, &#39;ate&#39;], [&#39;tan&#39;, &#39;nat&#39;], [&#39;bat&#39;]]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2: Group anagrams from a list with an empty string</span>
-<span class="n">strs2</span> <span class="o">=</span> <span class="p">[</span><span class="s2">&quot;&quot;</span><span class="p">]</span>
-<span class="n">result2</span> <span class="o">=</span> <span class="n">groupAnagrams</span><span class="p">(</span><span class="n">strs2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result2</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[&#39;&#39;]]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3: Group anagrams from a list with a single word</span>
-<span class="n">strs3</span> <span class="o">=</span> <span class="p">[</span><span class="s2">&quot;a&quot;</span><span class="p">]</span>
-<span class="n">result3</span> <span class="o">=</span> <span class="n">groupAnagrams</span><span class="p">(</span><span class="n">strs3</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result3</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[&#39;a&#39;]]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">groupAnagrams</span></code> method:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>First, we create an instance of the <code class="docutils literal notranslate"><span class="pre">Solution</span></code> class, which is an O(1) operation.</p></li>
-<li><p>Inside the <code class="docutils literal notranslate"><span class="pre">groupAnagrams</span></code> method, we iterate through the list of input strings <code class="docutils literal notranslate"><span class="pre">strs</span></code> once in a loop:</p>
-<ul class="simple">
-<li><p>For each word in <code class="docutils literal notranslate"><span class="pre">strs</span></code>, we perform character counting, which is done in O(K) time, where K is the maximum length of a word in the list.</p></li>
-</ul>
-</li>
-<li><p>The overall time complexity of the method is O(N * K), where N is the number of words in the input list <code class="docutils literal notranslate"><span class="pre">strs</span></code>, and K is the maximum length of a word in <code class="docutils literal notranslate"><span class="pre">strs</span></code>.</p></li>
-</ol>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>We use a <code class="docutils literal notranslate"><span class="pre">defaultdict</span></code> called <code class="docutils literal notranslate"><span class="pre">ans</span></code> to store the anagram groups. In the worst case, each word belongs to a distinct anagram group, so the space complexity for <code class="docutils literal notranslate"><span class="pre">ans</span></code> is O(N).</p></li>
-<li><p>Within the loop, we create a <code class="docutils literal notranslate"><span class="pre">count</span></code> list of length 26 to store character counts for each word. This is a fixed-size list and does not depend on the input size, so it has a constant space complexity of O(26), which is effectively O(1).</p></li>
-<li><p>The space used for other variables is also constant and does not depend on the input size.</p></li>
-<li><p>The final result, <code class="docutils literal notranslate"><span class="pre">result</span></code>, is a list of lists that contains the grouped anagrams. In the worst case, each word is a distinct anagram group, so the space complexity for <code class="docutils literal notranslate"><span class="pre">result</span></code> is O(N).</p></li>
-<li><p>The overall space complexity of the method is O(N) due to the space used by <code class="docutils literal notranslate"><span class="pre">ans</span></code> and <code class="docutils literal notranslate"><span class="pre">result</span></code>. The space used by the <code class="docutils literal notranslate"><span class="pre">count</span></code> list and other variables is constant and does not significantly affect the overall space complexity.</p></li>
-</ol>
-<p><strong>In summary,</strong> the time complexity of the <code class="docutils literal notranslate"><span class="pre">groupAnagrams</span></code> method is O(N * K), and the space complexity is O(N), where N is the number of words in the input list <code class="docutils literal notranslate"><span class="pre">strs</span></code>, and K is the maximum length of a word in <code class="docutils literal notranslate"><span class="pre">strs</span></code>.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Anagramic Palindromes</strong>: Given a list of strings, find all the groups of anagrams where each group contains words that can be rearranged into a palindrome. For example, in the list [“bat”, “tab”, “act”, “tac”, “dog”], there are two groups: [“bat”, “tab”, “act”, “tac”] and [“dog”].</p></li>
-<li><p><strong>Largest Anagram Group</strong>: Given a list of strings, find the largest group of anagrams. Return the list of anagrams in that group. If there are multiple largest groups, return all of them.</p></li>
-</ol>
-</section>
-</section>
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-<h1>659· Encode and Decode Strings<a class="headerlink" href="#encode-and-decode-strings" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://www.lintcode.com/problem/659/">To see the Encode and Decode Strings problem on LintCode, click here!</a></p>
-<hr class="docutils" />
-<p>Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.</p>
-<p>Please implement <code class="docutils literal notranslate"><span class="pre">encode</span></code> and <code class="docutils literal notranslate"><span class="pre">decode</span></code>.</p>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">Codec</span><span class="p">:</span>
-
-    <span class="k">def</span> <span class="nf">encode</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">strs</span><span class="p">):</span>
-<span class="w">        </span><span class="sd">&quot;&quot;&quot;Encodes a list of strings to a single string.</span>
-<span class="sd">        </span>
-<span class="sd">        :type strs: List[str]</span>
-<span class="sd">        :rtype: str</span>
-<span class="sd">        &quot;&quot;&quot;</span>
-        <span class="n">encoded</span> <span class="o">=</span> <span class="s2">&quot;&quot;</span>
-        <span class="k">for</span> <span class="n">s</span> <span class="ow">in</span> <span class="n">strs</span><span class="p">:</span>
-            <span class="n">encoded</span> <span class="o">+=</span> <span class="n">s</span><span class="o">.</span><span class="n">replace</span><span class="p">(</span><span class="s1">&#39;:&#39;</span><span class="p">,</span> <span class="s1">&#39;::&#39;</span><span class="p">)</span> <span class="o">+</span> <span class="s1">&#39;:;&#39;</span>
-        <span class="k">return</span> <span class="n">encoded</span>
-
-    <span class="k">def</span> <span class="nf">decode</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">s</span><span class="p">):</span>
-<span class="w">        </span><span class="sd">&quot;&quot;&quot;Decodes a single string to a list of strings.</span>
-<span class="sd">        </span>
-<span class="sd">        :type s: str</span>
-<span class="sd">        :rtype: List[str]</span>
-<span class="sd">        &quot;&quot;&quot;</span>
-        <span class="n">decoded</span> <span class="o">=</span> <span class="p">[]</span>
-        <span class="n">i</span> <span class="o">=</span> <span class="mi">0</span>
-        <span class="k">while</span> <span class="n">i</span> <span class="o">&lt;</span> <span class="nb">len</span><span class="p">(</span><span class="n">s</span><span class="p">):</span>
-            <span class="n">end</span> <span class="o">=</span> <span class="n">s</span><span class="o">.</span><span class="n">find</span><span class="p">(</span><span class="s1">&#39;:;&#39;</span><span class="p">,</span> <span class="n">i</span><span class="p">)</span>
-            <span class="k">if</span> <span class="n">end</span> <span class="o">==</span> <span class="o">-</span><span class="mi">1</span><span class="p">:</span>
-                <span class="n">end</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">s</span><span class="p">)</span>
-            <span class="n">decoded</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">s</span><span class="p">[</span><span class="n">i</span><span class="p">:</span><span class="n">end</span><span class="p">]</span><span class="o">.</span><span class="n">replace</span><span class="p">(</span><span class="s1">&#39;::&#39;</span><span class="p">,</span> <span class="s1">&#39;:&#39;</span><span class="p">))</span>
-            <span class="n">i</span> <span class="o">=</span> <span class="n">end</span> <span class="o">+</span> <span class="mi">2</span>
-        <span class="k">return</span> <span class="n">decoded</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We define a class called <code class="docutils literal notranslate"><span class="pre">Codec</span></code> to encapsulate the encoding and decoding operations for a list of strings.</p></li>
-<li><p>In the <code class="docutils literal notranslate"><span class="pre">encode</span></code> method, we take a list of strings (<code class="docutils literal notranslate"><span class="pre">strs</span></code>) as input and return a single encoded string. The encoding process involves concatenating the strings together with a delimiter <code class="docutils literal notranslate"><span class="pre">:;</span></code>. We iterate through each string in the input list, replace any occurrence of <code class="docutils literal notranslate"><span class="pre">:</span></code> with <code class="docutils literal notranslate"><span class="pre">::</span></code> (to avoid conflicts with the delimiter), and then append <code class="docutils literal notranslate"><span class="pre">:;</span></code> to indicate the end of that string.</p></li>
-<li><p>In the <code class="docutils literal notranslate"><span class="pre">decode</span></code> method, we take an encoded string (<code class="docutils literal notranslate"><span class="pre">s</span></code>) as input and return a list of strings. The decoding process involves splitting the encoded string using the <code class="docutils literal notranslate"><span class="pre">:;</span></code> delimiter. We start from the beginning of the encoded string (<code class="docutils literal notranslate"><span class="pre">i</span> <span class="pre">=</span> <span class="pre">0</span></code>) and repeatedly find the next occurrence of <code class="docutils literal notranslate"><span class="pre">:;</span></code>. We extract the substring between the current position (<code class="docutils literal notranslate"><span class="pre">i</span></code>) and the next delimiter (<code class="docutils literal notranslate"><span class="pre">end</span></code>). Before adding it to the result list, we replace any <code class="docutils literal notranslate"><span class="pre">::</span></code> with <code class="docutils literal notranslate"><span class="pre">:</span></code> to revert the encoding. We then update the current position <code class="docutils literal notranslate"><span class="pre">i</span></code> to be after the delimiter, so we can find the next substring in the next iteration.</p></li>
-<li><p>Finally, we create an instance of the <code class="docutils literal notranslate"><span class="pre">Codec</span></code> class and test it with two examples:</p></li>
-</ol>
-<p>By using the <code class="docutils literal notranslate"><span class="pre">:;</span></code> delimiter and handling the <code class="docutils literal notranslate"><span class="pre">::</span></code> escaping, this code can encode and decode lists of strings, preserving the original content even if it contains colons or the delimiter itself.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-<span class="n">codec</span> <span class="o">=</span> <span class="n">Codec</span><span class="p">()</span>
-
-<span class="n">input1</span> <span class="o">=</span> <span class="p">[</span><span class="s2">&quot;lint&quot;</span><span class="p">,</span> <span class="s2">&quot;code&quot;</span><span class="p">,</span> <span class="s2">&quot;love&quot;</span><span class="p">,</span> <span class="s2">&quot;you&quot;</span><span class="p">]</span>
-<span class="n">encoded1</span> <span class="o">=</span> <span class="n">codec</span><span class="o">.</span><span class="n">encode</span><span class="p">(</span><span class="n">input1</span><span class="p">)</span>
-<span class="n">decoded1</span> <span class="o">=</span> <span class="n">codec</span><span class="o">.</span><span class="n">decode</span><span class="p">(</span><span class="n">encoded1</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">decoded1</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[&#39;lint&#39;, &#39;code&#39;, &#39;love&#39;, &#39;you&#39;]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">codec</span> <span class="o">=</span> <span class="n">Codec</span><span class="p">()</span> 
-
-<span class="n">input2</span> <span class="o">=</span> <span class="p">[</span><span class="s2">&quot;we&quot;</span><span class="p">,</span> <span class="s2">&quot;say&quot;</span><span class="p">,</span> <span class="s2">&quot;:&quot;</span><span class="p">,</span> <span class="s2">&quot;yes&quot;</span><span class="p">]</span>
-<span class="n">encoded2</span> <span class="o">=</span> <span class="n">codec</span><span class="o">.</span><span class="n">encode</span><span class="p">(</span><span class="n">input2</span><span class="p">)</span>
-<span class="n">decoded2</span> <span class="o">=</span> <span class="n">codec</span><span class="o">.</span><span class="n">decode</span><span class="p">(</span><span class="n">encoded2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">decoded2</span><span class="p">)</span>  
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[&#39;we&#39;, &#39;say&#39;, &#39;:&#39;, &#39;yes&#39;]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">encode</span></code> and <code class="docutils literal notranslate"><span class="pre">decode</span></code> methods in the provided code.</p>
-<p><strong>Encode Method (<code class="docutils literal notranslate"><span class="pre">encode</span></code>):</strong></p>
-<ul class="simple">
-<li><p>Time Complexity: O(n * m)</p>
-<ul>
-<li><p>Here, <code class="docutils literal notranslate"><span class="pre">n</span></code> is the number of strings in the input list <code class="docutils literal notranslate"><span class="pre">strs</span></code>, and <code class="docutils literal notranslate"><span class="pre">m</span></code> is the average length of these strings.</p></li>
-<li><p>In the worst case, for each string in the list, we iterate over its characters once to replace any colons (<code class="docutils literal notranslate"><span class="pre">:</span></code>) with double colons (<code class="docutils literal notranslate"><span class="pre">::</span></code>) and then append <code class="docutils literal notranslate"><span class="pre">:;</span></code>. This is done for each string in the list.</p></li>
-</ul>
-</li>
-<li><p>Space Complexity: O(n * m)</p>
-<ul>
-<li><p>The space complexity is also O(n * m) because we build the encoded string by concatenating the modified strings. In the worst case, the encoded string can be of the same length as the original strings.</p></li>
-</ul>
-</li>
-</ul>
-<p><strong>Decode Method (<code class="docutils literal notranslate"><span class="pre">decode</span></code>):</strong></p>
-<ul class="simple">
-<li><p>Time Complexity: O(n * m)</p>
-<ul>
-<li><p>Similar to the encode method, we iterate through the encoded string in the decode method. In the worst case, we may have to scan each character in the encoded string to find the <code class="docutils literal notranslate"><span class="pre">:;</span></code> delimiter.</p></li>
-</ul>
-</li>
-<li><p>Space Complexity: O(n * m)</p>
-<ul>
-<li><p>The space complexity of the decode method is also O(n * m) because we build the list of decoded strings, which can be of the same length as the encoded string.</p></li>
-</ul>
-</li>
-</ul>
-<p>Overall, both the <code class="docutils literal notranslate"><span class="pre">encode</span></code> and <code class="docutils literal notranslate"><span class="pre">decode</span></code> methods have time and space complexities of O(n * m), where <code class="docutils literal notranslate"><span class="pre">n</span></code> is the number of strings in the input list, and <code class="docutils literal notranslate"><span class="pre">m</span></code> is the average length of these strings. The space complexity arises from storing the encoded or decoded strings, and the time complexity arises from iterating through the characters in these strings.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Multiple Delimiters</strong>: Modify the code to allow for multiple delimiters, not just one. Each string could have its own delimiter, and the code should be able to handle this complexity.</p></li>
-<li><p><strong>Optimize Encoding</strong>: Modify the <code class="docutils literal notranslate"><span class="pre">encode</span></code> method to achieve a more efficient encoding algorithm. Try to minimize the length of the encoded string while ensuring that it can be correctly decoded.</p></li>
-</ol>
-</section>
-</section>
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-<h1>Array &amp; Hashing Solutions - Blind 75 LeetCode Problems<a class="headerlink" href="#array-hashing-solutions-blind-75-leetcode-problems" title="Permalink to this heading">#</a></h1>
-<p><img alt="Array &amp; Hashing Solutions" src="https://img.shields.io/badge/Array_&amp;_Hashing_Solutions-blue" /><img alt="Blind 75 LeetCode" src="https://img.shields.io/badge/Blind_75_LeetCode-Problems-blue?labelColor=red" /></p>
-<p>This repository contains my solutions to Array &amp; Hashing Problems from Blind 75 LeetCode problems. I’ve organized the solutions by categories for easier navigation and reference. Each problem solution is presented in Jupyter Notebook format (<code class="docutils literal notranslate"><span class="pre">.ipynb</span></code>).</p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/217.%20Contains%20Duplicate.ipynb">217. Contains Duplicate</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/242.%20Valid%20Anagram.ipynb">242. Valid Anagram</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/1.%20Two%20Sum.ipynb">1. Two Sum</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/49.%20Group%20Anagrams.ipynb">49. Group Anagrams</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.ipynb">347. Top K Frequent Elements</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.ipynb">238. Product of Array Except Self</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.ipynb">659. Encode and Decode Strings</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.ipynb">128. Longest Consecutive Sequence</a></p></td>
-<td><p>Hard</p></td>
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-<li><p><a class="reference external" href="https://leetcode.com/discuss/">LeetCode Discuss</a>: LeetCode’s discussion forum for each problem. You can find helpful hints and discussions here.</p></li>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="container-with-most-water">
-<h1>11. Container With Most Water<a class="headerlink" href="#container-with-most-water" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/container-with-most-water/">To see the Container With Most Water problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>You are given an integer array <code class="docutils literal notranslate"><span class="pre">height</span></code> of length <code class="docutils literal notranslate"><span class="pre">n</span></code>. There are <code class="docutils literal notranslate"><span class="pre">n</span></code> vertical lines drawn such that the two endpoints of the <span class="math notranslate nohighlight">\(i^{th}\)</span> line are <code class="docutils literal notranslate"><span class="pre">(i,</span> <span class="pre">0)</span></code> and <code class="docutils literal notranslate"><span class="pre">(i,</span> <span class="pre">height[i])</span></code>.</p>
-<p>Find two lines that together with the x-axis form a container, such that the container contains the most water.</p>
-<p>Return <em>the maximum amount of water a container can store.</em></p>
-<p><strong>Notice</strong> that you may not slant the container.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">==</span> <span class="pre">height.length</span></code></p></li>
-<li><p>2 &lt;= <code class="docutils literal notranslate"><span class="pre">n</span></code> &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-<li><p>0 &lt;= <code class="docutils literal notranslate"><span class="pre">height[i]</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">maxArea</span><span class="p">(</span><span class="n">height</span><span class="p">):</span>
-    <span class="c1"># Initialize the maximum area to 0.</span>
-    <span class="n">max_area</span> <span class="o">=</span> <span class="mi">0</span>
-    <span class="c1"># Initialize two pointers, one at the beginning (left) and one at the end (right) of the array.</span>
-    <span class="n">left</span> <span class="o">=</span> <span class="mi">0</span>
-    <span class="n">right</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">height</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span>
-
-    <span class="c1"># Iterate until the left pointer is less than the right pointer.</span>
-    <span class="k">while</span> <span class="n">left</span> <span class="o">&lt;</span> <span class="n">right</span><span class="p">:</span>
-        <span class="c1"># Calculate the height of the container, which is the minimum height of the two lines.</span>
-        <span class="n">h</span> <span class="o">=</span> <span class="nb">min</span><span class="p">(</span><span class="n">height</span><span class="p">[</span><span class="n">left</span><span class="p">],</span> <span class="n">height</span><span class="p">[</span><span class="n">right</span><span class="p">])</span>
-        <span class="c1"># Calculate the width of the container, which is the distance between the two pointers.</span>
-        <span class="n">width</span> <span class="o">=</span> <span class="n">right</span> <span class="o">-</span> <span class="n">left</span>
-        <span class="c1"># Calculate the area of the container using height and width and update max_area if it&#39;s greater.</span>
-        <span class="n">max_area</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">max_area</span><span class="p">,</span> <span class="n">h</span> <span class="o">*</span> <span class="n">width</span><span class="p">)</span>
-
-        <span class="c1"># Move the pointer with the shorter line inward.</span>
-        <span class="k">if</span> <span class="n">height</span><span class="p">[</span><span class="n">left</span><span class="p">]</span> <span class="o">&lt;</span> <span class="n">height</span><span class="p">[</span><span class="n">right</span><span class="p">]:</span>
-            <span class="n">left</span> <span class="o">+=</span> <span class="mi">1</span>
-        <span class="k">else</span><span class="p">:</span>
-            <span class="n">right</span> <span class="o">-=</span> <span class="mi">1</span>
-
-    <span class="c1"># Continue this process until left &lt; right, and then return the maximum area.</span>
-    <span class="k">return</span> <span class="n">max_area</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We start with initializing <code class="docutils literal notranslate"><span class="pre">max_area</span></code> to 0, which will store the maximum area of water that can be contained by the lines.</p></li>
-<li><p>Two pointers, <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code>, are initialized to the beginning and end of the input <code class="docutils literal notranslate"><span class="pre">height</span></code> array.</p></li>
-<li><p>The main loop runs while <code class="docutils literal notranslate"><span class="pre">left</span></code> is less than <code class="docutils literal notranslate"><span class="pre">right</span></code>. This ensures that we are checking all possible pairs of lines in a systematic way.</p></li>
-<li><p>Inside the loop:</p>
-<ul class="simple">
-<li><p>We calculate the height of the container <code class="docutils literal notranslate"><span class="pre">h</span></code> as the minimum of the heights at the <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> pointers. This represents the height of the water the container can hold.</p></li>
-<li><p>We calculate the width of the container as the difference between <code class="docutils literal notranslate"><span class="pre">right</span></code> and <code class="docutils literal notranslate"><span class="pre">left</span></code>. This represents the distance between the two lines.</p></li>
-<li><p>We calculate the area of the container using <code class="docutils literal notranslate"><span class="pre">h</span> <span class="pre">*</span> <span class="pre">width</span></code> and update <code class="docutils literal notranslate"><span class="pre">max_area</span></code> if the calculated area is greater than the current <code class="docutils literal notranslate"><span class="pre">max_area</span></code>.</p></li>
-</ul>
-</li>
-<li><p>After calculating the area and updating <code class="docutils literal notranslate"><span class="pre">max_area</span></code>, we move one of the pointers. We move the pointer pointing to the shorter line inward because moving the pointer pointing to the taller line won’t increase the height and will only decrease the width, which will result in a smaller area.</p></li>
-<li><p>We continue this process until <code class="docutils literal notranslate"><span class="pre">left</span></code> is no longer less than <code class="docutils literal notranslate"><span class="pre">right</span></code>, indicating that we have checked all possible pairs of lines.</p></li>
-<li><p>Finally, we return the <code class="docutils literal notranslate"><span class="pre">max_area</span></code> which contains the maximum area of water that can be contained by two lines from the input array.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="n">height1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">8</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">5</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">8</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">7</span><span class="p">]</span>
-<span class="n">result1</span> <span class="o">=</span> <span class="n">maxArea</span><span class="p">(</span><span class="n">height1</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="s2">&quot;Output:&quot;</span><span class="p">,</span> <span class="n">result1</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Output: 49
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">height2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
-<span class="n">result2</span> <span class="o">=</span> <span class="n">maxArea</span><span class="p">(</span><span class="n">height2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="s2">&quot;Output:&quot;</span><span class="p">,</span> <span class="n">result2</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Output: 1
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Additional Example</span>
-<span class="n">height3</span> <span class="o">=</span> <span class="p">[</span><span class="mi">3</span><span class="p">,</span> <span class="mi">9</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">7</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">12</span><span class="p">,</span> <span class="mi">6</span><span class="p">]</span>
-<span class="n">result3</span> <span class="o">=</span> <span class="n">maxArea</span><span class="p">(</span><span class="n">height3</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="s2">&quot;Output:&quot;</span><span class="p">,</span> <span class="n">result3</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Output: 45
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">maxArea</span></code> function:</p>
-<p><strong>Time Complexity:</strong>
-The function uses a two-pointer approach to find the maximum area of water. In the worst case, both the left and right pointers traverse the entire input array once. Therefore, the time complexity is O(n), where n is the number of elements in the <code class="docutils literal notranslate"><span class="pre">height</span></code> array.</p>
-<p><strong>Space Complexity:</strong>
-The function uses a constant amount of extra space for variables like <code class="docutils literal notranslate"><span class="pre">max_area</span></code>, <code class="docutils literal notranslate"><span class="pre">left</span></code>, <code class="docutils literal notranslate"><span class="pre">right</span></code>, <code class="docutils literal notranslate"><span class="pre">h</span></code>, and <code class="docutils literal notranslate"><span class="pre">width</span></code>. It does not use any data structures whose space consumption depends on the input size. As a result, the space complexity is O(1), which means it has constant space complexity.</p>
-<p><strong>In summary,</strong> the <code class="docutils literal notranslate"><span class="pre">maxArea</span></code> function has a time complexity of O(n) and a space complexity of O(1), making it an efficient algorithm for finding the maximum area of water that can be contained by two lines in the input array.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Optimization Challenge:</strong> Modify the maxArea function to not only return the maximum area but also the indices of the two lines that form the container with the maximum area.</p></li>
-<li><p><strong>Variant Problem:</strong> Instead of finding the maximum area of water, find the minimum amount of water required to fill all the gaps between the lines. You’ll need to return the total water volume needed.</p></li>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="valid-palindrome">
-<h1>125. Valid Palindrome<a class="headerlink" href="#valid-palindrome" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/valid-palindrome/">To see the Valid Palindrome problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>A phrase is a <strong>palindrome</strong> if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.</p>
-<p>Given a string <code class="docutils literal notranslate"><span class="pre">s</span></code>, <em>return <code class="docutils literal notranslate"><span class="pre">true</span></code> if it is a <strong>palindrome</strong>, or <code class="docutils literal notranslate"><span class="pre">false</span></code> otherwise.</em></p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">s.length</span></code> &lt;= <span class="math notranslate nohighlight">\(2*10^5\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">s</span></code> consists only of printable ASCII characters.</p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">isPalindrome</span><span class="p">(</span><span class="n">s</span><span class="p">):</span>
-    <span class="c1"># Step 1: Convert the input string to lowercase</span>
-    <span class="n">s</span> <span class="o">=</span> <span class="n">s</span><span class="o">.</span><span class="n">lower</span><span class="p">()</span>
-    
-    <span class="c1"># Step 2: Remove all non-alphanumeric characters from the string</span>
-    <span class="n">cleaned_s</span> <span class="o">=</span> <span class="s1">&#39;&#39;</span><span class="o">.</span><span class="n">join</span><span class="p">(</span><span class="n">c</span> <span class="k">for</span> <span class="n">c</span> <span class="ow">in</span> <span class="n">s</span> <span class="k">if</span> <span class="n">c</span><span class="o">.</span><span class="n">isalnum</span><span class="p">())</span>
-    
-    <span class="c1"># Step 3: Initialize two pointers, one at the beginning and one at the end of the cleaned string</span>
-    <span class="n">left</span><span class="p">,</span> <span class="n">right</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">cleaned_s</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span>
-    
-    <span class="c1"># Step 4: Compare characters using the two pointers while moving them towards each other</span>
-    <span class="k">while</span> <span class="n">left</span> <span class="o">&lt;</span> <span class="n">right</span><span class="p">:</span>
-        <span class="k">if</span> <span class="n">cleaned_s</span><span class="p">[</span><span class="n">left</span><span class="p">]</span> <span class="o">!=</span> <span class="n">cleaned_s</span><span class="p">[</span><span class="n">right</span><span class="p">]:</span>
-            <span class="k">return</span> <span class="kc">False</span>  <span class="c1"># If characters don&#39;t match, it&#39;s not a palindrome</span>
-        <span class="n">left</span> <span class="o">+=</span> <span class="mi">1</span>
-        <span class="n">right</span> <span class="o">-=</span> <span class="mi">1</span>
-    
-    <span class="c1"># Step 5: If the loop completes without returning False, it&#39;s a palindrome</span>
-    <span class="k">return</span> <span class="kc">True</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">isPalindrome</span></code> function takes an input string <code class="docutils literal notranslate"><span class="pre">s</span></code>.</p></li>
-<li><p>It starts by converting the input string to lowercase using <code class="docutils literal notranslate"><span class="pre">s.lower()</span></code>. This step ensures that the function is case-insensitive when checking for palindromes.</p></li>
-<li><p>Next, it removes all non-alphanumeric characters from the string and stores the result in the <code class="docutils literal notranslate"><span class="pre">cleaned_s</span></code> variable. It does this by iterating through each character in the input string and only keeping characters that are alphanumeric (letters and digits). This step removes spaces, punctuation, and other non-alphanumeric characters.</p></li>
-<li><p>The function then initializes two pointers, <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code>, which will be used to compare characters from the beginning and end of the cleaned string.</p></li>
-<li><p>In a while loop, it compares characters using the two pointers while moving them towards each other. The loop continues until the <code class="docutils literal notranslate"><span class="pre">left</span></code> pointer is less than the <code class="docutils literal notranslate"><span class="pre">right</span></code> pointer. This is done to check all characters in the cleaned string.</p></li>
-<li><p>Inside the loop, it checks if the characters at the <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> pointers don’t match. If they don’t match, it means the input is not a palindrome, so the function returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-<li><p>If the characters at the <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> pointers match, the pointers are moved closer to each other by incrementing <code class="docutils literal notranslate"><span class="pre">left</span></code> and decrementing <code class="docutils literal notranslate"><span class="pre">right</span></code>.</p></li>
-<li><p>The loop continues to compare characters until it either finds a mismatch (returning <code class="docutils literal notranslate"><span class="pre">False</span></code>) or until the <code class="docutils literal notranslate"><span class="pre">left</span></code> pointer is greater than or equal to the <code class="docutils literal notranslate"><span class="pre">right</span></code> pointer. If the loop completes without finding a mismatch, it means the input is a palindrome, so the function returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-</ol>
-<p>In summary, the code converts the input string to lowercase, removes non-alphanumeric characters, and then uses two pointers to compare characters from the cleaned string’s beginning and end. If all characters match during the comparison, the function returns <code class="docutils literal notranslate"><span class="pre">True</span></code>, indicating that the input is a palindrome. If any characters do not match, it returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="n">input1</span> <span class="o">=</span> <span class="s2">&quot;A man, a plan, a canal: Panama&quot;</span>
-<span class="n">output1</span> <span class="o">=</span> <span class="n">isPalindrome</span><span class="p">(</span><span class="n">input1</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">output1</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">input2</span> <span class="o">=</span> <span class="s2">&quot;race a car&quot;</span>
-<span class="n">output2</span> <span class="o">=</span> <span class="n">isPalindrome</span><span class="p">(</span><span class="n">input2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">output2</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3</span>
-<span class="n">input3</span> <span class="o">=</span> <span class="s2">&quot;&quot;</span>
-<span class="n">output3</span> <span class="o">=</span> <span class="n">isPalindrome</span><span class="p">(</span><span class="n">input3</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">output3</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">isPalindrome</span></code> function that uses two pointers:</p>
-<ol class="arabic">
-<li><p><strong>Time Complexity:</strong></p>
-<ul class="simple">
-<li><p>Converting the input string to lowercase using <code class="docutils literal notranslate"><span class="pre">s.lower()</span></code> takes O(n) time, where ‘n’ is the length of the input string.</p></li>
-<li><p>Removing non-alphanumeric characters using the list comprehension <code class="docutils literal notranslate"><span class="pre">''.join(c</span> <span class="pre">for</span> <span class="pre">c</span> <span class="pre">in</span> <span class="pre">s</span> <span class="pre">if</span> <span class="pre">c.isalnum())</span></code> also takes O(n) time because it iterates through each character in the string once.</p></li>
-<li><p>The comparison of characters using the two pointers (<code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code>) is done in a loop that runs until <code class="docutils literal notranslate"><span class="pre">left</span></code> is less than <code class="docutils literal notranslate"><span class="pre">right</span></code>. In the worst case, the loop iterates through half of the string, so it takes O(n/2) time, which is still considered O(n).</p></li>
-</ul>
-<p>Therefore, the overall time complexity of the <code class="docutils literal notranslate"><span class="pre">isPalindrome</span></code> function is O(n), where ‘n’ is the length of the input string.</p>
-</li>
-<li><p><strong>Space Complexity:</strong></p>
-<ul class="simple">
-<li><p>The space complexity is primarily determined by the additional string <code class="docutils literal notranslate"><span class="pre">cleaned_s</span></code> created to store the cleaned version of the input string. In the worst case, if all characters in the input string are alphanumeric, this cleaned string can be as large as the original input string.</p></li>
-<li><p>Additionally, the function uses two pointers, <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code>, which consume constant space and do not depend on the size of the input.</p></li>
-</ul>
-</li>
-</ol>
-<p><strong>In summary,</strong> the overall space complexity of the <code class="docutils literal notranslate"><span class="pre">isPalindrome</span></code> function is O(n) in the worst case, where ‘n’ is the length of the input string. The space complexity is dominated by the space required for <code class="docutils literal notranslate"><span class="pre">cleaned_s</span></code>.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Longest Palindromic Substring:</strong> Given a string <code class="docutils literal notranslate"><span class="pre">s</span></code>, find and return the longest palindromic substring within <code class="docutils literal notranslate"><span class="pre">s</span></code>. For example, if <code class="docutils literal notranslate"><span class="pre">s</span> <span class="pre">=</span> <span class="pre">&quot;babad&quot;</span></code>, the function should return <code class="docutils literal notranslate"><span class="pre">&quot;bab&quot;</span></code> or <code class="docutils literal notranslate"><span class="pre">&quot;aba&quot;</span></code>.</p></li>
-<li><p><strong>Palindrome with Limited Characters:</strong> Given a string <code class="docutils literal notranslate"><span class="pre">s</span></code> and a list of characters <code class="docutils literal notranslate"><span class="pre">allowed_chars</span></code>, check if <code class="docutils literal notranslate"><span class="pre">s</span></code> is a palindrome considering only the characters from <code class="docutils literal notranslate"><span class="pre">allowed_chars</span></code>. Ignore all other characters. For example, if <code class="docutils literal notranslate"><span class="pre">s</span> <span class="pre">=</span> <span class="pre">&quot;A</span> <span class="pre">man,</span> <span class="pre">a</span> <span class="pre">plan,</span> <span class="pre">a</span> <span class="pre">canal:</span> <span class="pre">Panama&quot;</span></code> and <code class="docutils literal notranslate"><span class="pre">allowed_chars</span> <span class="pre">=</span> <span class="pre">['a',</span> <span class="pre">'m',</span> <span class="pre">'n',</span> <span class="pre">'p']</span></code>, the function should return <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
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-    <h1>15. 3Sum</h1>
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-                <h2> Contents </h2>
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-                <ul class="visible nav section-nav flex-column">
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="sum">
-<h1>15. 3Sum<a class="headerlink" href="#sum" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/3sum/">To see the 3Sum problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given an integer array nums, <em>return all the triplets <code class="docutils literal notranslate"><span class="pre">[nums[i],</span> <span class="pre">nums[j],</span> <span class="pre">nums[k]]</span></code> such that <code class="docutils literal notranslate"><span class="pre">i</span> <span class="pre">!=</span> <span class="pre">j</span></code>, <code class="docutils literal notranslate"><span class="pre">i</span> <span class="pre">!=</span> <span class="pre">k</span></code>, and <code class="docutils literal notranslate"><span class="pre">j</span> <span class="pre">!=</span> <span class="pre">k</span></code>, and <code class="docutils literal notranslate"><span class="pre">nums[i]</span> <span class="pre">+</span> <span class="pre">nums[j]</span> <span class="pre">+</span> <span class="pre">nums[k]</span> <span class="pre">==</span> <span class="pre">0</span></code>.</em></p>
-<p><strong>Notice</strong> that the solution set must not contain duplicate triplets.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">3</span> <span class="pre">&lt;=</span> <span class="pre">nums.length</span> <span class="pre">&lt;=</span> <span class="pre">3000</span></code></p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^5\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">nums[i]</span></code> &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">threeSum</span><span class="p">(</span><span class="n">nums</span><span class="p">):</span>
-    <span class="c1"># Sort the input array in ascending order</span>
-    <span class="n">nums</span><span class="o">.</span><span class="n">sort</span><span class="p">()</span>
-    <span class="n">triplets</span> <span class="o">=</span> <span class="p">[]</span>
-    
-    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span> <span class="o">-</span> <span class="mi">2</span><span class="p">):</span>
-        <span class="c1"># Skip duplicates to avoid duplicate triplets</span>
-        <span class="k">if</span> <span class="n">i</span> <span class="o">&gt;</span> <span class="mi">0</span> <span class="ow">and</span> <span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">==</span> <span class="n">nums</span><span class="p">[</span><span class="n">i</span> <span class="o">-</span> <span class="mi">1</span><span class="p">]:</span>
-            <span class="k">continue</span>
-        
-        <span class="n">left</span><span class="p">,</span> <span class="n">right</span> <span class="o">=</span> <span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span>
-        
-        <span class="k">while</span> <span class="n">left</span> <span class="o">&lt;</span> <span class="n">right</span><span class="p">:</span>
-            <span class="c1"># Calculate the total sum of the current triplet</span>
-            <span class="n">total</span> <span class="o">=</span> <span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">+</span> <span class="n">nums</span><span class="p">[</span><span class="n">left</span><span class="p">]</span> <span class="o">+</span> <span class="n">nums</span><span class="p">[</span><span class="n">right</span><span class="p">]</span>
-            
-            <span class="k">if</span> <span class="n">total</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
-                <span class="c1"># If the total is zero, we found a valid triplet</span>
-                <span class="n">triplets</span><span class="o">.</span><span class="n">append</span><span class="p">([</span><span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="n">nums</span><span class="p">[</span><span class="n">left</span><span class="p">],</span> <span class="n">nums</span><span class="p">[</span><span class="n">right</span><span class="p">]])</span>
-                
-                <span class="c1"># Skip duplicates of left and right pointers</span>
-                <span class="k">while</span> <span class="n">left</span> <span class="o">&lt;</span> <span class="n">right</span> <span class="ow">and</span> <span class="n">nums</span><span class="p">[</span><span class="n">left</span><span class="p">]</span> <span class="o">==</span> <span class="n">nums</span><span class="p">[</span><span class="n">left</span> <span class="o">+</span> <span class="mi">1</span><span class="p">]:</span>
-                    <span class="n">left</span> <span class="o">+=</span> <span class="mi">1</span>
-                <span class="k">while</span> <span class="n">left</span> <span class="o">&lt;</span> <span class="n">right</span> <span class="ow">and</span> <span class="n">nums</span><span class="p">[</span><span class="n">right</span><span class="p">]</span> <span class="o">==</span> <span class="n">nums</span><span class="p">[</span><span class="n">right</span> <span class="o">-</span> <span class="mi">1</span><span class="p">]:</span>
-                    <span class="n">right</span> <span class="o">-=</span> <span class="mi">1</span>
-                
-                <span class="c1"># Move the pointers to the next unique elements</span>
-                <span class="n">left</span> <span class="o">+=</span> <span class="mi">1</span>
-                <span class="n">right</span> <span class="o">-=</span> <span class="mi">1</span>
-            <span class="k">elif</span> <span class="n">total</span> <span class="o">&lt;</span> <span class="mi">0</span><span class="p">:</span>
-                <span class="c1"># If the total is negative, we need to increase the sum by moving the left pointer to the right</span>
-                <span class="n">left</span> <span class="o">+=</span> <span class="mi">1</span>
-            <span class="k">else</span><span class="p">:</span>
-                <span class="c1"># If the total is positive, we need to decrease the sum by moving the right pointer to the left</span>
-                <span class="n">right</span> <span class="o">-=</span> <span class="mi">1</span>
-    
-    <span class="k">return</span> <span class="n">triplets</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>It starts by sorting the input array <code class="docutils literal notranslate"><span class="pre">nums</span></code> in ascending order. Sorting helps in efficiently finding triplets that sum up to zero.</p></li>
-<li><p>It initializes an empty list <code class="docutils literal notranslate"><span class="pre">triplets</span></code> to store the unique triplets that meet the given conditions.</p></li>
-<li><p>The code then iterates through the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array using a loop with index <code class="docutils literal notranslate"><span class="pre">i</span></code>. This loop will consider each element of the array as a potential starting point for a triplet.</p></li>
-<li><p>Inside the loop, it checks for duplicates and skips them. This is done to ensure that the solution set does not contain duplicate triplets. If <code class="docutils literal notranslate"><span class="pre">nums[i]</span></code> is the same as the previous element <code class="docutils literal notranslate"><span class="pre">nums[i-1]</span></code>, it continues to the next iteration of the loop.</p></li>
-<li><p>Two pointers, <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code>, are initialized. <code class="docutils literal notranslate"><span class="pre">left</span></code> starts just after the current element <code class="docutils literal notranslate"><span class="pre">nums[i]</span></code>, and <code class="docutils literal notranslate"><span class="pre">right</span></code> starts at the end of the array.</p></li>
-<li><p>The code enters another loop with <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> pointers, trying to find a triplet that sums up to zero.</p></li>
-<li><p>It calculates the <code class="docutils literal notranslate"><span class="pre">total</span></code> sum of the current triplet as <code class="docutils literal notranslate"><span class="pre">nums[i]</span> <span class="pre">+</span> <span class="pre">nums[left]</span> <span class="pre">+</span> <span class="pre">nums[right]</span></code>.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">total</span></code> is zero, it means a valid triplet is found, so it appends <code class="docutils literal notranslate"><span class="pre">[nums[i],</span> <span class="pre">nums[left],</span> <span class="pre">nums[right]]</span></code> to the <code class="docutils literal notranslate"><span class="pre">triplets</span></code> list. Then, it moves both the <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> pointers to their next unique elements while skipping duplicates.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">total</span></code> is less than zero, it means the sum is negative, so it increments the <code class="docutils literal notranslate"><span class="pre">left</span></code> pointer to move towards a larger sum.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">total</span></code> is greater than zero, it means the sum is positive, so it decrements the <code class="docutils literal notranslate"><span class="pre">right</span></code> pointer to move towards a smaller sum.</p></li>
-<li><p>This process continues until all possible triplets have been considered.</p></li>
-<li><p>Finally, the function returns the <code class="docutils literal notranslate"><span class="pre">triplets</span></code> list, which contains all the unique triplets that sum up to zero in the sorted <code class="docutils literal notranslate"><span class="pre">nums</span></code> array.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="o">-</span><span class="mi">4</span><span class="p">]</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">threeSum</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[-1, -1, 2], [-1, 0, 1]]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">threeSum</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3</span>
-<span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">0</span><span class="p">]</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">threeSum</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[0, 0, 0]]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">threeSum</span></code> function:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>Sorting the input array <code class="docutils literal notranslate"><span class="pre">nums</span></code> takes O(n log n) time, where n is the length of the array.</p></li>
-<li><p>The outer loop runs for each element of the array, so it iterates O(n) times.</p></li>
-<li><p>Inside the outer loop, we have a while loop with two pointers (<code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code>). In the worst case, the while loop can iterate O(n) times (when all elements are unique).</p></li>
-<li><p>Inside the while loop, we have constant time operations (additions, comparisons, and list appends).</p></li>
-</ol>
-<p>Overall, the time complexity of the <code class="docutils literal notranslate"><span class="pre">threeSum</span></code> function is dominated by the sorting step, so it is O(n log n) due to the sorting. The other operations inside the loops contribute linearly but are dominated by the sorting step.</p>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The space used by the <code class="docutils literal notranslate"><span class="pre">triplets</span></code> list to store the output triplets is O(k), where k is the number of unique triplets that sum up to zero. In the worst case, this can be O(n^2/3) because there can be roughly O(n^2) possible triplets, and many of them may sum up to zero.</p></li>
-<li><p>Other than the <code class="docutils literal notranslate"><span class="pre">triplets</span></code> list, the function uses only a constant amount of additional space for variables like <code class="docutils literal notranslate"><span class="pre">i</span></code>, <code class="docutils literal notranslate"><span class="pre">left</span></code>, <code class="docutils literal notranslate"><span class="pre">right</span></code>, and <code class="docutils literal notranslate"><span class="pre">total</span></code>.</p></li>
-</ol>
-<p>Overall, the space complexity of the <code class="docutils literal notranslate"><span class="pre">threeSum</span></code> function is O(k), where k is the number of unique triplets that meet the criteria. It’s important to note that the space complexity does not depend on the size of the input array but rather on the number of valid triplets found.</p>
-<p><strong>In summary:</strong></p>
-<ul class="simple">
-<li><p>Time Complexity: O(n log n) due to sorting (dominant factor) and O(n^2) in the worst case for the nested loops.</p></li>
-<li><p>Space Complexity: O(k) for storing the output triplets, where k is the number of unique triplets that meet the criteria.</p></li>
-</ul>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>K-Sum Problem:</strong> Generalize the problem to the K-Sum problem, where instead of finding triplets that sum to zero, you need to find K elements that sum to a target value. Implement a function for this generalization.</p></li>
-<li><p><strong>Count the Number of Unique Triplets:</strong> Modify the algorithm to count the number of unique triplets that satisfy the conditions instead of returning the triplets themselves. This will require some modifications to handle duplicates efficiently.</p></li>
-</ol>
-</section>
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-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
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-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/valid-palindrome/">125. Valid Palindrome</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/3sum/">15. 3Sum</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/container-with-most-water/">11. Container With Most Water</a></p></td>
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-<li><p><a class="reference external" href="https://leetcode.com/discuss/">LeetCode Discuss</a>: LeetCode’s discussion forum for each problem. You can find helpful hints and discussions here.</p></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="best-time-to-buy-and-sell-stock">
-<h1>121. Best Time to Buy and Sell Stock<a class="headerlink" href="#best-time-to-buy-and-sell-stock" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/best-time-to-buy-and-sell-stock/">To see the Best Time to Buy and Sell Stock problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>You are given an array <code class="docutils literal notranslate"><span class="pre">prices</span></code> where <code class="docutils literal notranslate"><span class="pre">prices[i]</span></code> is the price of a given stock on the <span class="math notranslate nohighlight">\(i^{th}\)</span> day.</p>
-<p>You want to maximize your profit by choosing a <strong>single day</strong> to buy one stock and choosing a <strong>different day in the future</strong> to sell that stock.</p>
-<p><em>Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return <code class="docutils literal notranslate"><span class="pre">0</span></code>.</em></p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">prices.length</span></code> &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-<li><p>0 &lt;= <code class="docutils literal notranslate"><span class="pre">prices[i]</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">maxProfit</span><span class="p">(</span><span class="n">prices</span><span class="p">):</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">prices</span><span class="p">:</span>
-        <span class="k">return</span> <span class="mi">0</span>
-    
-    <span class="n">min_price</span> <span class="o">=</span> <span class="n">prices</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
-    <span class="n">max_profit</span> <span class="o">=</span> <span class="mi">0</span>
-    
-    <span class="k">for</span> <span class="n">price</span> <span class="ow">in</span> <span class="n">prices</span><span class="p">:</span>
-        <span class="c1"># Update the minimum price if a lower price is encountered</span>
-        <span class="n">min_price</span> <span class="o">=</span> <span class="nb">min</span><span class="p">(</span><span class="n">min_price</span><span class="p">,</span> <span class="n">price</span><span class="p">)</span>
-        <span class="c1"># Calculate the profit if selling at the current price</span>
-        <span class="n">max_profit</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">max_profit</span><span class="p">,</span> <span class="n">price</span> <span class="o">-</span> <span class="n">min_price</span><span class="p">)</span>
-    
-    <span class="k">return</span> <span class="n">max_profit</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">maxProfit</span></code> function takes a single argument, <code class="docutils literal notranslate"><span class="pre">prices</span></code>, which is a list of stock prices.</p></li>
-<li><p>The first line of the function checks if the <code class="docutils literal notranslate"><span class="pre">prices</span></code> list is empty. If it’s empty, there are no prices to analyze, so the function returns 0 (no profit can be made).</p></li>
-<li><p>Then, we initialize two variables:</p>
-<ul>
-<li><p><code class="docutils literal notranslate"><span class="pre">min_price</span></code>: This variable will keep track of the minimum price seen so far. We initialize it with the price of the first day in the list (<code class="docutils literal notranslate"><span class="pre">prices[0]</span></code>).</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">max_profit</span></code>: This variable will keep track of the maximum profit we can achieve. We initialize it to 0 because initially, we haven’t made any profit.</p></li>
-</ul>
-</li>
-<li><p>We start iterating through the <code class="docutils literal notranslate"><span class="pre">prices</span></code> list, where <code class="docutils literal notranslate"><span class="pre">price</span></code> represents the stock price for the current day.</p></li>
-<li><p>In each iteration, we check if the current <code class="docutils literal notranslate"><span class="pre">price</span></code> is lower than the <code class="docutils literal notranslate"><span class="pre">min_price</span></code> we’ve seen so far. If it is, we update <code class="docutils literal notranslate"><span class="pre">min_price</span></code> to the current <code class="docutils literal notranslate"><span class="pre">price</span></code>. This is because we want to buy the stock at the lowest possible price.</p></li>
-<li><p>Next, we calculate the profit we would make if we sold the stock at the current <code class="docutils literal notranslate"><span class="pre">price</span></code> (assuming we bought it at <code class="docutils literal notranslate"><span class="pre">min_price</span></code>). We do this by subtracting <code class="docutils literal notranslate"><span class="pre">min_price</span></code> from the current <code class="docutils literal notranslate"><span class="pre">price</span></code>. This represents the profit for the current day.</p></li>
-<li><p>We also use the <code class="docutils literal notranslate"><span class="pre">max</span></code> function to keep track of the maximum profit seen so far. If the profit calculated for the current day is greater than the <code class="docutils literal notranslate"><span class="pre">max_profit</span></code> we’ve seen previously, we update <code class="docutils literal notranslate"><span class="pre">max_profit</span></code> with this higher value.</p></li>
-<li><p>Finally, after iterating through all the days in the <code class="docutils literal notranslate"><span class="pre">prices</span></code> list, we return <code class="docutils literal notranslate"><span class="pre">max_profit</span></code>, which represents the maximum profit that can be achieved by buying and selling a single stock.</p></li>
-</ul>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="n">prices1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">7</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">5</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">4</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">maxProfit</span><span class="p">(</span><span class="n">prices1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>5
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">prices2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">7</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">maxProfit</span><span class="p">(</span><span class="n">prices2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>0
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong></p>
-<ul class="simple">
-<li><p>The code iterates through the <code class="docutils literal notranslate"><span class="pre">prices</span></code> array once, examining each price exactly once.</p></li>
-<li><p>Within the loop, there are constant-time operations such as comparisons, additions, and subtractions.</p></li>
-<li><p>Therefore, the time complexity of this code is O(n), where n is the length of the <code class="docutils literal notranslate"><span class="pre">prices</span></code> array. It performs a linear number of operations relative to the size of the input.</p></li>
-</ul>
-<p><strong>Space Complexity:</strong></p>
-<ul class="simple">
-<li><p>The space complexity of the code is constant, O(1).</p></li>
-<li><p>Regardless of the size of the input <code class="docutils literal notranslate"><span class="pre">prices</span></code> array, the code only uses a fixed amount of additional memory to store two variables (<code class="docutils literal notranslate"><span class="pre">min_price</span></code> and <code class="docutils literal notranslate"><span class="pre">max_profit</span></code>) and a few loop variables.</p></li>
-<li><p>The memory usage does not depend on the size of the input, so it is considered constant space complexity.</p></li>
-</ul>
-<p><strong>In summary:</strong></p>
-<ul class="simple">
-<li><p>Time Complexity: O(n) where n is the length of the <code class="docutils literal notranslate"><span class="pre">prices</span></code> array.</p></li>
-<li><p>Space Complexity: O(1) (constant space usage).</p></li>
-</ul>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Multiple Transactions Allowed:</strong> Modify the problem to allow multiple transactions (buy and sell multiple times), but you must sell before buying again. Find the maximum profit in this scenario.</p></li>
-<li><p><strong>K Transactions Allowed:</strong> Extend the problem to allow at most k transactions. Find the maximum profit considering this constraint.</p></li>
-<li><p><strong>With Transaction Fee:</strong> Introduce a transaction fee for each buy/sell operation. Modify the code to maximize profit while considering the transaction fee.</p></li>
-</ol>
-</section>
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-    <h1>3. Longest Substring Without Repeating Characters</h1>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="longest-substring-without-repeating-characters">
-<h1>3. Longest Substring Without Repeating Characters<a class="headerlink" href="#longest-substring-without-repeating-characters" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/longest-substring-without-repeating-characters/">To see the Longest Substring Without Repeating Characters problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given a string <code class="docutils literal notranslate"><span class="pre">s</span></code>, find the length of the <strong>longest substring</strong> without repeating characters.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>0 &lt;= <code class="docutils literal notranslate"><span class="pre">s.length</span></code> &lt;= <span class="math notranslate nohighlight">\(5 * 10^4\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">s</span></code> consists of English letters, digits, symbols and spaces.</p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">length_of_longest_substring</span><span class="p">(</span><span class="n">s</span><span class="p">):</span>
-    <span class="c1"># Create a dictionary to store the index of each character&#39;s last occurrence.</span>
-    <span class="n">char_index</span> <span class="o">=</span> <span class="p">{}</span>
-    
-    <span class="c1"># Initialize variables to keep track of the start and end of the current substring.</span>
-    <span class="n">start</span> <span class="o">=</span> <span class="mi">0</span>
-    <span class="n">max_length</span> <span class="o">=</span> <span class="mi">0</span>
-    
-    <span class="k">for</span> <span class="n">end</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">s</span><span class="p">)):</span>
-        <span class="c1"># If the character is in the dictionary and its last occurrence is after the start of the current substring,</span>
-        <span class="c1"># update the start of the substring to the next character after its last occurrence.</span>
-        <span class="k">if</span> <span class="n">s</span><span class="p">[</span><span class="n">end</span><span class="p">]</span> <span class="ow">in</span> <span class="n">char_index</span> <span class="ow">and</span> <span class="n">char_index</span><span class="p">[</span><span class="n">s</span><span class="p">[</span><span class="n">end</span><span class="p">]]</span> <span class="o">&gt;=</span> <span class="n">start</span><span class="p">:</span>
-            <span class="n">start</span> <span class="o">=</span> <span class="n">char_index</span><span class="p">[</span><span class="n">s</span><span class="p">[</span><span class="n">end</span><span class="p">]]</span> <span class="o">+</span> <span class="mi">1</span>
-        
-        <span class="c1"># Update the last occurrence index of the current character.</span>
-        <span class="n">char_index</span><span class="p">[</span><span class="n">s</span><span class="p">[</span><span class="n">end</span><span class="p">]]</span> <span class="o">=</span> <span class="n">end</span>
-        
-        <span class="c1"># Calculate the length of the current substring and update the maximum length if necessary.</span>
-        <span class="n">current_length</span> <span class="o">=</span> <span class="n">end</span> <span class="o">-</span> <span class="n">start</span> <span class="o">+</span> <span class="mi">1</span>
-        <span class="n">max_length</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">max_length</span><span class="p">,</span> <span class="n">current_length</span><span class="p">)</span>
-    
-    <span class="k">return</span> <span class="n">max_length</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We start by defining a function called <code class="docutils literal notranslate"><span class="pre">length_of_longest_substring</span></code> that takes a single argument <code class="docutils literal notranslate"><span class="pre">s</span></code>, which is the input string for which we want to find the length of the longest substring without repeating characters.</p></li>
-<li><p>We create a dictionary called <code class="docutils literal notranslate"><span class="pre">char_index</span></code> to store the index of the last occurrence of each character in the input string. This dictionary will help us keep track of where each character was last seen.</p></li>
-<li><p>We initialize two variables, <code class="docutils literal notranslate"><span class="pre">start</span></code> and <code class="docutils literal notranslate"><span class="pre">max_length</span></code>.</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">start</span></code> represents the start index of the current substring without repeating characters. It starts at 0.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">max_length</span></code> is used to keep track of the maximum length found so far and is initially set to 0.</p></li>
-</ul>
-</li>
-<li><p>We then iterate through the input string <code class="docutils literal notranslate"><span class="pre">s</span></code> using a for loop, where the loop variable <code class="docutils literal notranslate"><span class="pre">end</span></code> represents the current end index of the substring we are considering.</p></li>
-<li><p>Inside the loop:</p>
-<ul class="simple">
-<li><p>We check if the character <code class="docutils literal notranslate"><span class="pre">s[end]</span></code> is already in the <code class="docutils literal notranslate"><span class="pre">char_index</span></code> dictionary and if its last occurrence is within or after the current substring. If so, it means that we’ve encountered a repeating character, and we need to update the <code class="docutils literal notranslate"><span class="pre">start</span></code> of the substring to the next character after the last occurrence of <code class="docutils literal notranslate"><span class="pre">s[end]</span></code>. This ensures that we have a new valid substring without repeating characters.</p></li>
-<li><p>We then update the <code class="docutils literal notranslate"><span class="pre">char_index</span></code> dictionary by storing the current index <code class="docutils literal notranslate"><span class="pre">end</span></code> as the last occurrence index of the character <code class="docutils literal notranslate"><span class="pre">s[end]</span></code>.</p></li>
-</ul>
-</li>
-<li><p>Next, we calculate the length of the current substring without repeating characters, which is <code class="docutils literal notranslate"><span class="pre">current_length</span> <span class="pre">=</span> <span class="pre">end</span> <span class="pre">-</span> <span class="pre">start</span> <span class="pre">+</span> <span class="pre">1</span></code>. We add 1 to account for the fact that the indices are zero-based.</p></li>
-<li><p>We update the <code class="docutils literal notranslate"><span class="pre">max_length</span></code> with the maximum of its current value and the <code class="docutils literal notranslate"><span class="pre">current_length</span></code>. This step ensures that we keep track of the longest valid substring we have encountered so far.</p></li>
-<li><p>The loop continues to iterate through the input string, and at the end, we return the <code class="docutils literal notranslate"><span class="pre">max_length</span></code>, which represents the length of the longest substring without repeating characters.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">length_of_longest_substring</span><span class="p">(</span><span class="s2">&quot;abcabcbb&quot;</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>3
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">length_of_longest_substring</span><span class="p">(</span><span class="s2">&quot;bbbbb&quot;</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>1
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3:</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">length_of_longest_substring</span><span class="p">(</span><span class="s2">&quot;pwwkew&quot;</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>3
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">length_of_longest_substring</span></code> function:</p>
-<p><strong>Time Complexity:</strong></p>
-<ul class="simple">
-<li><p>The function iterates through the input string <code class="docutils literal notranslate"><span class="pre">s</span></code> using a single for loop. The loop runs from the beginning of the string to the end once.</p></li>
-<li><p>Inside the loop, we perform constant-time operations such as dictionary lookups and updates, comparisons, and arithmetic operations.</p></li>
-<li><p>Therefore, the time complexity of the function is O(n), where n is the length of the input string <code class="docutils literal notranslate"><span class="pre">s</span></code>.</p></li>
-</ul>
-<p><strong>Space Complexity:</strong></p>
-<ul class="simple">
-<li><p>The primary data structure that consumes space in this function is the <code class="docutils literal notranslate"><span class="pre">char_index</span></code> dictionary.</p></li>
-<li><p>In the worst case, if there are no repeating characters in the input string, the dictionary can store all unique characters in the string.</p></li>
-<li><p>Therefore, the space complexity is O(min(n, m)), where n is the length of the input string <code class="docutils literal notranslate"><span class="pre">s</span></code>, and m is the number of unique characters in the string. In the worst case, when all characters are unique, m is equal to n, so the space complexity is O(n).</p></li>
-<li><p>Additionally, there are a few integer variables used for indices and lengths, which consume constant space.</p></li>
-<li><p>Overall, the space complexity of the function is O(min(n, m)) or simply O(n) in the worst case.</p></li>
-</ul>
-<p><strong>In summary,</strong> the time complexity of the <code class="docutils literal notranslate"><span class="pre">length_of_longest_substring</span></code> function is O(n), and the space complexity is O(n) in the worst case due to the <code class="docutils literal notranslate"><span class="pre">char_index</span></code> dictionary. This algorithm provides an efficient way to find the length of the longest substring without repeating characters in linear time.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Longest Substring with K Distinct Characters:</strong> Modify the problem to find the length of the longest substring with exactly K distinct characters. For example, given the input string “abcabcbb” and K = 2, the answer would be 4 because the longest substring with two distinct characters is “abca.”</p></li>
-<li><p><strong>Longest Substring with Unique Characters:</strong> Write a function to find the length of the longest substring in a given string where all characters are unique. For example, given the input string “abcabcbb,” the answer would be 4 because “abcd” is the longest substring with unique characters.</p></li>
-</ol>
-</section>
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-    <h1>424. Longest Repeating Character Replacement</h1>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="longest-repeating-character-replacement">
-<h1>424. Longest Repeating Character Replacement<a class="headerlink" href="#longest-repeating-character-replacement" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/longest-repeating-character-replacement/">To see the Longest Repeating Character Replacement problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>You are given a string <code class="docutils literal notranslate"><span class="pre">s</span></code> and an integer <code class="docutils literal notranslate"><span class="pre">k</span></code>. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most <code class="docutils literal notranslate"><span class="pre">k</span></code> times.</p>
-<p><em>Return the length of the longest substring containing the same letter you can get after performing the above operations.</em></p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">s.length</span></code> &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">s</span></code> consists of only uppercase English letters.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">0</span> <span class="pre">&lt;=</span> <span class="pre">k</span> <span class="pre">&lt;=</span> <span class="pre">s.length</span></code></p></li>
-</ul>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">characterReplacement</span><span class="p">(</span><span class="n">s</span><span class="p">,</span> <span class="n">k</span><span class="p">):</span>
-    <span class="n">max_length</span> <span class="o">=</span> <span class="mi">0</span>  <span class="c1"># Initialize the maximum length</span>
-    <span class="n">max_count</span> <span class="o">=</span> <span class="mi">0</span>   <span class="c1"># Initialize the maximum count of repeating characters</span>
-    <span class="n">start</span> <span class="o">=</span> <span class="mi">0</span>       <span class="c1"># Initialize the start of the sliding window</span>
-    <span class="n">char_count</span> <span class="o">=</span> <span class="p">{}</span> <span class="c1"># Dictionary to store the count of each character</span>
-
-    <span class="k">for</span> <span class="n">end</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">s</span><span class="p">)):</span>
-        <span class="c1"># Update the count of the current character in the dictionary</span>
-        <span class="n">char_count</span><span class="p">[</span><span class="n">s</span><span class="p">[</span><span class="n">end</span><span class="p">]]</span> <span class="o">=</span> <span class="n">char_count</span><span class="o">.</span><span class="n">get</span><span class="p">(</span><span class="n">s</span><span class="p">[</span><span class="n">end</span><span class="p">],</span> <span class="mi">0</span><span class="p">)</span> <span class="o">+</span> <span class="mi">1</span>
-        
-        <span class="c1"># Update the maximum count of repeating characters</span>
-        <span class="n">max_count</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">max_count</span><span class="p">,</span> <span class="n">char_count</span><span class="p">[</span><span class="n">s</span><span class="p">[</span><span class="n">end</span><span class="p">]])</span>
-        
-        <span class="c1"># Check if the current window size is greater than k</span>
-        <span class="k">if</span> <span class="p">(</span><span class="n">end</span> <span class="o">-</span> <span class="n">start</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span> <span class="o">-</span> <span class="n">max_count</span> <span class="o">&gt;</span> <span class="n">k</span><span class="p">:</span>
-            <span class="c1"># Move the start of the window to the right</span>
-            <span class="n">char_count</span><span class="p">[</span><span class="n">s</span><span class="p">[</span><span class="n">start</span><span class="p">]]</span> <span class="o">-=</span> <span class="mi">1</span>
-            <span class="n">start</span> <span class="o">+=</span> <span class="mi">1</span>
-        
-        <span class="c1"># Update the maximum length</span>
-        <span class="n">max_length</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">max_length</span><span class="p">,</span> <span class="n">end</span> <span class="o">-</span> <span class="n">start</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span>
-
-    <span class="k">return</span> <span class="n">max_length</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We initialize some variables:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">max_length</span></code> to keep track of the maximum length of the substring containing the same letter.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">max_count</span></code> to keep track of the maximum count of repeating characters within the current window.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">start</span></code> to represent the start index of the sliding window.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">char_count</span></code> is a dictionary that will store the count of each character within the current window.</p></li>
-</ul>
-</li>
-<li><p>We use a <code class="docutils literal notranslate"><span class="pre">for</span></code> loop to iterate through the characters of the string <code class="docutils literal notranslate"><span class="pre">s</span></code>.</p></li>
-<li><p>Inside the loop, we do the following for each character at position <code class="docutils literal notranslate"><span class="pre">end</span></code>:</p>
-<ul class="simple">
-<li><p>Update the count of the current character in the <code class="docutils literal notranslate"><span class="pre">char_count</span></code> dictionary.</p></li>
-<li><p>Update the <code class="docutils literal notranslate"><span class="pre">max_count</span></code> to be the maximum of the current <code class="docutils literal notranslate"><span class="pre">max_count</span></code> and the count of the current character. This keeps track of the maximum count of repeating characters within the current window.</p></li>
-</ul>
-</li>
-<li><p>We check if the current window size (the difference between <code class="docutils literal notranslate"><span class="pre">end</span></code> and <code class="docutils literal notranslate"><span class="pre">start</span></code> plus one) minus the <code class="docutils literal notranslate"><span class="pre">max_count</span></code> is greater than <code class="docutils literal notranslate"><span class="pre">k</span></code>. This condition checks whether we have exceeded the allowed number of replacements (<code class="docutils literal notranslate"><span class="pre">k</span></code>) within the current window.</p>
-<ul class="simple">
-<li><p>If we have exceeded the allowed replacements, it means we need to shrink the window from the left side. We do this by moving the <code class="docutils literal notranslate"><span class="pre">start</span></code> of the window to the right and decrementing the count of the character at <code class="docutils literal notranslate"><span class="pre">s[start]</span></code> in the <code class="docutils literal notranslate"><span class="pre">char_count</span></code> dictionary. This effectively removes characters from the left side of the window until we have a valid window again.</p></li>
-<li><p>By doing this, we ensure that the difference between the current window size and the <code class="docutils literal notranslate"><span class="pre">max_count</span></code> is always less than or equal to <code class="docutils literal notranslate"><span class="pre">k</span></code>.</p></li>
-</ul>
-</li>
-<li><p>After each character, we update the <code class="docutils literal notranslate"><span class="pre">max_length</span></code> to be the maximum of the current <code class="docutils literal notranslate"><span class="pre">max_length</span></code> and the size of the current window (<code class="docutils literal notranslate"><span class="pre">end</span> <span class="pre">-</span> <span class="pre">start</span> <span class="pre">+</span> <span class="pre">1</span></code>).</p></li>
-<li><p>Finally, we return <code class="docutils literal notranslate"><span class="pre">max_length</span></code>, which holds the length of the longest substring containing the same letter with at most <code class="docutils literal notranslate"><span class="pre">k</span></code> replacements.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="n">s1</span> <span class="o">=</span> <span class="s2">&quot;ABAB&quot;</span>
-<span class="n">k1</span> <span class="o">=</span> <span class="mi">2</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">characterReplacement</span><span class="p">(</span><span class="n">s1</span><span class="p">,</span> <span class="n">k1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>4
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">s2</span> <span class="o">=</span> <span class="s2">&quot;AABABBA&quot;</span>
-<span class="n">k2</span> <span class="o">=</span> <span class="mi">1</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">characterReplacement</span><span class="p">(</span><span class="n">s2</span><span class="p">,</span> <span class="n">k2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>4
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong></p>
-<ul class="simple">
-<li><p>The code uses a single <code class="docutils literal notranslate"><span class="pre">for</span></code> loop to iterate through the characters of the input string <code class="docutils literal notranslate"><span class="pre">s</span></code>. The loop runs from the beginning to the end of the string.</p></li>
-<li><p>Inside the loop, we perform constant-time operations such as updating the <code class="docutils literal notranslate"><span class="pre">char_count</span></code> dictionary and updating variables like <code class="docutils literal notranslate"><span class="pre">max_count</span></code> and <code class="docutils literal notranslate"><span class="pre">max_length</span></code>.</p></li>
-<li><p>The code’s time complexity is primarily determined by the loop, which iterates through each character in the string once. Therefore, the time complexity is O(n), where n is the length of the input string <code class="docutils literal notranslate"><span class="pre">s</span></code>.</p></li>
-</ul>
-<p><strong>Space Complexity:</strong></p>
-<ul class="simple">
-<li><p>The code uses several variables to store information, but the space they consume is constant and does not depend on the size of the input string. These variables include <code class="docutils literal notranslate"><span class="pre">max_length</span></code>, <code class="docutils literal notranslate"><span class="pre">max_count</span></code>, <code class="docutils literal notranslate"><span class="pre">start</span></code>, and <code class="docutils literal notranslate"><span class="pre">char_count</span></code>. Therefore, the space complexity is O(1) or constant space.</p></li>
-</ul>
-<p><strong>In summary</strong>, the time complexity of the code is O(n), where n is the length of the input string, and the space complexity is O(1), as it uses a constant amount of additional space regardless of the input size. This algorithm efficiently solves the problem with a linear time complexity.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Optimization Challenge:</strong> Modify the code to find the longest substring containing the same letter with at most k replacements in O(n) time complexity and O(1) space complexity. Hint: You may need to update the approach to achieve this.</p></li>
-<li><p><strong>Variation with Lowercase Letters:</strong> Extend the problem to include both uppercase and lowercase English letters in the input string s. Write a function that can handle this extended input and still find the longest substring with at most k replacements efficiently.</p></li>
-</ol>
-</section>
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-    <h1>76. Minimum Window Substring</h1>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="minimum-window-substring">
-<h1>76. Minimum Window Substring<a class="headerlink" href="#minimum-window-substring" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Hard</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/minimum-window-substring/">To see the Minimum Window Substring problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given two strings <code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code> of lengths <code class="docutils literal notranslate"><span class="pre">m</span></code> and <code class="docutils literal notranslate"><span class="pre">n</span></code> respectively, return <em>the <strong>minimum window substring</strong> of <code class="docutils literal notranslate"><span class="pre">s</span></code> such that every character in <code class="docutils literal notranslate"><span class="pre">t</span></code> (<strong>including duplicates</strong>) is included in the window. If there is no such substring, return the empty string <code class="docutils literal notranslate"><span class="pre">&quot;&quot;</span></code>.</em></p>
-<p>The testcases will be generated such that the answer is <strong>unique</strong>.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">==</span> <span class="pre">s.length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">==</span> <span class="pre">t.length</span></code></p></li>
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">m,</span> <span class="pre">n</span></code> &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">s</span></code> and <code class="docutils literal notranslate"><span class="pre">t</span></code> consist of uppercase and lowercase English letters.</p></li>
-</ul>
-<p><strong>Follow up</strong>: Could you find an algorithm that runs in <span class="math notranslate nohighlight">\(O(m + n)\)</span> time?</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">min_window_substring</span><span class="p">(</span><span class="n">s</span><span class="p">,</span> <span class="n">t</span><span class="p">):</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">s</span> <span class="ow">or</span> <span class="ow">not</span> <span class="n">t</span><span class="p">:</span>
-        <span class="k">return</span> <span class="s2">&quot;&quot;</span>
-
-    <span class="c1"># Initialize dictionaries to keep track of character counts for t and the current window in s.</span>
-    <span class="n">t_dict</span> <span class="o">=</span> <span class="p">{}</span>
-    <span class="n">current_window_dict</span> <span class="o">=</span> <span class="p">{}</span>
-    
-    <span class="c1"># Populate t_dict with character counts for string t.</span>
-    <span class="k">for</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">t</span><span class="p">:</span>
-        <span class="n">t_dict</span><span class="p">[</span><span class="n">char</span><span class="p">]</span> <span class="o">=</span> <span class="n">t_dict</span><span class="o">.</span><span class="n">get</span><span class="p">(</span><span class="n">char</span><span class="p">,</span> <span class="mi">0</span><span class="p">)</span> <span class="o">+</span> <span class="mi">1</span>
-
-    <span class="c1"># Initialize pointers for the sliding window.</span>
-    <span class="n">left</span> <span class="o">=</span> <span class="mi">0</span>
-    <span class="n">min_len</span> <span class="o">=</span> <span class="nb">float</span><span class="p">(</span><span class="s1">&#39;inf&#39;</span><span class="p">)</span>
-    <span class="n">min_window</span> <span class="o">=</span> <span class="s2">&quot;&quot;</span>
-    <span class="n">required_chars</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">t_dict</span><span class="p">)</span>
-
-    <span class="c1"># Initialize a variable to keep track of how many required characters have been found in the current window.</span>
-    <span class="n">found_chars</span> <span class="o">=</span> <span class="mi">0</span>
-
-    <span class="c1"># Iterate over the string s using the right pointer.</span>
-    <span class="k">for</span> <span class="n">right</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">s</span><span class="p">)):</span>
-        <span class="n">char</span> <span class="o">=</span> <span class="n">s</span><span class="p">[</span><span class="n">right</span><span class="p">]</span>
-
-        <span class="c1"># Update the current_window_dict.</span>
-        <span class="n">current_window_dict</span><span class="p">[</span><span class="n">char</span><span class="p">]</span> <span class="o">=</span> <span class="n">current_window_dict</span><span class="o">.</span><span class="n">get</span><span class="p">(</span><span class="n">char</span><span class="p">,</span> <span class="mi">0</span><span class="p">)</span> <span class="o">+</span> <span class="mi">1</span>
-
-        <span class="c1"># Check if the current character is a required character and if its count matches the required count.</span>
-        <span class="k">if</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">t_dict</span> <span class="ow">and</span> <span class="n">current_window_dict</span><span class="p">[</span><span class="n">char</span><span class="p">]</span> <span class="o">==</span> <span class="n">t_dict</span><span class="p">[</span><span class="n">char</span><span class="p">]:</span>
-            <span class="n">found_chars</span> <span class="o">+=</span> <span class="mi">1</span>
-
-        <span class="c1"># Try to minimize the window by moving the left pointer.</span>
-        <span class="k">while</span> <span class="n">left</span> <span class="o">&lt;=</span> <span class="n">right</span> <span class="ow">and</span> <span class="n">found_chars</span> <span class="o">==</span> <span class="n">required_chars</span><span class="p">:</span>
-            <span class="c1"># Calculate the current window size.</span>
-            <span class="n">window_size</span> <span class="o">=</span> <span class="n">right</span> <span class="o">-</span> <span class="n">left</span> <span class="o">+</span> <span class="mi">1</span>
-
-            <span class="c1"># If the current window is smaller than the minimum found so far, update min_len and min_window.</span>
-            <span class="k">if</span> <span class="n">window_size</span> <span class="o">&lt;</span> <span class="n">min_len</span><span class="p">:</span>
-                <span class="n">min_len</span> <span class="o">=</span> <span class="n">window_size</span>
-                <span class="n">min_window</span> <span class="o">=</span> <span class="n">s</span><span class="p">[</span><span class="n">left</span><span class="p">:</span><span class="n">right</span><span class="o">+</span><span class="mi">1</span><span class="p">]</span>
-
-            <span class="c1"># Move the left pointer to the right to shrink the window.</span>
-            <span class="n">left_char</span> <span class="o">=</span> <span class="n">s</span><span class="p">[</span><span class="n">left</span><span class="p">]</span>
-            <span class="n">current_window_dict</span><span class="p">[</span><span class="n">left_char</span><span class="p">]</span> <span class="o">-=</span> <span class="mi">1</span>
-
-            <span class="c1"># Check if the character removed from the window was a required character.</span>
-            <span class="k">if</span> <span class="n">left_char</span> <span class="ow">in</span> <span class="n">t_dict</span> <span class="ow">and</span> <span class="n">current_window_dict</span><span class="p">[</span><span class="n">left_char</span><span class="p">]</span> <span class="o">&lt;</span> <span class="n">t_dict</span><span class="p">[</span><span class="n">left_char</span><span class="p">]:</span>
-                <span class="n">found_chars</span> <span class="o">-=</span> <span class="mi">1</span>
-
-            <span class="c1"># Move the left pointer further to continue shrinking the window.</span>
-            <span class="n">left</span> <span class="o">+=</span> <span class="mi">1</span>
-
-    <span class="k">return</span> <span class="n">min_window</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We start by checking if either of the input strings <code class="docutils literal notranslate"><span class="pre">s</span></code> or <code class="docutils literal notranslate"><span class="pre">t</span></code> is empty. If either of them is empty, we return an empty string since there can’t be any valid substring in this case.</p></li>
-<li><p>We initialize two dictionaries: <code class="docutils literal notranslate"><span class="pre">t_dict</span></code> and <code class="docutils literal notranslate"><span class="pre">current_window_dict</span></code>. These dictionaries will be used to keep track of character counts in the string <code class="docutils literal notranslate"><span class="pre">t</span></code> and the current window in string <code class="docutils literal notranslate"><span class="pre">s</span></code>, respectively.</p></li>
-<li><p>We populate the <code class="docutils literal notranslate"><span class="pre">t_dict</span></code> dictionary by iterating through string <code class="docutils literal notranslate"><span class="pre">t</span></code>. For each character, we increment its count in the dictionary using <code class="docutils literal notranslate"><span class="pre">t_dict.get(char,</span> <span class="pre">0)</span> <span class="pre">+</span> <span class="pre">1</span></code>. This allows us to count the occurrences of each character in <code class="docutils literal notranslate"><span class="pre">t</span></code>.</p></li>
-<li><p>We initialize two pointers: <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code>. The <code class="docutils literal notranslate"><span class="pre">left</span></code> pointer will represent the start of the current window, and the <code class="docutils literal notranslate"><span class="pre">right</span></code> pointer will represent the end of the current window. We also initialize <code class="docutils literal notranslate"><span class="pre">min_len</span></code> to store the length of the minimum window found so far, and <code class="docutils literal notranslate"><span class="pre">min_window</span></code> to store the actual minimum window substring.</p></li>
-<li><p>We determine the number of required characters in <code class="docutils literal notranslate"><span class="pre">t</span></code> (i.e., the number of distinct characters in <code class="docutils literal notranslate"><span class="pre">t</span></code>) and store it in the variable <code class="docutils literal notranslate"><span class="pre">required_chars</span></code>.</p></li>
-<li><p>We initialize a variable <code class="docutils literal notranslate"><span class="pre">found_chars</span></code> to keep track of how many required characters have been found in the current window. Initially, it is set to 0.</p></li>
-<li><p>We iterate over the string <code class="docutils literal notranslate"><span class="pre">s</span></code> using the <code class="docutils literal notranslate"><span class="pre">right</span></code> pointer. In each iteration, we do the following:</p>
-<ul class="simple">
-<li><p>Update the <code class="docutils literal notranslate"><span class="pre">current_window_dict</span></code> by incrementing the count of the current character.</p></li>
-<li><p>Check if the current character is a required character (present in <code class="docutils literal notranslate"><span class="pre">t_dict</span></code>) and if its count in the <code class="docutils literal notranslate"><span class="pre">current_window_dict</span></code> matches the required count from <code class="docutils literal notranslate"><span class="pre">t_dict</span></code>. If so, we increment <code class="docutils literal notranslate"><span class="pre">found_chars</span></code>.</p></li>
-</ul>
-</li>
-<li><p>After updating the window, we attempt to minimize the window size by moving the <code class="docutils literal notranslate"><span class="pre">left</span></code> pointer to the right. In this step, we:</p>
-<ul class="simple">
-<li><p>Calculate the size of the current window.</p></li>
-<li><p>If the current window size is smaller than the minimum found so far (<code class="docutils literal notranslate"><span class="pre">min_len</span></code>), we update <code class="docutils literal notranslate"><span class="pre">min_len</span></code> and <code class="docutils literal notranslate"><span class="pre">min_window</span></code> to store the current window substring.</p></li>
-<li><p>Move the <code class="docutils literal notranslate"><span class="pre">left</span></code> pointer to the right to shrink the window.</p></li>
-<li><p>Check if the character being removed from the window was a required character. If it was, we decrement <code class="docutils literal notranslate"><span class="pre">found_chars</span></code>.</p></li>
-<li><p>Continue moving the <code class="docutils literal notranslate"><span class="pre">left</span></code> pointer further to continue shrinking the window if the window still contains all required characters.</p></li>
-</ul>
-</li>
-<li><p>Finally, we return <code class="docutils literal notranslate"><span class="pre">min_window</span></code>, which will contain the minimum window substring that contains all characters from <code class="docutils literal notranslate"><span class="pre">t</span></code>.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">min_window_substring</span><span class="p">(</span><span class="s2">&quot;ADOBECODEBANC&quot;</span><span class="p">,</span> <span class="s2">&quot;ABC&quot;</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>BANC
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">min_window_substring</span><span class="p">(</span><span class="s2">&quot;a&quot;</span><span class="p">,</span> <span class="s2">&quot;a&quot;</span><span class="p">))</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>a
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3:</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">min_window_substring</span><span class="p">(</span><span class="s2">&quot;a&quot;</span><span class="p">,</span> <span class="s2">&quot;aa&quot;</span><span class="p">))</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">min_window_substring</span></code> function:</p>
-<p><strong>Time Complexity:</strong></p>
-<ul class="simple">
-<li><p>The main loop iterates through the string <code class="docutils literal notranslate"><span class="pre">s</span></code> from left to right using the <code class="docutils literal notranslate"><span class="pre">right</span></code> pointer. This loop runs in O(m) time, where ‘m’ is the length of string <code class="docutils literal notranslate"><span class="pre">s</span></code>.</p></li>
-<li><p>Inside the loop, we have a while loop that moves the <code class="docutils literal notranslate"><span class="pre">left</span></code> pointer to the right to shrink the window as needed. In the worst case, this while loop can also run in O(m) time because in the worst case, we may have to move the <code class="docutils literal notranslate"><span class="pre">left</span></code> pointer all the way to the end of the string.</p></li>
-<li><p>Within each iteration of the while loop, we perform constant time operations, such as updating dictionaries and comparing character counts.</p></li>
-<li><p>The initialization of the <code class="docutils literal notranslate"><span class="pre">t_dict</span></code> dictionary takes O(n) time, where ‘n’ is the length of string <code class="docutils literal notranslate"><span class="pre">t</span></code>.</p></li>
-</ul>
-<p>Therefore, the overall time complexity of the function is O(m + n) because the dominant factor is the length of string <code class="docutils literal notranslate"><span class="pre">s</span></code>, and the length of string <code class="docutils literal notranslate"><span class="pre">t</span></code> has a smaller impact.</p>
-<p><strong>Space Complexity:</strong></p>
-<ul class="simple">
-<li><p>The space complexity is determined by the space used by the <code class="docutils literal notranslate"><span class="pre">t_dict</span></code> dictionary, the <code class="docutils literal notranslate"><span class="pre">current_window_dict</span></code> dictionary, and a few variables.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">t_dict</span></code> dictionary stores character counts for string <code class="docutils literal notranslate"><span class="pre">t</span></code>. In the worst case, when all characters in <code class="docutils literal notranslate"><span class="pre">t</span></code> are distinct, this dictionary can have a maximum of ‘n’ key-value pairs. So, the space complexity for <code class="docutils literal notranslate"><span class="pre">t_dict</span></code> is O(n).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">current_window_dict</span></code> dictionary stores character counts for the current window. In the worst case, it can have a maximum of ‘m’ key-value pairs. So, the space complexity for <code class="docutils literal notranslate"><span class="pre">current_window_dict</span></code> is also O(m).</p></li>
-<li><p>Other variables used in the function, such as <code class="docutils literal notranslate"><span class="pre">left</span></code>, <code class="docutils literal notranslate"><span class="pre">right</span></code>, <code class="docutils literal notranslate"><span class="pre">min_len</span></code>, <code class="docutils literal notranslate"><span class="pre">min_window</span></code>, <code class="docutils literal notranslate"><span class="pre">required_chars</span></code>, and <code class="docutils literal notranslate"><span class="pre">found_chars</span></code>, are all of constant size and do not depend on the input sizes.</p></li>
-</ul>
-<p><strong>In summary</strong>, the overall space complexity of the function is O(max(m, n)), which means it is determined by the larger of the two input strings’ lengths. This is because the space used for <code class="docutils literal notranslate"><span class="pre">t_dict</span></code> and <code class="docutils literal notranslate"><span class="pre">current_window_dict</span></code> dominates the space complexity.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Smallest Distinct Substring:</strong> Instead of finding the minimum window substring containing all characters, find the smallest distinct (unique) substring of s that contains all characters from t. This variation adds complexity because you must find a substring with distinct characters.</p></li>
-<li><p><strong>No Extra Space:</strong> Solve the problem without using any extra space, such as dictionaries or arrays, other than a constant amount of space. This is a significant optimization challenge.</p></li>
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-    <h1>20. Valid Parentheses</h1>
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-<h1>20. Valid Parentheses<a class="headerlink" href="#valid-parentheses" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/valid-parentheses/">To see the Valid Parentheses problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given a string s containing just the characters <code class="docutils literal notranslate"><span class="pre">'('</span></code>, <code class="docutils literal notranslate"><span class="pre">')'</span></code>, <code class="docutils literal notranslate"><span class="pre">'{'</span></code>, <code class="docutils literal notranslate"><span class="pre">'}'</span></code>, <code class="docutils literal notranslate"><span class="pre">'['</span></code> and <code class="docutils literal notranslate"><span class="pre">']'</span></code>, determine if the input string is valid.</p>
-<p>An input string is valid if:</p>
-<ol class="arabic simple">
-<li><p>Open brackets must be closed by the same type of brackets.</p></li>
-<li><p>Open brackets must be closed in the correct order.</p></li>
-<li><p>Every close bracket has a corresponding open bracket of the same type.</p></li>
-</ol>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">s.length</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">s</span></code> consists of parentheses only <code class="docutils literal notranslate"><span class="pre">'()[]{}'</span></code>.</p></li>
-</ul>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">isValid</span><span class="p">(</span><span class="n">s</span><span class="p">):</span>
-    <span class="c1"># Create an empty stack to store opening brackets</span>
-    <span class="n">stack</span> <span class="o">=</span> <span class="p">[]</span>
-    
-    <span class="c1"># Define a mapping for matching brackets</span>
-    <span class="n">bracket_mapping</span> <span class="o">=</span> <span class="p">{</span><span class="s1">&#39;)&#39;</span><span class="p">:</span> <span class="s1">&#39;(&#39;</span><span class="p">,</span> <span class="s1">&#39;}&#39;</span><span class="p">:</span> <span class="s1">&#39;{&#39;</span><span class="p">,</span> <span class="s1">&#39;]&#39;</span><span class="p">:</span> <span class="s1">&#39;[&#39;</span><span class="p">}</span>
-    
-    <span class="c1"># Iterate through the characters in the input string</span>
-    <span class="k">for</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">s</span><span class="p">:</span>
-        <span class="c1"># If the character is an opening bracket, push it onto the stack</span>
-        <span class="k">if</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">bracket_mapping</span><span class="o">.</span><span class="n">values</span><span class="p">():</span>
-            <span class="n">stack</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">char</span><span class="p">)</span>
-        <span class="c1"># If the character is a closing bracket</span>
-        <span class="k">elif</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">bracket_mapping</span><span class="o">.</span><span class="n">keys</span><span class="p">():</span>
-            <span class="c1"># Pop the top element from the stack if it exists, or use a dummy value &#39;#&#39;</span>
-            <span class="n">top_element</span> <span class="o">=</span> <span class="n">stack</span><span class="o">.</span><span class="n">pop</span><span class="p">()</span> <span class="k">if</span> <span class="n">stack</span> <span class="k">else</span> <span class="s1">&#39;#&#39;</span>
-            <span class="c1"># If the popped element does not match the corresponding opening bracket, return False</span>
-            <span class="k">if</span> <span class="n">bracket_mapping</span><span class="p">[</span><span class="n">char</span><span class="p">]</span> <span class="o">!=</span> <span class="n">top_element</span><span class="p">:</span>
-                <span class="k">return</span> <span class="kc">False</span>
-        <span class="c1"># If the character is not a bracket, ignore it</span>
-        
-    <span class="c1"># After processing the entire string, if there are any unmatched opening brackets left in the stack, return False</span>
-    <span class="k">return</span> <span class="nb">len</span><span class="p">(</span><span class="n">stack</span><span class="p">)</span> <span class="o">==</span> <span class="mi">0</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We start by defining the <code class="docutils literal notranslate"><span class="pre">isValid</span></code> function that takes a single argument <code class="docutils literal notranslate"><span class="pre">s</span></code>, which is the input string containing parentheses and brackets.</p></li>
-<li><p>Inside the function, we create an empty stack, which is a list in Python, to store opening brackets as we encounter them in the input string. The stack will help us keep track of the brackets and their order.</p></li>
-<li><p>We define a <code class="docutils literal notranslate"><span class="pre">bracket_mapping</span></code> dictionary that maps each closing bracket to its corresponding opening bracket. This mapping will be used to check if a closing bracket matches the most recent opening bracket in the stack.</p></li>
-<li><p>We iterate through each character <code class="docutils literal notranslate"><span class="pre">char</span></code> in the input string <code class="docutils literal notranslate"><span class="pre">s</span></code> using a <code class="docutils literal notranslate"><span class="pre">for</span></code> loop.</p></li>
-<li><p>If the character <code class="docutils literal notranslate"><span class="pre">char</span></code> is an opening bracket (i.e., it exists in the <code class="docutils literal notranslate"><span class="pre">values()</span></code> of <code class="docutils literal notranslate"><span class="pre">bracket_mapping</span></code>), we push it onto the stack using the <code class="docutils literal notranslate"><span class="pre">append()</span></code> method.</p></li>
-<li><p>If the character <code class="docutils literal notranslate"><span class="pre">char</span></code> is a closing bracket (i.e., it exists in the <code class="docutils literal notranslate"><span class="pre">keys()</span></code> of <code class="docutils literal notranslate"><span class="pre">bracket_mapping</span></code>), we need to check if it matches the corresponding opening bracket. To do this, we pop the top element from the stack (if it exists) and store it in <code class="docutils literal notranslate"><span class="pre">top_element</span></code>. We use a dummy value <code class="docutils literal notranslate"><span class="pre">'#'</span></code> if the stack is empty to avoid errors.</p></li>
-<li><p>We then compare <code class="docutils literal notranslate"><span class="pre">top_element</span></code> with the corresponding opening bracket using <code class="docutils literal notranslate"><span class="pre">bracket_mapping[char]</span></code>. If they do not match, it means the string is not valid, and we return <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-<li><p>If the character <code class="docutils literal notranslate"><span class="pre">char</span></code> is not a bracket, we simply ignore it and continue the loop.</p></li>
-<li><p>After processing the entire string, we check if there are any unmatched opening brackets left in the stack. If the stack is empty, it means all opening brackets have been properly matched and closed, and we return <code class="docutils literal notranslate"><span class="pre">True</span></code>. Otherwise, we return <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-<li><p>Finally, we provide some test cases at the bottom of the code to demonstrate how the function works for different input strings.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isValid</span><span class="p">(</span><span class="s2">&quot;()&quot;</span><span class="p">))</span>    
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isValid</span><span class="p">(</span><span class="s2">&quot;()[]</span><span class="si">{}</span><span class="s2">&quot;</span><span class="p">))</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3:</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isValid</span><span class="p">(</span><span class="s2">&quot;(]&quot;</span><span class="p">))</span>     
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">isValid</span></code> function:</p>
-<p><strong>Time Complexity:</strong></p>
-<ul class="simple">
-<li><p>The function iterates through each character in the input string <code class="docutils literal notranslate"><span class="pre">s</span></code> once, performing constant-time operations for each character.</p></li>
-<li><p>Pushing and popping elements from the stack (list) also takes constant time in most cases.</p></li>
-<li><p>Therefore, the overall time complexity of the function is <span class="math notranslate nohighlight">\(O(n)\)</span>, where <span class="math notranslate nohighlight">\(n\)</span> is the length of the input string <code class="docutils literal notranslate"><span class="pre">s</span></code>.</p></li>
-</ul>
-<p><strong>Space Complexity:</strong></p>
-<ul class="simple">
-<li><p>The space complexity of the function is determined by the space used by the stack and the <code class="docutils literal notranslate"><span class="pre">bracket_mapping</span></code> dictionary.</p></li>
-<li><p>In the worst case, if the input string <code class="docutils literal notranslate"><span class="pre">s</span></code> consists entirely of opening brackets, the stack can potentially contain all of these opening brackets, resulting in a space complexity of <span class="math notranslate nohighlight">\(O(n)\)</span> in terms of the stack.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">bracket_mapping</span></code> dictionary has a constant number of key-value pairs (3 pairs in this case).</p></li>
-<li><p>Therefore, the overall space complexity of the function is <span class="math notranslate nohighlight">\(O(n)\)</span>, where <span class="math notranslate nohighlight">\(n\)</span> is the length of the input string <code class="docutils literal notranslate"><span class="pre">s</span></code>, mainly due to the stack space.</p></li>
-</ul>
-<p><strong>In summary:</strong></p>
-<ul class="simple">
-<li><p>The time complexity is <span class="math notranslate nohighlight">\(O(n)\)</span> because we iterate through the string once.</p></li>
-<li><p>The space complexity is <span class="math notranslate nohighlight">\(O(n)\)</span> due to the stack used to keep track of opening brackets.</p></li>
-</ul>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Valid Expressions:</strong> Modify the problem to validate not only brackets but also arithmetic expressions containing parentheses, such as “<span class="math notranslate nohighlight">\(2 * (3 + 5) / (4 - 2)\)</span>”.</p></li>
-<li><p><strong>Valid Parentheses Combinations:</strong>
-Write a function to generate all valid combinations of n pairs of parentheses, where <span class="math notranslate nohighlight">\(n\)</span> is a positive integer. For example, for <span class="math notranslate nohighlight">\(n = 3\)</span>, the valid combinations are [“((()))”, “(()())”, “(())()”, “()(())”, “()()()”].</p></li>
-</ol>
-</section>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="find-minimum-in-rotated-sorted-array">
-<h1>153. Find Minimum in Rotated Sorted Array<a class="headerlink" href="#find-minimum-in-rotated-sorted-array" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/">To see the Find Minimum in Rotated Sorted Array problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Suppose an array of length <code class="docutils literal notranslate"><span class="pre">n</span></code> sorted in ascending order is <strong>rotated</strong> between <code class="docutils literal notranslate"><span class="pre">1</span></code> and <code class="docutils literal notranslate"><span class="pre">n</span></code> times. For example, the array <code class="docutils literal notranslate"><span class="pre">nums</span> <span class="pre">=</span> <span class="pre">[0,1,2,4,5,6,7]</span></code> might become:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">[4,5,6,7,0,1,2]</span></code> if it was rotated <code class="docutils literal notranslate"><span class="pre">4</span></code> times.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">[0,1,2,4,5,6,7]</span></code> if it was rotated <code class="docutils literal notranslate"><span class="pre">7</span></code> times.</p></li>
-</ul>
-<p>Notice that <strong>rotating</strong> an array <code class="docutils literal notranslate"><span class="pre">[a[0],</span> <span class="pre">a[1],</span> <span class="pre">a[2],</span> <span class="pre">...,</span> <span class="pre">a[n-1]]</span></code> 1 time results in the array <code class="docutils literal notranslate"><span class="pre">[a[n-1],</span> <span class="pre">a[0],</span> <span class="pre">a[1],</span> <span class="pre">a[2],</span> <span class="pre">...,</span> <span class="pre">a[n-2]]</span></code>.</p>
-<p>Given the sorted rotated array <code class="docutils literal notranslate"><span class="pre">nums</span></code> of <strong>unique</strong> elements, return <em>the minimum element of this array.</em></p>
-<p>You must write an algorithm that runs in <span class="math notranslate nohighlight">\(O(log\ n)\)</span> time.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">==</span> <span class="pre">nums.length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">n</span> <span class="pre">&lt;=</span> <span class="pre">5000</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-5000</span> <span class="pre">&lt;=</span> <span class="pre">nums[i]</span> <span class="pre">&lt;=</span> <span class="pre">5000</span></code></p></li>
-<li><p>All the integers of <code class="docutils literal notranslate"><span class="pre">nums</span></code> are <strong>unique</strong>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">nums</span></code> is sorted and rotated between <code class="docutils literal notranslate"><span class="pre">1</span></code> and <code class="docutils literal notranslate"><span class="pre">n</span></code> times.</p></li>
-</ul>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">findMin</span><span class="p">(</span><span class="n">nums</span><span class="p">):</span>
-    <span class="n">left</span><span class="p">,</span> <span class="n">right</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span>
-    
-    <span class="k">while</span> <span class="n">left</span> <span class="o">&lt;</span> <span class="n">right</span><span class="p">:</span>
-        <span class="n">mid</span> <span class="o">=</span> <span class="n">left</span> <span class="o">+</span> <span class="p">(</span><span class="n">right</span> <span class="o">-</span> <span class="n">left</span><span class="p">)</span> <span class="o">//</span> <span class="mi">2</span>
-        
-        <span class="c1"># If the mid element is greater than the rightmost element,</span>
-        <span class="c1"># it means the minimum element is in the right half.</span>
-        <span class="k">if</span> <span class="n">nums</span><span class="p">[</span><span class="n">mid</span><span class="p">]</span> <span class="o">&gt;</span> <span class="n">nums</span><span class="p">[</span><span class="n">right</span><span class="p">]:</span>
-            <span class="n">left</span> <span class="o">=</span> <span class="n">mid</span> <span class="o">+</span> <span class="mi">1</span>
-        <span class="c1"># If the mid element is less than or equal to the rightmost element,</span>
-        <span class="c1"># it means the minimum element is in the left half or at the mid itself.</span>
-        <span class="k">else</span><span class="p">:</span>
-            <span class="n">right</span> <span class="o">=</span> <span class="n">mid</span>
-    
-    <span class="c1"># The loop will break when left and right converge to the minimum element.</span>
-    <span class="c1"># At this point, left (or right) will be pointing to the minimum element.</span>
-    <span class="k">return</span> <span class="n">nums</span><span class="p">[</span><span class="n">left</span><span class="p">]</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>The function <code class="docutils literal notranslate"><span class="pre">findMin(nums)</span></code> takes an input list <code class="docutils literal notranslate"><span class="pre">nums</span></code>, which represents the rotated sorted array.</p></li>
-<li><p>Initialize two pointers <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> to track the current search range. Initially, <code class="docutils literal notranslate"><span class="pre">left</span></code> is set to 0 (the beginning of the array), and <code class="docutils literal notranslate"><span class="pre">right</span></code> is set to <code class="docutils literal notranslate"><span class="pre">len(nums)</span> <span class="pre">-</span> <span class="pre">1</span></code> (the end of the array).</p></li>
-<li><p>Enter a <code class="docutils literal notranslate"><span class="pre">while</span></code> loop with the condition <code class="docutils literal notranslate"><span class="pre">left</span> <span class="pre">&lt;</span> <span class="pre">right</span></code>. This loop continues until <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> converge to a single element, which will be the minimum element in the rotated array.</p></li>
-<li><p>Inside the loop, calculate the middle index <code class="docutils literal notranslate"><span class="pre">mid</span></code> using integer division. This helps in finding the middle element of the current search range.</p></li>
-<li><p>Check if the element at the middle index (<code class="docutils literal notranslate"><span class="pre">nums[mid]</span></code>) is greater than the element at the rightmost index (<code class="docutils literal notranslate"><span class="pre">nums[right]</span></code>). If this condition is true, it means that the minimum element must be in the right half of the current search range. So, update <code class="docutils literal notranslate"><span class="pre">left</span></code> to <code class="docutils literal notranslate"><span class="pre">mid</span> <span class="pre">+</span> <span class="pre">1</span></code>, effectively eliminating the left half of the search range.</p></li>
-<li><p>If the condition from step 5 is not met, it implies that the minimum element can be in the left half of the current search range or could be the element at the <code class="docutils literal notranslate"><span class="pre">mid</span></code> index itself. In this case, update <code class="docutils literal notranslate"><span class="pre">right</span></code> to <code class="docutils literal notranslate"><span class="pre">mid</span></code>, effectively eliminating the right half of the search range.</p></li>
-<li><p>Repeat steps 4-6 until <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> converge to the minimum element.</p></li>
-<li><p>When the loop exits, it means that <code class="docutils literal notranslate"><span class="pre">left</span></code> (or <code class="docutils literal notranslate"><span class="pre">right</span></code>) points to the minimum element in the array.</p></li>
-<li><p>Return <code class="docutils literal notranslate"><span class="pre">nums[left]</span></code> (or <code class="docutils literal notranslate"><span class="pre">nums[right]</span></code>) as the result, which is the minimum element in the rotated sorted array.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 1:</span>
-<span class="n">nums1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">3</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">5</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">findMin</span><span class="p">(</span><span class="n">nums1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>1
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">nums2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">4</span><span class="p">,</span> <span class="mi">5</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">7</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">findMin</span><span class="p">(</span><span class="n">nums2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>0
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3:</span>
-<span class="n">nums3</span> <span class="o">=</span> <span class="p">[</span><span class="mi">11</span><span class="p">,</span> <span class="mi">13</span><span class="p">,</span> <span class="mi">15</span><span class="p">,</span> <span class="mi">17</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">findMin</span><span class="p">(</span><span class="n">nums3</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>11
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong>
-The time complexity of this code is <span class="math notranslate nohighlight">\(O(log\ n)\)</span>, where ‘<span class="math notranslate nohighlight">\(n\)</span>’ is the length of the input array <code class="docutils literal notranslate"><span class="pre">nums</span></code>. This is because the binary search algorithm is employed, which continually divides the search range in half with each iteration.</p>
-<p>In each iteration of the <code class="docutils literal notranslate"><span class="pre">while</span></code> loop, the search range is halved, and this process continues until the <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> pointers converge to the minimum element. In the worst case, it will take logarithmic time to reduce the search range to a single element.</p>
-<p>Therefore, the binary search used in this code runs in <span class="math notranslate nohighlight">\(O(log\ n)\)</span> time complexity.</p>
-<p><strong>Space Complexity:</strong>
-The space complexity of this code is <span class="math notranslate nohighlight">\(O(1)\)</span>, which means it uses a constant amount of additional space that does not depend on the size of the input array.</p>
-<p>The algorithm uses a fixed number of variables (<code class="docutils literal notranslate"><span class="pre">left</span></code>, <code class="docutils literal notranslate"><span class="pre">right</span></code>, <code class="docutils literal notranslate"><span class="pre">mid</span></code>) to keep track of the search range and indices, but the number of these variables does not grow with the size of the input array. Hence, the space complexity is constant.</p>
-<p><strong>In summary:</strong></p>
-<ul class="simple">
-<li><p>Time Complexity: <span class="math notranslate nohighlight">\(O(log\ n)\)</span></p></li>
-<li><p>Space Complexity: <span class="math notranslate nohighlight">\(O(1)\)</span></p></li>
-</ul>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>Solve the problem with a variation where the input array may contain duplicates.</p></li>
-<li><p>Modify the problem to return the index of the minimum element in the rotated sorted array, rather than the element itself. The algorithm should still run in <span class="math notranslate nohighlight">\(O(log\ n)\)</span> time.</p></li>
-<li><p>Implement a function that finds the maximum element in a rotated sorted array. How would you adapt the binary search algorithm to solve this problem efficiently in <span class="math notranslate nohighlight">\(O(log\ n)\)</span> time?</p></li>
-<li><p>Implement a function that can find the kth smallest element in a rotated sorted array. This is an extension of the original problem, and the algorithm should still run in <span class="math notranslate nohighlight">\(O(log\ n)\)</span> time.</p></li>
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-    <h1>33. Search in Rotated Sorted Array</h1>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-<h1>33. Search in Rotated Sorted Array<a class="headerlink" href="#search-in-rotated-sorted-array" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/search-in-rotated-sorted-array/">To see the Search in Rotated Sorted Array problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>There is an integer array <code class="docutils literal notranslate"><span class="pre">nums</span></code> sorted in ascending order (with <strong>distinct</strong> values).</p>
-<p>Prior to being passed to your function, <code class="docutils literal notranslate"><span class="pre">nums</span></code> is <strong>possibly rotated</strong> at an unknown pivot index <code class="docutils literal notranslate"><span class="pre">k</span> <span class="pre">(1</span> <span class="pre">&lt;=</span> <span class="pre">k</span> <span class="pre">&lt;</span> <span class="pre">nums.length)</span></code> such that the resulting array is <code class="docutils literal notranslate"><span class="pre">[nums[k],</span> <span class="pre">nums[k+1],</span> <span class="pre">...,</span> <span class="pre">nums[n-1],</span> <span class="pre">nums[0],</span> <span class="pre">nums[1],</span> <span class="pre">...,</span> <span class="pre">nums[k-1]]</span></code> (<strong>0-indexed</strong>). For example, <code class="docutils literal notranslate"><span class="pre">[0,1,2,4,5,6,7]</span></code> might be rotated at pivot index 3 and become <code class="docutils literal notranslate"><span class="pre">[4,5,6,7,0,1,2]</span></code>.</p>
-<p>Given the array <code class="docutils literal notranslate"><span class="pre">nums</span></code> <strong>after</strong> the possible rotation and an integer <code class="docutils literal notranslate"><span class="pre">target</span></code>, return <em>the index of <code class="docutils literal notranslate"><span class="pre">target</span></code> if it is in <code class="docutils literal notranslate"><span class="pre">nums</span></code>, or <code class="docutils literal notranslate"><span class="pre">-1</span></code> if it is not in <code class="docutils literal notranslate"><span class="pre">nums</span></code>.</em></p>
-<p>You must write an algorithm with <span class="math notranslate nohighlight">\(O(log\ n)\)</span> runtime complexity.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">nums.length</span> <span class="pre">&lt;=</span> <span class="pre">5000</span></code></p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^4\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">nums[i]</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p>All values of <code class="docutils literal notranslate"><span class="pre">nums</span></code> are <strong>unique</strong>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">nums</span></code> is an ascending array that is possibly rotated.</p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^4\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">target</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-</ul>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">search</span><span class="p">(</span><span class="n">nums</span><span class="p">,</span> <span class="n">target</span><span class="p">):</span>
-    <span class="n">left</span><span class="p">,</span> <span class="n">right</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span>
-
-    <span class="c1"># Find the pivot index using binary search</span>
-    <span class="k">while</span> <span class="n">left</span> <span class="o">&lt;</span> <span class="n">right</span><span class="p">:</span>
-        <span class="n">mid</span> <span class="o">=</span> <span class="n">left</span> <span class="o">+</span> <span class="p">(</span><span class="n">right</span> <span class="o">-</span> <span class="n">left</span><span class="p">)</span> <span class="o">//</span> <span class="mi">2</span>
-
-        <span class="k">if</span> <span class="n">nums</span><span class="p">[</span><span class="n">mid</span><span class="p">]</span> <span class="o">&gt;</span> <span class="n">nums</span><span class="p">[</span><span class="n">right</span><span class="p">]:</span>
-            <span class="n">left</span> <span class="o">=</span> <span class="n">mid</span> <span class="o">+</span> <span class="mi">1</span>
-        <span class="k">else</span><span class="p">:</span>
-            <span class="n">right</span> <span class="o">=</span> <span class="n">mid</span>
-
-    <span class="n">pivot</span> <span class="o">=</span> <span class="n">left</span>
-    <span class="n">left</span><span class="p">,</span> <span class="n">right</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span>
-
-    <span class="c1"># Determine which part of the array to search in</span>
-    <span class="k">if</span> <span class="n">target</span> <span class="o">&gt;=</span> <span class="n">nums</span><span class="p">[</span><span class="n">pivot</span><span class="p">]</span> <span class="ow">and</span> <span class="n">target</span> <span class="o">&lt;=</span> <span class="n">nums</span><span class="p">[</span><span class="n">right</span><span class="p">]:</span>
-        <span class="n">left</span> <span class="o">=</span> <span class="n">pivot</span>
-    <span class="k">else</span><span class="p">:</span>
-        <span class="n">right</span> <span class="o">=</span> <span class="n">pivot</span>
-
-    <span class="c1"># Perform binary search to find the target</span>
-    <span class="k">while</span> <span class="n">left</span> <span class="o">&lt;=</span> <span class="n">right</span><span class="p">:</span>
-        <span class="n">mid</span> <span class="o">=</span> <span class="n">left</span> <span class="o">+</span> <span class="p">(</span><span class="n">right</span> <span class="o">-</span> <span class="n">left</span><span class="p">)</span> <span class="o">//</span> <span class="mi">2</span>
-
-        <span class="k">if</span> <span class="n">nums</span><span class="p">[</span><span class="n">mid</span><span class="p">]</span> <span class="o">==</span> <span class="n">target</span><span class="p">:</span>
-            <span class="k">return</span> <span class="n">mid</span>
-        <span class="k">elif</span> <span class="n">nums</span><span class="p">[</span><span class="n">mid</span><span class="p">]</span> <span class="o">&lt;</span> <span class="n">target</span><span class="p">:</span>
-            <span class="n">left</span> <span class="o">=</span> <span class="n">mid</span> <span class="o">+</span> <span class="mi">1</span>
-        <span class="k">else</span><span class="p">:</span>
-            <span class="n">right</span> <span class="o">=</span> <span class="n">mid</span> <span class="o">-</span> <span class="mi">1</span>
-
-    <span class="k">return</span> <span class="o">-</span><span class="mi">1</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">search</span></code> function takes two arguments: <code class="docutils literal notranslate"><span class="pre">nums</span></code>, which is the rotated sorted array, and <code class="docutils literal notranslate"><span class="pre">target</span></code>, which is the value we want to find in the array.</p></li>
-<li><p>We initialize two pointers, <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code>, which represent the range in which we are going to perform binary search. Initially, <code class="docutils literal notranslate"><span class="pre">left</span></code> is set to 0, and <code class="docutils literal notranslate"><span class="pre">right</span></code> is set to the index of the last element in the array (<code class="docutils literal notranslate"><span class="pre">len(nums)</span> <span class="pre">-</span> <span class="pre">1</span></code>).</p></li>
-<li><p>We start by finding the pivot index using binary search. The pivot index is the index at which the array is rotated. We use a <code class="docutils literal notranslate"><span class="pre">while</span></code> loop to continue the search until <code class="docutils literal notranslate"><span class="pre">left</span></code> is less than <code class="docutils literal notranslate"><span class="pre">right</span></code>.</p></li>
-<li><p>Inside the loop, we calculate the middle index <code class="docutils literal notranslate"><span class="pre">mid</span></code> using integer division. We then compare the value at <code class="docutils literal notranslate"><span class="pre">mid</span></code> with the value at <code class="docutils literal notranslate"><span class="pre">right</span></code>. If <code class="docutils literal notranslate"><span class="pre">nums[mid]</span></code> is greater than <code class="docutils literal notranslate"><span class="pre">nums[right]</span></code>, it means the pivot point lies to the right of <code class="docutils literal notranslate"><span class="pre">mid</span></code>, so we update <code class="docutils literal notranslate"><span class="pre">left</span></code> to <code class="docutils literal notranslate"><span class="pre">mid</span> <span class="pre">+</span> <span class="pre">1</span></code>. Otherwise, the pivot point lies to the left of <code class="docutils literal notranslate"><span class="pre">mid</span></code>, so we update <code class="docutils literal notranslate"><span class="pre">right</span></code> to <code class="docutils literal notranslate"><span class="pre">mid</span></code>. This process continues until we find the pivot index.</p></li>
-<li><p>After finding the pivot index, we have divided the array into two parts: one part is sorted in ascending order, and the other part is also sorted in ascending order but rotated. We need to determine which part of the array contains the target value.</p></li>
-<li><p>We reset <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> pointers. If the target value is within the range <code class="docutils literal notranslate"><span class="pre">[nums[pivot],</span> <span class="pre">nums[right]]</span></code>, we set <code class="docutils literal notranslate"><span class="pre">left</span></code> to <code class="docutils literal notranslate"><span class="pre">pivot</span></code> (the start of the rotated part), indicating that we should search in the rotated part. Otherwise, we set <code class="docutils literal notranslate"><span class="pre">right</span></code> to <code class="docutils literal notranslate"><span class="pre">pivot</span></code> (the end of the sorted part), indicating that we should search in the sorted part.</p></li>
-<li><p>We then perform binary search again within the chosen range (<code class="docutils literal notranslate"><span class="pre">left</span></code> to <code class="docutils literal notranslate"><span class="pre">right</span></code>) to find the target element. The binary search continues until <code class="docutils literal notranslate"><span class="pre">left</span></code> is less than or equal to <code class="docutils literal notranslate"><span class="pre">right</span></code>.</p></li>
-<li><p>Inside the binary search loop, we calculate the middle index <code class="docutils literal notranslate"><span class="pre">mid</span></code> and compare <code class="docutils literal notranslate"><span class="pre">nums[mid]</span></code> with the target value. If they are equal, we return <code class="docutils literal notranslate"><span class="pre">mid</span></code> as the index of the target element. If <code class="docutils literal notranslate"><span class="pre">nums[mid]</span></code> is less than the target, we update <code class="docutils literal notranslate"><span class="pre">left</span></code> to <code class="docutils literal notranslate"><span class="pre">mid</span> <span class="pre">+</span> <span class="pre">1</span></code> to search in the right half. If <code class="docutils literal notranslate"><span class="pre">nums[mid]</span></code> is greater than the target, we update <code class="docutils literal notranslate"><span class="pre">right</span></code> to <code class="docutils literal notranslate"><span class="pre">mid</span> <span class="pre">-</span> <span class="pre">1</span></code> to search in the left half.</p></li>
-<li><p>If we exit the binary search loop without finding the target, we return -1 to indicate that the target is not present in the array.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">nums1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">4</span><span class="p">,</span> <span class="mi">5</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">7</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>
-<span class="n">target1</span> <span class="o">=</span> <span class="mi">0</span>
-<span class="n">result1</span> <span class="o">=</span> <span class="n">search</span><span class="p">(</span><span class="n">nums1</span><span class="p">,</span> <span class="n">target1</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result1</span><span class="p">)</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>4
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">nums2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">4</span><span class="p">,</span> <span class="mi">5</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">7</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>
-<span class="n">target2</span> <span class="o">=</span> <span class="mi">3</span>
-<span class="n">result2</span> <span class="o">=</span> <span class="n">search</span><span class="p">(</span><span class="n">nums2</span><span class="p">,</span> <span class="n">target2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result2</span><span class="p">)</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>-1
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3:</span>
-<span class="n">nums3</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">]</span>
-<span class="n">target3</span> <span class="o">=</span> <span class="mi">0</span>
-<span class="n">result3</span> <span class="o">=</span> <span class="n">search</span><span class="p">(</span><span class="n">nums3</span><span class="p">,</span> <span class="n">target3</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result3</span><span class="p">)</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>-1
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>Finding the pivot index using binary search takes <span class="math notranslate nohighlight">\(O(log\ n)\)</span> time, where ‘<span class="math notranslate nohighlight">\(n\)</span>’ is the number of elements in the array.</p></li>
-<li><p>After finding the pivot index, performing binary search to find the target element also takes <span class="math notranslate nohighlight">\(O(log\ n)\)</span> time in the worst case.</p></li>
-</ol>
-<p>The dominant factor in the time complexity is the binary search, and since we perform two binary searches sequentially, the overall time complexity of the code is <span class="math notranslate nohighlight">\(O(log\ n)\)</span>.</p>
-<p><strong>Space Complexity:</strong></p>
-<p>The space complexity of the code is very minimal:</p>
-<ol class="arabic simple">
-<li><p>We use a constant amount of additional space for variables such as <code class="docutils literal notranslate"><span class="pre">left</span></code>, <code class="docutils literal notranslate"><span class="pre">right</span></code>, <code class="docutils literal notranslate"><span class="pre">pivot</span></code>, and <code class="docutils literal notranslate"><span class="pre">mid</span></code>. These variables do not depend on the input size, so they contribute <span class="math notranslate nohighlight">\(O(1)\)</span> space complexity.</p></li>
-<li><p>The function does not use any additional data structures that scale with the input size.</p></li>
-</ol>
-<p>Therefore, the space complexity of the code is <span class="math notranslate nohighlight">\(O(1)\)</span>, which means it has a constant space complexity and does not depend on the size of the input array.</p>
-<p><strong>In summary:</strong></p>
-<ul class="simple">
-<li><p>Time Complexity: <span class="math notranslate nohighlight">\(O(log\ n)\)</span></p></li>
-<li><p>Space Complexity: <span class="math notranslate nohighlight">\(O(1)\)</span></p></li>
-</ul>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Find Minimum Element in Rotated Sorted Array</strong>: Write an algorithm to find the minimum element in a rotated sorted array. This is a variation of the problem where you don’t need to search for a target value but instead find the smallest element.</p></li>
-<li><p><strong>Search in a Circularly Sorted Array</strong>: Consider an array that is sorted in a circular manner (e.g., [4, 5, 6, 7, 0, 1, 2, 3]). Adapt the search algorithm to work for circularly sorted arrays while maintaining <span class="math notranslate nohighlight">\(O(log\ n)\)</span> complexity.</p></li>
-</ol>
-</section>
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-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/">153. Find Minimum In Rotated Sorted Array</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/search-in-rotated-sorted-array/">33. Search In Rotated Sorted Array</a></p></td>
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-<li><p><a class="reference external" href="https://leetcode.com/discuss/">LeetCode Discuss</a>: LeetCode’s discussion forum for each problem. You can find helpful hints and discussions here.</p></li>
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-    <h1>141. Linked List Cycle</h1>
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-                <h2> Contents </h2>
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-            <nav aria-label="Page">
-                <ul class="visible nav section-nav flex-column">
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="linked-list-cycle">
-<h1>141. Linked List Cycle<a class="headerlink" href="#linked-list-cycle" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/linked-list-cycle/">To see the Linked List Cycle problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given <code class="docutils literal notranslate"><span class="pre">head</span></code>, the head of a linked list, determine if the linked list has a cycle in it.</p>
-<p>There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the <code class="docutils literal notranslate"><span class="pre">next</span></code> pointer. Internally, <code class="docutils literal notranslate"><span class="pre">pos</span></code> is used to denote the index of the node that tail’s <code class="docutils literal notranslate"><span class="pre">next</span></code> pointer is connected to. <strong>Note that <code class="docutils literal notranslate"><span class="pre">pos</span></code> is not passed as a parameter.</strong></p>
-<p>Return <code class="docutils literal notranslate"><span class="pre">true</span></code> if there is a cycle in the linked list. Otherwise, return <code class="docutils literal notranslate"><span class="pre">false</span></code>.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of the nodes in the list is in the range <span class="math notranslate nohighlight">\([0, 10^4]\)</span>.</p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^5\)</span> &lt;= Node.val &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">pos</span></code> is <code class="docutils literal notranslate"><span class="pre">-1</span></code> or a <strong>valid index</strong> in the linked-list.</p></li>
-</ul>
-<p><strong>Follow up:</strong> Can you solve it using <span class="math notranslate nohighlight">\(O(1)\)</span> (i.e. constant) memory?</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Definition for singly-linked list.</span>
-<span class="k">class</span> <span class="nc">ListNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="kc">None</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">hasCycle</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">head</span><span class="p">:</span> <span class="n">ListNode</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="nb">bool</span><span class="p">:</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">head</span><span class="p">:</span>
-            <span class="k">return</span> <span class="kc">False</span>
-        
-        <span class="n">slow</span> <span class="o">=</span> <span class="n">head</span>
-        <span class="n">fast</span> <span class="o">=</span> <span class="n">head</span>
-        
-        <span class="k">while</span> <span class="n">fast</span> <span class="ow">and</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span><span class="p">:</span>
-            <span class="n">slow</span> <span class="o">=</span> <span class="n">slow</span><span class="o">.</span><span class="n">next</span>
-            <span class="n">fast</span> <span class="o">=</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span><span class="o">.</span><span class="n">next</span>
-            
-            <span class="k">if</span> <span class="n">slow</span> <span class="o">==</span> <span class="n">fast</span><span class="p">:</span>
-                <span class="k">return</span> <span class="kc">True</span>
-        
-        <span class="k">return</span> <span class="kc">False</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The given code defines a class <code class="docutils literal notranslate"><span class="pre">Solution</span></code> with a method <code class="docutils literal notranslate"><span class="pre">hasCycle</span></code> for determining if a linked list has a cycle.</p>
-<p>The <code class="docutils literal notranslate"><span class="pre">ListNode</span></code> class represents nodes in the linked list, where each node has a <code class="docutils literal notranslate"><span class="pre">val</span></code> (a value associated with the node) and a <code class="docutils literal notranslate"><span class="pre">next</span></code> attribute (a reference to the next node in the list).</p>
-<p>The <code class="docutils literal notranslate"><span class="pre">hasCycle</span></code> method takes the <code class="docutils literal notranslate"><span class="pre">head</span></code> of a linked list as input and returns <code class="docutils literal notranslate"><span class="pre">True</span></code> if there is a cycle in the linked list, and <code class="docutils literal notranslate"><span class="pre">False</span></code> otherwise. It uses two pointers, <code class="docutils literal notranslate"><span class="pre">slow</span></code> and <code class="docutils literal notranslate"><span class="pre">fast</span></code>, to traverse the linked list. If there is a cycle, the <code class="docutils literal notranslate"><span class="pre">fast</span></code> pointer will eventually catch up to the <code class="docutils literal notranslate"><span class="pre">slow</span></code> pointer.</p>
-<p>The algorithm works as follows:</p>
-<ol class="arabic simple">
-<li><p>It checks if the input <code class="docutils literal notranslate"><span class="pre">head</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code> (an empty list). If it is, the method returns <code class="docutils literal notranslate"><span class="pre">False</span></code> because an empty list can’t have a cycle.</p></li>
-<li><p>Two pointers, <code class="docutils literal notranslate"><span class="pre">slow</span></code> and <code class="docutils literal notranslate"><span class="pre">fast</span></code>, initially point to the <code class="docutils literal notranslate"><span class="pre">head</span></code> of the list.</p></li>
-<li><p>The code enters a <code class="docutils literal notranslate"><span class="pre">while</span></code> loop where <code class="docutils literal notranslate"><span class="pre">fast</span></code> moves two steps at a time, and <code class="docutils literal notranslate"><span class="pre">slow</span></code> moves one step at a time.</p></li>
-<li><p>If there is a cycle, <code class="docutils literal notranslate"><span class="pre">fast</span></code> will eventually catch up to <code class="docutils literal notranslate"><span class="pre">slow</span></code>, and the method returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>If the loop completes without finding a cycle, the method returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-</ol>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Helper function to create a linked list with a cycle</span>
-<span class="k">def</span> <span class="nf">create_linked_list_with_cycle</span><span class="p">(</span><span class="n">values</span><span class="p">,</span> <span class="n">pos</span><span class="p">):</span>
-    <span class="n">dummy</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">()</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">dummy</span>
-    <span class="n">cycle_node</span> <span class="o">=</span> <span class="kc">None</span>
-    
-    <span class="k">for</span> <span class="n">i</span><span class="p">,</span> <span class="n">val</span> <span class="ow">in</span> <span class="nb">enumerate</span><span class="p">(</span><span class="n">values</span><span class="p">):</span>
-        <span class="n">current</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="n">val</span><span class="p">)</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-        
-        <span class="k">if</span> <span class="n">i</span> <span class="o">==</span> <span class="n">pos</span><span class="p">:</span>
-            <span class="n">cycle_node</span> <span class="o">=</span> <span class="n">current</span>
-    
-    <span class="k">if</span> <span class="n">cycle_node</span><span class="p">:</span>
-        <span class="n">current</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">cycle_node</span>
-    
-    <span class="k">return</span> <span class="n">dummy</span><span class="o">.</span><span class="n">next</span>
-</pre></div>
-</div>
-</div>
-</div>
-<p>This helper function is designed to create a linked list with a cycle for testing purposes. It takes two arguments:</p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">values</span></code>: A list of values representing the nodes in the linked list.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">pos</span></code>: An integer representing the index of the node where the cycle begins.</p></li>
-</ol>
-<p>Here’s how the function works:</p>
-<ul class="simple">
-<li><p>It starts by creating an empty <code class="docutils literal notranslate"><span class="pre">ListNode</span></code> called <code class="docutils literal notranslate"><span class="pre">dummy</span></code>. This <code class="docutils literal notranslate"><span class="pre">dummy</span></code> node is used to simplify the creation of the linked list.</p></li>
-<li><p>It initializes a <code class="docutils literal notranslate"><span class="pre">current</span></code> pointer to the <code class="docutils literal notranslate"><span class="pre">dummy</span></code> node. This pointer will be used to traverse the linked list.</p></li>
-<li><p>It initializes a <code class="docutils literal notranslate"><span class="pre">cycle_node</span></code> variable to <code class="docutils literal notranslate"><span class="pre">None</span></code>. This variable will hold the reference to the node where the cycle begins, if any.</p></li>
-<li><p>It then iterates through the <code class="docutils literal notranslate"><span class="pre">values</span></code> list, creating a new <code class="docutils literal notranslate"><span class="pre">ListNode</span></code> for each value and appending it to the linked list.</p></li>
-<li><p>Inside the loop, it checks if the current index <code class="docutils literal notranslate"><span class="pre">i</span></code> is equal to the specified <code class="docutils literal notranslate"><span class="pre">pos</span></code>. If they match, it sets <code class="docutils literal notranslate"><span class="pre">cycle_node</span></code> to the current node. This simulates the creation of a cycle in the linked list.</p></li>
-<li><p>After the loop, if <code class="docutils literal notranslate"><span class="pre">cycle_node</span></code> is not <code class="docutils literal notranslate"><span class="pre">None</span></code> (indicating a cycle should be created), it connects the last node in the linked list to <code class="docutils literal notranslate"><span class="pre">cycle_node</span></code>, effectively creating a cycle.</p></li>
-<li><p>Finally, it returns the reference to the first node of the linked list (not the <code class="docutils literal notranslate"><span class="pre">dummy</span></code> node).</p></li>
-</ul>
-<p>This helper function allows you to easily create test cases where you can specify the values for the linked list and the position where the cycle starts. It’s particularly useful for testing the code’s ability to detect cycles in linked lists.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 1</span>
-<span class="n">head</span> <span class="o">=</span> <span class="p">[</span><span class="mi">3</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">0</span><span class="p">,</span><span class="o">-</span><span class="mi">4</span><span class="p">]</span>
-<span class="n">pos</span> <span class="o">=</span> <span class="mi">1</span>
-
-<span class="n">head</span> <span class="o">=</span> <span class="n">create_linked_list_with_cycle</span><span class="p">(</span><span class="n">head</span><span class="p">,</span> <span class="n">pos</span><span class="p">)</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">hasCycle</span><span class="p">(</span><span class="n">head</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 2</span>
-<span class="n">head</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">]</span>
-<span class="n">pos</span> <span class="o">=</span> <span class="mi">0</span>
-
-<span class="n">head</span> <span class="o">=</span> <span class="n">create_linked_list_with_cycle</span><span class="p">(</span><span class="n">head</span><span class="p">,</span> <span class="n">pos</span><span class="p">)</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">hasCycle</span><span class="p">(</span><span class="n">head</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
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-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 3</span>
-<span class="n">head</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">]</span>
-<span class="n">pos</span> <span class="o">=</span> <span class="o">-</span><span class="mi">1</span>
-
-<span class="n">head</span> <span class="o">=</span> <span class="n">create_linked_list_with_cycle</span><span class="p">(</span><span class="n">head</span><span class="p">,</span> <span class="n">pos</span><span class="p">)</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">hasCycle</span><span class="p">(</span><span class="n">head</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s discuss the time and space complexity of the code for detecting a cycle in a linked list:</p>
-<p><strong>Time Complexity:</strong></p>
-<p>The time complexity of this code is <span class="math notranslate nohighlight">\(O(n)\)</span>, where “<span class="math notranslate nohighlight">\(n\)</span>” is the number of nodes in the linked list.</p>
-<p>The reason for this is that both the <code class="docutils literal notranslate"><span class="pre">slow</span></code> and <code class="docutils literal notranslate"><span class="pre">fast</span></code> pointers traverse the linked list, and they move at different speeds. In the worst case, when there is no cycle, the <code class="docutils literal notranslate"><span class="pre">fast</span></code> pointer will reach the end of the list after going through approximately <span class="math notranslate nohighlight">\(n/2\)</span> nodes. In the case of a cycle, it may take some extra iterations for the <code class="docutils literal notranslate"><span class="pre">fast</span></code> pointer to catch up to the <code class="docutils literal notranslate"><span class="pre">slow</span></code> pointer. However, the total number of iterations is still proportional to the number of nodes in the linked list, making it <span class="math notranslate nohighlight">\(O(n)\)</span>.</p>
-<p><strong>Space Complexity:</strong></p>
-<p>The space complexity of this code is <span class="math notranslate nohighlight">\(O(1)\)</span>, which means it uses constant space.</p>
-<p>The reason for this is that regardless of the size of the linked list, the code only uses two additional pointers (<code class="docutils literal notranslate"><span class="pre">slow</span></code> and <code class="docutils literal notranslate"><span class="pre">fast</span></code>) to traverse the list. These pointers do not depend on the size of the input linked list, so the space complexity remains constant.</p>
-<p><strong>In summary</strong>, this code efficiently detects cycles in a linked list with a time complexity of <span class="math notranslate nohighlight">\(O(n)\)</span> and a space complexity of <span class="math notranslate nohighlight">\(O(1)\)</span>. It is an example of an algorithm that solves a complex problem with minimal memory usage.</p>
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-    <h1>143. Reorder List</h1>
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-                <h2> Contents </h2>
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-                <ul class="visible nav section-nav flex-column">
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="reorder-list">
-<h1>143. Reorder List<a class="headerlink" href="#reorder-list" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/reorder-list/">To see the Reorder List problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>You are given the head of a singly linked-list. The list can be represented as:</p>
-<p><span class="math notranslate nohighlight">\(L_0 → L_1 → … → L_{n - 1} → L_n\)</span></p>
-<p>Reorder the list to be on the following form:</p>
-<p><span class="math notranslate nohighlight">\(L_0 → L_n → L_1 → L_{n - 1} → L_2 → L_{n - 2} → …\)</span></p>
-<p>You may not modify the values in the list’s nodes. Only nodes themselves may be changed.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the list is in the range <span class="math notranslate nohighlight">\([1, 5 * 10^4]\)</span>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">1000</span></code></p></li>
-</ul>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">ListNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="nb">next</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="nb">next</span>
-
-<span class="k">def</span> <span class="nf">reorderList</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">head</span> <span class="ow">or</span> <span class="ow">not</span> <span class="n">head</span><span class="o">.</span><span class="n">next</span><span class="p">:</span>
-        <span class="k">return</span> <span class="n">head</span>
-
-    <span class="c1"># Step 1: Find the middle of the linked list</span>
-    <span class="n">slow</span><span class="p">,</span> <span class="n">fast</span> <span class="o">=</span> <span class="n">head</span><span class="p">,</span> <span class="n">head</span>
-    <span class="k">while</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span> <span class="ow">and</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span><span class="o">.</span><span class="n">next</span><span class="p">:</span>
-        <span class="n">slow</span> <span class="o">=</span> <span class="n">slow</span><span class="o">.</span><span class="n">next</span>
-        <span class="n">fast</span> <span class="o">=</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span><span class="o">.</span><span class="n">next</span>
-
-    <span class="c1"># Split the list into two halves</span>
-    <span class="n">first_half</span> <span class="o">=</span> <span class="n">head</span>
-    <span class="n">second_half</span> <span class="o">=</span> <span class="n">slow</span><span class="o">.</span><span class="n">next</span>
-    <span class="n">slow</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="kc">None</span>
-
-    <span class="c1"># Step 2: Reverse the second half of the linked list</span>
-    <span class="n">prev</span> <span class="o">=</span> <span class="kc">None</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">second_half</span>
-    <span class="k">while</span> <span class="n">current</span><span class="p">:</span>
-        <span class="n">next_node</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-        <span class="n">current</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">prev</span>
-        <span class="n">prev</span> <span class="o">=</span> <span class="n">current</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">next_node</span>
-    <span class="n">second_half</span> <span class="o">=</span> <span class="n">prev</span>
-
-    <span class="c1"># Step 3: Merge the first and reversed second halves alternately</span>
-    <span class="n">p1</span><span class="p">,</span> <span class="n">p2</span> <span class="o">=</span> <span class="n">first_half</span><span class="p">,</span> <span class="n">second_half</span>
-    <span class="k">while</span> <span class="n">p2</span><span class="p">:</span>
-        <span class="n">next_p1</span> <span class="o">=</span> <span class="n">p1</span><span class="o">.</span><span class="n">next</span>
-        <span class="n">next_p2</span> <span class="o">=</span> <span class="n">p2</span><span class="o">.</span><span class="n">next</span>
-        <span class="n">p1</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">p2</span>
-        <span class="n">p2</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">next_p1</span>
-        <span class="n">p1</span> <span class="o">=</span> <span class="n">next_p1</span>
-        <span class="n">p2</span> <span class="o">=</span> <span class="n">next_p2</span>
-
-    <span class="k">return</span> <span class="n">head</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The <code class="docutils literal notranslate"><span class="pre">reorderList</span></code> function is designed to reorder a singly linked list following the specific pattern described in the problem statement. Here’s a detailed explanation of how the function works:</p>
-<ol class="arabic simple">
-<li><p><strong>Base Cases Handling</strong>:</p>
-<ul class="simple">
-<li><p>The function first checks if the input linked list is empty (<code class="docutils literal notranslate"><span class="pre">not</span> <span class="pre">head</span></code>) or contains only one element (<code class="docutils literal notranslate"><span class="pre">not</span> <span class="pre">head.next</span></code>). If either of these conditions is met, the list cannot be reordered, so the function returns the original list as-is.</p></li>
-</ul>
-</li>
-<li><p><strong>Finding the Middle of the Linked List</strong>:</p>
-<ul class="simple">
-<li><p>To reorder the list, we first need to find the middle point so that we can split the list into two halves. This is done using two pointers, <code class="docutils literal notranslate"><span class="pre">slow</span></code> and <code class="docutils literal notranslate"><span class="pre">fast</span></code>, initialized to the head of the list.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">slow</span></code> moves one step at a time while <code class="docutils literal notranslate"><span class="pre">fast</span></code> moves two steps at a time. When <code class="docutils literal notranslate"><span class="pre">fast</span></code> reaches the end of the list or the second-to-last node, <code class="docutils literal notranslate"><span class="pre">slow</span></code> will be at the middle node.</p></li>
-</ul>
-</li>
-<li><p><strong>Splitting the List</strong>:</p>
-<ul class="simple">
-<li><p>After finding the middle node, we split the list into two halves:</p>
-<ul>
-<li><p>The first half, <code class="docutils literal notranslate"><span class="pre">first_half</span></code>, contains nodes from the beginning up to the middle.</p></li>
-<li><p>The second half, <code class="docutils literal notranslate"><span class="pre">second_half</span></code>, contains nodes from the middle to the end.</p></li>
-</ul>
-</li>
-<li><p>To split the list, we set the <code class="docutils literal notranslate"><span class="pre">next</span></code> pointer of the node before the middle node to <code class="docutils literal notranslate"><span class="pre">None</span></code>.</p></li>
-</ul>
-</li>
-<li><p><strong>Reversing the Second Half</strong>:</p>
-<ul class="simple">
-<li><p>We reverse the second half of the linked list using the <code class="docutils literal notranslate"><span class="pre">prev</span></code>, <code class="docutils literal notranslate"><span class="pre">current</span></code>, and <code class="docutils literal notranslate"><span class="pre">next_node</span></code> pointers.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">prev</span></code> is initially <code class="docutils literal notranslate"><span class="pre">None</span></code>, and we iterate through the second half. For each node, we:</p>
-<ul>
-<li><p>Set the <code class="docutils literal notranslate"><span class="pre">next</span></code> of the current node to <code class="docutils literal notranslate"><span class="pre">prev</span></code>, effectively reversing the next pointer direction.</p></li>
-<li><p>Update <code class="docutils literal notranslate"><span class="pre">prev</span></code> to the current node.</p></li>
-<li><p>Move to the next node using the <code class="docutils literal notranslate"><span class="pre">next_node</span></code>.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p><strong>Merging the Two Halves Alternately</strong>:</p>
-<ul class="simple">
-<li><p>We now have two linked lists: <code class="docutils literal notranslate"><span class="pre">first_half</span></code> and the reversed <code class="docutils literal notranslate"><span class="pre">second_half</span></code>.</p></li>
-<li><p>We merge these two lists alternately by adjusting the <code class="docutils literal notranslate"><span class="pre">next</span></code> pointers of the nodes.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">p1</span></code> and <code class="docutils literal notranslate"><span class="pre">p2</span></code> are pointers to the current nodes in <code class="docutils literal notranslate"><span class="pre">first_half</span></code> and <code class="docutils literal notranslate"><span class="pre">second_half</span></code>, respectively.</p></li>
-<li><p>We iterate through both lists while reordering nodes:</p>
-<ul>
-<li><p>Set the <code class="docutils literal notranslate"><span class="pre">next</span></code> of <code class="docutils literal notranslate"><span class="pre">p1</span></code> to <code class="docutils literal notranslate"><span class="pre">p2</span></code> to link a node from the first half to a node from the reversed second half.</p></li>
-<li><p>Update <code class="docutils literal notranslate"><span class="pre">p1</span></code> and <code class="docutils literal notranslate"><span class="pre">p2</span></code> to their respective next nodes.</p></li>
-<li><p>Repeat this process until we have processed all nodes in both halves.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p><strong>Returning the Reordered List</strong>:</p>
-<ul class="simple">
-<li><p>After the merging process is complete, the linked list is reordered as specified.</p></li>
-<li><p>The function returns the <code class="docutils literal notranslate"><span class="pre">head</span></code> of the reordered list.</p></li>
-</ul>
-</li>
-</ol>
-<p>The overall result is a singly linked list that has been reordered according to the given pattern. The time complexity of this algorithm is O(N), where N is the number of nodes in the linked list, as we traverse the list once to find the middle, once to reverse the second half, and once to merge the two halves.</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Helper function to print the linked list</span>
-<span class="k">def</span> <span class="nf">printLinkedList</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
-    <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
-    <span class="k">while</span> <span class="n">head</span><span class="p">:</span>
-        <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">head</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-        <span class="n">head</span> <span class="o">=</span> <span class="n">head</span><span class="o">.</span><span class="n">next</span>
-    <span class="k">return</span> <span class="n">result</span>
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s explain the helper functions:
-<strong>printLinkedList Function</strong>:</p>
-<ul class="simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">printLinkedList</span></code> function takes the <code class="docutils literal notranslate"><span class="pre">head</span></code> of a linked list as input and returns a list of values representing the nodes in the linked list.</p></li>
-<li><p>Inside the function, a <code class="docutils literal notranslate"><span class="pre">result</span></code> list is initialized to store the values of the linked list nodes.</p></li>
-<li><p>The function then iterates through the linked list starting from the <code class="docutils literal notranslate"><span class="pre">head</span></code> node and appends the <code class="docutils literal notranslate"><span class="pre">val</span></code> attribute of each node to the <code class="docutils literal notranslate"><span class="pre">result</span></code> list.</p></li>
-<li><p>As it iterates through the list, it moves to the next node using the <code class="docutils literal notranslate"><span class="pre">next</span></code> attribute of each node.</p></li>
-<li><p>Finally, the function returns the <code class="docutils literal notranslate"><span class="pre">result</span></code> list containing the values of the linked list nodes in the order they appear in the linked list.</p></li>
-</ul>
-<p>This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1</span>
-<span class="n">head1</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">4</span><span class="p">))))</span>
-<span class="n">reorderList</span><span class="p">(</span><span class="n">head1</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">printLinkedList</span><span class="p">(</span><span class="n">head1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[1, 4, 2, 3]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">head2</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">4</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">5</span><span class="p">)))))</span>
-<span class="n">reorderList</span><span class="p">(</span><span class="n">head2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">printLinkedList</span><span class="p">(</span><span class="n">head2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[1, 5, 2, 4, 3]
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">reorderList</span></code> function:</p>
-<p><strong>Time Complexity</strong>:</p>
-<ol class="arabic simple">
-<li><p><strong>Finding the Middle of the Linked List</strong>: In this step, we use two pointers, one moving one step at a time (<code class="docutils literal notranslate"><span class="pre">slow</span></code>) and the other moving two steps at a time (<code class="docutils literal notranslate"><span class="pre">fast</span></code>). This process takes <span class="math notranslate nohighlight">\(O(N/2)\)</span> time, where <span class="math notranslate nohighlight">\(N\)</span> is the number of nodes in the linked list.</p></li>
-<li><p><strong>Splitting the List</strong>: After finding the middle, we split the list into two halves by setting the <code class="docutils literal notranslate"><span class="pre">next</span></code> pointer of the node before the middle node to <code class="docutils literal notranslate"><span class="pre">None</span></code>. This step takes <span class="math notranslate nohighlight">\(O(1)\)</span> time.</p></li>
-<li><p><strong>Reversing the Second Half</strong>: We reverse the second half of the linked list using a while loop that iterates through the second half of the list once. Therefore, this step also takes <span class="math notranslate nohighlight">\(O(N/2)\)</span> time.</p></li>
-<li><p><strong>Merging the Two Halves Alternately</strong>: In this step, we merge the two halves alternately by adjusting the <code class="docutils literal notranslate"><span class="pre">next</span></code> pointers of the nodes. We iterate through both halves once, so this step takes <span class="math notranslate nohighlight">\(O(N/2)\)</span> time.</p></li>
-</ol>
-<p>Overall, the time complexity of the <code class="docutils literal notranslate"><span class="pre">reorderList</span></code> function is dominated by the steps involving reversing and merging the two halves, both of which take <span class="math notranslate nohighlight">\(O(N/2)\)</span> time. Therefore, the total time complexity is <span class="math notranslate nohighlight">\(O(N)\)</span>.</p>
-<p><strong>Space Complexity</strong>:
-The space complexity of the <code class="docutils literal notranslate"><span class="pre">reorderList</span></code> function is primarily determined by the variables and data structures used within the function. Let’s break down the space complexity components:</p>
-<ol class="arabic simple">
-<li><p><strong>Constant Space Variables</strong>: The variables <code class="docutils literal notranslate"><span class="pre">slow</span></code>, <code class="docutils literal notranslate"><span class="pre">fast</span></code>, <code class="docutils literal notranslate"><span class="pre">prev</span></code>, <code class="docutils literal notranslate"><span class="pre">current</span></code>, <code class="docutils literal notranslate"><span class="pre">next_node</span></code>, <code class="docutils literal notranslate"><span class="pre">p1</span></code>, and <code class="docutils literal notranslate"><span class="pre">p2</span></code> are used to traverse and manipulate the linked list. These variables occupy constant space, regardless of the input size. Therefore, they contribute <span class="math notranslate nohighlight">\(O(1)\)</span> to the space complexity.</p></li>
-<li><p><strong>Split Linked Lists</strong>: In the splitting step, we create two new linked list segments (<code class="docutils literal notranslate"><span class="pre">first_half</span></code> and <code class="docutils literal notranslate"><span class="pre">second_half</span></code>) that store references to nodes from the original linked list. These segments occupy space proportional to half of the input list, <span class="math notranslate nohighlight">\(O(N/2)\)</span>.</p></li>
-<li><p><strong>Reversed Second Half</strong>: During the reversal step, we reverse the second half of the linked list in-place without creating any additional data structures. Therefore, it doesn’t contribute to additional space complexity.</p></li>
-<li><p><strong>Overall</strong>: Combining the above components, the total space complexity is <span class="math notranslate nohighlight">\(O(N/2)\)</span>, which simplifies to <span class="math notranslate nohighlight">\(O(N)\)</span> in terms of space complexity.</p></li>
-</ol>
-<p><strong>In summary</strong>, the <code class="docutils literal notranslate"><span class="pre">reorderList</span></code> function has a time complexity of <span class="math notranslate nohighlight">\(O(N)\)</span> and a space complexity of <span class="math notranslate nohighlight">\(O(N)\)</span> due to the creation of two new linked list segments while splitting the list. The constant space variables used for traversal do not significantly impact the overall space complexity.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Reorder by K Elements:</strong> Generalize the reorderList function to reorder the list in a different pattern, where you reorder the list by <span class="math notranslate nohighlight">\(K\)</span> elements at a time. For example, for <span class="math notranslate nohighlight">\(K=3\)</span>, you would reorder it as (<span class="math notranslate nohighlight">\(L_0 → L_1 → L_2 → L_n → L_{n-1} → L_{n-2} → ...\)</span>).
-You may need to handle cases where the number of nodes is not a multiple of <span class="math notranslate nohighlight">\(K\)</span>.</p></li>
-<li><p><strong>Reorder by Odd and Even Nodes:</strong> Modify the reorderList function to reorder the list in a pattern where odd-indexed nodes (1-based index) come before even-indexed nodes. For example, (<span class="math notranslate nohighlight">\(L_0 → L_n → L_1 → L_{n-1} → L_2 → L_{n-2} → ...\)</span>).</p></li>
-</ol>
-</section>
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-<h1>19. Remove Nth Node From End of List<a class="headerlink" href="#remove-nth-node-from-end-of-list" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/remove-nth-node-from-end-of-list/">To see the Remove Nth Node From End of List problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given the head of a linked list, remove the <span class="math notranslate nohighlight">\(n^{th}\)</span> node from the end of the list and return its head.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the list is <code class="docutils literal notranslate"><span class="pre">sz</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">sz</span> <span class="pre">&lt;=</span> <span class="pre">30</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">0</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">100</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">n</span> <span class="pre">&lt;=</span> <span class="pre">sz</span></code></p></li>
-</ul>
-<p><strong>Follow up:</strong> Could you do this in one pass?</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">ListNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="nb">next</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="nb">next</span>
-
-<span class="k">def</span> <span class="nf">removeNthFromEnd</span><span class="p">(</span><span class="n">head</span><span class="p">,</span> <span class="n">n</span><span class="p">):</span>
-    <span class="c1"># Create a dummy node to handle the case of removing the head node.</span>
-    <span class="n">dummy</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span>
-    <span class="n">dummy</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">head</span>
-    <span class="n">first</span> <span class="o">=</span> <span class="n">dummy</span>
-    <span class="n">second</span> <span class="o">=</span> <span class="n">dummy</span>
-    
-    <span class="c1"># Advance the first pointer by n+1 nodes.</span>
-    <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">):</span>
-        <span class="n">first</span> <span class="o">=</span> <span class="n">first</span><span class="o">.</span><span class="n">next</span>
-    
-    <span class="c1"># Move both pointers simultaneously until first reaches the end.</span>
-    <span class="k">while</span> <span class="n">first</span> <span class="ow">is</span> <span class="ow">not</span> <span class="kc">None</span><span class="p">:</span>
-        <span class="n">first</span> <span class="o">=</span> <span class="n">first</span><span class="o">.</span><span class="n">next</span>
-        <span class="n">second</span> <span class="o">=</span> <span class="n">second</span><span class="o">.</span><span class="n">next</span>
-    
-    <span class="c1"># Remove the nth node by updating the next pointer of the previous node.</span>
-    <span class="n">second</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">second</span><span class="o">.</span><span class="n">next</span><span class="o">.</span><span class="n">next</span>
-    
-    <span class="k">return</span> <span class="n">dummy</span><span class="o">.</span><span class="n">next</span>  <span class="c1"># Return the modified head of the linked list</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We start by defining a <code class="docutils literal notranslate"><span class="pre">ListNode</span></code> class to represent individual nodes in a linked list. Each node has two attributes: <code class="docutils literal notranslate"><span class="pre">val</span></code> (the value of the node) and <code class="docutils literal notranslate"><span class="pre">next</span></code> (a reference to the next node in the list).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">removeNthFromEnd</span></code> function takes the head of a linked list (<code class="docutils literal notranslate"><span class="pre">head</span></code>) and the value <code class="docutils literal notranslate"><span class="pre">n</span></code>, which represents the position from the end of the list of the node to be removed. It returns the modified head of the linked list after the removal.</p></li>
-<li><p>We create a dummy node (<code class="docutils literal notranslate"><span class="pre">dummy</span></code>) at the beginning of the list. This dummy node simplifies the code by handling cases where the head of the list needs to be removed.</p></li>
-<li><p>We initialize two pointers, <code class="docutils literal notranslate"><span class="pre">first</span></code> and <code class="docutils literal notranslate"><span class="pre">second</span></code>, both initially pointing to the dummy node.</p></li>
-<li><p>To advance the <code class="docutils literal notranslate"><span class="pre">first</span></code> pointer by <code class="docutils literal notranslate"><span class="pre">n+1</span></code> nodes, we use a <code class="docutils literal notranslate"><span class="pre">for</span></code> loop. This positions the <code class="docutils literal notranslate"><span class="pre">first</span></code> pointer <code class="docutils literal notranslate"><span class="pre">n</span></code> nodes ahead of the <code class="docutils literal notranslate"><span class="pre">second</span></code> pointer.</p></li>
-<li><p>We then move both <code class="docutils literal notranslate"><span class="pre">first</span></code> and <code class="docutils literal notranslate"><span class="pre">second</span></code> pointers simultaneously until <code class="docutils literal notranslate"><span class="pre">first</span></code> reaches the end of the list. This ensures that the <code class="docutils literal notranslate"><span class="pre">second</span></code> pointer ends up pointing to the node that needs to be removed.</p></li>
-<li><p>To remove the nth node from the end, we update the <code class="docutils literal notranslate"><span class="pre">next</span></code> pointer of the node pointed to by the <code class="docutils literal notranslate"><span class="pre">second</span></code> pointer to skip over the node to be removed.</p></li>
-<li><p>Finally, we return <code class="docutils literal notranslate"><span class="pre">dummy.next</span></code>, which is the modified head of the linked list without the removed node.</p></li>
-<li><p>Additionally, there are two helper functions:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">createLinkedList(values)</span></code> creates a linked list from a list of values.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">convertToList(head)</span></code> converts a linked list back into a list of values.</p></li>
-</ul>
-</li>
-</ol>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Helper function to create a linked list from a list of values.</span>
-<span class="k">def</span> <span class="nf">createLinkedList</span><span class="p">(</span><span class="n">values</span><span class="p">):</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">values</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">None</span>
-    <span class="n">head</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="n">values</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">head</span>
-    <span class="k">for</span> <span class="n">val</span> <span class="ow">in</span> <span class="n">values</span><span class="p">[</span><span class="mi">1</span><span class="p">:]:</span>
-        <span class="n">current</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="n">val</span><span class="p">)</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-    <span class="k">return</span> <span class="n">head</span>
-
-<span class="c1"># Helper function to convert a linked list to a list of values.</span>
-<span class="k">def</span> <span class="nf">convertToList</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
-    <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">head</span>
-    <span class="k">while</span> <span class="n">current</span><span class="p">:</span>
-        <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">current</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-    <span class="k">return</span> <span class="n">result</span>
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s explain the two helper functions:</p>
-<ol class="arabic">
-<li><p><code class="docutils literal notranslate"><span class="pre">createLinkedList(values)</span></code>: This function creates a linked list from a list of values.</p>
-<ul class="simple">
-<li><p>Input: <code class="docutils literal notranslate"><span class="pre">values</span></code> is a list containing values that you want to insert into the linked list.</p></li>
-<li><p>Output: The function returns the head of the created linked list.</p></li>
-</ul>
-<p>Explanation:</p>
-<ul class="simple">
-<li><p>The function starts by checking if the <code class="docutils literal notranslate"><span class="pre">values</span></code> list is empty. If it is, it returns <code class="docutils literal notranslate"><span class="pre">None</span></code> to indicate an empty linked list.</p></li>
-<li><p>If the <code class="docutils literal notranslate"><span class="pre">values</span></code> list is not empty, it creates the first node of the linked list with the value from the first element of <code class="docutils literal notranslate"><span class="pre">values</span></code>.</p></li>
-<li><p>It then iterates through the remaining elements of <code class="docutils literal notranslate"><span class="pre">values</span></code> and creates new nodes for each value, connecting them together using the <code class="docutils literal notranslate"><span class="pre">next</span></code> attribute to form a linked list.</p></li>
-<li><p>Finally, it returns the head of the linked list, which is the first node created.</p></li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">convertToList(head)</span></code>: This function converts a linked list into a list of values.</p>
-<ul class="simple">
-<li><p>Input: <code class="docutils literal notranslate"><span class="pre">head</span></code> is the head node of the linked list that you want to convert.</p></li>
-<li><p>Output: The function returns a list containing the values of the nodes in the linked list in order.</p></li>
-</ul>
-<p>Explanation:</p>
-<ul class="simple">
-<li><p>The function starts by initializing an empty list, <code class="docutils literal notranslate"><span class="pre">result</span></code>, to store the values of the linked list nodes.</p></li>
-<li><p>It then iterates through the linked list starting from the <code class="docutils literal notranslate"><span class="pre">head</span></code> node and appends the <code class="docutils literal notranslate"><span class="pre">val</span></code> attribute of each node to the <code class="docutils literal notranslate"><span class="pre">result</span></code> list.</p></li>
-<li><p>The iteration continues until it reaches the end of the linked list (i.e., the <code class="docutils literal notranslate"><span class="pre">next</span></code> attribute of the current node becomes <code class="docutils literal notranslate"><span class="pre">None</span></code>).</p></li>
-<li><p>After the iteration is complete, the function returns the <code class="docutils literal notranslate"><span class="pre">result</span></code> list, which now contains the values of all the nodes in the linked list in the same order as they appear in the linked list.</p></li>
-</ul>
-</li>
-</ol>
-<p>These helper functions are useful for creating linked lists from lists of values and converting linked lists back into lists of values, making it easier to work with linked list examples and test cases in a more familiar list format.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1</span>
-<span class="n">head1</span> <span class="o">=</span> <span class="n">createLinkedList</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">5</span><span class="p">])</span>
-<span class="n">n1</span> <span class="o">=</span> <span class="mi">2</span>
-<span class="n">new_head1</span> <span class="o">=</span> <span class="n">removeNthFromEnd</span><span class="p">(</span><span class="n">head1</span><span class="p">,</span> <span class="n">n1</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">convertToList</span><span class="p">(</span><span class="n">new_head1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[1, 2, 3, 5]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">head2</span> <span class="o">=</span> <span class="n">createLinkedList</span><span class="p">([</span><span class="mi">1</span><span class="p">])</span>
-<span class="n">n2</span> <span class="o">=</span> <span class="mi">1</span>
-<span class="n">new_head2</span> <span class="o">=</span> <span class="n">removeNthFromEnd</span><span class="p">(</span><span class="n">head2</span><span class="p">,</span> <span class="n">n2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">convertToList</span><span class="p">(</span><span class="n">new_head2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3</span>
-<span class="n">head3</span> <span class="o">=</span> <span class="n">createLinkedList</span><span class="p">([</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">])</span>
-<span class="n">n3</span> <span class="o">=</span> <span class="mi">1</span>
-<span class="n">new_head3</span> <span class="o">=</span> <span class="n">removeNthFromEnd</span><span class="p">(</span><span class="n">head3</span><span class="p">,</span> <span class="n">n3</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">convertToList</span><span class="p">(</span><span class="n">new_head3</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[1]
-</pre></div>
-</div>
-</div>
-</div>
-<p>Certainly! Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">removeNthFromEnd</span></code> function in isolation:</p>
-<p><strong>Time Complexity:</strong></p>
-<ul class="simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">removeNthFromEnd</span></code> function performs two passes through the linked list. In the worst case, it needs to traverse the entire linked list of length <span class="math notranslate nohighlight">\(N\)</span>.</p></li>
-<li><p>The first pass positions the <code class="docutils literal notranslate"><span class="pre">first</span></code> pointer <span class="math notranslate nohighlight">\(n\)</span> nodes ahead of the <code class="docutils literal notranslate"><span class="pre">second</span></code> pointer. This pass takes <span class="math notranslate nohighlight">\(O(N)\)</span> time because it involves iterating through all N nodes.</p></li>
-<li><p>The second pass moves both <code class="docutils literal notranslate"><span class="pre">first</span></code> and <code class="docutils literal notranslate"><span class="pre">second</span></code> pointers simultaneously until <code class="docutils literal notranslate"><span class="pre">first</span></code> reaches the end of the list. This also takes <span class="math notranslate nohighlight">\(O(N)\)</span> time in the worst case.</p></li>
-<li><p>Therefore, the overall time complexity of the <code class="docutils literal notranslate"><span class="pre">removeNthFromEnd</span></code> function is <span class="math notranslate nohighlight">\(O(N)\)</span>.</p></li>
-</ul>
-<p><strong>Space Complexity:</strong></p>
-<ul class="simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">removeNthFromEnd</span></code> function uses a constant amount of extra space for its variables (e.g., <code class="docutils literal notranslate"><span class="pre">first</span></code>, <code class="docutils literal notranslate"><span class="pre">second</span></code>, and <code class="docutils literal notranslate"><span class="pre">dummy</span></code>). This space usage does not depend on the size of the linked list.</p></li>
-<li><p>The space complexity is <span class="math notranslate nohighlight">\(O(1)\)</span>, indicating that the space used by the function is constant and independent of the size of the input linked list.</p></li>
-</ul>
-<p><strong>In summary</strong>, the <code class="docutils literal notranslate"><span class="pre">removeNthFromEnd</span></code> function has a time complexity of <span class="math notranslate nohighlight">\(O(N)\)</span> and a space complexity of <span class="math notranslate nohighlight">\(O(1)\)</span>. It efficiently removes the nth node from the end of the linked list with a single pass through the list and constant extra space usage.</p>
-</section>
-</section>
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-    <h1>206. Reverse Linked List</h1>
-    <!-- Table of contents -->
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-            <div>
-                <h2> Contents </h2>
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-            <nav aria-label="Page">
-                <ul class="visible nav section-nav flex-column">
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="reverse-linked-list">
-<h1>206. Reverse Linked List<a class="headerlink" href="#reverse-linked-list" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/reverse-linked-list/">To see the Reverse Linked List problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given the <code class="docutils literal notranslate"><span class="pre">head</span></code> of a singly linked list, reverse the list, and return the reversed list.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the list is the range <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">5000]</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-5000</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">5000</span></code></p></li>
-</ul>
-<p><strong>Follow up:</strong> A linked list can be reversed either iteratively or recursively. Could you implement both?</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Definition for singly-linked list.</span>
-<span class="k">class</span> <span class="nc">ListNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="nb">next</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="nb">next</span>
-
-<span class="c1"># Function to reverse a linked list iteratively</span>
-<span class="k">def</span> <span class="nf">reverseListIterative</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
-    <span class="n">prev</span> <span class="o">=</span> <span class="kc">None</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">head</span>
-    
-    <span class="k">while</span> <span class="n">current</span><span class="p">:</span>
-        <span class="n">next_node</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-        <span class="n">current</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">prev</span>
-        <span class="n">prev</span> <span class="o">=</span> <span class="n">current</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">next_node</span>
-    
-    <span class="k">return</span> <span class="n">prev</span>
-
-<span class="c1"># Function to reverse a linked list recursively</span>
-<span class="k">def</span> <span class="nf">reverseListRecursive</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">head</span> <span class="ow">or</span> <span class="ow">not</span> <span class="n">head</span><span class="o">.</span><span class="n">next</span><span class="p">:</span>
-        <span class="k">return</span> <span class="n">head</span>
-
-    <span class="n">new_head</span> <span class="o">=</span> <span class="n">reverseListRecursive</span><span class="p">(</span><span class="n">head</span><span class="o">.</span><span class="n">next</span><span class="p">)</span>
-    <span class="n">head</span><span class="o">.</span><span class="n">next</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">head</span>
-    <span class="n">head</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="kc">None</span>
-
-    <span class="k">return</span> <span class="n">new_head</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">reverseListIterative(head)</span></code>: This function reverses a linked list iteratively using a loop. Here’s a step-by-step explanation:</p>
-<ul class="simple">
-<li><p>It takes the <code class="docutils literal notranslate"><span class="pre">head</span></code> of the input linked list as a parameter.</p></li>
-<li><p>It initializes two pointers, <code class="docutils literal notranslate"><span class="pre">prev</span></code> and <code class="docutils literal notranslate"><span class="pre">current</span></code>, initially set to <code class="docutils literal notranslate"><span class="pre">None</span></code> and the <code class="docutils literal notranslate"><span class="pre">head</span></code> of the list, respectively.</p></li>
-<li><p>It enters a <code class="docutils literal notranslate"><span class="pre">while</span></code> loop that continues until <code class="docutils literal notranslate"><span class="pre">current</span></code> reaches the end of the list (i.e., becomes <code class="docutils literal notranslate"><span class="pre">None</span></code>).</p></li>
-<li><p>Inside the loop:</p>
-<ul>
-<li><p>It stores the <code class="docutils literal notranslate"><span class="pre">next_node</span></code> which is the next node after <code class="docutils literal notranslate"><span class="pre">current</span></code>.</p></li>
-<li><p>It updates the <code class="docutils literal notranslate"><span class="pre">next</span></code> pointer of <code class="docutils literal notranslate"><span class="pre">current</span></code> to point to the <code class="docutils literal notranslate"><span class="pre">prev</span></code> node. This effectively reverses the link direction.</p></li>
-<li><p>It moves <code class="docutils literal notranslate"><span class="pre">prev</span></code> to <code class="docutils literal notranslate"><span class="pre">current</span></code> and <code class="docutils literal notranslate"><span class="pre">current</span></code> to <code class="docutils literal notranslate"><span class="pre">next_node</span></code>, advancing the pointers one step further in the list.</p></li>
-</ul>
-</li>
-<li><p>Once the loop completes, <code class="docutils literal notranslate"><span class="pre">prev</span></code> will be pointing to the new head of the reversed list (which was the last node of the original list), so it returns <code class="docutils literal notranslate"><span class="pre">prev</span></code>.</p></li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">reverseListRecursive(head)</span></code>: This function reverses a linked list recursively. Here’s how it works:</p>
-<ul class="simple">
-<li><p>It takes the <code class="docutils literal notranslate"><span class="pre">head</span></code> of the input linked list as a parameter.</p></li>
-<li><p>The base case is checked: if <code class="docutils literal notranslate"><span class="pre">head</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code> or <code class="docutils literal notranslate"><span class="pre">head.next</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code>, meaning the list is empty or has only one node, it returns <code class="docutils literal notranslate"><span class="pre">head</span></code> as there’s no need to reverse a list with zero or one element.</p></li>
-<li><p>In the recursive case:</p>
-<ul>
-<li><p>It calls itself with <code class="docutils literal notranslate"><span class="pre">head.next</span></code>, effectively moving down the list until it reaches the end.</p></li>
-<li><p>Once it reaches the end, it starts reversing the links:</p>
-<ul>
-<li><p><code class="docutils literal notranslate"><span class="pre">head.next.next</span></code> is set to <code class="docutils literal notranslate"><span class="pre">head</span></code>, reversing the link between <code class="docutils literal notranslate"><span class="pre">head</span></code> and the next node.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">head.next</span></code> is set to <code class="docutils literal notranslate"><span class="pre">None</span></code> to avoid cycles.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p>Finally, it returns the new head of the reversed list, which will be the last node of the original list.</p></li>
-</ul>
-</li>
-</ol>
-<p><strong>In summary,</strong> <code class="docutils literal notranslate"><span class="pre">reverseListIterative</span></code> reverses the linked list by iterating through it and changing the next pointers, while <code class="docutils literal notranslate"><span class="pre">reverseListRecursive</span></code> reverses the linked list by recursively reaching the end and then reversing the links on the way back up the recursion stack. Both functions achieve the same result: reversing the linked list.</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Helper function to create a linked list from a list of values</span>
-<span class="k">def</span> <span class="nf">createLinkedList</span><span class="p">(</span><span class="n">values</span><span class="p">):</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">values</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">None</span>
-    <span class="n">head</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="n">values</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">head</span>
-    <span class="k">for</span> <span class="n">val</span> <span class="ow">in</span> <span class="n">values</span><span class="p">[</span><span class="mi">1</span><span class="p">:]:</span>
-        <span class="n">current</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="n">val</span><span class="p">)</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-    <span class="k">return</span> <span class="n">head</span>
-
-<span class="c1"># Helper function to convert a linked list to a list for testing</span>
-<span class="k">def</span> <span class="nf">linkedListToList</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
-    <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">head</span>
-    <span class="k">while</span> <span class="n">current</span><span class="p">:</span>
-        <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">current</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-    <span class="k">return</span> <span class="n">result</span>
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s explain the two helper functions used in the provided code:</p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">createLinkedList(values)</span></code>:</p>
-<ul class="simple">
-<li><p>Purpose: This function is used to create a linked list from a list of values. It takes a list of values as input and returns the head of the linked list.</p></li>
-<li><p>How it works:</p>
-<ul>
-<li><p>It first checks if the input list <code class="docutils literal notranslate"><span class="pre">values</span></code> is empty. If it’s empty, it returns <code class="docutils literal notranslate"><span class="pre">None</span></code> to indicate an empty linked list.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">values</span></code> is not empty, it initializes a <code class="docutils literal notranslate"><span class="pre">head</span></code> node with the value of the first element in <code class="docutils literal notranslate"><span class="pre">values</span></code>.</p></li>
-<li><p>It then iterates through the remaining values in <code class="docutils literal notranslate"><span class="pre">values</span></code>, creating new nodes for each value and linking them together to form a linked list.</p></li>
-<li><p>Finally, it returns the <code class="docutils literal notranslate"><span class="pre">head</span></code> node of the newly created linked list.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">linkedListToList(head)</span></code>:</p>
-<ul class="simple">
-<li><p>Purpose: This function is used to convert a linked list back into a Python list for testing and output purposes.</p></li>
-<li><p>How it works:</p>
-<ul>
-<li><p>It takes the <code class="docutils literal notranslate"><span class="pre">head</span></code> of the linked list as input.</p></li>
-<li><p>It initializes an empty list called <code class="docutils literal notranslate"><span class="pre">result</span></code>.</p></li>
-<li><p>It then iterates through the linked list, starting from <code class="docutils literal notranslate"><span class="pre">head</span></code>, and appends each node’s <code class="docutils literal notranslate"><span class="pre">val</span></code> (value) to the <code class="docutils literal notranslate"><span class="pre">result</span></code> list.</p></li>
-<li><p>This process continues until the end of the linked list is reached.</p></li>
-<li><p>Finally, it returns the <code class="docutils literal notranslate"><span class="pre">result</span></code> list, which contains the values from the linked list in the same order.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-</ol>
-<p>These helper functions are used to simplify the process of creating linked lists from lists of values and converting linked lists back into lists, making it easier to work with linked lists in the provided examples and test cases.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: Iterative</span>
-<span class="n">head1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">4</span><span class="p">,</span><span class="mi">5</span><span class="p">]</span>
-<span class="n">head1</span> <span class="o">=</span> <span class="n">createLinkedList</span><span class="p">(</span><span class="n">head1</span><span class="p">)</span>
-<span class="n">reversed_head1</span> <span class="o">=</span> <span class="n">reverseListIterative</span><span class="p">(</span><span class="n">head1</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">linkedListToList</span><span class="p">(</span><span class="n">reversed_head1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[5, 4, 3, 2, 1]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2: Recursive</span>
-<span class="n">head2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">]</span>
-<span class="n">head2</span> <span class="o">=</span> <span class="n">createLinkedList</span><span class="p">(</span><span class="n">head2</span><span class="p">)</span>
-<span class="n">reversed_head2</span> <span class="o">=</span> <span class="n">reverseListRecursive</span><span class="p">(</span><span class="n">head2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">linkedListToList</span><span class="p">(</span><span class="n">reversed_head2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[2, 1]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3: Empty list</span>
-<span class="n">head3</span> <span class="o">=</span> <span class="p">[]</span>
-<span class="n">head3</span> <span class="o">=</span> <span class="n">createLinkedList</span><span class="p">(</span><span class="n">head3</span><span class="p">)</span>
-<span class="n">reversed_head3</span> <span class="o">=</span> <span class="n">reverseListIterative</span><span class="p">(</span><span class="n">head3</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">linkedListToList</span><span class="p">(</span><span class="n">reversed_head3</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[]
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the two functions:</p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">reverseListIterative(head)</span></code>:</p>
-<ul class="simple">
-<li><p>Time Complexity: <span class="math notranslate nohighlight">\(O(n)\)</span></p>
-<ul>
-<li><p>The iterative function visits each node in the linked list exactly once in a single pass, where “<span class="math notranslate nohighlight">\(n\)</span>” is the number of nodes in the list. Therefore, the time complexity is linear in the number of nodes.</p></li>
-</ul>
-</li>
-<li><p>Space Complexity: <span class="math notranslate nohighlight">\(O(1)\)</span></p>
-<ul>
-<li><p>This function uses a constant amount of extra space for the <code class="docutils literal notranslate"><span class="pre">prev</span></code>, <code class="docutils literal notranslate"><span class="pre">current</span></code>, and <code class="docutils literal notranslate"><span class="pre">next_node</span></code> pointers. Regardless of the size of the input list, the amount of additional memory used remains the same, so the space complexity is constant.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">reverseListRecursive(head)</span></code>:</p>
-<ul class="simple">
-<li><p>Time Complexity: <span class="math notranslate nohighlight">\(O(n)\)</span></p>
-<ul>
-<li><p>The recursive function also visits each node in the linked list exactly once. It recursively traverses the list from the head to the tail, reversing the links on the way back. Therefore, like the iterative approach, the time complexity is <span class="math notranslate nohighlight">\(O(n)\)</span>, where “<span class="math notranslate nohighlight">\(n\)</span>” is the number of nodes.</p></li>
-</ul>
-</li>
-<li><p>Space Complexity: <span class="math notranslate nohighlight">\(O(n)\)</span></p>
-<ul>
-<li><p>The recursive function uses space on the call stack for each recursive call. In the worst case, when the list has “<span class="math notranslate nohighlight">\(n\)</span>” nodes, it will create “<span class="math notranslate nohighlight">\(n\)</span>” recursive function calls on the stack, resulting in a space complexity of <span class="math notranslate nohighlight">\(O(n)\)</span>. This is because each function call stores information about its state, including the <code class="docutils literal notranslate"><span class="pre">head</span></code> pointer, until it reaches the base case and starts returning.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-</ol>
-<p><strong>In summary:</strong></p>
-<ul class="simple">
-<li><p>The time complexity of both functions is <span class="math notranslate nohighlight">\(O(n)\)</span> because they both visit each node once.</p></li>
-<li><p>The space complexity of <code class="docutils literal notranslate"><span class="pre">reverseListIterative</span></code> is <span class="math notranslate nohighlight">\(O(1)\)</span> because it uses a constant amount of extra space.</p></li>
-<li><p>The space complexity of <code class="docutils literal notranslate"><span class="pre">reverseListRecursive</span></code> is <span class="math notranslate nohighlight">\(O(n)\)</span> due to the recursive function calls and the associated call stack space.</p></li>
-</ul>
-<p>Both functions are efficient in terms of time complexity, but <code class="docutils literal notranslate"><span class="pre">reverseListIterative</span></code> is more memory-efficient because it uses a constant amount of additional space, while <code class="docutils literal notranslate"><span class="pre">reverseListRecursive</span></code> uses space on the call stack proportional to the number of nodes in the list.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>K-Group Reverse:</strong> Modify the reverseListIterative function to reverse the linked list in groups of <span class="math notranslate nohighlight">\(K\)</span> elements. For example, if the input linked list is [1, 2, 3, 4, 5, 6], and <span class="math notranslate nohighlight">\(K\)</span> is 3, the output should be [3, 2, 1, 6, 5, 4].</p></li>
-<li><p><strong>Rotate Linked List:</strong> Write a function to rotate a linked list to the right by <span class="math notranslate nohighlight">\(K\)</span> places. For example, if the input is [1, 2, 3, 4, 5] and <span class="math notranslate nohighlight">\(K\)</span> is 2, the output should be [4, 5, 1, 2, 3].</p></li>
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-    <h1>21. Merge Two Sorted Lists</h1>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
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-<h1>21. Merge Two Sorted Lists<a class="headerlink" href="#merge-two-sorted-lists" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/merge-two-sorted-lists/">To see the Merge Two Sorted Lists problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>You are given the heads of two sorted linked lists <code class="docutils literal notranslate"><span class="pre">list1</span></code> and <code class="docutils literal notranslate"><span class="pre">list2</span></code>.</p>
-<p>Merge the two lists into one <strong>sorted</strong> list. The list should be made by splicing together the nodes of the first two lists.</p>
-<p><em>Return the head of the merged linked list.</em></p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in both lists is in the range <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">50]</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-100</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">100</span></code></p></li>
-<li><p>Both <code class="docutils literal notranslate"><span class="pre">list1</span></code> and <code class="docutils literal notranslate"><span class="pre">list2</span></code> are sorted in non-decreasing order.</p></li>
-</ul>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">ListNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="nb">next</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="nb">next</span>
-
-<span class="k">def</span> <span class="nf">mergeTwoLists</span><span class="p">(</span><span class="n">list1</span><span class="p">,</span> <span class="n">list2</span><span class="p">):</span>
-    <span class="c1"># Create a dummy node to simplify the code.</span>
-    <span class="n">dummy</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">()</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">dummy</span>
-
-    <span class="k">while</span> <span class="n">list1</span> <span class="ow">and</span> <span class="n">list2</span><span class="p">:</span>
-        <span class="k">if</span> <span class="n">list1</span><span class="o">.</span><span class="n">val</span> <span class="o">&lt;</span> <span class="n">list2</span><span class="o">.</span><span class="n">val</span><span class="p">:</span>
-            <span class="n">current</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">list1</span>
-            <span class="n">list1</span> <span class="o">=</span> <span class="n">list1</span><span class="o">.</span><span class="n">next</span>
-        <span class="k">else</span><span class="p">:</span>
-            <span class="n">current</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">list2</span>
-            <span class="n">list2</span> <span class="o">=</span> <span class="n">list2</span><span class="o">.</span><span class="n">next</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-
-    <span class="c1"># Append the remaining elements from either list.</span>
-    <span class="k">if</span> <span class="n">list1</span><span class="p">:</span>
-        <span class="n">current</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">list1</span>
-    <span class="k">else</span><span class="p">:</span>
-        <span class="n">current</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">list2</span>
-
-    <span class="k">return</span> <span class="n">dummy</span><span class="o">.</span><span class="n">next</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We define a class <code class="docutils literal notranslate"><span class="pre">ListNode</span></code> to represent the nodes of a singly-linked list. Each node has a <code class="docutils literal notranslate"><span class="pre">val</span></code> attribute to store the value of the node and a <code class="docutils literal notranslate"><span class="pre">next</span></code> attribute to point to the next node in the list.</p></li>
-<li><p>We define the <code class="docutils literal notranslate"><span class="pre">mergeTwoLists</span></code> function that takes two linked list heads, <code class="docutils literal notranslate"><span class="pre">list1</span></code> and <code class="docutils literal notranslate"><span class="pre">list2</span></code>, as input.</p></li>
-<li><p>We create a <code class="docutils literal notranslate"><span class="pre">dummy</span></code> node at the beginning of the merged list. The <code class="docutils literal notranslate"><span class="pre">dummy</span></code> node simplifies the code by serving as a placeholder, and its <code class="docutils literal notranslate"><span class="pre">next</span></code> attribute will point to the actual merged list.</p></li>
-<li><p>We initialize a <code class="docutils literal notranslate"><span class="pre">current</span></code> pointer to point to the <code class="docutils literal notranslate"><span class="pre">dummy</span></code> node initially. This <code class="docutils literal notranslate"><span class="pre">current</span></code> pointer helps us traverse and build the merged list.</p></li>
-<li><p>We enter a <code class="docutils literal notranslate"><span class="pre">while</span></code> loop that continues until either <code class="docutils literal notranslate"><span class="pre">list1</span></code> or <code class="docutils literal notranslate"><span class="pre">list2</span></code> becomes empty. Inside the loop, we compare the values of the nodes at the current positions of <code class="docutils literal notranslate"><span class="pre">list1</span></code> and <code class="docutils literal notranslate"><span class="pre">list2</span></code>.</p></li>
-<li><p>If the <code class="docutils literal notranslate"><span class="pre">val</span></code> of the node in <code class="docutils literal notranslate"><span class="pre">list1</span></code> is smaller than the <code class="docutils literal notranslate"><span class="pre">val</span></code> of the node in <code class="docutils literal notranslate"><span class="pre">list2</span></code>, we attach the node from <code class="docutils literal notranslate"><span class="pre">list1</span></code> to the merged list by updating the <code class="docutils literal notranslate"><span class="pre">next</span></code> attribute of the <code class="docutils literal notranslate"><span class="pre">current</span></code> node to point to the node in <code class="docutils literal notranslate"><span class="pre">list1</span></code>. We then move the <code class="docutils literal notranslate"><span class="pre">list1</span></code> pointer to the next node.</p></li>
-<li><p>If the <code class="docutils literal notranslate"><span class="pre">val</span></code> of the node in <code class="docutils literal notranslate"><span class="pre">list2</span></code> is smaller than or equal to the <code class="docutils literal notranslate"><span class="pre">val</span></code> of the node in <code class="docutils literal notranslate"><span class="pre">list1</span></code>, we attach the node from <code class="docutils literal notranslate"><span class="pre">list2</span></code> to the merged list in a similar manner. We move the <code class="docutils literal notranslate"><span class="pre">list2</span></code> pointer to the next node.</p></li>
-<li><p>After attaching a node to the merged list, we move the <code class="docutils literal notranslate"><span class="pre">current</span></code> pointer to the newly added node. This step is essential for keeping track of the end of the merged list.</p></li>
-<li><p>The loop continues until either <code class="docutils literal notranslate"><span class="pre">list1</span></code> or <code class="docutils literal notranslate"><span class="pre">list2</span></code> becomes empty.</p></li>
-<li><p>Once the loop exits, we check if there are any remaining elements in <code class="docutils literal notranslate"><span class="pre">list1</span></code> or <code class="docutils literal notranslate"><span class="pre">list2</span></code>. If <code class="docutils literal notranslate"><span class="pre">list1</span></code> is not empty, we attach the remaining elements of <code class="docutils literal notranslate"><span class="pre">list1</span></code> to the merged list. If <code class="docutils literal notranslate"><span class="pre">list2</span></code> is not empty, we attach the remaining elements of <code class="docutils literal notranslate"><span class="pre">list2</span></code>.</p></li>
-<li><p>Finally, we return the <code class="docutils literal notranslate"><span class="pre">next</span></code> attribute of the <code class="docutils literal notranslate"><span class="pre">dummy</span></code> node as the head of the merged list, which represents the sorted merged list.</p></li>
-</ol>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Helper function to print the linked list</span>
-<span class="k">def</span> <span class="nf">printLinkedList</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
-    <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
-    <span class="k">while</span> <span class="n">head</span><span class="p">:</span>
-        <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">head</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-        <span class="n">head</span> <span class="o">=</span> <span class="n">head</span><span class="o">.</span><span class="n">next</span>
-    <span class="k">return</span> <span class="n">result</span>
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s explain the helper functions:
-<strong>printLinkedList Function</strong>:</p>
-<ul class="simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">printLinkedList</span></code> function takes the <code class="docutils literal notranslate"><span class="pre">head</span></code> of a linked list as input and returns a list of values representing the nodes in the linked list.</p></li>
-<li><p>Inside the function, a <code class="docutils literal notranslate"><span class="pre">result</span></code> list is initialized to store the values of the linked list nodes.</p></li>
-<li><p>The function then iterates through the linked list starting from the <code class="docutils literal notranslate"><span class="pre">head</span></code> node and appends the <code class="docutils literal notranslate"><span class="pre">val</span></code> attribute of each node to the <code class="docutils literal notranslate"><span class="pre">result</span></code> list.</p></li>
-<li><p>As it iterates through the list, it moves to the next node using the <code class="docutils literal notranslate"><span class="pre">next</span></code> attribute of each node.</p></li>
-<li><p>Finally, the function returns the <code class="docutils literal notranslate"><span class="pre">result</span></code> list containing the values of the linked list nodes in the order they appear in the linked list.</p></li>
-</ul>
-<p>This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="c1"># Input: list1 = [1,2,4], list2 = [1,3,4]</span>
-<span class="n">list1</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)))</span>
-<span class="n">list2</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)))</span>
-<span class="n">merged_list</span> <span class="o">=</span> <span class="n">mergeTwoLists</span><span class="p">(</span><span class="n">list1</span><span class="p">,</span> <span class="n">list2</span><span class="p">)</span>
-<span class="n">printLinkedList</span><span class="p">(</span><span class="n">merged_list</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output text_plain highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[1, 1, 2, 3, 4, 4]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="c1"># Input: list1 = [], list2 = []</span>
-<span class="n">list1</span> <span class="o">=</span> <span class="kc">None</span>
-<span class="n">list2</span> <span class="o">=</span> <span class="kc">None</span>
-<span class="n">merged_list</span> <span class="o">=</span> <span class="n">mergeTwoLists</span><span class="p">(</span><span class="n">list1</span><span class="p">,</span> <span class="n">list2</span><span class="p">)</span>
-<span class="n">printLinkedList</span><span class="p">(</span><span class="n">merged_list</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output text_plain highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3:</span>
-<span class="c1"># list1 = [], list2 = [0]</span>
-<span class="n">list1</span> <span class="o">=</span> <span class="kc">None</span>
-<span class="n">list2</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span>
-<span class="n">merged_list</span> <span class="o">=</span> <span class="n">mergeTwoLists</span><span class="p">(</span><span class="n">list1</span><span class="p">,</span> <span class="n">list2</span><span class="p">)</span>
-<span class="n">printLinkedList</span><span class="p">(</span><span class="n">merged_list</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output text_plain highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[0]
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">mergeTwoLists</span></code> function.</p>
-<p><strong>Time Complexity:</strong></p>
-<p>The time complexity of this function is <span class="math notranslate nohighlight">\(O(N + M\)</span>), where <span class="math notranslate nohighlight">\(N\)</span> and <span class="math notranslate nohighlight">\(M\)</span> are the lengths of <code class="docutils literal notranslate"><span class="pre">list1</span></code> and <code class="docutils literal notranslate"><span class="pre">list2</span></code>, respectively. Here’s why:</p>
-<ol class="arabic simple">
-<li><p>In the worst case, we need to traverse both <code class="docutils literal notranslate"><span class="pre">list1</span></code> and <code class="docutils literal notranslate"><span class="pre">list2</span></code> completely. This involves iterating through all the nodes in both lists once.</p></li>
-<li><p>The while loop runs until either <code class="docutils literal notranslate"><span class="pre">list1</span></code> or <code class="docutils literal notranslate"><span class="pre">list2</span></code> becomes empty. The number of iterations depends on the total number of nodes in both lists, which is N + M in the worst case.</p></li>
-<li><p>Inside the loop, we perform constant-time operations for each iteration, such as comparisons and updating pointers.</p></li>
-</ol>
-<p>Therefore, the dominant factor in the time complexity is the combined length of both input lists, resulting in a linear time complexity of <span class="math notranslate nohighlight">\(O(N + M)\)</span>.</p>
-<p><strong>Space Complexity:</strong></p>
-<p>The space complexity of this function is <span class="math notranslate nohighlight">\(O(1)\)</span>, which means it uses a constant amount of extra space regardless of the input sizes. Here’s why:</p>
-<ol class="arabic simple">
-<li><p>We create a few extra pointers (<code class="docutils literal notranslate"><span class="pre">dummy</span></code>, <code class="docutils literal notranslate"><span class="pre">current</span></code>) and temporary variables (<code class="docutils literal notranslate"><span class="pre">val</span></code>) to manage the merging process. These consume a fixed amount of memory, regardless of the input sizes. The number of these extra variables is independent of the lengths of <code class="docutils literal notranslate"><span class="pre">list1</span></code> and <code class="docutils literal notranslate"><span class="pre">list2</span></code>.</p></li>
-<li><p>We do not create a new data structure or allocate memory for the merged list. Instead, we rearrange the existing nodes from <code class="docutils literal notranslate"><span class="pre">list1</span></code> and <code class="docutils literal notranslate"><span class="pre">list2</span></code> to form the merged list. This operation does not consume additional memory proportional to the input sizes.</p></li>
-</ol>
-<p><strong>In summary</strong>, the <code class="docutils literal notranslate"><span class="pre">mergeTwoLists</span></code> function has a time complexity of <span class="math notranslate nohighlight">\(O(N + M)\)</span> and a space complexity of <span class="math notranslate nohighlight">\(O(1)\)</span>, making it an efficient solution for merging two sorted linked lists.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Merge K Sorted Lists:</strong> Extend the problem to merge <span class="math notranslate nohighlight">\(K\)</span> sorted linked lists instead of just two. You need to efficiently merge <span class="math notranslate nohighlight">\(K\)</span> lists into one sorted list.</p></li>
-<li><p><strong>Merge Lists in a Zigzag Pattern:</strong> Merge two sorted lists in a zigzag pattern. For example, given [1, 2, 4] and [1, 3, 5], the merged list should be [1, 1, 2, 3, 4, 5].</p></li>
-</ol>
-</section>
-</section>
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-<h1>23. Merge k Sorted Lists<a class="headerlink" href="#merge-k-sorted-lists" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Hard</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/merge-k-sorted-lists/">To see the Merge k Sorted Lists problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>You are given an array of <span class="math notranslate nohighlight">\(k\)</span> linked-lists lists, each linked-list is sorted in ascending order.</p>
-<p>Merge all the linked-lists into one sorted linked-list and return it.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">k</span> <span class="pre">==</span> <span class="pre">lists.length</span></code></p></li>
-<li><p>0 &lt;= k &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">0</span> <span class="pre">&lt;=</span> <span class="pre">lists[i].length</span> <span class="pre">&lt;=</span> <span class="pre">500</span></code></p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^4\)</span> &lt;= lists[i][j] &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">lists[i]</span></code> is sorted in ascending order.</p></li>
-<li><p>The sum of <code class="docutils literal notranslate"><span class="pre">lists[i].length</span></code> will not exceed <span class="math notranslate nohighlight">\(10^4\)</span>.</p></li>
-</ul>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Definition for singly-linked list.</span>
-<span class="k">class</span> <span class="nc">ListNode</span><span class="p">:</span>
-     <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="nb">next</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-         <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-         <span class="bp">self</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="nb">next</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">mergeKLists</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">lists</span><span class="p">:</span> <span class="p">[[</span><span class="n">ListNode</span><span class="p">]])</span> <span class="o">-&gt;</span> <span class="n">ListNode</span><span class="p">:</span>
-        <span class="c1"># Check if the input list of linked lists is empty</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">lists</span> <span class="ow">or</span> <span class="nb">len</span><span class="p">(</span><span class="n">lists</span><span class="p">)</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
-            <span class="k">return</span> <span class="kc">None</span>
-
-        <span class="c1"># Loop until there is only one merged list left</span>
-        <span class="k">while</span> <span class="nb">len</span><span class="p">(</span><span class="n">lists</span><span class="p">)</span> <span class="o">&gt;</span> <span class="mi">1</span><span class="p">:</span>
-            <span class="n">mergedLists</span> <span class="o">=</span> <span class="p">[]</span>
-
-            <span class="c1"># Merge pairs of lists</span>
-            <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">lists</span><span class="p">),</span> <span class="mi">2</span><span class="p">):</span>
-                <span class="n">l1</span> <span class="o">=</span> <span class="n">lists</span><span class="p">[</span><span class="n">i</span><span class="p">]</span>
-                <span class="n">l2</span> <span class="o">=</span> <span class="n">lists</span><span class="p">[</span><span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">]</span> <span class="k">if</span> <span class="p">(</span><span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span> <span class="o">&lt;</span> <span class="nb">len</span><span class="p">(</span><span class="n">lists</span><span class="p">)</span> <span class="k">else</span> <span class="kc">None</span>
-                <span class="n">mergedLists</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">mergeList</span><span class="p">(</span><span class="n">l1</span><span class="p">,</span> <span class="n">l2</span><span class="p">))</span>
-
-            <span class="c1"># Update the list of lists with merged results</span>
-            <span class="n">lists</span> <span class="o">=</span> <span class="n">mergedLists</span>
-        
-        <span class="c1"># Return the final merged list</span>
-        <span class="k">return</span> <span class="n">lists</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
-
-    <span class="k">def</span> <span class="nf">mergeList</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">l1</span><span class="p">,</span> <span class="n">l2</span><span class="p">):</span>
-        <span class="c1"># Create a dummy node to simplify merging</span>
-        <span class="n">dummy</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">()</span>
-        <span class="n">tail</span> <span class="o">=</span> <span class="n">dummy</span>
-
-        <span class="c1"># Merge the two sorted lists</span>
-        <span class="k">while</span> <span class="n">l1</span> <span class="ow">and</span> <span class="n">l2</span><span class="p">:</span>
-            <span class="k">if</span> <span class="n">l1</span><span class="o">.</span><span class="n">val</span> <span class="o">&lt;</span> <span class="n">l2</span><span class="o">.</span><span class="n">val</span><span class="p">:</span>
-                <span class="n">tail</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">l1</span>
-                <span class="n">l1</span> <span class="o">=</span> <span class="n">l1</span><span class="o">.</span><span class="n">next</span>
-            <span class="k">else</span><span class="p">:</span>
-                <span class="n">tail</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">l2</span>
-                <span class="n">l2</span> <span class="o">=</span> <span class="n">l2</span><span class="o">.</span><span class="n">next</span>
-            <span class="n">tail</span> <span class="o">=</span> <span class="n">tail</span><span class="o">.</span><span class="n">next</span>
-
-        <span class="c1"># Append any remaining elements from l1 or l2 (if any)</span>
-        <span class="k">if</span> <span class="n">l1</span><span class="p">:</span>
-            <span class="n">tail</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">l1</span>
-        <span class="k">if</span> <span class="n">l2</span><span class="p">:</span>
-            <span class="n">tail</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">l2</span>
-
-        <span class="c1"># Return the merged result starting from the next of the dummy node</span>
-        <span class="k">return</span> <span class="n">dummy</span><span class="o">.</span><span class="n">next</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The provided code defines a Python class <code class="docutils literal notranslate"><span class="pre">Solution</span></code> with two methods for merging k sorted linked lists:</p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">mergeKLists(self,</span> <span class="pre">lists:</span> <span class="pre">List[ListNode])</span> <span class="pre">-&gt;</span> <span class="pre">ListNode</span></code>: This method takes a list of k sorted linked lists as input and returns a single merged sorted linked list. It uses a divide-and-conquer approach to repeatedly merge pairs of lists until only one merged list remains.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">mergeList(self,</span> <span class="pre">l1,</span> <span class="pre">l2)</span></code>: This method takes two sorted linked lists, <code class="docutils literal notranslate"><span class="pre">l1</span></code> and <code class="docutils literal notranslate"><span class="pre">l2</span></code>, as input and merges them into a single sorted linked list. It uses a dummy node to simplify the merging process.</p></li>
-</ol>
-<p>Here’s a high-level overview of how the code works:</p>
-<ul class="simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">mergeKLists</span></code> method checks if the input list of linked lists is empty or contains no lists. If there are no lists, it returns <code class="docutils literal notranslate"><span class="pre">None</span></code>.</p></li>
-<li><p>Inside a <code class="docutils literal notranslate"><span class="pre">while</span></code> loop, the code repeatedly merges pairs of linked lists until only one merged list remains in the <code class="docutils literal notranslate"><span class="pre">lists</span></code> array. It does this by iterating through the input lists in pairs and calling the <code class="docutils literal notranslate"><span class="pre">mergeList</span></code> method to merge each pair.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">mergeList</span></code> method takes two sorted linked lists, <code class="docutils literal notranslate"><span class="pre">l1</span></code> and <code class="docutils literal notranslate"><span class="pre">l2</span></code>, and merges them into a single sorted linked list. It uses a dummy node (<code class="docutils literal notranslate"><span class="pre">dummy</span></code>) and a <code class="docutils literal notranslate"><span class="pre">tail</span></code> pointer to keep track of the merged list while comparing and merging elements from <code class="docutils literal notranslate"><span class="pre">l1</span></code> and <code class="docutils literal notranslate"><span class="pre">l2</span></code>.</p></li>
-<li><p>After merging all pairs of lists and updating the <code class="docutils literal notranslate"><span class="pre">lists</span></code> array with the merged results, the loop continues until only one merged list remains in the <code class="docutils literal notranslate"><span class="pre">lists</span></code> array.</p></li>
-<li><p>Finally, the <code class="docutils literal notranslate"><span class="pre">mergeKLists</span></code> method returns the merged list.</p></li>
-</ul>
-<p>Overall, this code efficiently merges k sorted linked lists using a divide-and-conquer strategy, resulting in a single merged sorted linked list as the output.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 1</span>
-<span class="c1"># Input: lists = [[1,4,5],[1,3,4],[2,6]]</span>
-
-<span class="n">lists1</span> <span class="o">=</span> <span class="p">[</span>
-    <span class="n">ListNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">4</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">5</span><span class="p">))),</span>
-    <span class="n">ListNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">4</span><span class="p">))),</span>
-    <span class="n">ListNode</span><span class="p">(</span><span class="mi">2</span><span class="p">,</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">6</span><span class="p">))</span>
-<span class="p">]</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="n">result1</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">mergeKLists</span><span class="p">(</span><span class="n">lists1</span><span class="p">)</span>
-
-<span class="c1"># Print the result</span>
-<span class="k">if</span> <span class="n">result1</span><span class="p">:</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">result1</span>
-    <span class="k">while</span> <span class="n">current</span><span class="p">:</span>
-        <span class="nb">print</span><span class="p">(</span><span class="n">current</span><span class="o">.</span><span class="n">val</span><span class="p">,</span> <span class="n">end</span><span class="o">=</span><span class="s2">&quot; -&gt; &quot;</span><span class="p">)</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-<span class="k">else</span><span class="p">:</span>
-    <span class="nb">print</span><span class="p">(</span><span class="s2">&quot;None&quot;</span><span class="p">)</span>  <span class="c1"># Print &quot;None&quot; for input with a single None element</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>1 -&gt; 1 -&gt; 2 -&gt; 3 -&gt; 4 -&gt; 4 -&gt; 5 -&gt; 6 -&gt; 
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 2</span>
-<span class="c1"># Input: lists = []</span>
-
-<span class="n">lists2</span> <span class="o">=</span> <span class="p">[]</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="n">result2</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">mergeKLists</span><span class="p">(</span><span class="n">lists2</span><span class="p">)</span>
-
-<span class="c1"># Print the result</span>
-<span class="k">if</span> <span class="n">result2</span><span class="p">:</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">result2</span>
-    <span class="k">while</span> <span class="n">current</span><span class="p">:</span>
-        <span class="nb">print</span><span class="p">(</span><span class="n">current</span><span class="o">.</span><span class="n">val</span><span class="p">,</span> <span class="n">end</span><span class="o">=</span><span class="s2">&quot; -&gt; &quot;</span><span class="p">)</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-<span class="k">else</span><span class="p">:</span>
-    <span class="nb">print</span><span class="p">(</span><span class="s2">&quot;None&quot;</span><span class="p">)</span>  <span class="c1"># Print &quot;None&quot; for input with a single None element</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>None
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 3</span>
-<span class="c1"># Input: lists = [[]]</span>
-
-<span class="n">lists3</span> <span class="o">=</span> <span class="p">[</span><span class="kc">None</span><span class="p">]</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="n">result3</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">mergeKLists</span><span class="p">(</span><span class="n">lists3</span><span class="p">)</span>
-
-<span class="c1"># Print the result</span>
-<span class="k">if</span> <span class="n">result3</span><span class="p">:</span>
-    <span class="n">current</span> <span class="o">=</span> <span class="n">result3</span>
-    <span class="k">while</span> <span class="n">current</span><span class="p">:</span>
-        <span class="nb">print</span><span class="p">(</span><span class="n">current</span><span class="o">.</span><span class="n">val</span><span class="p">,</span> <span class="n">end</span><span class="o">=</span><span class="s2">&quot; -&gt; &quot;</span><span class="p">)</span>
-        <span class="n">current</span> <span class="o">=</span> <span class="n">current</span><span class="o">.</span><span class="n">next</span>
-<span class="k">else</span><span class="p">:</span>
-    <span class="nb">print</span><span class="p">(</span><span class="s2">&quot;None&quot;</span><span class="p">)</span>  <span class="c1"># Print &quot;None&quot; for input with a single None element</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>None
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic">
-<li><p><strong>Heap Initialization:</strong> The code does not use a heap. Instead, it uses a divide-and-conquer approach. The initial check for empty input lists takes <span class="math notranslate nohighlight">\(O(1)\)</span> time.</p></li>
-<li><p><strong>Merging:</strong> The merging operation is performed in a divide-and-conquer fashion. In each iteration of the <code class="docutils literal notranslate"><span class="pre">while</span></code> loop, we merge pairs of linked lists, and the number of comparisons made is proportional to the total number of nodes across all the linked lists (<span class="math notranslate nohighlight">\(n\)</span>). In each merge step, we effectively process each node once. The number of iterations required to reduce k lists to 1 is <span class="math notranslate nohighlight">\(O(log\ k)\)</span>.</p>
-<p>Therefore, the overall time complexity of the code is <span class="math notranslate nohighlight">\(O(n * log\ k)\)</span>, where n is the total number of nodes across all lists, and <span class="math notranslate nohighlight">\(k\)</span> is the number of linked lists.</p>
-</li>
-</ol>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic">
-<li><p><strong>Heap Space:</strong> The code doesn’t use a heap data structure, so there’s no additional space complexity due to a heap.</p></li>
-<li><p><strong>Merged Lists:</strong> In the <code class="docutils literal notranslate"><span class="pre">mergeList</span></code> method, we create a new merged list. However, this list is not stored in memory for all lists; it’s replaced with each merged pair. The space used for these merged lists is proportional to the size of the largest merged list, which is <span class="math notranslate nohighlight">\(O(n)\)</span> in the worst case.</p></li>
-<li><p><strong>Additional Variables:</strong> The code uses a few additional variables, such as <code class="docutils literal notranslate"><span class="pre">dummy</span></code> and <code class="docutils literal notranslate"><span class="pre">tail</span></code>, but these occupy a constant amount of space and don’t depend on the input size.</p>
-<p>Therefore, the overall space complexity of the code is <span class="math notranslate nohighlight">\(O(n)\)</span>, where n is the total number of nodes across all lists.</p>
-</li>
-</ol>
-<p>In summary, the code’s time complexity is <span class="math notranslate nohighlight">\(O(n * log(k))\)</span>, and its space complexity is <span class="math notranslate nohighlight">\(O(n)\)</span>. This code efficiently merges <span class="math notranslate nohighlight">\(k\)</span> sorted linked lists using a divide-and-conquer approach with a relatively low space overhead.</p>
-</section>
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-<h1>100: Same Tree<a class="headerlink" href="#same-tree" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/same-tree/">To see the Same Tree problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given the roots of two binary trees <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code>, write a function to check if they are the same or not.</p>
-<p>Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.</p>
-<p><strong>Constraints</strong></p>
-<ol class="arabic simple">
-<li><p>The number of nodes in both trees is in the range <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">100]</span></code>.</p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^4\)</span> &lt;= Node.val &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-</ol>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Definition for a binary tree node.</span>
-<span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="c1"># Initialize a TreeNode with a value (val), left child, and right child.</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">def</span> <span class="nf">isSameTree</span><span class="p">(</span><span class="n">p</span><span class="p">,</span> <span class="n">q</span><span class="p">):</span>
-    <span class="c1"># Base case: If both p and q are None, the trees are the same.</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">p</span> <span class="ow">and</span> <span class="ow">not</span> <span class="n">q</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">True</span>
-    
-    <span class="c1"># Base case: If either p or q is None (but not both), the trees are different.</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">p</span> <span class="ow">or</span> <span class="ow">not</span> <span class="n">q</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">False</span>
-    
-    <span class="c1"># Check if the values of the current nodes (p.val and q.val) are equal.</span>
-    <span class="k">if</span> <span class="n">p</span><span class="o">.</span><span class="n">val</span> <span class="o">!=</span> <span class="n">q</span><span class="o">.</span><span class="n">val</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">False</span>
-    
-    <span class="c1"># Recursively check the left and right subtrees of p and q.</span>
-    <span class="c1"># If both subtrees are the same, the entire trees are the same.</span>
-    <span class="k">return</span> <span class="n">isSameTree</span><span class="p">(</span><span class="n">p</span><span class="o">.</span><span class="n">left</span><span class="p">,</span> <span class="n">q</span><span class="o">.</span><span class="n">left</span><span class="p">)</span> <span class="ow">and</span> <span class="n">isSameTree</span><span class="p">(</span><span class="n">p</span><span class="o">.</span><span class="n">right</span><span class="p">,</span> <span class="n">q</span><span class="o">.</span><span class="n">right</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-</div>
-<p>In this code, we define a <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class to represent binary tree nodes and a <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function to check if two binary trees are the same. The function uses recursive traversal to compare the trees’ structures and values.</p>
-<ol class="arabic simple">
-<li><p>We start by defining a <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class, which represents a node in a binary tree. Each node has a <code class="docutils literal notranslate"><span class="pre">val</span></code> (the node’s value), a <code class="docutils literal notranslate"><span class="pre">left</span></code> child, and a right child. This class will help us create and work with binary trees.</p></li>
-<li><p>Next, we define the <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function, which checks if two binary trees (<code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code>) are the same.</p>
-<ul class="simple">
-<li><p>The base case for the recursion is when both <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code> are <code class="docutils literal notranslate"><span class="pre">None</span></code>. In this case, they are considered the same, so we return <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>If either <code class="docutils literal notranslate"><span class="pre">p</span></code> or <code class="docutils literal notranslate"><span class="pre">q</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code> (but not both), they cannot be the same, so we return <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-<li><p>If the values of the current nodes <code class="docutils literal notranslate"><span class="pre">p.val</span></code> and <code class="docutils literal notranslate"><span class="pre">q.val</span></code> are not equal, we return <code class="docutils literal notranslate"><span class="pre">False</span></code> because the trees cannot be the same.</p></li>
-<li><p>Finally, we recursively check the left and right subtrees of <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code> to see if they are the same.</p></li>
-</ul>
-</li>
-</ol>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 1</span>
-
-<span class="c1">#Input: `p = [1,2,3]`, `q = [1,2,3]`</span>
-
-<span class="n">p1</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">),</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">))</span>
-<span class="n">q1</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">),</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">))</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isSameTree</span><span class="p">(</span><span class="n">p1</span><span class="p">,</span> <span class="n">q1</span><span class="p">))</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 2</span>
-
-<span class="c1">#Input: `p = [1,2]`, `q = [1,null,2]`</span>
-
-<span class="n">p2</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">),</span> <span class="kc">None</span><span class="p">)</span>
-<span class="n">q2</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="kc">None</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">))</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isSameTree</span><span class="p">(</span><span class="n">p2</span><span class="p">,</span> <span class="n">q2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 3</span>
-
-<span class="c1">#Input: p = [1,2,1], q = [1,1,2]</span>
-
-<span class="n">p3</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">),</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">))</span>
-<span class="n">q3</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">),</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">))</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">isSameTree</span><span class="p">(</span><span class="n">p3</span><span class="p">,</span> <span class="n">q3</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-complexity">
-<h2>Time Complexity<a class="headerlink" href="#time-complexity" title="Permalink to this heading">#</a></h2>
-<p>The time complexity of the <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function can be analyzed as follows:</p>
-<p>In the worst case, the function needs to visit every node in both trees once to determine if they are the same.
-Since each node is visited exactly once, the time complexity is <span class="math notranslate nohighlight">\(O(n)\)</span>, where <span class="math notranslate nohighlight">\(n\)</span> is the total number of nodes in the input trees.</p>
-</section>
-<section id="space-complexity">
-<h2>Space Complexity<a class="headerlink" href="#space-complexity" title="Permalink to this heading">#</a></h2>
-<p>The space complexity of the <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function can be analyzed as follows:</p>
-<p>The space used by the function’s call stack during recursion is proportional to the maximum depth of the binary trees.
-In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be <span class="math notranslate nohighlight">\(n\)</span>, where <span class="math notranslate nohighlight">\(n\)</span> is the total number of nodes in the input trees.
-Therefore, the space complexity is <span class="math notranslate nohighlight">\(O(n)\)</span> due to the recursive call stack.
-In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.</p>
-</section>
-<section id="in-summary">
-<h2>In summary:<a class="headerlink" href="#in-summary" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p><strong>Time Complexity:</strong> <span class="math notranslate nohighlight">\(O(n)\)</span> where <span class="math notranslate nohighlight">\(n\)</span> is the total number of nodes in the input trees.</p></li>
-<li><p><strong>Space Complexity:</strong> <span class="math notranslate nohighlight">\(O(n)\)</span> due to the recursive call stack.</p></li>
-</ul>
-</section>
-</section>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
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-<h1>102. Binary Tree Level Order Traversal<a class="headerlink" href="#binary-tree-level-order-traversal" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/binary-tree-level-order-traversal/">To see the Binary Tree Level Order Traversal problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given the <code class="docutils literal notranslate"><span class="pre">root</span></code> of a binary tree, return <em>the level order traversal of its nodes’ values</em>. (i.e., from left to right, level by level).</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the tree is in the range <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">2000]</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-1000</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">1000</span></code></p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">def</span> <span class="nf">levelOrder</span><span class="p">(</span><span class="n">root</span><span class="p">):</span>
-    <span class="c1"># Check if the tree is empty, if so, return an empty list.</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">root</span><span class="p">:</span>
-        <span class="k">return</span> <span class="p">[]</span>
-
-    <span class="c1"># Initialize an empty list to store the result.</span>
-    <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
-    
-    <span class="c1"># Initialize a queue with the root node to perform BFS.</span>
-    <span class="n">queue</span> <span class="o">=</span> <span class="p">[</span><span class="n">root</span><span class="p">]</span>
-
-    <span class="k">while</span> <span class="n">queue</span><span class="p">:</span>
-        <span class="c1"># Initialize an empty list to store the values of nodes at the current level.</span>
-        <span class="n">level_values</span> <span class="o">=</span> <span class="p">[]</span>
-        
-        <span class="c1"># Get the number of nodes at the current level.</span>
-        <span class="n">level_size</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">queue</span><span class="p">)</span>
-
-        <span class="c1"># Iterate through the nodes at the current level.</span>
-        <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">level_size</span><span class="p">):</span>
-            <span class="c1"># Dequeue the front node from the queue.</span>
-            <span class="n">node</span> <span class="o">=</span> <span class="n">queue</span><span class="o">.</span><span class="n">pop</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span>
-            
-            <span class="c1"># Append the value of the current node to the level_values list.</span>
-            <span class="n">level_values</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-
-            <span class="c1"># Enqueue the left and right children of the current node if they exist.</span>
-            <span class="k">if</span> <span class="n">node</span><span class="o">.</span><span class="n">left</span><span class="p">:</span>
-                <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">left</span><span class="p">)</span>
-            <span class="k">if</span> <span class="n">node</span><span class="o">.</span><span class="n">right</span><span class="p">:</span>
-                <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">right</span><span class="p">)</span>
-
-        <span class="c1"># Append the level_values list (values at the current level) to the result list.</span>
-        <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">level_values</span><span class="p">)</span>
-
-    <span class="c1"># Return the final result, which is a list of lists representing level order traversal.</span>
-    <span class="k">return</span> <span class="n">result</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We define a <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class that represents a node in the binary tree. Each <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> object has a value (<code class="docutils literal notranslate"><span class="pre">val</span></code>) and two child nodes: <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code>.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">levelOrder</span></code> function takes the root of the binary tree as its input and returns the level order traversal of the tree as a list of lists.</p></li>
-<li><p>We start by checking if the input <code class="docutils literal notranslate"><span class="pre">root</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code>, which indicates an empty tree. If the tree is empty, we return an empty list because there are no nodes to traverse.</p></li>
-<li><p>We initialize an empty list called <code class="docutils literal notranslate"><span class="pre">result</span></code> to store the final result, which will be a list of lists containing node values at each level.</p></li>
-<li><p>We initialize a queue called <code class="docutils literal notranslate"><span class="pre">queue</span></code> with the root node. This queue will be used for breadth-first traversal of the tree.</p></li>
-<li><p>We enter a while loop that continues until the <code class="docutils literal notranslate"><span class="pre">queue</span></code> is empty. Inside the loop, we perform the following steps:</p>
-<ul class="simple">
-<li><p>We initialize an empty list called <code class="docutils literal notranslate"><span class="pre">level_values</span></code> to store the values of nodes at the current level.</p></li>
-<li><p>We determine the number of nodes at the current level by getting the length of the <code class="docutils literal notranslate"><span class="pre">queue</span></code>. This is done to process nodes level by level.</p></li>
-<li><p>We iterate through the nodes at the current level using a for loop. For each node in the current level:</p>
-<ul>
-<li><p>We dequeue (remove) the front node from the <code class="docutils literal notranslate"><span class="pre">queue</span></code>.</p></li>
-<li><p>We append the value of the dequeued node to the <code class="docutils literal notranslate"><span class="pre">level_values</span></code> list, effectively collecting the values of nodes at the current level.</p></li>
-<li><p>If the dequeued node has a left child, we enqueue the left child to the <code class="docutils literal notranslate"><span class="pre">queue</span></code>.</p></li>
-<li><p>If the dequeued node has a right child, we enqueue the right child to the <code class="docutils literal notranslate"><span class="pre">queue</span></code>.</p></li>
-</ul>
-</li>
-<li><p>After processing all nodes at the current level, we append the <code class="docutils literal notranslate"><span class="pre">level_values</span></code> list to the <code class="docutils literal notranslate"><span class="pre">result</span></code> list. This represents the values at the current level.</p></li>
-</ul>
-</li>
-<li><p>The loop continues until all levels have been traversed, and the <code class="docutils literal notranslate"><span class="pre">queue</span></code> becomes empty.</p></li>
-<li><p>Finally, we return the <code class="docutils literal notranslate"><span class="pre">result</span></code> list, which contains lists of node values at each level, representing the level order traversal of the binary tree.</p></li>
-</ol>
-<p>The code effectively performs a breadth-first traversal of the binary tree, processing nodes level by level, and constructs the result list that represents the level order traversal.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Construct the tree</span>
-<span class="n">root</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">9</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">20</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">15</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">7</span><span class="p">)</span>
-
-<span class="n">result</span> <span class="o">=</span> <span class="n">levelOrder</span><span class="p">(</span><span class="n">root</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[3], [9, 20], [15, 7]]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-
-<span class="c1"># Using the same tree as before</span>
-<span class="n">root</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">levelOrder</span><span class="p">(</span><span class="n">root</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[1]]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3:</span>
-<span class="c1"># Creating a new tree for this example</span>
-<span class="n">root</span> <span class="o">=</span> <span class="kc">None</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">levelOrder</span><span class="p">(</span><span class="n">root</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">levelOrder</span></code> function:</p>
-<p><strong>Time Complexity:</strong></p>
-<p>The time complexity of this function is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the breadth-first traversal.</p>
-<p>In the worst case, we have to enqueue and dequeue all nodes in the binary tree, which is proportional to the number of nodes (N). In each level, we process all nodes in that level, and since there are a total of log(N) levels in a balanced binary tree, the time complexity can also be approximated as O(N) for unbalanced trees.</p>
-<p><strong>Space Complexity:</strong></p>
-<p>The space complexity of this function is O(N), where N is the number of nodes in the binary tree. Here’s how the space complexity breaks down:</p>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">result</span></code> list stores the level order traversal, and in the worst case, it contains N/2 levels (for a completely unbalanced binary tree). So, the space used by <code class="docutils literal notranslate"><span class="pre">result</span></code> is O(N).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">queue</span></code> data structure is used for BFS traversal. In the worst case, it can store all nodes at the last level of the tree. In a balanced binary tree, the maximum number of nodes at any level is 2^(log(N)), which is still O(N). In the case of an unbalanced tree, it can be even worse. So, the space used by <code class="docutils literal notranslate"><span class="pre">queue</span></code> is O(N).</p></li>
-</ol>
-<p>Overall, the dominant factor in terms of space complexity is the <code class="docutils literal notranslate"><span class="pre">queue</span></code>, and the space complexity is O(N).</p>
-<p>In summary, the function’s time complexity is O(N), and its space complexity is also O(N). It is an efficient and optimal solution for performing a level order traversal of a binary tree.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Reverse Level Order Traversal:</strong> Modify the <code class="docutils literal notranslate"><span class="pre">levelOrder</span></code> function to return the level order traversal in reverse order (from bottom to top). For example, if the input tree is <code class="docutils literal notranslate"><span class="pre">[3,</span> <span class="pre">9,</span> <span class="pre">20,</span> <span class="pre">null,</span> <span class="pre">null,</span> <span class="pre">15,</span> <span class="pre">7]</span></code>, the output should be <code class="docutils literal notranslate"><span class="pre">[[15,</span> <span class="pre">7],</span> <span class="pre">[9,</span> <span class="pre">20],</span> <span class="pre">[3]]</span></code>.</p></li>
-<li><p><strong>Zigzag Level Order Traversal:</strong> Write a function that performs a level order traversal of a binary tree in a zigzag pattern. In a zigzag traversal, the nodes at even levels are traversed from left to right, and nodes at odd levels are traversed from right to left. For example, if the input tree is <code class="docutils literal notranslate"><span class="pre">[3,</span> <span class="pre">9,</span> <span class="pre">20,</span> <span class="pre">null,</span> <span class="pre">null,</span> <span class="pre">15,</span> <span class="pre">7]</span></code>, the output should be <code class="docutils literal notranslate"><span class="pre">[[3],</span> <span class="pre">[20,</span> <span class="pre">9],</span> <span class="pre">[15,</span> <span class="pre">7]]</span></code>.</p></li>
-</ol>
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-<h1>104. Maximum Depth of Binary Tree<a class="headerlink" href="#maximum-depth-of-binary-tree" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/invert-binary-tree/">To see the Invert Binary Tree problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given the <code class="docutils literal notranslate"><span class="pre">root</span></code> of a binary tree, return its maximum depth.</p>
-<p>A binary tree’s <strong>maximum depth</strong> is the number of nodes along the longest path from the root node down to the farthest leaf node.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the tree is in the range <span class="math notranslate nohighlight">\([0, 10^4]\)</span>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-100</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">100</span></code></p></li>
-</ul>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Definition for a binary tree node.</span>
-<span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">def</span> <span class="nf">maxDepth</span><span class="p">(</span><span class="n">root</span><span class="p">):</span>
-    <span class="c1"># Base case: If the root is None, the depth is 0.</span>
-    <span class="k">if</span> <span class="n">root</span> <span class="ow">is</span> <span class="kc">None</span><span class="p">:</span>
-        <span class="k">return</span> <span class="mi">0</span>
-    
-    <span class="c1"># Recursively calculate the maximum depth of the left and right subtrees.</span>
-    <span class="n">left_depth</span> <span class="o">=</span> <span class="n">maxDepth</span><span class="p">(</span><span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="p">)</span>
-    <span class="n">right_depth</span> <span class="o">=</span> <span class="n">maxDepth</span><span class="p">(</span><span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="p">)</span>
-    
-    <span class="c1"># Return the maximum depth of the tree by adding 1 to the maximum depth of the subtrees.</span>
-    <span class="k">return</span> <span class="nb">max</span><span class="p">(</span><span class="n">left_depth</span><span class="p">,</span> <span class="n">right_depth</span><span class="p">)</span> <span class="o">+</span> <span class="mi">1</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>Let’s go through the code step by step to understand how it works:</p>
-<ol class="arabic simple">
-<li><p>We start by defining a <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class to represent the nodes of the binary tree. Each node has three attributes:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">val</span></code>: the value stored in the node.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">left</span></code>: a reference to the left child node.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">right</span></code>: a reference to the right child node.</p></li>
-</ul>
-</li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">maxDepth</span></code> function is the main function that calculates the maximum depth of the binary tree. It takes a single argument, <code class="docutils literal notranslate"><span class="pre">root</span></code>, which is the root node of the binary tree.</p></li>
-<li><p>In the <code class="docutils literal notranslate"><span class="pre">maxDepth</span></code> function, we have a base case:</p>
-<ul class="simple">
-<li><p>If the <code class="docutils literal notranslate"><span class="pre">root</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code>, it means we have reached the end of a branch of the tree (a leaf node or an empty subtree). In this case, the depth is 0 because there are no nodes to count.</p></li>
-</ul>
-</li>
-<li><p>If the <code class="docutils literal notranslate"><span class="pre">root</span></code> is not <code class="docutils literal notranslate"><span class="pre">None</span></code>, we recursively calculate the maximum depth of the left and right subtrees:</p>
-<ul class="simple">
-<li><p>We call <code class="docutils literal notranslate"><span class="pre">maxDepth</span></code> on the <code class="docutils literal notranslate"><span class="pre">root.left</span></code> to calculate the maximum depth of the left subtree and store it in the variable <code class="docutils literal notranslate"><span class="pre">left_depth</span></code>.</p></li>
-<li><p>We call <code class="docutils literal notranslate"><span class="pre">maxDepth</span></code> on the <code class="docutils literal notranslate"><span class="pre">root.right</span></code> to calculate the maximum depth of the right subtree and store it in the variable <code class="docutils literal notranslate"><span class="pre">right_depth</span></code>.</p></li>
-</ul>
-</li>
-<li><p>Finally, we return the maximum depth of the tree by taking the maximum of <code class="docutils literal notranslate"><span class="pre">left_depth</span></code> and <code class="docutils literal notranslate"><span class="pre">right_depth</span></code> and adding 1 to it. This is because we are counting the current level as well.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 1</span>
-<span class="c1"># Input: root = [3,9,20,null,null,15,7]</span>
-
-<span class="n">root</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">9</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">20</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">15</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">7</span><span class="p">)</span>
-
-<span class="nb">print</span><span class="p">(</span><span class="n">maxDepth</span><span class="p">(</span><span class="n">root</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>3
-</pre></div>
-</div>
-</div>
-</div>
-<p>In this example, we create a binary tree based on the given input <code class="docutils literal notranslate"><span class="pre">[3,9,20,null,null,15,7]</span></code> and calculate its maximum depth, which is <code class="docutils literal notranslate"><span class="pre">3</span></code>. The tree structure is as follows:</p>
-<div class="highlight-default notranslate"><div class="highlight"><pre><span></span>    <span class="mi">3</span>
-   <span class="o">/</span> \
-  <span class="mi">9</span>  <span class="mi">20</span>
-    <span class="o">/</span>  \
-   <span class="mi">15</span>   <span class="mi">7</span>
-</pre></div>
-</div>
-<p>So, the maximum depth is the length of the longest path from the root to a leaf node, which is 3 in this case.</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 2</span>
-<span class="c1"># Input: root = [1,null,2]</span>
-
-<span class="n">root</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-
-<span class="c1"># Calculate the maximum depth of the tree and print the result</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">maxDepth</span><span class="p">(</span><span class="n">root</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>2
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong></p>
-<ul class="simple">
-<li><p>The time complexity of the <code class="docutils literal notranslate"><span class="pre">maxDepth</span></code> function is <span class="math notranslate nohighlight">\(O(N)\)</span>, where <span class="math notranslate nohighlight">\(N\)</span> is the number of nodes in the binary tree.</p></li>
-<li><p>This is because in the worst case, the function visits every node exactly once in a depth-first manner.</p></li>
-<li><p>The recursion explores all nodes of the tree, so the time complexity is linear with respect to the number of nodes.</p></li>
-</ul>
-<p><strong>Space Complexity:</strong></p>
-<ul class="simple">
-<li><p>The space complexity of the <code class="docutils literal notranslate"><span class="pre">maxDepth</span></code> function is <span class="math notranslate nohighlight">\(O(H)\)</span>, where H is the height of the binary tree.</p></li>
-<li><p>In the worst case, if the binary tree is completely unbalanced (skewed), the recursion stack can go as deep as the height of the tree.</p></li>
-<li><p>In the best case, if the binary tree is perfectly balanced, the height is <span class="math notranslate nohighlight">\(O(log\ N)\)</span>, where <span class="math notranslate nohighlight">\(N\)</span> is the number of nodes.</p></li>
-<li><p>Therefore, the space complexity can vary from <span class="math notranslate nohighlight">\(O(log\ N)\)</span> to <span class="math notranslate nohighlight">\(O(N)\)</span> depending on the shape of the tree.</p></li>
-<li><p>In addition to the recursion stack, there is a small constant amount of space used for variables and function call overhead.</p></li>
-</ul>
-<p><strong>In summary:</strong></p>
-<ul class="simple">
-<li><p>The time complexity of the code is <span class="math notranslate nohighlight">\(O(N)\)</span> as it visits each node once.</p></li>
-<li><p>The space complexity is <span class="math notranslate nohighlight">\(O(H)\)</span>, where H is the height of the tree, which can vary from <span class="math notranslate nohighlight">\(O(log\ N)\)</span> to <span class="math notranslate nohighlight">\(O(N)\)</span> depending on the tree’s shape.</p></li>
-</ul>
-</section>
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-<h1>105. Construct Binary Tree from Preorder and Inorder Traversal<a class="headerlink" href="#construct-binary-tree-from-preorder-and-inorder-traversal" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/">To see the Construct Binary Tree from Preorder and Inorder Traversal problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given two integer arrays <code class="docutils literal notranslate"><span class="pre">preorder</span></code> and <code class="docutils literal notranslate"><span class="pre">inorder</span></code> where <code class="docutils literal notranslate"><span class="pre">preorder</span></code> is the preorder traversal of a binary tree and <code class="docutils literal notranslate"><span class="pre">inorder</span></code> is the inorder traversal of the same tree, construct and return <em>the binary tree.</em></p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">preorder.length</span> <span class="pre">&lt;=</span> <span class="pre">3000</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">inorder.length</span> <span class="pre">==</span> <span class="pre">preorder.length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-3000</span> <span class="pre">&lt;=</span> <span class="pre">preorder[i],</span> <span class="pre">inorder[i]</span> <span class="pre">&lt;=</span> <span class="pre">3000</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">preorder</span></code> and <code class="docutils literal notranslate"><span class="pre">inorder</span></code> consist of <strong>unique</strong> values.</p></li>
-<li><p>Each value of <code class="docutils literal notranslate"><span class="pre">inorder</span></code> also appears in <code class="docutils literal notranslate"><span class="pre">preorder</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">preorder</span></code> is <strong>guaranteed</strong> to be the preorder traversal of the tree.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">inorder</span></code> is <strong>guaranteed</strong> to be the inorder traversal of the tree.</p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">def</span> <span class="nf">buildTree</span><span class="p">(</span><span class="n">preorder</span><span class="p">,</span> <span class="n">inorder</span><span class="p">):</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">preorder</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">None</span>
-
-    <span class="c1"># The first element in the preorder traversal is the root of the current subtree</span>
-    <span class="n">root_val</span> <span class="o">=</span> <span class="n">preorder</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
-    <span class="n">root</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="n">root_val</span><span class="p">)</span>
-
-    <span class="c1"># Find the index of the root value in the inorder traversal</span>
-    <span class="n">root_idx_inorder</span> <span class="o">=</span> <span class="n">inorder</span><span class="o">.</span><span class="n">index</span><span class="p">(</span><span class="n">root_val</span><span class="p">)</span>
-
-    <span class="c1"># Recursively build left and right subtrees</span>
-    <span class="n">root</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">buildTree</span><span class="p">(</span><span class="n">preorder</span><span class="p">[</span><span class="mi">1</span><span class="p">:</span><span class="mi">1</span> <span class="o">+</span> <span class="n">root_idx_inorder</span><span class="p">],</span> <span class="n">inorder</span><span class="p">[:</span><span class="n">root_idx_inorder</span><span class="p">])</span>
-    <span class="n">root</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">buildTree</span><span class="p">(</span><span class="n">preorder</span><span class="p">[</span><span class="mi">1</span> <span class="o">+</span> <span class="n">root_idx_inorder</span><span class="p">:],</span> <span class="n">inorder</span><span class="p">[</span><span class="n">root_idx_inorder</span> <span class="o">+</span> <span class="mi">1</span><span class="p">:])</span>
-
-    <span class="k">return</span> <span class="n">root</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic">
-<li><p><code class="docutils literal notranslate"><span class="pre">class</span> <span class="pre">TreeNode:</span></code>: This is a class definition for a binary tree node. It has three attributes:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">val</span></code>: The value of the node.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">left</span></code>: A reference to the left child node.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">right</span></code>: A reference to the right child node.</p></li>
-</ul>
-<p>This class is used to create instances of binary tree nodes, which will be used to build the binary tree.</p>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">def</span> <span class="pre">buildTree(preorder,</span> <span class="pre">inorder):</span></code>: This is the main function that constructs a binary tree from its preorder and inorder traversals.</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">preorder</span></code>: A list representing the preorder traversal of the binary tree.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">inorder</span></code>: A list representing the inorder traversal of the binary tree.</p></li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">if</span> <span class="pre">not</span> <span class="pre">preorder:</span> <span class="pre">return</span> <span class="pre">None</span></code>: This line checks if the <code class="docutils literal notranslate"><span class="pre">preorder</span></code> list is empty. If it is, it means there are no nodes to construct, so the function returns <code class="docutils literal notranslate"><span class="pre">None</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">root_val</span> <span class="pre">=</span> <span class="pre">preorder[0]</span></code>: The value of the root node is extracted from the first element of the <code class="docutils literal notranslate"><span class="pre">preorder</span></code> list.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">root</span> <span class="pre">=</span> <span class="pre">TreeNode(root_val)</span></code>: A new <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> object is created with the <code class="docutils literal notranslate"><span class="pre">root_val</span></code> as its value. This represents the root of the current subtree.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">root_idx_inorder</span> <span class="pre">=</span> <span class="pre">inorder.index(root_val)</span></code>: The index of the <code class="docutils literal notranslate"><span class="pre">root_val</span></code> in the <code class="docutils literal notranslate"><span class="pre">inorder</span></code> list is found. This index indicates the position of the root node in the inorder traversal.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">root.left</span></code>: The left subtree is constructed recursively by calling the <code class="docutils literal notranslate"><span class="pre">buildTree</span></code> function with the appropriate sublists of <code class="docutils literal notranslate"><span class="pre">preorder</span></code> and <code class="docutils literal notranslate"><span class="pre">inorder</span></code>. The left subtree’s preorder and inorder traversals are the slices of <code class="docutils literal notranslate"><span class="pre">preorder</span></code> and <code class="docutils literal notranslate"><span class="pre">inorder</span></code> lists up to the <code class="docutils literal notranslate"><span class="pre">root_idx_inorder</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">root.right</span></code>: The right subtree is constructed recursively by calling the <code class="docutils literal notranslate"><span class="pre">buildTree</span></code> function with the appropriate sublists of <code class="docutils literal notranslate"><span class="pre">preorder</span></code> and <code class="docutils literal notranslate"><span class="pre">inorder</span></code>. The right subtree’s preorder and inorder traversals are the slices of <code class="docutils literal notranslate"><span class="pre">preorder</span></code> and <code class="docutils literal notranslate"><span class="pre">inorder</span></code> lists starting from <code class="docutils literal notranslate"><span class="pre">root_idx_inorder</span> <span class="pre">+</span> <span class="pre">1</span></code>.</p></li>
-<li><p>Finally, the function returns the <code class="docutils literal notranslate"><span class="pre">root</span></code> of the subtree it just constructed.</p></li>
-</ol>
-<p>The recursive approach used here divides the problem of constructing the entire binary tree into smaller subproblems, starting with the root node and then recursively building the left and right subtrees until the entire tree is constructed.</p>
-</section>
-<section id="helper-function">
-<h2>Helper Function<a class="headerlink" href="#helper-function" title="Permalink to this heading">#</a></h2>
-<p>Here is a helper function to represent the binary tree elements.</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">treeToList</span><span class="p">(</span><span class="n">root</span><span class="p">):</span>
-    <span class="c1"># Initialize an empty list to store the elements of the binary tree</span>
-    <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
-    
-    <span class="c1"># Initialize a queue for level-order traversal starting with the root</span>
-    <span class="n">queue</span> <span class="o">=</span> <span class="p">[</span><span class="n">root</span><span class="p">]</span>
-
-    <span class="k">while</span> <span class="n">queue</span><span class="p">:</span>
-        <span class="c1"># Pop the front node from the queue</span>
-        <span class="n">node</span> <span class="o">=</span> <span class="n">queue</span><span class="o">.</span><span class="n">pop</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span>
-        
-        <span class="c1"># If the node is not None (i.e., a valid node), add its value to the result list</span>
-        <span class="k">if</span> <span class="n">node</span><span class="p">:</span>
-            <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-            
-            <span class="c1"># Add the left and right children of the node to the queue</span>
-            <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">left</span><span class="p">)</span>
-            <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">right</span><span class="p">)</span>
-        <span class="k">else</span><span class="p">:</span>
-            <span class="c1"># If the node is None, add None to the result to represent an empty node</span>
-            <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="kc">None</span><span class="p">)</span>
-
-    <span class="c1"># Remove any trailing None values from the result list</span>
-    <span class="k">while</span> <span class="n">result</span> <span class="ow">and</span> <span class="n">result</span><span class="p">[</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="ow">is</span> <span class="kc">None</span><span class="p">:</span>
-        <span class="n">result</span><span class="o">.</span><span class="n">pop</span><span class="p">()</span>
-
-    <span class="c1"># Return the resulting list representing the binary tree elements</span>
-    <span class="k">return</span> <span class="n">result</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="id1">
-<h2>Explanation:<a class="headerlink" href="#id1" title="Permalink to this heading">#</a></h2>
-<p>Let’s break down the <code class="docutils literal notranslate"><span class="pre">treeToList</span></code> helper function step by step:</p>
-<ol class="arabic">
-<li><p><code class="docutils literal notranslate"><span class="pre">def</span> <span class="pre">treeToList(root):</span></code>: This function takes the root of a binary tree as input and converts it into a list.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">result</span> <span class="pre">=</span> <span class="pre">[]</span></code>: Initialize an empty list called <code class="docutils literal notranslate"><span class="pre">result</span></code> to store the elements of the binary tree in list format.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">queue</span> <span class="pre">=</span> <span class="pre">[root]</span></code>: Initialize a queue data structure for a level-order traversal, starting with the root node.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">while</span> <span class="pre">queue:</span></code>: This starts a loop that continues until the queue is empty, indicating that all nodes have been processed.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">node</span> <span class="pre">=</span> <span class="pre">queue.pop(0)</span></code>: Dequeue the first node from the queue, effectively processing it.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">if</span> <span class="pre">node:</span></code>: Check if the <code class="docutils literal notranslate"><span class="pre">node</span></code> is not None, which means it’s a valid node in the binary tree.</p>
-<p>a. <code class="docutils literal notranslate"><span class="pre">result.append(node.val)</span></code>: If the node is valid, append its value (<code class="docutils literal notranslate"><span class="pre">node.val</span></code>) to the <code class="docutils literal notranslate"><span class="pre">result</span></code> list. This represents the value of the current node in the binary tree.</p>
-<p>b. <code class="docutils literal notranslate"><span class="pre">queue.append(node.left)</span></code>: Enqueue the left child of the current node if it exists. This adds the left child to the queue for processing in the next iteration.</p>
-<p>c. <code class="docutils literal notranslate"><span class="pre">queue.append(node.right)</span></code>: Enqueue the right child of the current node if it exists. This adds the right child to the queue for processing in the next iteration.</p>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">else:</span></code>: If the <code class="docutils literal notranslate"><span class="pre">node</span></code> is None, it represents an empty node in the binary tree.</p>
-<p>a. <code class="docutils literal notranslate"><span class="pre">result.append(None)</span></code>: Append <code class="docutils literal notranslate"><span class="pre">None</span></code> to the <code class="docutils literal notranslate"><span class="pre">result</span></code> list to indicate an empty node.</p>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">while</span> <span class="pre">result</span> <span class="pre">and</span> <span class="pre">result[-1]</span> <span class="pre">is</span> <span class="pre">None:</span></code>: After the traversal is complete, there might be trailing <code class="docutils literal notranslate"><span class="pre">None</span></code> values in the <code class="docutils literal notranslate"><span class="pre">result</span></code> list. This loop removes any such trailing <code class="docutils literal notranslate"><span class="pre">None</span></code> values to ensure a clean representation of the tree’s elements.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">return</span> <span class="pre">result</span></code>: Return the <code class="docutils literal notranslate"><span class="pre">result</span></code> list, which now contains the elements of the binary tree in a format where <code class="docutils literal notranslate"><span class="pre">None</span></code> represents empty nodes and the order of elements reflects a level-order traversal of the tree.</p></li>
-</ol>
-<p>In summary, the <code class="docutils literal notranslate"><span class="pre">treeToList</span></code> function performs a level-order traversal of the binary tree using a queue, constructing a list where each element corresponds to a node’s value or represents an empty node with <code class="docutils literal notranslate"><span class="pre">None</span></code>. This list represents the binary tree’s elements in a structured format.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Example usage:</span>
-<span class="n">preorder</span> <span class="o">=</span> <span class="p">[</span><span class="mi">3</span><span class="p">,</span> <span class="mi">9</span><span class="p">,</span> <span class="mi">20</span><span class="p">,</span> <span class="mi">15</span><span class="p">,</span> <span class="mi">7</span><span class="p">]</span>
-<span class="n">inorder</span> <span class="o">=</span> <span class="p">[</span><span class="mi">9</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">15</span><span class="p">,</span> <span class="mi">20</span><span class="p">,</span> <span class="mi">7</span><span class="p">]</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">buildTree</span><span class="p">(</span><span class="n">preorder</span><span class="p">,</span> <span class="n">inorder</span><span class="p">)</span>
-
-<span class="nb">print</span><span class="p">(</span><span class="n">treeToList</span><span class="p">(</span><span class="n">result</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[3, 9, 20, None, None, 15, 7]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-
-<span class="n">preorder</span> <span class="o">=</span> <span class="p">[</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span>
-<span class="n">inorder</span> <span class="o">=</span> <span class="p">[</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">buildTree</span><span class="p">(</span><span class="n">preorder</span><span class="p">,</span> <span class="n">inorder</span><span class="p">)</span>
-
-<span class="nb">print</span><span class="p">(</span><span class="n">treeToList</span><span class="p">(</span><span class="n">result</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[-1]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">buildTree</span></code> function:</p>
-<p><strong>Time Complexity:</strong>
-The time complexity of the <code class="docutils literal notranslate"><span class="pre">buildTree</span></code> function can be analyzed in terms of the number of nodes in the binary tree. Let’s assume there are <code class="docutils literal notranslate"><span class="pre">n</span></code> nodes in the binary tree.</p>
-<ol class="arabic simple">
-<li><p>Finding the root value in the <code class="docutils literal notranslate"><span class="pre">preorder</span></code> traversal takes O(1) time as it’s just extracting the first element from the list.</p></li>
-<li><p>Finding the index of the root value in the <code class="docutils literal notranslate"><span class="pre">inorder</span></code> traversal using <code class="docutils literal notranslate"><span class="pre">inorder.index(root_val)</span></code> takes O(n) time in the worst case because in the worst case, it might have to search through all <code class="docutils literal notranslate"><span class="pre">n</span></code> elements in the <code class="docutils literal notranslate"><span class="pre">inorder</span></code> list to find the index.</p></li>
-<li><p>The recursive calls to <code class="docutils literal notranslate"><span class="pre">buildTree</span></code> for the left and right subtrees are made once for each node in the tree. Therefore, the recursive calls have a combined time complexity of O(n) as they process each node once.</p></li>
-</ol>
-<p>The total time complexity of the <code class="docutils literal notranslate"><span class="pre">buildTree</span></code> function is O(n) due to the recursive calls and finding the root’s index in the <code class="docutils literal notranslate"><span class="pre">inorder</span></code> list.</p>
-<p><strong>Space Complexity:</strong>
-The space complexity is determined by the space used by the function’s call stack during recursion and any additional data structures used.</p>
-<ol class="arabic simple">
-<li><p>The space used by the function call stack during recursion depends on the height of the binary tree. In the worst case, where the tree is highly unbalanced (e.g., a skewed tree), the space complexity for the call stack is O(n) as it can go as deep as the number of nodes in the tree.</p></li>
-<li><p>Additionally, the function creates TreeNode objects for each node in the binary tree. Therefore, the space complexity for these objects is also O(n).</p></li>
-</ol>
-<p>Overall, the space complexity of the <code class="docutils literal notranslate"><span class="pre">buildTree</span></code> function is O(n) due to the space used by the call stack and the TreeNode objects.</p>
-<p>In summary, the time complexity of <code class="docutils literal notranslate"><span class="pre">buildTree</span></code> is O(n), and the space complexity is O(n), where ‘n’ is the number of nodes in the binary tree.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Construct Binary Tree from Postorder and Inorder Traversal:</strong>
-Modify the problem to construct a binary tree from its postorder and inorder traversals. Implement a function similar to <code class="docutils literal notranslate"><span class="pre">buildTree</span></code> but for postorder and inorder traversals.</p></li>
-<li><p><strong>Reconstruct Binary Tree with Duplicate Values:</strong>
-Extend the problem to handle binary trees with duplicate values. Ensure that your solution correctly handles scenarios where there are duplicate values in the preorder and inorder traversals.</p></li>
-</ol>
-</section>
-</section>
-
-    <script type="text/x-thebe-config">
-    {
-        requestKernel: true,
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-    <h1>124. Binary Tree Maximum Path Sum</h1>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="binary-tree-maximum-path-sum">
-<h1>124. Binary Tree Maximum Path Sum<a class="headerlink" href="#binary-tree-maximum-path-sum" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Hard</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/binary-tree-maximum-path-sum/">To see the Binary Tree Maximum Path Sum problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>A <strong>path</strong> in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence <strong>at most once</strong>. Note that the path does not need to pass through the root.</p>
-<p>The <strong>path sum</strong> of a path is the sum of the node’s values in the path.</p>
-<p>Given the <code class="docutils literal notranslate"><span class="pre">root</span></code> of a binary tree, return <em>the maximum <strong>path sum</strong> of any <strong>non-empty</strong> path</em>.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the tree is in the range <span class="math notranslate nohighlight">\([1, 3*10^4]\)</span>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-1000</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">1000</span></code></p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem is to find the maximum path sum in a binary tree. A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. The path sum of a path is the sum of the node values in that path.</p>
-<p>In this problem, you are given the root of a binary tree, and you need to find the maximum path sum among all possible paths in the tree. Note that the path does not need to start at the root node or end at a leaf node; it can start and end at any nodes in the tree.</p>
-<p>To find the maximum path sum, you need to consider both left and right subtrees of each node while traversing the tree. At each node, you have two choices:</p>
-<ol class="arabic simple">
-<li><p>Include the current node in the path: In this case, you add the value of the current node to the sum and continue exploring both the left and right subtrees for possible extensions of the path.</p></li>
-<li><p>Start a new path from the current node: In this case, you do not include the current node in the path sum, and you choose either the left subtree or the right subtree to start a new path.</p></li>
-</ol>
-<p>To solve this problem, you can use a recursive approach to traverse the binary tree. For each node, you calculate the maximum path sum that can pass through that node (option 1) and also return the maximum path sum that can be extended from that node (option 2). You keep track of the global maximum path sum encountered during the traversal.</p>
-<p>Ultimately, the maximum path sum among all paths in the tree will be the maximum of the global maximum path sum and the maximum path sum calculated for each node.</p>
-<p>The constraints for this problem include:</p>
-<ul class="simple">
-<li><p>The number of nodes in the tree is in the range <span class="math notranslate nohighlight">\([1, 3 * 10^4]\)</span>.</p></li>
-<li><p>Node values are in the range [-1000, 1000].</p></li>
-</ul>
-<p>By considering all possible paths through the tree, the algorithm aims to find the most significant sum of node values in any path in the binary tree.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">maxPathSum</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">root</span><span class="p">):</span>
-        <span class="k">def</span> <span class="nf">max_path_sum_helper</span><span class="p">(</span><span class="n">node</span><span class="p">):</span>
-            <span class="k">if</span> <span class="ow">not</span> <span class="n">node</span><span class="p">:</span>
-                <span class="k">return</span> <span class="mi">0</span>
-            
-            <span class="c1"># Recursively calculate the maximum path sum for left and right subtrees</span>
-            <span class="n">left_max</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">max_path_sum_helper</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">left</span><span class="p">))</span>
-            <span class="n">right_max</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">max_path_sum_helper</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">right</span><span class="p">))</span>
-            
-            <span class="c1"># Update the global maximum path sum</span>
-            <span class="bp">self</span><span class="o">.</span><span class="n">max_sum</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">max_sum</span><span class="p">,</span> <span class="n">left_max</span> <span class="o">+</span> <span class="n">right_max</span> <span class="o">+</span> <span class="n">node</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-            
-            <span class="c1"># Return the maximum path sum starting from the current node</span>
-            <span class="k">return</span> <span class="nb">max</span><span class="p">(</span><span class="n">left_max</span><span class="p">,</span> <span class="n">right_max</span><span class="p">)</span> <span class="o">+</span> <span class="n">node</span><span class="o">.</span><span class="n">val</span>
-        
-        <span class="bp">self</span><span class="o">.</span><span class="n">max_sum</span> <span class="o">=</span> <span class="nb">float</span><span class="p">(</span><span class="s1">&#39;-inf&#39;</span><span class="p">)</span>  <span class="c1"># Initialize the global maximum to negative infinity</span>
-        <span class="n">max_path_sum_helper</span><span class="p">(</span><span class="n">root</span><span class="p">)</span>  <span class="c1"># Start the recursive traversal from the root</span>
-        <span class="k">return</span> <span class="bp">self</span><span class="o">.</span><span class="n">max_sum</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>The code starts by defining a <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class to represent nodes in the binary tree. Each <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> has a <code class="docutils literal notranslate"><span class="pre">val</span></code> attribute representing its value, a <code class="docutils literal notranslate"><span class="pre">left</span></code> attribute pointing to the left child, and a <code class="docutils literal notranslate"><span class="pre">right</span></code> attribute pointing to the right child.</p></li>
-<li><p>Next, a <code class="docutils literal notranslate"><span class="pre">Solution</span></code> class is defined to contain the <code class="docutils literal notranslate"><span class="pre">maxPathSum</span></code> method, which will be used to find the maximum path sum in the binary tree.</p></li>
-<li><p>Within the <code class="docutils literal notranslate"><span class="pre">maxPathSum</span></code> method, a helper function called <code class="docutils literal notranslate"><span class="pre">max_path_sum_helper</span></code> is defined. This recursive function takes a single argument, <code class="docutils literal notranslate"><span class="pre">node</span></code>, which is the current node being considered during traversal.</p></li>
-<li><p>Inside the <code class="docutils literal notranslate"><span class="pre">max_path_sum_helper</span></code> function:</p>
-<ul class="simple">
-<li><p>If <code class="docutils literal notranslate"><span class="pre">node</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code> (i.e., the current node is a null node), it returns 0. This is the base case of the recursion.</p></li>
-<li><p>It calculates the maximum path sum starting from the left subtree (<code class="docutils literal notranslate"><span class="pre">left_max</span></code>) and the right subtree (<code class="docutils literal notranslate"><span class="pre">right_max</span></code>). Importantly, it uses <code class="docutils literal notranslate"><span class="pre">max(0,</span> <span class="pre">...)</span></code> to ensure that negative values (which would make the path sum smaller) are not considered. If a subtree’s path sum is negative, it’s better to not include it in the path.</p></li>
-<li><p>The code then updates the global maximum path sum (<code class="docutils literal notranslate"><span class="pre">self.max_sum</span></code>) by checking if the sum of <code class="docutils literal notranslate"><span class="pre">left_max</span></code>, <code class="docutils literal notranslate"><span class="pre">right_max</span></code>, and the current node’s value is greater than the current maximum.</p></li>
-<li><p>Finally, it returns the maximum path sum starting from the current node, which is the maximum of <code class="docutils literal notranslate"><span class="pre">left_max</span></code> and <code class="docutils literal notranslate"><span class="pre">right_max</span></code> plus the current node’s value.</p></li>
-</ul>
-</li>
-<li><p>Before calling the <code class="docutils literal notranslate"><span class="pre">max_path_sum_helper</span></code> function, the <code class="docutils literal notranslate"><span class="pre">max_sum</span></code> attribute of the <code class="docutils literal notranslate"><span class="pre">Solution</span></code> class is initialized to negative infinity (<code class="docutils literal notranslate"><span class="pre">float('-inf')</span></code>). This is done to ensure that any valid path sum encountered during traversal will be greater than the initial value of <code class="docutils literal notranslate"><span class="pre">max_sum</span></code>.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">max_path_sum_helper</span></code> function is called with the root of the binary tree, effectively starting the traversal from the root node.</p></li>
-<li><p>Once the traversal is complete, the method returns the value of <code class="docutils literal notranslate"><span class="pre">self.max_sum</span></code>, which contains the maximum path sum found in the binary tree.</p></li>
-<li><p>Example usage at the bottom of the code demonstrates how to create binary trees and use the <code class="docutils literal notranslate"><span class="pre">maxPathSum</span></code> method to find the maximum path sum for two different tree structures.</p></li>
-</ol>
-<p>In summary, the code uses a recursive approach to explore all possible paths in the binary tree, ensuring that negative path sums are not considered. It keeps track of the maximum path sum encountered and returns that maximum value as the result.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Example usage:</span>
-<span class="n">root1</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">solution</span><span class="o">.</span><span class="n">maxPathSum</span><span class="p">(</span><span class="n">root1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>6
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-
-<span class="n">root2</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="o">-</span><span class="mi">10</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">9</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">20</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">15</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">7</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">solution</span><span class="o">.</span><span class="n">maxPathSum</span><span class="p">(</span><span class="n">root2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>42
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the given code:</p>
-<p>Time Complexity:</p>
-<ul class="simple">
-<li><p>The code uses a recursive depth-first traversal to explore each node in the binary tree exactly once.</p></li>
-<li><p>During the traversal, for each node, we perform constant time operations, such as updating the <code class="docutils literal notranslate"><span class="pre">max_sum</span></code> variable and calculating the maximum path sum starting from that node.</p></li>
-<li><p>Therefore, the time complexity of the code is O(N), where N is the number of nodes in the binary tree.</p></li>
-</ul>
-<p>Space Complexity:</p>
-<ul class="simple">
-<li><p>The space complexity of the code is determined by the space used in the call stack during the recursive traversal.</p></li>
-<li><p>In the worst case, the depth of the call stack can be equal to the height of the binary tree. In an unbalanced tree, this could be O(N), but in a balanced binary tree, the height is O(log N).</p></li>
-<li><p>Additionally, the code uses a constant amount of space for variables like <code class="docutils literal notranslate"><span class="pre">left_max</span></code>, <code class="docutils literal notranslate"><span class="pre">right_max</span></code>, and <code class="docutils literal notranslate"><span class="pre">max_sum</span></code>.</p></li>
-<li><p>Therefore, the overall space complexity is O(H), where H is the height of the binary tree. In the worst case, it’s O(N), and in the best case (a balanced binary tree), it’s O(log N).</p></li>
-</ul>
-<p>In summary:</p>
-<ul class="simple">
-<li><p>Time Complexity: O(N)</p></li>
-<li><p>Space Complexity: O(H), where H is the height of the binary tree, ranging from O(log N) to O(N) depending on the tree’s balance.</p></li>
-</ul>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Print the Maximum Path:</strong>
-Modify the code to not only find the maximum path sum but also print the nodes involved in the maximum path. You can print the nodes in the correct order from the root to the leaf.</p></li>
-<li><p><strong>Path Sum Count:</strong>
-Instead of finding the maximum path sum, find the count of unique paths that sum to a given target value. Each path can start and end anywhere in the tree.</p></li>
-</ol>
-</section>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
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-<h1>199. Binary Tree Right Side View<a class="headerlink" href="#binary-tree-right-side-view" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/binary-tree-right-side-view/">To see the Binary Tree Right Side View problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given the <code class="docutils literal notranslate"><span class="pre">root</span></code> of a binary tree, imagine yourself standing on the <strong>right side</strong> of it, return <em>the values of the nodes you can see ordered from top to bottom</em>.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the tree is in the range <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">2000]</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-100</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">100</span></code></p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">def</span> <span class="nf">rightSideView</span><span class="p">(</span><span class="n">root</span><span class="p">):</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">root</span><span class="p">:</span>
-        <span class="k">return</span> <span class="p">[]</span>  <span class="c1"># If the tree is empty, return an empty list.</span>
-
-    <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>  <span class="c1"># Initialize a list to store the right side view values.</span>
-    <span class="n">queue</span> <span class="o">=</span> <span class="p">[</span><span class="n">root</span><span class="p">]</span>  <span class="c1"># Initialize a queue for level-order traversal with the root node.</span>
-
-    <span class="k">while</span> <span class="n">queue</span><span class="p">:</span>
-        <span class="c1"># Get the number of nodes at the current level.</span>
-        <span class="n">level_size</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">queue</span><span class="p">)</span>
-
-        <span class="c1"># Traverse all nodes at the current level and add the rightmost node to the result.</span>
-        <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">level_size</span><span class="p">):</span>
-            <span class="n">node</span> <span class="o">=</span> <span class="n">queue</span><span class="o">.</span><span class="n">pop</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span>  <span class="c1"># Dequeue the first node from the queue.</span>
-
-            <span class="c1"># If it&#39;s the rightmost node at this level, add its value to the result.</span>
-            <span class="k">if</span> <span class="n">i</span> <span class="o">==</span> <span class="n">level_size</span> <span class="o">-</span> <span class="mi">1</span><span class="p">:</span>
-                <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-
-            <span class="c1"># Add the children of the current node to the queue.</span>
-            <span class="k">if</span> <span class="n">node</span><span class="o">.</span><span class="n">left</span><span class="p">:</span>
-                <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">left</span><span class="p">)</span>
-            <span class="k">if</span> <span class="n">node</span><span class="o">.</span><span class="n">right</span><span class="p">:</span>
-                <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">right</span><span class="p">)</span>
-
-    <span class="k">return</span> <span class="n">result</span>  <span class="c1"># Return the list of right side view values.</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic">
-<li><p>We start with a binary tree represented using a TreeNode class. Each node has a value (<code class="docutils literal notranslate"><span class="pre">val</span></code>) and can have a left child (<code class="docutils literal notranslate"><span class="pre">left</span></code>) and a right child (<code class="docutils literal notranslate"><span class="pre">right</span></code>).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">rightSideView</span></code> function takes the root node of the binary tree as input and returns a list of values representing the nodes you can see when standing on the right side of the tree.</p></li>
-<li><p>We check if the <code class="docutils literal notranslate"><span class="pre">root</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code>, which means the tree is empty. If it’s empty, we return an empty list because there are no nodes to see.</p></li>
-<li><p>We initialize an empty list called <code class="docutils literal notranslate"><span class="pre">result</span></code> to store the right side view values.</p></li>
-<li><p>We initialize a queue called <code class="docutils literal notranslate"><span class="pre">queue</span></code> with the root node. This queue is used for a level-order traversal of the tree.</p></li>
-<li><p>We enter a while loop that continues until the <code class="docutils literal notranslate"><span class="pre">queue</span></code> is empty. Inside this loop, we perform the following steps for each level of the tree:</p>
-<p>a. We determine the number of nodes at the current level by getting the length of the <code class="docutils literal notranslate"><span class="pre">queue</span></code>. This is important for processing nodes at the same level together.</p>
-<p>b. We then iterate through the nodes at the current level using a for loop. For each node in the level, we dequeue it from the <code class="docutils literal notranslate"><span class="pre">queue</span></code> using <code class="docutils literal notranslate"><span class="pre">queue.pop(0)</span></code>.</p>
-<p>c. If the node we dequeue is the rightmost node at the current level (determined by <code class="docutils literal notranslate"><span class="pre">i</span> <span class="pre">==</span> <span class="pre">level_size</span> <span class="pre">-</span> <span class="pre">1</span></code>), we add its <code class="docutils literal notranslate"><span class="pre">val</span></code> to the <code class="docutils literal notranslate"><span class="pre">result</span></code> list. This is because, when standing on the right side of the tree, you can see the rightmost node at each level.</p>
-<p>d. We enqueue the left and right children of the current node (if they exist) to the <code class="docutils literal notranslate"><span class="pre">queue</span></code>. This ensures that we process the next level in the subsequent iterations of the while loop.</p>
-</li>
-<li><p>After processing all levels of the tree, the <code class="docutils literal notranslate"><span class="pre">result</span></code> list contains the values of the rightmost nodes in each level, ordered from top to bottom.</p></li>
-<li><p>Finally, we return the <code class="docutils literal notranslate"><span class="pre">result</span></code> list as the output of the function, which represents the right side view of the binary tree.</p></li>
-</ol>
-<p>The code effectively performs a level-order traversal of the binary tree while keeping track of the rightmost nodes at each level and adding them to the result list. This ensures that we obtain the correct order of nodes visible from the right side of the tree.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Construct the tree</span>
-<span class="n">root1</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">rightSideView</span><span class="p">(</span><span class="n">root1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[1, 3, 4]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-
-<span class="n">root2</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">rightSideView</span><span class="p">(</span><span class="n">root2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[1, 3]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3:</span>
-<span class="c1"># Creating a new tree for this example</span>
-<span class="n">root</span> <span class="o">=</span> <span class="kc">None</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">rightSideView</span><span class="p">(</span><span class="n">root</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the given code:</p>
-<p><strong>Time Complexity:</strong></p>
-<p>The time complexity of the code is O(N), where N is the number of nodes in the binary tree. Here’s why:</p>
-<ol class="arabic simple">
-<li><p>We perform a level-order traversal of the tree using a queue. In the worst case, we visit all nodes once, which takes O(N) time since we visit each node exactly once.</p></li>
-<li><p>For each node, we perform constant-time operations like dequeuing it from the queue, checking if it’s the rightmost node at the current level, and enqueuing its children (if they exist). These operations do not depend on the size of the tree, so they do not contribute to the overall time complexity.</p></li>
-</ol>
-<p>Therefore, the dominant factor in the time complexity is the level-order traversal, making it O(N).</p>
-<p><strong>Space Complexity:</strong></p>
-<p>The space complexity of the code is also O(N). Here’s why:</p>
-<ol class="arabic simple">
-<li><p>We use a queue to perform level-order traversal. In the worst case, the queue can contain all nodes at the maximum width of the tree, which can be up to N/2 nodes for a completely unbalanced tree. Therefore, the space required for the queue is O(N).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">result</span></code> list stores the values of the rightmost nodes. In the worst case, when the binary tree is a complete binary tree, it can have roughly N/2 rightmost nodes. Therefore, the space required for the <code class="docutils literal notranslate"><span class="pre">result</span></code> list is also O(N).</p></li>
-<li><p>Other auxiliary variables like <code class="docutils literal notranslate"><span class="pre">level_size</span></code>, <code class="docutils literal notranslate"><span class="pre">i</span></code>, and <code class="docutils literal notranslate"><span class="pre">node</span></code> require only constant space and do not depend on the size of the tree.</p></li>
-</ol>
-<p>Combining the space used by the queue and the <code class="docutils literal notranslate"><span class="pre">result</span></code> list, we get a space complexity of O(N).</p>
-<p>In summary, the code has a time complexity of O(N) and a space complexity of O(N), where N is the number of nodes in the binary tree. These complexities are efficient and scale linearly with the size of the input tree.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Variant with Left Side View</strong>: Modify the code to find and return the values of nodes visible from the left side of the binary tree instead of the right side.</p></li>
-<li><p><strong>Zigzag Right Side View</strong>: Extend the code to return the right side view values in a zigzag order. In a zigzag order, you alternate between starting from the rightmost node at level 0, then the leftmost at level 1, rightmost at level 2, and so on.</p></li>
-</ol>
-</section>
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-    <h1>226: Invert Binary Tree</h1>
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-                <h2> Contents </h2>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
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-<h1>226: Invert Binary Tree<a class="headerlink" href="#invert-binary-tree" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/invert-binary-tree/">To see the Invert Binary Tree problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given the root of a binary tree, invert the tree, and return its root.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the tree is in the range <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">100]</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-100</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">100</span></code></p></li>
-</ul>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Definition for a binary tree node.</span>
-<span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">def</span> <span class="nf">invertTree</span><span class="p">(</span><span class="n">root</span><span class="p">):</span>
-    <span class="c1"># Base case: If the root is None or the tree is empty, return None.</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">root</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">None</span>
-
-    <span class="c1"># Swap the left and right subtrees of the current node.</span>
-    <span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="p">,</span> <span class="n">root</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="p">,</span> <span class="n">root</span><span class="o">.</span><span class="n">left</span>
-
-    <span class="c1"># Recursively invert the left and right subtrees.</span>
-    <span class="n">invertTree</span><span class="p">(</span><span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="p">)</span>
-    <span class="n">invertTree</span><span class="p">(</span><span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="p">)</span>
-
-    <span class="k">return</span> <span class="n">root</span>
-</pre></div>
-</div>
-</div>
-</div>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We start by defining a TreeNode class, which represents a node in a binary tree. Each node has a value (<code class="docutils literal notranslate"><span class="pre">val</span></code>), a left child (<code class="docutils literal notranslate"><span class="pre">left</span></code>), and a right child (<code class="docutils literal notranslate"><span class="pre">right</span></code>).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">invertTree</span></code> function is defined to invert the binary tree. It takes the root node of the tree as an argument.</p></li>
-<li><p>In the <code class="docutils literal notranslate"><span class="pre">invertTree</span></code> function, we have a base case to handle the scenario when the root is <code class="docutils literal notranslate"><span class="pre">None</span></code> (empty tree). In such cases, we return <code class="docutils literal notranslate"><span class="pre">None</span></code> because there’s nothing to invert.</p></li>
-<li><p>For non-empty trees, we swap the left and right subtrees of the current node. This effectively inverts the tree at the current node.</p></li>
-<li><p>We then recursively call <code class="docutils literal notranslate"><span class="pre">invertTree</span></code> on the left and right subtrees to invert them.</p></li>
-<li><p>Finally, we return the root of the inverted tree.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 1</span>
-<span class="c1"># Input: root = [4,2,7,1,3,6,9]</span>
-
-<span class="n">root1</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">7</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">6</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">9</span><span class="p">)</span>
-
-<span class="n">inverted_root1</span> <span class="o">=</span> <span class="n">invertTree</span><span class="p">(</span><span class="n">root1</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="s2">&quot;Inverted Tree (Example 1):&quot;</span><span class="p">,</span> <span class="n">inverted_root1</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Inverted Tree (Example 1): &lt;__main__.TreeNode object at 0x7faa5c445ab0&gt;
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 2</span>
-<span class="c1"># Input: root = [2,1,3]</span>
-
-<span class="n">root2</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-
-<span class="n">inverted_root2</span> <span class="o">=</span> <span class="n">invertTree</span><span class="p">(</span><span class="n">root2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="s2">&quot;Inverted Tree (Example 2):&quot;</span><span class="p">,</span> <span class="n">inverted_root2</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Inverted Tree (Example 2): &lt;__main__.TreeNode object at 0x7faa5c444310&gt;
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">### Example 3</span>
-<span class="c1"># Input: root = []</span>
-
-<span class="n">root3</span> <span class="o">=</span> <span class="kc">None</span>  <span class="c1"># Empty tree</span>
-
-<span class="n">inverted_root3</span> <span class="o">=</span> <span class="n">invertTree</span><span class="p">(</span><span class="n">root3</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="s2">&quot;Inverted Tree (Example 3):&quot;</span><span class="p">,</span> <span class="n">inverted_root3</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Inverted Tree (Example 3): None
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s discuss the time and space complexity of the provided Python code to invert a binary tree.</p>
-<p><strong>Time Complexity:</strong></p>
-<p>The time complexity of the <code class="docutils literal notranslate"><span class="pre">invertTree</span></code> function is <span class="math notranslate nohighlight">\(O(n)\)</span>, where <span class="math notranslate nohighlight">\(n\)</span> is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal of the tree. In the worst case, we have to visit all nodes in the tree.</p>
-<p>The reason for this time complexity is the depth-first traversal of the tree, where we recursively process the left and right subtrees of each node.</p>
-<p><strong>Space Complexity:</strong></p>
-<p>The space complexity of the code is determined by the function call stack during the recursion. In the worst case, the space complexity is <span class="math notranslate nohighlight">\(O(h)\)</span>, where <span class="math notranslate nohighlight">\(h\)</span> is the height of the binary tree.</p>
-<p>In a completely unbalanced binary tree (essentially a linked list), the height <span class="math notranslate nohighlight">\(h\)</span> is equal to the number of nodes <span class="math notranslate nohighlight">\(n\)</span>, resulting in a space complexity of <span class="math notranslate nohighlight">\(O(n)\)</span>. This occurs when the tree is skewed to one side.</p>
-<p>In a balanced binary tree, such as a full binary tree, the height <span class="math notranslate nohighlight">\(h\)</span> is <span class="math notranslate nohighlight">\(O(log\ n)\)</span>, and the space complexity is <span class="math notranslate nohighlight">\(O(log\ n)\)</span>.</p>
-<p>The space complexity depends on how balanced the tree is. In practical scenarios, binary trees are often approximately balanced, so the space complexity is typically closer to <span class="math notranslate nohighlight">\(O(log\ n)\)</span>.</p>
-<p><strong>In summary:</strong></p>
-<ul class="simple">
-<li><p>Time Complexity: <span class="math notranslate nohighlight">\(O(n)\)</span> where <span class="math notranslate nohighlight">\(n\)</span> is the number of nodes.</p></li>
-<li><p>Space Complexity: <span class="math notranslate nohighlight">\(O(h)\)</span>, where <span class="math notranslate nohighlight">\(h\)</span> is the height of the binary tree. In the worst case, it can be <span class="math notranslate nohighlight">\(O(n)\)</span>, and in a balanced tree, it is <span class="math notranslate nohighlight">\(O(log\ n)\)</span>.</p></li>
-</ul>
-<p>Keep in mind that these complexities are based on the recursive implementation provided. Iterative solutions can achieve the same task with <span class="math notranslate nohighlight">\(O(1)\)</span> space complexity, using auxiliary data structures like stacks or queues to mimic the recursion stack.</p>
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-<h1>230. Kth Smallest Element in a BST<a class="headerlink" href="#kth-smallest-element-in-a-bst" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/kth-smallest-element-in-a-bst/">To see the Kth Smallest Element in a BST problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given the <code class="docutils literal notranslate"><span class="pre">root</span></code> of a binary search tree, and an integer <code class="docutils literal notranslate"><span class="pre">k</span></code>, return <em>the <span class="math notranslate nohighlight">\(k^{th}\)</span> smallest value (<strong>1-indexed</strong>) of all the values of the nodes in the tree.</em></p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the tree is <code class="docutils literal notranslate"><span class="pre">n</span></code>.</p></li>
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">k</span> <span class="pre">&lt;=</span> <span class="pre">n</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p>0 &lt;= <code class="docutils literal notranslate"><span class="pre">Node.val</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-</ul>
-<p><strong>Follow up</strong>: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?</p>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">def</span> <span class="nf">kthSmallest</span><span class="p">(</span><span class="n">root</span><span class="p">,</span> <span class="n">k</span><span class="p">):</span>
-    <span class="n">stack</span> <span class="o">=</span> <span class="p">[]</span>  <span class="c1"># Initialize an empty stack to simulate the traversal</span>
-    <span class="k">while</span> <span class="n">root</span> <span class="ow">or</span> <span class="n">stack</span><span class="p">:</span>
-        <span class="k">while</span> <span class="n">root</span><span class="p">:</span>
-            <span class="n">stack</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">root</span><span class="p">)</span>  <span class="c1"># Push the current node onto the stack</span>
-            <span class="n">root</span> <span class="o">=</span> <span class="n">root</span><span class="o">.</span><span class="n">left</span>  <span class="c1"># Move to the left child</span>
-        <span class="n">root</span> <span class="o">=</span> <span class="n">stack</span><span class="o">.</span><span class="n">pop</span><span class="p">()</span>  <span class="c1"># Pop a node from the stack</span>
-        <span class="n">k</span> <span class="o">-=</span> <span class="mi">1</span>  <span class="c1"># Decrement k since we&#39;ve visited a node</span>
-        <span class="k">if</span> <span class="n">k</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
-            <span class="k">return</span> <span class="n">root</span><span class="o">.</span><span class="n">val</span>  <span class="c1"># If k becomes 0, return the current node&#39;s value as the kth smallest</span>
-        <span class="n">root</span> <span class="o">=</span> <span class="n">root</span><span class="o">.</span><span class="n">right</span>  <span class="c1"># Move to the right child to continue the traversal</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>We start by defining a <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class, which represents a node in the binary search tree (BST). Each node has a value (<code class="docutils literal notranslate"><span class="pre">val</span></code>), a left child (<code class="docutils literal notranslate"><span class="pre">left</span></code>), and a right child (<code class="docutils literal notranslate"><span class="pre">right</span></code>).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">kthSmallest</span></code> function takes two parameters: <code class="docutils literal notranslate"><span class="pre">root</span></code>, which is the root of the BST, and <code class="docutils literal notranslate"><span class="pre">k</span></code>, which represents the kth smallest element we want to find.</p></li>
-<li><p>We initialize an empty stack (<code class="docutils literal notranslate"><span class="pre">stack</span></code>) to simulate the traversal of the BST. The stack will be used to keep track of nodes as we traverse the tree in an iterative manner.</p></li>
-<li><p>We enter a while loop that continues until either <code class="docutils literal notranslate"><span class="pre">root</span></code> becomes <code class="docutils literal notranslate"><span class="pre">None</span></code> (indicating we have traversed the entire tree) or the <code class="docutils literal notranslate"><span class="pre">stack</span></code> is empty.</p></li>
-<li><p>Within the loop, we start another while loop to traverse as far left as possible in the BST. We repeatedly push nodes onto the <code class="docutils literal notranslate"><span class="pre">stack</span></code> and move to their left children until we reach the leftmost leaf node. This is the smallest node in the BST.</p></li>
-<li><p>Once we have reached the leftmost leaf node (the smallest node), we pop nodes from the <code class="docutils literal notranslate"><span class="pre">stack</span></code> one by one. As we pop each node, we decrement <code class="docutils literal notranslate"><span class="pre">k</span></code> by 1 to keep track of the number of nodes we have visited.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">k</span></code> becomes 0 after decrementing, it means we have found the kth smallest element. In this case, we return the <code class="docutils literal notranslate"><span class="pre">val</span></code> of the current node as the result.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">k</span></code> is still greater than 0, it means we haven’t found the kth smallest element yet. In this case, we move to the right child of the current node to continue the traversal, as the kth smallest element, if it exists, will be in the right subtree of the current node.</p></li>
-<li><p>The process continues until we find the kth smallest element or traverse the entire tree.</p></li>
-<li><p>Finally, we return the kth smallest element found.</p></li>
-</ol>
-<p>The key idea in this code is to perform an in-order traversal of the BST while keeping track of the kth smallest element. By visiting nodes in ascending order, we can efficiently find the kth smallest element in O(h + k) time, where h is the height of the BST and k is the desired kth element. This approach is both concise and efficient for this problem.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Example usage:</span>
-<span class="c1"># Create the tree for the first example: root = [3,1,4,null,2], k = 1</span>
-<span class="n">root1</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">k1</span> <span class="o">=</span> <span class="mi">1</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">kthSmallest</span><span class="p">(</span><span class="n">root1</span><span class="p">,</span> <span class="n">k1</span><span class="p">))</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>1
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-
-<span class="c1"># Create the tree for the second example: root = [5,3,6,2,4,null,null,1], k = 3</span>
-<span class="n">root2</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">6</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">k2</span> <span class="o">=</span> <span class="mi">3</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">kthSmallest</span><span class="p">(</span><span class="n">root2</span><span class="p">,</span> <span class="n">k2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>3
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the given code:</p>
-<p><strong>Time Complexity:</strong></p>
-<p>The time complexity of this code is O(h + k), where:</p>
-<ul class="simple">
-<li><p>h is the height of the binary search tree (BST).</p></li>
-<li><p>k is the desired kth smallest element we want to find.</p></li>
-</ul>
-<ol class="arabic simple">
-<li><p>In the worst case, where the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the time complexity is O(n + k).</p></li>
-<li><p>In the best case, where the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the time complexity is O(log(n) + k).</p></li>
-</ol>
-<p>So, the time complexity can vary from O(log(n) + k) in the best-case scenario to O(n + k) in the worst-case scenario. Typically, for balanced BSTs, the time complexity is closer to O(log(n) + k).</p>
-<p><strong>Space Complexity:</strong></p>
-<p>The space complexity of this code is O(h) due to the stack used for the iterative in-order traversal, where h is the height of the BST.</p>
-<ol class="arabic simple">
-<li><p>In the worst case, when the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the space complexity is O(n) because the stack can potentially store all n nodes.</p></li>
-<li><p>In the best case, when the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the space complexity is O(log(n)) because the stack will have at most log(n) nodes.</p></li>
-</ol>
-<p>So, the space complexity depends on the height of the BST and can vary from O(log(n)) in the best-case scenario to O(n) in the worst-case scenario.</p>
-<p>In practical terms, for balanced BSTs or moderately unbalanced BSTs, the space complexity is usually close to O(log(n)), and the code is efficient in terms of space usage.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Kth Largest Element</strong>: Modify the code to find the kth largest element in the BST instead of the kth smallest element.</p></li>
-<li><p><strong>Kth Smallest Element in Two BSTs</strong>: Given two BSTs, find the kth smallest element when considering both BSTs as a single sorted list. This involves merging the two BSTs while finding the kth element efficiently.</p></li>
-</ol>
-</section>
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-<h1>235. Lowest Common Ancestor of a Binary Search Tree<a class="headerlink" href="#lowest-common-ancestor-of-a-binary-search-tree" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/">To see the Lowest Common Ancestor of a Binary Search Tree problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.</p>
-<p>According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code> as the lowest node in <code class="docutils literal notranslate"><span class="pre">T</span></code> that has both <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code> as descendants (where we allow <strong>a node to be a descendant of itself</strong>).”</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the tree is in the range <span class="math notranslate nohighlight">\([2, 10^5]\)</span>.</p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^9\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">Node.val</span></code> &lt;= <span class="math notranslate nohighlight">\(10^9\)</span></p></li>
-<li><p>All <code class="docutils literal notranslate"><span class="pre">Node.val</span></code> are <strong>unique</strong>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">p</span> <span class="pre">!=</span> <span class="pre">q</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code> will exist in the BST.</p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">def</span> <span class="nf">lowestCommonAncestor</span><span class="p">(</span><span class="n">root</span><span class="p">,</span> <span class="n">p</span><span class="p">,</span> <span class="n">q</span><span class="p">):</span>
-    <span class="c1"># Ensure p is less than q</span>
-    <span class="k">if</span> <span class="n">p</span><span class="o">.</span><span class="n">val</span> <span class="o">&gt;</span> <span class="n">q</span><span class="o">.</span><span class="n">val</span><span class="p">:</span>
-        <span class="n">p</span><span class="p">,</span> <span class="n">q</span> <span class="o">=</span> <span class="n">q</span><span class="p">,</span> <span class="n">p</span>
-
-    <span class="k">while</span> <span class="n">root</span><span class="p">:</span>
-        <span class="c1"># If the current node value is greater than both p and q, move left</span>
-        <span class="k">if</span> <span class="n">root</span><span class="o">.</span><span class="n">val</span> <span class="o">&gt;</span> <span class="n">q</span><span class="o">.</span><span class="n">val</span><span class="p">:</span>
-            <span class="n">root</span> <span class="o">=</span> <span class="n">root</span><span class="o">.</span><span class="n">left</span>
-        <span class="c1"># If the current node value is less than both p and q, move right</span>
-        <span class="k">elif</span> <span class="n">root</span><span class="o">.</span><span class="n">val</span> <span class="o">&lt;</span> <span class="n">p</span><span class="o">.</span><span class="n">val</span><span class="p">:</span>
-            <span class="n">root</span> <span class="o">=</span> <span class="n">root</span><span class="o">.</span><span class="n">right</span>
-        <span class="c1"># If the current node value is between p and q (inclusive), or it matches either p or q, it&#39;s the LCA</span>
-        <span class="k">else</span><span class="p">:</span>
-            <span class="k">return</span> <span class="n">root</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">class</span> <span class="pre">TreeNode:</span></code> defines a simple class for representing nodes in a binary tree. Each <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> has three attributes:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">val</span></code>: The value of the node.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">left</span></code>: A reference to the left child node.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">right</span></code>: A reference to the right child node.</p></li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">def</span> <span class="pre">lowestCommonAncestor(root,</span> <span class="pre">p,</span> <span class="pre">q):</span></code> is a function that takes three arguments:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">root</span></code>: The root node of the BST.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">p</span></code>: A TreeNode representing one of the nodes for which we want to find the LCA.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">q</span></code>: A TreeNode representing the other node for which we want to find the LCA.</p></li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">if</span> <span class="pre">p.val</span> <span class="pre">&gt;</span> <span class="pre">q.val:</span></code> checks if the value of <code class="docutils literal notranslate"><span class="pre">p</span></code> is greater than the value of <code class="docutils literal notranslate"><span class="pre">q</span></code>. In a BST, it’s essential to ensure that <code class="docutils literal notranslate"><span class="pre">p</span></code> represents the smaller value, and <code class="docutils literal notranslate"><span class="pre">q</span></code> represents the larger value. If <code class="docutils literal notranslate"><span class="pre">p</span></code> is greater than <code class="docutils literal notranslate"><span class="pre">q</span></code>, the code swaps their values.</p></li>
-<li><p>The main logic is inside the <code class="docutils literal notranslate"><span class="pre">while</span></code> loop, which runs until <code class="docutils literal notranslate"><span class="pre">root</span></code> becomes <code class="docutils literal notranslate"><span class="pre">None</span></code>. It performs the following steps to find the LCA:</p>
-<ul class="simple">
-<li><p>If the current <code class="docutils literal notranslate"><span class="pre">root</span></code> node’s value is greater than the value of <code class="docutils literal notranslate"><span class="pre">q</span></code>, it means both <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code> are on the left subtree of the current node. So, we move to the left child of the current node by setting <code class="docutils literal notranslate"><span class="pre">root</span> <span class="pre">=</span> <span class="pre">root.left</span></code>.</p></li>
-<li><p>If the current <code class="docutils literal notranslate"><span class="pre">root</span></code> node’s value is less than the value of <code class="docutils literal notranslate"><span class="pre">p</span></code>, it means both <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code> are on the right subtree of the current node. So, we move to the right child of the current node by setting <code class="docutils literal notranslate"><span class="pre">root</span> <span class="pre">=</span> <span class="pre">root.right</span></code>.</p></li>
-<li><p>If neither of the above conditions is met, it means the current <code class="docutils literal notranslate"><span class="pre">root</span></code> node’s value is between <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code>, or it matches either <code class="docutils literal notranslate"><span class="pre">p</span></code> or <code class="docutils literal notranslate"><span class="pre">q</span></code>. In this case, the current node is the lowest common ancestor (LCA), so we return <code class="docutils literal notranslate"><span class="pre">root</span></code>.</p></li>
-</ul>
-</li>
-</ol>
-<p>The algorithm is efficient because it takes advantage of the properties of a BST. It eliminates subtrees that cannot contain the LCA by comparing the values of <code class="docutils literal notranslate"><span class="pre">p</span></code>, <code class="docutils literal notranslate"><span class="pre">q</span></code>, and the current <code class="docutils literal notranslate"><span class="pre">root</span></code> node. Eventually, it reaches the LCA node, and that node is returned as the result.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Construct the tree</span>
-<span class="n">root</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">6</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">8</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">7</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">9</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span>
-
-<span class="n">p</span> <span class="o">=</span> <span class="n">root</span><span class="o">.</span><span class="n">left</span>  <span class="c1"># Node 2</span>
-<span class="n">q</span> <span class="o">=</span> <span class="n">root</span><span class="o">.</span><span class="n">right</span>  <span class="c1"># Node 8</span>
-
-<span class="n">result</span> <span class="o">=</span> <span class="n">lowestCommonAncestor</span><span class="p">(</span><span class="n">root</span><span class="p">,</span> <span class="n">p</span><span class="p">,</span> <span class="n">q</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>6
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-
-<span class="c1"># Using the same tree as before</span>
-<span class="n">p</span> <span class="o">=</span> <span class="n">root</span><span class="o">.</span><span class="n">left</span>  <span class="c1"># Node 2</span>
-<span class="n">q</span> <span class="o">=</span> <span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span>  <span class="c1"># Node 4</span>
-
-<span class="n">result</span> <span class="o">=</span> <span class="n">lowestCommonAncestor</span><span class="p">(</span><span class="n">root</span><span class="p">,</span> <span class="n">p</span><span class="p">,</span> <span class="n">q</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="o">.</span><span class="n">val</span><span class="p">)</span> 
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>2
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3:</span>
-<span class="c1"># Creating a new tree for this example</span>
-<span class="n">root2</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-
-<span class="n">p</span> <span class="o">=</span> <span class="n">root2</span>  <span class="c1"># Node 2</span>
-<span class="n">q</span> <span class="o">=</span> <span class="n">root2</span><span class="o">.</span><span class="n">left</span>  <span class="c1"># Node 1</span>
-
-<span class="n">result</span> <span class="o">=</span> <span class="n">lowestCommonAncestor</span><span class="p">(</span><span class="n">root2</span><span class="p">,</span> <span class="n">p</span><span class="p">,</span> <span class="n">q</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">result</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>2
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided algorithm for finding the lowest common ancestor (LCA) in a Binary Search Tree (BST).</p>
-<p><strong>Time Complexity:</strong>
-The time complexity of this algorithm is O(h), where “h” is the height of the BST. In the worst case, where the tree is completely unbalanced (essentially a linked list), the height of the tree is O(n), where “n” is the number of nodes in the tree. However, in a well-balanced BST, the height is logarithmic, which is O(log n).</p>
-<p>The reason for this time complexity is that the algorithm efficiently narrows down the search space by traversing either left or right subtrees based on the values of the target nodes <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code>. It eliminates entire subtrees that cannot contain the LCA, leading to a relatively quick search.</p>
-<p><strong>Space Complexity:</strong>
-The space complexity of the algorithm is O(1). This is because it uses a constant amount of extra space, regardless of the size of the input BST. The only variables used are <code class="docutils literal notranslate"><span class="pre">root</span></code>, <code class="docutils literal notranslate"><span class="pre">p</span></code>, and <code class="docutils literal notranslate"><span class="pre">q</span></code>, and there are no data structures like stacks or queues used for additional space. The algorithm performs a simple traversal without recursion, so it does not consume extra memory as the tree depth increases.</p>
-<p><strong>In summary</strong>, the provided algorithm for finding the LCA in a BST is both time and space-efficient. Its time complexity is O(h), where “h” is the height of the tree, and its space complexity is O(1), making it suitable for practical use even in large BSTs.</p>
-<section id="challenging-exercises">
-<h3>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h3>
-<ol class="arabic simple">
-<li><p><strong>Multiple LCAs:</strong> Extend the algorithm to find all the lowest common ancestors of two given nodes <code class="docutils literal notranslate"><span class="pre">p</span></code> and <code class="docutils literal notranslate"><span class="pre">q</span></code> in a BST. In some cases, there can be multiple LCAs.</p></li>
-<li><p><strong>LCA with k Nodes:</strong> Given a BST and k nodes, find the lowest common ancestor of these k nodes. This is an extension of the problem where you must find the LCA of more than two nodes.</p></li>
-</ol>
-</section>
-</section>
-</section>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
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-    <h1>297. Serialize and Deserialize Binary Tree</h1>
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-                <ul class="visible nav section-nav flex-column">
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="serialize-and-deserialize-binary-tree">
-<h1>297. Serialize and Deserialize Binary Tree<a class="headerlink" href="#serialize-and-deserialize-binary-tree" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Hard</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/serialize-and-deserialize-binary-tree/">To see the Serialize and Deserialize Binary Tree problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.</p>
-<p>Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.</p>
-<p><strong>Clarification:</strong> The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the tree is in the range <span class="math notranslate nohighlight">\([0, 10^4]\)</span>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">-1000</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">1000</span></code></p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>TThe problem is to design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of characters or bytes so that it can be easily stored in a file, transmitted over a network, or otherwise persisted. Deserialization is the reverse process of converting the serialized data back into the original data structure.</p>
-<p>In the context of this problem:</p>
-<ol class="arabic simple">
-<li><p><strong>Serialization</strong>: You are given a binary tree, and your task is to convert it into a string representation such that you can later recreate the same binary tree from this string. The format of serialization is flexible, but it should allow you to reconstruct the original binary tree accurately.</p></li>
-<li><p><strong>Deserialization</strong>: Given a serialized string, you need to reconstruct the binary tree it represents, ensuring that it is identical to the original tree.</p></li>
-</ol>
-<p>Here are some key points to consider in solving this problem:</p>
-<ul class="simple">
-<li><p>The input can include any valid binary tree, including trees with nodes having integer values within the range [-1000, 1000].</p></li>
-<li><p>You don’t necessarily need to follow a specific format for serialization, but it should be designed in a way that allows unambiguous deserialization.</p></li>
-<li><p>The goal is to serialize the tree structure and its values and then deserialize it back into the same structure and values.</p></li>
-</ul>
-<p>A common approach for serialization is to use a traversal method like preorder traversal, where you visit nodes in the order: root, left subtree, right subtree. This way, you can serialize the tree into a string, and during deserialization, you can reconstruct the tree by parsing the string in the same order.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Definition for a binary tree node.</span>
-<span class="k">class</span> <span class="nc">TreeNode</span><span class="p">(</span><span class="nb">object</span><span class="p">):</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">x</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">x</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="kc">None</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="kc">None</span>
-
-<span class="k">class</span> <span class="nc">Codec</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">serialize</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">root</span><span class="p">):</span>
-        <span class="n">serialized</span>  <span class="o">=</span> <span class="p">[]</span>
-
-        <span class="k">def</span> <span class="nf">dfs</span><span class="p">(</span><span class="n">node</span><span class="p">):</span>
-            <span class="k">if</span> <span class="ow">not</span> <span class="n">node</span><span class="p">:</span>
-                <span class="c1"># If the node is None, represent it as &quot;Null&quot; in the serialized string</span>
-                <span class="n">serialized</span> <span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="s2">&quot;Null&quot;</span><span class="p">)</span>
-                <span class="k">return</span>
-            <span class="c1"># Convert the node&#39;s value to a string and add it to the serialized string</span>
-            <span class="n">serialized</span> <span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="nb">str</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">val</span><span class="p">))</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">left</span><span class="p">)</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">right</span><span class="p">)</span>
-
-        <span class="n">dfs</span><span class="p">(</span><span class="n">root</span><span class="p">)</span>
-        <span class="c1"># Join the serialized values with &quot;,&quot;</span>
-        <span class="k">return</span> <span class="s2">&quot;,&quot;</span><span class="o">.</span><span class="n">join</span><span class="p">(</span><span class="n">serialized</span> <span class="p">)</span>
-
-    <span class="k">def</span> <span class="nf">deserialize</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">data</span><span class="p">):</span>
-        <span class="n">vals</span> <span class="o">=</span> <span class="n">data</span><span class="o">.</span><span class="n">split</span><span class="p">(</span><span class="s2">&quot;,&quot;</span><span class="p">)</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">i</span> <span class="o">=</span> <span class="mi">0</span>
-
-        <span class="k">def</span> <span class="nf">dfs</span><span class="p">():</span>
-            <span class="k">if</span> <span class="n">vals</span><span class="p">[</span><span class="bp">self</span><span class="o">.</span><span class="n">i</span><span class="p">]</span> <span class="o">==</span> <span class="s2">&quot;Null&quot;</span><span class="p">:</span>
-                <span class="bp">self</span><span class="o">.</span><span class="n">i</span> <span class="o">+=</span> <span class="mi">1</span>
-                <span class="k">return</span> <span class="kc">None</span>
-            <span class="c1"># Convert the value to an integer to create a new node</span>
-            <span class="n">node</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="nb">int</span><span class="p">(</span><span class="n">vals</span><span class="p">[</span><span class="bp">self</span><span class="o">.</span><span class="n">i</span><span class="p">]))</span>
-            <span class="bp">self</span><span class="o">.</span><span class="n">i</span> <span class="o">+=</span> <span class="mi">1</span>
-            <span class="n">node</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">dfs</span><span class="p">()</span>
-            <span class="n">node</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">dfs</span><span class="p">()</span>
-            <span class="k">return</span> <span class="n">node</span>
-
-        <span class="k">return</span> <span class="n">dfs</span><span class="p">()</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>Here’s a step-by-step explanation of the code:</p>
-<ol class="arabic simple">
-<li><p><strong>Definition of TreeNode</strong>:</p>
-<ul class="simple">
-<li><p>The code defines a simple <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class that represents a node in a binary tree. Each <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> has three attributes:</p>
-<ul>
-<li><p><code class="docutils literal notranslate"><span class="pre">val</span></code>: The integer value stored in the node.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">left</span></code>: A reference to the left child node (or None if there is no left child).</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">right</span></code>: A reference to the right child node (or None if there is no right child).</p></li>
-</ul>
-</li>
-<li><p>This class allows you to create binary tree nodes with integer values and left and right child references.</p></li>
-</ul>
-</li>
-<li><p><strong>Serialization (serialize method)</strong>:</p>
-<ul class="simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">serialize</span></code> method takes a <code class="docutils literal notranslate"><span class="pre">root</span></code> node as input, which is the root of the binary tree that needs to be serialized.</p></li>
-<li><p>It initializes an empty list <code class="docutils literal notranslate"><span class="pre">serialized</span></code> to store the serialized elements.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">dfs</span></code> (depth-first search) function is defined within the <code class="docutils literal notranslate"><span class="pre">serialize</span></code> method to perform a preorder traversal of the binary tree.</p></li>
-<li><p>In the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function:</p>
-<ul>
-<li><p>If the current <code class="docutils literal notranslate"><span class="pre">node</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code> (i.e., a leaf node or a child of a leaf node), it appends the string “Null” to the <code class="docutils literal notranslate"><span class="pre">serialized</span></code> list to represent the absence of a node.</p></li>
-<li><p>If the current <code class="docutils literal notranslate"><span class="pre">node</span></code> is not <code class="docutils literal notranslate"><span class="pre">None</span></code>, it appends the string representation of the <code class="docutils literal notranslate"><span class="pre">node.val</span></code> to the <code class="docutils literal notranslate"><span class="pre">serialized</span></code> list and then recursively calls <code class="docutils literal notranslate"><span class="pre">dfs</span></code> on the left and right children.</p></li>
-</ul>
-</li>
-<li><p>After the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> traversal, the <code class="docutils literal notranslate"><span class="pre">serialized</span></code> list contains the serialized binary tree elements.</p></li>
-<li><p>The method returns a string obtained by joining the elements in the <code class="docutils literal notranslate"><span class="pre">serialized</span></code> list with commas.</p></li>
-</ul>
-</li>
-<li><p><strong>Deserialization (deserialize method)</strong>:</p>
-<ul class="simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">deserialize</span></code> method takes a serialized string <code class="docutils literal notranslate"><span class="pre">data</span></code> as input, which represents a binary tree in the custom format.</p></li>
-<li><p>It splits the <code class="docutils literal notranslate"><span class="pre">data</span></code> string by commas to obtain a list of elements called <code class="docutils literal notranslate"><span class="pre">vals</span></code>.</p></li>
-<li><p>The method initializes an index <code class="docutils literal notranslate"><span class="pre">self.i</span></code> to 0. This index keeps track of the current position in the <code class="docutils literal notranslate"><span class="pre">vals</span></code> list during deserialization.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function is defined within the <code class="docutils literal notranslate"><span class="pre">deserialize</span></code> method to perform the deserialization process using a recursive approach.</p></li>
-<li><p>In the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function:</p>
-<ul>
-<li><p>If the current element in <code class="docutils literal notranslate"><span class="pre">vals</span></code> is “Null,” it means there is no node at this position in the binary tree, so it returns <code class="docutils literal notranslate"><span class="pre">None</span></code>.</p></li>
-<li><p>If the current element is not “Null,” it converts the element to an integer to create a new <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> with that value.</p></li>
-<li><p>It then recursively calls <code class="docutils literal notranslate"><span class="pre">dfs</span></code> to set the left and right children of the current node.</p></li>
-</ul>
-</li>
-<li><p>The method returns the root node of the reconstructed binary tree.</p></li>
-</ul>
-</li>
-</ol>
-<p>Overall, this code demonstrates a way to serialize a binary tree into a string format and then deserialize it back into the original tree structure, allowing for the representation of both the tree structure and the values stored in the nodes.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Example usage:</span>
-<span class="c1"># Serialize a binary tree</span>
-<span class="n">root</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span>
-
-<span class="n">codec</span> <span class="o">=</span> <span class="n">Codec</span><span class="p">()</span>
-<span class="n">serialized_tree</span> <span class="o">=</span> <span class="n">codec</span><span class="o">.</span><span class="n">serialize</span><span class="p">(</span><span class="n">root</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">serialized_tree</span><span class="p">)</span>
-
-<span class="c1"># Deserialize the serialized tree</span>
-<span class="n">new_root</span> <span class="o">=</span> <span class="n">codec</span><span class="o">.</span><span class="n">deserialize</span><span class="p">(</span><span class="n">serialized_tree</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>1,2,Null,Null,3,4,Null,Null,5,Null,Null
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-
-<span class="c1"># Example usage:</span>
-<span class="c1"># Serialize a binary tree</span>
-<span class="n">root</span> <span class="o">=</span> <span class="kc">None</span>
-
-<span class="n">codec</span> <span class="o">=</span> <span class="n">Codec</span><span class="p">()</span>
-<span class="n">serialized_tree</span> <span class="o">=</span> <span class="n">codec</span><span class="o">.</span><span class="n">serialize</span><span class="p">(</span><span class="n">root</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">serialized_tree</span><span class="p">)</span>
-
-<span class="c1"># Deserialize the serialized tree</span>
-<span class="n">new_root</span> <span class="o">=</span> <span class="n">codec</span><span class="o">.</span><span class="n">deserialize</span><span class="p">(</span><span class="n">serialized_tree</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Null
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code for serializing and deserializing a binary tree:</p>
-<ol class="arabic simple">
-<li><p><strong>Serialization (serialize method)</strong>:</p>
-<ul class="simple">
-<li><p>Time Complexity:</p>
-<ul>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">serialize</span></code> method uses a depth-first traversal (preorder) of the binary tree.</p></li>
-<li><p>It visits each node exactly once.</p></li>
-<li><p>At each node, it performs constant-time operations (string conversion and list append).</p></li>
-<li><p>Therefore, the time complexity for serialization is O(N), where N is the number of nodes in the binary tree.</p></li>
-</ul>
-</li>
-<li><p>Space Complexity:</p>
-<ul>
-<li><p>The space complexity for serialization includes the space used by the <code class="docutils literal notranslate"><span class="pre">serialized</span></code> list and the recursive call stack.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">serialized</span></code> list stores the serialized elements, and its space is proportional to the number of nodes.</p></li>
-<li><p>The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).</p></li>
-<li><p>Therefore, the space complexity for serialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p><strong>Deserialization (deserialize method)</strong>:</p>
-<ul class="simple">
-<li><p>Time Complexity:</p>
-<ul>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">deserialize</span></code> method splits the serialized string into a list of elements, which takes O(N) time.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function is a recursive function that visits each element in the list exactly once.</p></li>
-<li><p>At each element, it performs constant-time operations (string comparison, integer conversion, and recursive function calls).</p></li>
-<li><p>Therefore, the time complexity for deserialization is O(N).</p></li>
-</ul>
-</li>
-<li><p>Space Complexity:</p>
-<ul>
-<li><p>The space complexity for deserialization includes the space used by the <code class="docutils literal notranslate"><span class="pre">vals</span></code> list and the recursive call stack.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">vals</span></code> list stores the elements from the serialized string, and its space is proportional to the number of nodes.</p></li>
-<li><p>The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).</p></li>
-<li><p>Therefore, the space complexity for deserialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).</p></li>
-</ul>
-</li>
-</ul>
-</li>
-</ol>
-<p>Overall, the provided code has a time complexity of O(N) for both serialization and deserialization and a space complexity of O(N) in the majority of cases (unless the tree is severely unbalanced, in which case the height H dominates the space complexity). It efficiently serializes and deserializes a binary tree while using a linear amount of memory.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Efficient Serialization</strong>:
-Optimize the serialization process to minimize the length of the serialized string. Consider using techniques like binary encoding to represent node values more efficiently.</p></li>
-<li><p><strong>Custom Serialization Format</strong>:
-Modify the serialization and deserialization methods to use a custom format different from the one provided. Ensure that the new format allows for accurate reconstruction of the original binary tree.</p></li>
-</ol>
-</section>
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-    <h1>572: Subtree of Another Tree</h1>
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-                <h2> Contents </h2>
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-                <ul class="visible nav section-nav flex-column">
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#here-s-a-detailed-explanation-of-the-code">Here’s a detailed explanation of the code:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
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-<h1>572: Subtree of Another Tree<a class="headerlink" href="#subtree-of-another-tree" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/subtree-of-another-tree/">To see the Subtree of Another Tree problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p><strong>Problem Description:</strong></p>
-<p>Given the roots of two binary trees <code class="docutils literal notranslate"><span class="pre">root</span></code> and <code class="docutils literal notranslate"><span class="pre">subRoot</span></code>, return <code class="docutils literal notranslate"><span class="pre">True</span></code> if there is a subtree of <code class="docutils literal notranslate"><span class="pre">root</span></code> with the same structure and node values as <code class="docutils literal notranslate"><span class="pre">subRoot</span></code>, and <code class="docutils literal notranslate"><span class="pre">False</span></code> otherwise.</p>
-<p>A subtree of a binary tree <code class="docutils literal notranslate"><span class="pre">tree</span></code> is a tree that consists of a node in <code class="docutils literal notranslate"><span class="pre">tree</span></code> and all of this node’s descendants. The tree <code class="docutils literal notranslate"><span class="pre">tree</span></code> could also be considered as a subtree of itself.</p>
-<p><strong>Constraints:</strong></p>
-<ol class="arabic simple">
-<li><p>The number of nodes in the <code class="docutils literal notranslate"><span class="pre">root</span></code> tree is in the range <span class="math notranslate nohighlight">\([1, 2000]\)</span>.</p></li>
-<li><p>The number of nodes in the <code class="docutils literal notranslate"><span class="pre">subRoot</span></code> tree is in the range <span class="math notranslate nohighlight">\([1, 1000]\)</span>.</p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^4\)</span> &lt;= root.val &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^4\)</span> &lt;= subRoot.val &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-</ol>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Definition for a binary tree node.</span>
-<span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">isSubtree</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">root</span><span class="p">,</span> <span class="n">subRoot</span><span class="p">):</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">root</span><span class="p">:</span>
-            <span class="k">return</span> <span class="kc">False</span>
-        
-        <span class="c1"># Check if the current subtree is equal to the subRoot</span>
-        <span class="k">if</span> <span class="bp">self</span><span class="o">.</span><span class="n">isSameTree</span><span class="p">(</span><span class="n">root</span><span class="p">,</span> <span class="n">subRoot</span><span class="p">):</span>
-            <span class="k">return</span> <span class="kc">True</span>
-        
-        <span class="c1"># Recursively check the left and right subtrees</span>
-        <span class="k">return</span> <span class="bp">self</span><span class="o">.</span><span class="n">isSubtree</span><span class="p">(</span><span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="p">,</span> <span class="n">subRoot</span><span class="p">)</span> <span class="ow">or</span> <span class="bp">self</span><span class="o">.</span><span class="n">isSubtree</span><span class="p">(</span><span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="p">,</span> <span class="n">subRoot</span><span class="p">)</span>
-    
-    <span class="k">def</span> <span class="nf">isSameTree</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">tree1</span><span class="p">,</span> <span class="n">tree2</span><span class="p">):</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">tree1</span> <span class="ow">and</span> <span class="ow">not</span> <span class="n">tree2</span><span class="p">:</span>
-            <span class="k">return</span> <span class="kc">True</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">tree1</span> <span class="ow">or</span> <span class="ow">not</span> <span class="n">tree2</span><span class="p">:</span>
-            <span class="k">return</span> <span class="kc">False</span>
-        
-        <span class="k">return</span> <span class="p">(</span>
-            <span class="n">tree1</span><span class="o">.</span><span class="n">val</span> <span class="o">==</span> <span class="n">tree2</span><span class="o">.</span><span class="n">val</span> <span class="ow">and</span>
-            <span class="bp">self</span><span class="o">.</span><span class="n">isSameTree</span><span class="p">(</span><span class="n">tree1</span><span class="o">.</span><span class="n">left</span><span class="p">,</span> <span class="n">tree2</span><span class="o">.</span><span class="n">left</span><span class="p">)</span> <span class="ow">and</span>
-            <span class="bp">self</span><span class="o">.</span><span class="n">isSameTree</span><span class="p">(</span><span class="n">tree1</span><span class="o">.</span><span class="n">right</span><span class="p">,</span> <span class="n">tree2</span><span class="o">.</span><span class="n">right</span><span class="p">)</span>
-        <span class="p">)</span>
-</pre></div>
-</div>
-</div>
-</div>
-<p>This code defines a TreeNode class for the binary tree nodes and a Solution class with two methods:</p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">isSubtree</span></code>: This method checks if there is a subtree of the <code class="docutils literal notranslate"><span class="pre">root</span></code> tree with the same structure and node values as the <code class="docutils literal notranslate"><span class="pre">subRoot</span></code> tree. It uses a helper function <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> to compare two trees for equality.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">isSameTree</span></code>: This helper method recursively compares two trees to check if they are the same in structure and node values.</p></li>
-</ol>
-<section id="here-s-a-detailed-explanation-of-the-code">
-<h2>Here’s a detailed explanation of the code:<a class="headerlink" href="#here-s-a-detailed-explanation-of-the-code" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>TreeNode Class:</strong></p></li>
-</ol>
-<p>The <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class is defined to represent nodes in a binary tree. Each node has a <code class="docutils literal notranslate"><span class="pre">val</span></code> (the node’s value) and may have a <code class="docutils literal notranslate"><span class="pre">left</span></code> and <code class="docutils literal notranslate"><span class="pre">right</span></code> child.</p>
-<ol class="arabic simple" start="2">
-<li><p><strong>Solution Class:</strong></p>
-<ul class="simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">Solution</span></code> class contains the solution for the problem and defines two important methods:</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">isSubtree(self,</span> <span class="pre">root,</span> <span class="pre">subRoot)</span></code>:</p>
-<ul>
-<li><p>This method checks whether <code class="docutils literal notranslate"><span class="pre">subRoot</span></code> is a subtree of <code class="docutils literal notranslate"><span class="pre">root</span></code>. It takes two tree nodes, <code class="docutils literal notranslate"><span class="pre">root</span></code> and <code class="docutils literal notranslate"><span class="pre">subRoot</span></code>, as input arguments.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">root</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code>, it returns <code class="docutils literal notranslate"><span class="pre">False</span></code> because there is no subtree to search.</p></li>
-<li><p>It then checks if the current subtree with <code class="docutils literal notranslate"><span class="pre">root</span></code> as its root is equal to <code class="docutils literal notranslate"><span class="pre">subRoot</span></code> using the <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> method. If they are the same, it returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>If the current subtree is not the same as <code class="docutils literal notranslate"><span class="pre">subRoot</span></code>, it recursively checks the left and right subtrees of <code class="docutils literal notranslate"><span class="pre">root</span></code> to see if <code class="docutils literal notranslate"><span class="pre">subRoot</span></code> is a subtree of any of them.</p></li>
-<li><p>It returns <code class="docutils literal notranslate"><span class="pre">True</span></code> if <code class="docutils literal notranslate"><span class="pre">subRoot</span></code> is found in either the left or right subtree; otherwise, it returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">isSameTree(self,</span> <span class="pre">tree1,</span> <span class="pre">tree2)</span></code>:</p>
-<ul>
-<li><p>This method checks whether two trees, <code class="docutils literal notranslate"><span class="pre">tree1</span></code> and <code class="docutils literal notranslate"><span class="pre">tree2</span></code>, are the same.</p></li>
-<li><p>If both <code class="docutils literal notranslate"><span class="pre">tree1</span></code> and <code class="docutils literal notranslate"><span class="pre">tree2</span></code> are <code class="docutils literal notranslate"><span class="pre">None</span></code>, they are considered the same tree, so it returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>If only one of them is <code class="docutils literal notranslate"><span class="pre">None</span></code> (but not both), they are different trees, so it returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-<li><p>If both <code class="docutils literal notranslate"><span class="pre">tree1</span></code> and <code class="docutils literal notranslate"><span class="pre">tree2</span></code> have values, it checks if their values are equal and recursively checks if their left and right subtrees are the same.</p></li>
-<li><p>It returns <code class="docutils literal notranslate"><span class="pre">True</span></code> if the trees are the same; otherwise, it returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-</ol>
-<p>The code effectively uses recursion to traverse the binary trees and check for subtree equality. The <code class="docutils literal notranslate"><span class="pre">isSubtree</span></code> method starts the recursive search, and the <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> method is used to compare individual subtrees. The approach is efficient and avoids unnecessary checks when possible.</p>
-<p>This solution allows you to determine if there exists a subtree within the <code class="docutils literal notranslate"><span class="pre">root</span></code> tree that matches the structure and node values of <code class="docutils literal notranslate"><span class="pre">subRoot</span></code>.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">#root = [3,4,5,1,2]</span>
-<span class="c1">#subRoot = [4,1,2]</span>
-
-<span class="c1"># Example input</span>
-<span class="n">root</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-
-<span class="n">subRoot</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="n">subRoot</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">subRoot</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-
-<span class="c1"># Create an instance of the Solution class</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-
-<span class="c1"># Test the isSubtree method</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">isSubtree</span><span class="p">(</span><span class="n">root</span><span class="p">,</span> <span class="n">subRoot</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="s2">&quot;Is subRoot a subtree of root?&quot;</span><span class="p">,</span> <span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Is subRoot a subtree of root? True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1">#root = [3,4,5,1,2,null,null,null,null,0] </span>
-<span class="c1">#subRoot = [4,1,2]</span>
-
-
-<span class="c1"># Example input</span>
-<span class="n">root</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span>
-
-<span class="n">subRoot</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="n">subRoot</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">subRoot</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-
-<span class="c1"># Create an instance of the Solution class</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-
-<span class="c1"># Test the isSubtree method</span>
-<span class="n">result</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">isSubtree</span><span class="p">(</span><span class="n">root</span><span class="p">,</span> <span class="n">subRoot</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="s2">&quot;Is subRoot a subtree of root?&quot;</span><span class="p">,</span> <span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Is subRoot a subtree of root? False
-</pre></div>
-</div>
-</div>
-</div>
-<p>Let’s analyze the time and space complexity of the provided code for the “Subtree of Another Tree” problem:</p>
-<p><strong>Time Complexity</strong></p>
-<p>The time complexity of the code primarily depends on the recursive traversal of the binary tree. In both the <code class="docutils literal notranslate"><span class="pre">isSubtree</span></code> and <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> functions, we visit each node in the binary trees once. Let’s break down the time complexity:</p>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">isSubtree</span></code> function:</p>
-<ul class="simple">
-<li><p>In the worst case, we need to visit every node in the <code class="docutils literal notranslate"><span class="pre">root</span></code> tree.</p></li>
-<li><p>For each node in the <code class="docutils literal notranslate"><span class="pre">root</span></code> tree, we call the <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function, which has its own traversal.</p></li>
-<li><p>So, the total time complexity is <span class="math notranslate nohighlight">\(O(n * m)\)</span>, where <span class="math notranslate nohighlight">\(n\)</span> is the number of nodes in the <code class="docutils literal notranslate"><span class="pre">root</span></code> tree, and <span class="math notranslate nohighlight">\(m\)</span> is the number of nodes in the <code class="docutils literal notranslate"><span class="pre">subRoot</span></code> tree.</p></li>
-</ul>
-</li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function:</p>
-<ul class="simple">
-<li><p>In the worst case, we visit every node in both <code class="docutils literal notranslate"><span class="pre">tree1</span></code> and <code class="docutils literal notranslate"><span class="pre">tree2</span></code>.</p></li>
-<li><p>The number of recursive calls made is proportional to the number of nodes in the trees.</p></li>
-<li><p>So, the time complexity of this function is <span class="math notranslate nohighlight">\(O(max(n, m))\)</span>, where <span class="math notranslate nohighlight">\(n\)</span> and <span class="math notranslate nohighlight">\(m\)</span> are the numbers of nodes in <code class="docutils literal notranslate"><span class="pre">tree1</span></code> and <code class="docutils literal notranslate"><span class="pre">tree2</span></code>, respectively.</p></li>
-</ul>
-</li>
-</ol>
-<p>Overall, the time complexity of the entire code is <span class="math notranslate nohighlight">\(O(n * m)\)</span>, where <span class="math notranslate nohighlight">\(n\)</span> is the number of nodes in the <code class="docutils literal notranslate"><span class="pre">root</span></code> tree, and <span class="math notranslate nohighlight">\(m\)</span> is the number of nodes in the <code class="docutils literal notranslate"><span class="pre">subRoot</span></code> tree. In practice, it may be less than <span class="math notranslate nohighlight">\(O(n * m)\)</span> if a subtree mismatch is detected early during the traversal.</p>
-<p><strong>Space Complexity</strong></p>
-<p>The space complexity of the code is determined by the function call stack during recursion and the space used by the recursive functions. Let’s analyze the space complexity:</p>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">isSubtree</span></code> function:</p>
-<ul class="simple">
-<li><p>It uses the call stack for recursion.</p></li>
-<li><p>The maximum depth of the recursion is equal to the height of the <code class="docutils literal notranslate"><span class="pre">root</span></code> tree, which can be <span class="math notranslate nohighlight">\(O(n)\)</span> in the worst case (unbalanced tree).</p></li>
-<li><p>Additionally, the function doesn’t use any significant extra space other than the recursion stack.</p></li>
-</ul>
-</li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">isSameTree</span></code> function:</p>
-<ul class="simple">
-<li><p>It also uses the call stack for recursion.</p></li>
-<li><p>The maximum depth of the recursion is equal to the height of the <code class="docutils literal notranslate"><span class="pre">tree1</span></code> or <code class="docutils literal notranslate"><span class="pre">tree2</span></code>, whichever is greater.</p></li>
-<li><p>So, the maximum space used for the call stack is <span class="math notranslate nohighlight">\(O(max(n, m))\)</span>.</p></li>
-</ul>
-</li>
-</ol>
-<p>In summary, the space complexity of the code is <span class="math notranslate nohighlight">\(O(max(n, m))\)</span> due to the function call stack. It scales with the maximum height of the trees being compared.</p>
-<p>Overall, the code is efficient and works well for trees with moderate sizes. However, it’s important to keep in mind that the worst-case time complexity is <span class="math notranslate nohighlight">\(O(n * m)\)</span>, so for very large trees, the performance may degrade.</p>
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-    <h1>98. Validate Binary Search Tree</h1>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
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-<h1>98. Validate Binary Search Tree<a class="headerlink" href="#validate-binary-search-tree" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/validate-binary-search-tree/">To see the Validate Binary Search Tree problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given the <code class="docutils literal notranslate"><span class="pre">root</span></code> of a binary tree, <em>determine if it is a valid binary search tree (BST)</em>.</p>
-<p>A <strong>valid BST</strong> is defined as follows:</p>
-<ul class="simple">
-<li><p>The left subtree of a node contains only nodes with keys <strong>less than</strong> the node’s key.</p></li>
-<li><p>The right subtree of a node contains only nodes with keys <strong>greater than</strong> the node’s key.</p></li>
-<li><p>Both the left and right subtrees must also be binary search trees.</p></li>
-</ul>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the tree is in the range <span class="math notranslate nohighlight">\([1, 10^4]\)</span>.</p></li>
-<li><p><span class="math notranslate nohighlight">\(-2^{31}\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">Node.val</span></code> &lt;= <span class="math notranslate nohighlight">\(2^{31} - 1\)</span></p></li>
-</ul>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to solve this problem:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">TreeNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">left</span><span class="o">=</span><span class="kc">None</span><span class="p">,</span> <span class="n">right</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">left</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">right</span>
-
-<span class="k">def</span> <span class="nf">is_valid_BST</span><span class="p">(</span><span class="n">root</span><span class="p">):</span>
-    <span class="k">def</span> <span class="nf">is_valid</span><span class="p">(</span><span class="n">node</span><span class="p">,</span> <span class="n">min_val</span><span class="p">,</span> <span class="n">max_val</span><span class="p">):</span>
-        <span class="c1"># Base case: If the node is None, it&#39;s a valid BST.</span>
-        <span class="k">if</span> <span class="n">node</span> <span class="ow">is</span> <span class="kc">None</span><span class="p">:</span>
-            <span class="k">return</span> <span class="kc">True</span>
-        
-        <span class="c1"># Check if the current node&#39;s value is within the valid range.</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="p">(</span><span class="n">min_val</span> <span class="o">&lt;</span> <span class="n">node</span><span class="o">.</span><span class="n">val</span> <span class="o">&lt;</span> <span class="n">max_val</span><span class="p">):</span>
-            <span class="k">return</span> <span class="kc">False</span>
-        
-        <span class="c1"># Recursively check the left and right subtrees with updated ranges.</span>
-        <span class="k">return</span> <span class="p">(</span><span class="n">is_valid</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">left</span><span class="p">,</span> <span class="n">min_val</span><span class="p">,</span> <span class="n">node</span><span class="o">.</span><span class="n">val</span><span class="p">)</span> <span class="ow">and</span>
-                <span class="n">is_valid</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">right</span><span class="p">,</span> <span class="n">node</span><span class="o">.</span><span class="n">val</span><span class="p">,</span> <span class="n">max_val</span><span class="p">))</span>
-    
-    <span class="c1"># Call the helper function starting with a wide range for root node.</span>
-    <span class="k">return</span> <span class="n">is_valid</span><span class="p">(</span><span class="n">root</span><span class="p">,</span> <span class="nb">float</span><span class="p">(</span><span class="s1">&#39;-inf&#39;</span><span class="p">),</span> <span class="nb">float</span><span class="p">(</span><span class="s1">&#39;inf&#39;</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p>It defines a <code class="docutils literal notranslate"><span class="pre">TreeNode</span></code> class to represent nodes in the binary tree.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">is_valid_BST</span></code> function takes the root node of the binary tree as input and returns <code class="docutils literal notranslate"><span class="pre">True</span></code> if the tree is a valid BST, and <code class="docutils literal notranslate"><span class="pre">False</span></code> otherwise.</p></li>
-<li><p>Inside the <code class="docutils literal notranslate"><span class="pre">is_valid_BST</span></code> function, there is a helper function called <code class="docutils literal notranslate"><span class="pre">is_valid</span></code> that performs the actual validation using a recursive approach.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">is_valid</span></code> function checks each node in the tree to ensure that it satisfies the properties of a BST:</p>
-<ul>
-<li><p>The value of the current node must be within a valid range defined by <code class="docutils literal notranslate"><span class="pre">min_val</span></code> and <code class="docutils literal notranslate"><span class="pre">max_val</span></code>.</p></li>
-<li><p>The left subtree of the current node should contain values less than the current node’s value.</p></li>
-<li><p>The right subtree of the current node should contain values greater than the current node’s value.</p></li>
-</ul>
-</li>
-<li><p>If any of these conditions are violated, the function returns <code class="docutils literal notranslate"><span class="pre">False</span></code>, indicating that the tree is not a valid BST.</p></li>
-<li><p>If all nodes satisfy these conditions, the function returns <code class="docutils literal notranslate"><span class="pre">True</span></code>, indicating that the tree is a valid BST.</p></li>
-<li><p>The code calls the <code class="docutils literal notranslate"><span class="pre">is_valid</span></code> function with the root node and initial range values of negative infinity to positive infinity to start the validation.</p></li>
-<li><p>The time complexity of the code is O(N), where N is the number of nodes in the tree, as it traverses each node once.</p></li>
-<li><p>The space complexity depends on the height of the tree. In the average case for a balanced BST, the space complexity is O(log N), but in the worst case (skewed tree), it can be O(N) due to the recursive call stack.</p></li>
-</ul>
-<p>In summary, this code checks whether a given binary tree is a valid BST by recursively validating each node’s value and its left and right subtrees while maintaining valid value ranges. If all nodes satisfy the BST properties, the tree is considered a valid BST.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Example usage:</span>
-<span class="c1"># Create the tree for the first example: [2, 1, 3]</span>
-<span class="n">root1</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root1</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">is_valid_BST</span><span class="p">(</span><span class="n">root1</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-
-<span class="c1"># Create the tree for the second example: [5, 1, 4, None, None, 3, 6]</span>
-<span class="n">root2</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">5</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">4</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="n">root2</span><span class="o">.</span><span class="n">right</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="mi">6</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">is_valid_BST</span><span class="p">(</span><span class="n">root2</span><span class="p">))</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">is_valid_BST</span></code> function:</p>
-<p><strong>Time Complexity:</strong></p>
-<ul class="simple">
-<li><p>The time complexity of the function primarily depends on the number of nodes in the binary tree.</p></li>
-<li><p>In the worst case, we may have to visit every node in the tree once to validate whether it’s part of a valid BST.</p></li>
-<li><p>Since we are performing a depth-first search (DFS) traversal of the tree, the time complexity is O(N), where N is the number of nodes in the tree.</p></li>
-</ul>
-<p><strong>Space Complexity:</strong></p>
-<ul class="simple">
-<li><p>The space complexity is determined by the space used by the recursive call stack during the DFS traversal.</p></li>
-<li><p>In the worst case, if the tree is completely unbalanced (a skewed tree), the maximum depth of the call stack would be equal to the height of the tree.</p></li>
-<li><p>In a balanced BST, the height is approximately log2(N), where N is the number of nodes.</p></li>
-<li><p>Therefore, the space complexity of the call stack is O(log N) for a balanced BST.</p></li>
-<li><p>In the worst case (skewed tree), the space complexity can be O(N) as the height of the tree can be equal to N.</p></li>
-</ul>
-<p><strong>Overall:</strong></p>
-<ul class="simple">
-<li><p>Time Complexity: O(N)</p></li>
-<li><p>Space Complexity: O(log N) on average for a balanced BST, and O(N) in the worst case for a skewed tree.</p></li>
-</ul>
-<p>The space complexity is typically dominated by the recursive call stack, and it varies depending on the shape of the binary tree. In practice, for balanced binary trees, the space complexity is often close to O(log N), which is quite efficient.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Complex Constraints</strong>: Modify the problem to have more complex constraints on the tree. For example, consider a binary tree with constraints like “The left subtree of a node contains only nodes with keys less than or equal to the node’s key,” and “The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.” How would you adapt the code to handle these constraints?</p></li>
-<li><p><strong>Handling Duplicates</strong>: Modify the code to handle binary search trees that allow duplicate values. In a standard BST, each value is unique, but in this case, multiple nodes can have the same value. The tree should still be considered valid if it follows the BST property.</p></li>
-</ol>
-</section>
-</section>
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-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/invert-binary-tree/">226. Invert Binary Tree</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/maximum-depth-of-binary-tree/">104. Maximum Depth of Binary Tree</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/same-tree/">100. Same Tree</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/subtree-of-another-tree/">572. Subtree of Another Tree</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/">235. Lowest Common Ancestor of a Binary Search Tree</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/binary-tree-level-order-traversal/">102. Binary Tree Level Order Traversal</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/validate-binary-search-tree/">98. Validate Binary Search Tree</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/kth-smallest-element-in-a-bst/">230. Kth Smallest Element In a BST</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/">105. Construct Binary Tree From Preorder And Inorder Traversal</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/binary-tree-maximum-path-sum/">124. Binary Tree Maximum Path Sum</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/serialize-and-deserialize-binary-tree/">297. Serialize And Deserialize Binary Tree</a></p></td>
-<td><p>Hard</p></td>
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-</tbody>
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-<h2>Contribute<a class="headerlink" href="#contribute" title="Permalink to this heading">#</a></h2>
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-<li><p><a class="reference external" href="https://leetcode.com/discuss/">LeetCode Discuss</a>: LeetCode’s discussion forum for each problem. You can find helpful hints and discussions here.</p></li>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
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-<h1>208. Implement Trie (Prefix Tree)<a class="headerlink" href="#implement-trie-prefix-tree" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/implement-trie-prefix-tree/">To see the Implement Trie problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>A <strong>trie</strong> (pronounced as “try”) or <strong>prefix tree</strong> is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.</p>
-<p>Implement the Trie class:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">Trie()</span></code> Initializes the trie object.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">void</span> <span class="pre">insert(String</span> <span class="pre">word)</span></code> Inserts the string <code class="docutils literal notranslate"><span class="pre">word</span></code> into the trie.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">boolean</span> <span class="pre">search(String</span> <span class="pre">word)</span></code> Returns <code class="docutils literal notranslate"><span class="pre">true</span></code> if the string <code class="docutils literal notranslate"><span class="pre">word</span></code> is in the trie (i.e., was inserted before), and <code class="docutils literal notranslate"><span class="pre">false</span></code> otherwise.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">boolean</span> <span class="pre">startsWith(String</span> <span class="pre">prefix)</span></code> Returns <code class="docutils literal notranslate"><span class="pre">true</span></code> if there is a previously inserted string <code class="docutils literal notranslate"><span class="pre">word</span></code> that has the prefix <code class="docutils literal notranslate"><span class="pre">prefix</span></code>, and <code class="docutils literal notranslate"><span class="pre">false</span></code> otherwise.</p></li>
-</ul>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">word.length,</span> <span class="pre">prefix.length</span> <span class="pre">&lt;=</span> <span class="pre">2000</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">word</span></code> and <code class="docutils literal notranslate"><span class="pre">prefix</span></code> consist only of lowercase English letters.</p></li>
-<li><p>At most <span class="math notranslate nohighlight">\(3 * 10^4\)</span> calls in total will be made to <code class="docutils literal notranslate"><span class="pre">insert</span></code>, <code class="docutils literal notranslate"><span class="pre">search</span></code>, and <code class="docutils literal notranslate"><span class="pre">startsWith</span></code>.</p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem at hand is to implement a data structure known as a Trie (pronounced “try”) or Prefix Tree. A Trie is a tree-like data structure that is used to efficiently store and retrieve a set of strings or words. It’s particularly useful for tasks involving string manipulation, such as autocomplete and spell-checking. The problem statement defines the following operations for implementing the Trie class:</p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">Trie()</span></code>: This is the constructor that initializes the Trie object.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">void</span> <span class="pre">insert(String</span> <span class="pre">word)</span></code>: This method inserts the given string <code class="docutils literal notranslate"><span class="pre">word</span></code> into the Trie. It effectively adds the characters of the word to the Trie’s structure, creating nodes for each character if they don’t already exist. At the end of the word, a special flag is set to indicate that this node represents the end of a valid word.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">boolean</span> <span class="pre">search(String</span> <span class="pre">word)</span></code>: This method checks if the given string <code class="docutils literal notranslate"><span class="pre">word</span></code> is present in the Trie. It starts from the root of the Trie and traverses the Trie by following the characters in the word. If it successfully traverses the Trie and reaches the end of the word, it returns <code class="docutils literal notranslate"><span class="pre">true</span></code>, indicating that the word exists in the Trie; otherwise, it returns <code class="docutils literal notranslate"><span class="pre">false</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">boolean</span> <span class="pre">startsWith(String</span> <span class="pre">prefix)</span></code>: This method checks if there is any previously inserted word in the Trie that has the given <code class="docutils literal notranslate"><span class="pre">prefix</span></code>. It’s similar to the <code class="docutils literal notranslate"><span class="pre">search</span></code> method but does not require that the prefix be a complete word. If there is any word in the Trie that starts with the given prefix, it returns <code class="docutils literal notranslate"><span class="pre">true</span></code>; otherwise, it returns <code class="docutils literal notranslate"><span class="pre">false</span></code>.</p></li>
-</ol>
-<p>The problem also provides an example scenario where these operations are called, demonstrating the expected output for each operation.</p>
-<p>In summary, the goal is to create a data structure (Trie) that efficiently stores a set of strings and provides methods to insert new strings, search for complete words, and check if a given prefix exists in the stored set of strings.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">TrieNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">children</span> <span class="o">=</span> <span class="p">{}</span>           <span class="c1"># A dictionary to store child nodes</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">is_end_of_word</span> <span class="o">=</span> <span class="kc">False</span>  <span class="c1"># Indicates if a word ends at this node</span>
-
-<span class="k">class</span> <span class="nc">Trie</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">root</span> <span class="o">=</span> <span class="n">TrieNode</span><span class="p">()</span>       <span class="c1"># Initialize the Trie with a root node</span>
-
-    <span class="k">def</span> <span class="nf">insert</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">word</span><span class="p">):</span>
-        <span class="n">node</span> <span class="o">=</span> <span class="bp">self</span><span class="o">.</span><span class="n">root</span>             <span class="c1"># Start from the root</span>
-        <span class="k">for</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">word</span><span class="p">:</span>
-            <span class="k">if</span> <span class="n">char</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">:</span>
-                <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">[</span><span class="n">char</span><span class="p">]</span> <span class="o">=</span> <span class="n">TrieNode</span><span class="p">()</span>  <span class="c1"># Create a new node if the character is not present</span>
-            <span class="n">node</span> <span class="o">=</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">[</span><span class="n">char</span><span class="p">]</span>           <span class="c1"># Move to the child node</span>
-        <span class="n">node</span><span class="o">.</span><span class="n">is_end_of_word</span> <span class="o">=</span> <span class="kc">True</span>                <span class="c1"># Mark the end of the inserted word</span>
-
-    <span class="k">def</span> <span class="nf">search</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">word</span><span class="p">):</span>
-        <span class="n">node</span> <span class="o">=</span> <span class="bp">self</span><span class="o">.</span><span class="n">root</span>             <span class="c1"># Start from the root</span>
-        <span class="k">for</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">word</span><span class="p">:</span>
-            <span class="k">if</span> <span class="n">char</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">:</span>
-                <span class="k">return</span> <span class="kc">False</span>         <span class="c1"># If the character is not found, the word is not in the Trie</span>
-            <span class="n">node</span> <span class="o">=</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">[</span><span class="n">char</span><span class="p">]</span>  <span class="c1"># Move to the child node</span>
-        <span class="k">return</span> <span class="n">node</span><span class="o">.</span><span class="n">is_end_of_word</span>    <span class="c1"># Check if the node represents the end of a valid word</span>
-
-    <span class="k">def</span> <span class="nf">startsWith</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">prefix</span><span class="p">):</span>
-        <span class="n">node</span> <span class="o">=</span> <span class="bp">self</span><span class="o">.</span><span class="n">root</span>             <span class="c1"># Start from the root</span>
-        <span class="k">for</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">prefix</span><span class="p">:</span>
-            <span class="k">if</span> <span class="n">char</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">:</span>
-                <span class="k">return</span> <span class="kc">False</span>         <span class="c1"># If the character is not found, no word starts with the prefix</span>
-            <span class="n">node</span> <span class="o">=</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">[</span><span class="n">char</span><span class="p">]</span>  <span class="c1"># Move to the child node</span>
-        <span class="k">return</span> <span class="kc">True</span>                  <span class="c1"># Prefix found in the Trie</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>Let’s explain how the code works step by step:</p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">TrieNode</span></code> Class:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">TrieNode</span></code> is a class that represents nodes in the Trie. Each node has two attributes:</p>
-<ul>
-<li><p><code class="docutils literal notranslate"><span class="pre">children</span></code>: This is a dictionary that stores child nodes. Each key in the dictionary represents a character, and the corresponding value is the child node for that character.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">is_end_of_word</span></code>: This boolean flag indicates whether the node represents the end of a complete word. Initially, it’s set to <code class="docutils literal notranslate"><span class="pre">False</span></code> for all nodes.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">Trie</span></code> Class:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">Trie</span></code> is the main class that implements the Trie data structure. It has the following methods:</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">__init__(self)</span></code>: The constructor initializes the Trie object by creating a root node, which serves as the starting point for all Trie operations.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">insert(self,</span> <span class="pre">word)</span></code>: This method inserts a string <code class="docutils literal notranslate"><span class="pre">word</span></code> into the Trie. It starts from the root node and iterates through each character in the word.</p>
-<ul>
-<li><p>If a character is not present as a child node, it creates a new node for that character.</p></li>
-<li><p>It then moves to the child node and continues the process until the entire word is inserted.</p></li>
-<li><p>Finally, it sets the <code class="docutils literal notranslate"><span class="pre">is_end_of_word</span></code> flag to <code class="docutils literal notranslate"><span class="pre">True</span></code> for the last node to mark the end of the inserted word.</p></li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">search(self,</span> <span class="pre">word)</span></code>: This method checks if a complete word (string) exists in the Trie. It starts from the root node and iterates through each character in the word.</p>
-<ul>
-<li><p>If a character is not found as a child node, it immediately returns <code class="docutils literal notranslate"><span class="pre">False</span></code> because the word is not in the Trie.</p></li>
-<li><p>It continues to move to the child node for each character.</p></li>
-<li><p>After reaching the end of the word, it checks if the <code class="docutils literal notranslate"><span class="pre">is_end_of_word</span></code> flag is <code class="docutils literal notranslate"><span class="pre">True</span></code> for the last node to confirm the presence of the word.</p></li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">startsWith(self,</span> <span class="pre">prefix)</span></code>: This method checks if there is any previously inserted word in the Trie that starts with a given <code class="docutils literal notranslate"><span class="pre">prefix</span></code>. It follows the same logic as the <code class="docutils literal notranslate"><span class="pre">search</span></code> method but does not require the entire word to be present.</p>
-<ul>
-<li><p>If the prefix is found in the Trie, it returns <code class="docutils literal notranslate"><span class="pre">True</span></code>; otherwise, it returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p>Example Usage:</p>
-<ul class="simple">
-<li><p>The code demonstrates how to create a Trie object, insert words (“apple” and “app”), and perform operations on the Trie:</p>
-<ul>
-<li><p>It inserts the word “apple” into the Trie.</p></li>
-<li><p>It checks if “apple” is in the Trie, which returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>It checks if “app” is in the Trie, which returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-<li><p>It checks if there is any word in the Trie that starts with “app,” which returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>It inserts the word “app” into the Trie.</p></li>
-<li><p>It checks if “app” is in the Trie again, which now returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-</ol>
-<p>In summary, the code efficiently implements a Trie data structure with the ability to insert words, search for complete words, and check for the existence of words with a given prefix. The Trie is organized as a tree of nodes, with each node representing a character in the words being stored.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Example usage:</span>
-
-<span class="n">trie</span> <span class="o">=</span> <span class="n">Trie</span><span class="p">()</span>
-<span class="n">trie</span><span class="o">.</span><span class="n">insert</span><span class="p">(</span><span class="s2">&quot;apple&quot;</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">trie</span><span class="o">.</span><span class="n">search</span><span class="p">(</span><span class="s2">&quot;apple&quot;</span><span class="p">))</span>   <span class="c1"># Output: True</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">trie</span><span class="o">.</span><span class="n">search</span><span class="p">(</span><span class="s2">&quot;app&quot;</span><span class="p">))</span>     <span class="c1"># Output: False</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">trie</span><span class="o">.</span><span class="n">startsWith</span><span class="p">(</span><span class="s2">&quot;app&quot;</span><span class="p">))</span> <span class="c1"># Output: True</span>
-<span class="n">trie</span><span class="o">.</span><span class="n">insert</span><span class="p">(</span><span class="s2">&quot;app&quot;</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">trie</span><span class="o">.</span><span class="n">search</span><span class="p">(</span><span class="s2">&quot;app&quot;</span><span class="p">))</span>     <span class="c1"># Output: True</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-False
-True
-True
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the Trie implementation:</p>
-<ol class="arabic">
-<li><p>Time Complexity:</p>
-<ul class="simple">
-<li><p>Insertion (<code class="docutils literal notranslate"><span class="pre">insert</span></code> method): The time complexity of inserting a word into the Trie is O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to insert it into the Trie.</p></li>
-<li><p>Search (<code class="docutils literal notranslate"><span class="pre">search</span></code> method): The time complexity of searching for a word in the Trie is also O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to determine if it exists in the Trie.</p></li>
-<li><p>Starts with (<code class="docutils literal notranslate"><span class="pre">startsWith</span></code> method): The time complexity of checking if a prefix exists in the Trie is O(p), where p is the length of the prefix. In the worst case, you may need to traverse the entire prefix to determine its presence.</p></li>
-</ul>
-<p>Therefore, for each operation (insertion, search, or starts with), the time complexity is O(m) or O(p), where m is the length of the word being operated on, and p is the length of the prefix.</p>
-</li>
-<li><p>Space Complexity:</p>
-<ul class="simple">
-<li><p>The space complexity of the Trie is determined by the number of nodes and characters stored in it. In the worst case, where none of the inserted words share common prefixes, the space complexity is O(N), where N is the total number of characters across all inserted words.</p></li>
-<li><p>Each node in the Trie represents a character, and the number of nodes is directly related to the total number of characters.</p></li>
-</ul>
-<p>In practice, the space complexity can be less than O(N) because common prefixes are shared among words in the Trie, leading to space optimization.</p>
-</li>
-</ol>
-<p>In summary:</p>
-<ul class="simple">
-<li><p>Time complexity for each operation (insertion, search, starts with) is O(m) or O(p), where m is the length of the word and p is the length of the prefix.</p></li>
-<li><p>Space complexity is O(N) in the worst case, where N is the total number of characters across all inserted words. However, space optimization occurs due to common prefix sharing in practice, potentially reducing the actual space used.</p></li>
-</ul>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Count Prefixes</strong>:
-Implement a method to count the number of words in the Trie that have a specific prefix. This exercise requires you to maintain additional data in the Trie nodes to keep track of the count.</p></li>
-<li><p><strong>Auto-Complete Suggestions</strong>:
-Implement an auto-complete feature using the Trie. Given a prefix, the program should return a list of suggested words based on the words previously inserted into the Trie.</p></li>
-</ol>
-</section>
-</section>
-
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-    <h1>211. Design Add and Search Words Data Structure</h1>
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-                <ul class="visible nav section-nav flex-column">
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="design-add-and-search-words-data-structure">
-<h1>211. Design Add and Search Words Data Structure<a class="headerlink" href="#design-add-and-search-words-data-structure" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/design-add-and-search-words-data-structure/">To see the Design Add and Search Words Data Structure problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Design a data structure that supports adding new words and finding if a string matches any previously added string.</p>
-<p>Implement the <code class="docutils literal notranslate"><span class="pre">WordDictionary</span></code> class:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">WordDictionary()</span></code> Initializes the object.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">void</span> <span class="pre">addWord(word)</span></code> Adds <code class="docutils literal notranslate"><span class="pre">word</span></code> to the data structure, it can be matched later.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">bool</span> <span class="pre">search(word)</span></code> Returns <code class="docutils literal notranslate"><span class="pre">true</span></code> if there is any string in the data structure that matches <code class="docutils literal notranslate"><span class="pre">word</span></code> or <code class="docutils literal notranslate"><span class="pre">false</span></code> otherwise. <code class="docutils literal notranslate"><span class="pre">word</span></code> may contain dots <code class="docutils literal notranslate"><span class="pre">'.'</span></code> where dots can be matched with any letter.</p></li>
-</ul>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">word.length</span> <span class="pre">&lt;=</span> <span class="pre">25</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">word</span></code> in <code class="docutils literal notranslate"><span class="pre">addWord</span></code> consists of lowercase English letters.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">word</span></code> in <code class="docutils literal notranslate"><span class="pre">search</span></code> consist of <code class="docutils literal notranslate"><span class="pre">'.'</span></code> or lowercase English letters.</p></li>
-<li><p>There will be at most <code class="docutils literal notranslate"><span class="pre">2</span></code> dots in <code class="docutils literal notranslate"><span class="pre">word</span></code> for <code class="docutils literal notranslate"><span class="pre">search</span></code> queries.</p></li>
-<li><p>At most <span class="math notranslate nohighlight">\(10^4\)</span> calls will be made to <code class="docutils literal notranslate"><span class="pre">addWord</span></code> and <code class="docutils literal notranslate"><span class="pre">search</span></code>.</p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem is to design a data structure called <code class="docutils literal notranslate"><span class="pre">WordDictionary</span></code> that allows you to add words and search for words efficiently. The key feature of this data structure is that it should support wildcard search, where a dot (‘.’) can match any letter.</p>
-<p>Here are the specific requirements and explanations for the problem:</p>
-<ol class="arabic simple">
-<li><p>Initialize the <code class="docutils literal notranslate"><span class="pre">WordDictionary</span></code> object using the constructor <code class="docutils literal notranslate"><span class="pre">WordDictionary()</span></code>.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">addWord(word)</span></code> method allows you to add a word to the data structure. Once added, you should be able to search for this word later. The words consist of lowercase English letters, and each word has a length between 1 and 25 characters.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">search(word)</span></code> method allows you to search for a word in the data structure. This method returns <code class="docutils literal notranslate"><span class="pre">True</span></code> if there is any string in the data structure that matches the given word or <code class="docutils literal notranslate"><span class="pre">False</span></code> otherwise. The word you are searching for may contain dots (‘.’), where each dot can match any letter.</p></li>
-<li><p>The wildcard character (‘.’) in the <code class="docutils literal notranslate"><span class="pre">search</span></code> method allows for partial or fuzzy searching. For example, if the word dictionary contains the words “bad,” “dad,” and “mad,” then:</p>
-<ul class="simple">
-<li><p>Searching for “pad” should return <code class="docutils literal notranslate"><span class="pre">False</span></code> because there is no exact match.</p></li>
-<li><p>Searching for “bad” should return <code class="docutils literal notranslate"><span class="pre">True</span></code> because “bad” is in the dictionary.</p></li>
-<li><p>Searching for “.ad” should return <code class="docutils literal notranslate"><span class="pre">True</span></code> because it can match “bad,” “dad,” or “mad.”</p></li>
-<li><p>Searching for “b..” should return <code class="docutils literal notranslate"><span class="pre">True</span></code> because it can match “bad,” “bed,” “bet,” etc.</p></li>
-</ul>
-</li>
-<li><p>The problem specifies that there will be at most two dots (‘.’) in search queries, and there will be at most 10^4 calls to both <code class="docutils literal notranslate"><span class="pre">addWord</span></code> and <code class="docutils literal notranslate"><span class="pre">search</span></code>.</p></li>
-</ol>
-<p>In summary, you need to implement the <code class="docutils literal notranslate"><span class="pre">WordDictionary</span></code> class that efficiently supports adding words and searching for words, where the search can include wildcard characters (‘.’) that match any letter.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">TrieNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">):</span>
-        <span class="c1"># Initialize a TrieNode with children and a flag to indicate the end of a word.</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">children</span> <span class="o">=</span> <span class="p">{}</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">is_end_of_word</span> <span class="o">=</span> <span class="kc">False</span>
-
-<span class="k">class</span> <span class="nc">WordDictionary</span><span class="p">:</span>
-
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">):</span>
-        <span class="c1"># Initialize the WordDictionary with a root node.</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">root</span> <span class="o">=</span> <span class="n">TrieNode</span><span class="p">()</span>
-
-    <span class="k">def</span> <span class="nf">addWord</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">word</span><span class="p">:</span> <span class="nb">str</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="kc">None</span><span class="p">:</span>
-        <span class="n">node</span> <span class="o">=</span> <span class="bp">self</span><span class="o">.</span><span class="n">root</span>
-        <span class="c1"># Iterate through the characters in the word.</span>
-        <span class="k">for</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">word</span><span class="p">:</span>
-            <span class="k">if</span> <span class="n">char</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">:</span>
-                <span class="c1"># Create a new node for the character if it doesn&#39;t exist.</span>
-                <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">[</span><span class="n">char</span><span class="p">]</span> <span class="o">=</span> <span class="n">TrieNode</span><span class="p">()</span>
-            <span class="c1"># Move to the next node.</span>
-            <span class="n">node</span> <span class="o">=</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">[</span><span class="n">char</span><span class="p">]</span>
-        <span class="c1"># Mark the end of the word by setting the flag to True.</span>
-        <span class="n">node</span><span class="o">.</span><span class="n">is_end_of_word</span> <span class="o">=</span> <span class="kc">True</span>
-
-    <span class="k">def</span> <span class="nf">search_helper</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">node</span><span class="p">:</span> <span class="n">TrieNode</span><span class="p">,</span> <span class="n">word</span><span class="p">:</span> <span class="nb">str</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="nb">bool</span><span class="p">:</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">word</span><span class="p">:</span>
-            <span class="c1"># If there are no characters left in the word, check if we reached the end of a word in the trie.</span>
-            <span class="k">return</span> <span class="n">node</span><span class="o">.</span><span class="n">is_end_of_word</span>
-        
-        <span class="n">char</span> <span class="o">=</span> <span class="n">word</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
-        <span class="k">if</span> <span class="n">char</span> <span class="o">==</span> <span class="s1">&#39;.&#39;</span><span class="p">:</span>
-            <span class="c1"># If the character is a dot, explore all possible child nodes.</span>
-            <span class="k">for</span> <span class="n">child</span> <span class="ow">in</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="o">.</span><span class="n">values</span><span class="p">():</span>
-                <span class="k">if</span> <span class="bp">self</span><span class="o">.</span><span class="n">search_helper</span><span class="p">(</span><span class="n">child</span><span class="p">,</span> <span class="n">word</span><span class="p">[</span><span class="mi">1</span><span class="p">:]):</span>
-                    <span class="k">return</span> <span class="kc">True</span>
-        <span class="k">elif</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">:</span>
-            <span class="c1"># If the character is in the children of the current node, move to the next node.</span>
-            <span class="k">return</span> <span class="bp">self</span><span class="o">.</span><span class="n">search_helper</span><span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">[</span><span class="n">char</span><span class="p">],</span> <span class="n">word</span><span class="p">[</span><span class="mi">1</span><span class="p">:])</span>
-        
-        <span class="c1"># If the character is not found and is not a dot, the word is not in the trie.</span>
-        <span class="k">return</span> <span class="kc">False</span>
-
-    <span class="k">def</span> <span class="nf">search</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">word</span><span class="p">:</span> <span class="nb">str</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="nb">bool</span><span class="p">:</span>
-        <span class="c1"># Start the search from the root of the trie.</span>
-        <span class="k">return</span> <span class="bp">self</span><span class="o">.</span><span class="n">search_helper</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">root</span><span class="p">,</span> <span class="n">word</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>This code defines a <code class="docutils literal notranslate"><span class="pre">WordDictionary</span></code> class that uses a Trie data structure to efficiently store and search for words. Here’s a step-by-step explanation:</p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">TrieNode</span></code> class: This class represents nodes in the Trie data structure. Each node has two attributes: <code class="docutils literal notranslate"><span class="pre">children</span></code> (a dictionary to store child nodes) and <code class="docutils literal notranslate"><span class="pre">is_end_of_word</span></code> (a flag to indicate whether the node marks the end of a word).</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">WordDictionary</span></code> class: This class initializes with an empty Trie by creating a root node.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">addWord</span></code> method: This method adds a word to the Trie. It iterates through each character in the word and creates Trie nodes as necessary. It sets the <code class="docutils literal notranslate"><span class="pre">is_end_of_word</span></code> flag to <code class="docutils literal notranslate"><span class="pre">True</span></code> for the final character of the word to mark the end of the word.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">search_helper</span></code> method: This is a recursive helper function for searching words in the Trie. It takes a Trie node (<code class="docutils literal notranslate"><span class="pre">node</span></code>) and a word to search (<code class="docutils literal notranslate"><span class="pre">word</span></code>) as input. If there are no characters left in the word (<code class="docutils literal notranslate"><span class="pre">not</span> <span class="pre">word</span></code>), it checks if the current node marks the end of a word. If it does, it returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>If the first character of the word is a dot (‘.’), the function explores all possible child nodes by iterating through <code class="docutils literal notranslate"><span class="pre">node.children.values()</span></code> and recursively calls <code class="docutils literal notranslate"><span class="pre">search_helper</span></code> for each child with the remaining part of the word (<code class="docutils literal notranslate"><span class="pre">word[1:]</span></code>).</p></li>
-<li><p>If the first character of the word is not a dot and is found in the children of the current node, the function recursively moves to the next node by calling <code class="docutils literal notranslate"><span class="pre">search_helper</span></code> on the child node and the remaining part of the word.</p></li>
-<li><p>If the character is not found in the children and is not a dot, the word is not in the Trie, so the function returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">search</span></code> method: This is the public method for searching words. It initiates the search process by calling <code class="docutils literal notranslate"><span class="pre">search_helper</span></code> starting from the root node of the Trie.</p></li>
-</ol>
-<p>Overall, this code efficiently implements a Trie-based data structure that allows adding and searching for words, including support for wildcard characters represented by dots (‘.’).</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Example usage:</span>
-
-<span class="n">wordDictionary</span> <span class="o">=</span> <span class="n">WordDictionary</span><span class="p">()</span>
-<span class="n">wordDictionary</span><span class="o">.</span><span class="n">addWord</span><span class="p">(</span><span class="s2">&quot;bad&quot;</span><span class="p">)</span>
-<span class="n">wordDictionary</span><span class="o">.</span><span class="n">addWord</span><span class="p">(</span><span class="s2">&quot;dad&quot;</span><span class="p">)</span>
-<span class="n">wordDictionary</span><span class="o">.</span><span class="n">addWord</span><span class="p">(</span><span class="s2">&quot;mad&quot;</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">wordDictionary</span><span class="o">.</span><span class="n">search</span><span class="p">(</span><span class="s2">&quot;pad&quot;</span><span class="p">))</span>  <span class="c1"># Output: False</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">wordDictionary</span><span class="o">.</span><span class="n">search</span><span class="p">(</span><span class="s2">&quot;bad&quot;</span><span class="p">))</span>  <span class="c1"># Output: True</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">wordDictionary</span><span class="o">.</span><span class="n">search</span><span class="p">(</span><span class="s2">&quot;.ad&quot;</span><span class="p">))</span>  <span class="c1"># Output: True</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">wordDictionary</span><span class="o">.</span><span class="n">search</span><span class="p">(</span><span class="s2">&quot;b..&quot;</span><span class="p">))</span>  <span class="c1"># Output: True</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-True
-True
-True
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">WordDictionary</span></code> class methods:</p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">addWord</span></code> Method:</p>
-<ul class="simple">
-<li><p>Time Complexity: O(L)</p>
-<ul>
-<li><p>L is the length of the word being added.</p></li>
-<li><p>In the worst case, we need to iterate through each character in the word and create nodes in the Trie. This takes O(L) time.</p></li>
-</ul>
-</li>
-<li><p>Space Complexity: O(L)</p>
-<ul>
-<li><p>The space complexity for storing the word in the Trie is also O(L) because we create nodes for each character in the word.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">search_helper</span></code> Method (called by <code class="docutils literal notranslate"><span class="pre">search</span></code>):</p>
-<ul class="simple">
-<li><p>Time Complexity: O(2^M * L)</p>
-<ul>
-<li><p>M is the maximum number of dots (‘.’) in the search word.</p></li>
-<li><p>In the worst case, when there are M dots in the search word, we may explore all possible child nodes at each dot. This results in a branching factor of 26 (for lowercase letters) at each dot.</p></li>
-<li><p>So, the time complexity is exponential in the number of dots and linear in the length of the word being searched (L).</p></li>
-</ul>
-</li>
-<li><p>Space Complexity: O(L)</p>
-<ul>
-<li><p>The space complexity for the recursive call stack is O(L) in the worst case because we may have to recurse to a depth equal to the length of the search word.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">search</span></code> Method:</p>
-<ul class="simple">
-<li><p>Time Complexity: O(2^M * L)</p>
-<ul>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">search</span></code> method calls the <code class="docutils literal notranslate"><span class="pre">search_helper</span></code> method, which has the same time complexity explained above.</p></li>
-</ul>
-</li>
-<li><p>Space Complexity: O(L)</p>
-<ul>
-<li><p>The space complexity for the recursive call stack is O(L) in the worst case, as explained earlier.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-</ol>
-<p>Overall, the time complexity of the <code class="docutils literal notranslate"><span class="pre">search</span></code> method is mainly affected by the number of dots (‘.’) in the search word. In the worst case, when there are multiple dots and many possibilities to explore at each dot, the time complexity can be exponential. However, for most practical cases, it performs reasonably well.</p>
-<p>The space complexity primarily depends on the space required to store the words in the Trie. It is proportional to the total number of characters in all added words and is linear with respect to the length of the words added.</p>
-<p>In summary:</p>
-<ul class="simple">
-<li><p>Time Complexity for <code class="docutils literal notranslate"><span class="pre">addWord</span></code>: O(L)</p></li>
-<li><p>Time Complexity for <code class="docutils literal notranslate"><span class="pre">search</span></code>: O(2^M * L)</p></li>
-<li><p>Space Complexity: O(L) for storing words in the Trie, O(L) for the recursive call stack during search.</p></li>
-</ul>
-<p>Note: The space complexity analysis assumes that the dictionary contains a reasonable number of words and does not account for the overhead of Python objects or system-level memory allocation.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Dictionary Auto-Completion</strong>: Extend the <code class="docutils literal notranslate"><span class="pre">WordDictionary</span></code> class to provide auto-completion suggestions based on the prefix of a word. Implement a method that returns a list of words that match the given prefix.</p></li>
-<li><p><strong>Support Word Deletion</strong>: Extend the <code class="docutils literal notranslate"><span class="pre">WordDictionary</span></code> class to support word deletion. Add a method <code class="docutils literal notranslate"><span class="pre">deleteWord(word)</span></code> that removes a word from the data structure. Ensure that the search operation still works correctly after deletion.</p></li>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="word-search-ii">
-<h1>212. Word Search II<a class="headerlink" href="#word-search-ii" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Hard</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/word-search-ii/">To see the Word Search II problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given an <code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">x</span> <span class="pre">n</span></code> <code class="docutils literal notranslate"><span class="pre">board</span></code> of characters and a list of strings <code class="docutils literal notranslate"><span class="pre">words</span></code>, the task is to return <em>all words on the board</em>.</p>
-<p>Each word must be constructed from letters of sequentially <strong>adjacent cells</strong>, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">==</span> <span class="pre">board.length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">==</span> <span class="pre">board[i].length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">m,</span> <span class="pre">n</span> <span class="pre">&lt;=</span> <span class="pre">12</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">board[i][j]</span></code> is a lowercase English letter.</p></li>
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">words.length</span></code> &lt;= <span class="math notranslate nohighlight">\(3 * 10^4\)</span></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">words[i].length</span> <span class="pre">&lt;=</span> <span class="pre">10</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">words[i]</span></code> consists of lowercase English letters.</p></li>
-<li><p>All the strings of <code class="docutils literal notranslate"><span class="pre">words</span></code> are unique.</p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>To solve the “Word Search II” problem, you can use a Trie data structure and perform a depth-first search (DFS) on the board. Here’s a step-by-step approach to solve the problem:</p>
-<ol class="arabic simple">
-<li><p>Build a Trie from the given list of words:</p>
-<ul class="simple">
-<li><p>Create a Trie data structure to store all the words from the input list.</p></li>
-<li><p>For each word in the input list, insert it into the Trie, character by character. Make sure to mark the end of each word in the Trie.</p></li>
-</ul>
-</li>
-<li><p>Perform DFS on the board:</p>
-<ul class="simple">
-<li><p>Iterate through each cell (i, j) on the board.</p></li>
-<li><p>For each cell, start a DFS traversal from that cell to search for words.</p></li>
-<li><p>During the DFS traversal, maintain a current path string.</p></li>
-<li><p>At each cell, check if the current path string, concatenated with the character in the cell, matches any word prefix in the Trie. If it does, continue the DFS.</p></li>
-<li><p>If the current path string matches a word in the Trie, add it to the result.</p></li>
-</ul>
-</li>
-<li><p>Implement DFS with backtracking:</p>
-<ul class="simple">
-<li><p>During the DFS traversal, you will explore neighboring cells (horizontally and vertically) if they are within bounds and haven’t been visited before.</p></li>
-<li><p>To prevent revisiting the same cell within the same word, mark the cell as visited (e.g., change its character to a special character like ‘#’) before exploring it and restore its original character after the DFS traversal.</p></li>
-</ul>
-</li>
-<li><p>Return the result:</p>
-<ul class="simple">
-<li><p>After completing the DFS traversal for all cells on the board, you will have collected all the valid words in the result set.</p></li>
-<li><p>Convert the result set to a list and return it as the final output.</p></li>
-</ul>
-</li>
-</ol>
-<p>This approach efficiently finds all words on the board by utilizing the Trie data structure to prune unnecessary DFS paths and avoiding duplicate visits to the same cell within the same word. It ensures that each word is constructed from letters of sequentially adjacent cells without using the same letter cell more than once in a word.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Define a TrieNode class to represent nodes in the Trie.</span>
-<span class="k">class</span> <span class="nc">TrieNode</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">children</span> <span class="o">=</span> <span class="p">{}</span>  <span class="c1"># Dictionary to store child nodes.</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">is_end_of_word</span> <span class="o">=</span> <span class="kc">False</span>  <span class="c1"># Flag to mark the end of a word.</span>
-
-<span class="c1"># Main function to find words on the board.</span>
-<span class="k">def</span> <span class="nf">findWords</span><span class="p">(</span><span class="n">board</span><span class="p">,</span> <span class="n">words</span><span class="p">):</span>
-    <span class="c1"># DFS function to search for words starting from a given cell.</span>
-    <span class="k">def</span> <span class="nf">dfs</span><span class="p">(</span><span class="n">node</span><span class="p">,</span> <span class="n">i</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="n">path</span><span class="p">):</span>
-        <span class="n">char</span> <span class="o">=</span> <span class="n">board</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">]</span>  <span class="c1"># Get the character at the current cell.</span>
-        <span class="n">curr_node</span> <span class="o">=</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="o">.</span><span class="n">get</span><span class="p">(</span><span class="n">char</span><span class="p">)</span>  <span class="c1"># Find the corresponding Trie node.</span>
-
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">curr_node</span><span class="p">:</span>
-            <span class="k">return</span>  <span class="c1"># If no matching Trie node, stop the search.</span>
-
-        <span class="n">path</span> <span class="o">+=</span> <span class="n">char</span>  <span class="c1"># Add the character to the current path.</span>
-
-        <span class="k">if</span> <span class="n">curr_node</span><span class="o">.</span><span class="n">is_end_of_word</span><span class="p">:</span>
-            <span class="n">result</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="n">path</span><span class="p">)</span>  <span class="c1"># If the path matches a word, add it to the result.</span>
-
-        <span class="n">temp</span><span class="p">,</span> <span class="n">board</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">]</span> <span class="o">=</span> <span class="n">board</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">],</span> <span class="s2">&quot;#&quot;</span>  <span class="c1"># Mark the cell as visited.</span>
-
-        <span class="c1"># Explore neighboring cells (up, down, left, right).</span>
-        <span class="k">if</span> <span class="n">i</span> <span class="o">&gt;</span> <span class="mi">0</span><span class="p">:</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">curr_node</span><span class="p">,</span> <span class="n">i</span> <span class="o">-</span> <span class="mi">1</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="n">path</span><span class="p">)</span>
-        <span class="k">if</span> <span class="n">i</span> <span class="o">&lt;</span> <span class="n">m</span> <span class="o">-</span> <span class="mi">1</span><span class="p">:</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">curr_node</span><span class="p">,</span> <span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="n">path</span><span class="p">)</span>
-        <span class="k">if</span> <span class="n">j</span> <span class="o">&gt;</span> <span class="mi">0</span><span class="p">:</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">curr_node</span><span class="p">,</span> <span class="n">i</span><span class="p">,</span> <span class="n">j</span> <span class="o">-</span> <span class="mi">1</span><span class="p">,</span> <span class="n">path</span><span class="p">)</span>
-        <span class="k">if</span> <span class="n">j</span> <span class="o">&lt;</span> <span class="n">n</span> <span class="o">-</span> <span class="mi">1</span><span class="p">:</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">curr_node</span><span class="p">,</span> <span class="n">i</span><span class="p">,</span> <span class="n">j</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">path</span><span class="p">)</span>
-
-        <span class="n">board</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">]</span> <span class="o">=</span> <span class="n">temp</span>  <span class="c1"># Restore the cell&#39;s original character.</span>
-
-    <span class="c1"># Function to build a Trie from the list of words.</span>
-    <span class="k">def</span> <span class="nf">buildTrie</span><span class="p">():</span>
-        <span class="n">root</span> <span class="o">=</span> <span class="n">TrieNode</span><span class="p">()</span>  <span class="c1"># Create a root node.</span>
-        <span class="k">for</span> <span class="n">word</span> <span class="ow">in</span> <span class="n">words</span><span class="p">:</span>
-            <span class="n">node</span> <span class="o">=</span> <span class="n">root</span>
-            <span class="k">for</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">word</span><span class="p">:</span>
-                <span class="k">if</span> <span class="n">char</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">:</span>
-                    <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">[</span><span class="n">char</span><span class="p">]</span> <span class="o">=</span> <span class="n">TrieNode</span><span class="p">()</span>  <span class="c1"># Create a new node if needed.</span>
-                <span class="n">node</span> <span class="o">=</span> <span class="n">node</span><span class="o">.</span><span class="n">children</span><span class="p">[</span><span class="n">char</span><span class="p">]</span>
-            <span class="n">node</span><span class="o">.</span><span class="n">is_end_of_word</span> <span class="o">=</span> <span class="kc">True</span>  <span class="c1"># Mark the end of the word.</span>
-        <span class="k">return</span> <span class="n">root</span>
-
-    <span class="n">m</span><span class="p">,</span> <span class="n">n</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">board</span><span class="p">),</span> <span class="nb">len</span><span class="p">(</span><span class="n">board</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span>  <span class="c1"># Get the dimensions of the board.</span>
-    <span class="n">root</span> <span class="o">=</span> <span class="n">buildTrie</span><span class="p">()</span>  <span class="c1"># Build the Trie from the list of words.</span>
-    <span class="n">result</span> <span class="o">=</span> <span class="nb">set</span><span class="p">()</span>  <span class="c1"># Use a set to store unique words found on the board.</span>
-
-    <span class="c1"># Iterate through each cell on the board and start DFS from there.</span>
-    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">m</span><span class="p">):</span>
-        <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">root</span><span class="p">,</span> <span class="n">i</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="s2">&quot;&quot;</span><span class="p">)</span>  <span class="c1"># Start DFS from each cell on the board.</span>
-
-    <span class="k">return</span> <span class="nb">list</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>  <span class="c1"># Convert the set to a list and return the result.</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The code solves the “Word Search II” problem, where you have a grid of characters and a list of words. The goal is to find all words from the list that can be constructed by traversing adjacent cells (horizontally or vertically) in the grid, without using the same cell more than once for a single word.</p>
-<p>The code does the following:</p>
-<ol class="arabic simple">
-<li><p>Defines a <code class="docutils literal notranslate"><span class="pre">TrieNode</span></code> class to represent nodes in a Trie data structure. Each node has children (other characters in the word) and a flag to mark the end of a word.</p></li>
-<li><p>Defines the <code class="docutils literal notranslate"><span class="pre">findWords</span></code> function, which takes the grid (<code class="docutils literal notranslate"><span class="pre">board</span></code>) and a list of words (<code class="docutils literal notranslate"><span class="pre">words</span></code>) as input.</p></li>
-<li><p>Inside the <code class="docutils literal notranslate"><span class="pre">findWords</span></code> function, there is a <code class="docutils literal notranslate"><span class="pre">dfs</span></code> (depth-first search) function. This function is used to explore the grid starting from a specific cell. It checks if the current path matches any word in the Trie and adds it to the result if found.</p></li>
-<li><p>The code also includes a <code class="docutils literal notranslate"><span class="pre">buildTrie</span></code> function that constructs a Trie data structure from the list of words. It iterates through each word, creating nodes for each character in the Trie and marking the end of words.</p></li>
-<li><p>The dimensions of the grid are obtained (m x n).</p></li>
-<li><p>A Trie is built using the <code class="docutils literal notranslate"><span class="pre">buildTrie</span></code> function.</p></li>
-<li><p>A set called <code class="docutils literal notranslate"><span class="pre">result</span></code> is used to store unique words found in the grid.</p></li>
-<li><p>The code iterates through each cell on the grid (using nested loops) and starts a DFS search from each cell.</p></li>
-<li><p>During the DFS, it explores neighboring cells (up, down, left, right) if they haven’t been visited before and if the current path matches a prefix in the Trie.</p></li>
-<li><p>When a word is found during the DFS traversal, it is added to the <code class="docutils literal notranslate"><span class="pre">result</span></code> set.</p></li>
-<li><p>The grid cell is marked as visited during the DFS by changing its character to “#”, and it is restored to its original character after the DFS traversal.</p></li>
-<li><p>Finally, the set of unique words found is converted to a list and returned as the output.</p></li>
-</ol>
-<p>The code effectively uses a Trie data structure to optimize the search and ensures that each word is constructed from adjacent cells without reusing the same cell for the same word.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="n">board1</span> <span class="o">=</span> <span class="p">[[</span><span class="s2">&quot;o&quot;</span><span class="p">,</span><span class="s2">&quot;a&quot;</span><span class="p">,</span><span class="s2">&quot;a&quot;</span><span class="p">,</span><span class="s2">&quot;n&quot;</span><span class="p">],[</span><span class="s2">&quot;e&quot;</span><span class="p">,</span><span class="s2">&quot;t&quot;</span><span class="p">,</span><span class="s2">&quot;a&quot;</span><span class="p">,</span><span class="s2">&quot;e&quot;</span><span class="p">],[</span><span class="s2">&quot;i&quot;</span><span class="p">,</span><span class="s2">&quot;h&quot;</span><span class="p">,</span><span class="s2">&quot;k&quot;</span><span class="p">,</span><span class="s2">&quot;r&quot;</span><span class="p">],[</span><span class="s2">&quot;i&quot;</span><span class="p">,</span><span class="s2">&quot;f&quot;</span><span class="p">,</span><span class="s2">&quot;l&quot;</span><span class="p">,</span><span class="s2">&quot;v&quot;</span><span class="p">]]</span>
-<span class="n">words1</span> <span class="o">=</span> <span class="p">[</span><span class="s2">&quot;oath&quot;</span><span class="p">,</span><span class="s2">&quot;pea&quot;</span><span class="p">,</span><span class="s2">&quot;eat&quot;</span><span class="p">,</span><span class="s2">&quot;rain&quot;</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">findWords</span><span class="p">(</span><span class="n">board1</span><span class="p">,</span> <span class="n">words1</span><span class="p">))</span>  <span class="c1"># Output: [&quot;eat&quot;, &quot;oath&quot;]</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[&#39;eat&#39;, &#39;oath&#39;]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-
-<span class="n">board2</span> <span class="o">=</span> <span class="p">[[</span><span class="s2">&quot;a&quot;</span><span class="p">,</span><span class="s2">&quot;b&quot;</span><span class="p">],[</span><span class="s2">&quot;c&quot;</span><span class="p">,</span><span class="s2">&quot;d&quot;</span><span class="p">]]</span>
-<span class="n">words2</span> <span class="o">=</span> <span class="p">[</span><span class="s2">&quot;abcb&quot;</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">findWords</span><span class="p">(</span><span class="n">board2</span><span class="p">,</span> <span class="n">words2</span><span class="p">))</span>  <span class="c1"># Output: []</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p><strong>Building the Trie (<code class="docutils literal notranslate"><span class="pre">buildTrie</span></code> function):</strong></p>
-<ul class="simple">
-<li><p>Suppose there are <code class="docutils literal notranslate"><span class="pre">N</span></code> words in the input list, and the average length of a word is <code class="docutils literal notranslate"><span class="pre">L</span></code>.</p></li>
-<li><p>Building the Trie takes O(N * L) time, as we iterate through each word and insert each character into the Trie.</p></li>
-</ul>
-</li>
-<li><p><strong>DFS Traversal (<code class="docutils literal notranslate"><span class="pre">dfs</span></code> function):</strong></p>
-<ul class="simple">
-<li><p>In the worst case, the DFS function explores all possible paths in the grid.</p></li>
-<li><p>The maximum number of recursive calls for each cell is 4 (up, down, left, right).</p></li>
-<li><p>The maximum depth of the DFS is limited by the length of the longest word in the Trie, which is at most <code class="docutils literal notranslate"><span class="pre">L</span></code>.</p></li>
-<li><p>Therefore, the DFS traversal for each cell takes <span class="math notranslate nohighlight">\(O(4^L)\)</span> time.</p></li>
-</ul>
-</li>
-<li><p><strong>Iterating Through the Grid (<code class="docutils literal notranslate"><span class="pre">findWords</span></code> function):</strong></p>
-<ul class="simple">
-<li><p>In the worst case, we iterate through all <code class="docutils literal notranslate"><span class="pre">m</span></code> rows and <code class="docutils literal notranslate"><span class="pre">n</span></code> columns in the grid.</p></li>
-<li><p>For each cell, we start a DFS traversal.</p></li>
-<li><p>Therefore, iterating through the entire grid takes <span class="math notranslate nohighlight">\(O(m * n * 4^L)\)</span> time.</p></li>
-</ul>
-</li>
-</ol>
-<p>Overall, the time complexity of the code is <span class="math notranslate nohighlight">\(O(N * L + m * n * 4^L)\)</span>.</p>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p><strong>Trie Data Structure:</strong></p>
-<ul class="simple">
-<li><p>The space required to store the Trie depends on the number of unique characters in all the words.</p></li>
-<li><p>In the worst case, where all words are unique and have no common prefixes, the space complexity of the Trie is O(N * L).</p></li>
-<li><p>In practice, it may be less if there are common prefixes among words.</p></li>
-</ul>
-</li>
-<li><p><strong>DFS Stack (Recursive Calls):</strong></p>
-<ul class="simple">
-<li><p>The depth of the recursive call stack during DFS is at most <code class="docutils literal notranslate"><span class="pre">L</span></code>, where <code class="docutils literal notranslate"><span class="pre">L</span></code> is the length of the longest word in the Trie.</p></li>
-<li><p>Therefore, the space complexity for the recursive call stack is O(L).</p></li>
-</ul>
-</li>
-<li><p><strong>Result Set (<code class="docutils literal notranslate"><span class="pre">result</span></code> set):</strong></p>
-<ul class="simple">
-<li><p>The space used to store the result set depends on the number of valid words found in the grid.</p></li>
-<li><p>In the worst case, when all words are found, the space complexity of the result set is O(N * L).</p></li>
-</ul>
-</li>
-</ol>
-<p>Overall, the space complexity of the code is O(N * L) for the Trie and O(L) for the recursive call stack.</p>
-<p>In summary, the time complexity is dominated by the DFS traversal and is influenced by the number of words, their lengths, and the size of the grid. The space complexity is mainly determined by the Trie structure and the recursive call stack depth during DFS.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Word Search Paths:</strong>
-Extend the basic word search exercise to find all possible paths (sequences of cells) that spell out a given word on the board. Return a list of paths if the word can be formed, otherwise an empty list.</p></li>
-<li><p><strong>Reverse Word Search:</strong>
-Modify the code to search for words in reverse order on the board. Given a 2D grid of characters and a list of words, find all words from the list that can be formed by traversing the board in reverse (right to left, bottom to top, etc.).</p></li>
-</ol>
-</section>
-</section>
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-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/design-add-and-search-words-data-structure/">211. Design Add And Search Words Data Structure</a></p></td>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="last-stone-weight">
-<h1>1046. Last Stone Weight<a class="headerlink" href="#last-stone-weight" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/last-stone-weight/">To see the Last Stone Weight problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>You are given an array of integers <code class="docutils literal notranslate"><span class="pre">stones</span></code> where <code class="docutils literal notranslate"><span class="pre">stones[i]</span></code> is the weight of the <code class="docutils literal notranslate"><span class="pre">i</span></code>th stone.</p>
-<p>We are playing a game with the stones. On each turn, you choose the <strong>heaviest two stones</strong> and smash them together. Suppose the heaviest two stones have weights <code class="docutils literal notranslate"><span class="pre">x</span></code> and <code class="docutils literal notranslate"><span class="pre">y</span></code> with <code class="docutils literal notranslate"><span class="pre">x</span> <span class="pre">&lt;=</span> <span class="pre">y</span></code>. The result of this smash is:</p>
-<ol class="arabic simple">
-<li><p>If <code class="docutils literal notranslate"><span class="pre">x</span> <span class="pre">==</span> <span class="pre">y</span></code>, both stones are destroyed.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">x</span> <span class="pre">!=</span> <span class="pre">y</span></code>, the stone of weight <code class="docutils literal notranslate"><span class="pre">x</span></code> is destroyed, and the stone of weight <code class="docutils literal notranslate"><span class="pre">y</span></code> has new weight <code class="docutils literal notranslate"><span class="pre">y</span> <span class="pre">-</span> <span class="pre">x</span></code>.</p></li>
-</ol>
-<p>At the end of the game, there is <strong>at most one</strong> stone left.</p>
-<p>Return <em>the weight of the last remaining stone</em>. If there are no stones left, return <code class="docutils literal notranslate"><span class="pre">0</span></code>.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">stones.length</span> <span class="pre">&lt;=</span> <span class="pre">30</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">stones[i]</span> <span class="pre">&lt;=</span> <span class="pre">1000</span></code></p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The “Last Stone Weight” problem involves a game played with a collection of stones, each stone having a certain weight. The goal is to find the weight of the last remaining stone after applying a specific set of rules.</p>
-<p>Here’s how the game works:</p>
-<ol class="arabic simple">
-<li><p>You start with an array of stones, where each stone is represented by its weight. The weights of the stones are given in the input as an array.</p></li>
-<li><p>In each turn of the game, you select the two heaviest stones from the remaining stones. If there is only one stone left, you’ve found the answer.</p></li>
-<li><p>If there are two stones with weights <code class="docutils literal notranslate"><span class="pre">x</span></code> and <code class="docutils literal notranslate"><span class="pre">y</span></code>, where <code class="docutils literal notranslate"><span class="pre">x</span></code> is less than or equal to <code class="docutils literal notranslate"><span class="pre">y</span></code>, the following happens:</p>
-<ul class="simple">
-<li><p>If <code class="docutils literal notranslate"><span class="pre">x</span></code> is equal to <code class="docutils literal notranslate"><span class="pre">y</span></code>, both stones are completely destroyed, and they are removed from the array of stones.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">x</span></code> is not equal to <code class="docutils literal notranslate"><span class="pre">y</span></code>, the stone with weight <code class="docutils literal notranslate"><span class="pre">x</span></code> is destroyed, and the stone with weight <code class="docutils literal notranslate"><span class="pre">y</span></code> now has a new weight of <code class="docutils literal notranslate"><span class="pre">y</span> <span class="pre">-</span> <span class="pre">x</span></code>. The stone with weight <code class="docutils literal notranslate"><span class="pre">x</span></code> is removed from the array, and the modified stone with weight <code class="docutils literal notranslate"><span class="pre">y</span> <span class="pre">-</span> <span class="pre">x</span></code> remains in the array.</p></li>
-</ul>
-</li>
-<li><p>The game continues with the modified array of stones until there is at most one stone left.</p></li>
-<li><p>The goal is to find the weight of the last remaining stone, or if there are no stones left, return <code class="docutils literal notranslate"><span class="pre">0</span></code>.</p></li>
-</ol>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">heapq</span>
-
-<span class="k">def</span> <span class="nf">lastStoneWeight</span><span class="p">(</span><span class="n">stones</span><span class="p">):</span>
-    <span class="c1"># Convert the input list to a max heap (negate the values to simulate a max heap)</span>
-    <span class="n">max_heap</span> <span class="o">=</span> <span class="p">[</span><span class="o">-</span><span class="n">stone</span> <span class="k">for</span> <span class="n">stone</span> <span class="ow">in</span> <span class="n">stones</span><span class="p">]</span>
-    <span class="n">heapq</span><span class="o">.</span><span class="n">heapify</span><span class="p">(</span><span class="n">max_heap</span><span class="p">)</span>
-
-    <span class="c1"># Continue as long as there are more than 1 stone in the heap</span>
-    <span class="k">while</span> <span class="nb">len</span><span class="p">(</span><span class="n">max_heap</span><span class="p">)</span> <span class="o">&gt;</span> <span class="mi">1</span><span class="p">:</span>
-        <span class="c1"># Get the two heaviest stones</span>
-        <span class="n">y</span> <span class="o">=</span> <span class="o">-</span><span class="n">heapq</span><span class="o">.</span><span class="n">heappop</span><span class="p">(</span><span class="n">max_heap</span><span class="p">)</span>  <span class="c1"># Get the heaviest stone and negate it back to positive</span>
-        <span class="n">x</span> <span class="o">=</span> <span class="o">-</span><span class="n">heapq</span><span class="o">.</span><span class="n">heappop</span><span class="p">(</span><span class="n">max_heap</span><span class="p">)</span>  <span class="c1"># Get the second heaviest stone and negate it back to positive</span>
-
-        <span class="c1"># Calculate the new stone&#39;s weight after smashing</span>
-        <span class="k">if</span> <span class="n">x</span> <span class="o">!=</span> <span class="n">y</span><span class="p">:</span>
-            <span class="n">new_stone</span> <span class="o">=</span> <span class="n">y</span> <span class="o">-</span> <span class="n">x</span>
-            <span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="n">max_heap</span><span class="p">,</span> <span class="o">-</span><span class="n">new_stone</span><span class="p">)</span>  <span class="c1"># Negate the value and push it back to the heap</span>
-
-    <span class="c1"># If there is one stone left, return its weight (negate it back to positive)</span>
-    <span class="k">if</span> <span class="n">max_heap</span><span class="p">:</span>
-        <span class="k">return</span> <span class="o">-</span><span class="n">max_heap</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
-    <span class="k">else</span><span class="p">:</span>
-        <span class="k">return</span> <span class="mi">0</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The code defines a Python function called <code class="docutils literal notranslate"><span class="pre">lastStoneWeight</span></code> that takes a list of stone weights as input. It aims to find the weight of the last remaining stone after playing the stone smashing game described in the problem statement.</p>
-<p>Here’s a step-by-step explanation of the code:</p>
-<ol class="arabic simple">
-<li><p>We import the <code class="docutils literal notranslate"><span class="pre">heapq</span></code> library, which allows us to create and manipulate a heap (priority queue).</p></li>
-<li><p>Inside the <code class="docutils literal notranslate"><span class="pre">lastStoneWeight</span></code> function:</p>
-<ul class="simple">
-<li><p>We create a max heap (negating stone weights to simulate a max heap) using the <code class="docutils literal notranslate"><span class="pre">heapq.heapify</span></code> function. This heap will keep track of the heaviest stones.</p></li>
-<li><p>We enter a loop that continues as long as there are more than 1 stone in the heap.</p></li>
-<li><p>Inside the loop, we:</p>
-<ul>
-<li><p>Retrieve and negate the heaviest stone (<code class="docutils literal notranslate"><span class="pre">y</span></code>) from the heap.</p></li>
-<li><p>Retrieve and negate the second heaviest stone (<code class="docutils literal notranslate"><span class="pre">x</span></code>) from the heap.</p></li>
-<li><p>Calculate the new stone’s weight after smashing (<code class="docutils literal notranslate"><span class="pre">y</span> <span class="pre">-</span> <span class="pre">x</span></code>), and negate it back to simulate a max heap.</p></li>
-<li><p>Add the new stone back to the heap using <code class="docutils literal notranslate"><span class="pre">heapq.heappush</span></code>.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p>After the loop, if there is one stone left in the heap, we retrieve and negate it to get its actual weight, and return it as the result.</p></li>
-<li><p>If there are no stones left (the heap is empty), we return <code class="docutils literal notranslate"><span class="pre">0</span></code>, indicating that no stones remain.</p></li>
-<li><p>Finally, we provide two test cases to demonstrate the function’s usage and correctness.</p></li>
-</ol>
-<p>In summary, the code efficiently implements the stone smashing game by maintaining a max heap, selecting the heaviest stones in each turn, smashing them, and updating the heap until there is at most one stone left. It then returns the weight of the last remaining stone or 0 if there are no stones left.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="n">stones1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">7</span><span class="p">,</span> <span class="mi">4</span><span class="p">,</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">8</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">lastStoneWeight</span><span class="p">(</span><span class="n">stones1</span><span class="p">))</span>  <span class="c1"># Output: 1</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>1
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2: </span>
-
-<span class="n">stones2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">lastStoneWeight</span><span class="p">(</span><span class="n">stones2</span><span class="p">))</span>  <span class="c1"># Output: 1</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>1
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided Python code for the “Last Stone Weight” problem:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">heapq.heapify</span></code> operation to create the max heap from the input stone weights has a time complexity of O(n), where n is the number of stones.</p></li>
-<li><p>The loop that continues until there are more than 1 stone in the heap performs the following operations in each iteration:</p>
-<ul class="simple">
-<li><p>Retrieving and negating the heaviest stone (<code class="docutils literal notranslate"><span class="pre">y</span></code>) from the heap, which takes O(log n) time.</p></li>
-<li><p>Retrieving and negating the second heaviest stone (<code class="docutils literal notranslate"><span class="pre">x</span></code>) from the heap, also taking O(log n) time.</p></li>
-<li><p>Calculating the new stone’s weight and pushing it back into the heap using <code class="docutils literal notranslate"><span class="pre">heapq.heappush</span></code>, which takes O(log n) time.</p></li>
-</ul>
-</li>
-<li><p>In the worst case, the loop runs until there is only one stone left, which requires approximately (n-1) iterations.</p></li>
-</ol>
-<p>Overall, the dominant time complexity is O(n) for the initial heap creation, and within the loop, each iteration takes O(log n) time. Therefore, the overall time complexity of the algorithm is O(n log n).</p>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The space complexity is primarily determined by the space required for the max heap. In the worst case, this heap can contain all the stones, so the space complexity for the heap is O(n).</p></li>
-<li><p>The rest of the variables used in the function (e.g., <code class="docutils literal notranslate"><span class="pre">x</span></code>, <code class="docutils literal notranslate"><span class="pre">y</span></code>, <code class="docutils literal notranslate"><span class="pre">new_stone</span></code>) require only constant space, so they do not significantly contribute to the space complexity.</p></li>
-</ol>
-<p>Therefore, the overall space complexity of the algorithm is O(n) due to the space used for the max heap.</p>
-<p>In summary:</p>
-<ul class="simple">
-<li><p>Time Complexity: <span class="math notranslate nohighlight">\(O(n\ log\ n)\)</span></p></li>
-<li><p>Space Complexity: <span class="math notranslate nohighlight">\(O(n)\)</span></p></li>
-</ul>
-<p>The algorithm is efficient enough to handle the problem’s constraints, as the worst-case time and space complexities are both linearithmic in terms of the number of stones.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Kth Last Stone Weight</strong>: Modify the problem to find the weight of the Kth last remaining stone, where K is an integer input. Your function should work efficiently for large values of K.</p></li>
-<li><p><strong>Multiple Stones Smash</strong>: Modify the problem so that instead of smashing two stones at a time, you can smash up to K stones at each turn, where K is an integer input. Determine the weight of the last remaining stone in this variation.</p></li>
-</ol>
-</section>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
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-<h1>295. Find Median from Data Stream<a class="headerlink" href="#find-median-from-data-stream" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Hard</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/find-median-from-data-stream/">To see the Find Median from Data Stream problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>The <strong>median</strong> is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.</p>
-<p>Implement the MedianFinder class:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">MedianFinder()</span></code> initializes the <code class="docutils literal notranslate"><span class="pre">MedianFinder</span></code> object.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">void</span> <span class="pre">addNum(int</span> <span class="pre">num)</span></code> adds the integer <code class="docutils literal notranslate"><span class="pre">num</span></code> from the data stream to the data structure.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">double</span> <span class="pre">findMedian()</span></code> returns the median of all elements so far. Answers within <span class="math notranslate nohighlight">\(10^{-5}\)</span> of the actual answer will be accepted.</p></li>
-</ul>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><span class="math notranslate nohighlight">\(-10^{5}\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">num</span></code> &lt;= <span class="math notranslate nohighlight">\(10^{5}\)</span></p></li>
-<li><p>There will be at least one element in the data structure before calling <code class="docutils literal notranslate"><span class="pre">findMedian</span></code>.</p></li>
-<li><p>At most <span class="math notranslate nohighlight">\(5 * 10^4\)</span> calls will be made to <code class="docutils literal notranslate"><span class="pre">addNum</span></code> and <code class="docutils literal notranslate"><span class="pre">findMedian</span></code>.</p></li>
-</ul>
-<p><strong>Follow-up:</strong></p>
-<ol class="arabic simple">
-<li><p>If all integer numbers from the stream are in the range <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">100]</span></code>, how would you optimize your solution?</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">99%</span></code> of all integer numbers from the stream are in the range <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">100]</span></code>, how would you optimize your solution?</p></li>
-</ol>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem, “Find Median from Data Stream,” deals with efficiently computing the median of a sequence of numbers as they are incrementally provided. The median is the middle value in an ordered list of numbers. When the list has an even number of elements, the median is the average of the two middle values.</p>
-<p>Here’s a breakdown of the problem:</p>
-<ol class="arabic simple">
-<li><p><strong>Initialization</strong>: You are asked to implement a <code class="docutils literal notranslate"><span class="pre">MedianFinder</span></code> class that supports the following operations:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">MedianFinder()</span></code>: Initializes the MedianFinder object.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">addNum(int</span> <span class="pre">num)</span></code>: Adds an integer <code class="docutils literal notranslate"><span class="pre">num</span></code> from the data stream to the data structure.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">findMedian()</span></code>: Returns the median of all elements added to the data structure.</p></li>
-</ul>
-</li>
-<li><p><strong>Median Calculation</strong>: The <code class="docutils literal notranslate"><span class="pre">findMedian</span></code> operation should efficiently compute the median, and the result should have a precision of at least <span class="math notranslate nohighlight">\(10^{-5}\)</span> (i.e., answers within this range will be considered correct).</p></li>
-<li><p><strong>Examples</strong>:</p>
-<ul class="simple">
-<li><p>If you add numbers [1, 2], the median should be 1.5 because (1 + 2) / 2 = 1.5.</p></li>
-<li><p>If you add numbers [1, 2, 3], the median should be 2 because it’s the middle value.</p></li>
-</ul>
-</li>
-<li><p><strong>Constraints</strong>:</p>
-<ul class="simple">
-<li><p>The integers provided in the data stream are in the range from <span class="math notranslate nohighlight">\(-10^5\)</span> to <span class="math notranslate nohighlight">\(10^5\)</span>.</p></li>
-<li><p>There will be at least one element in the data structure before calling <code class="docutils literal notranslate"><span class="pre">findMedian</span></code>.</p></li>
-<li><p>At most, <span class="math notranslate nohighlight">\(5 * 10^4\)</span> calls will be made to <code class="docutils literal notranslate"><span class="pre">addNum</span></code> and <code class="docutils literal notranslate"><span class="pre">findMedian</span></code>.</p></li>
-</ul>
-</li>
-</ol>
-<p>The problem can be solved by using two priority queues (heaps): one max-heap to store the smaller half of the numbers and one min-heap to store the larger half of the numbers. The max-heap ensures that the largest number in the smaller half is at the top, and the min-heap ensures that the smallest number in the larger half is at the top. These heaps are balanced to efficiently find the median when requested.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">heapq</span>
-
-<span class="k">class</span> <span class="nc">MedianFinder</span><span class="p">:</span>
-
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">):</span>
-        <span class="c1"># Initialize two heaps: a max-heap (small_half) for the smaller half of numbers,</span>
-        <span class="c1"># and a min-heap (large_half) for the larger half of numbers.</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">small_half</span> <span class="o">=</span> <span class="p">[]</span>  <span class="c1"># Max-heap for the smaller half of the numbers</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">large_half</span> <span class="o">=</span> <span class="p">[]</span>  <span class="c1"># Min-heap for the larger half of the numbers</span>
-
-    <span class="k">def</span> <span class="nf">addNum</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">num</span><span class="p">:</span> <span class="nb">int</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="kc">None</span><span class="p">:</span>
-        <span class="c1"># Add the number to the appropriate heap based on its value</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="bp">self</span><span class="o">.</span><span class="n">small_half</span> <span class="ow">or</span> <span class="n">num</span> <span class="o">&lt;=</span> <span class="o">-</span><span class="bp">self</span><span class="o">.</span><span class="n">small_half</span><span class="p">[</span><span class="mi">0</span><span class="p">]:</span>
-            <span class="c1"># If the number is less than or equal to the current maximum in the smaller half,</span>
-            <span class="c1"># add it to the smaller half (max-heap)</span>
-            <span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">small_half</span><span class="p">,</span> <span class="o">-</span><span class="n">num</span><span class="p">)</span>
-        <span class="k">else</span><span class="p">:</span>
-            <span class="c1"># Otherwise, add it to the larger half (min-heap)</span>
-            <span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">large_half</span><span class="p">,</span> <span class="n">num</span><span class="p">)</span>
-
-        <span class="c1"># Balance the heaps if necessary to ensure the size difference is at most 1</span>
-        <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">small_half</span><span class="p">)</span> <span class="o">&gt;</span> <span class="nb">len</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">large_half</span><span class="p">)</span> <span class="o">+</span> <span class="mi">1</span><span class="p">:</span>
-            <span class="c1"># If the size of the smaller half is more than one greater than the size of the larger half,</span>
-            <span class="c1"># move the maximum from the smaller half to the larger half to balance them.</span>
-            <span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">large_half</span><span class="p">,</span> <span class="o">-</span><span class="n">heapq</span><span class="o">.</span><span class="n">heappop</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">small_half</span><span class="p">))</span>
-        <span class="k">elif</span> <span class="nb">len</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">large_half</span><span class="p">)</span> <span class="o">&gt;</span> <span class="nb">len</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">small_half</span><span class="p">):</span>
-            <span class="c1"># If the size of the larger half is greater than the size of the smaller half,</span>
-            <span class="c1"># move the minimum from the larger half to the smaller half to balance them.</span>
-            <span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">small_half</span><span class="p">,</span> <span class="o">-</span><span class="n">heapq</span><span class="o">.</span><span class="n">heappop</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">large_half</span><span class="p">))</span>
-
-    <span class="k">def</span> <span class="nf">findMedian</span><span class="p">(</span><span class="bp">self</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="nb">float</span><span class="p">:</span>
-        <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">small_half</span><span class="p">)</span> <span class="o">==</span> <span class="nb">len</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">large_half</span><span class="p">):</span>
-            <span class="c1"># If both halves have the same size, there&#39;s an even number of elements,</span>
-            <span class="c1"># so the median is the average of the top elements in both heaps.</span>
-            <span class="k">return</span> <span class="p">(</span><span class="o">-</span><span class="bp">self</span><span class="o">.</span><span class="n">small_half</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">+</span> <span class="bp">self</span><span class="o">.</span><span class="n">large_half</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span> <span class="o">/</span> <span class="mf">2.0</span>
-        <span class="k">else</span><span class="p">:</span>
-            <span class="c1"># If the smaller half has more elements, it contains the median (odd number of elements).</span>
-            <span class="k">return</span> <span class="o">-</span><span class="bp">self</span><span class="o">.</span><span class="n">small_half</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The code defines a <code class="docutils literal notranslate"><span class="pre">MedianFinder</span></code> class for efficiently finding the median of a stream of numbers. It uses two heaps: a max-heap (<code class="docutils literal notranslate"><span class="pre">small_half</span></code>) to store the smaller half of the numbers and a min-heap (<code class="docutils literal notranslate"><span class="pre">large_half</span></code>) to store the larger half. Here’s how the code works:</p>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">__init__</span></code> method initializes the two heaps.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">addNum</span></code> method adds a number to the appropriate heap, ensuring that the heaps remain balanced. It does this by comparing the number to the current maximum in the smaller half and adding it to the appropriate heap. Then, it checks the sizes of the two heaps and rebalances them if necessary.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">findMedian</span></code> method calculates and returns the median. If both halves have the same size (even number of elements), it computes the average of the tops of both heaps. If the smaller half has more elements (odd number of elements), it returns the top of the max-heap, which is the median.</p></li>
-</ol>
-<p>In summary, this code efficiently maintains two heaps to divide the elements into smaller and larger halves, allowing for quick median retrieval. It offers a time complexity of O(log N) for adding elements and O(1) for finding the median, where N is the total number of elements processed. The space complexity is O(1) as the space used by the two heaps is independent of the input size.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Example usage:</span>
-<span class="n">medianFinder</span> <span class="o">=</span> <span class="n">MedianFinder</span><span class="p">()</span>
-<span class="n">medianFinder</span><span class="o">.</span><span class="n">addNum</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-<span class="n">medianFinder</span><span class="o">.</span><span class="n">addNum</span><span class="p">(</span><span class="mi">2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">medianFinder</span><span class="o">.</span><span class="n">findMedian</span><span class="p">())</span>  <span class="c1"># Output: 1.5</span>
-<span class="n">medianFinder</span><span class="o">.</span><span class="n">addNum</span><span class="p">(</span><span class="mi">3</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">medianFinder</span><span class="o">.</span><span class="n">findMedian</span><span class="p">())</span>  <span class="c1"># Output: 2.0</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>1.5
-2
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">MedianFinder</span></code> class:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">MedianFinder()</span></code> Initialization: This operation has a time complexity of O(1). It simply initializes the two heaps.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">addNum(int</span> <span class="pre">num)</span></code>: The time complexity for adding a number is O(log N), where N is the total number of elements processed so far. This is because, in the worst case, you might need to perform logarithmic operations on the heaps to maintain their balance.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">findMedian()</span></code>: Finding the median has a time complexity of O(1). It simply involves accessing the tops of the two heaps, which takes constant time.</p></li>
-</ol>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">MedianFinder()</span></code> Initialization: The space complexity for initialization is O(1). It involves creating two empty heaps, which do not depend on the size of the data stream.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">addNum(int</span> <span class="pre">num)</span></code>: The space complexity for adding a number is also O(1). It involves adding the number to one of the two heaps, which do not significantly affect the overall space complexity.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">findMedian()</span></code>: The space complexity for finding the median is O(1). It doesn’t require any additional data structures that grow with the input size.</p></li>
-</ol>
-<p>In summary, the time complexity of the <code class="docutils literal notranslate"><span class="pre">addNum</span></code> operation is O(log N), and the time complexity of the <code class="docutils literal notranslate"><span class="pre">findMedian</span></code> operation is O(1). The space complexity is O(1) as well, as the space used by the two heaps is independent of the size of the data stream. This solution efficiently maintains and retrieves the median, making it suitable for the given problem constraints.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Generalized kth Element</strong>: Extend the <code class="docutils literal notranslate"><span class="pre">MedianFinder</span></code> class to support finding the kth element in the data stream efficiently. This means finding the kth smallest or kth largest element in the stream.</p></li>
-<li><p><strong>Sliding Window Median</strong>: Given an array and a sliding window size, find the median within the window as it slides over the array. This is an extension of the problem to a moving window scenario.</p></li>
-</ol>
-</section>
-</section>
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-  <section class="tex2jax_ignore mathjax_ignore" id="kth-largest-element-in-a-stream">
-<h1>703. Kth Largest Element in a Stream<a class="headerlink" href="#kth-largest-element-in-a-stream" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/kth-largest-element-in-a-stream/">To see the Kth Largest Element in a Stream problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Design a class to find the <code class="docutils literal notranslate"><span class="pre">kth</span></code> largest element in a stream. Note that it is the <code class="docutils literal notranslate"><span class="pre">kth</span></code> largest element in the sorted order, not the <code class="docutils literal notranslate"><span class="pre">kth</span></code> distinct element.</p>
-<p>Implement the <code class="docutils literal notranslate"><span class="pre">KthLargest</span></code> class with the following methods:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">KthLargest(int</span> <span class="pre">k,</span> <span class="pre">int[]</span> <span class="pre">nums)</span></code>: Initializes the object with the integer <code class="docutils literal notranslate"><span class="pre">k</span></code> and the stream of integers <code class="docutils literal notranslate"><span class="pre">nums</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">int</span> <span class="pre">add(int</span> <span class="pre">val)</span></code>: Appends the integer <code class="docutils literal notranslate"><span class="pre">val</span></code> to the stream and returns the element representing the <code class="docutils literal notranslate"><span class="pre">kth</span></code> largest element in the stream.</p></li>
-</ul>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">k</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p>0 &lt;= <code class="docutils literal notranslate"><span class="pre">nums.length</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^4\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">nums[i]</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p><span class="math notranslate nohighlight">\(-10^4\)</span> &lt;= <code class="docutils literal notranslate"><span class="pre">val</span></code> &lt;= <span class="math notranslate nohighlight">\(10^4\)</span></p></li>
-<li><p>At most <span class="math notranslate nohighlight">\(10^4\)</span> calls will be made to <code class="docutils literal notranslate"><span class="pre">add</span></code>.</p></li>
-<li><p>It is guaranteed that there will be at least <code class="docutils literal notranslate"><span class="pre">k</span></code> elements in the array when you search for the <code class="docutils literal notranslate"><span class="pre">kth</span></code> element.</p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem you’re trying to solve is to design a class called <code class="docutils literal notranslate"><span class="pre">KthLargest</span></code> that can efficiently find the kth largest element in a stream of integers. You need to support two main operations:</p>
-<p>To solve this problem efficiently, you can use a min-heap data structure. Here’s how the approach works:</p>
-<ol class="arabic simple">
-<li><p>Initialize a min-heap to store the k largest elements. Initially, it’s empty.</p></li>
-<li><p>In the <code class="docutils literal notranslate"><span class="pre">__init__</span></code> method, populate the min-heap with the first <code class="docutils literal notranslate"><span class="pre">k</span></code> elements from <code class="docutils literal notranslate"><span class="pre">nums</span></code>. This ensures that you have the k largest elements seen so far.</p></li>
-<li><p>Whenever you add a new element to the stream using the <code class="docutils literal notranslate"><span class="pre">add</span></code> method, follow these steps:</p>
-<ul class="simple">
-<li><p>Add the new element to the min-heap.</p></li>
-<li><p>If the size of the min-heap exceeds <code class="docutils literal notranslate"><span class="pre">k</span></code>, remove the smallest element from the min-heap. This ensures that you always have the k largest elements in the min-heap.</p></li>
-<li><p>The smallest element in the min-heap (the root) will always represent the kth largest element in the stream.</p></li>
-</ul>
-</li>
-</ol>
-<p>Here’s why this approach works efficiently:</p>
-<ul class="simple">
-<li><p>By using a min-heap, you can quickly maintain the k largest elements, and finding the smallest element in the heap (the root) takes constant time.</p></li>
-<li><p>When you add a new element, the min-heap’s size is kept at most <code class="docutils literal notranslate"><span class="pre">k</span></code>, which ensures that you only track the k largest elements and discard the smaller ones.</p></li>
-<li><p>The time complexity for adding an element is O(log k), which is very efficient compared to sorting the entire stream, which would be O(n log n).</p></li>
-<li><p>This approach meets the constraints of the problem, including handling large streams and having a low time complexity for both initialization and adding elements.</p></li>
-</ul>
-<p>In summary, the min-heap approach efficiently tracks the kth largest element in the stream by maintaining a heap of the k largest elements seen so far, updating it as new elements are added. This approach provides a fast and scalable solution to the problem.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">heapq</span>
-<span class="kn">from</span> <span class="nn">typing</span> <span class="kn">import</span> <span class="n">List</span>  <span class="c1"># Import the List type from the typing module</span>
-
-<span class="k">class</span> <span class="nc">KthLargest</span><span class="p">:</span>
-
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">k</span><span class="p">:</span> <span class="nb">int</span><span class="p">,</span> <span class="n">nums</span><span class="p">:</span> <span class="n">List</span><span class="p">[</span><span class="nb">int</span><span class="p">]):</span>
-        <span class="c1"># Initialize the KthLargest object with k and nums.</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">k</span> <span class="o">=</span> <span class="n">k</span>
-        <span class="c1"># Create a min-heap to store the k largest elements.</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">min_heap</span> <span class="o">=</span> <span class="p">[]</span>
-        <span class="c1"># Populate the min-heap with the first k elements from nums.</span>
-        <span class="k">for</span> <span class="n">num</span> <span class="ow">in</span> <span class="n">nums</span><span class="p">:</span>
-            <span class="bp">self</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="n">num</span><span class="p">)</span>
-
-    <span class="k">def</span> <span class="nf">add</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="p">:</span> <span class="nb">int</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="nb">int</span><span class="p">:</span>
-        <span class="c1"># Add val to the min-heap.</span>
-        <span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">min_heap</span><span class="p">,</span> <span class="n">val</span><span class="p">)</span>
-        <span class="c1"># If the size of the min-heap exceeds k, remove the smallest element.</span>
-        <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">min_heap</span><span class="p">)</span> <span class="o">&gt;</span> <span class="bp">self</span><span class="o">.</span><span class="n">k</span><span class="p">:</span>
-            <span class="n">heapq</span><span class="o">.</span><span class="n">heappop</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">min_heap</span><span class="p">)</span>
-        <span class="c1"># The root of the min-heap is the kth largest element.</span>
-        <span class="k">return</span> <span class="bp">self</span><span class="o">.</span><span class="n">min_heap</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p>We start by importing the necessary modules:</p>
-<ul>
-<li><p><code class="docutils literal notranslate"><span class="pre">heapq</span></code>: This module provides functions to create and manipulate heaps.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">List</span></code> from the <code class="docutils literal notranslate"><span class="pre">typing</span></code> module: This is used to specify the type of the <code class="docutils literal notranslate"><span class="pre">nums</span></code> parameter.</p></li>
-</ul>
-</li>
-<li><p>We define the <code class="docutils literal notranslate"><span class="pre">KthLargest</span></code> class:</p>
-<ul>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">__init__</span></code> method initializes the object with the integer <code class="docutils literal notranslate"><span class="pre">k</span></code> and the list of integers <code class="docutils literal notranslate"><span class="pre">nums</span></code>.</p></li>
-<li><p>It also creates an empty min-heap called <code class="docutils literal notranslate"><span class="pre">self.min_heap</span></code> to store the k largest elements.</p></li>
-<li><p>It populates the min-heap with the first <code class="docutils literal notranslate"><span class="pre">k</span></code> elements from <code class="docutils literal notranslate"><span class="pre">nums</span></code> by calling the <code class="docutils literal notranslate"><span class="pre">add</span></code> method.</p></li>
-</ul>
-</li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">add</span></code> method:</p>
-<ul>
-<li><p>Adds the new integer <code class="docutils literal notranslate"><span class="pre">val</span></code> to the min-heap using <code class="docutils literal notranslate"><span class="pre">heapq.heappush</span></code>. This maintains the min-heap property.</p></li>
-<li><p>Checks if the size of the min-heap exceeds <code class="docutils literal notranslate"><span class="pre">k</span></code>. If it does, it removes the smallest element (the k+1th largest) using <code class="docutils literal notranslate"><span class="pre">heapq.heappop</span></code>.</p></li>
-<li><p>Finally, it returns the smallest element in the min-heap, which is always the kth largest element.</p></li>
-</ul>
-</li>
-</ul>
-<p>Overall, this code implements a class that efficiently finds the kth largest element in a stream of integers by maintaining a min-heap of the k largest elements seen so far.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Example usage:</span>
-<span class="n">kthLargest</span> <span class="o">=</span> <span class="n">KthLargest</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span> <span class="p">[</span><span class="mi">4</span><span class="p">,</span> <span class="mi">5</span><span class="p">,</span> <span class="mi">8</span><span class="p">,</span> <span class="mi">2</span><span class="p">])</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">kthLargest</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="mi">3</span><span class="p">))</span>  <span class="c1"># Output: 4</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">kthLargest</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="mi">5</span><span class="p">))</span>  <span class="c1"># Output: 5</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">kthLargest</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="mi">10</span><span class="p">))</span> <span class="c1"># Output: 5</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">kthLargest</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="mi">9</span><span class="p">))</span>  <span class="c1"># Output: 8</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">kthLargest</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="mi">4</span><span class="p">))</span>  <span class="c1"># Output: 8</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>4
-5
-5
-8
-8
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">KthLargest</span></code> class using the min-heap approach:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">__init__</span></code> method:</p>
-<ul class="simple">
-<li><p>In the <code class="docutils literal notranslate"><span class="pre">__init__</span></code> method, you iterate over the first <code class="docutils literal notranslate"><span class="pre">k</span></code> elements in <code class="docutils literal notranslate"><span class="pre">nums</span></code> and add them to the min-heap. Each <code class="docutils literal notranslate"><span class="pre">heapq.heappush</span></code> operation takes O(log k) time.</p></li>
-<li><p>Therefore, the time complexity of the <code class="docutils literal notranslate"><span class="pre">__init__</span></code> method is O(k * log k).</p></li>
-</ul>
-</li>
-<li><p><code class="docutils literal notranslate"><span class="pre">add</span></code> method:</p>
-<ul class="simple">
-<li><p>In the <code class="docutils literal notranslate"><span class="pre">add</span></code> method, you perform the following operations:</p>
-<ul>
-<li><p><code class="docutils literal notranslate"><span class="pre">heapq.heappush</span></code>: O(log k) to add the new element to the min-heap.</p></li>
-<li><p>If the size of the min-heap exceeds <code class="docutils literal notranslate"><span class="pre">k</span></code>, you perform <code class="docutils literal notranslate"><span class="pre">heapq.heappop</span></code>, which is also O(log k).</p></li>
-<li><p>Finally, you return the smallest element from the min-heap, which is O(1) because it’s always the root.</p></li>
-</ul>
-</li>
-<li><p>Overall, the time complexity of the <code class="docutils literal notranslate"><span class="pre">add</span></code> method is O(log k).</p></li>
-</ul>
-</li>
-<li><p>Overall, if you make <code class="docutils literal notranslate"><span class="pre">n</span></code> calls to the <code class="docutils literal notranslate"><span class="pre">add</span></code> method, the total time complexity is O(n * log k), where <code class="docutils literal notranslate"><span class="pre">n</span></code> is the total number of elements added to the stream.</p></li>
-</ol>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The space complexity is determined by the space used to store the min-heap and the instance variables.</p></li>
-<li><p>The min-heap stores at most <code class="docutils literal notranslate"><span class="pre">k</span></code> elements at any given time, so its space complexity is O(k).</p></li>
-<li><p>The instance variables (such as <code class="docutils literal notranslate"><span class="pre">self.k</span></code> and <code class="docutils literal notranslate"><span class="pre">self.min_heap</span></code>) have constant space requirements.</p></li>
-<li><p>Therefore, the overall space complexity of the <code class="docutils literal notranslate"><span class="pre">KthLargest</span></code> class is O(k).</p></li>
-</ol>
-<p>In summary, the time complexity of the <code class="docutils literal notranslate"><span class="pre">KthLargest</span></code> class is O(n * log k) for <code class="docutils literal notranslate"><span class="pre">n</span></code> add operations, and the space complexity is O(k), where <code class="docutils literal notranslate"><span class="pre">k</span></code> is the parameter passed during initialization. This implementation efficiently maintains the kth largest element in the stream while meeting the problem’s constraints.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Dynamic k:</strong> Modify the <code class="docutils literal notranslate"><span class="pre">KthLargest</span></code> class to support dynamic changes in the value of <code class="docutils literal notranslate"><span class="pre">k</span></code>. Implement a method <code class="docutils literal notranslate"><span class="pre">update_k</span></code> that allows changing the value of <code class="docutils literal notranslate"><span class="pre">k</span></code> during the lifetime of the object. Ensure that the object can still correctly find the kth largest element based on the updated value of <code class="docutils literal notranslate"><span class="pre">k</span></code>.</p></li>
-<li><p><strong>Implement kth Smallest:</strong> Create a new class called <code class="docutils literal notranslate"><span class="pre">KthSmallest</span></code> that finds the kth smallest element in a stream instead of the kth largest. You may need to modify the data structure used in the implementation.</p></li>
-</ol>
-</section>
-</section>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases:</a></li>
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-<td><p>Easy</p></td>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="combination-sum">
-<h1>39. Combination Sum<a class="headerlink" href="#combination-sum" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/combination-sum/">To see the Combination Sum problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given an array of <strong>distinct</strong> integers <code class="docutils literal notranslate"><span class="pre">candidates</span></code> and a target integer <code class="docutils literal notranslate"><span class="pre">target</span></code>, return <em>a list of all <strong>unique combinations</strong> of <code class="docutils literal notranslate"><span class="pre">candidates</span></code> where the chosen numbers sum to the <code class="docutils literal notranslate"><span class="pre">target</span></code></em>. You may return the combinations in <strong>any order</strong>.</p>
-<p>The <strong>same</strong> number may be chosen from <code class="docutils literal notranslate"><span class="pre">candidates</span></code> an <strong>unlimited number</strong> of times. Two combinations are unique if the <strong>frequency</strong> of at least one of the chosen numbers is different.</p>
-<p>The test cases are generated such that the number of unique combinations that sum up to <code class="docutils literal notranslate"><span class="pre">target</span></code> is less than <code class="docutils literal notranslate"><span class="pre">150</span></code> combinations for the given input.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">candidates.length</span> <span class="pre">&lt;=</span> <span class="pre">30</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">2</span> <span class="pre">&lt;=</span> <span class="pre">candidates[i]</span> <span class="pre">&lt;=</span> <span class="pre">40</span></code></p></li>
-<li><p>All elements of <code class="docutils literal notranslate"><span class="pre">candidates</span></code> are <strong>distinct</strong>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">target</span> <span class="pre">&lt;=</span> <span class="pre">40</span></code></p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem is to find all unique combinations of numbers from a given array (candidates) such that their sum equals a target number. Here are the details of the problem:</p>
-<ul class="simple">
-<li><p>You are given an array of distinct integers called “candidates.”</p></li>
-<li><p>You are also given a target integer called “target.”</p></li>
-</ul>
-<p>The goal is to find all unique combinations of numbers from the candidates array where the sum of the selected numbers is equal to the target. You can use the same number from the candidates array an unlimited number of times. A combination is considered unique if it has a different frequency (i.e., a different number of occurrences) of at least one chosen number compared to other combinations.</p>
-<p>For example:</p>
-<p><strong>Example 1:</strong></p>
-<div class="highlight-python notranslate"><div class="highlight"><pre><span></span><span class="n">Input</span><span class="p">:</span> <span class="n">candidates</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">7</span><span class="p">],</span> <span class="n">target</span> <span class="o">=</span> <span class="mi">7</span>
-<span class="n">Output</span><span class="p">:</span> <span class="p">[[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">],</span> <span class="p">[</span><span class="mi">7</span><span class="p">]]</span>
-</pre></div>
-</div>
-<p>In this example, there are two unique combinations that sum up to the target:</p>
-<ul class="simple">
-<li><p>2 + 2 + 3 = 7</p></li>
-<li><p>7 = 7</p></li>
-</ul>
-<p><strong>Example 2:</strong></p>
-<div class="highlight-python notranslate"><div class="highlight"><pre><span></span><span class="n">Input</span><span class="p">:</span> <span class="n">candidates</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">5</span><span class="p">],</span> <span class="n">target</span> <span class="o">=</span> <span class="mi">8</span>
-<span class="n">Output</span><span class="p">:</span> <span class="p">[[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">2</span><span class="p">],</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">3</span><span class="p">],</span> <span class="p">[</span><span class="mi">3</span><span class="p">,</span> <span class="mi">5</span><span class="p">]]</span>
-</pre></div>
-</div>
-<p>Here, there are three unique combinations that sum up to the target.</p>
-<p><strong>Example 3:</strong></p>
-<div class="highlight-python notranslate"><div class="highlight"><pre><span></span><span class="n">Input</span><span class="p">:</span> <span class="n">candidates</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">],</span> <span class="n">target</span> <span class="o">=</span> <span class="mi">1</span>
-<span class="n">Output</span><span class="p">:</span> <span class="p">[]</span>
-</pre></div>
-</div>
-<p>In this case, there are no combinations that can be formed from the candidates to reach the target of 1, so the output is an empty list.</p>
-<p>The problem asks you to find and return these combinations in any order.</p>
-<p>The constraints for this problem include the length of the candidates array, the values of candidates, and the target value.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">combinationSum</span><span class="p">(</span><span class="n">candidates</span><span class="p">,</span> <span class="n">target</span><span class="p">):</span>
-    <span class="c1"># Define a recursive DFS function to find combinations</span>
-    <span class="k">def</span> <span class="nf">dfs</span><span class="p">(</span><span class="n">remaining</span><span class="p">,</span> <span class="n">start</span><span class="p">,</span> <span class="n">path</span><span class="p">):</span>
-        <span class="c1"># If the remaining target is 0, we found a valid combination</span>
-        <span class="k">if</span> <span class="n">remaining</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
-            <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">path</span><span class="p">)</span>
-            <span class="k">return</span>
-        <span class="c1"># If the remaining target is negative, this path is invalid</span>
-        <span class="k">if</span> <span class="n">remaining</span> <span class="o">&lt;</span> <span class="mi">0</span><span class="p">:</span>
-            <span class="k">return</span>
-        <span class="c1"># Iterate through candidates starting from &#39;start&#39; to avoid duplicates</span>
-        <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">start</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">candidates</span><span class="p">)):</span>
-            <span class="c1"># Explore the current candidate by subtracting it from the remaining target</span>
-            <span class="c1"># Add the current candidate to the path</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">remaining</span> <span class="o">-</span> <span class="n">candidates</span><span class="p">[</span><span class="n">i</span><span class="p">],</span> <span class="n">i</span><span class="p">,</span> <span class="n">path</span> <span class="o">+</span> <span class="p">[</span><span class="n">candidates</span><span class="p">[</span><span class="n">i</span><span class="p">]])</span>
-
-    <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>  <span class="c1"># Initialize an empty list to store the result</span>
-    <span class="n">candidates</span><span class="o">.</span><span class="n">sort</span><span class="p">()</span>  <span class="c1"># Sort the candidates for deduplication and early stopping</span>
-    <span class="n">dfs</span><span class="p">(</span><span class="n">target</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="p">[])</span>  <span class="c1"># Start the DFS from the target value with an empty path</span>
-    <span class="k">return</span> <span class="n">result</span>  <span class="c1"># Return the list of unique combinations</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The <code class="docutils literal notranslate"><span class="pre">combinationSum</span></code> function takes an array of distinct integers <code class="docutils literal notranslate"><span class="pre">candidates</span></code> and a target integer <code class="docutils literal notranslate"><span class="pre">target</span></code>. It returns a list of all unique combinations of <code class="docutils literal notranslate"><span class="pre">candidates</span></code> where the chosen numbers sum to the <code class="docutils literal notranslate"><span class="pre">target</span></code>. Each number in the <code class="docutils literal notranslate"><span class="pre">candidates</span></code> array can be used an unlimited number of times.</p>
-<p>Here’s an overview of how the code works:</p>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">combinationSum</span></code> function is defined, and it takes two arguments: <code class="docutils literal notranslate"><span class="pre">candidates</span></code> and <code class="docutils literal notranslate"><span class="pre">target</span></code>.</p></li>
-<li><p>Inside the function, there is a nested helper function called <code class="docutils literal notranslate"><span class="pre">backtrack</span></code>. This function is responsible for the actual combination generation using a recursive approach.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">backtrack</span></code> function is called with three arguments: <code class="docutils literal notranslate"><span class="pre">start</span></code>, <code class="docutils literal notranslate"><span class="pre">target</span></code>, and <code class="docutils literal notranslate"><span class="pre">path</span></code>. The <code class="docutils literal notranslate"><span class="pre">start</span></code> variable helps keep track of the current index in the <code class="docutils literal notranslate"><span class="pre">candidates</span></code> array, <code class="docutils literal notranslate"><span class="pre">target</span></code> keeps track of the remaining sum to be achieved, and <code class="docutils literal notranslate"><span class="pre">path</span></code> is a list that stores the current combination.</p></li>
-<li><p>Within the <code class="docutils literal notranslate"><span class="pre">backtrack</span></code> function, there are three main conditional statements:</p>
-<ul class="simple">
-<li><p>If <code class="docutils literal notranslate"><span class="pre">target</span></code> becomes zero, it means we have found a combination that sums up to the target, so we add <code class="docutils literal notranslate"><span class="pre">path</span></code> to the <code class="docutils literal notranslate"><span class="pre">result</span></code> list.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">target</span></code> becomes negative, it means the current combination doesn’t work, so we return without doing anything.</p></li>
-<li><p>Otherwise, we enter a loop that iterates over the <code class="docutils literal notranslate"><span class="pre">candidates</span></code> array, starting from the current index <code class="docutils literal notranslate"><span class="pre">start</span></code>.</p></li>
-</ul>
-</li>
-<li><p>In the loop, the <code class="docutils literal notranslate"><span class="pre">backtrack</span></code> function is called recursively with the updated <code class="docutils literal notranslate"><span class="pre">target</span></code> and <code class="docutils literal notranslate"><span class="pre">path</span></code> after including the current candidate. This process explores different combinations by considering the current candidate or moving to the next candidate.</p></li>
-<li><p>After the loop completes, the <code class="docutils literal notranslate"><span class="pre">result</span></code> list contains all unique combinations that sum to the target.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">candidates</span></code> array is sorted to optimize the search. Sorting helps in avoiding unnecessary recursive branches and reduces the number of explored combinations.</p></li>
-<li><p>Finally, the <code class="docutils literal notranslate"><span class="pre">backtrack</span></code> function is initially called from the <code class="docutils literal notranslate"><span class="pre">combinationSum</span></code> function with <code class="docutils literal notranslate"><span class="pre">start</span></code> set to 0, <code class="docutils literal notranslate"><span class="pre">target</span></code> set to the original target value, and an empty <code class="docutils literal notranslate"><span class="pre">path</span></code>. The result is returned as a list of unique combinations.</p></li>
-</ol>
-<p>The code includes a few example cases to demonstrate how the function works with different inputs.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Example 1</span>
-<span class="n">candidates1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">6</span><span class="p">,</span> <span class="mi">7</span><span class="p">]</span>
-<span class="n">target1</span> <span class="o">=</span> <span class="mi">7</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">combinationSum</span><span class="p">(</span><span class="n">candidates1</span><span class="p">,</span> <span class="n">target1</span><span class="p">))</span>  <span class="c1"># Output: [[2, 2, 3], [7]]</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[2, 2, 3], [7]]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">candidates2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">5</span><span class="p">]</span>
-<span class="n">target2</span> <span class="o">=</span> <span class="mi">8</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">combinationSum</span><span class="p">(</span><span class="n">candidates2</span><span class="p">,</span> <span class="n">target2</span><span class="p">))</span>  <span class="c1"># Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[2, 2, 2, 2], [2, 3, 3], [3, 5]]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3</span>
-<span class="n">candidates3</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">]</span>
-<span class="n">target3</span> <span class="o">=</span> <span class="mi">1</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">combinationSum</span><span class="p">(</span><span class="n">candidates3</span><span class="p">,</span> <span class="n">target3</span><span class="p">))</span>  <span class="c1"># Output: []</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">combinationSum</span></code> function.</p>
-<p><strong>Time Complexity:</strong>
-The time complexity of this function is influenced by the number of recursive calls made by the <code class="docutils literal notranslate"><span class="pre">backtrack</span></code> function and the amount of work done within each call. In the worst case, the function will explore all possible combinations.</p>
-<p>The worst-case scenario occurs when we have many combinations to reach the target. The time complexity can be expressed as <span class="math notranslate nohighlight">\(O(2^n)\)</span>, where ‘n’ is the maximum number of recursive calls. In this context, ‘n’ corresponds to the number of elements in the <code class="docutils literal notranslate"><span class="pre">candidates</span></code> list, and it may be up to 30 (as per the constraints). However, the actual number of combinations explored is generally much smaller because we eliminate branches when the target becomes negative, and we skip over candidates that cannot lead to a solution. The sorting step also helps to optimize the search.</p>
-<p><strong>Space Complexity:</strong>
-The space complexity is determined by the auxiliary space used for storing the results and the stack space for the recursive calls.</p>
-<ol class="arabic simple">
-<li><p>The space required for the <code class="docutils literal notranslate"><span class="pre">result</span></code> list can be up to <span class="math notranslate nohighlight">\(O(2^n)\)</span> in the worst case because each combination is stored.</p></li>
-<li><p>The stack space for the recursive calls can also be up to O(n) in the worst case, where ‘n’ is the number of elements in the <code class="docutils literal notranslate"><span class="pre">candidates</span></code> list.</p></li>
-</ol>
-<p>Therefore, the overall space complexity of the function is <span class="math notranslate nohighlight">\(O(2^n + n)\)</span>. However, in practice, the space used for the results list is often the dominant factor, and the actual space used may be much smaller than 2^n because not all combinations are explored.</p>
-<p>In summary, the time complexity is exponential but typically smaller in practice due to optimization, and the space complexity is influenced by the number of results and the depth of the recursion.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Find the Number of Unique Combinations:</strong> Instead of listing all unique combinations, modify the function to return the count of unique combinations that sum to the target.</p></li>
-<li><p><strong>Combinations with a Maximum Value:</strong> Add a maximum value constraint for each combination, so they cannot exceed a certain value. Modify the code to find combinations respecting this constraint.</p></li>
-</ol>
-</section>
-</section>
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-  <section class="tex2jax_ignore mathjax_ignore" id="word-search">
-<h1>79. Word Search<a class="headerlink" href="#word-search" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/word-search/">To see the Word Search problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given an <code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">x</span> <span class="pre">n</span></code> grid of characters <code class="docutils literal notranslate"><span class="pre">board</span></code> and a string <code class="docutils literal notranslate"><span class="pre">word</span></code>, return <em><code class="docutils literal notranslate"><span class="pre">true</span></code> if <code class="docutils literal notranslate"><span class="pre">word</span></code> exists in the grid.</em></p>
-<p>The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">==</span> <span class="pre">board.length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">=</span> <span class="pre">board[i].length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">m,</span> <span class="pre">n</span> <span class="pre">&lt;=</span> <span class="pre">6</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">word.length</span> <span class="pre">&lt;=</span> <span class="pre">15</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">board</span></code> and <code class="docutils literal notranslate"><span class="pre">word</span></code> consists of only lowercase and uppercase English letters.</p></li>
-</ul>
-<p><strong>Follow up:</strong> Could you use search pruning to make your solution faster with a larger <code class="docutils literal notranslate"><span class="pre">board</span></code>?</p>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem is to determine whether a given word can be constructed from characters in an <code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">x</span> <span class="pre">n</span></code> grid of characters. The word can be formed by moving sequentially through horizontally or vertically neighboring cells in the grid, and you cannot use the same cell more than once.</p>
-<p>Here are the specifics of the problem:</p>
-<ul class="simple">
-<li><p>You are given an <code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">x</span> <span class="pre">n</span></code> grid, which is a 2D board of characters.</p></li>
-<li><p>You are also given a string <code class="docutils literal notranslate"><span class="pre">word</span></code> consisting of lowercase and uppercase English letters.</p></li>
-<li><p>Your task is to determine if the word can be found in the grid by moving from one cell to an adjacent cell (horizontally or vertically) and collecting the characters sequentially to form the word.</p></li>
-<li><p>You can’t use the same cell more than once when forming the word.</p></li>
-<li><p>If it’s possible to form the word in the grid, return <code class="docutils literal notranslate"><span class="pre">True</span></code>; otherwise, return <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-</ul>
-<p>For example, consider the following grid and word:</p>
-<div class="highlight-python notranslate"><div class="highlight"><pre><span></span><span class="n">board</span> <span class="o">=</span> <span class="p">[[</span><span class="s2">&quot;A&quot;</span><span class="p">,</span> <span class="s2">&quot;B&quot;</span><span class="p">,</span> <span class="s2">&quot;C&quot;</span><span class="p">,</span> <span class="s2">&quot;E&quot;</span><span class="p">],</span>
-         <span class="p">[</span><span class="s2">&quot;S&quot;</span><span class="p">,</span> <span class="s2">&quot;F&quot;</span><span class="p">,</span> <span class="s2">&quot;C&quot;</span><span class="p">,</span> <span class="s2">&quot;S&quot;</span><span class="p">],</span>
-         <span class="p">[</span><span class="s2">&quot;A&quot;</span><span class="p">,</span> <span class="s2">&quot;D&quot;</span><span class="p">,</span> <span class="s2">&quot;E&quot;</span><span class="p">,</span> <span class="s2">&quot;E&quot;</span><span class="p">]]</span>         
-<span class="n">word</span> <span class="o">=</span> <span class="s2">&quot;ABCCED&quot;</span>
-</pre></div>
-</div>
-<p>In this case, the word “ABCCED” can be formed by moving through the cells (0,0) -&gt; (0,1) -&gt; (0,2) -&gt; (1,2) -&gt; (2,2) -&gt; (2,3) in the grid. Therefore, the function should return <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p>
-<p>Conversely, if the word is impossible to construct, such as in the case of the word “ABCB” for the same grid, where there’s no valid path that forms the word, the function should return <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p>
-<p>The problem involves searching for a path through the grid that matches the characters in the word while respecting the constraints mentioned. It can be solved using depth-first search (DFS) or backtracking.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">from</span> <span class="nn">typing</span> <span class="kn">import</span> <span class="n">List</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">exist</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">board</span><span class="p">:</span> <span class="n">List</span><span class="p">[</span><span class="n">List</span><span class="p">[</span><span class="nb">str</span><span class="p">]],</span> <span class="n">word</span><span class="p">:</span> <span class="nb">str</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="nb">bool</span><span class="p">:</span>
-        <span class="k">def</span> <span class="nf">dfs</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="n">k</span><span class="p">):</span>
-            <span class="c1"># Check if the current position is out of bounds or doesn&#39;t match the character in the word.</span>
-            <span class="k">if</span> <span class="ow">not</span> <span class="p">(</span><span class="mi">0</span> <span class="o">&lt;=</span> <span class="n">i</span> <span class="o">&lt;</span> <span class="nb">len</span><span class="p">(</span><span class="n">board</span><span class="p">))</span> <span class="ow">or</span> <span class="ow">not</span> <span class="p">(</span><span class="mi">0</span> <span class="o">&lt;=</span> <span class="n">j</span> <span class="o">&lt;</span> <span class="nb">len</span><span class="p">(</span><span class="n">board</span><span class="p">[</span><span class="mi">0</span><span class="p">]))</span> <span class="ow">or</span> <span class="n">board</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">]</span> <span class="o">!=</span> <span class="n">word</span><span class="p">[</span><span class="n">k</span><span class="p">]:</span>
-                <span class="k">return</span> <span class="kc">False</span>
-            
-            <span class="c1"># If we have matched all characters in the word, return True.</span>
-            <span class="k">if</span> <span class="n">k</span> <span class="o">==</span> <span class="nb">len</span><span class="p">(</span><span class="n">word</span><span class="p">)</span> <span class="o">-</span> <span class="mi">1</span><span class="p">:</span>
-                <span class="k">return</span> <span class="kc">True</span>
-
-            <span class="c1"># Temporarily mark the current cell to prevent revisiting.</span>
-            <span class="n">tmp</span><span class="p">,</span> <span class="n">board</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">]</span> <span class="o">=</span> <span class="n">board</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">],</span> <span class="s2">&quot;/&quot;</span>
-
-            <span class="c1"># Explore adjacent cells by making recursive calls in all four directions.</span>
-            <span class="n">found</span> <span class="o">=</span> <span class="n">dfs</span><span class="p">(</span><span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="n">k</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span> <span class="ow">or</span> <span class="n">dfs</span><span class="p">(</span><span class="n">i</span> <span class="o">-</span> <span class="mi">1</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="n">k</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span> <span class="ow">or</span> <span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">j</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">k</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span> <span class="ow">or</span> <span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">j</span> <span class="o">-</span> <span class="mi">1</span><span class="p">,</span> <span class="n">k</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span>
-
-            <span class="c1"># Restore the original character in the cell.</span>
-            <span class="n">board</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">]</span> <span class="o">=</span> <span class="n">tmp</span>
-
-            <span class="k">return</span> <span class="n">found</span>
-
-        <span class="c1"># Iterate through all cells in the board.</span>
-        <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">board</span><span class="p">)):</span>
-            <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">board</span><span class="p">[</span><span class="mi">0</span><span class="p">])):</span>
-                <span class="c1"># Check if the &#39;dfs&#39; function starting from this cell returns True.</span>
-                <span class="k">if</span> <span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="mi">0</span><span class="p">):</span>
-                    <span class="k">return</span> <span class="kc">True</span>
-
-        <span class="c1"># If no match is found after exploring all cells, return False.</span>
-        <span class="k">return</span> <span class="kc">False</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>Import the necessary module:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">from</span> <span class="pre">typing</span> <span class="pre">import</span> <span class="pre">List</span></code>: This line imports the <code class="docutils literal notranslate"><span class="pre">List</span></code> type from the <code class="docutils literal notranslate"><span class="pre">typing</span></code> module, which is used to specify the type of the <code class="docutils literal notranslate"><span class="pre">board</span></code> parameter in the method signature.</p></li>
-</ul>
-</li>
-<li><p>Define the <code class="docutils literal notranslate"><span class="pre">Solution</span></code> class:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">class</span> <span class="pre">Solution:</span></code>: This line defines a class called <code class="docutils literal notranslate"><span class="pre">Solution</span></code> to encapsulate the solution for the word search problem.</p></li>
-</ul>
-</li>
-<li><p>Define the <code class="docutils literal notranslate"><span class="pre">exist</span></code> method:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">def</span> <span class="pre">exist(self,</span> <span class="pre">board:</span> <span class="pre">List[List[str]],</span> <span class="pre">word:</span> <span class="pre">str)</span> <span class="pre">-&gt;</span> <span class="pre">bool:</span></code>: This method is a member of the <code class="docutils literal notranslate"><span class="pre">Solution</span></code> class and takes two parameters: <code class="docutils literal notranslate"><span class="pre">board</span></code>, which is a 2D list of characters, and <code class="docutils literal notranslate"><span class="pre">word</span></code>, which is a string. It returns a boolean value, indicating whether the word exists in the board.</p></li>
-</ul>
-</li>
-<li><p>Implement the inner <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">def</span> <span class="pre">dfs(i,</span> <span class="pre">j,</span> <span class="pre">k):</span></code>: This is a helper function used for depth-first search (DFS). It is called recursively to explore the board and search for the word. It takes three parameters: <code class="docutils literal notranslate"><span class="pre">i</span></code> and <code class="docutils literal notranslate"><span class="pre">j</span></code> representing the current position on the board, and <code class="docutils literal notranslate"><span class="pre">k</span></code> representing the current index in the <code class="docutils literal notranslate"><span class="pre">word</span></code> string.</p></li>
-<li><p>Inside <code class="docutils literal notranslate"><span class="pre">dfs</span></code>, it checks if the current position is out of bounds or if the character at the current position on the board does not match the character at the current index in the <code class="docutils literal notranslate"><span class="pre">word</span></code>. If either of these conditions is met, it returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">k</span></code> is equal to the length of the <code class="docutils literal notranslate"><span class="pre">word</span></code> minus one, it means we have successfully matched all characters in the <code class="docutils literal notranslate"><span class="pre">word</span></code>, so it returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>The function then temporarily replaces the character at the current position with a marker, “/”, to prevent revisiting the same cell. It then explores adjacent cells by making recursive calls in all four directions: up, down, left, and right. If any of these recursive calls return <code class="docutils literal notranslate"><span class="pre">True</span></code>, it propagates the <code class="docutils literal notranslate"><span class="pre">True</span></code> result up the call stack. After the exploration is complete, it restores the original character in the cell.</p></li>
-</ul>
-</li>
-<li><p>Iterate through the board:</p>
-<ul class="simple">
-<li><p>The outer loop iterates through all cells in the <code class="docutils literal notranslate"><span class="pre">board</span></code> using two nested loops. For each cell, it checks if the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function starting from that cell (with <code class="docutils literal notranslate"><span class="pre">k</span></code> initially set to 0) returns <code class="docutils literal notranslate"><span class="pre">True</span></code>. If it does, it means the word exists, so the method returns <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-</ul>
-</li>
-<li><p>Return <code class="docutils literal notranslate"><span class="pre">False</span></code>:</p>
-<ul class="simple">
-<li><p>If the loop completes without finding the word in the board, the method returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-</ul>
-</li>
-</ol>
-<p>To use this code, you can create an instance of the <code class="docutils literal notranslate"><span class="pre">Solution</span></code> class and call the <code class="docutils literal notranslate"><span class="pre">exist</span></code> method on that instance, passing the <code class="docutils literal notranslate"><span class="pre">board</span></code> and the <code class="docutils literal notranslate"><span class="pre">word</span></code> as arguments to check if the word exists in the board.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1: </span>
-
-<span class="c1"># Create an instance of the Solution class</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-
-<span class="c1"># Test case 1: Word &quot;ABCCED&quot; can be found in the grid</span>
-<span class="n">board1</span> <span class="o">=</span> <span class="p">[[</span><span class="s2">&quot;A&quot;</span><span class="p">,</span><span class="s2">&quot;B&quot;</span><span class="p">,</span><span class="s2">&quot;C&quot;</span><span class="p">,</span><span class="s2">&quot;E&quot;</span><span class="p">],</span>
-          <span class="p">[</span><span class="s2">&quot;S&quot;</span><span class="p">,</span><span class="s2">&quot;F&quot;</span><span class="p">,</span><span class="s2">&quot;C&quot;</span><span class="p">,</span><span class="s2">&quot;S&quot;</span><span class="p">],</span>
-          <span class="p">[</span><span class="s2">&quot;A&quot;</span><span class="p">,</span><span class="s2">&quot;D&quot;</span><span class="p">,</span><span class="s2">&quot;E&quot;</span><span class="p">,</span><span class="s2">&quot;E&quot;</span><span class="p">]]</span>
-<span class="n">word1</span> <span class="o">=</span> <span class="s2">&quot;ABCCED&quot;</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">solution</span><span class="o">.</span><span class="n">exist</span><span class="p">(</span><span class="n">board1</span><span class="p">,</span> <span class="n">word1</span><span class="p">))</span>  <span class="c1"># Expected output: True</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-
-<span class="c1"># Test case 2: Word &quot;SEE&quot; can be found in the grid</span>
-<span class="n">word2</span> <span class="o">=</span> <span class="s2">&quot;SEE&quot;</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">solution</span><span class="o">.</span><span class="n">exist</span><span class="p">(</span><span class="n">board1</span><span class="p">,</span> <span class="n">word2</span><span class="p">))</span>  <span class="c1"># Expected output: True</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 3</span>
-
-<span class="c1"># Test case 3: Word &quot;ABCB&quot; cannot be found in the grid</span>
-<span class="n">word3</span> <span class="o">=</span> <span class="s2">&quot;ABCB&quot;</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">solution</span><span class="o">.</span><span class="n">exist</span><span class="p">(</span><span class="n">board1</span><span class="p">,</span> <span class="n">word3</span><span class="p">))</span>  <span class="c1"># Expected output: False</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">exist</span></code> method in the given code.</p>
-<p><strong>Time Complexity:</strong></p>
-<p>The primary work in this code is done by the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function, which performs a depth-first search through the grid. The time complexity of the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function is <span class="math notranslate nohighlight">\(O(4^(n))\)</span>, where “n” is the length of the <code class="docutils literal notranslate"><span class="pre">word</span></code>. This is because, in the worst case, the function explores all possible directions (up, down, left, right) for each character in the <code class="docutils literal notranslate"><span class="pre">word</span></code>.</p>
-<p>The outer loops iterate through all cells in the <code class="docutils literal notranslate"><span class="pre">board</span></code>, which has dimensions <code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">x</span> <span class="pre">n</span></code>. So, the total number of iterations of <code class="docutils literal notranslate"><span class="pre">dfs</span></code> is O(m * n). However, since the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function has a higher time complexity, the overall time complexity is still dominated by the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function.</p>
-<p>Therefore, the overall time complexity of the <code class="docutils literal notranslate"><span class="pre">exist</span></code> method is <span class="math notranslate nohighlight">\(O(m * n * 4^n)\)</span>, where “m” and “n” are the dimensions of the board, and “n” is the length of the word.</p>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The primary space usage comes from the recursive calls in the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function. The depth of the recursion can be at most equal to the length of the <code class="docutils literal notranslate"><span class="pre">word</span></code>, which is O(n).</p></li>
-<li><p>Additionally, the <code class="docutils literal notranslate"><span class="pre">tmp</span></code> variable and the recursive calls on the call stack consume a small amount of additional memory.</p></li>
-<li><p>The marking of visited cells with the “/” character temporarily modifies the <code class="docutils literal notranslate"><span class="pre">board</span></code> in-place but doesn’t significantly contribute to the space complexity.</p></li>
-</ol>
-<p>Therefore, the overall space complexity is O(n) due to the recursive call stack.</p>
-<p>In summary, the time complexity is <span class="math notranslate nohighlight">\(O(m * n * 4^n)\)</span>, and the space complexity is O(n), where “m” and “n” are the dimensions of the board, and “n” is the length of the word. The time complexity is exponential in the worst case due to the depth-first search through all possible paths in the grid.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Optimization Challenge:</strong> Modify the code to find all unique words that can be constructed from the grid. Instead of returning <code class="docutils literal notranslate"><span class="pre">True</span></code> or <code class="docutils literal notranslate"><span class="pre">False</span></code>, return a list of words found. Ensure that the same cell is not used more than once in constructing each word.</p></li>
-<li><p><strong>Board Generation:</strong> Create a function to generate random grids of characters and then use the <code class="docutils literal notranslate"><span class="pre">exist</span></code> method to solve them. Test the efficiency and correctness of your code by generating larger boards and words.</p></li>
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-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/combination-sum/">39. Combination Sum</a></p></td>
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-    <h1>133. Clone Graph</h1>
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-                <h2> Contents </h2>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="clone-graph">
-<h1>133. Clone Graph<a class="headerlink" href="#clone-graph" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/clone-graph/">To see the Clone Graph problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given a reference of a node in a <strong>connected</strong> undirected graph.</p>
-<p>Return a <strong>deep copy</strong> (clone) of the graph.</p>
-<p>Each node in the graph contains a value (<code class="docutils literal notranslate"><span class="pre">int</span></code>) and a list (<code class="docutils literal notranslate"><span class="pre">List[Node]</span></code>) of its neighbors.</p>
-<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">Node</span> <span class="p">{</span>
-    <span class="n">public</span> <span class="nb">int</span> <span class="n">val</span><span class="p">;</span>
-    <span class="n">public</span> <span class="n">List</span><span class="o">&lt;</span><span class="n">Node</span><span class="o">&gt;</span> <span class="n">neighbors</span><span class="p">;</span>
-<span class="p">}</span>
-</pre></div>
-</div>
-<p><strong>Test case format:</strong></p>
-<p>For simplicity, each node’s value is the same as the node’s index (1-indexed). For example, the first node with <code class="docutils literal notranslate"><span class="pre">val</span> <span class="pre">==</span> <span class="pre">1</span></code>, the second node with <code class="docutils literal notranslate"><span class="pre">val</span> <span class="pre">==</span> <span class="pre">2</span></code>, and so on. The graph is represented in the test case using an adjacency list.</p>
-<p><strong>An adjacency list</strong> is a collection of unordered <strong>lists</strong> used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.</p>
-<p>The given node will always be the first node with <code class="docutils literal notranslate"><span class="pre">val</span> <span class="pre">=</span> <span class="pre">1</span></code>. You must return the <strong>copy of the given node</strong> as a reference to the cloned graph.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>The number of nodes in the graph is in the range <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">100]</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">Node.val</span> <span class="pre">&lt;=</span> <span class="pre">100</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">Node.val</span></code> is unique for each node.</p></li>
-<li><p>There are no repeated edges and no self-loops in the graph.</p></li>
-<li><p>The Graph is connected and all nodes can be visited starting from the given node.</p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem at hand is to clone a connected undirected graph, which means we need to create a deep copy of the original graph while preserving its structure and relationships between nodes. The graph is represented using adjacency lists, where each node has a value (an integer) and a list of its neighbors.</p>
-<p>Here are the key aspects of the problem:</p>
-<ol class="arabic simple">
-<li><p><strong>Input</strong>: You are given a reference to one of the nodes in the original graph. This reference node serves as the entry point to the graph. The entire graph can be explored starting from this node.</p></li>
-<li><p><strong>Output</strong>: The goal is to create a deep copy of the graph and return a reference to the cloned graph. The cloned graph should have the same structure and relationships as the original one but should be a separate instance in memory.</p></li>
-<li><p><strong>Graph Structure</strong>: The graph is composed of nodes (vertices) and edges. Nodes have values (integers) and are connected to other nodes through edges (undirected connections). Each node has a list of neighbors, which are other nodes it is connected to.</p></li>
-<li><p><strong>Connected Graph</strong>: The problem assumes that the graph is connected, meaning that you can start from the reference node provided and traverse the entire graph by following the edges. This ensures that there are no isolated components in the graph.</p></li>
-</ol>
-<p>To solve this problem, you typically need to perform a graph traversal (e.g., depth-first search or breadth-first search) to visit all the nodes in the original graph, create corresponding nodes in the clone, and establish the same connections between nodes in the cloned graph as in the original one. To avoid duplication of nodes, you also need to keep track of visited nodes to ensure that the cloning process is efficient and that nodes are not duplicated in the clone.</p>
-<p>The problem can be challenging because you need to create a deep copy of the graph while handling cyclic dependencies between nodes and ensuring that the structure and relationships of the original graph are preserved in the clone. The solution requires recursion or a data structure to maintain the mapping between original nodes and their clones to achieve an efficient and accurate cloning process.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Definition for a Node.</span>
-<span class="k">class</span> <span class="nc">Node</span><span class="p">:</span>
-    <span class="k">def</span> <span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="n">neighbors</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
-        <span class="bp">self</span><span class="o">.</span><span class="n">neighbors</span> <span class="o">=</span> <span class="n">neighbors</span> <span class="k">if</span> <span class="n">neighbors</span> <span class="ow">is</span> <span class="ow">not</span> <span class="kc">None</span> <span class="k">else</span> <span class="p">[]</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">cloneGraph</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">node</span><span class="p">:</span> <span class="s1">&#39;Node&#39;</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="s1">&#39;Node&#39;</span><span class="p">:</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">node</span><span class="p">:</span>
-            <span class="k">return</span> <span class="kc">None</span>
-
-        <span class="n">visited</span> <span class="o">=</span> <span class="p">{}</span>  <span class="c1"># Dictionary to keep track of visited nodes</span>
-
-        <span class="k">def</span> <span class="nf">dfs</span><span class="p">(</span><span class="n">original_node</span><span class="p">):</span>
-            <span class="k">if</span> <span class="n">original_node</span> <span class="ow">in</span> <span class="n">visited</span><span class="p">:</span>
-                <span class="k">return</span> <span class="n">visited</span><span class="p">[</span><span class="n">original_node</span><span class="p">]</span>
-
-            <span class="c1"># Create a copy of the original node</span>
-            <span class="n">copy_node</span> <span class="o">=</span> <span class="n">Node</span><span class="p">(</span><span class="n">original_node</span><span class="o">.</span><span class="n">val</span><span class="p">)</span>
-            <span class="n">visited</span><span class="p">[</span><span class="n">original_node</span><span class="p">]</span> <span class="o">=</span> <span class="n">copy_node</span>
-
-            <span class="c1"># Recursively clone the neighbors of the original node</span>
-            <span class="k">for</span> <span class="n">neighbor</span> <span class="ow">in</span> <span class="n">original_node</span><span class="o">.</span><span class="n">neighbors</span><span class="p">:</span>
-                <span class="n">copy_node</span><span class="o">.</span><span class="n">neighbors</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">dfs</span><span class="p">(</span><span class="n">neighbor</span><span class="p">))</span>
-
-            <span class="k">return</span> <span class="n">copy_node</span>
-
-        <span class="k">return</span> <span class="n">dfs</span><span class="p">(</span><span class="n">node</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The given code defines a Python class, <code class="docutils literal notranslate"><span class="pre">Node</span></code>, and a solution class, <code class="docutils literal notranslate"><span class="pre">Solution</span></code>, to clone a connected undirected graph represented by nodes and their neighbors. Here’s an explanation of the code:</p>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">Node</span></code> class defines a node in the graph. Each node has a value (<code class="docutils literal notranslate"><span class="pre">val</span></code>) which is an integer and a list of neighbors (<code class="docutils literal notranslate"><span class="pre">neighbors</span></code>) that are other nodes in the graph.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">Solution</span></code> class contains the method <code class="docutils literal notranslate"><span class="pre">cloneGraph</span></code>, which takes a reference to the first node of the graph (<code class="docutils literal notranslate"><span class="pre">node</span></code>) as input and returns a deep copy of the entire graph, preserving the structure and relationships between nodes.</p></li>
-<li><p>Inside the <code class="docutils literal notranslate"><span class="pre">cloneGraph</span></code> method:</p>
-<ul class="simple">
-<li><p>It first checks if the input <code class="docutils literal notranslate"><span class="pre">node</span></code> is <code class="docutils literal notranslate"><span class="pre">None</span></code>. If it’s <code class="docutils literal notranslate"><span class="pre">None</span></code>, it means there is no graph to clone, and the method returns <code class="docutils literal notranslate"><span class="pre">None</span></code>.</p></li>
-<li><p>It initializes a <code class="docutils literal notranslate"><span class="pre">visited</span></code> dictionary to keep track of visited nodes. This dictionary is used to ensure that each node is cloned only once to avoid duplication.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">dfs</span></code> (depth-first search) function is defined within the <code class="docutils literal notranslate"><span class="pre">cloneGraph</span></code> method. It takes an original node as input and returns its corresponding clone. This function performs the actual cloning of the graph.</p></li>
-<li><p>Inside the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function:</p>
-<ul>
-<li><p>It checks if the original node has been visited before by looking it up in the <code class="docutils literal notranslate"><span class="pre">visited</span></code> dictionary. If it has been visited, it returns the corresponding clone from the <code class="docutils literal notranslate"><span class="pre">visited</span></code> dictionary.</p></li>
-<li><p>If the original node is not visited, it creates a new node, <code class="docutils literal notranslate"><span class="pre">copy_node</span></code>, with the same value as the original node. This is the first step in creating the clone of the original node.</p></li>
-<li><p>It then recursively clones the neighbors of the original node by iterating through the <code class="docutils literal notranslate"><span class="pre">neighbors</span></code> list of the original node and appending the corresponding cloned neighbors to the <code class="docutils literal notranslate"><span class="pre">neighbors</span></code> list of the <code class="docutils literal notranslate"><span class="pre">copy_node</span></code>. This step ensures that the relationships between nodes are preserved in the clone.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">copy_node</span></code> is added to the <code class="docutils literal notranslate"><span class="pre">visited</span></code> dictionary with the original node as the key and the clone as the value.</p></li>
-<li><p>Finally, the <code class="docutils literal notranslate"><span class="pre">copy_node</span></code> is returned as the result of the DFS for the original node.</p></li>
-</ul>
-</li>
-</ul>
-</li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">cloneGraph</span></code> method returns the result of the DFS on the input <code class="docutils literal notranslate"><span class="pre">node</span></code>, which is the clone of the entire graph with its structure and relationships intact.</p></li>
-</ol>
-<p>In summary, the code uses depth-first search (DFS) to traverse the original graph, creating a clone for each node and its neighbors. It ensures that each node is cloned only once to avoid duplication and returns a reference to the cloned graph.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">get_adjacency_list</span><span class="p">(</span><span class="n">node</span><span class="p">):</span>
-    <span class="n">visited</span> <span class="o">=</span> <span class="p">{}</span>  <span class="c1"># Dictionary to keep track of visited nodes</span>
-
-    <span class="k">def</span> <span class="nf">dfs</span><span class="p">(</span><span class="n">node</span><span class="p">):</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">node</span><span class="p">:</span>
-            <span class="k">return</span> <span class="p">[]</span>
-
-        <span class="k">if</span> <span class="n">node</span> <span class="ow">in</span> <span class="n">visited</span><span class="p">:</span>
-            <span class="k">return</span> <span class="n">visited</span><span class="p">[</span><span class="n">node</span><span class="p">]</span>
-
-        <span class="n">neighbors</span> <span class="o">=</span> <span class="p">[</span><span class="n">n</span><span class="o">.</span><span class="n">val</span> <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="n">node</span><span class="o">.</span><span class="n">neighbors</span><span class="p">]</span>
-        <span class="n">visited</span><span class="p">[</span><span class="n">node</span><span class="p">]</span> <span class="o">=</span> <span class="n">neighbors</span>
-
-        <span class="k">for</span> <span class="n">neighbor</span> <span class="ow">in</span> <span class="n">node</span><span class="o">.</span><span class="n">neighbors</span><span class="p">:</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">neighbor</span><span class="p">)</span>
-
-    <span class="n">dfs</span><span class="p">(</span><span class="n">node</span><span class="p">)</span>
-    <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
-
-    <span class="k">def</span> <span class="nf">get_node_and_neighbors</span><span class="p">(</span><span class="n">node</span><span class="p">):</span>
-        <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">([</span><span class="n">node</span><span class="o">.</span><span class="n">val</span><span class="p">]</span> <span class="o">+</span> <span class="n">visited</span><span class="p">[</span><span class="n">node</span><span class="p">])</span>
-
-    <span class="n">get_node_and_neighbors</span><span class="p">(</span><span class="n">node</span><span class="p">)</span>
-    <span class="k">return</span> <span class="n">result</span>
-
-
-<span class="c1"># Testing the code with the provided examples</span>
-<span class="n">adjList1</span> <span class="o">=</span> <span class="p">[[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">4</span><span class="p">],</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">3</span><span class="p">],</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">4</span><span class="p">],</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">3</span><span class="p">]]</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-
-<span class="c1"># Create the graph from adjList</span>
-<span class="n">nodes</span> <span class="o">=</span> <span class="p">[</span><span class="n">Node</span><span class="p">(</span><span class="n">i</span><span class="p">)</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">adjList1</span><span class="p">)</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)]</span>
-<span class="k">for</span> <span class="n">i</span><span class="p">,</span> <span class="n">adj</span> <span class="ow">in</span> <span class="nb">enumerate</span><span class="p">(</span><span class="n">adjList1</span><span class="p">):</span>
-    <span class="n">nodes</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">.</span><span class="n">neighbors</span> <span class="o">=</span> <span class="p">[</span><span class="n">nodes</span><span class="p">[</span><span class="n">j</span> <span class="o">-</span> <span class="mi">1</span><span class="p">]</span> <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="n">adj</span><span class="p">]</span>
-
-<span class="c1"># Clone the graph and get its adjacency list</span>
-<span class="n">cloned_node1</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">cloneGraph</span><span class="p">(</span><span class="n">nodes</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span>
-<span class="n">cloned_adjList1</span> <span class="o">=</span> <span class="n">get_adjacency_list</span><span class="p">(</span><span class="n">cloned_node1</span><span class="p">)</span>
-
-<span class="c1"># Print the cloned graph as an adjacency list</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">cloned_adjList1</span><span class="p">)</span>
-
-<span class="c1"># Output should match adjList1</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[1, 2, 4]]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="c1"># Input: adjList = [[]]</span>
-
-<span class="n">adjList2</span> <span class="o">=</span> <span class="p">[[]]</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-
-<span class="c1"># Create the graph from adjList</span>
-<span class="n">node1</span> <span class="o">=</span> <span class="n">Node</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span>
-
-<span class="c1"># Clone the graph and get its adjacency list</span>
-<span class="n">cloned_node1</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">cloneGraph</span><span class="p">(</span><span class="n">node1</span><span class="p">)</span>
-<span class="n">cloned_adjList2</span> <span class="o">=</span> <span class="n">get_adjacency_list</span><span class="p">(</span><span class="n">cloned_node1</span><span class="p">)</span>
-
-<span class="c1"># Print the cloned graph as an adjacency list</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">cloned_adjList2</span><span class="p">)</span>
-
-<span class="c1"># Output should be [[]], which represents an empty graph</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[1]]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code for cloning a graph using depth-first search (DFS).</p>
-<p>Time Complexity:</p>
-<ol class="arabic simple">
-<li><p>Creating the graph: The time complexity to create the original graph from <code class="docutils literal notranslate"><span class="pre">adjList</span></code> is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. We iterate through each node and its neighbors in <code class="docutils literal notranslate"><span class="pre">adjList</span></code>.</p></li>
-<li><p>Cloning the graph (DFS): The depth-first search (DFS) is used to traverse the original graph and create a clone. The time complexity of DFS is O(V + E), as we visit each node and each edge once.</p></li>
-</ol>
-<p>Overall, the time complexity is O(V + E) for both creating the original graph and cloning it.</p>
-<p>Space Complexity:</p>
-<ol class="arabic simple">
-<li><p>Storage for the original graph: We need to store the original graph, which includes all nodes and their adjacency lists. In the worst case, this requires O(V + E) space, where V is the number of nodes, and E is the number of edges.</p></li>
-<li><p>Storage for the visited dictionary: The <code class="docutils literal notranslate"><span class="pre">visited</span></code> dictionary stores mappings from original nodes to their corresponding clones. In the worst case, this requires O(V) space because each node is visited once.</p></li>
-<li><p>Recursive stack for DFS: The space used for the recursive call stack during DFS can be up to O(V) in the worst case, where V is the number of nodes.</p></li>
-</ol>
-<p>Overall, the space complexity is O(V + E) for storing the original graph, and O(V) for auxiliary data structures, including the <code class="docutils literal notranslate"><span class="pre">visited</span></code> dictionary and the DFS call stack. Therefore, the overall space complexity is O(V + E).</p>
-<p>In summary, the time and space complexities of the provided code for cloning a graph are both O(V + E), where V is the number of nodes, and E is the number of edges in the graph.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Cyclic Graph</strong>: Design a graph with cycles (nodes connected in a cycle). Test if the code can correctly clone graphs with cycles and doesn’t fall into an infinite loop during traversal.</p></li>
-<li><p><strong>Partial Graph Cloning</strong>: Modify the code to clone only a subset of the graph starting from a given reference node, rather than the entire graph. Implement this as an extension to the problem.</p></li>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="graph-valid-tree">
-<h1>178. Graph Valid Tree<a class="headerlink" href="#graph-valid-tree" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://www.lintcode.com/problem/178/">To see the Graph Valid Tree problem on LintCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given <code class="docutils literal notranslate"><span class="pre">n</span></code> nodes labeled from <code class="docutils literal notranslate"><span class="pre">0</span></code> to <code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">-</span> <span class="pre">1</span></code> and a list of <code class="docutils literal notranslate"><span class="pre">undirected</span></code> edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>You can assume that no duplicate edges will appear in edges. Since all edges are <code class="docutils literal notranslate"><span class="pre">undirected</span></code>, <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">1]</span></code> is the same as <code class="docutils literal notranslate"><span class="pre">[1,</span> <span class="pre">0]</span></code> and thus will not appear together in edges.</p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem is to determine whether a given set of undirected edges can form a valid tree. In this context, a “tree” is a specific type of graph with the following properties:</p>
-<ol class="arabic simple">
-<li><p>It is a connected graph, meaning there is a path between any pair of nodes in the graph.</p></li>
-<li><p>It has no cycles, which means there are no closed loops or circuits in the graph.</p></li>
-<li><p>It has exactly <code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">-</span> <span class="pre">1</span></code> edges, where <code class="docutils literal notranslate"><span class="pre">n</span></code> is the number of nodes in the tree.</p></li>
-</ol>
-<p>The problem is typically given with two pieces of information:</p>
-<ol class="arabic simple">
-<li><p>The number of nodes (<code class="docutils literal notranslate"><span class="pre">n</span></code>), which are usually labeled from 0 to <code class="docutils literal notranslate"><span class="pre">n-1</span></code>.</p></li>
-<li><p>A list of undirected edges, where each edge is represented as a pair of nodes.</p></li>
-</ol>
-<p>The task is to write a function or algorithm to determine whether the provided set of edges can be arranged to form a valid tree based on the properties mentioned above.</p>
-<p>To solve this problem, you need to check the following conditions:</p>
-<ol class="arabic simple">
-<li><p>There should be exactly <code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">-</span> <span class="pre">1</span></code> edges to ensure that the graph is connected and tree-like.</p></li>
-<li><p>There should be no cycles in the graph. If there are any cycles, the set of edges cannot form a tree.</p></li>
-<li><p>The graph should be connected, which means you can reach any node from any other node through a series of edges.</p></li>
-</ol>
-<p>The solution typically involves using graph traversal algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the graph and check these conditions. If all conditions are met, the given edges form a valid tree; otherwise, they do not.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">validTree</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">edges</span><span class="p">):</span>
-    <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="n">edges</span><span class="p">)</span> <span class="o">!=</span> <span class="n">n</span> <span class="o">-</span> <span class="mi">1</span><span class="p">:</span>
-        <span class="k">return</span> <span class="kc">False</span>  <span class="c1"># A tree with n nodes must have n-1 edges.</span>
-
-    <span class="c1"># Create an adjacency list to represent the graph.</span>
-    <span class="n">adj_list</span> <span class="o">=</span> <span class="p">{</span><span class="n">i</span><span class="p">:</span> <span class="p">[]</span> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">)}</span>
-    <span class="k">for</span> <span class="n">u</span><span class="p">,</span> <span class="n">v</span> <span class="ow">in</span> <span class="n">edges</span><span class="p">:</span>
-        <span class="n">adj_list</span><span class="p">[</span><span class="n">u</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">v</span><span class="p">)</span>
-        <span class="n">adj_list</span><span class="p">[</span><span class="n">v</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">u</span><span class="p">)</span>
-
-    <span class="n">visited</span> <span class="o">=</span> <span class="nb">set</span><span class="p">()</span>
-
-    <span class="c1"># Define a recursive DFS function.</span>
-    <span class="k">def</span> <span class="nf">dfs</span><span class="p">(</span><span class="n">node</span><span class="p">,</span> <span class="n">parent</span><span class="p">):</span>
-        <span class="k">if</span> <span class="n">node</span> <span class="ow">in</span> <span class="n">visited</span><span class="p">:</span>
-            <span class="k">return</span> <span class="kc">False</span>  <span class="c1"># If we encounter a visited node, there&#39;s a cycle.</span>
-        <span class="n">visited</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="n">node</span><span class="p">)</span>
-
-        <span class="k">for</span> <span class="n">neighbor</span> <span class="ow">in</span> <span class="n">adj_list</span><span class="p">[</span><span class="n">node</span><span class="p">]:</span>
-            <span class="k">if</span> <span class="n">neighbor</span> <span class="o">!=</span> <span class="n">parent</span><span class="p">:</span>
-                <span class="k">if</span> <span class="ow">not</span> <span class="n">dfs</span><span class="p">(</span><span class="n">neighbor</span><span class="p">,</span> <span class="n">node</span><span class="p">):</span>
-                    <span class="k">return</span> <span class="kc">False</span>
-
-        <span class="k">return</span> <span class="kc">True</span>
-
-    <span class="c1"># Start DFS from any node (e.g., node 0).</span>
-    <span class="k">if</span> <span class="ow">not</span> <span class="n">dfs</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="o">-</span><span class="mi">1</span><span class="p">):</span>
-        <span class="k">return</span> <span class="kc">False</span>  <span class="c1"># If there&#39;s a cycle, it&#39;s not a valid tree.</span>
-
-    <span class="k">return</span> <span class="nb">len</span><span class="p">(</span><span class="n">visited</span><span class="p">)</span> <span class="o">==</span> <span class="n">n</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The <code class="docutils literal notranslate"><span class="pre">validTree</span></code> function checks if a given set of undirected edges forms a valid tree. A valid tree must satisfy these conditions:</p>
-<ol class="arabic simple">
-<li><p>It has exactly <code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">-</span> <span class="pre">1</span></code> edges, where <code class="docutils literal notranslate"><span class="pre">n</span></code> is the number of nodes.</p></li>
-<li><p>There are no cycles in the graph.</p></li>
-<li><p>The graph is connected, meaning all nodes are reachable from any starting node.</p></li>
-</ol>
-<p>The code accomplishes this as follows:</p>
-<ul class="simple">
-<li><p>It first checks if the number of edges is equal to <code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">-</span> <span class="pre">1</span></code>. If not, it immediately returns <code class="docutils literal notranslate"><span class="pre">False</span></code> since a tree with <code class="docutils literal notranslate"><span class="pre">n</span></code> nodes must have <code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">-</span> <span class="pre">1</span></code> edges to be connected.</p></li>
-<li><p>It then constructs an adjacency list (<code class="docutils literal notranslate"><span class="pre">adj_list</span></code>) to represent the graph efficiently. This data structure stores the neighbors of each node.</p></li>
-<li><p>The code uses a <code class="docutils literal notranslate"><span class="pre">visited</span></code> set to keep track of nodes visited during depth-first search (DFS).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function is a recursive function that checks for cycles in the graph. It does this by visiting nodes and marking them as visited. If it encounters a node that has already been visited (indicating a cycle), it returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-<li><p>The DFS function explores the neighbors of each node, avoiding revisiting the parent node to prevent cycles.</p></li>
-<li><p>After the DFS is complete, the code checks if the number of visited nodes is equal to <code class="docutils literal notranslate"><span class="pre">n</span></code>, ensuring the graph is connected.</p></li>
-<li><p>Finally, if all conditions are met, the code returns <code class="docutils literal notranslate"><span class="pre">True</span></code>, indicating that the given edges form a valid tree. Otherwise, it returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-</ul>
-<p>The code provides examples to demonstrate how to use the <code class="docutils literal notranslate"><span class="pre">validTree</span></code> function to check whether a set of edges forms a valid tree or not.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1</span>
-<span class="n">n1</span> <span class="o">=</span> <span class="mi">5</span>
-<span class="n">edges1</span> <span class="o">=</span> <span class="p">[[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">],</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">2</span><span class="p">],</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">3</span><span class="p">],</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">4</span><span class="p">]]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">validTree</span><span class="p">(</span><span class="n">n1</span><span class="p">,</span> <span class="n">edges1</span><span class="p">))</span>  <span class="c1"># Output: true</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">n2</span> <span class="o">=</span> <span class="mi">5</span>
-<span class="n">edges2</span> <span class="o">=</span> <span class="p">[[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">],</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">],</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">],</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">3</span><span class="p">],</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">4</span><span class="p">]]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">validTree</span><span class="p">(</span><span class="n">n2</span><span class="p">,</span> <span class="n">edges2</span><span class="p">))</span>  <span class="c1"># Output: false</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexities of the <code class="docutils literal notranslate"><span class="pre">validTree</span></code> function:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>Constructing the adjacency list (<code class="docutils literal notranslate"><span class="pre">adj_list</span></code>) takes O(E) time, where E is the number of edges in the input graph (|edges|).</p></li>
-<li><p>The DFS function, when considering all nodes, has a time complexity of O(V + E), where V is the number of nodes in the graph. This is because, in the worst case, the DFS function visits each node and each edge once.</p></li>
-<li><p>Checking if the number of visited nodes is equal to <code class="docutils literal notranslate"><span class="pre">n</span></code> takes O(1) time.</p></li>
-</ol>
-<p>Overall, the time complexity of the <code class="docutils literal notranslate"><span class="pre">validTree</span></code> function is O(E + V) in the worst case.</p>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>The adjacency list (<code class="docutils literal notranslate"><span class="pre">adj_list</span></code>) consumes additional space to store the graph structure, taking O(V + E) space. This is because it stores information about each node and its adjacent nodes.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">visited</span></code> set keeps track of visited nodes, and its space complexity is O(V) in the worst case, as there can be at most V unique nodes in the set.</p></li>
-<li><p>The depth of the function call stack during DFS can be at most V, which contributes O(V) space to the space complexity.</p></li>
-</ol>
-<p>Overall, the space complexity of the <code class="docutils literal notranslate"><span class="pre">validTree</span></code> function is O(V + E) due to the adjacency list and O(V) due to the <code class="docutils literal notranslate"><span class="pre">visited</span></code> set and DFS call stack.</p>
-<p>In summary, the time complexity is O(E + V), and the space complexity is O(V + E). These complexities are based on the worst-case scenario, where all nodes and edges are considered. In practical cases, the actual complexities may be smaller, depending on the specific structure of the graph.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Minimum Spanning Tree (MST)</strong>: Implement an algorithm to find the minimum spanning tree of a connected graph using either Kruskal’s or Prim’s algorithm. Compare the MST to the original graph to determine if the original graph is a tree. This exercise reinforces the concept of tree properties.</p></li>
-<li><p><strong>Detecting Cycles</strong>: Modify the code to not only check if the input forms a tree but also to identify and print out any cycles present in the graph if it’s not a tree. This exercise enhances your ability to detect and visualize cycles in a graph.</p></li>
-</ol>
-</section>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="number-of-islands">
-<h1>200. Number of Islands<a class="headerlink" href="#number-of-islands" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/number-of-islands/">To see the Number of Islands problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>Given an <code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">x</span> <span class="pre">n</span></code> 2D binary grid <code class="docutils literal notranslate"><span class="pre">grid</span></code> which represents a map of <code class="docutils literal notranslate"><span class="pre">'1'</span></code>s (land) and <code class="docutils literal notranslate"><span class="pre">'0'</span></code>s (water), return <em>the number of islands</em>.</p>
-<p>An <strong>island</strong> is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">==</span> <span class="pre">grid.length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">==</span> <span class="pre">grid[i].length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">m,</span> <span class="pre">n</span> <span class="pre">&lt;=</span> <span class="pre">300</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">grid[i][j]</span></code> is <code class="docutils literal notranslate"><span class="pre">'0'</span></code> or <code class="docutils literal notranslate"><span class="pre">'1'</span></code>.</p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem is about counting the number of islands in a 2D binary grid. In this grid, ‘1’ represents land, and ‘0’ represents water. An island is defined as a group of ‘1’ cells that are adjacent to each other either horizontally or vertically. It’s important to note that diagonal connections do not count. Additionally, it’s assumed that all four edges of the grid are surrounded by water, meaning the grid is finite and doesn’t “wrap around.”</p>
-<p>The objective is to determine how many distinct islands exist in the grid. To clarify, an island consists of connected ‘1’ cells, and these connections can only occur horizontally or vertically. If there are no ‘1’ cells in the grid, the number of islands would be zero.</p>
-<p>The problem often involves using graph traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to identify and count these islands. You start at one ‘1’ cell and explore its adjacent cells to determine the extent of the island. Once you’ve visited all the ‘1’ cells in an island, you move on to the next unvisited ‘1’ cell and repeat the process until you’ve counted all the islands in the grid.</p>
-<p>In essence, the problem asks you to analyze the grid to find groups of connected ‘1’ cells and count them as individual islands. The goal is to return the total count of islands in the given grid.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">from</span> <span class="nn">typing</span> <span class="kn">import</span> <span class="n">List</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">numIslands</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">grid</span><span class="p">:</span> <span class="n">List</span><span class="p">[</span><span class="n">List</span><span class="p">[</span><span class="nb">str</span><span class="p">]])</span> <span class="o">-&gt;</span> <span class="nb">int</span><span class="p">:</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">grid</span><span class="p">:</span>
-            <span class="k">return</span> <span class="mi">0</span>
-
-        <span class="n">m</span><span class="p">,</span> <span class="n">n</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">grid</span><span class="p">),</span> <span class="nb">len</span><span class="p">(</span><span class="n">grid</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span>  <span class="c1"># Get the dimensions of the grid</span>
-        <span class="n">num_islands</span> <span class="o">=</span> <span class="mi">0</span>  <span class="c1"># Initialize the count of islands</span>
-
-        <span class="k">def</span> <span class="nf">dfs</span><span class="p">(</span><span class="n">row</span><span class="p">,</span> <span class="n">col</span><span class="p">):</span>
-            <span class="k">if</span> <span class="n">row</span> <span class="o">&lt;</span> <span class="mi">0</span> <span class="ow">or</span> <span class="n">row</span> <span class="o">&gt;=</span> <span class="n">m</span> <span class="ow">or</span> <span class="n">col</span> <span class="o">&lt;</span> <span class="mi">0</span> <span class="ow">or</span> <span class="n">col</span> <span class="o">&gt;=</span> <span class="n">n</span> <span class="ow">or</span> <span class="n">grid</span><span class="p">[</span><span class="n">row</span><span class="p">][</span><span class="n">col</span><span class="p">]</span> <span class="o">==</span> <span class="s1">&#39;0&#39;</span><span class="p">:</span>
-                <span class="k">return</span>
-
-            <span class="n">grid</span><span class="p">[</span><span class="n">row</span><span class="p">][</span><span class="n">col</span><span class="p">]</span> <span class="o">=</span> <span class="s1">&#39;0&#39;</span>  <span class="c1"># Mark the current cell as visited by changing it to &#39;0&#39;</span>
-            <span class="n">directions</span> <span class="o">=</span> <span class="p">[(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">),</span> <span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">),</span> <span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">),</span> <span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="o">-</span><span class="mi">1</span><span class="p">)]</span>  <span class="c1"># Define four possible directions to move</span>
-
-            <span class="k">for</span> <span class="n">dr</span><span class="p">,</span> <span class="n">dc</span> <span class="ow">in</span> <span class="n">directions</span><span class="p">:</span>
-                <span class="c1"># Explore adjacent cells</span>
-                <span class="n">dfs</span><span class="p">(</span><span class="n">row</span> <span class="o">+</span> <span class="n">dr</span><span class="p">,</span> <span class="n">col</span> <span class="o">+</span> <span class="n">dc</span><span class="p">)</span>
-
-        <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">m</span><span class="p">):</span>
-            <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
-                <span class="k">if</span> <span class="n">grid</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">]</span> <span class="o">==</span> <span class="s1">&#39;1&#39;</span><span class="p">:</span>
-                    <span class="n">num_islands</span> <span class="o">+=</span> <span class="mi">1</span>  <span class="c1"># Increment the island count</span>
-                    <span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">j</span><span class="p">)</span>  <span class="c1"># Start a DFS from the current &#39;1&#39; cell to explore the entire island</span>
-
-        <span class="k">return</span> <span class="n">num_islands</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The provided code defines a Python class called <code class="docutils literal notranslate"><span class="pre">Solution</span></code> with a method called <code class="docutils literal notranslate"><span class="pre">numIslands</span></code>. This class is designed to count the number of islands in a 2D binary grid.</p>
-<p>The <code class="docutils literal notranslate"><span class="pre">numIslands</span></code> method takes a 2D grid as input, where ‘1’ represents land, and ‘0’ represents water. It uses a depth-first search (DFS) approach to identify and count islands. The key steps in the code include:</p>
-<ol class="arabic simple">
-<li><p>Checking if the input grid is empty, and if so, returning 0 (indicating there are no islands).</p></li>
-<li><p>Determining the dimensions of the grid (number of rows and columns).</p></li>
-<li><p>Initializing a variable <code class="docutils literal notranslate"><span class="pre">num_islands</span></code> to keep track of the count of islands.</p></li>
-<li><p>Defining a nested function <code class="docutils literal notranslate"><span class="pre">dfs</span></code> to perform depth-first search. This function recursively explores adjacent land cells connected to the current cell and marks them as visited by changing ‘1’ to ‘0’.</p></li>
-<li><p>In the <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function, it checks for boundary conditions and whether the current cell is ‘0’. If these conditions are met, it returns without further exploration.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function explores neighboring cells in all four directions (up, down, left, right) using a loop.</p></li>
-<li><p>The main loop iterates through each cell of the grid. When a ‘1’ cell is encountered, it increments the <code class="docutils literal notranslate"><span class="pre">num_islands</span></code> count and starts a DFS from that cell to explore the entire island.</p></li>
-<li><p>Finally, the method returns the total count of islands found in the grid.</p></li>
-</ol>
-<p>This code encapsulates the solution in a class, allowing you to create an instance of the <code class="docutils literal notranslate"><span class="pre">Solution</span></code> class and call the <code class="docutils literal notranslate"><span class="pre">numIslands</span></code> method on it to count the number of islands in a given grid.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">grid1</span> <span class="o">=</span> <span class="p">[</span>
-  <span class="p">[</span><span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">],</span>
-  <span class="p">[</span><span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">],</span>
-  <span class="p">[</span><span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">],</span>
-  <span class="p">[</span><span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">]</span>
-<span class="p">]</span>
-
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="n">result1</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">numIslands</span><span class="p">(</span><span class="n">grid1</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="s2">&quot;Example 1 Output:&quot;</span><span class="p">,</span> <span class="n">result1</span><span class="p">)</span>  <span class="c1"># Expected output: 1</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Example 1 Output: 1
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="n">grid2</span> <span class="o">=</span> <span class="p">[</span>
-  <span class="p">[</span><span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">],</span>
-  <span class="p">[</span><span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">],</span>
-  <span class="p">[</span><span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">],</span>
-  <span class="p">[</span><span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;0&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">,</span> <span class="s2">&quot;1&quot;</span><span class="p">]</span>
-<span class="p">]</span>
-
-<span class="n">result2</span> <span class="o">=</span> <span class="n">solution</span><span class="o">.</span><span class="n">numIslands</span><span class="p">(</span><span class="n">grid2</span><span class="p">)</span>
-<span class="nb">print</span><span class="p">(</span><span class="s2">&quot;Example 2 Output:&quot;</span><span class="p">,</span> <span class="n">result2</span><span class="p">)</span>  <span class="c1"># Expected output: 3</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Example 2 Output: 3
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<p>Time Complexity:</p>
-<ul class="simple">
-<li><p>The code uses a depth-first search (DFS) to explore the grid and count the number of islands.</p></li>
-<li><p>In the worst case, it visits every cell in the grid exactly once.</p></li>
-<li><p>The DFS function explores neighboring cells in four directions (up, down, left, right), so it has a time complexity of <span class="math notranslate nohighlight">\(O(4^{max(m, n)})\)</span>, where m is the number of rows and n is the number of columns.</p></li>
-<li><p>Therefore, the overall time complexity of the code is O(m * n), where m is the number of rows, and n is the number of columns.</p></li>
-</ul>
-<p>Space Complexity:</p>
-<ul class="simple">
-<li><p>The space complexity is determined by the recursive calls and the depth of the DFS stack.</p></li>
-<li><p>In the worst case, when the entire grid consists of ‘1’s, the depth of the DFS recursion can be as deep as max(m, n).</p></li>
-<li><p>Therefore, the space complexity of the code is O(max(m, n)), representing the maximum depth of the DFS stack.</p></li>
-</ul>
-<p>In practical terms, for reasonably sized grids, the time and space complexity is linear, and the code should perform well. However, it’s important to keep in mind that the worst-case time complexity is O(m * n), and the space complexity can be O(max(m, n)).</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Identify Largest Island:</strong> In addition to counting the number of islands, modify the code to identify the largest island in the grid and return its size.</p></li>
-<li><p><strong>Optimize for Space:</strong> Modify the solution to reduce its space complexity to O(1) without modifying the input grid. This is a more memory-efficient version of the problem.</p></li>
-</ol>
-</section>
-</section>
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-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/course-schedule/">To see the Course Schedule problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>There are a total of <code class="docutils literal notranslate"><span class="pre">numCourses</span></code> courses to take, labeled from 0 to <code class="docutils literal notranslate"><span class="pre">numCourses</span> <span class="pre">-</span> <span class="pre">1</span></code>. You are given an array <code class="docutils literal notranslate"><span class="pre">prerequisites</span></code> where <code class="docutils literal notranslate"><span class="pre">prerequisites[i]</span> <span class="pre">=</span> <span class="pre">[ai,</span> <span class="pre">bi]</span></code> indicates that you <strong>must</strong> take course <code class="docutils literal notranslate"><span class="pre">bi</span></code> first if you want to take course <code class="docutils literal notranslate"><span class="pre">ai</span></code>.</p>
-<ul class="simple">
-<li><p>For example, the pair <code class="docutils literal notranslate"><span class="pre">[0,</span> <span class="pre">1]</span></code>, indicates that to take course <code class="docutils literal notranslate"><span class="pre">0</span></code> you have to first take course <code class="docutils literal notranslate"><span class="pre">1</span></code>.</p></li>
-</ul>
-<p>Return <code class="docutils literal notranslate"><span class="pre">true</span></code> if you can finish all courses. Otherwise, return <code class="docutils literal notranslate"><span class="pre">false</span></code>.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>1 &lt;= <code class="docutils literal notranslate"><span class="pre">numCourses</span></code> &lt;= 2000</p></li>
-<li><p>0 &lt;= <code class="docutils literal notranslate"><span class="pre">prerequisites.length</span></code> &lt;= 5000</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">prerequisites[i].length</span></code> == 2</p></li>
-<li><p>0 &lt;= <code class="docutils literal notranslate"><span class="pre">ai</span></code>, <code class="docutils literal notranslate"><span class="pre">bi</span></code> &lt; <code class="docutils literal notranslate"><span class="pre">numCourses</span></code></p></li>
-<li><p>All the pairs <code class="docutils literal notranslate"><span class="pre">prerequisites[i]</span></code> are <strong>unique</strong>.</p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem is to determine whether it’s possible to complete a set of courses given a list of prerequisites. You are given the following:</p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">numCourses</span></code>: The total number of courses to be taken, labeled from 0 to <code class="docutils literal notranslate"><span class="pre">numCourses</span> <span class="pre">-</span> <span class="pre">1</span></code>.</p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">prerequisites</span></code>: A list of prerequisite courses, where each element <code class="docutils literal notranslate"><span class="pre">prerequisites[i]</span></code> is a pair <code class="docutils literal notranslate"><span class="pre">[ai,</span> <span class="pre">bi]</span></code>, indicating that you must take course <code class="docutils literal notranslate"><span class="pre">bi</span></code> before you can take course <code class="docutils literal notranslate"><span class="pre">ai</span></code>.</p></li>
-</ul>
-<p>The goal is to check whether it’s possible to complete all the courses while respecting the prerequisite requirements. In other words, you need to determine if there are any circular dependencies in the course prerequisites that would prevent you from taking all the courses.</p>
-<p>For example:</p>
-<ul class="simple">
-<li><p>If <code class="docutils literal notranslate"><span class="pre">numCourses</span></code> is 2, and <code class="docutils literal notranslate"><span class="pre">prerequisites</span></code> is <code class="docutils literal notranslate"><span class="pre">[[1,</span> <span class="pre">0]]</span></code>, it means you have two courses, and you must finish course 0 before you can take course 1. In this case, it’s possible to complete all courses, so the function should return <code class="docutils literal notranslate"><span class="pre">True</span></code>.</p></li>
-<li><p>If <code class="docutils literal notranslate"><span class="pre">numCourses</span></code> is 2, and <code class="docutils literal notranslate"><span class="pre">prerequisites</span></code> is <code class="docutils literal notranslate"><span class="pre">[[1,</span> <span class="pre">0],</span> <span class="pre">[0,</span> <span class="pre">1]]</span></code>, it means you have two courses, but there’s a circular dependency where course 1 requires course 0 and course 0 requires course 1. In this case, it’s impossible to complete all courses, so the function should return <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-</ul>
-<p>The problem is essentially about checking for the presence of cycles in a directed graph, where the courses are nodes, and the prerequisites represent directed edges between them. If there are no cycles, it’s possible to complete all courses; otherwise, it’s not possible.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">from</span> <span class="nn">collections</span> <span class="kn">import</span> <span class="n">defaultdict</span><span class="p">,</span> <span class="n">deque</span>
-<span class="kn">from</span> <span class="nn">typing</span> <span class="kn">import</span> <span class="n">List</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">canFinish</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">numCourses</span><span class="p">:</span> <span class="nb">int</span><span class="p">,</span> <span class="n">prerequisites</span><span class="p">:</span> <span class="n">List</span><span class="p">[</span><span class="n">List</span><span class="p">[</span><span class="nb">int</span><span class="p">]])</span> <span class="o">-&gt;</span> <span class="nb">bool</span><span class="p">:</span>
-        <span class="c1"># Create a graph using an adjacency list to represent the courses and their prerequisites.</span>
-        <span class="n">graph</span> <span class="o">=</span> <span class="n">defaultdict</span><span class="p">(</span><span class="nb">list</span><span class="p">)</span>  <span class="c1"># Initialize an empty adjacency list.</span>
-        <span class="n">in_degree</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="n">numCourses</span>  <span class="c1"># Initialize an array to store in-degrees of courses.</span>
-
-        <span class="c1"># Populate the graph and calculate in-degrees.</span>
-        <span class="k">for</span> <span class="n">course</span><span class="p">,</span> <span class="n">prerequisite</span> <span class="ow">in</span> <span class="n">prerequisites</span><span class="p">:</span>
-            <span class="n">graph</span><span class="p">[</span><span class="n">prerequisite</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">course</span><span class="p">)</span>  <span class="c1"># Add the course as a neighbor to its prerequisite.</span>
-            <span class="n">in_degree</span><span class="p">[</span><span class="n">course</span><span class="p">]</span> <span class="o">+=</span> <span class="mi">1</span>  <span class="c1"># Increment the in-degree of the course.</span>
-
-        <span class="c1"># Initialize a queue with courses that have no prerequisites.</span>
-        <span class="n">queue</span> <span class="o">=</span> <span class="n">deque</span><span class="p">()</span>
-        <span class="k">for</span> <span class="n">course</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">numCourses</span><span class="p">):</span>
-            <span class="k">if</span> <span class="n">in_degree</span><span class="p">[</span><span class="n">course</span><span class="p">]</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
-                <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">course</span><span class="p">)</span>
-
-        <span class="c1"># Perform topological sorting.</span>
-        <span class="k">while</span> <span class="n">queue</span><span class="p">:</span>
-            <span class="n">course</span> <span class="o">=</span> <span class="n">queue</span><span class="o">.</span><span class="n">popleft</span><span class="p">()</span>  <span class="c1"># Take a course with no prerequisites.</span>
-            <span class="n">numCourses</span> <span class="o">-=</span> <span class="mi">1</span>  <span class="c1"># Decrement the count of remaining courses.</span>
-
-            <span class="k">for</span> <span class="n">neighbor</span> <span class="ow">in</span> <span class="n">graph</span><span class="p">[</span><span class="n">course</span><span class="p">]:</span>
-                <span class="n">in_degree</span><span class="p">[</span><span class="n">neighbor</span><span class="p">]</span> <span class="o">-=</span> <span class="mi">1</span>  <span class="c1"># Remove the prerequisite relationship.</span>
-                <span class="k">if</span> <span class="n">in_degree</span><span class="p">[</span><span class="n">neighbor</span><span class="p">]</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
-                    <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">neighbor</span><span class="p">)</span>  <span class="c1"># If no more prerequisites, add to the queue.</span>
-
-        <span class="c1"># If all courses were successfully taken (numCourses becomes 0), return True.</span>
-        <span class="k">return</span> <span class="n">numCourses</span> <span class="o">==</span> <span class="mi">0</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>This Python code is designed to determine whether it is possible to complete all the required courses given a set of prerequisites. It uses a topological sorting algorithm to accomplish this. Here’s a brief explanation of the code:</p>
-<ol class="arabic simple">
-<li><p>The code defines a <code class="docutils literal notranslate"><span class="pre">Solution</span></code> class with a <code class="docutils literal notranslate"><span class="pre">canFinish</span></code> method that takes two parameters: <code class="docutils literal notranslate"><span class="pre">numCourses</span></code> (the total number of courses) and <code class="docutils literal notranslate"><span class="pre">prerequisites</span></code> (a list of pairs indicating which course must be taken before another).</p></li>
-<li><p>It initializes an empty graph as an adjacency list using a <code class="docutils literal notranslate"><span class="pre">defaultdict</span></code> to represent the courses and their prerequisites. It also initializes a list called <code class="docutils literal notranslate"><span class="pre">in_degree</span></code> to keep track of the in-degrees of courses, which is used in the topological sorting process.</p></li>
-<li><p>The code then populates the graph by iterating through the <code class="docutils literal notranslate"><span class="pre">prerequisites</span></code> list. For each pair <code class="docutils literal notranslate"><span class="pre">(ai,</span> <span class="pre">bi)</span></code>, it adds course <code class="docutils literal notranslate"><span class="pre">ai</span></code> as a neighbor to its prerequisite course <code class="docutils literal notranslate"><span class="pre">bi</span></code> in the graph and increments the in-degree of course <code class="docutils literal notranslate"><span class="pre">ai</span></code>.</p></li>
-<li><p>Next, it initializes a queue with courses that have no prerequisites. It does this by iterating through all the courses and adding those with an in-degree of 0 to the queue.</p></li>
-<li><p>The code performs a topological sorting of the courses using a while loop. In each iteration of the loop, it dequeues a course from the front of the queue (a course with no prerequisites).</p></li>
-<li><p>For each neighbor (course that depends on the dequeued course), it decrements the in-degree of the neighbor. If the neighbor’s in-degree becomes 0, it means all of its prerequisites have been taken, so the neighbor is added to the queue.</p></li>
-<li><p>The loop continues until there are no more courses with no prerequisites to dequeue, and the in-degrees are updated accordingly.</p></li>
-<li><p>After the loop, if all courses were successfully taken (i.e., <code class="docutils literal notranslate"><span class="pre">numCourses</span></code> becomes 0), it means that there were no circular dependencies, and it’s possible to finish all courses. In this case, the method returns <code class="docutils literal notranslate"><span class="pre">True</span></code>. Otherwise, if there are remaining courses (indicating a circular dependency), it returns <code class="docutils literal notranslate"><span class="pre">False</span></code>.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1</span>
-<span class="n">numCourses1</span> <span class="o">=</span> <span class="mi">2</span>
-<span class="n">prerequisites1</span> <span class="o">=</span> <span class="p">[[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">]]</span>
-<span class="n">sol</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">sol</span><span class="o">.</span><span class="n">canFinish</span><span class="p">(</span><span class="n">numCourses1</span><span class="p">,</span> <span class="n">prerequisites1</span><span class="p">))</span>  <span class="c1"># Output: True</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>True
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">numCourses2</span> <span class="o">=</span> <span class="mi">2</span>
-<span class="n">prerequisites2</span> <span class="o">=</span> <span class="p">[[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">],</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">]]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">sol</span><span class="o">.</span><span class="n">canFinish</span><span class="p">(</span><span class="n">numCourses2</span><span class="p">,</span> <span class="n">prerequisites2</span><span class="p">))</span>  <span class="c1"># Output: False</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>False
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the <code class="docutils literal notranslate"><span class="pre">canFinish</span></code> method:</p>
-<p><strong>Time Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>Constructing the graph: In the first loop where we iterate through the <code class="docutils literal notranslate"><span class="pre">prerequisites</span></code> list, we populate the graph with prerequisites. This loop has a time complexity of O(E), where E is the number of prerequisites (edges).</p></li>
-<li><p>Initializing in-degrees and finding courses with no prerequisites: In the worst case, we iterate through all <code class="docutils literal notranslate"><span class="pre">numCourses</span></code> courses to initialize in-degrees and find the courses with no prerequisites. This has a time complexity of O(V), where V is the number of courses (vertices).</p></li>
-<li><p>Topological Sorting: In the while loop, we perform topological sorting, processing each course once and its outgoing edges. In the worst case, each course is processed once. This has a time complexity of O(V + E).</p></li>
-</ol>
-<p>The overall time complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.</p>
-<p><strong>Space Complexity:</strong></p>
-<ol class="arabic simple">
-<li><p>Graph Representation: The space complexity for storing the graph as an adjacency list is O(E), where E is the number of prerequisites.</p></li>
-<li><p>In-Degree Array: The space complexity for storing the in-degrees of courses is O(V), where V is the number of courses.</p></li>
-<li><p>Queue: The space complexity for the queue is O(V), as it may contain all the courses.</p></li>
-</ol>
-<p>The overall space complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.</p>
-<p>In most practical cases, the number of prerequisites (E) is much smaller than the number of courses (V), so the time and space complexity can often be considered as O(V).</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Optimizing Course Scheduling</strong>: Given a list of courses with their durations and prerequisites, find the most efficient way to schedule these courses to minimize the time it takes to complete all of them.</p></li>
-<li><p><strong>Detecting Cycles in Course Dependencies</strong>: Modify the original problem to not just determine if you can finish all courses but also to identify the specific courses that form a cycle in the prerequisite dependencies.</p></li>
-</ol>
-</section>
-</section>
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-  <section class="tex2jax_ignore mathjax_ignore" id="number-of-connected-components-in-an-undirected-graph">
-<h1>3651. Number of Connected Components in an Undirected Graph<a class="headerlink" href="#number-of-connected-components-in-an-undirected-graph" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://www.lintcode.com/problem/3651/">To see the Number of Connected Components in an Undirected Graph problem on LintCode, click here!</a></p>
-<hr class="docutils" />
-<p>In this problem, there is an undirected graph with <code class="docutils literal notranslate"><span class="pre">n</span></code> nodes. There is also an <code class="docutils literal notranslate"><span class="pre">edges</span></code> array. Where <code class="docutils literal notranslate"><span class="pre">edges[i]</span> <span class="pre">=</span> <span class="pre">[a,</span> <span class="pre">b]</span></code> means that there is an edge between <em>node a</em> and <em>node b</em> in the graph.</p>
-<p>You need to return <em>the number of connected components in that graph</em>.</p>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem at hand is to determine the number of connected components in an undirected graph. Here’s a more detailed explanation of the problem:</p>
-<ul class="simple">
-<li><p>You are given an undirected graph, which consists of nodes and edges. The graph may have multiple distinct connected components.</p></li>
-<li><p>A connected component is a group of nodes within the graph where each node can be reached from every other node in the same component. In other words, there is a path between any two nodes in the same connected component, but there are no paths connecting nodes from different components.</p></li>
-<li><p>Your task is to write a function that takes as input the number of nodes <code class="docutils literal notranslate"><span class="pre">n</span></code> and a list of edges <code class="docutils literal notranslate"><span class="pre">edges</span></code>, where each edge is represented as a pair of nodes <code class="docutils literal notranslate"><span class="pre">[a,</span> <span class="pre">b]</span></code> indicating an edge between node <code class="docutils literal notranslate"><span class="pre">a</span></code> and node <code class="docutils literal notranslate"><span class="pre">b</span></code>.</p></li>
-<li><p>The goal is to determine the number of distinct connected components in the graph.</p></li>
-<li><p>You need to implement an algorithm that can identify these connected components and count how many there are in the given graph.</p></li>
-</ul>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">countConnectedComponents</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">edges</span><span class="p">):</span>
-    <span class="c1"># Define a helper function for depth-first search (DFS).</span>
-    <span class="k">def</span> <span class="nf">dfs</span><span class="p">(</span><span class="n">node</span><span class="p">):</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">visited</span><span class="p">[</span><span class="n">node</span><span class="p">]:</span>  <span class="c1"># If the node has not been visited yet:</span>
-            <span class="n">visited</span><span class="p">[</span><span class="n">node</span><span class="p">]</span> <span class="o">=</span> <span class="kc">True</span>  <span class="c1"># Mark it as visited.</span>
-            <span class="k">for</span> <span class="n">neighbor</span> <span class="ow">in</span> <span class="n">graph</span><span class="p">[</span><span class="n">node</span><span class="p">]:</span>  <span class="c1"># Explore all neighbors of the current node.</span>
-                <span class="n">dfs</span><span class="p">(</span><span class="n">neighbor</span><span class="p">)</span>  <span class="c1"># Recursively call DFS on the neighbor.</span>
-
-    <span class="c1"># Create an empty adjacency list to represent the graph.</span>
-    <span class="n">graph</span> <span class="o">=</span> <span class="p">[[]</span> <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">)]</span>
-
-    <span class="c1"># Initialize a boolean list to keep track of visited nodes.</span>
-    <span class="n">visited</span> <span class="o">=</span> <span class="p">[</span><span class="kc">False</span><span class="p">]</span> <span class="o">*</span> <span class="n">n</span>
-
-    <span class="c1"># Populate the adjacency list with edges from the input.</span>
-    <span class="k">for</span> <span class="n">a</span><span class="p">,</span> <span class="n">b</span> <span class="ow">in</span> <span class="n">edges</span><span class="p">:</span>
-        <span class="n">graph</span><span class="p">[</span><span class="n">a</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">b</span><span class="p">)</span>
-        <span class="n">graph</span><span class="p">[</span><span class="n">b</span><span class="p">]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">a</span><span class="p">)</span>  <span class="c1"># Since the graph is undirected, we add edges in both directions.</span>
-
-    <span class="c1"># Initialize a counter to keep track of the number of connected components.</span>
-    <span class="n">count</span> <span class="o">=</span> <span class="mi">0</span>
-
-    <span class="c1"># Iterate through all nodes in the graph.</span>
-    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">visited</span><span class="p">[</span><span class="n">i</span><span class="p">]:</span>  <span class="c1"># If the node has not been visited:</span>
-            <span class="n">count</span> <span class="o">+=</span> <span class="mi">1</span>  <span class="c1"># Increment the count since we&#39;ve found a new connected component.</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="p">)</span>  <span class="c1"># Start a depth-first search from the unvisited node to explore the component.</span>
-
-    <span class="k">return</span> <span class="n">count</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The provided Python code calculates the number of connected components in an undirected graph. Here’s a high-level explanation of how the code works:</p>
-<ol class="arabic simple">
-<li><p>The <code class="docutils literal notranslate"><span class="pre">countConnectedComponents</span></code> function takes two arguments: <code class="docutils literal notranslate"><span class="pre">n</span></code> (the number of nodes in the graph) and <code class="docutils literal notranslate"><span class="pre">edges</span></code> (a list of edge pairs representing connections between nodes).</p></li>
-<li><p>Inside the function, there is a helper function called <code class="docutils literal notranslate"><span class="pre">dfs</span></code> (depth-first search), which is used to explore and mark nodes as visited.</p></li>
-<li><p>An empty adjacency list called <code class="docutils literal notranslate"><span class="pre">graph</span></code> is created to represent the graph. The <code class="docutils literal notranslate"><span class="pre">graph</span></code> is a list of lists where each element represents a node, and the list associated with each node contains its neighboring nodes.</p></li>
-<li><p>Another list called <code class="docutils literal notranslate"><span class="pre">visited</span></code> is created to keep track of visited nodes. Initially, all nodes are marked as not visited (<code class="docutils literal notranslate"><span class="pre">False</span></code>).</p></li>
-<li><p>The code populates the <code class="docutils literal notranslate"><span class="pre">graph</span></code> by iterating through the <code class="docutils literal notranslate"><span class="pre">edges</span></code> list and adding each edge to the adjacency list. Since the graph is undirected, both directions of each edge are added.</p></li>
-<li><p>A counter variable called <code class="docutils literal notranslate"><span class="pre">count</span></code> is initialized to keep track of the number of connected components.</p></li>
-<li><p>The code iterates through all nodes from <code class="docutils literal notranslate"><span class="pre">0</span></code> to <code class="docutils literal notranslate"><span class="pre">n-1</span></code>. For each node, it checks whether it has been visited yet. If it hasn’t been visited, it means a new connected component is found. The counter <code class="docutils literal notranslate"><span class="pre">count</span></code> is incremented, and a depth-first search (DFS) is initiated from this unvisited node to explore the connected component.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">dfs</span></code> function recursively visits all nodes in the connected component, marking them as visited. It does this by checking each neighbor of the current node and recursively calling <code class="docutils literal notranslate"><span class="pre">dfs</span></code> on unvisited neighbors.</p></li>
-<li><p>After the loop through all nodes, the <code class="docutils literal notranslate"><span class="pre">count</span></code> variable contains the total number of connected components in the graph.</p></li>
-<li><p>The function returns the value of <code class="docutils literal notranslate"><span class="pre">count</span></code>, which represents the number of connected components.</p></li>
-<li><p>Two examples are provided at the end to demonstrate how to use the function with different inputs.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1</span>
-<span class="n">n1</span> <span class="o">=</span> <span class="mi">3</span>
-<span class="n">edges1</span> <span class="o">=</span> <span class="p">[[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">],</span> <span class="p">[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">2</span><span class="p">]]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">countConnectedComponents</span><span class="p">(</span><span class="n">n1</span><span class="p">,</span> <span class="n">edges1</span><span class="p">))</span>  <span class="c1"># Output: 1</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>1
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">n2</span> <span class="o">=</span> <span class="mi">6</span>
-<span class="n">edges2</span> <span class="o">=</span> <span class="p">[[</span><span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">],</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">],</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">],</span> <span class="p">[</span><span class="mi">4</span><span class="p">,</span> <span class="mi">5</span><span class="p">]]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">countConnectedComponents</span><span class="p">(</span><span class="n">n2</span><span class="p">,</span> <span class="n">edges2</span><span class="p">))</span>  <span class="c1"># Output: 2</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>2
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the code for counting the number of connected components in an undirected graph using depth-first search (DFS).</p>
-<p><strong>Time Complexity:</strong></p>
-<ul class="simple">
-<li><p>The code iterates through all nodes in the graph (from 0 to <code class="docutils literal notranslate"><span class="pre">n-1</span></code>).</p></li>
-<li><p>For each unvisited node, it initiates a depth-first search (DFS) from that node to explore the connected component.</p></li>
-<li><p>In the worst case, every node may belong to a separate connected component, and you perform a DFS for each node.</p></li>
-<li><p>The time complexity of the DFS for a single connected component is O(V + E), where V is the number of nodes in the component, and E is the number of edges.</p></li>
-<li><p>In the worst case, you perform O(n) DFS calls, each with its own O(V + E) complexity.</p></li>
-</ul>
-<p>So, the overall time complexity of the code is O(n * (V + E)). In the worst case, if the graph is connected (a single connected component), this simplifies to O(n * (n + m)), where m is the number of edges.</p>
-<p><strong>Space Complexity:</strong></p>
-<ul class="simple">
-<li><p>The main data structures used are the <code class="docutils literal notranslate"><span class="pre">graph</span></code> (adjacency list), <code class="docutils literal notranslate"><span class="pre">visited</span></code> (boolean array), and the recursive call stack for DFS.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">graph</span></code> has a space complexity of O(n + m), where n is the number of nodes, and m is the number of edges.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">visited</span></code> boolean array also has a space complexity of O(n) to keep track of whether nodes have been visited.</p></li>
-<li><p>The depth of the recursive call stack for DFS can go up to O(n) in the worst case, when all nodes are part of a single connected component.</p></li>
-</ul>
-<p>So, the overall space complexity of the code is O(n + m) for data structures and O(n) for the call stack. In the worst case, this simplifies to O(n + m).</p>
-<p>In summary, the time complexity is O(n * (V + E)), and the space complexity is O(n + m).</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Disconnected Components with Multiple Queries:</strong> You are given an undirected graph with <code class="docutils literal notranslate"><span class="pre">n</span></code> nodes and <code class="docutils literal notranslate"><span class="pre">m</span></code> edges. Initially, the graph is disconnected, and you need to process <code class="docutils literal notranslate"><span class="pre">q</span></code> queries. Each query consists of adding an edge to connect two nodes. After each query, you need to output the number of connected components in the updated graph.</p></li>
-<li><p><strong>Maximum Size of Connected Components:</strong> Given an undirected graph with <code class="docutils literal notranslate"><span class="pre">n</span></code> nodes and <code class="docutils literal notranslate"><span class="pre">m</span></code> edges, find the size of the largest connected component in the graph. Additionally, identify the nodes that belong to this largest component. You can assume there is at least one connected component in the graph.</p></li>
-</ol>
-</section>
-</section>
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-    <h1>417. Pacific Atlantic Water Flow</h1>
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-                <h2> Contents </h2>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="pacific-atlantic-water-flow">
-<h1>417. Pacific Atlantic Water Flow<a class="headerlink" href="#pacific-atlantic-water-flow" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/pacific-atlantic-water-flow/">To see the Pacific Atlantic Water Flow problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>There is an <code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">x</span> <span class="pre">n</span></code> rectangular island that borders both the <strong>Pacific Ocean</strong> and <strong>Atlantic Ocean</strong>. The <strong>Pacific Ocea</strong>n touches the island’s left and top edges, and the <strong>Atlantic Ocean</strong> touches the island’s right and bottom edges.</p>
-<p>The island is partitioned into a grid of square cells. You are given an <code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">x</span> <span class="pre">n</span></code> integer matrix <code class="docutils literal notranslate"><span class="pre">heights</span></code> where <code class="docutils literal notranslate"><span class="pre">heights[r][c]</span></code> represents the <strong>height above sea level</strong> of the cell at coordinate <code class="docutils literal notranslate"><span class="pre">(r,</span> <span class="pre">c)</span></code>.</p>
-<p>The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell’s height is <strong>less than or equal to</strong> the current cell’s height. Water can flow from any cell adjacent to an ocean into the ocean.</p>
-<p>Return <em>a <strong>2D list</strong> of grid coordinates <code class="docutils literal notranslate"><span class="pre">result</span></code> where <span class="math notranslate nohighlight">\(result[i] = [r_i, c_i]\)</span> denotes that rain water can flow from cell <span class="math notranslate nohighlight">\((r_i, c_i)\)</span> to <strong>both</strong> the Pacific and Atlantic oceans</em>.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">m</span> <span class="pre">==</span> <span class="pre">heights.length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">==</span> <span class="pre">heights[r].length</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">m,</span> <span class="pre">n</span> <span class="pre">&lt;=</span> <span class="pre">200</span></code></p></li>
-<li><p>0 &lt;= <code class="docutils literal notranslate"><span class="pre">heights[r][c]</span></code> &lt;= <span class="math notranslate nohighlight">\(10^5\)</span></p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem at hand involves a scenario with a rectangular island surrounded by both the Pacific Ocean and the Atlantic Ocean. The Pacific Ocean is in contact with the island’s left and top edges, while the Atlantic Ocean is in contact with the island’s right and bottom edges. The goal is to determine which cells on the island can allow rainwater to flow to both the Pacific and Atlantic Oceans.</p>
-<p>Here are the key details and constraints of the problem:</p>
-<ol class="arabic simple">
-<li><p>The island is divided into a grid of square cells.</p></li>
-<li><p>Each cell is represented by an integer value in a matrix <code class="docutils literal notranslate"><span class="pre">heights</span></code>. The value at <code class="docutils literal notranslate"><span class="pre">heights[r][c]</span></code> represents the height above sea level of the cell at coordinates <code class="docutils literal notranslate"><span class="pre">(r,</span> <span class="pre">c)</span></code> on the island.</p></li>
-<li><p>Rainwater can flow from a cell to neighboring cells directly north, south, east, or west only if the neighboring cell’s height is less than or equal to the current cell’s height.</p></li>
-<li><p>Water can flow from any cell adjacent to the ocean into the ocean. This means water can flow from cells on the edge of the island to the Pacific Ocean and the Atlantic Ocean.</p></li>
-</ol>
-<p>The task is to find and return a 2D list of grid coordinates where rainwater can flow from a cell to both the Pacific and Atlantic Oceans. In other words, you need to identify the cells that can send water to both the left and top edges (Pacific) and the right and bottom edges (Atlantic) of the island.</p>
-<p>The problem is solved by performing a depth-first search (DFS) starting from the ocean edges and marking the cells that can flow to each ocean. Then, the algorithm identifies cells that are reachable from both oceans and returns their coordinates as the result.</p>
-<p>The problem involves both traversal of the island and backtracking to find the solution, which makes it a classic graph traversal problem.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">from</span> <span class="nn">typing</span> <span class="kn">import</span> <span class="n">List</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">pacificAtlantic</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">heights</span><span class="p">:</span> <span class="n">List</span><span class="p">[</span><span class="n">List</span><span class="p">[</span><span class="nb">int</span><span class="p">]])</span> <span class="o">-&gt;</span> <span class="n">List</span><span class="p">[</span><span class="n">List</span><span class="p">[</span><span class="nb">int</span><span class="p">]]:</span>
-        <span class="c1"># Check if the input matrix is empty</span>
-        <span class="k">if</span> <span class="ow">not</span> <span class="n">heights</span><span class="p">:</span>
-            <span class="k">return</span> <span class="p">[]</span>
-        
-        <span class="n">m</span><span class="p">,</span> <span class="n">n</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">heights</span><span class="p">),</span> <span class="nb">len</span><span class="p">(</span><span class="n">heights</span><span class="p">[</span><span class="mi">0</span><span class="p">])</span>
-        
-        <span class="c1"># Initialize boolean matrices to keep track of cells reachable from each ocean</span>
-        <span class="n">pacific_reachable</span> <span class="o">=</span> <span class="p">[[</span><span class="kc">False</span><span class="p">]</span> <span class="o">*</span> <span class="n">n</span> <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">m</span><span class="p">)]</span>
-        <span class="n">atlantic_reachable</span> <span class="o">=</span> <span class="p">[[</span><span class="kc">False</span><span class="p">]</span> <span class="o">*</span> <span class="n">n</span> <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">m</span><span class="p">)]</span>
-        
-        <span class="c1"># Depth-First Search (DFS) function to mark cells that can be reached from an ocean</span>
-        <span class="k">def</span> <span class="nf">dfs</span><span class="p">(</span><span class="n">x</span><span class="p">,</span> <span class="n">y</span><span class="p">,</span> <span class="n">reachable</span><span class="p">):</span>
-            <span class="k">if</span> <span class="n">reachable</span><span class="p">[</span><span class="n">x</span><span class="p">][</span><span class="n">y</span><span class="p">]:</span>
-                <span class="k">return</span>
-            <span class="n">reachable</span><span class="p">[</span><span class="n">x</span><span class="p">][</span><span class="n">y</span><span class="p">]</span> <span class="o">=</span> <span class="kc">True</span>
-            <span class="k">for</span> <span class="n">dx</span><span class="p">,</span> <span class="n">dy</span> <span class="ow">in</span> <span class="p">[(</span><span class="mi">0</span><span class="p">,</span> <span class="mi">1</span><span class="p">),</span> <span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="o">-</span><span class="mi">1</span><span class="p">),</span> <span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">),</span> <span class="p">(</span><span class="o">-</span><span class="mi">1</span><span class="p">,</span> <span class="mi">0</span><span class="p">)]:</span>
-                <span class="n">new_x</span><span class="p">,</span> <span class="n">new_y</span> <span class="o">=</span> <span class="n">x</span> <span class="o">+</span> <span class="n">dx</span><span class="p">,</span> <span class="n">y</span> <span class="o">+</span> <span class="n">dy</span>
-                <span class="k">if</span> <span class="mi">0</span> <span class="o">&lt;=</span> <span class="n">new_x</span> <span class="o">&lt;</span> <span class="n">m</span> <span class="ow">and</span> <span class="mi">0</span> <span class="o">&lt;=</span> <span class="n">new_y</span> <span class="o">&lt;</span> <span class="n">n</span> <span class="ow">and</span> <span class="n">heights</span><span class="p">[</span><span class="n">new_x</span><span class="p">][</span><span class="n">new_y</span><span class="p">]</span> <span class="o">&gt;=</span> <span class="n">heights</span><span class="p">[</span><span class="n">x</span><span class="p">][</span><span class="n">y</span><span class="p">]:</span>
-                    <span class="n">dfs</span><span class="p">(</span><span class="n">new_x</span><span class="p">,</span> <span class="n">new_y</span><span class="p">,</span> <span class="n">reachable</span><span class="p">)</span>
-        
-        <span class="c1"># Start DFS from the Pacific Ocean (left and top edges)</span>
-        <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">m</span><span class="p">):</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="mi">0</span><span class="p">,</span> <span class="n">pacific_reachable</span><span class="p">)</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">n</span> <span class="o">-</span> <span class="mi">1</span><span class="p">,</span> <span class="n">atlantic_reachable</span><span class="p">)</span>
-        
-        <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="n">pacific_reachable</span><span class="p">)</span>
-            <span class="n">dfs</span><span class="p">(</span><span class="n">m</span> <span class="o">-</span> <span class="mi">1</span><span class="p">,</span> <span class="n">j</span><span class="p">,</span> <span class="n">atlantic_reachable</span><span class="p">)</span>
-        
-        <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
-        
-        <span class="c1"># Find cells that can flow to both the Pacific and Atlantic oceans</span>
-        <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">m</span><span class="p">):</span>
-            <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
-                <span class="k">if</span> <span class="n">pacific_reachable</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">]</span> <span class="ow">and</span> <span class="n">atlantic_reachable</span><span class="p">[</span><span class="n">i</span><span class="p">][</span><span class="n">j</span><span class="p">]:</span>
-                    <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">([</span><span class="n">i</span><span class="p">,</span> <span class="n">j</span><span class="p">])</span>
-        
-        <span class="k">return</span> <span class="n">result</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The code defines a class called <code class="docutils literal notranslate"><span class="pre">Solution</span></code> with a method named <code class="docutils literal notranslate"><span class="pre">pacificAtlantic</span></code>. This method takes a 2D matrix <code class="docutils literal notranslate"><span class="pre">heights</span></code>, where each element represents the height above sea level of a cell on a rectangular island. The goal of the method is to find and return a list of grid coordinates (cells) from which rainwater can flow into both the Pacific and Atlantic Oceans.</p>
-<p>The code uses a Depth-First Search (DFS) algorithm to traverse the island, starting from the ocean edges, and marks cells that are reachable from either the Pacific or Atlantic Ocean. It does this by creating two boolean matrices, <code class="docutils literal notranslate"><span class="pre">pacific_reachable</span></code> and <code class="docutils literal notranslate"><span class="pre">atlantic_reachable</span></code>, to keep track of cells that can be reached from each ocean.</p>
-<p>The code iterates through the entire island, initiating DFS searches from the edges of the island. If a cell is reachable from an ocean (either Pacific or Atlantic), it is marked as reachable in the corresponding boolean matrix. The DFS function ensures that the traversal follows the rule that water can flow to a neighboring cell if the neighboring cell’s height is less than or equal to the current cell’s height.</p>
-<p>After marking all reachable cells from both oceans, the code then looks for cells that are reachable from both the Pacific and Atlantic Oceans. If a cell satisfies this condition, it is added to the <code class="docutils literal notranslate"><span class="pre">result</span></code> list.</p>
-<p>Finally, the code returns the <code class="docutils literal notranslate"><span class="pre">result</span></code> list, which contains the grid coordinates of cells from which rainwater can flow into both the Pacific and Atlantic Oceans.</p>
-<p>The code provides a solution to the problem of finding cells on the island where water can reach both oceans, demonstrating a traversal algorithm that explores the connectivity of cells on the island.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1</span>
-<span class="n">heights1</span> <span class="o">=</span> <span class="p">[</span>
-    <span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">5</span><span class="p">],</span>
-    <span class="p">[</span><span class="mi">3</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">4</span><span class="p">,</span><span class="mi">4</span><span class="p">],</span>
-    <span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">4</span><span class="p">,</span><span class="mi">5</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">1</span><span class="p">],</span>
-    <span class="p">[</span><span class="mi">6</span><span class="p">,</span><span class="mi">7</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">4</span><span class="p">,</span><span class="mi">5</span><span class="p">],</span>
-    <span class="p">[</span><span class="mi">5</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">4</span><span class="p">]</span>
-<span class="p">]</span>
-
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">solution</span><span class="o">.</span><span class="n">pacificAtlantic</span><span class="p">(</span><span class="n">heights1</span><span class="p">))</span> <span class="c1"># Expected Output: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">heights2</span> <span class="o">=</span> <span class="p">[[</span><span class="mi">1</span><span class="p">]]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">solution</span><span class="o">.</span><span class="n">pacificAtlantic</span><span class="p">(</span><span class="n">heights2</span><span class="p">))</span> <span class="c1"># Expected Output: [[0, 0]]</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[[0, 0]]
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Time Complexity:
-The time complexity of this solution is O(m * n), where ‘m’ is the number of rows in the input matrix and ‘n’ is the number of columns. Here’s a breakdown of the time complexity:</p>
-<ol class="arabic simple">
-<li><p>Creating the <code class="docutils literal notranslate"><span class="pre">pacific_reachable</span></code> and <code class="docutils literal notranslate"><span class="pre">atlantic_reachable</span></code> matrices with dimensions m x n takes O(m * n) time.</p></li>
-<li><p>Initiating Depth-First Search (DFS) from the edges of the island takes O(m * n) time because it processes each cell once.</p></li>
-<li><p>The DFS function visits each cell at most once, and its time complexity is O(1) for each cell. The number of cells visited by the DFS in the worst case is m * n.</p></li>
-<li><p>The final step of identifying and appending cells to the result list takes O(m * n) time in the worst case.</p></li>
-</ol>
-<p>Overall, the time complexity is dominated by the DFS and is O(m * n).</p>
-<p>Space Complexity:
-The space complexity of this solution is also O(m * n). Here’s how the space is used:</p>
-<ol class="arabic simple">
-<li><p>Two boolean matrices, <code class="docutils literal notranslate"><span class="pre">pacific_reachable</span></code> and <code class="docutils literal notranslate"><span class="pre">atlantic_reachable</span></code>, are created with dimensions m x n. Each matrix takes O(m * n) space, resulting in O(2 * m * n) space usage.</p></li>
-<li><p>The depth-first search (DFS) stack space during the recursive calls can go as deep as the diagonal of the grid, which is at most min(m, n). This additional space for the call stack is relatively small compared to the boolean matrices and can be considered O(min(m, n)).</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">result</span></code> list stores the grid coordinates, and its size is determined by the number of cells that can flow to both oceans. In the worst case, this list can contain all m * n cells, so it takes O(m * n) space.</p></li>
-</ol>
-<p>The overall space complexity is the sum of these components, which is O(m * n) + O(min(m, n)) + O(m * n). In big O notation, we can simplify this to O(m * n).</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Optimization Challenge:</strong> Modify the algorithm to find the maximum flow of water from any cell to both the Pacific and Atlantic Oceans. In other words, find the cell(s) that can send the most water to both oceans.</p></li>
-<li><p><strong>Efficiency Challenge:</strong> Optimize the time and space complexity of the algorithm while maintaining correctness. Aim to reduce both time and space complexity as much as possible.</p></li>
-</ol>
-</section>
-</section>
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-    <h1>892. Alien Dictionary</h1>
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-                <h2> Contents </h2>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="alien-dictionary">
-<h1>892. Alien Dictionary<a class="headerlink" href="#alien-dictionary" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Hard</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://www.lintcode.com/problem/892/">To see the Alien Dictionary problem on LintCode, click here!</a></p>
-<hr class="docutils" />
-<p>There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of <strong>non-empty</strong> words from the dictionary, where words are <strong>sorted lexicographically by the rules of this new language</strong>. Derive the order of letters in this language.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p>You may assume all letters are in lowercase.</p></li>
-<li><p>The dictionary is invalid, if string a is prefix of string b and b is appear before a.</p></li>
-<li><p>If the order is invalid, return an empty string.</p></li>
-<li><p>There may be multiple valid order of letters, return the smallest in normal lexicographical order.</p></li>
-<li><p>The letters in <strong>one</strong> string are of the same rank by default and are sorted in Human dictionary order.</p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem at hand involves determining the order of letters in a new alien language that uses the Latin alphabet. However, the order of letters in this alien language is unknown. To solve this problem, we are given a list of non-empty words from the dictionary, where the words are sorted lexicographically based on the rules of the alien language. The goal is to derive the correct order of letters in this new language.</p>
-<p>Here’s a more detailed explanation of the problem:</p>
-<ol class="arabic simple">
-<li><p>The alphabet: The new alien language uses the Latin alphabet, which consists of lowercase letters.</p></li>
-<li><p>Dictionary: We are provided with a list of non-empty words. These words are sorted in lexicographical order based on the rules of the alien language. The order of letters in the words reflects the correct order in the new language.</p></li>
-<li><p>Invalid order: The problem specifies that the dictionary is considered invalid if there exists a situation where one string ‘a’ is a prefix of another string ‘b,’ and ‘b’ appears before ‘a’ in the dictionary. This condition ensures that there are no inconsistencies in the order of letters.</p></li>
-<li><p>Multiple valid orders: It’s possible that there are multiple valid orders of letters that satisfy the provided dictionary. However, we are instructed to return the smallest valid order in normal lexicographical order.</p></li>
-</ol>
-<p>To solve this problem, you need to analyze the given list of words and determine the correct order of letters in the alien language while adhering to the specified constraints. You should return the order as a string in lexicographical order if it’s valid, and an empty string if the order is invalid.</p>
-<p>The key to solving this problem is to construct a directed graph of letter relationships based on the order information provided by the words in the dictionary. Then, perform a topological sort on this graph to derive the order of letters, ensuring that the order is consistent and valid according to the given constraints.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">from</span> <span class="nn">collections</span> <span class="kn">import</span> <span class="n">defaultdict</span><span class="p">,</span> <span class="n">deque</span>
-
-<span class="k">def</span> <span class="nf">alienOrder</span><span class="p">(</span><span class="n">words</span><span class="p">):</span>
-    <span class="c1"># Create an adjacency list to represent the graph</span>
-    <span class="n">graph</span> <span class="o">=</span> <span class="n">defaultdict</span><span class="p">(</span><span class="nb">list</span><span class="p">)</span>
-    
-    <span class="c1"># Create a dictionary to store the in-degrees of each letter</span>
-    <span class="n">in_degree</span> <span class="o">=</span> <span class="p">{}</span>
-    
-    <span class="c1"># Initialize in-degrees to 0 for all letters</span>
-    <span class="k">for</span> <span class="n">word</span> <span class="ow">in</span> <span class="n">words</span><span class="p">:</span>
-        <span class="k">for</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">word</span><span class="p">:</span>
-            <span class="k">if</span> <span class="n">char</span> <span class="ow">not</span> <span class="ow">in</span> <span class="n">in_degree</span><span class="p">:</span>
-                <span class="n">in_degree</span><span class="p">[</span><span class="n">char</span><span class="p">]</span> <span class="o">=</span> <span class="mi">0</span>
-    
-    <span class="c1"># Build the graph and update in-degrees</span>
-    <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">words</span><span class="p">)):</span>
-        <span class="n">word1</span><span class="p">,</span> <span class="n">word2</span> <span class="o">=</span> <span class="n">words</span><span class="p">[</span><span class="n">i</span><span class="o">-</span><span class="mi">1</span><span class="p">],</span> <span class="n">words</span><span class="p">[</span><span class="n">i</span><span class="p">]</span>
-        <span class="n">min_length</span> <span class="o">=</span> <span class="nb">min</span><span class="p">(</span><span class="nb">len</span><span class="p">(</span><span class="n">word1</span><span class="p">),</span> <span class="nb">len</span><span class="p">(</span><span class="n">word2</span><span class="p">))</span>
-        
-        <span class="c1"># Compare characters in the two words</span>
-        <span class="n">j</span> <span class="o">=</span> <span class="mi">0</span>
-        <span class="k">while</span> <span class="n">j</span> <span class="o">&lt;</span> <span class="n">min_length</span> <span class="ow">and</span> <span class="n">word1</span><span class="p">[</span><span class="n">j</span><span class="p">]</span> <span class="o">==</span> <span class="n">word2</span><span class="p">[</span><span class="n">j</span><span class="p">]:</span>
-            <span class="n">j</span> <span class="o">+=</span> <span class="mi">1</span>
-        
-        <span class="k">if</span> <span class="n">j</span> <span class="o">&lt;</span> <span class="n">min_length</span><span class="p">:</span>
-            <span class="c1"># If word1[j] is lexicographically before word2[j], add an edge</span>
-            <span class="n">graph</span><span class="p">[</span><span class="n">word1</span><span class="p">[</span><span class="n">j</span><span class="p">]]</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">word2</span><span class="p">[</span><span class="n">j</span><span class="p">])</span>
-            <span class="n">in_degree</span><span class="p">[</span><span class="n">word2</span><span class="p">[</span><span class="n">j</span><span class="p">]]</span> <span class="o">+=</span> <span class="mi">1</span>
-            
-            <span class="c1"># Break the loop to avoid further comparisons</span>
-            <span class="k">break</span>
-    
-    <span class="c1"># Perform topological sorting using Kahn&#39;s algorithm</span>
-    <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
-    <span class="n">queue</span> <span class="o">=</span> <span class="n">deque</span><span class="p">([</span><span class="n">char</span> <span class="k">for</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">in_degree</span> <span class="k">if</span> <span class="n">in_degree</span><span class="p">[</span><span class="n">char</span><span class="p">]</span> <span class="o">==</span> <span class="mi">0</span><span class="p">])</span>
-    
-    <span class="k">while</span> <span class="n">queue</span><span class="p">:</span>
-        <span class="n">char</span> <span class="o">=</span> <span class="n">queue</span><span class="o">.</span><span class="n">popleft</span><span class="p">()</span>
-        <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">char</span><span class="p">)</span>
-        
-        <span class="k">for</span> <span class="n">neighbor</span> <span class="ow">in</span> <span class="n">graph</span><span class="p">[</span><span class="n">char</span><span class="p">]:</span>
-            <span class="n">in_degree</span><span class="p">[</span><span class="n">neighbor</span><span class="p">]</span> <span class="o">-=</span> <span class="mi">1</span>
-            <span class="k">if</span> <span class="n">in_degree</span><span class="p">[</span><span class="n">neighbor</span><span class="p">]</span> <span class="o">==</span> <span class="mi">0</span><span class="p">:</span>
-                <span class="n">queue</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">neighbor</span><span class="p">)</span>
-    
-    <span class="c1"># Check if there is a valid order</span>
-    <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="n">result</span><span class="p">)</span> <span class="o">&lt;</span> <span class="nb">len</span><span class="p">(</span><span class="n">in_degree</span><span class="p">):</span>
-        <span class="k">return</span> <span class="s2">&quot;&quot;</span>
-    
-    <span class="k">return</span> <span class="s2">&quot;&quot;</span><span class="o">.</span><span class="n">join</span><span class="p">(</span><span class="n">result</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The code is designed to determine the order of letters in an unknown alien language that uses the Latin alphabet. It takes a list of non-empty words from a dictionary as input, where the words are sorted lexicographically based on the rules of this alien language.</p>
-<p>The code follows these main steps:</p>
-<ol class="arabic simple">
-<li><p>Create a data structure to represent the graph of letters and their relationships in the alien language. It uses a defaultdict to store a list of letters that come after each letter. Additionally, it uses a dictionary called <code class="docutils literal notranslate"><span class="pre">in_degree</span></code> to keep track of the in-degrees (the number of letters that precede a letter) for each letter.</p></li>
-<li><p>Initialize the in-degrees of all letters to 0. This is done by iterating through all the words and characters in the words, adding each character to the <code class="docutils literal notranslate"><span class="pre">in_degree</span></code> dictionary with an initial in-degree of 0.</p></li>
-<li><p>Build the graph by comparing adjacent words in the dictionary. For each pair of adjacent words, the code finds the first differing character between the two words. If such a character exists, it means there’s an order relationship between the letters represented by these characters. The code updates the graph and in-degrees accordingly.</p></li>
-<li><p>After constructing the graph, the code performs a topological sorting of the letters using Kahn’s algorithm. It starts by initializing an empty result list and a queue containing letters with in-degrees of 0. The algorithm repeatedly removes a letter with an in-degree of 0 from the queue, adds it to the result list, and updates the in-degrees of its neighbors. This process continues until the queue is empty.</p></li>
-<li><p>Finally, the code checks if the topological sorting was successful by comparing the length of the result list with the number of unique letters in the input. If the lengths differ, it means there’s a cycle in the graph, and the order is invalid. In such a case, an empty string is returned. Otherwise, the result list is joined together to form the smallest order of letters in normal lexicographical order, and it is returned as the output.</p></li>
-</ol>
-<p>The code provides a way to determine the order of letters in the alien language while handling various edge cases, including checking for invalid order conditions.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1</span>
-<span class="n">words1</span> <span class="o">=</span> <span class="p">[</span><span class="s2">&quot;wrt&quot;</span><span class="p">,</span> <span class="s2">&quot;wrf&quot;</span><span class="p">,</span> <span class="s2">&quot;er&quot;</span><span class="p">,</span> <span class="s2">&quot;ett&quot;</span><span class="p">,</span> <span class="s2">&quot;rftt&quot;</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">alienOrder</span><span class="p">(</span><span class="n">words1</span><span class="p">))</span>  <span class="c1"># Output: &quot;wertf&quot;</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>wrtef
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2</span>
-<span class="n">words2</span> <span class="o">=</span> <span class="p">[</span><span class="s2">&quot;z&quot;</span><span class="p">,</span> <span class="s2">&quot;x&quot;</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">alienOrder</span><span class="p">(</span><span class="n">words2</span><span class="p">))</span>  <span class="c1"># Output: &quot;zx&quot;</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>zx
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code for determining the order of letters in an alien language.</p>
-<p>Time Complexity:</p>
-<ol class="arabic simple">
-<li><p>Building the graph: The code iterates through the list of words and compares adjacent words. In the worst case, where all words have the same prefix, it takes O(N) time to build the graph, where N is the total number of characters in all the words.</p></li>
-<li><p>Topological sorting: Performing a topological sort on the graph takes O(V + E) time, where V is the number of unique letters (vertices) and E is the number of relationships between letters (edges) in the graph.</p></li>
-</ol>
-<p>Overall, the time complexity is O(N + V + E), where N is the total number of characters in the words, V is the number of unique letters, and E is the number of order relationships between letters.</p>
-<p>Space Complexity:</p>
-<ol class="arabic simple">
-<li><p>Graph and In-degrees: The code uses data structures to represent the graph and in-degrees. The space complexity for these data structures is O(V + E), where V is the number of unique letters, and E is the number of order relationships between letters in the words.</p></li>
-<li><p>Result List: The result list stores the order of letters, which can have a maximum size of V, where V is the number of unique letters.</p></li>
-<li><p>Queue: The space used by the queue during topological sorting is also O(V).</p></li>
-</ol>
-<p>Overall, the space complexity is O(V + E) due to the data structures used for graph representation and the space needed for storing the result and the queue.</p>
-<p>In the worst case, where there are many unique letters and many order relationships, the space complexity is dominated by the number of unique letters and their relationships.</p>
-<p>In summary, the time complexity of the code is O(N + V + E), and the space complexity is O(V + E). The actual performance will depend on the specific input data, such as the number of unique letters and the relationships between them in the alien language.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Multiple Valid Orders</strong>: Extend the problem to handle cases where there are multiple valid orders of letters in the alien language. Modify the code to return all valid orders instead of just one. Be mindful of performance.</p></li>
-<li><p><strong>Detect Invalid Dictionary</strong>: Given a dictionary of words, write a function to detect whether the dictionary is valid or not based on the constraints mentioned in the problem statement.</p></li>
-</ol>
-</section>
-</section>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
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-<h1>House Robber<a class="headerlink" href="#house-robber" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<hr class="docutils" />
-<p>You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and <strong>it will automatically contact the police if two adjacent houses were broken into on the same night.</strong></p>
-<p>Given an integer array <code class="docutils literal notranslate"><span class="pre">nums</span></code> representing the amount of money of each house, return <em>the maximum amount of money you can rob tonight <strong>without alerting the police</strong>.</em></p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">nums.length</span> <span class="pre">&lt;=</span> <span class="pre">100</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">0</span> <span class="pre">&lt;=</span> <span class="pre">nums[i]</span> <span class="pre">&lt;=</span> <span class="pre">400</span></code></p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The problem is a classic dynamic programming challenge known as the “House Robber” problem. The scenario is that you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, but the constraint is that adjacent houses have security systems connected. If two adjacent houses are broken into on the same night, the security systems will automatically contact the police.</p>
-<p>The goal is to determine the maximum amount of money you can rob tonight without alerting the police. You need to find a strategy to maximize your earnings while avoiding robbing two adjacent houses.</p>
-<p>The input to the problem is an array <code class="docutils literal notranslate"><span class="pre">nums</span></code>, where <code class="docutils literal notranslate"><span class="pre">nums[i]</span></code> represents the amount of money in the <code class="docutils literal notranslate"><span class="pre">i</span></code>-th house. The task is to return the maximum amount of money that can be robbed without triggering the security system.</p>
-<p>For example:</p>
-<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">Input</span><span class="p">:</span> <span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span><span class="mi">2</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">1</span><span class="p">]</span>
-<span class="n">Output</span><span class="p">:</span> <span class="mi">4</span>
-<span class="n">Explanation</span><span class="p">:</span> <span class="n">Rob</span> <span class="n">house</span> <span class="mi">1</span> <span class="p">(</span><span class="n">money</span> <span class="o">=</span> <span class="mi">1</span><span class="p">)</span> <span class="ow">and</span> <span class="n">then</span> <span class="n">rob</span> <span class="n">house</span> <span class="mi">3</span> <span class="p">(</span><span class="n">money</span> <span class="o">=</span> <span class="mi">3</span><span class="p">)</span><span class="o">.</span>
-<span class="n">Total</span> <span class="n">amount</span> <span class="n">you</span> <span class="n">can</span> <span class="n">rob</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">+</span> <span class="mi">3</span> <span class="o">=</span> <span class="mf">4.</span>
-
-<span class="n">Input</span><span class="p">:</span> <span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span><span class="mi">7</span><span class="p">,</span><span class="mi">9</span><span class="p">,</span><span class="mi">3</span><span class="p">,</span><span class="mi">1</span><span class="p">]</span>
-<span class="n">Output</span><span class="p">:</span> <span class="mi">12</span>
-<span class="n">Explanation</span><span class="p">:</span> <span class="n">Rob</span> <span class="n">house</span> <span class="mi">1</span> <span class="p">(</span><span class="n">money</span> <span class="o">=</span> <span class="mi">2</span><span class="p">),</span> <span class="n">rob</span> <span class="n">house</span> <span class="mi">3</span> <span class="p">(</span><span class="n">money</span> <span class="o">=</span> <span class="mi">9</span><span class="p">),</span> <span class="ow">and</span> <span class="n">rob</span> <span class="n">house</span> <span class="mi">5</span> <span class="p">(</span><span class="n">money</span> <span class="o">=</span> <span class="mi">1</span><span class="p">)</span><span class="o">.</span>
-<span class="n">Total</span> <span class="n">amount</span> <span class="n">you</span> <span class="n">can</span> <span class="n">rob</span> <span class="o">=</span> <span class="mi">2</span> <span class="o">+</span> <span class="mi">9</span> <span class="o">+</span> <span class="mi">1</span> <span class="o">=</span> <span class="mf">12.</span>
-</pre></div>
-</div>
-<p>The challenge involves designing an algorithm to efficiently solve this problem and find the optimal strategy for robbing houses to maximize the stolen money while adhering to the constraint of not robbing adjacent houses. The provided solution utilizes dynamic programming to achieve this.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">from</span> <span class="nn">typing</span> <span class="kn">import</span> <span class="n">List</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">rob</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">nums</span><span class="p">:</span> <span class="n">List</span><span class="p">[</span><span class="nb">int</span><span class="p">])</span> <span class="o">-&gt;</span> <span class="nb">int</span><span class="p">:</span>
-        <span class="c1"># Initialize variables to keep track of the maximum amount of money robbed at the previous and current houses</span>
-        <span class="n">prev</span><span class="p">,</span> <span class="n">curr</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">0</span>
-
-        <span class="c1"># Iterate through the houses</span>
-        <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="n">nums</span><span class="p">:</span>
-            <span class="c1"># Calculate the maximum amount of money that can be robbed at the current house</span>
-            <span class="c1"># It is the maximum of either robbing the current house and the money robbed at the house before the previous one (prev),</span>
-            <span class="c1"># or skipping the current house and taking the maximum amount from the previous house (curr)</span>
-            <span class="n">temp</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">n</span> <span class="o">+</span> <span class="n">prev</span><span class="p">,</span> <span class="n">curr</span><span class="p">)</span>
-
-            <span class="c1"># Update the variables for the next iteration</span>
-            <span class="n">prev</span><span class="p">,</span> <span class="n">curr</span> <span class="o">=</span> <span class="n">curr</span><span class="p">,</span> <span class="n">temp</span>
-
-        <span class="c1"># The final result is the maximum amount of money that can be robbed at the last house</span>
-        <span class="k">return</span> <span class="n">curr</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The provided code is a Python implementation of the “House Robber” problem using a dynamic programming approach. It’s part of a class <code class="docutils literal notranslate"><span class="pre">Solution</span></code> and includes a method <code class="docutils literal notranslate"><span class="pre">rob</span></code> that takes a list of integers <code class="docutils literal notranslate"><span class="pre">nums</span></code> as input and returns the maximum amount of money that can be robbed without alerting the police.</p>
-<p>The key idea of the algorithm is to maintain two variables, <code class="docutils literal notranslate"><span class="pre">prev</span></code> and <code class="docutils literal notranslate"><span class="pre">curr</span></code>, which represent the maximum amount of money that can be robbed up to the previous house and the current house, respectively. The algorithm iterates through the houses, updating these variables based on the maximum amount that can be robbed at the current house, considering the constraint of not robbing adjacent houses.</p>
-<p>The loop calculates a temporary variable <code class="docutils literal notranslate"><span class="pre">temp</span></code> that represents the maximum amount of money that can be robbed at the current house. It considers two scenarios: either robbing the current house and adding the money robbed at the house before the previous one (<code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">+</span> <span class="pre">prev</span></code>), or skipping the current house and taking the maximum amount from the previous house (<code class="docutils literal notranslate"><span class="pre">curr</span></code>). The maximum of these two scenarios is assigned to <code class="docutils literal notranslate"><span class="pre">temp</span></code>.</p>
-<p>The <code class="docutils literal notranslate"><span class="pre">prev</span></code> and <code class="docutils literal notranslate"><span class="pre">curr</span></code> variables are then updated for the next iteration. <code class="docutils literal notranslate"><span class="pre">prev</span></code> is set to the previous value of <code class="docutils literal notranslate"><span class="pre">curr</span></code>, and <code class="docutils literal notranslate"><span class="pre">curr</span></code> is set to the calculated <code class="docutils literal notranslate"><span class="pre">temp</span></code>.</p>
-<p>The final result is the value stored in the <code class="docutils literal notranslate"><span class="pre">curr</span></code> variable, representing the maximum amount of money that can be robbed at the last house. This value is returned as the result of the <code class="docutils literal notranslate"><span class="pre">rob</span></code> method.</p>
-<p>The provided test cases at the end demonstrate the usage of the <code class="docutils literal notranslate"><span class="pre">rob</span></code> method with different input arrays.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="n">sol</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-
-<span class="n">nums1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">sol</span><span class="o">.</span><span class="n">rob</span><span class="p">(</span><span class="n">nums1</span><span class="p">))</span>  <span class="c1"># Output: 4</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>4
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">nums2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">7</span><span class="p">,</span> <span class="mi">9</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">sol</span><span class="o">.</span><span class="n">rob</span><span class="p">(</span><span class="n">nums2</span><span class="p">))</span>  <span class="c1"># Output: 12</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>12
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code.</p>
-<section id="time-complexity">
-<h3>Time Complexity:<a class="headerlink" href="#time-complexity" title="Permalink to this heading">#</a></h3>
-<p>The time complexity of the code is O(n), where n is the number of houses. The algorithm iterates through the array once, and in each iteration, it performs constant time operations. The loop runs for each house, and the number of operations inside the loop is independent of the input size. Therefore, the overall time complexity is linear, O(n).</p>
-</section>
-<section id="space-complexity">
-<h3>Space Complexity:<a class="headerlink" href="#space-complexity" title="Permalink to this heading">#</a></h3>
-<p>The space complexity is O(1), constant space. The algorithm uses only a constant amount of extra space regardless of the input size. The variables <code class="docutils literal notranslate"><span class="pre">prev</span></code>, <code class="docutils literal notranslate"><span class="pre">curr</span></code>, and <code class="docutils literal notranslate"><span class="pre">temp</span></code> are the only additional variables used, and they do not depend on the size of the input array. Thus, the space complexity is constant, O(1).</p>
-<p>In summary:</p>
-<ul class="simple">
-<li><p>Time Complexity: O(n)</p></li>
-<li><p>Space Complexity: O(1)</p></li>
-</ul>
-</section>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Variation with Maximum Number of Houses to Rob:</strong> Modify the problem to include an additional constraint: you are allowed to rob at most ‘k’ houses without triggering the alarm. Find the maximum amount you can rob under this new constraint.</p></li>
-<li><p><strong>Robbing in a Circular Street:</strong> Extend the problem to a circular street where the first and last houses are adjacent. You cannot rob both the first and last houses simultaneously. Find the maximum amount you can rob in this circular setting.</p></li>
-</ol>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/house-robber/description/">To see the House Robber problem on LeetCode, click here!</a></p>
-</section>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a><ul class="nav section-nav flex-column">
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#time-complexity">Time Complexity:</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#space-complexity">Space Complexity:</a></li>
-</ul>
-</li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
-</ul>
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-  <section class="tex2jax_ignore mathjax_ignore" id="house-robber-ii">
-<h1>House Robber II<a class="headerlink" href="#house-robber-ii" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Medium</p>
-<hr class="docutils" />
-<p>You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are <strong>arranged in a circle</strong>. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and <strong>it will automatically contact the police if two adjacent houses were broken into on the same night</strong>.</p>
-<p>Given an integer array <code class="docutils literal notranslate"><span class="pre">nums</span></code> representing the amount of money of each house, return <em>the maximum amount of money you can rob tonight without alerting the police</em>.</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">nums.length</span> <span class="pre">&lt;=</span> <span class="pre">100</span></code></p></li>
-<li><p><code class="docutils literal notranslate"><span class="pre">0</span> <span class="pre">&lt;=</span> <span class="pre">nums[i]</span> <span class="pre">&lt;=</span> <span class="pre">1000</span></code></p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>Certainly! The problem statement describes a scenario where you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. However, there is a security system in place: adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.</p>
-<p>The twist in this problem is that the houses are arranged in a circle. That means the first house is the neighbor of the last one. So, robbing the first house has implications for the last house, and vice versa.</p>
-<p>The goal is to find the maximum amount of money you can rob tonight without alerting the police. You need to implement a function that takes an integer array <code class="docutils literal notranslate"><span class="pre">nums</span></code> representing the amount of money in each house and returns the maximum amount of money that can be robbed without triggering the security system.</p>
-<p>Here are a few examples to illustrate the problem:</p>
-<ol class="arabic">
-<li><p><strong>Example 1:</strong></p>
-<div class="highlight-python notranslate"><div class="highlight"><pre><span></span><span class="n">Input</span><span class="p">:</span> <span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>
-<span class="n">Output</span><span class="p">:</span> <span class="mi">3</span>
-<span class="n">Explanation</span><span class="p">:</span> <span class="n">You</span> <span class="n">cannot</span> <span class="n">rob</span> <span class="n">house</span> <span class="mi">1</span> <span class="p">(</span><span class="n">money</span> <span class="o">=</span> <span class="mi">2</span><span class="p">)</span> <span class="ow">and</span> <span class="n">then</span> <span class="n">rob</span> <span class="n">house</span> <span class="mi">3</span> <span class="p">(</span><span class="n">money</span> <span class="o">=</span> <span class="mi">2</span><span class="p">)</span> <span class="n">because</span> <span class="n">they</span> <span class="n">are</span> <span class="n">adjacent</span> <span class="n">houses</span><span class="o">.</span> <span class="n">The</span> <span class="n">optimal</span> <span class="n">strategy</span> <span class="ow">is</span> <span class="n">to</span> <span class="n">rob</span> <span class="n">house</span> <span class="mi">2</span> <span class="p">(</span><span class="n">money</span> <span class="o">=</span> <span class="mi">3</span><span class="p">)</span><span class="o">.</span>
-</pre></div>
-</div>
-</li>
-<li><p><strong>Example 2:</strong></p>
-<div class="highlight-python notranslate"><div class="highlight"><pre><span></span><span class="n">Input</span><span class="p">:</span> <span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
-<span class="n">Output</span><span class="p">:</span> <span class="mi">4</span>
-<span class="n">Explanation</span><span class="p">:</span> <span class="n">Robbing</span> <span class="n">house</span> <span class="mi">1</span> <span class="p">(</span><span class="n">money</span> <span class="o">=</span> <span class="mi">1</span><span class="p">)</span> <span class="ow">and</span> <span class="n">then</span> <span class="n">robbing</span> <span class="n">house</span> <span class="mi">3</span> <span class="p">(</span><span class="n">money</span> <span class="o">=</span> <span class="mi">3</span><span class="p">)</span> <span class="n">results</span> <span class="ow">in</span> <span class="n">a</span> <span class="n">total</span> <span class="n">amount</span> <span class="n">of</span> <span class="mf">4.</span> <span class="n">Robbing</span> <span class="n">house</span> <span class="mi">2</span> <span class="ow">is</span> <span class="n">skipped</span> <span class="n">to</span> <span class="n">avoid</span> <span class="n">alerting</span> <span class="n">the</span> <span class="n">police</span><span class="o">.</span>
-</pre></div>
-</div>
-</li>
-<li><p><strong>Example 3:</strong></p>
-<div class="highlight-python notranslate"><div class="highlight"><pre><span></span><span class="n">Input</span><span class="p">:</span> <span class="n">nums</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">]</span>
-<span class="n">Output</span><span class="p">:</span> <span class="mi">3</span>
-<span class="n">Explanation</span><span class="p">:</span> <span class="n">Since</span> <span class="n">the</span> <span class="n">houses</span> <span class="n">are</span> <span class="ow">in</span> <span class="n">a</span> <span class="n">circle</span><span class="p">,</span> <span class="n">you</span> <span class="n">can</span> <span class="n">choose</span> <span class="n">either</span> <span class="n">house</span> <span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="ow">or</span> <span class="mf">3.</span> <span class="n">The</span> <span class="n">optimal</span> <span class="n">strategy</span> <span class="ow">is</span> <span class="n">to</span> <span class="n">rob</span> <span class="n">house</span> <span class="mi">3</span> <span class="k">for</span> <span class="n">a</span> <span class="n">total</span> <span class="n">amount</span> <span class="n">of</span> <span class="mf">3.</span>
-</pre></div>
-</div>
-</li>
-</ol>
-<p>The constraints for the problem are that the length of the <code class="docutils literal notranslate"><span class="pre">nums</span></code> array is between 1 and 100, and the amount of money in each house (<code class="docutils literal notranslate"><span class="pre">nums[i]</span></code>) is between 0 and 1000.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="kn">from</span> <span class="nn">typing</span> <span class="kn">import</span> <span class="n">List</span>
-
-<span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">rob</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">nums</span><span class="p">:</span> <span class="n">List</span><span class="p">[</span><span class="nb">int</span><span class="p">])</span> <span class="o">-&gt;</span> <span class="nb">int</span><span class="p">:</span>
-        <span class="k">def</span> <span class="nf">simple_rob</span><span class="p">(</span><span class="n">nums</span><span class="p">):</span>
-            <span class="c1"># Helper function for basic house robbery problem</span>
-            <span class="n">prev</span><span class="p">,</span> <span class="n">curr</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span> <span class="mi">0</span>
-            <span class="k">for</span> <span class="n">num</span> <span class="ow">in</span> <span class="n">nums</span><span class="p">:</span>
-                <span class="c1"># Calculate the maximum amount of money that can be robbed</span>
-                <span class="c1"># without alerting the police for each house</span>
-                <span class="n">prev</span><span class="p">,</span> <span class="n">curr</span> <span class="o">=</span> <span class="n">curr</span><span class="p">,</span> <span class="nb">max</span><span class="p">(</span><span class="n">prev</span> <span class="o">+</span> <span class="n">num</span><span class="p">,</span> <span class="n">curr</span><span class="p">)</span>
-            <span class="k">return</span> <span class="n">curr</span>
-
-        <span class="c1"># Check the base cases where the length of nums is 1 or 2</span>
-        <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span> <span class="o">==</span> <span class="mi">1</span><span class="p">:</span>
-            <span class="k">return</span> <span class="n">nums</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
-        <span class="k">elif</span> <span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span> <span class="o">==</span> <span class="mi">2</span><span class="p">:</span>
-            <span class="k">return</span> <span class="nb">max</span><span class="p">(</span><span class="n">nums</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span> <span class="n">nums</span><span class="p">[</span><span class="mi">1</span><span class="p">])</span>
-
-        <span class="c1"># Case 1: First house is robbed, last house is not</span>
-        <span class="n">case1</span> <span class="o">=</span> <span class="n">simple_rob</span><span class="p">(</span><span class="n">nums</span><span class="p">[:</span><span class="o">-</span><span class="mi">1</span><span class="p">])</span>
-
-        <span class="c1"># Case 2: First house is not robbed, last house is</span>
-        <span class="n">case2</span> <span class="o">=</span> <span class="n">simple_rob</span><span class="p">(</span><span class="n">nums</span><span class="p">[</span><span class="mi">1</span><span class="p">:])</span>
-
-        <span class="c1"># Maximum of the two cases</span>
-        <span class="k">return</span> <span class="nb">max</span><span class="p">(</span><span class="n">case1</span><span class="p">,</span> <span class="n">case2</span><span class="p">)</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The provided code defines a class <code class="docutils literal notranslate"><span class="pre">Solution</span></code> with a method <code class="docutils literal notranslate"><span class="pre">rob</span></code>, which aims to determine the maximum amount of money a professional robber can steal from houses arranged in a circle without alerting the police. Each house contains a certain amount of money, and adjacent houses have a security system that triggers if both are robbed on the same night.</p>
-<p>The code uses dynamic programming to solve the problem efficiently. It contains a helper function <code class="docutils literal notranslate"><span class="pre">simple_rob</span></code> that computes the maximum amount of money that can be robbed from a linear arrangement of houses. The main method <code class="docutils literal notranslate"><span class="pre">rob</span></code> handles the circular arrangement by considering two cases: robbing the first house and not robbing the last, and not robbing the first house and robbing the last. The result is the maximum of these two cases.</p>
-<p>The class is instantiated as <code class="docutils literal notranslate"><span class="pre">solution</span></code>, and the <code class="docutils literal notranslate"><span class="pre">rob</span></code> method is applied to three different scenarios (test cases) to demonstrate its functionality. The final output represents the maximum amount of money the robber can steal without triggering the security system.</p>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-
-<span class="c1"># Create an instance of the Solution class</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-
-<span class="n">nums1</span> <span class="o">=</span> <span class="p">[</span><span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">solution</span><span class="o">.</span><span class="n">rob</span><span class="p">(</span><span class="n">nums1</span><span class="p">))</span>  <span class="c1"># Output: 3</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>3
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">nums2</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">,</span> <span class="mi">1</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">solution</span><span class="o">.</span><span class="n">rob</span><span class="p">(</span><span class="n">nums2</span><span class="p">))</span>  <span class="c1"># Output: 4</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>4
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">nums3</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">,</span> <span class="mi">3</span><span class="p">]</span>
-<span class="nb">print</span><span class="p">(</span><span class="n">solution</span><span class="o">.</span><span class="n">rob</span><span class="p">(</span><span class="n">nums3</span><span class="p">))</span>  <span class="c1"># Output: 3</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>3
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Let’s analyze the time and space complexity of the provided code:</p>
-<section id="time-complexity">
-<h3>Time Complexity:<a class="headerlink" href="#time-complexity" title="Permalink to this heading">#</a></h3>
-<p>The time complexity is primarily determined by the <code class="docutils literal notranslate"><span class="pre">simple_rob</span></code> helper function, which has a time complexity of O(N), where N is the number of houses in the input array. The main <code class="docutils literal notranslate"><span class="pre">rob</span></code> function calls <code class="docutils literal notranslate"><span class="pre">simple_rob</span></code> twice, once for the case of robbing the first house and not robbing the last, and once for the case of not robbing the first house and robbing the last. Therefore, the overall time complexity is O(N).</p>
-</section>
-<section id="space-complexity">
-<h3>Space Complexity:<a class="headerlink" href="#space-complexity" title="Permalink to this heading">#</a></h3>
-<p>The space complexity is O(1) because the space used by the algorithm is constant, regardless of the size of the input array. The only variables used are <code class="docutils literal notranslate"><span class="pre">prev</span></code> and <code class="docutils literal notranslate"><span class="pre">curr</span></code> in the <code class="docutils literal notranslate"><span class="pre">simple_rob</span></code> function, which represent the previous and current maximum amounts of money that can be robbed without alerting the police. These variables are updated iteratively, and no additional data structures are used that scale with the input size. Therefore, the space complexity is constant, or O(1).</p>
-<p>In summary:</p>
-<ul class="simple">
-<li><p>Time Complexity: O(N)</p></li>
-<li><p>Space Complexity: O(1)</p></li>
-</ul>
-</section>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Optimization Challenge:</strong> Optimize the space complexity of the solution. Can you solve the problem with O(1) space complexity without sacrificing the time complexity?</p></li>
-<li><p><strong>Circular Robbery Variations:</strong> Consider variations of the circular house robbery problem, such as allowing the robber to skip a certain number of houses or restricting the number of consecutive houses that can be robbed. Modify the solution to accommodate these variations.</p></li>
-</ol>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/house-robber-ii/">To see the House Robber II problem on LeetCode, click here!</a></p>
-</section>
-</section>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#probelm-explanation">Probelm Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#solution">Solution:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#explanation">Explanation:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#test-cases">Test cases:</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#time-and-space-complexity-analysis">Time and Space Complexity Analysis</a><ul class="nav section-nav flex-column">
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#time-complexity">Time Complexity:</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#space-complexity">Space Complexity:</a></li>
-</ul>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#challenging-exercises">Challenging Exercises:</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="climbing-stairs">
-<h1>70. Climbing Stairs<a class="headerlink" href="#climbing-stairs" title="Permalink to this heading">#</a></h1>
-<p><strong>Difficulty:</strong> Easy</p>
-<p><strong>Link to Problem:</strong> <a class="reference external" href="https://leetcode.com/problems/climbing-stairs/">To see the Climbing Stairs problem on LeetCode, click here!</a></p>
-<hr class="docutils" />
-<p>You are climbing a staircase. It takes <code class="docutils literal notranslate"><span class="pre">n</span></code> steps to reach the top.</p>
-<p>Each time you can either climb <code class="docutils literal notranslate"><span class="pre">1</span></code> or <code class="docutils literal notranslate"><span class="pre">2</span></code> steps. In how many distinct ways can you climb to the top?</p>
-<p><strong>Constraints:</strong></p>
-<ul class="simple">
-<li><p><code class="docutils literal notranslate"><span class="pre">1</span> <span class="pre">&lt;=</span> <span class="pre">n</span> <span class="pre">&lt;=</span> <span class="pre">45</span></code></p></li>
-</ul>
-<section id="probelm-explanation">
-<h2>Probelm Explanation:<a class="headerlink" href="#probelm-explanation" title="Permalink to this heading">#</a></h2>
-<p>The “Climbing Stairs” problem is a classic algorithmic problem that asks you to find the number of distinct ways to climb a staircase with a given number of steps. Here’s the problem statement:</p>
-<p>You are given a staircase with ‘n’ steps. Each time you can either climb 1 step or 2 steps. You need to find out how many distinct ways there are to reach the top of the staircase.</p>
-<p>For example, if ‘n’ is 2, there are two ways to climb to the top:</p>
-<ol class="arabic simple">
-<li><p>Climbing two steps at once.</p></li>
-<li><p>Climbing one step, and then another step.</p></li>
-</ol>
-<p>If ‘n’ is 3, there are three ways:</p>
-<ol class="arabic simple">
-<li><p>Climbing one step, then one step, and finally one step.</p></li>
-<li><p>Climbing one step, then two steps.</p></li>
-<li><p>Climbing two steps, then one step.</p></li>
-</ol>
-<p>The problem essentially asks you to find a solution that counts the possible combinations of 1-step and 2-step climbs that can be taken to reach the top of the staircase. It’s a classic example of a dynamic programming problem where you can break it down into smaller subproblems and build a solution incrementally, as shown in the code I provided earlier.</p>
-</section>
-<section id="solution">
-<h2>Solution:<a class="headerlink" href="#solution" title="Permalink to this heading">#</a></h2>
-<p>Here’s a Python function to implement this algorithm:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">class</span> <span class="nc">Solution</span><span class="p">:</span>
-    <span class="k">def</span> <span class="nf">climbStairs</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">n</span><span class="p">:</span> <span class="nb">int</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="nb">int</span><span class="p">:</span>
-        <span class="c1"># If there are 1 or 2 steps, there are exactly n ways to reach the top.</span>
-        <span class="k">if</span> <span class="n">n</span> <span class="o">&lt;=</span> <span class="mi">2</span><span class="p">:</span>
-            <span class="k">return</span> <span class="n">n</span>
-
-        <span class="c1"># Create an array to store the number of ways to reach each step.</span>
-        <span class="n">dp</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="p">(</span><span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span>
-
-        <span class="c1"># There is one way to reach the first step (base case).</span>
-        <span class="n">dp</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span> <span class="o">=</span> <span class="mi">1</span>
-
-        <span class="c1"># There are two ways to reach the second step (base case).</span>
-        <span class="n">dp</span><span class="p">[</span><span class="mi">2</span><span class="p">]</span> <span class="o">=</span> <span class="mi">2</span>
-
-        <span class="c1"># Calculate the number of ways to reach each step from the previous steps.</span>
-        <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span> <span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">):</span>
-            <span class="c1"># The number of ways to reach step &#39;i&#39; is the sum of the ways to reach</span>
-            <span class="c1"># the previous two steps, as you can take 1 or 2 steps at a time.</span>
-            <span class="n">dp</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">=</span> <span class="n">dp</span><span class="p">[</span><span class="n">i</span> <span class="o">-</span> <span class="mi">1</span><span class="p">]</span> <span class="o">+</span> <span class="n">dp</span><span class="p">[</span><span class="n">i</span> <span class="o">-</span> <span class="mi">2</span><span class="p">]</span>
-
-        <span class="c1"># The result is the number of ways to reach the top, which is stored in dp[n].</span>
-        <span class="k">return</span> <span class="n">dp</span><span class="p">[</span><span class="n">n</span><span class="p">]</span>
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="explanation">
-<h2>Explanation:<a class="headerlink" href="#explanation" title="Permalink to this heading">#</a></h2>
-<p>The code defines a class <code class="docutils literal notranslate"><span class="pre">Solution</span></code> with a method <code class="docutils literal notranslate"><span class="pre">climbStairs</span></code> that takes an integer <code class="docutils literal notranslate"><span class="pre">n</span></code> as input and returns the number of distinct ways to climb a staircase with <code class="docutils literal notranslate"><span class="pre">n</span></code> steps.</p>
-<ol class="arabic simple">
-<li><p>If <code class="docutils literal notranslate"><span class="pre">n</span></code> is 1 or 2, the function returns <code class="docutils literal notranslate"><span class="pre">n</span></code> directly because there are only 1 or 2 distinct ways to reach the top in those cases.</p></li>
-<li><p>For <code class="docutils literal notranslate"><span class="pre">n</span></code> greater than 2, the code uses dynamic programming to calculate the number of ways to reach each step. It creates an array <code class="docutils literal notranslate"><span class="pre">dp</span></code> of length <code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">+</span> <span class="pre">1</span></code> to store these values.</p></li>
-<li><p>The <code class="docutils literal notranslate"><span class="pre">dp[1]</span></code> and <code class="docutils literal notranslate"><span class="pre">dp[2]</span></code> are initialized to 1 and 2, respectively. This is because there is only one way to reach the first step (by taking 1 step), and there are two ways to reach the second step (by taking 2 steps or two 1-step climbs).</p></li>
-<li><p>The code then uses a loop to fill in the <code class="docutils literal notranslate"><span class="pre">dp</span></code> array for steps 3 and above. For each step <code class="docutils literal notranslate"><span class="pre">i</span></code> from 3 to <code class="docutils literal notranslate"><span class="pre">n</span></code>, it calculates the number of ways to reach that step by adding the number of ways to reach the previous two steps, as you can either take 1 step or 2 steps at a time.</p></li>
-<li><p>Finally, the function returns the value stored in <code class="docutils literal notranslate"><span class="pre">dp[n]</span></code>, which represents the total number of distinct ways to climb the staircase with <code class="docutils literal notranslate"><span class="pre">n</span></code> steps.</p></li>
-</ol>
-</section>
-<section id="test-cases">
-<h2>Test cases:<a class="headerlink" href="#test-cases" title="Permalink to this heading">#</a></h2>
-<p>Here’s how you can use this solution:</p>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 1:</span>
-<span class="n">n</span> <span class="o">=</span> <span class="mi">2</span>
-<span class="n">solution</span> <span class="o">=</span> <span class="n">Solution</span><span class="p">()</span>
-<span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;Example 1 - Input: n = </span><span class="si">{</span><span class="n">n</span><span class="si">}</span><span class="s2">, Output: </span><span class="si">{</span><span class="n">solution</span><span class="o">.</span><span class="n">climbStairs</span><span class="p">(</span><span class="n">n</span><span class="p">)</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>  <span class="c1"># Output should be 2</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Example 1 - Input: n = 2, Output: 2
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="c1"># Example 2:</span>
-<span class="n">n</span> <span class="o">=</span> <span class="mi">3</span>
-<span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;Example 2 - Input: n = </span><span class="si">{</span><span class="n">n</span><span class="si">}</span><span class="s2">, Output: </span><span class="si">{</span><span class="n">solution</span><span class="o">.</span><span class="n">climbStairs</span><span class="p">(</span><span class="n">n</span><span class="p">)</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>  <span class="c1"># Output should be 3</span>
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Example 2 - Input: n = 3, Output: 3
-</pre></div>
-</div>
-</div>
-</div>
-</section>
-<section id="time-and-space-complexity-analysis">
-<h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-space-complexity-analysis" title="Permalink to this heading">#</a></h2>
-<p>Time Complexity:
-The time complexity of the solution is O(n) because we use a single for loop that iterates from 3 to n to fill the <code class="docutils literal notranslate"><span class="pre">dp</span></code> array. In each iteration, we perform constant time operations (addition and assignment). Therefore, the time complexity is linear with respect to the input value <code class="docutils literal notranslate"><span class="pre">n</span></code>.</p>
-<p>Space Complexity:
-The space complexity of the solution is O(n) as well. We use an array <code class="docutils literal notranslate"><span class="pre">dp</span></code> of length <code class="docutils literal notranslate"><span class="pre">n</span> <span class="pre">+</span> <span class="pre">1</span></code> to store the number of ways to reach each step. The size of this array is directly proportional to the input value <code class="docutils literal notranslate"><span class="pre">n</span></code>, which makes the space complexity linear in terms of <code class="docutils literal notranslate"><span class="pre">n</span></code>. The rest of the variables used in the function (e.g., <code class="docutils literal notranslate"><span class="pre">i</span></code>) occupy constant space and do not contribute significantly to the overall space complexity.</p>
-<p>In summary, the time complexity is O(n), and the space complexity is O(n) for this dynamic programming solution to the “Climbing Stairs” problem.</p>
-</section>
-<section id="challenging-exercises">
-<h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p><strong>Generalized Step Sizes</strong>: Instead of being limited to 1 step or 2 steps, consider a scenario where you can take steps of sizes <code class="docutils literal notranslate"><span class="pre">[1,</span> <span class="pre">2,</span> <span class="pre">3]</span></code>. Implement a function that calculates the number of ways to climb the staircase with these different step sizes.</p></li>
-<li><p><strong>Minimize Steps</strong>: Given an integer <code class="docutils literal notranslate"><span class="pre">n</span></code>, find the minimum number of steps required to reach the top of the staircase. Return both the minimum number of steps and the specific sequence of steps taken.</p></li>
-</ol>
-</section>
-</section>
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-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
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-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/climbing-stairs/">70. Climbing Stairs</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/house-robber/">198. House Robber</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/house-robber-ii/">213. House Robber II</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/longest-palindromic-substring/">5. Longest Palindromic Substring</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/palindromic-substrings/">647. Palindromic Substrings</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/decode-ways/">91. Decode Ways</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/coin-change/">322. Coin Change</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/maximum-product-subarray/">152. Maximum Product Subarray</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/word-break/">139. Word Break</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/longest-increasing-subsequence/">300. Longest Increasing Subsequence</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-<section id="contribute">
-<h2>Contribute<a class="headerlink" href="#contribute" title="Permalink to this heading">#</a></h2>
-<p>If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.</p>
-</section>
-<section id="resources">
-<h2>Resources<a class="headerlink" href="#resources" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p><a class="reference external" href="https://leetcode.com/">LeetCode</a>: The official LeetCode website.</p></li>
-<li><p><a class="reference external" href="https://leetcode.com/discuss/">LeetCode Discuss</a>: LeetCode’s discussion forum for each problem. You can find helpful hints and discussions here.</p></li>
-</ul>
-<p>Happy coding!</p>
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-<h1>Blind 75 LeetCode Problems<a class="headerlink" href="#blind-75-leetcode-problems" title="Permalink to this heading">#</a></h1>
-<p>This repository contains a curated list of 75 LeetCode problems that are commonly known as the “Blind 75.” These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.</p>
-<section id="problem-list">
-<h2>Problem List<a class="headerlink" href="#problem-list" title="Permalink to this heading">#</a></h2>
-<section id="d-dynamic-programming">
-<h3>2-D Dynamic Programming<a class="headerlink" href="#d-dynamic-programming" title="Permalink to this heading">#</a></h3>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/unique-paths/">62. Unique Paths</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/longest-common-subsequence/">1143. Longest Common Subsequence</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-</section>
-<section id="how-to-use">
-<h2>How to Use<a class="headerlink" href="#how-to-use" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>Clone or download this repository to your local machine.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span>git<span class="w"> </span>clone<span class="w"> </span>https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-</pre></div>
-</div>
-<ol class="arabic simple" start="2">
-<li><p>Navigate to the directory of the specific problem you want to solve.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span><span class="nb">cd</span><span class="w"> </span>Blind-75-LeetCode/Problem-Directory
-</pre></div>
-</div>
-<ol class="arabic simple" start="3">
-<li><p>Open the problem’s directory to find its description, code template, and solution(s).</p></li>
-<li><p>Solve the problem using your preferred programming language.</p></li>
-<li><p>Check the solution(s) provided in the directory for reference and to compare your solution.</p></li>
-</ol>
-</section>
-<section id="contribute">
-<h2>Contribute<a class="headerlink" href="#contribute" title="Permalink to this heading">#</a></h2>
-<p>If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.</p>
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-<h1>Blind 75 LeetCode Problems<a class="headerlink" href="#blind-75-leetcode-problems" title="Permalink to this heading">#</a></h1>
-<p>This repository contains a curated list of 75 LeetCode problems that are commonly known as the “Blind 75.” These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.</p>
-<section id="problem-list">
-<h2>Problem List<a class="headerlink" href="#problem-list" title="Permalink to this heading">#</a></h2>
-<section id="greedy">
-<h3>Greedy<a class="headerlink" href="#greedy" title="Permalink to this heading">#</a></h3>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
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-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/maximum-subarray/">53. Maximum Subarray</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/jump-game/">55. Jump Game</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
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-<h2>How to Use<a class="headerlink" href="#how-to-use" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>Clone or download this repository to your local machine.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span>git<span class="w"> </span>clone<span class="w"> </span>https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-</pre></div>
-</div>
-<ol class="arabic simple" start="2">
-<li><p>Navigate to the directory of the specific problem you want to solve.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span><span class="nb">cd</span><span class="w"> </span>Blind-75-LeetCode/Problem-Directory
-</pre></div>
-</div>
-<ol class="arabic simple" start="3">
-<li><p>Open the problem’s directory to find its description, code template, and solution(s).</p></li>
-<li><p>Solve the problem using your preferred programming language.</p></li>
-<li><p>Check the solution(s) provided in the directory for reference and to compare your solution.</p></li>
-</ol>
-</section>
-<section id="contribute">
-<h2>Contribute<a class="headerlink" href="#contribute" title="Permalink to this heading">#</a></h2>
-<p>If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.</p>
-</section>
-<section id="resources">
-<h2>Resources<a class="headerlink" href="#resources" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p><a class="reference external" href="https://leetcode.com/">LeetCode</a>: The official LeetCode website.</p></li>
-<li><p><a class="reference external" href="https://leetcode.com/discuss/">LeetCode Discuss</a>: LeetCode’s discussion forum for each problem. You can find helpful hints and discussions here.</p></li>
-</ul>
-</section>
-<section id="disclaimer">
-<h2>Disclaimer<a class="headerlink" href="#disclaimer" title="Permalink to this heading">#</a></h2>
-<p>This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode’s terms of use and guidelines when using this repository.</p>
-<p>Happy coding!</p>
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-<h1>Blind 75 LeetCode Problems<a class="headerlink" href="#blind-75-leetcode-problems" title="Permalink to this heading">#</a></h1>
-<p>This repository contains a curated list of 75 LeetCode problems that are commonly known as the “Blind 75.” These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.</p>
-<section id="problem-list">
-<h2>Problem List<a class="headerlink" href="#problem-list" title="Permalink to this heading">#</a></h2>
-<section id="intervals">
-<h3>Intervals<a class="headerlink" href="#intervals" title="Permalink to this heading">#</a></h3>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
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-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/insert-interval/">57. Insert Interval</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/merge-intervals/">56. Merge Intervals</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/non-overlapping-intervals/">435. Non Overlapping Intervals</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/meeting-rooms/">252. Meeting Rooms</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/meeting-rooms-ii/">253. Meeting Rooms II</a></p></td>
-<td><p>Medium</p></td>
-</tr>
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-<h2>How to Use<a class="headerlink" href="#how-to-use" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>Clone or download this repository to your local machine.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span>git<span class="w"> </span>clone<span class="w"> </span>https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-</pre></div>
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-<ol class="arabic simple" start="2">
-<li><p>Navigate to the directory of the specific problem you want to solve.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span><span class="nb">cd</span><span class="w"> </span>Blind-75-LeetCode/Problem-Directory
-</pre></div>
-</div>
-<ol class="arabic simple" start="3">
-<li><p>Open the problem’s directory to find its description, code template, and solution(s).</p></li>
-<li><p>Solve the problem using your preferred programming language.</p></li>
-<li><p>Check the solution(s) provided in the directory for reference and to compare your solution.</p></li>
-</ol>
-</section>
-<section id="contribute">
-<h2>Contribute<a class="headerlink" href="#contribute" title="Permalink to this heading">#</a></h2>
-<p>If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.</p>
-</section>
-<section id="resources">
-<h2>Resources<a class="headerlink" href="#resources" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p><a class="reference external" href="https://leetcode.com/">LeetCode</a>: The official LeetCode website.</p></li>
-<li><p><a class="reference external" href="https://leetcode.com/discuss/">LeetCode Discuss</a>: LeetCode’s discussion forum for each problem. You can find helpful hints and discussions here.</p></li>
-</ul>
-</section>
-<section id="disclaimer">
-<h2>Disclaimer<a class="headerlink" href="#disclaimer" title="Permalink to this heading">#</a></h2>
-<p>This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode’s terms of use and guidelines when using this repository.</p>
-<p>Happy coding!</p>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#problem-list">Problem List</a><ul class="nav section-nav flex-column">
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#math-geometry">Math &amp; Geometry</a></li>
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-<h1>Blind 75 LeetCode Problems<a class="headerlink" href="#blind-75-leetcode-problems" title="Permalink to this heading">#</a></h1>
-<p>This repository contains a curated list of 75 LeetCode problems that are commonly known as the “Blind 75.” These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.</p>
-<section id="problem-list">
-<h2>Problem List<a class="headerlink" href="#problem-list" title="Permalink to this heading">#</a></h2>
-<section id="math-geometry">
-<h3>Math &amp; Geometry<a class="headerlink" href="#math-geometry" title="Permalink to this heading">#</a></h3>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/rotate-image/">48. Rotate Image</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/spiral-matrix/">54. Spiral Matrix</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/set-matrix-zeroes/">73. Set Matrix Zeroes</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-</section>
-<section id="how-to-use">
-<h2>How to Use<a class="headerlink" href="#how-to-use" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>Clone or download this repository to your local machine.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span>git<span class="w"> </span>clone<span class="w"> </span>https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-</pre></div>
-</div>
-<ol class="arabic simple" start="2">
-<li><p>Navigate to the directory of the specific problem you want to solve.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span><span class="nb">cd</span><span class="w"> </span>Blind-75-LeetCode/Problem-Directory
-</pre></div>
-</div>
-<ol class="arabic simple" start="3">
-<li><p>Open the problem’s directory to find its description, code template, and solution(s).</p></li>
-<li><p>Solve the problem using your preferred programming language.</p></li>
-<li><p>Check the solution(s) provided in the directory for reference and to compare your solution.</p></li>
-</ol>
-</section>
-<section id="contribute">
-<h2>Contribute<a class="headerlink" href="#contribute" title="Permalink to this heading">#</a></h2>
-<p>If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.</p>
-</section>
-<section id="resources">
-<h2>Resources<a class="headerlink" href="#resources" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p><a class="reference external" href="https://leetcode.com/">LeetCode</a>: The official LeetCode website.</p></li>
-<li><p><a class="reference external" href="https://leetcode.com/discuss/">LeetCode Discuss</a>: LeetCode’s discussion forum for each problem. You can find helpful hints and discussions here.</p></li>
-</ul>
-</section>
-<section id="disclaimer">
-<h2>Disclaimer<a class="headerlink" href="#disclaimer" title="Permalink to this heading">#</a></h2>
-<p>This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode’s terms of use and guidelines when using this repository.</p>
-<p>Happy coding!</p>
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-    <h1>Blind 75 LeetCode Problems</h1>
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-                <h2> Contents </h2>
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-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#problem-list">Problem List</a><ul class="nav section-nav flex-column">
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-<h1>Blind 75 LeetCode Problems<a class="headerlink" href="#blind-75-leetcode-problems" title="Permalink to this heading">#</a></h1>
-<p>This repository contains a curated list of 75 LeetCode problems that are commonly known as the “Blind 75.” These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.</p>
-<section id="problem-list">
-<h2>Problem List<a class="headerlink" href="#problem-list" title="Permalink to this heading">#</a></h2>
-<section id="bit-manupulation">
-<h3>Bit Manupulation<a class="headerlink" href="#bit-manupulation" title="Permalink to this heading">#</a></h3>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/number-of-1-bits/">191. Number of 1 Bits</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/counting-bits/">338. Counting Bits</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/reverse-bits/">190. Reverse Bits</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/missing-number/">268. Missing Number</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/sum-of-two-integers/">371. Sum of Two Integers</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-</section>
-<section id="how-to-use">
-<h2>How to Use<a class="headerlink" href="#how-to-use" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>Clone or download this repository to your local machine.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span>git<span class="w"> </span>clone<span class="w"> </span>https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-</pre></div>
-</div>
-<ol class="arabic simple" start="2">
-<li><p>Navigate to the directory of the specific problem you want to solve.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span><span class="nb">cd</span><span class="w"> </span>Blind-75-LeetCode/Problem-Directory
-</pre></div>
-</div>
-<ol class="arabic simple" start="3">
-<li><p>Open the problem’s directory to find its description, code template, and solution(s).</p></li>
-<li><p>Solve the problem using your preferred programming language.</p></li>
-<li><p>Check the solution(s) provided in the directory for reference and to compare your solution.</p></li>
-</ol>
-</section>
-<section id="contribute">
-<h2>Contribute<a class="headerlink" href="#contribute" title="Permalink to this heading">#</a></h2>
-<p>If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.</p>
-</section>
-<section id="resources">
-<h2>Resources<a class="headerlink" href="#resources" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p><a class="reference external" href="https://leetcode.com/">LeetCode</a>: The official LeetCode website.</p></li>
-<li><p><a class="reference external" href="https://leetcode.com/discuss/">LeetCode Discuss</a>: LeetCode’s discussion forum for each problem. You can find helpful hints and discussions here.</p></li>
-</ul>
-</section>
-<section id="disclaimer">
-<h2>Disclaimer<a class="headerlink" href="#disclaimer" title="Permalink to this heading">#</a></h2>
-<p>This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode’s terms of use and guidelines when using this repository.</p>
-<p>Happy coding!</p>
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-Requirement already satisfied: jsonschema&lt;5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (4.19.1)
-Requirement already satisfied: linkify-it-py~=2.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (2.0.2)
-Requirement already satisfied: myst-nb~=0.17.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.17.2)
-Requirement already satisfied: pyyaml in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (6.0.1)
-Requirement already satisfied: sphinx&lt;6,&gt;=4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (5.0.2)
-Requirement already satisfied: sphinx-comments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.0.3)
-Requirement already satisfied: sphinx-copybutton in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.5.2)
-Requirement already satisfied: sphinx-external-toc~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.1)
-Requirement already satisfied: sphinx-jupyterbook-latex~=0.5.2 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.5.2)
-Requirement already satisfied: sphinx-design~=0.3.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.0)
-Requirement already satisfied: sphinx-thebe~=0.2.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.2.1)
-Requirement already satisfied: sphinx-book-theme~=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (1.0.1)
-Requirement already satisfied: sphinx_togglebutton in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.2)
-Requirement already satisfied: sphinxcontrib-bibtex&lt;=2.5.0,&gt;=2.2.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (2.5.0)
-Requirement already satisfied: sphinx-multitoc-numbering~=0.1.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.1.3)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: attrs&gt;=22.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema&lt;5-&gt;jupyter-book) (23.1.0)
-Requirement already satisfied: jsonschema-specifications&gt;=2023.03.6 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema&lt;5-&gt;jupyter-book) (2023.7.1)
-Requirement already satisfied: referencing&gt;=0.28.4 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema&lt;5-&gt;jupyter-book) (0.30.2)
-Requirement already satisfied: rpds-py&gt;=0.7.1 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema&lt;5-&gt;jupyter-book) (0.10.6)
-Requirement already satisfied: uc-micro-py in /usr/local/python/3.10.8/lib/python3.10/site-packages (from linkify-it-py~=2.0.0-&gt;jupyter-book) (1.0.2)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: importlib_metadata in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (6.8.0)
-Requirement already satisfied: ipython in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (8.16.1)
-Requirement already satisfied: jupyter-cache&lt;0.7,&gt;=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (0.6.1)
-Requirement already satisfied: nbclient in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (0.7.4)
-Requirement already satisfied: myst-parser~=0.18.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (0.18.1)
-Requirement already satisfied: nbformat~=5.0 in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (5.9.2)
-Requirement already satisfied: typing-extensions in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (4.8.0)
-Requirement already satisfied: ipykernel in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (6.25.2)
-Requirement already satisfied: sphinxcontrib-applehelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.0.7)
-Requirement already satisfied: sphinxcontrib-devhelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.0.5)
-Requirement already satisfied: sphinxcontrib-jsmath in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.0.1)
-Requirement already satisfied: sphinxcontrib-htmlhelp&gt;=2.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (2.0.4)
-Requirement already satisfied: sphinxcontrib-serializinghtml&gt;=1.1.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.1.9)
-Requirement already satisfied: sphinxcontrib-qthelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.0.6)
-Requirement already satisfied: Pygments&gt;=2.0 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (2.16.1)
-Requirement already satisfied: snowballstemmer&gt;=1.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (2.2.0)
-Requirement already satisfied: babel&gt;=1.3 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (2.13.0)
-Requirement already satisfied: alabaster&lt;0.8,&gt;=0.7 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (0.7.13)
-Requirement already satisfied: imagesize in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.4.1)
-Requirement already satisfied: requests&gt;=2.5.0 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (2.31.0)
-Requirement already satisfied: packaging in /home/codespace/.local/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (23.2)
-Requirement already satisfied: MarkupSafe&gt;=2.0 in /home/codespace/.local/lib/python3.10/site-packages (from Jinja2-&gt;jupyter-book) (2.1.3)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: pydata-sphinx-theme&gt;=0.13.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx-book-theme~=1.0.0-&gt;jupyter-book) (0.14.3)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: pybtex&gt;=0.24 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex&lt;=2.5.0,&gt;=2.2.0-&gt;jupyter-book) (0.24.0)
-Requirement already satisfied: pybtex-docutils&gt;=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex&lt;=2.5.0,&gt;=2.2.0-&gt;jupyter-book) (1.0.3)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: setuptools in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx_togglebutton-&gt;jupyter-book) (68.0.0)
-Requirement already satisfied: wheel in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx_togglebutton-&gt;jupyter-book) (0.41.3)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: sqlalchemy&lt;3,&gt;=1.3.12 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-cache&lt;0.7,&gt;=0.5-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.0.23)
-Requirement already satisfied: tabulate in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-cache&lt;0.7,&gt;=0.5-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.9.0)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: markdown-it-py&lt;3.0.0,&gt;=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.2.0)
-Requirement already satisfied: mdit-py-plugins~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.3.5)
-Requirement already satisfied: jupyter-client&gt;=6.1.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (8.4.0)
-Requirement already satisfied: jupyter-core!=5.0.*,&gt;=4.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (5.4.0)
-Requirement already satisfied: traitlets&gt;=5.3 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (5.11.2)
-Requirement already satisfied: fastjsonschema in /home/codespace/.local/lib/python3.10/site-packages (from nbformat~=5.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.18.1)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: latexcodec&gt;=1.0.4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pybtex&gt;=0.24-&gt;sphinxcontrib-bibtex&lt;=2.5.0,&gt;=2.2.0-&gt;jupyter-book) (2.0.1)
-Requirement already satisfied: six in /home/codespace/.local/lib/python3.10/site-packages (from pybtex&gt;=0.24-&gt;sphinxcontrib-bibtex&lt;=2.5.0,&gt;=2.2.0-&gt;jupyter-book) (1.16.0)
-Requirement already satisfied: beautifulsoup4 in /home/codespace/.local/lib/python3.10/site-packages (from pydata-sphinx-theme&gt;=0.13.3-&gt;sphinx-book-theme~=1.0.0-&gt;jupyter-book) (4.12.2)
-Requirement already satisfied: accessible-pygments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pydata-sphinx-theme&gt;=0.13.3-&gt;sphinx-book-theme~=1.0.0-&gt;jupyter-book) (0.0.4)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: charset-normalizer&lt;4,&gt;=2 in /home/codespace/.local/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (3.3.0)
-Requirement already satisfied: idna&lt;4,&gt;=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (3.4)
-Requirement already satisfied: urllib3&lt;3,&gt;=1.21.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.26.18)
-Requirement already satisfied: certifi&gt;=2017.4.17 in /home/codespace/.local/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (2023.7.22)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: zipp&gt;=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from importlib_metadata-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (3.17.0)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: comm&gt;=0.1.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.1.4)
-Requirement already satisfied: debugpy&gt;=1.6.5 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (1.8.0)
-Requirement already satisfied: matplotlib-inline&gt;=0.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.1.6)
-Requirement already satisfied: nest-asyncio in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (1.5.8)
-Requirement already satisfied: psutil in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (5.9.6)
-Requirement already satisfied: pyzmq&gt;=20 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (25.1.1)
-Requirement already satisfied: tornado&gt;=6.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (6.3.3)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: backcall in /home/codespace/.local/lib/python3.10/site-packages (from ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.2.0)
-Requirement already satisfied: decorator in /home/codespace/.local/lib/python3.10/site-packages (from ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (5.1.1)
-Requirement already satisfied: jedi&gt;=0.16 in /home/codespace/.local/lib/python3.10/site-packages (from ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.19.1)
-Requirement already satisfied: pickleshare in /home/codespace/.local/lib/python3.10/site-packages (from ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.7.5)
-Requirement already satisfied: prompt-toolkit!=3.0.37,&lt;3.1.0,&gt;=3.0.30 in /home/codespace/.local/lib/python3.10/site-packages (from ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (3.0.39)
-Requirement already satisfied: stack-data in /home/codespace/.local/lib/python3.10/site-packages (from ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.6.3)
-Requirement already satisfied: exceptiongroup in /home/codespace/.local/lib/python3.10/site-packages (from ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (1.1.3)
-Requirement already satisfied: pexpect&gt;4.3 in /home/codespace/.local/lib/python3.10/site-packages (from ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (4.8.0)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: parso&lt;0.9.0,&gt;=0.8.3 in /home/codespace/.local/lib/python3.10/site-packages (from jedi&gt;=0.16-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.8.3)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: python-dateutil&gt;=2.8.2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-client&gt;=6.1.12-&gt;nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.8.2)
-Requirement already satisfied: platformdirs&gt;=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-core!=5.0.*,&gt;=4.12-&gt;nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (3.11.0)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py&lt;3.0.0,&gt;=1.0.0-&gt;myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.1.2)
-Requirement already satisfied: ptyprocess&gt;=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect&gt;4.3-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.7.0)
-Requirement already satisfied: wcwidth in /home/codespace/.local/lib/python3.10/site-packages (from prompt-toolkit!=3.0.37,&lt;3.1.0,&gt;=3.0.30-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.2.8)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: greenlet!=0.4.17 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sqlalchemy&lt;3,&gt;=1.3.12-&gt;jupyter-cache&lt;0.7,&gt;=0.5-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (3.0.1)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: soupsieve&gt;1.2 in /home/codespace/.local/lib/python3.10/site-packages (from beautifulsoup4-&gt;pydata-sphinx-theme&gt;=0.13.3-&gt;sphinx-book-theme~=1.0.0-&gt;jupyter-book) (2.5)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: executing&gt;=1.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.0.0)
-Requirement already satisfied: asttokens&gt;=2.1.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.4.0)
-Requirement already satisfied: pure-eval in /home/codespace/.local/lib/python3.10/site-packages (from stack-data-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.2.2)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Note: you may need to restart the kernel to use updated packages.
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="o">%</span><span class="k">cd</span> /workspaces/LeetCode-Solutions
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="o">!</span>jupyter-book<span class="w"> </span>toc<span class="w"> </span>migrate<span class="w"> </span>/workspaces/LeetCode-Solutions/_toc.yml<span class="w"> </span>-o<span class="w"> </span>/workspaces/LeetCode-Solutions/_toc.yml
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Traceback (most recent call last):
-  File &quot;/home/codespace/.python/current/bin/jupyter-book&quot;, line 8, in &lt;module&gt;
-    sys.exit(main())
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1157, in __call__
-    return self.main(*args, **kwargs)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1078, in main
-    rv = self.invoke(ctx)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1688, in invoke
-    return _process_result(sub_ctx.command.invoke(sub_ctx))
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1688, in invoke
-    return _process_result(sub_ctx.command.invoke(sub_ctx))
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1434, in invoke
-    return ctx.invoke(self.callback, **ctx.params)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 783, in invoke
-    return __callback(*args, **kwargs)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx_external_toc/cli.py&quot;, line 139, in migrate_toc
-    toc = migrate_jupyter_book(Path(toc_file))
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx_external_toc/tools.py&quot;, line 327, in migrate_jupyter_book
-    raise MalformedError(&quot;no top-level &#39;file&#39; key found&quot;)
-sphinx_external_toc.parsing.MalformedError: no top-level &#39;file&#39; key found
-</pre></div>
-</div>
-</div>
-</div>
-<div class="cell docutils container">
-<div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="o">!</span>jupyter-book<span class="w"> </span>build<span class="w"> </span>.
-</pre></div>
-</div>
-</div>
-<div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold -Color-Bold-Green">Running Jupyter-Book v0.15.1</span>
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold -Color-Bold-Blue">Source Folder: </span>/workspaces/LeetCode-Solutions
-<span class=" -Color -Color-Bold -Color-Bold-Blue">Config Path: </span>/workspaces/LeetCode-Solutions/_config.yml
-<span class=" -Color -Color-Bold -Color-Bold-Blue">Output Path: </span>/workspaces/LeetCode-Solutions/_build/html
-<span class=" -Color -Color-Bold">Running Sphinx v5.0.2</span>
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[etoc] Changing master_doc to &#39;README&#39;
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">myst v0.18.1:</span> MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=[&#39;colon_fence&#39;, &#39;dollarmath&#39;, &#39;linkify&#39;, &#39;substitution&#39;, &#39;tasklist&#39;], disable_syntax=[], all_links_external=False, url_schemes=[&#39;mailto&#39;, &#39;http&#39;, &#39;https&#39;], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=(&#39;{&#39;, &#39;}&#39;), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes=&#39;tex2jax_process|mathjax_process|math|output_area&#39;)
-<span class=" -Color -Color-Bold">myst-nb v0.17.2:</span> NbParserConfig(custom_formats={}, metadata_key=&#39;mystnb&#39;, cell_metadata_key=&#39;mystnb&#39;, kernel_rgx_aliases={}, execution_mode=&#39;auto&#39;, execution_cache_path=&#39;&#39;, execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin=&#39;default&#39;, remove_code_source=False, remove_code_outputs=False, code_prompt_show=&#39;Show code cell {type}&#39;, code_prompt_hide=&#39;Hide code cell {type}&#39;, number_source_lines=False, output_stderr=&#39;show&#39;, render_text_lexer=&#39;myst-ansi&#39;, render_error_lexer=&#39;ipythontb&#39;, render_image_options={}, render_figure_options={}, render_markdown_format=&#39;commonmark&#39;, output_folder=&#39;build&#39;, append_css=True, metadata_to_fm=False)
-Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">building [mo]: </span>targets for 0 po files that are out of date
-<span class=" -Color -Color-Bold">building [html]: </span>targets for 77 source files that are out of date
-<span class=" -Color -Color-Bold">updating environment: </span>[new config] 77 added, 0 changed, 0 removed
-<span class=" -Color -Color-Bold">reading sources... </span>[  1%] <span class=" -Color -Color-Magenta">01. Array Hashing/1. Two Sum</span>                          
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.07 seconds [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[  2%] <span class=" -Color -Color-Magenta">01. Array Hashing/128. Longest Consecutive Sequence</span>   
-/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.17 seconds [mystnb]
-<span class=" -Color -Color-Bold">reading sources... </span>[  3%] <span class=" -Color -Color-Magenta">01. Array Hashing/217. Contains Duplicate</span>             
-/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.19 seconds [mystnb]
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-/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 1.26 seconds [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[  6%] <span class=" -Color -Color-Magenta">01. Array Hashing/242. Valid Anagram</span>                  
-/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 1.18 seconds [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[  7%] <span class=" -Color -Color-Magenta">01. Array Hashing/347. Top K Frequent Elements</span>        
-/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.13 seconds [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[  9%] <span class=" -Color -Color-Magenta">01. Array Hashing/49. Group Anagrams</span>                  
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-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 4.02 seconds [mystnb]
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-</pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.05 seconds [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 11%] <span class=" -Color -Color-Magenta">01. Array Hashing/README</span>                              
-<span class=" -Color -Color-Bold">reading sources... </span>[ 12%] <span class=" -Color -Color-Magenta">02. Two Pointers/11. Container With Most Water</span>        
-/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.26 seconds [mystnb]
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-/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 1.07 seconds [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 15%] <span class=" -Color -Color-Magenta">02. Two Pointers/15. 3Sum</span>                             
-/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.00 seconds [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 16%] <span class=" -Color -Color-Magenta">02. Two Pointers/README</span>                               
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-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.89 seconds [mystnb]
-<span class=" -Color -Color-Bold">reading sources... </span>[ 19%] <span class=" -Color -Color-Magenta">03. Sliding Window/3. Longest Substring Without Repeating Characters</span>
-</pre></div>
-</div>
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-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 1.16 seconds [mystnb]
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 20%] <span class=" -Color -Color-Magenta">03. Sliding Window/424. Longest Repeating Character Replacement</span>
-/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.05 seconds [mystnb]
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-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 0.87 seconds [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 23%] <span class=" -Color -Color-Magenta">03. Sliding Window/README</span>                             
-<span class=" -Color -Color-Bold">reading sources... </span>[ 24%] <span class=" -Color -Color-Magenta">04. Stack/20. Valid Parentheses</span>                       
-/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 4.02 seconds [mystnb]
-<span class=" -Color -Color-Bold">reading sources... </span>[ 25%] <span class=" -Color -Color-Magenta">04. Stack/README</span>                                      
-<span class=" -Color -Color-Bold">reading sources... </span>[ 27%] <span class=" -Color -Color-Magenta">05. Binary Search/153. Find Minimum in Rotated Sorted Array</span>
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 0.98 seconds [mystnb]
-<span class=" -Color -Color-Bold">reading sources... </span>[ 28%] <span class=" -Color -Color-Magenta">05. Binary Search/33. Search in Rotated Sorted Array</span>  
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.05 seconds [mystnb]
-<span class=" -Color -Color-Bold">reading sources... </span>[ 29%] <span class=" -Color -Color-Magenta">05. Binary Search/README</span>                              
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 31%] <span class=" -Color -Color-Magenta">06. Linked List/141. Linked List Cycle</span>                
-/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 0.96 seconds [mystnb]
-<span class=" -Color -Color-Bold">reading sources... </span>[ 32%] <span class=" -Color -Color-Magenta">06. Linked List/143. Reorder List</span>                     
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>^C
-
-Exception occurred:
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 512, in _async_cleanup_kernel
-    await ensure_async(self.km.cleanup_resources())
-AttributeError: &#39;NoneType&#39; object has no attribute &#39;cleanup_resources&#39;
-The full traceback has been saved in /tmp/sphinx-err-tl19fjln.log, if you want to report the issue to the developers.
-Please also report this if it was a user error, so that a better error message can be provided next time.
-A bug report can be filed in the tracker at &lt;https://github.com/sphinx-doc/sphinx/issues&gt;. Thanks!
-Traceback (most recent call last):
-  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py&quot;, line 89, in wrapper
-    self._ready.set_result(None)
-asyncio.exceptions.InvalidStateError: invalid state
-
-During handling of the above exception, another exception occurred:
-
-Traceback (most recent call last):
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 505, in _async_cleanup_kernel
-    await ensure_async(self.km.shutdown_kernel(now=now))
-  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py&quot;, line 189, in ensure_async
-    result = await obj
-  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py&quot;, line 92, in wrapper
-    self._ready.set_exception(e)
-asyncio.exceptions.InvalidStateError: invalid state
-
-During handling of the above exception, another exception occurred:
-
-Traceback (most recent call last):
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/sphinx.py&quot;, line 171, in build_sphinx
-    app.build(force_all, filenames)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/application.py&quot;, line 329, in build
-    self.builder.build_update()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 288, in build_update
-    self.build(to_build,
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 302, in build
-    updated_docnames = set(self.read())
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 409, in read
-    self._read_serial(docnames)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 430, in _read_serial
-    self.read_doc(docname)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 483, in read_doc
-    publisher.publish()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/core.py&quot;, line 217, in publish
-    self.document = self.reader.read(self.source, self.parser,
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/io.py&quot;, line 103, in read
-    self.parse()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/readers/__init__.py&quot;, line 78, in parse
-    self.parser.parse(self.input, document)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/sphinx_.py&quot;, line 150, in parse
-    with create_client(
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/base.py&quot;, line 83, in __enter__
-    self.start_client()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/direct.py&quot;, line 40, in start_client
-    result = single_nb_execution(
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_cache/executors/utils.py&quot;, line 58, in single_nb_execution
-    executenb(
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 1305, in execute
-    return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute()
-  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py&quot;, line 173, in wrapped
-    return loop.run_until_complete(inner)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/asyncio/base_events.py&quot;, line 649, in run_until_complete
-    return future.result()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 689, in async_execute
-    async with self.async_setup_kernel(**kwargs):
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/contextlib.py&quot;, line 206, in __aexit__
-    await anext(self.gen)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 656, in async_setup_kernel
-    await self._async_cleanup_kernel()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 512, in _async_cleanup_kernel
-    await ensure_async(self.km.cleanup_resources())
-AttributeError: &#39;NoneType&#39; object has no attribute &#39;cleanup_resources&#39;
-
-The above exception was the direct cause of the following exception:
-
-Traceback (most recent call last):
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-<span class=" -Color -Color-Bold">reading sources... </span>[100%] <span class=" -Color -Color-Magenta">intro</span>                                                 
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-[etoc] missing index.html written as redirect to &#39;intro.html&#39;
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-<h1>Contributing Guidelines<a class="headerlink" href="#contributing-guidelines" title="Permalink to this heading">#</a></h1>
-<p>Thank you for considering contributing to the Blind 75 LeetCode repository! We welcome contributions from the community to help improve this resource and make it even more valuable.</p>
-<p>Please take a moment to review this document to understand how you can contribute.</p>
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-<li><p><span class="xref myst">Ways to Contribute</span></p></li>
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-<h2>Ways to Contribute<a class="headerlink" href="#ways-to-contribute" title="Permalink to this heading">#</a></h2>
-<p>There are several ways you can contribute to this repository:</p>
-<ol class="arabic simple">
-<li><p><strong>Suggest Improvements</strong>: If you have ideas for making the Blind 75 LeetCode repository better, please open an issue to discuss your suggestions.</p></li>
-<li><p><strong>Report Issues</strong>: If you encounter any issues with the existing content or code, please open an issue and provide as much detail as possible.</p></li>
-<li><p><strong>Contribute Solutions</strong>: You can contribute solutions to the problems listed in the repository. Please follow the guidelines mentioned below when submitting your solutions.</p></li>
-<li><p><strong>Enhance Documentation</strong>: Improve documentation, including explanations and time/space complexity analysis, for existing solutions.</p></li>
-<li><p><strong>Fix Bugs</strong>: If you find and fix any bugs in the existing code or documentation, you are welcome to submit a pull request.</p></li>
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-<h2>Getting Started<a class="headerlink" href="#getting-started" title="Permalink to this heading">#</a></h2>
-<p>Before you start contributing, please ensure you have the following prerequisites:</p>
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-<li><p>A GitHub account.</p></li>
-<li><p>Familiarity with Git and GitHub.</p></li>
-<li><p>Code editor or IDE for making changes.</p></li>
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-<h2>Contributing Process<a class="headerlink" href="#contributing-process" title="Permalink to this heading">#</a></h2>
-<p>Here’s the general process for contributing:</p>
-<ol class="arabic">
-<li><p><strong>Fork the Repository</strong>: Click the “Fork” button at the top of this repository to create your copy.</p></li>
-<li><p><strong>Clone Your Fork</strong>: Clone your fork to your local machine.</p>
-<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">git</span> <span class="n">clone</span> <span class="n">https</span><span class="p">:</span><span class="o">//</span><span class="n">github</span><span class="o">.</span><span class="n">com</span><span class="o">/</span><span class="n">your</span><span class="o">-</span><span class="n">username</span><span class="o">/</span><span class="n">Blind</span><span class="o">-</span><span class="mi">75</span><span class="o">-</span><span class="n">LeetCode</span><span class="o">.</span><span class="n">git</span>
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-<li><p><strong>Create a Branch</strong>: Create a new branch for your contribution.</p>
-<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">git</span> <span class="n">checkout</span> <span class="o">-</span><span class="n">b</span> <span class="n">feature</span><span class="o">/</span><span class="n">your</span><span class="o">-</span><span class="n">feature</span><span class="o">-</span><span class="n">name</span>
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-<li><p><strong>Make Changes</strong>: Implement your changes, following the guidelines provided.</p></li>
-<li><p><strong>Test Your Changes</strong>: Ensure that your changes do not introduce new issues and are consistent with the repository’s style and structure.</p></li>
-<li><p><strong>Commit Changes</strong>: Commit your changes with a meaningful commit message.</p>
-<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">git</span> <span class="n">commit</span> <span class="o">-</span><span class="n">m</span> <span class="s2">&quot;Add solution to problem X&quot;</span>
-</pre></div>
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-<li><p><strong>Push Changes</strong>: Push your changes to your fork on GitHub.</p>
-<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="n">git</span> <span class="n">push</span> <span class="n">origin</span> <span class="n">feature</span><span class="o">/</span><span class="n">your</span><span class="o">-</span><span class="n">feature</span><span class="o">-</span><span class="n">name</span>
-</pre></div>
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-<li><p><strong>Create a Pull Request</strong>: Go to the original repository and create a pull request. Please provide a detailed description of your changes.</p></li>
-<li><p><strong>Review and Collaboration</strong>: Your pull request will be reviewed, and you may need to make further changes. Be responsive to comments and feedback.</p></li>
-<li><p><strong>Merge</strong>: Once your pull request is approved, it will be merged into the main branch.</p></li>
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-<h2>Code of Conduct<a class="headerlink" href="#code-of-conduct" title="Permalink to this heading">#</a></h2>
-<p>Please note that we have a <a class="reference internal" href="CODE_OF_CONDUCT.html"><span class="doc std std-doc">Code of Conduct</span></a> that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.</p>
-<p>We appreciate your contributions to the Blind 75 LeetCode repository and look forward to working together to make it an even better resource for the community!</p>
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-    <h1>Blind 75 LeetCode Solutions</h1>
-    <!-- Table of contents -->
-    <div id="print-main-content">
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-            <div>
-                <h2> Contents </h2>
-            </div>
-            <nav aria-label="Page">
-                <ul class="visible nav section-nav flex-column">
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#problem-list">Problem List</a><ul class="nav section-nav flex-column">
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#array-hashing">Array &amp; Hashing</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#two-pointers">Two Pointers</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#sliding-window">Sliding Window</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#stack">Stack</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#binary-search">Binary Search</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#linked-list">Linked List</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#trees">Trees</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#tries">Tries</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#heap-priority-queue">Heap / Priority Queue</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#backtracking">Backtracking</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#graphs">Graphs</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#advanced-graphs">Advanced Graphs</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#d-dynamic-programming">1-D Dynamic Programming</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#id1">2-D Dynamic Programming</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#greedy">Greedy</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#intervals">Intervals</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#math-geometry">Math &amp; Geometry</a></li>
-<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#bit-manipulation">Bit Manipulation</a></li>
-</ul>
-</li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#how-to-use">How to Use</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#contribute">Contribute</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#resources">Resources</a></li>
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-  <section class="tex2jax_ignore mathjax_ignore" id="blind-75-leetcode-solutions">
-<h1>Blind 75 LeetCode Solutions<a class="headerlink" href="#blind-75-leetcode-solutions" title="Permalink to this heading">#</a></h1>
-<p><img alt="Blind 75 LeetCode" src="https://img.shields.io/badge/Blind_75_LeetCode-Solutions-blue?labelColor=red" /></p>
-<p>This repository contains a curated list of 75 LeetCode problems that are commonly known as the “Blind 75.” These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.</p>
-<section id="problem-list">
-<h2>Problem List<a class="headerlink" href="#problem-list" title="Permalink to this heading">#</a></h2>
-<section id="array-hashing">
-<h3>Array &amp; Hashing<a class="headerlink" href="#array-hashing" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Array_&amp;_Hashing-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/contains-duplicate/">217. Contains Duplicate</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/valid-anagram/">242. Valid Anagram</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/two-sum/">1. Two Sum</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/group-anagrams/">49. Group Anagrams</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/top-k-frequent-elements/">347. Top K Frequent Elements</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/product-of-array-except-self/">238. Product of Array Except Self</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/encode-and-decode-strings/">659. Encode and Decode Strings</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/longest-consecutive-sequence/">128. Longest Consecutive Sequence</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="two-pointers">
-<h3>Two Pointers<a class="headerlink" href="#two-pointers" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Two_Pointers-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/valid-palindrome/">125. Valid Palindrome</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/3sum/">15. 3Sum</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/container-with-most-water/">11. Container With Most Water</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="sliding-window">
-<h3>Sliding Window<a class="headerlink" href="#sliding-window" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Sliding_Window-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/best-time-to-buy-and-sell-stock/">121. Best Time to Buy And Sell Stock</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/longest-substring-without-repeating-characters/">3. Longest Substring Without Repeating Characters</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/longest-repeating-character-replacement/">424. Longest Repeating Character Replacement</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/minimum-window-substring/">76. Minimum Window Substring</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="stack">
-<h3>Stack<a class="headerlink" href="#stack" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Stack-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/valid-parentheses/">20. Valid Parentheses</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="binary-search">
-<h3>Binary Search<a class="headerlink" href="#binary-search" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Binary_Search-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/">153. Find Minimum In Rotated Sorted Array</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/search-in-rotated-sorted-array/">33. Search In Rotated Sorted Array</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="linked-list">
-<h3>Linked List<a class="headerlink" href="#linked-list" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Linked_List-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/reverse-linked-list/">206. Reverse Linked List</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/merge-two-sorted-lists/">21. Merge Two Sorted Lists</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/reorder-list/">143. Reorder List</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/remove-nth-node-from-end-of-list/">19. Remove Nth Node From End of List</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/linked-list-cycle/">141. Linked List Cycle</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/merge-k-sorted-lists/">23. Merge K Sorted Lists</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="trees">
-<h3>Trees<a class="headerlink" href="#trees" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Trees-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/invert-binary-tree/">226. Invert Binary Tree</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/maximum-depth-of-binary-tree/">104. Maximum Depth of Binary Tree</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/same-tree/">100. Same Tree</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/subtree-of-another-tree/">572. Subtree of Another Tree</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/">235. Lowest Common Ancestor of a Binary Search Tree</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/binary-tree-level-order-traversal/">102. Binary Tree Level Order Traversal</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/validate-binary-search-tree/">98. Validate Binary Search Tree</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/kth-smallest-element-in-a-bst/">230. Kth Smallest Element In a BST</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/">105. Construct Binary Tree From Preorder And Inorder Traversal</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/binary-tree-maximum-path-sum/">124. Binary Tree Maximum Path Sum</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/serialize-and-deserialize-binary-tree/">297. Serialize And Deserialize Binary Tree</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="tries">
-<h3>Tries<a class="headerlink" href="#tries" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Tries-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/implement-trie-prefix-tree/">208. Implement Trie Prefix Tree</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/design-add-and-search-words-data-structure/">211. Design Add And Search Words Data Structure</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/word-search-ii/">212. Word Search II</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="heap-priority-queue">
-<h3>Heap / Priority Queue<a class="headerlink" href="#heap-priority-queue" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Heap_/_Priority_Queue-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/find-median-from-data-stream/">295. Find Median From Data Stream</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="backtracking">
-<h3>Backtracking<a class="headerlink" href="#backtracking" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Backtracking-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/combination-sum/">39. Combination Sum</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/word-search/">79. Word Search</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="graphs">
-<h3>Graphs<a class="headerlink" href="#graphs" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Graphs-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/number-of-islands/">200. Number of Islands</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/clone-graph/">133. Clone Graph</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/pacific-atlantic-water-flow/">417. Pacific Atlantic Water Flow</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/course-schedule/">207. Course Schedule</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/">323. Number of Connected Components In An Undirected Graph</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/graph-valid-tree/">261. Graph Valid Tree</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="advanced-graphs">
-<h3>Advanced Graphs<a class="headerlink" href="#advanced-graphs" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Advanced_Graphs-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/alien-dictionary/">269. Alien Dictionary</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="d-dynamic-programming">
-<h3>1-D Dynamic Programming<a class="headerlink" href="#d-dynamic-programming" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/1_D_Dynamic_Programming-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/climbing-stairs/">70. Climbing Stairs</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/house-robber/">198. House Robber</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/house-robber-ii/">213. House Robber II</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/longest-palindromic-substring/">5. Longest Palindromic Substring</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/palindromic-substrings/">647. Palindromic Substrings</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/decode-ways/">91. Decode Ways</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/coin-change/">322. Coin Change</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/maximum-product-subarray/">152. Maximum Product Subarray</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/word-break/">139. Word Break</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/longest-increasing-subsequence/">300. Longest Increasing Subsequence</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="id1">
-<h3>2-D Dynamic Programming<a class="headerlink" href="#id1" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/2_D_Dynamic_Programming-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/unique-paths/">62. Unique Paths</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/longest-common-subsequence/">1143. Longest Common Subsequence</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="greedy">
-<h3>Greedy<a class="headerlink" href="#greedy" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Greedy-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/maximum-subarray/">53. Maximum Subarray</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/jump-game/">55. Jump Game</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="intervals">
-<h3>Intervals<a class="headerlink" href="#intervals" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Intervals-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/insert-interval/">57. Insert Interval</a></p></td>
-<td><p>Hard</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/merge-intervals/">56. Merge Intervals</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/non-overlapping-intervals/">435. Non Overlapping Intervals</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/meeting-rooms/">252. Meeting Rooms</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/meeting-rooms-ii/">253. Meeting Rooms II</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="math-geometry">
-<h3>Math &amp; Geometry<a class="headerlink" href="#math-geometry" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Math_&amp;_Geometry-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/rotate-image/">48. Rotate Image</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/spiral-matrix/">54. Spiral Matrix</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/set-matrix-zeroes/">73. Set Matrix Zeroes</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-<section id="bit-manipulation">
-<h3>Bit Manipulation<a class="headerlink" href="#bit-manipulation" title="Permalink to this heading">#</a></h3>
-<p><img alt="Array &amp; Hashing" src="https://img.shields.io/badge/Bit_Manipulation-Solutions-blue?labelColor=red" /></p>
-<table class="table">
-<thead>
-<tr class="row-odd"><th class="head"><p>Problem Name</p></th>
-<th class="head"><p>Difficulty</p></th>
-</tr>
-</thead>
-<tbody>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/number-of-1-bits/">191. Number of 1 Bits</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/counting-bits/">338. Counting Bits</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/reverse-bits/">190. Reverse Bits</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-odd"><td><p><a class="reference external" href="https://leetcode.com/problems/missing-number/">268. Missing Number</a></p></td>
-<td><p>Easy</p></td>
-</tr>
-<tr class="row-even"><td><p><a class="reference external" href="https://leetcode.com/problems/sum-of-two-integers/">371. Sum of Two Integers</a></p></td>
-<td><p>Medium</p></td>
-</tr>
-</tbody>
-</table>
-</section>
-</section>
-<section id="how-to-use">
-<h2>How to Use<a class="headerlink" href="#how-to-use" title="Permalink to this heading">#</a></h2>
-<ol class="arabic simple">
-<li><p>Clone or download this repository to your local machine.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span>git<span class="w"> </span>clone<span class="w"> </span>https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-</pre></div>
-</div>
-<ol class="arabic simple" start="2">
-<li><p>Navigate to the directory of the specific problem you want to solve.</p></li>
-</ol>
-<div class="highlight-bash notranslate"><div class="highlight"><pre><span></span><span class="nb">cd</span><span class="w"> </span>Blind-75-LeetCode/Problem-Directory
-</pre></div>
-</div>
-<ol class="arabic simple" start="3">
-<li><p>Open the problem’s directory to find its description, code template, and solution(s).</p></li>
-<li><p>Solve the problem using your preferred programming language.</p></li>
-<li><p>Check the solution(s) provided in the directory for reference and to compare your solution.</p></li>
-</ol>
-</section>
-<section id="contribute">
-<h2>Contribute<a class="headerlink" href="#contribute" title="Permalink to this heading">#</a></h2>
-<p>If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.</p>
-</section>
-<section id="resources">
-<h2>Resources<a class="headerlink" href="#resources" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p><a class="reference external" href="https://leetcode.com/">LeetCode</a>: The official LeetCode website.</p></li>
-<li><p><a class="reference external" href="https://leetcode.com/discuss/">LeetCode Discuss</a>: LeetCode’s discussion forum for each problem. You can find helpful hints and discussions here.</p></li>
-</ul>
-</section>
-<section id="disclaimer">
-<h2>Disclaimer<a class="headerlink" href="#disclaimer" title="Permalink to this heading">#</a></h2>
-<p>This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode’s terms of use and guidelines when using this repository.</p>
-</section>
-<section id="code-of-conduct">
-<h2>Code of Conduct<a class="headerlink" href="#code-of-conduct" title="Permalink to this heading">#</a></h2>
-<p>We have a <a class="reference internal" href="CODE_OF_CONDUCT.html"><span class="doc std std-doc">Code of Conduct</span></a> that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.</p>
-</section>
-<section id="security-policy">
-<h2>Security Policy<a class="headerlink" href="#security-policy" title="Permalink to this heading">#</a></h2>
-<p>We take security seriously. Please review our <a class="reference internal" href="SECURITY.html"><span class="doc std std-doc">Security Policy</span></a> for information on reporting security vulnerabilities.</p>
-<p>Happy coding!</p>
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-  <section class="tex2jax_ignore mathjax_ignore" id="security-policy">
-<h1>Security Policy<a class="headerlink" href="#security-policy" title="Permalink to this heading">#</a></h1>
-<section id="reporting-a-vulnerability">
-<h2>Reporting a Vulnerability<a class="headerlink" href="#reporting-a-vulnerability" title="Permalink to this heading">#</a></h2>
-<p>We take the security of the Blind 75 LeetCode repository seriously. If you believe you have found a security vulnerability, please help us by following these steps:</p>
-<ol class="arabic">
-<li><p><strong>Do Not Disclose Publicly</strong>: Please do not disclose the issue publicly until it has been addressed by our team.</p></li>
-<li><p><strong>Report Privately</strong>: Create a GitHub issue with the title “Security Vulnerability Report” and provide a detailed description of the vulnerability. You can use a sample report template as follows:</p>
-<div class="highlight-default notranslate"><div class="highlight"><pre><span></span><span class="o">**</span><span class="n">Description</span><span class="o">**</span>
-<span class="n">A</span> <span class="n">clear</span> <span class="ow">and</span> <span class="n">concise</span> <span class="n">description</span> <span class="n">of</span> <span class="n">the</span> <span class="n">security</span> <span class="n">vulnerability</span><span class="o">.</span>
-
-<span class="o">**</span><span class="n">Steps</span> <span class="n">to</span> <span class="n">Reproduce</span><span class="o">**</span>
-<span class="n">Provide</span> <span class="n">detailed</span> <span class="n">steps</span> <span class="n">on</span> <span class="n">how</span> <span class="n">to</span> <span class="n">reproduce</span> <span class="n">the</span> <span class="n">vulnerability</span><span class="o">.</span>
-
-<span class="o">**</span><span class="n">Expected</span> <span class="n">Behavior</span><span class="o">**</span>
-<span class="n">Describe</span> <span class="n">what</span> <span class="n">you</span> <span class="n">expected</span> <span class="n">to</span> <span class="n">happen</span><span class="o">.</span>
-
-<span class="o">**</span><span class="n">Actual</span> <span class="n">Behavior</span><span class="o">**</span>
-<span class="n">Describe</span> <span class="n">what</span> <span class="n">actually</span> <span class="n">happened</span><span class="o">.</span>
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-<span class="n">Any</span> <span class="n">additional</span> <span class="n">information</span> <span class="n">that</span> <span class="n">can</span> <span class="n">help</span> <span class="n">us</span> <span class="n">understand</span> <span class="ow">and</span> <span class="n">reproduce</span> <span class="n">the</span> <span class="n">issue</span><span class="p">,</span> <span class="n">such</span> <span class="k">as</span> <span class="n">environment</span> <span class="n">details</span><span class="p">,</span> <span class="n">configuration</span><span class="p">,</span> <span class="ow">or</span> <span class="n">screenshots</span><span class="o">.</span>
-</pre></div>
-</div>
-</li>
-<li><p><strong>Secure Communication</strong>: If the vulnerability involves sensitive data, please encrypt your communication with our PGP key. Contact us for our PGP key information.</p></li>
-<li><p><strong>Response</strong>: We will acknowledge your report within 5 business days and provide an estimated timeline for the resolution.</p></li>
-<li><p><strong>Fix and Disclosure</strong>: We will work to fix the issue promptly and privately. Once resolved, we will provide a security advisory and give credit to the reporter if desired.</p></li>
-<li><p><strong>Responsible Disclosure</strong>: We encourage responsible disclosure. Please do not take advantage of the vulnerability or attempt to exploit it for malicious purposes.</p></li>
-</ol>
-</section>
-<section id="security-updates">
-<h2>Security Updates<a class="headerlink" href="#security-updates" title="Permalink to this heading">#</a></h2>
-<p>We are committed to promptly addressing and resolving any security issues. We recommend that you stay up-to-date with the latest releases of this repository, as they may include security updates and fixes.</p>
-</section>
-<section id="contact">
-<h2>Contact<a class="headerlink" href="#contact" title="Permalink to this heading">#</a></h2>
-<p>If you have any questions or need further assistance regarding security, you can contact us at <a class="reference external" href="mailto:mohsentabibian&#37;&#52;&#48;gmail&#46;com">mohsentabibian<span>&#64;</span>gmail<span>&#46;</span>com</a>.</p>
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diff --git a/docs/_build/html/_sources/01. Array Hashing/1. Two Sum.ipynb b/docs/_build/html/_sources/01. Array Hashing/1. Two Sum.ipynb
deleted file mode 100644
index b3b517f..0000000
--- a/docs/_build/html/_sources/01. Array Hashing/1. Two Sum.ipynb	
+++ /dev/null
@@ -1,211 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 100: Same Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-        "\n",
-        "---\n",
-        "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-        "\n",
-        "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-        "\n",
-        "**Constraints**\n",
-        "\n",
-        "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-        "2. $-10^4$ <= `Node.val` <= $10^4$"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def isSameTree(p, q):\n",
-        "    # Base case: If both p and q are None, the trees are the same.\n",
-        "    if not p and not q:\n",
-        "        return True\n",
-        "    \n",
-        "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-        "    if not p or not q:\n",
-        "        return False\n",
-        "    \n",
-        "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-        "    if p.val != q.val:\n",
-        "        return False\n",
-        "    \n",
-        "    # Recursively check the left and right subtrees of p and q.\n",
-        "    # If both subtrees are the same, the entire trees are the same.\n",
-        "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-        "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-        "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-        "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-        "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-        "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "\n",
-        "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-        "\n",
-        "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "print(isSameTree(p1, q1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "\n",
-        "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-        "\n",
-        "p2 = TreeNode(1, TreeNode(2), None)\n",
-        "q2 = TreeNode(1, None, TreeNode(2))\n",
-        "print(isSameTree(p2, q2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "\n",
-        "#Input: p = [1,2,1], q = [1,1,2]\n",
-        "\n",
-        "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-        "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-        "print(isSameTree(p3, q3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "**Time Complexity**\n",
-        "\n",
-        "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-        "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-        "\n",
-        "**Space Complexity**\n",
-        "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-        "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-        "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-        "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "+ Time Complexity: $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-        "+ Space Complexity: $O(n)$ due to the recursive call stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb b/docs/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb
deleted file mode 100644
index a77a03d..0000000
--- a/docs/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 128. Longest Consecutive Sequence\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Consecutive Sequence problem on LeetCode, click here!](https://leetcode.com/problems/longest-consecutive-sequence/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an unsorted array of integers `nums`, return *the length of the longest consecutive elements sequence.*\n",
-        "\n",
-        "You must write an algorithm that runs in `O(n)` time.\n",
-        "\n",
-        "*Constraints:**\n",
-        "- 0 <= `nums.length` <= $10^5$\n",
-        "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 58,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def longestConsecutive(nums):\n",
-        "    if not nums:\n",
-        "        return 0\n",
-        "\n",
-        "    num_set = set(nums)\n",
-        "    max_length = 0\n",
-        "\n",
-        "    for num in num_set:\n",
-        "        if num - 1 not in num_set:\n",
-        "            current_num = num\n",
-        "            current_length = 1\n",
-        "\n",
-        "            while current_num + 1 in num_set:\n",
-        "                current_num += 1\n",
-        "                current_length += 1\n",
-        "\n",
-        "            max_length = max(max_length, current_length)\n",
-        "\n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining the `longestConsecutive` function that takes the `nums` array as input.\n",
-        "\n",
-        "2. We check if the `nums` array is empty. If it's empty, there are no consecutive elements, so we return 0.\n",
-        "\n",
-        "3. We create a Python set called `num_set` and insert all elements from the `nums` array into it. Using a set allows us to efficiently check for the existence of elements in O(1) time.\n",
-        "\n",
-        "4. We initialize a variable `max_length` to 0. This variable will keep track of the maximum length of consecutive elements sequence found.\n",
-        "\n",
-        "5. We iterate through the elements of the `num_set`. For each element `num`, we check if `num - 1` exists in the `num_set`. If it doesn't exist, it means `num` is the starting element of a potential consecutive sequence.\n",
-        "\n",
-        "6. Inside the loop, we initialize two variables: `current_num` to the current element `num` and `current_length` to 1. We start with a length of 1 because `num` itself is part of the sequence.\n",
-        "\n",
-        "7. We then enter a while loop that continues as long as `current_num + 1` exists in the `num_set`. This means we are incrementing the consecutive sequence.\n",
-        "\n",
-        "8. Inside the while loop, we increment `current_num` by 1 and also increment `current_length` by 1 to account for the next consecutive element.\n",
-        "\n",
-        "9. We compare `current_length` with the `max_length` and update `max_length` if the current sequence is longer.\n",
-        "\n",
-        "10. After the while loop, we move to the next element in the outer loop and repeat the process.\n",
-        "\n",
-        "11. Finally, we return the `max_length` as the result, which represents the length of the longest consecutive elements sequence in the `nums` array.\n",
-        "\n",
-        "The key idea here is to use a set to efficiently check for the existence of elements and to iterate through the elements, considering each element as the potential start of a consecutive sequence. By doing this, we can find the longest consecutive sequence in O(n) time complexity, where n is the number of elements in the array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 59,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "nums1 = [100, 4, 200, 1, 3, 2]\n",
-        "print(longestConsecutive(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 60,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "9\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "nums2 = [0, 3, 7, 2, 5, 8, 4, 6, 0, 1]\n",
-        "print(longestConsecutive(nums2)) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The code is designed to run in O(n) time complexity, where n is the number of elements in the `nums` array. Here's the breakdown:\n",
-        "\n",
-        "1. Constructing the `num_set` by inserting all elements from `nums` into it takes O(n) time because we perform an insertion operation for each element in `nums`.\n",
-        "\n",
-        "2. The main loop iterates through the elements in `num_set`. In the worst case, each element is visited only once, so the loop itself takes O(n) time.\n",
-        "\n",
-        "3. Within the loop, we have a while loop that may also take O(n) time in the worst case. However, this while loop is nested inside the main loop, so its overall time complexity remains O(n).\n",
-        "\n",
-        "Therefore, the overall time complexity of the code is O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of the code is determined by the space used by the `num_set` and a few additional variables. Here's the breakdown:\n",
-        "\n",
-        "1. `num_set` is a set that stores the unique elements from `nums`. In the worst case, it can store all n elements from `nums`, so the space complexity is O(n).\n",
-        "\n",
-        "2. The additional variables used, such as `max_length`, `current_num`, and `current_length`, have constant space requirements and do not depend on the size of the input array.\n",
-        "\n",
-        "Therefore, the overall space complexity of the code is O(n) due to the space used by the `num_set`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. Write an algorithm that not only returns the length of the longest consecutive elements sequence but also returns the actual consecutive sequence itself.\n",
-        "\n",
-        "2. Extend the problem to allow elements to be considered consecutive if they are within a certain absolute difference (e.g., less than or equal to k) instead of exactly 1. Write an algorithm that finds the longest such sequence and runs in O(n) time."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb b/docs/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb
deleted file mode 100644
index 57bb55d..0000000
--- a/docs/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb	
+++ /dev/null
@@ -1,218 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 217. Contains Duplicate\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Contains Duplicate problem on LeetCode, click here!](https://leetcode.com/problems/contains-duplicate/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `nums.length` <= $10^5$\n",
-        "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 19,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def containsDuplicate(nums):\n",
-        "    # Create an empty set to store unique elements\n",
-        "    unique_elements = set()\n",
-        "    \n",
-        "    # Iterate through the array\n",
-        "    for num in nums:\n",
-        "        # If the element is already in the set, return True\n",
-        "        if num in unique_elements:\n",
-        "            return True\n",
-        "        # Otherwise, add it to the set\n",
-        "        else:\n",
-        "            unique_elements.add(num)\n",
-        "    \n",
-        "    # If the loop completes without finding duplicates, return False\n",
-        "    return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `containsDuplicate` function takes a single argument, `nums`, which is the input integer array.\n",
-        "\n",
-        "2. Inside the function, an empty set called `unique_elements` is created. This set will be used to keep track of unique elements in the input array.\n",
-        "\n",
-        "3. The function then iterates through the input array `nums` using a `for` loop.\n",
-        "\n",
-        "4. For each element `num` in the array, it checks whether `num` is already in the `unique_elements` set using the `if num in unique_elements:` condition.\n",
-        "\n",
-        "5. If `num` is already in the set, it means there is a duplicate element in the array, and the function immediately returns `True`.\n",
-        "\n",
-        "6. If `num` is not in the set, it is added to the `unique_elements` set using `unique_elements.add(num)`.\n",
-        "\n",
-        "7. The loop continues to the next element, and the process repeats.\n",
-        "\n",
-        "8. If the loop completes without finding any duplicates, it means that all elements in the array are distinct, and the function returns `False`.\n",
-        "\n",
-        "The code efficiently utilizes a set data structure to keep track of unique elements while iterating through the array, allowing it to quickly detect duplicate elements. This code meets the problem's requirements and constraints, as explained earlier."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "nums1 = [1, 2, 3, 1]\n",
-        "print(containsDuplicate(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "nums2 = [1, 2, 3, 4]\n",
-        "print(containsDuplicate(nums2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 22,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "nums3 = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]\n",
-        "print(containsDuplicate(nums3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for the \"Contains Duplicate\" problem:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The primary operation that affects the time complexity is the `for` loop that iterates through the input array `nums`. In the worst case, the loop will iterate through all `n` elements of the array, where `n` is the length of the input array.\n",
-        "\n",
-        "- Iterating through the array: O(n)\n",
-        "\n",
-        "Inside the loop, we perform two operations:\n",
-        "\n",
-        "1. Checking whether an element exists in the `unique_elements` set (`if num in unique_elements`). This operation has an average time complexity of O(1) for a set.\n",
-        "2. Adding an element to the `unique_elements` set (`unique_elements.add(num)`), which also has an average time complexity of O(1) for a set.\n",
-        "\n",
-        "Since these operations are performed for each element in the array, the overall time complexity remains O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity is determined by the additional data structures used in the code, which are the `unique_elements` set.\n",
-        "\n",
-        "- `unique_elements` set: This set stores unique elements from the input array. In the worst case, if all elements in the input array are distinct, the set will store all `n` elements.\n",
-        "\n",
-        "Therefore, the space complexity is O(n) because, in the worst case, the set's size will grow linearly with the input array's size.\n",
-        "\n",
-        "**In summary**, the time complexity of the code is O(n), where `n` is the length of the input array, and the space complexity is also O(n) in the worst case. This code is efficient and meets the constraints of the problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K Duplicates:** Modify the problem to return `true` if there are exactly K duplicate elements in the array. Write a function that takes an additional integer parameter K and returns `true` if there are exactly K duplicates.\n",
-        "2. **Frequency Count:** Write a function that returns a list of all the elements that appear more than once in the array, along with their frequencies."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb b/docs/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb
deleted file mode 100644
index 1ff9a9c..0000000
--- a/docs/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb	
+++ /dev/null
@@ -1,197 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 238. Product of Array Except Self\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Product of Array Except Self problem on LeetCode, click here!](https://leetcode.com/problems/product-of-array-except-self/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array `nums`, return *an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.*\n",
-        "\n",
-        "The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-        "\n",
-        "You must write an algorithm that runs in O(n) time and without using the division operation.\n",
-        "\n",
-        "Constraints:\n",
-        "- 2 <= `nums.length` <= $10^5$\n",
-        "- `-30 <= nums[i] <= 30`\n",
-        "- The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-        "\n",
-        "**Follow-up:** Can you solve the problem in O(1) extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 50,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def productExceptSelf(nums):\n",
-        "    n = len(nums)\n",
-        "    \n",
-        "    # Initialize two arrays to store products to the left and right of each element\n",
-        "    left_products = [1] * n\n",
-        "    right_products = [1] * n\n",
-        "    \n",
-        "    # Calculate products to the left of each element\n",
-        "    left_product = 1\n",
-        "    for i in range(n):\n",
-        "        left_products[i] = left_product\n",
-        "        left_product *= nums[i]\n",
-        "    \n",
-        "    # Calculate products to the right of each element\n",
-        "    right_product = 1\n",
-        "    for i in range(n - 1, -1, -1):\n",
-        "        right_products[i] = right_product\n",
-        "        right_product *= nums[i]\n",
-        "    \n",
-        "    # Calculate the final answer using left and right products\n",
-        "    answer = [left_products[i] * right_products[i] for i in range(n)]\n",
-        "    \n",
-        "    return answer"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `n` is initialized as the length of the `nums` array.\n",
-        "2. Two arrays, `left_products` and `right_products`, are created, each with a length of `n` and filled with ones initially.\n",
-        "3. `left_product` is initialized to 1.\n",
-        "4. A loop iterates through the `nums` array, from left to right. For each element at index `i`, it stores the product of all elements to its left (including itself) in the `left_products` array and updates `left_product` accordingly.\n",
-        "5. `right_product` is initialized to 1.\n",
-        "6. Another loop iterates through the `nums` array in reverse order, from right to left. For each element at index `i`, it stores the product of all elements to its right (including itself) in the `right_products` array and updates `right_product`.\n",
-        "7. The final `answer` list is constructed by multiplying corresponding elements from the `left_products` and `right_products` arrays for each index `i`.\n",
-        "8. The function returns the `answer` list, which contains the desired results.\n",
-        "\n",
-        "Overall, this algorithm efficiently computes the product of all elements except the current element, as required. It runs in O(n) time complexity and uses O(n) extra space, which is within the problem's constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 52,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[24, 12, 8, 6]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "nums = [1,2,3,4]\n",
-        "result1 = productExceptSelf(nums)\n",
-        "print(result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 53,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[0, 0, 9, 0, 0]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums = [-1,1,0,-3,3]\n",
-        "result2 = productExceptSelf(nums)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The first loop iterates through the `nums` array to calculate the products to the left of each element. This loop runs in O(n) time, where n is the length of the `nums` array.\n",
-        "2. The second loop also iterates through the `nums` array in reverse order to calculate the products to the right of each element. This loop also runs in O(n) time.\n",
-        "3. The final loop constructs the `answer` list by multiplying elements from the `left_products` and `right_products` arrays. This loop runs in O(n) time.\n",
-        "\n",
-        "Since all the loops are independent and sequential, the overall time complexity is O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. Two additional arrays, `left_products` and `right_products`, are created with a length of `n`, where `n` is the length of the `nums` array. Therefore, these arrays consume O(n) extra space.\n",
-        "2. The `left_product` and `right_product` variables are used to keep track of the product to the left and right of each element, respectively. These variables occupy O(1) extra space.\n",
-        "3. The `answer` list, which contains the final results, also consumes O(n) space, but it's not counted toward the extra space complexity analysis as per the problem's constraints.\n",
-        "\n",
-        "Overall, the space complexity is O(n) for this solution."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Handle Zeros in the Input**: Modify the solution to handle cases where the input array contains zeros. For example, if `nums = [0, 1, 2, 3, 0]`, the output should be `[0, 0, 0, 0, 0]` because any element multiplied by zero results in zero.\n",
-        "\n",
-        "2. **In-Place Modification**: Attempt to solve the problem with in-place modification of the `nums` array, where the output is stored directly in the input array without using any additional space."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb b/docs/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb
deleted file mode 100644
index d165899..0000000
--- a/docs/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb	
+++ /dev/null
@@ -1,198 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 242. Valid Anagram\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Anagram problem on LeetCode, click here!](https://leetcode.com/problems/valid-anagram/)\n",
-        "\n",
-        "---\n",
-        "Given two strings `s` and `t`, return `true` if `t` is an anagram of `s`, and `false` otherwise.\n",
-        "\n",
-        "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `s.length`, `t.length` <= $5 * 10^4$\n",
-        "- `s` and `t` consist of lowercase English letters.\n",
-        "\n",
-        "**Follow-up:** What if the inputs contain Unicode characters? How would you adapt your solution to such a case?\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 29,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isAnagram(s: str, t: str) -> bool:\n",
-        "    # Check if the lengths of s and t are not equal\n",
-        "    if len(s) != len(t):\n",
-        "        return False\n",
-        "\n",
-        "    # Create dictionaries to store character frequencies for s and t\n",
-        "    countS, countT = {}, {}\n",
-        "\n",
-        "    # Count character frequencies in s\n",
-        "    for i in range(len(s)):\n",
-        "        countS[s[i]] = 1 + countS.get(s[i], 0)\n",
-        "\n",
-        "    # Count character frequencies in t\n",
-        "    for i in range(len(t)):\n",
-        "        countT[t[i]] = 1 + countT.get(t[i], 0)\n",
-        "\n",
-        "    # Check if the character frequencies in s and t are the same\n",
-        "    return countS == countT"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The code defines a class called `Solution` with a method `isAnagram` that takes two input strings, `s` and `t`, and returns a boolean (`True` or `False`) based on whether `t` is an anagram of `s`.\n",
-        "\n",
-        "2. The first check performed in the method is whether the lengths of `s` and `t` are equal. If they are not equal, it immediately returns `False` because strings of different lengths cannot be anagrams of each other.\n",
-        "\n",
-        "3. Two dictionaries, `countS` and `countT`, are created to store the frequency of characters in `s` and `t`, respectively. These dictionaries will be used to count the occurrences of each character in the strings.\n",
-        "\n",
-        "4. The code then enters a loop to iterate through each character in string `s` using a `for` loop. Inside the loop, it updates the `countS` dictionary. The line `countS[s[i]] = 1 + countS.get(s[i], 0)` increments the count of character `s[i]` in `countS` by 1. If the character is not already in the dictionary, it initializes the count to 1.\n",
-        "\n",
-        "5. Similarly, the code iterates through each character in string `t` and updates the `countT` dictionary in the same way.\n",
-        "\n",
-        "6. After counting the character frequencies in both strings, the code compares the two dictionaries using `countS == countT`. If the dictionaries are equal, it means that both strings have the same character frequencies, and therefore, `t` is an anagram of `s`. In this case, the method returns `True`.\n",
-        "\n",
-        "7. If the dictionaries are not equal, the method returns `False`, indicating that `t` is not an anagram of `s`.\n",
-        "\n",
-        "In summary, this code efficiently determines whether two input strings are anagrams by counting the character frequencies in each string using dictionaries and then comparing these counts. If the character frequencies match, the strings are considered anagrams, and the method returns `True`; otherwise, it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 30,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "s = \"anagram\"\n",
-        "t = \"nagaram\"\n",
-        "print(isAnagram(s, t))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "s = \"rat\"\n",
-        "t = \"car\"\n",
-        "print(isAnagram(s, t))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. The code first checks whether the lengths of the input strings `s` and `t` are equal, which takes constant time. This check has a time complexity of O(1).\n",
-        "\n",
-        "2. The code then iterates through both strings, `s` and `t`, once to count the character frequencies. Since both strings have a maximum length of 5 * 10^4, the worst-case scenario is that the code iterates through 5 * 10^4 characters in each string. This results in a linear time complexity of O(n), where n is the length of the longer of the two input strings (`s` or `t`).\n",
-        "\n",
-        "3. Finally, the code compares the two dictionaries `countS` and `countT` to check if the character frequencies match. This comparison takes O(n) time in the worst case, where n is the length of the longer string.\n",
-        "\n",
-        "Overall, the time complexity of the code is O(n), where n is the length of the longer input string.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The code uses two dictionaries, `countS` and `countT`, to store the character frequencies of the input strings `s` and `t`. In the worst case, both dictionaries can contain all unique characters from the input strings. Since the input strings can have a maximum length of 5 * 10^4, the space complexity for these dictionaries is O(5 * 10^4) or simply O(n), where n is the length of the longer input string.\n",
-        "\n",
-        "2. The code uses a few additional variables for bookkeeping, but these variables have constant space requirements and do not depend on the input size. Therefore, they do not significantly impact the overall space complexity.\n",
-        "\n",
-        "**In summary**, the space complexity of the code is O(n), where n is the length of the longer input string."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Anagram Chains**: Given a list of words, find the longest chain of anagrams, where each word in the chain is an anagram of the previous word. For example, given [\"bat\", \"tab\", \"cat\", \"tac\", \"dog\"], the longest anagram chain is [\"bat\", \"tab\", \"cat\", \"tac\"].\n",
-        "\n",
-        "2. **Minimum Deletions to Make Anagrams**: Given two strings `s` and `t`, find the minimum number of character deletions required in both strings to make them anagrams of each other. For example, if `s = \"hello\"` and `t = \"billion\"`, you can remove the characters \"heo\" from `s` and \"billi\" from `t` to make them anagrams (\"lo\" and \"on\" are left in the two strings)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb b/docs/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb
deleted file mode 100644
index ecfa494..0000000
--- a/docs/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 347. Top K Frequent Elements\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Top K Frequent Elements problem on LeetCode, click here!](https://leetcode.com/problems/top-k-frequent-elements/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array `nums` and an integer `k`, return the `k` most frequent elements. You may return the answer in **any order**.\n",
-        "\n",
-        "Constraints:\n",
-        "- 1 <= `nums.length` <= $10^5$\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- `k` is in the range `[1, the number of unique elements in the array]`.\n",
-        "- It is **guaranteed** that the answer is **unique**.\n",
-        "\n",
-        "**Follow-up:** Your algorithm's time complexity must be better than $O(n\\ log\\ n)$, where n is the array's size.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 46,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def topKFrequent(nums, k):\n",
-        "    # Create a dictionary to store the frequency of each element\n",
-        "    count = {}\n",
-        "    # Create a list of lists to store elements with the same frequency\n",
-        "    frequency = [[] for i in range(len(nums) + 1)]\n",
-        "\n",
-        "    # Count the frequency of each element in nums\n",
-        "    for n in nums:\n",
-        "        count[n] = 1 + count.get(n, 0)\n",
-        "\n",
-        "    # Place elements in the freq list according to their frequency\n",
-        "    for n, c in count.items():\n",
-        "        frequency[c].append(n)\n",
-        "\n",
-        "    res = []\n",
-        "    # Traverse freq list from the end (higher frequencies)\n",
-        "    for i in range(len(frequency) - 1, 0, -1):\n",
-        "        for n in frequency[i]:\n",
-        "            res.append(n)\n",
-        "            if len(res) == k:\n",
-        "                return res"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `def topKFrequent(nums, k):`: This is a function that takes two arguments: `nums`, which is the input array of integers, and `k`, which is the number of most frequent elements to return.\n",
-        "\n",
-        "2. `count = {}`: This dictionary `count` will be used to store the frequency of each unique element in the `nums` array. The keys are elements from the input array, and the values are their corresponding frequencies.\n",
-        "\n",
-        "3. `frequency = [[] for i in range(len(nums) + 1)]`: This creates a list of empty lists called `frequency`. It's used to store elements based on their frequencies, similar to the `freq` list in the previous code. The size of this list is set to be one greater than the length of the input `nums`.\n",
-        "\n",
-        "4. `for n in nums:`: This loop iterates through each element `n` in the input `nums` array.\n",
-        "\n",
-        "5. `count[n] = 1 + count.get(n, 0)`: This line counts the frequency of each element `n` in the `nums` array. It uses the `count.get(n, 0)` method to retrieve the current count of `n` from the `count` dictionary. If `n` is not in the dictionary, it defaults to 0. It then increments the count by 1.\n",
-        "\n",
-        "6. After the above loop, the `count` dictionary will contain counts of each unique element in the `nums` array.\n",
-        "\n",
-        "7. `for n, c in count.items():`: This loop iterates through the items (key-value pairs) of the `count` dictionary. `n` represents the element, and `c` represents its frequency.\n",
-        "\n",
-        "8. `frequency[c].append(n)`: This line places the element `n` into the bucket corresponding to its frequency `c`. Buckets are represented by the `frequency` list. For example, if an element `n` has a frequency of 3, it will be added to `frequency[3]`.\n",
-        "\n",
-        "9. After this loop, the `frequency` list will contain buckets of elements grouped by their frequencies.\n",
-        "\n",
-        "10. `res = []`: This list `res` will be used to store the k most frequent elements.\n",
-        "\n",
-        "11. `for i in range(len(freq) - 1, 0, -1):`: This loop iterates in reverse order through the `frequency` list, starting from the highest frequency and going down to 1. Note that there is a typo here; it should be `frequency` instead of `freq`.\n",
-        "\n",
-        "12. `for n in frequency[i]:`: For each element `n` in the current bucket (i.e., elements with frequency `i`), it appends `n` to the result list `res`.\n",
-        "\n",
-        "13. `if len(res) == k:`: Once `res` contains k elements, the code exits and returns the result.\n",
-        "\n",
-        "The code efficiently finds the k most frequent elements in the input array without using sorting, similar to the previous explanation, with the only difference being the variable names (e.g., `frequency` instead of `freq`)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 48,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 2]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "nums = [1,1,1,2,2,3]\n",
-        "k = 2\n",
-        "result1 = topKFrequent(nums, k)\n",
-        "print(result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 49,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums = [1]\n",
-        "k = 1\n",
-        "result2 = topKFrequent(nums, k)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The first loop that counts the frequency of elements by iterating through `nums` has a time complexity of O(n), where n is the number of elements in the input array `nums`.\n",
-        "\n",
-        "2. The second loop iterates through the keys in the `count` dictionary, which has at most `k` unique elements (as per the constraints). Therefore, the second loop also has a time complexity of O(k).\n",
-        "\n",
-        "3. The third loop iterates through the `frequency` list, which has a length equal to the maximum frequency of elements in `nums`. In the worst case, this loop can have a time complexity of O(n), where n is the number of elements in `nums`.\n",
-        "\n",
-        "Overall, the dominant factor in terms of time complexity is the loop that iterates through the `count` dictionary. So, the total time complexity of the code is O(n + k).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The `count` dictionary stores the frequency of each unique element in `nums`. In the worst case, it can have at most `k` unique elements, so the space complexity for `count` is O(k).\n",
-        "\n",
-        "2. The `frequency` list is used to store elements grouped by their frequencies. It has a length of `len(nums) + 1`, which can be at most 105 based on the constraints. Therefore, the space complexity for `frequency` is O(105).\n",
-        "\n",
-        "3. The `res` list stores the k most frequent elements, which can be at most `k` elements, so the space complexity for `res` is O(k).\n",
-        "\n",
-        "**In summary**, the space complexity is dominated by the `count` dictionary and the `frequency` list, both of which have a space complexity of O(k + 105)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimized k Most Frequent Elements:** Modify the code to find the k most frequent elements in an array while ensuring that the time complexity is O(n + klogk). You can use a priority queue (heap) to achieve this.\n",
-        "\n",
-        "2. **Handling Duplicate Frequencies:** Extend the code to handle cases where multiple elements have the same frequency and are among the k most frequent elements. Ensure that the output contains exactly k elements."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb b/docs/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb
deleted file mode 100644
index afe89ec..0000000
--- a/docs/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb	
+++ /dev/null
@@ -1,229 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 49. Group Anagrams\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Group Anagrams problem on LeetCode, click here!](https://leetcode.com/problems/group-anagrams/)\n",
-        "\n",
-        "---\n",
-        "Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.\n",
-        "\n",
-        "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `strs.length <= $10^4$\n",
-        "- `0 <= strs[i].length <= 100`\n",
-        "- `strs[i]` consists of lowercase English letters.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 36,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict\n",
-        "\n",
-        "\n",
-        "def groupAnagrams(strs):\n",
-        "    ans = defaultdict(list)\n",
-        "\n",
-        "    # Iterate through the list of input strings\n",
-        "    for s in strs:\n",
-        "        # Initialize a list to represent character counts for each character (a-z)\n",
-        "        count = [0] * 26\n",
-        "\n",
-        "        # Count the occurrences of each character in the current word\n",
-        "        for c in s:\n",
-        "            count[ord(c) - ord(\"a\")] += 1\n",
-        "\n",
-        "        # Use a tuple of character counts as the key and append the word to the anagram group\n",
-        "        ans[tuple(count)].append(s)\n",
-        "\n",
-        "    # Convert the values (lists of anagrams) to a list of lists\n",
-        "    return list(ans.values())"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. Inside the method, a `defaultdict` called `ans` is created to store anagram groups. This dictionary will have a list as its default value, meaning that each key in the dictionary will be associated with an empty list by default.\n",
-        "\n",
-        "2. The code then iterates through the list of input strings, `strs`, using a `for` loop. For each word `s` in `strs`, it performs the following steps:\n",
-        "\n",
-        "   - It initializes a list `count` of length 26, where each element represents the count of a specific character (a-z) in the word `s`.\n",
-        "\n",
-        "   - The code then iterates through the characters of `s`. For each character `c`, it increments the corresponding count in the `count` list based on its ASCII value.\n",
-        "\n",
-        "   - After counting the characters in `s`, it converts the `count` list into a tuple `tuple(count)` to use as a key for the `ans` dictionary.\n",
-        "\n",
-        "   - It appends the word `s` to the list associated with the key (tuple) in the `ans` dictionary. This groups all anagrams of the same word together under the same key.\n",
-        "\n",
-        "3. After processing all words in `strs`, the code converts the values of the `ans` dictionary (which are lists of anagrams) to a list of lists using the `list()` constructor.\n",
-        "\n",
-        "4. Finally, the code returns the list of anagram groups, which is the result of grouping the anagrams in the input list.\n",
-        "\n",
-        "In summary, the code efficiently groups anagrams together by counting the characters in each word and using a dictionary to store them under the same key. The result is a list of lists, where each inner list represents a group of anagrams."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 37,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: Group anagrams from a list of words\n",
-        "strs1 = [\"eat\", \"tea\", \"tan\", \"ate\", \"nat\", \"bat\"]\n",
-        "result1 = groupAnagrams(strs1)\n",
-        "print(result1)\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 38,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2: Group anagrams from a list with an empty string\n",
-        "strs2 = [\"\"]\n",
-        "result2 = groupAnagrams(strs2)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 40,
-      "id": "c8367e2a-e7d5-4e18-8235-2870e341a04b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['a']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3: Group anagrams from a list with a single word\n",
-        "strs3 = [\"a\"]\n",
-        "result3 = groupAnagrams(strs3)\n",
-        "print(result3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `groupAnagrams` method:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. First, we create an instance of the `Solution` class, which is an O(1) operation.\n",
-        "\n",
-        "2. Inside the `groupAnagrams` method, we iterate through the list of input strings `strs` once in a loop:\n",
-        "   - For each word in `strs`, we perform character counting, which is done in O(K) time, where K is the maximum length of a word in the list.\n",
-        "\n",
-        "3. The overall time complexity of the method is O(N * K), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. We use a `defaultdict` called `ans` to store the anagram groups. In the worst case, each word belongs to a distinct anagram group, so the space complexity for `ans` is O(N).\n",
-        "\n",
-        "2. Within the loop, we create a `count` list of length 26 to store character counts for each word. This is a fixed-size list and does not depend on the input size, so it has a constant space complexity of O(26), which is effectively O(1).\n",
-        "\n",
-        "3. The space used for other variables is also constant and does not depend on the input size.\n",
-        "\n",
-        "4. The final result, `result`, is a list of lists that contains the grouped anagrams. In the worst case, each word is a distinct anagram group, so the space complexity for `result` is O(N).\n",
-        "\n",
-        "5. The overall space complexity of the method is O(N) due to the space used by `ans` and `result`. The space used by the `count` list and other variables is constant and does not significantly affect the overall space complexity.\n",
-        "\n",
-        "**In summary,** the time complexity of the `groupAnagrams` method is O(N * K), and the space complexity is O(N), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Anagramic Palindromes**: Given a list of strings, find all the groups of anagrams where each group contains words that can be rearranged into a palindrome. For example, in the list [\"bat\", \"tab\", \"act\", \"tac\", \"dog\"], there are two groups: [\"bat\", \"tab\", \"act\", \"tac\"] and [\"dog\"].\n",
-        "\n",
-        "2. **Largest Anagram Group**: Given a list of strings, find the largest group of anagrams. Return the list of anagrams in that group. If there are multiple largest groups, return all of them."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb b/docs/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb
deleted file mode 100644
index 037c476..0000000
--- a/docs/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb	
+++ /dev/null
@@ -1,204 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 659· Encode and Decode Strings\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Encode and Decode Strings problem on LintCode, click here!](https://www.lintcode.com/problem/659/)\n",
-        "\n",
-        "---\n",
-        "Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.\n",
-        "\n",
-        "Please implement `encode` and `decode`.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 54,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class Codec:\n",
-        "\n",
-        "    def encode(self, strs):\n",
-        "        \"\"\"Encodes a list of strings to a single string.\n",
-        "        \n",
-        "        :type strs: List[str]\n",
-        "        :rtype: str\n",
-        "        \"\"\"\n",
-        "        encoded = \"\"\n",
-        "        for s in strs:\n",
-        "            encoded += s.replace(':', '::') + ':;'\n",
-        "        return encoded\n",
-        "\n",
-        "    def decode(self, s):\n",
-        "        \"\"\"Decodes a single string to a list of strings.\n",
-        "        \n",
-        "        :type s: str\n",
-        "        :rtype: List[str]\n",
-        "        \"\"\"\n",
-        "        decoded = []\n",
-        "        i = 0\n",
-        "        while i < len(s):\n",
-        "            end = s.find(':;', i)\n",
-        "            if end == -1:\n",
-        "                end = len(s)\n",
-        "            decoded.append(s[i:end].replace('::', ':'))\n",
-        "            i = end + 2\n",
-        "        return decoded"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a class called `Codec` to encapsulate the encoding and decoding operations for a list of strings.\n",
-        "\n",
-        "2. In the `encode` method, we take a list of strings (`strs`) as input and return a single encoded string. The encoding process involves concatenating the strings together with a delimiter `:;`. We iterate through each string in the input list, replace any occurrence of `:` with `::` (to avoid conflicts with the delimiter), and then append `:;` to indicate the end of that string.\n",
-        "\n",
-        "3. In the `decode` method, we take an encoded string (`s`) as input and return a list of strings. The decoding process involves splitting the encoded string using the `:;` delimiter. We start from the beginning of the encoded string (`i = 0`) and repeatedly find the next occurrence of `:;`. We extract the substring between the current position (`i`) and the next delimiter (`end`). Before adding it to the result list, we replace any `::` with `:` to revert the encoding. We then update the current position `i` to be after the delimiter, so we can find the next substring in the next iteration.\n",
-        "\n",
-        "4. Finally, we create an instance of the `Codec` class and test it with two examples:\n",
-        "\n",
-        "By using the `:;` delimiter and handling the `::` escaping, this code can encode and decode lists of strings, preserving the original content even if it contains colons or the delimiter itself."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 55,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['lint', 'code', 'love', 'you']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "codec = Codec()\n",
-        "\n",
-        "input1 = [\"lint\", \"code\", \"love\", \"you\"]\n",
-        "encoded1 = codec.encode(input1)\n",
-        "decoded1 = codec.decode(encoded1)\n",
-        "print(decoded1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 57,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['we', 'say', ':', 'yes']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "codec = Codec() \n",
-        "\n",
-        "input2 = [\"we\", \"say\", \":\", \"yes\"]\n",
-        "encoded2 = codec.encode(input2)\n",
-        "decoded2 = codec.decode(encoded2)\n",
-        "print(decoded2)  "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `encode` and `decode` methods in the provided code.\n",
-        "\n",
-        "**Encode Method (`encode`):**\n",
-        "\n",
-        "- Time Complexity: O(n * m)\n",
-        "  - Here, `n` is the number of strings in the input list `strs`, and `m` is the average length of these strings.\n",
-        "  - In the worst case, for each string in the list, we iterate over its characters once to replace any colons (`:`) with double colons (`::`) and then append `:;`. This is done for each string in the list.\n",
-        "\n",
-        "- Space Complexity: O(n * m)\n",
-        "  - The space complexity is also O(n * m) because we build the encoded string by concatenating the modified strings. In the worst case, the encoded string can be of the same length as the original strings.\n",
-        "\n",
-        "**Decode Method (`decode`):**\n",
-        "\n",
-        "- Time Complexity: O(n * m)\n",
-        "  - Similar to the encode method, we iterate through the encoded string in the decode method. In the worst case, we may have to scan each character in the encoded string to find the `:;` delimiter.\n",
-        "\n",
-        "- Space Complexity: O(n * m)\n",
-        "  - The space complexity of the decode method is also O(n * m) because we build the list of decoded strings, which can be of the same length as the encoded string.\n",
-        "\n",
-        "Overall, both the `encode` and `decode` methods have time and space complexities of O(n * m), where `n` is the number of strings in the input list, and `m` is the average length of these strings. The space complexity arises from storing the encoded or decoded strings, and the time complexity arises from iterating through the characters in these strings."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Delimiters**: Modify the code to allow for multiple delimiters, not just one. Each string could have its own delimiter, and the code should be able to handle this complexity.\n",
-        "\n",
-        "2. **Optimize Encoding**: Modify the `encode` method to achieve a more efficient encoding algorithm. Try to minimize the length of the encoded string while ensuring that it can be correctly decoded."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/01. Array Hashing/README.md b/docs/_build/html/_sources/01. Array Hashing/README.md
deleted file mode 100644
index a331dab..0000000
--- a/docs/_build/html/_sources/01. Array Hashing/README.md	
+++ /dev/null
@@ -1,27 +0,0 @@
-# Array & Hashing Solutions - Blind 75 LeetCode Problems
-![Array & Hashing Solutions](https://img.shields.io/badge/Array_&_Hashing_Solutions-blue)![Blind 75 LeetCode](https://img.shields.io/badge/Blind_75_LeetCode-Problems-blue?labelColor=red)
-
-This repository contains my solutions to Array & Hashing Problems from Blind 75 LeetCode problems. I've organized the solutions by categories for easier navigation and reference. Each problem solution is presented in Jupyter Notebook format (`.ipynb`).
-
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [217. Contains Duplicate](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/217.%20Contains%20Duplicate.ipynb)                            | Easy       |
-| [242. Valid Anagram](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/242.%20Valid%20Anagram.ipynb)                                      | Easy       |
-| [1. Two Sum](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/1.%20Two%20Sum.ipynb)                                                      | Easy       |
-| [49. Group Anagrams](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/49.%20Group%20Anagrams.ipynb)                                      | Medium     |
-| [347. Top K Frequent Elements](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.ipynb)              | Medium     |
-| [238. Product of Array Except Self](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.ipynb)  | Medium     |
-| [659. Encode and Decode Strings](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.ipynb)          | Medium     |
-| [128. Longest Consecutive Sequence](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.ipynb)      | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb b/docs/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb
deleted file mode 100644
index 284c832..0000000
--- a/docs/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 11. Container With Most Water\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Container With Most Water problem on LeetCode, click here!](https://leetcode.com/problems/container-with-most-water/)\n",
-        "\n",
-        "---\n",
-        "You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the $i^{th}$ line are `(i, 0)` and `(i, height[i])`.\n",
-        "\n",
-        "Find two lines that together with the x-axis form a container, such that the container contains the most water.\n",
-        "\n",
-        "Return *the maximum amount of water a container can store.*\n",
-        "\n",
-        "**Notice** that you may not slant the container.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `n == height.length`\n",
-        "- 2 <= `n` <= $10^5$\n",
-        "- 0 <= `height[i]` <= $10^4$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def maxArea(height):\n",
-        "    # Initialize the maximum area to 0.\n",
-        "    max_area = 0\n",
-        "    # Initialize two pointers, one at the beginning (left) and one at the end (right) of the array.\n",
-        "    left = 0\n",
-        "    right = len(height) - 1\n",
-        "\n",
-        "    # Iterate until the left pointer is less than the right pointer.\n",
-        "    while left < right:\n",
-        "        # Calculate the height of the container, which is the minimum height of the two lines.\n",
-        "        h = min(height[left], height[right])\n",
-        "        # Calculate the width of the container, which is the distance between the two pointers.\n",
-        "        width = right - left\n",
-        "        # Calculate the area of the container using height and width and update max_area if it's greater.\n",
-        "        max_area = max(max_area, h * width)\n",
-        "\n",
-        "        # Move the pointer with the shorter line inward.\n",
-        "        if height[left] < height[right]:\n",
-        "            left += 1\n",
-        "        else:\n",
-        "            right -= 1\n",
-        "\n",
-        "    # Continue this process until left < right, and then return the maximum area.\n",
-        "    return max_area"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start with initializing `max_area` to 0, which will store the maximum area of water that can be contained by the lines.\n",
-        "\n",
-        "2. Two pointers, `left` and `right`, are initialized to the beginning and end of the input `height` array.\n",
-        "\n",
-        "3. The main loop runs while `left` is less than `right`. This ensures that we are checking all possible pairs of lines in a systematic way.\n",
-        "\n",
-        "4. Inside the loop:\n",
-        "   - We calculate the height of the container `h` as the minimum of the heights at the `left` and `right` pointers. This represents the height of the water the container can hold.\n",
-        "   - We calculate the width of the container as the difference between `right` and `left`. This represents the distance between the two lines.\n",
-        "   - We calculate the area of the container using `h * width` and update `max_area` if the calculated area is greater than the current `max_area`.\n",
-        "\n",
-        "5. After calculating the area and updating `max_area`, we move one of the pointers. We move the pointer pointing to the shorter line inward because moving the pointer pointing to the taller line won't increase the height and will only decrease the width, which will result in a smaller area.\n",
-        "\n",
-        "6. We continue this process until `left` is no longer less than `right`, indicating that we have checked all possible pairs of lines.\n",
-        "\n",
-        "7. Finally, we return the `max_area` which contains the maximum area of water that can be contained by two lines from the input array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 49\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "height1 = [1, 8, 6, 2, 5, 4, 8, 3, 7]\n",
-        "result1 = maxArea(height1)\n",
-        "print(\"Output:\", result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "height2 = [1, 1]\n",
-        "result2 = maxArea(height2)\n",
-        "print(\"Output:\", result2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 45\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Additional Example\n",
-        "height3 = [3, 9, 3, 4, 7, 2, 12, 6]\n",
-        "result3 = maxArea(height3)\n",
-        "print(\"Output:\", result3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `maxArea` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The function uses a two-pointer approach to find the maximum area of water. In the worst case, both the left and right pointers traverse the entire input array once. Therefore, the time complexity is O(n), where n is the number of elements in the `height` array.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The function uses a constant amount of extra space for variables like `max_area`, `left`, `right`, `h`, and `width`. It does not use any data structures whose space consumption depends on the input size. As a result, the space complexity is O(1), which means it has constant space complexity.\n",
-        "\n",
-        "**In summary,** the `maxArea` function has a time complexity of O(n) and a space complexity of O(1), making it an efficient algorithm for finding the maximum area of water that can be contained by two lines in the input array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the maxArea function to not only return the maximum area but also the indices of the two lines that form the container with the maximum area.\n",
-        "2. **Variant Problem:** Instead of finding the maximum area of water, find the minimum amount of water required to fill all the gaps between the lines. You'll need to return the total water volume needed."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb b/docs/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb
deleted file mode 100644
index 8520b3b..0000000
--- a/docs/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb	
+++ /dev/null
@@ -1,218 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 125. Valid Palindrome\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Palindrome problem on LeetCode, click here!](https://leetcode.com/problems/valid-palindrome/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.\n",
-        "\n",
-        "Given a string `s`, *return `true` if it is a **palindrome**, or `false` otherwise.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $2*10^5$\n",
-        "- `s` consists only of printable ASCII characters.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isPalindrome(s):\n",
-        "    # Step 1: Convert the input string to lowercase\n",
-        "    s = s.lower()\n",
-        "    \n",
-        "    # Step 2: Remove all non-alphanumeric characters from the string\n",
-        "    cleaned_s = ''.join(c for c in s if c.isalnum())\n",
-        "    \n",
-        "    # Step 3: Initialize two pointers, one at the beginning and one at the end of the cleaned string\n",
-        "    left, right = 0, len(cleaned_s) - 1\n",
-        "    \n",
-        "    # Step 4: Compare characters using the two pointers while moving them towards each other\n",
-        "    while left < right:\n",
-        "        if cleaned_s[left] != cleaned_s[right]:\n",
-        "            return False  # If characters don't match, it's not a palindrome\n",
-        "        left += 1\n",
-        "        right -= 1\n",
-        "    \n",
-        "    # Step 5: If the loop completes without returning False, it's a palindrome\n",
-        "    return True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `isPalindrome` function takes an input string `s`.\n",
-        "\n",
-        "2. It starts by converting the input string to lowercase using `s.lower()`. This step ensures that the function is case-insensitive when checking for palindromes.\n",
-        "\n",
-        "3. Next, it removes all non-alphanumeric characters from the string and stores the result in the `cleaned_s` variable. It does this by iterating through each character in the input string and only keeping characters that are alphanumeric (letters and digits). This step removes spaces, punctuation, and other non-alphanumeric characters.\n",
-        "\n",
-        "4. The function then initializes two pointers, `left` and `right`, which will be used to compare characters from the beginning and end of the cleaned string.\n",
-        "\n",
-        "5. In a while loop, it compares characters using the two pointers while moving them towards each other. The loop continues until the `left` pointer is less than the `right` pointer. This is done to check all characters in the cleaned string.\n",
-        "\n",
-        "6. Inside the loop, it checks if the characters at the `left` and `right` pointers don't match. If they don't match, it means the input is not a palindrome, so the function returns `False`.\n",
-        "\n",
-        "7. If the characters at the `left` and `right` pointers match, the pointers are moved closer to each other by incrementing `left` and decrementing `right`.\n",
-        "\n",
-        "8. The loop continues to compare characters until it either finds a mismatch (returning `False`) or until the `left` pointer is greater than or equal to the `right` pointer. If the loop completes without finding a mismatch, it means the input is a palindrome, so the function returns `True`.\n",
-        "\n",
-        "In summary, the code converts the input string to lowercase, removes non-alphanumeric characters, and then uses two pointers to compare characters from the cleaned string's beginning and end. If all characters match during the comparison, the function returns `True`, indicating that the input is a palindrome. If any characters do not match, it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "input1 = \"A man, a plan, a canal: Panama\"\n",
-        "output1 = isPalindrome(input1)\n",
-        "print(output1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "input2 = \"race a car\"\n",
-        "output2 = isPalindrome(input2)\n",
-        "print(output2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "input3 = \"\"\n",
-        "output3 = isPalindrome(input3)\n",
-        "print(output3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `isPalindrome` function that uses two pointers:\n",
-        "\n",
-        "1. **Time Complexity:**\n",
-        "   - Converting the input string to lowercase using `s.lower()` takes O(n) time, where 'n' is the length of the input string.\n",
-        "   - Removing non-alphanumeric characters using the list comprehension `''.join(c for c in s if c.isalnum())` also takes O(n) time because it iterates through each character in the string once.\n",
-        "   - The comparison of characters using the two pointers (`left` and `right`) is done in a loop that runs until `left` is less than `right`. In the worst case, the loop iterates through half of the string, so it takes O(n/2) time, which is still considered O(n).\n",
-        "\n",
-        "   Therefore, the overall time complexity of the `isPalindrome` function is O(n), where 'n' is the length of the input string.\n",
-        "\n",
-        "2. **Space Complexity:**\n",
-        "   - The space complexity is primarily determined by the additional string `cleaned_s` created to store the cleaned version of the input string. In the worst case, if all characters in the input string are alphanumeric, this cleaned string can be as large as the original input string.\n",
-        "   - Additionally, the function uses two pointers, `left` and `right`, which consume constant space and do not depend on the size of the input.\n",
-        "\n",
-        "**In summary,** the overall space complexity of the `isPalindrome` function is O(n) in the worst case, where 'n' is the length of the input string. The space complexity is dominated by the space required for `cleaned_s`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Longest Palindromic Substring:** Given a string `s`, find and return the longest palindromic substring within `s`. For example, if `s = \"babad\"`, the function should return `\"bab\"` or `\"aba\"`.\n",
-        "2. **Palindrome with Limited Characters:** Given a string `s` and a list of characters `allowed_chars`, check if `s` is a palindrome considering only the characters from `allowed_chars`. Ignore all other characters. For example, if `s = \"A man, a plan, a canal: Panama\"` and `allowed_chars = ['a', 'm', 'n', 'p']`, the function should return `True`."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb b/docs/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb
deleted file mode 100644
index 06de549..0000000
--- a/docs/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb	
+++ /dev/null
@@ -1,250 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 15. 3Sum\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the 3Sum problem on LeetCode, click here!](https://leetcode.com/problems/3sum/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array nums, *return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.*\n",
-        "\n",
-        "**Notice** that the solution set must not contain duplicate triplets.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `3 <= nums.length <= 3000`\n",
-        "- $-10^5$ <= `nums[i]` <= $10^5$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def threeSum(nums):\n",
-        "    # Sort the input array in ascending order\n",
-        "    nums.sort()\n",
-        "    triplets = []\n",
-        "    \n",
-        "    for i in range(len(nums) - 2):\n",
-        "        # Skip duplicates to avoid duplicate triplets\n",
-        "        if i > 0 and nums[i] == nums[i - 1]:\n",
-        "            continue\n",
-        "        \n",
-        "        left, right = i + 1, len(nums) - 1\n",
-        "        \n",
-        "        while left < right:\n",
-        "            # Calculate the total sum of the current triplet\n",
-        "            total = nums[i] + nums[left] + nums[right]\n",
-        "            \n",
-        "            if total == 0:\n",
-        "                # If the total is zero, we found a valid triplet\n",
-        "                triplets.append([nums[i], nums[left], nums[right]])\n",
-        "                \n",
-        "                # Skip duplicates of left and right pointers\n",
-        "                while left < right and nums[left] == nums[left + 1]:\n",
-        "                    left += 1\n",
-        "                while left < right and nums[right] == nums[right - 1]:\n",
-        "                    right -= 1\n",
-        "                \n",
-        "                # Move the pointers to the next unique elements\n",
-        "                left += 1\n",
-        "                right -= 1\n",
-        "            elif total < 0:\n",
-        "                # If the total is negative, we need to increase the sum by moving the left pointer to the right\n",
-        "                left += 1\n",
-        "            else:\n",
-        "                # If the total is positive, we need to decrease the sum by moving the right pointer to the left\n",
-        "                right -= 1\n",
-        "    \n",
-        "    return triplets"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. It starts by sorting the input array `nums` in ascending order. Sorting helps in efficiently finding triplets that sum up to zero.\n",
-        "\n",
-        "2. It initializes an empty list `triplets` to store the unique triplets that meet the given conditions.\n",
-        "\n",
-        "3. The code then iterates through the `nums` array using a loop with index `i`. This loop will consider each element of the array as a potential starting point for a triplet.\n",
-        "\n",
-        "4. Inside the loop, it checks for duplicates and skips them. This is done to ensure that the solution set does not contain duplicate triplets. If `nums[i]` is the same as the previous element `nums[i-1]`, it continues to the next iteration of the loop.\n",
-        "\n",
-        "5. Two pointers, `left` and `right`, are initialized. `left` starts just after the current element `nums[i]`, and `right` starts at the end of the array.\n",
-        "\n",
-        "6. The code enters another loop with `left` and `right` pointers, trying to find a triplet that sums up to zero.\n",
-        "\n",
-        "7. It calculates the `total` sum of the current triplet as `nums[i] + nums[left] + nums[right]`.\n",
-        "\n",
-        "8. If `total` is zero, it means a valid triplet is found, so it appends `[nums[i], nums[left], nums[right]]` to the `triplets` list. Then, it moves both the `left` and `right` pointers to their next unique elements while skipping duplicates.\n",
-        "\n",
-        "9. If `total` is less than zero, it means the sum is negative, so it increments the `left` pointer to move towards a larger sum.\n",
-        "\n",
-        "10. If `total` is greater than zero, it means the sum is positive, so it decrements the `right` pointer to move towards a smaller sum.\n",
-        "\n",
-        "11. This process continues until all possible triplets have been considered.\n",
-        "\n",
-        "12. Finally, the function returns the `triplets` list, which contains all the unique triplets that sum up to zero in the sorted `nums` array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[-1, -1, 2], [-1, 0, 1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "nums = [-1, 0, 1, 2, -1, -4]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "nums = [0, 1, 1]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 0, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "nums = [0, 0, 0]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `threeSum` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. Sorting the input array `nums` takes O(n log n) time, where n is the length of the array.\n",
-        "\n",
-        "2. The outer loop runs for each element of the array, so it iterates O(n) times.\n",
-        "\n",
-        "3. Inside the outer loop, we have a while loop with two pointers (`left` and `right`). In the worst case, the while loop can iterate O(n) times (when all elements are unique).\n",
-        "\n",
-        "4. Inside the while loop, we have constant time operations (additions, comparisons, and list appends).\n",
-        "\n",
-        "Overall, the time complexity of the `threeSum` function is dominated by the sorting step, so it is O(n log n) due to the sorting. The other operations inside the loops contribute linearly but are dominated by the sorting step.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The space used by the `triplets` list to store the output triplets is O(k), where k is the number of unique triplets that sum up to zero. In the worst case, this can be O(n^2/3) because there can be roughly O(n^2) possible triplets, and many of them may sum up to zero.\n",
-        "\n",
-        "2. Other than the `triplets` list, the function uses only a constant amount of additional space for variables like `i`, `left`, `right`, and `total`.\n",
-        "\n",
-        "Overall, the space complexity of the `threeSum` function is O(k), where k is the number of unique triplets that meet the criteria. It's important to note that the space complexity does not depend on the size of the input array but rather on the number of valid triplets found.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: O(n log n) due to sorting (dominant factor) and O(n^2) in the worst case for the nested loops.\n",
-        "- Space Complexity: O(k) for storing the output triplets, where k is the number of unique triplets that meet the criteria."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K-Sum Problem:** Generalize the problem to the K-Sum problem, where instead of finding triplets that sum to zero, you need to find K elements that sum to a target value. Implement a function for this generalization.\n",
-        "2. **Count the Number of Unique Triplets:** Modify the algorithm to count the number of unique triplets that satisfy the conditions instead of returning the triplets themselves. This will require some modifications to handle duplicates efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/02. Two Pointers/README.md b/docs/_build/html/_sources/02. Two Pointers/README.md
deleted file mode 100644
index 0d4c7cf..0000000
--- a/docs/_build/html/_sources/02. Two Pointers/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Two Pointers Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)                            | Easy       |
-| [15. 3Sum](https://leetcode.com/problems/3sum/)                                                       | Medium     |
-| [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)           | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb b/docs/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb
deleted file mode 100644
index 554c940..0000000
--- a/docs/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb	
+++ /dev/null
@@ -1,191 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 121. Best Time to Buy and Sell Stock\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Best Time to Buy and Sell Stock problem on LeetCode, click here!](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given an array `prices` where `prices[i]` is the price of a given stock on the $i^{th}$ day.\n",
-        "\n",
-        "You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.\n",
-        "\n",
-        "*Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `prices.length` <= $10^5$\n",
-        "- 0 <= `prices[i]` <= $10^4$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 65,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def maxProfit(prices):\n",
-        "    if not prices:\n",
-        "        return 0\n",
-        "    \n",
-        "    min_price = prices[0]\n",
-        "    max_profit = 0\n",
-        "    \n",
-        "    for price in prices:\n",
-        "        # Update the minimum price if a lower price is encountered\n",
-        "        min_price = min(min_price, price)\n",
-        "        # Calculate the profit if selling at the current price\n",
-        "        max_profit = max(max_profit, price - min_price)\n",
-        "    \n",
-        "    return max_profit"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- The `maxProfit` function takes a single argument, `prices`, which is a list of stock prices.\n",
-        "- The first line of the function checks if the `prices` list is empty. If it's empty, there are no prices to analyze, so the function returns 0 (no profit can be made).\n",
-        "\n",
-        "- Then, we initialize two variables:\n",
-        "  - `min_price`: This variable will keep track of the minimum price seen so far. We initialize it with the price of the first day in the list (`prices[0]`).\n",
-        "  - `max_profit`: This variable will keep track of the maximum profit we can achieve. We initialize it to 0 because initially, we haven't made any profit.\n",
-        "\n",
-        "- We start iterating through the `prices` list, where `price` represents the stock price for the current day.\n",
-        "\n",
-        "- In each iteration, we check if the current `price` is lower than the `min_price` we've seen so far. If it is, we update `min_price` to the current `price`. This is because we want to buy the stock at the lowest possible price.\n",
-        "\n",
-        "- Next, we calculate the profit we would make if we sold the stock at the current `price` (assuming we bought it at `min_price`). We do this by subtracting `min_price` from the current `price`. This represents the profit for the current day.\n",
-        "\n",
-        "- We also use the `max` function to keep track of the maximum profit seen so far. If the profit calculated for the current day is greater than the `max_profit` we've seen previously, we update `max_profit` with this higher value.\n",
-        "\n",
-        "- Finally, after iterating through all the days in the `prices` list, we return `max_profit`, which represents the maximum profit that can be achieved by buying and selling a single stock."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 66,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "5\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "prices1 = [7, 1, 5, 3, 6, 4]\n",
-        "print(maxProfit(prices1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 67,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "0\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "prices2 = [7, 6, 4, 3, 1]\n",
-        "print(maxProfit(prices2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code iterates through the `prices` array once, examining each price exactly once.\n",
-        "- Within the loop, there are constant-time operations such as comparisons, additions, and subtractions.\n",
-        "- Therefore, the time complexity of this code is O(n), where n is the length of the `prices` array. It performs a linear number of operations relative to the size of the input.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the code is constant, O(1).\n",
-        "- Regardless of the size of the input `prices` array, the code only uses a fixed amount of additional memory to store two variables (`min_price` and `max_profit`) and a few loop variables.\n",
-        "- The memory usage does not depend on the size of the input, so it is considered constant space complexity.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: O(n) where n is the length of the `prices` array.\n",
-        "- Space Complexity: O(1) (constant space usage)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Transactions Allowed:** Modify the problem to allow multiple transactions (buy and sell multiple times), but you must sell before buying again. Find the maximum profit in this scenario.\n",
-        "2. **K Transactions Allowed:** Extend the problem to allow at most k transactions. Find the maximum profit considering this constraint.\n",
-        "3. **With Transaction Fee:** Introduce a transaction fee for each buy/sell operation. Modify the code to maximize profit while considering the transaction fee."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb b/docs/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb
deleted file mode 100644
index 4a516de..0000000
--- a/docs/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb	
+++ /dev/null
@@ -1,214 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 3. Longest Substring Without Repeating Characters\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Substring Without Repeating Characters problem on LeetCode, click here!](https://leetcode.com/problems/longest-substring-without-repeating-characters/)\n",
-        "\n",
-        "---\n",
-        "Given a string `s`, find the length of the **longest substring** without repeating characters.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 0 <= `s.length` <= $5 * 10^4$\n",
-        "- `s` consists of English letters, digits, symbols and spaces.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 61,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def length_of_longest_substring(s):\n",
-        "    # Create a dictionary to store the index of each character's last occurrence.\n",
-        "    char_index = {}\n",
-        "    \n",
-        "    # Initialize variables to keep track of the start and end of the current substring.\n",
-        "    start = 0\n",
-        "    max_length = 0\n",
-        "    \n",
-        "    for end in range(len(s)):\n",
-        "        # If the character is in the dictionary and its last occurrence is after the start of the current substring,\n",
-        "        # update the start of the substring to the next character after its last occurrence.\n",
-        "        if s[end] in char_index and char_index[s[end]] >= start:\n",
-        "            start = char_index[s[end]] + 1\n",
-        "        \n",
-        "        # Update the last occurrence index of the current character.\n",
-        "        char_index[s[end]] = end\n",
-        "        \n",
-        "        # Calculate the length of the current substring and update the maximum length if necessary.\n",
-        "        current_length = end - start + 1\n",
-        "        max_length = max(max_length, current_length)\n",
-        "    \n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a function called `length_of_longest_substring` that takes a single argument `s`, which is the input string for which we want to find the length of the longest substring without repeating characters.\n",
-        "\n",
-        "2. We create a dictionary called `char_index` to store the index of the last occurrence of each character in the input string. This dictionary will help us keep track of where each character was last seen.\n",
-        "\n",
-        "3. We initialize two variables, `start` and `max_length`. \n",
-        "   - `start` represents the start index of the current substring without repeating characters. It starts at 0.\n",
-        "   - `max_length` is used to keep track of the maximum length found so far and is initially set to 0.\n",
-        "\n",
-        "4. We then iterate through the input string `s` using a for loop, where the loop variable `end` represents the current end index of the substring we are considering.\n",
-        "\n",
-        "5. Inside the loop:\n",
-        "   - We check if the character `s[end]` is already in the `char_index` dictionary and if its last occurrence is within or after the current substring. If so, it means that we've encountered a repeating character, and we need to update the `start` of the substring to the next character after the last occurrence of `s[end]`. This ensures that we have a new valid substring without repeating characters.\n",
-        "   - We then update the `char_index` dictionary by storing the current index `end` as the last occurrence index of the character `s[end]`.\n",
-        "\n",
-        "6. Next, we calculate the length of the current substring without repeating characters, which is `current_length = end - start + 1`. We add 1 to account for the fact that the indices are zero-based.\n",
-        "\n",
-        "7. We update the `max_length` with the maximum of its current value and the `current_length`. This step ensures that we keep track of the longest valid substring we have encountered so far.\n",
-        "\n",
-        "8. The loop continues to iterate through the input string, and at the end, we return the `max_length`, which represents the length of the longest substring without repeating characters."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 62,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(length_of_longest_substring(\"abcabcbb\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 63,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(length_of_longest_substring(\"bbbbb\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 64,
-      "id": "cf04eba0-0ead-4e8b-ba33-4316ea64a9d3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(length_of_longest_substring(\"pwwkew\"))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "Let's analyze the time and space complexity of the `length_of_longest_substring` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The function iterates through the input string `s` using a single for loop. The loop runs from the beginning of the string to the end once.\n",
-        "- Inside the loop, we perform constant-time operations such as dictionary lookups and updates, comparisons, and arithmetic operations.\n",
-        "- Therefore, the time complexity of the function is O(n), where n is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The primary data structure that consumes space in this function is the `char_index` dictionary.\n",
-        "- In the worst case, if there are no repeating characters in the input string, the dictionary can store all unique characters in the string.\n",
-        "- Therefore, the space complexity is O(min(n, m)), where n is the length of the input string `s`, and m is the number of unique characters in the string. In the worst case, when all characters are unique, m is equal to n, so the space complexity is O(n).\n",
-        "- Additionally, there are a few integer variables used for indices and lengths, which consume constant space.\n",
-        "- Overall, the space complexity of the function is O(min(n, m)) or simply O(n) in the worst case.\n",
-        "\n",
-        "**In summary,** the time complexity of the `length_of_longest_substring` function is O(n), and the space complexity is O(n) in the worst case due to the `char_index` dictionary. This algorithm provides an efficient way to find the length of the longest substring without repeating characters in linear time."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Longest Substring with K Distinct Characters:** Modify the problem to find the length of the longest substring with exactly K distinct characters. For example, given the input string \"abcabcbb\" and K = 2, the answer would be 4 because the longest substring with two distinct characters is \"abca.\"\n",
-        "2. **Longest Substring with Unique Characters:** Write a function to find the length of the longest substring in a given string where all characters are unique. For example, given the input string \"abcabcbb,\" the answer would be 4 because \"abcd\" is the longest substring with unique characters."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb b/docs/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb
deleted file mode 100644
index 07e96a2..0000000
--- a/docs/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb	
+++ /dev/null
@@ -1,195 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 424. Longest Repeating Character Replacement\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Repeating Character Replacement problem on LeetCode, click here!](https://leetcode.com/problems/longest-repeating-character-replacement/)\n",
-        "\n",
-        "---\n",
-        "You are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.\n",
-        "\n",
-        "*Return the length of the longest substring containing the same letter you can get after performing the above operations.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $10^5$\n",
-        "- `s` consists of only uppercase English letters.\n",
-        "- `0 <= k <= s.length`"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 58,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def characterReplacement(s, k):\n",
-        "    max_length = 0  # Initialize the maximum length\n",
-        "    max_count = 0   # Initialize the maximum count of repeating characters\n",
-        "    start = 0       # Initialize the start of the sliding window\n",
-        "    char_count = {} # Dictionary to store the count of each character\n",
-        "\n",
-        "    for end in range(len(s)):\n",
-        "        # Update the count of the current character in the dictionary\n",
-        "        char_count[s[end]] = char_count.get(s[end], 0) + 1\n",
-        "        \n",
-        "        # Update the maximum count of repeating characters\n",
-        "        max_count = max(max_count, char_count[s[end]])\n",
-        "        \n",
-        "        # Check if the current window size is greater than k\n",
-        "        if (end - start + 1) - max_count > k:\n",
-        "            # Move the start of the window to the right\n",
-        "            char_count[s[start]] -= 1\n",
-        "            start += 1\n",
-        "        \n",
-        "        # Update the maximum length\n",
-        "        max_length = max(max_length, end - start + 1)\n",
-        "\n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We initialize some variables:\n",
-        "   - `max_length` to keep track of the maximum length of the substring containing the same letter.\n",
-        "   - `max_count` to keep track of the maximum count of repeating characters within the current window.\n",
-        "   - `start` to represent the start index of the sliding window.\n",
-        "   - `char_count` is a dictionary that will store the count of each character within the current window.\n",
-        "\n",
-        "2. We use a `for` loop to iterate through the characters of the string `s`.\n",
-        "\n",
-        "3. Inside the loop, we do the following for each character at position `end`:\n",
-        "   - Update the count of the current character in the `char_count` dictionary.\n",
-        "   - Update the `max_count` to be the maximum of the current `max_count` and the count of the current character. This keeps track of the maximum count of repeating characters within the current window.\n",
-        "\n",
-        "4. We check if the current window size (the difference between `end` and `start` plus one) minus the `max_count` is greater than `k`. This condition checks whether we have exceeded the allowed number of replacements (`k`) within the current window.\n",
-        "\n",
-        "   - If we have exceeded the allowed replacements, it means we need to shrink the window from the left side. We do this by moving the `start` of the window to the right and decrementing the count of the character at `s[start]` in the `char_count` dictionary. This effectively removes characters from the left side of the window until we have a valid window again.\n",
-        "\n",
-        "   - By doing this, we ensure that the difference between the current window size and the `max_count` is always less than or equal to `k`.\n",
-        "\n",
-        "5. After each character, we update the `max_length` to be the maximum of the current `max_length` and the size of the current window (`end - start + 1`).\n",
-        "\n",
-        "6. Finally, we return `max_length`, which holds the length of the longest substring containing the same letter with at most `k` replacements."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 59,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "s1 = \"ABAB\"\n",
-        "k1 = 2\n",
-        "print(characterReplacement(s1, k1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 60,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "s2 = \"AABABBA\"\n",
-        "k2 = 1\n",
-        "print(characterReplacement(s2, k2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code uses a single `for` loop to iterate through the characters of the input string `s`. The loop runs from the beginning to the end of the string.\n",
-        "- Inside the loop, we perform constant-time operations such as updating the `char_count` dictionary and updating variables like `max_count` and `max_length`.\n",
-        "- The code's time complexity is primarily determined by the loop, which iterates through each character in the string once. Therefore, the time complexity is O(n), where n is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The code uses several variables to store information, but the space they consume is constant and does not depend on the size of the input string. These variables include `max_length`, `max_count`, `start`, and `char_count`. Therefore, the space complexity is O(1) or constant space.\n",
-        "\n",
-        "**In summary**, the time complexity of the code is O(n), where n is the length of the input string, and the space complexity is O(1), as it uses a constant amount of additional space regardless of the input size. This algorithm efficiently solves the problem with a linear time complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the code to find the longest substring containing the same letter with at most k replacements in O(n) time complexity and O(1) space complexity. Hint: You may need to update the approach to achieve this.\n",
-        "2. **Variation with Lowercase Letters:** Extend the problem to include both uppercase and lowercase English letters in the input string s. Write a function that can handle this extended input and still find the longest substring with at most k replacements efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb b/docs/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb
deleted file mode 100644
index 33c2a44..0000000
--- a/docs/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb	
+++ /dev/null
@@ -1,273 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 76. Minimum Window Substring\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Minimum Window Substring problem on LeetCode, click here!](https://leetcode.com/problems/minimum-window-substring/)\n",
-        "\n",
-        "---\n",
-        "Given two strings `s` and `t` of lengths `m` and `n` respectively, return *the **minimum window substring** of `s` such that every character in `t` (**including duplicates**) is included in the window. If there is no such substring, return the empty string `\"\"`.*\n",
-        "\n",
-        "The testcases will be generated such that the answer is **unique**.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `m == s.length`\n",
-        "- `n == t.length`\n",
-        "- 1 <= `m, n` <= $10^5$\n",
-        "- `s` and `t` consist of uppercase and lowercase English letters.\n",
-        "\n",
-        "**Follow up**: Could you find an algorithm that runs in $O(m + n)$ time?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 54,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def min_window_substring(s, t):\n",
-        "    if not s or not t:\n",
-        "        return \"\"\n",
-        "\n",
-        "    # Initialize dictionaries to keep track of character counts for t and the current window in s.\n",
-        "    t_dict = {}\n",
-        "    current_window_dict = {}\n",
-        "    \n",
-        "    # Populate t_dict with character counts for string t.\n",
-        "    for char in t:\n",
-        "        t_dict[char] = t_dict.get(char, 0) + 1\n",
-        "\n",
-        "    # Initialize pointers for the sliding window.\n",
-        "    left = 0\n",
-        "    min_len = float('inf')\n",
-        "    min_window = \"\"\n",
-        "    required_chars = len(t_dict)\n",
-        "\n",
-        "    # Initialize a variable to keep track of how many required characters have been found in the current window.\n",
-        "    found_chars = 0\n",
-        "\n",
-        "    # Iterate over the string s using the right pointer.\n",
-        "    for right in range(len(s)):\n",
-        "        char = s[right]\n",
-        "\n",
-        "        # Update the current_window_dict.\n",
-        "        current_window_dict[char] = current_window_dict.get(char, 0) + 1\n",
-        "\n",
-        "        # Check if the current character is a required character and if its count matches the required count.\n",
-        "        if char in t_dict and current_window_dict[char] == t_dict[char]:\n",
-        "            found_chars += 1\n",
-        "\n",
-        "        # Try to minimize the window by moving the left pointer.\n",
-        "        while left <= right and found_chars == required_chars:\n",
-        "            # Calculate the current window size.\n",
-        "            window_size = right - left + 1\n",
-        "\n",
-        "            # If the current window is smaller than the minimum found so far, update min_len and min_window.\n",
-        "            if window_size < min_len:\n",
-        "                min_len = window_size\n",
-        "                min_window = s[left:right+1]\n",
-        "\n",
-        "            # Move the left pointer to the right to shrink the window.\n",
-        "            left_char = s[left]\n",
-        "            current_window_dict[left_char] -= 1\n",
-        "\n",
-        "            # Check if the character removed from the window was a required character.\n",
-        "            if left_char in t_dict and current_window_dict[left_char] < t_dict[left_char]:\n",
-        "                found_chars -= 1\n",
-        "\n",
-        "            # Move the left pointer further to continue shrinking the window.\n",
-        "            left += 1\n",
-        "\n",
-        "    return min_window"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by checking if either of the input strings `s` or `t` is empty. If either of them is empty, we return an empty string since there can't be any valid substring in this case.\n",
-        "\n",
-        "2. We initialize two dictionaries: `t_dict` and `current_window_dict`. These dictionaries will be used to keep track of character counts in the string `t` and the current window in string `s`, respectively.\n",
-        "\n",
-        "3. We populate the `t_dict` dictionary by iterating through string `t`. For each character, we increment its count in the dictionary using `t_dict.get(char, 0) + 1`. This allows us to count the occurrences of each character in `t`.\n",
-        "\n",
-        "4. We initialize two pointers: `left` and `right`. The `left` pointer will represent the start of the current window, and the `right` pointer will represent the end of the current window. We also initialize `min_len` to store the length of the minimum window found so far, and `min_window` to store the actual minimum window substring.\n",
-        "\n",
-        "5. We determine the number of required characters in `t` (i.e., the number of distinct characters in `t`) and store it in the variable `required_chars`.\n",
-        "\n",
-        "6. We initialize a variable `found_chars` to keep track of how many required characters have been found in the current window. Initially, it is set to 0.\n",
-        "\n",
-        "7. We iterate over the string `s` using the `right` pointer. In each iteration, we do the following:\n",
-        "\n",
-        "   - Update the `current_window_dict` by incrementing the count of the current character.\n",
-        "   \n",
-        "   - Check if the current character is a required character (present in `t_dict`) and if its count in the `current_window_dict` matches the required count from `t_dict`. If so, we increment `found_chars`.\n",
-        "\n",
-        "8. After updating the window, we attempt to minimize the window size by moving the `left` pointer to the right. In this step, we:\n",
-        "\n",
-        "   - Calculate the size of the current window.\n",
-        "   \n",
-        "   - If the current window size is smaller than the minimum found so far (`min_len`), we update `min_len` and `min_window` to store the current window substring.\n",
-        "   \n",
-        "   - Move the `left` pointer to the right to shrink the window.\n",
-        "   \n",
-        "   - Check if the character being removed from the window was a required character. If it was, we decrement `found_chars`.\n",
-        "   \n",
-        "   - Continue moving the `left` pointer further to continue shrinking the window if the window still contains all required characters.\n",
-        "\n",
-        "9. Finally, we return `min_window`, which will contain the minimum window substring that contains all characters from `t`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 55,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "BANC\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(min_window_substring(\"ADOBECODEBANC\", \"ABC\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 56,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "a\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(min_window_substring(\"a\", \"a\")) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 57,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": []
-        },
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(min_window_substring(\"a\", \"aa\")) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `min_window_substring` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The main loop iterates through the string `s` from left to right using the `right` pointer. This loop runs in O(m) time, where 'm' is the length of string `s`.\n",
-        "\n",
-        "- Inside the loop, we have a while loop that moves the `left` pointer to the right to shrink the window as needed. In the worst case, this while loop can also run in O(m) time because in the worst case, we may have to move the `left` pointer all the way to the end of the string.\n",
-        "\n",
-        "- Within each iteration of the while loop, we perform constant time operations, such as updating dictionaries and comparing character counts.\n",
-        "\n",
-        "- The initialization of the `t_dict` dictionary takes O(n) time, where 'n' is the length of string `t`.\n",
-        "\n",
-        "Therefore, the overall time complexity of the function is O(m + n) because the dominant factor is the length of string `s`, and the length of string `t` has a smaller impact.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity is determined by the space used by the `t_dict` dictionary, the `current_window_dict` dictionary, and a few variables.\n",
-        "\n",
-        "- The `t_dict` dictionary stores character counts for string `t`. In the worst case, when all characters in `t` are distinct, this dictionary can have a maximum of 'n' key-value pairs. So, the space complexity for `t_dict` is O(n).\n",
-        "\n",
-        "- The `current_window_dict` dictionary stores character counts for the current window. In the worst case, it can have a maximum of 'm' key-value pairs. So, the space complexity for `current_window_dict` is also O(m).\n",
-        "\n",
-        "- Other variables used in the function, such as `left`, `right`, `min_len`, `min_window`, `required_chars`, and `found_chars`, are all of constant size and do not depend on the input sizes.\n",
-        "\n",
-        "**In summary**, the overall space complexity of the function is O(max(m, n)), which means it is determined by the larger of the two input strings' lengths. This is because the space used for `t_dict` and `current_window_dict` dominates the space complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Smallest Distinct Substring:** Instead of finding the minimum window substring containing all characters, find the smallest distinct (unique) substring of s that contains all characters from t. This variation adds complexity because you must find a substring with distinct characters.\n",
-        "2. **No Extra Space:** Solve the problem without using any extra space, such as dictionaries or arrays, other than a constant amount of space. This is a significant optimization challenge."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/03. Sliding Window/README.md b/docs/_build/html/_sources/03. Sliding Window/README.md
deleted file mode 100644
index e1294ff..0000000
--- a/docs/_build/html/_sources/03. Sliding Window/README.md	
+++ /dev/null
@@ -1,19 +0,0 @@
-# Sliding Window Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [121. Best Time to Buy And Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | Easy       |
-| [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) | Medium |
-| [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) | Medium |
-| [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)             | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb b/docs/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb
deleted file mode 100644
index c47292f..0000000
--- a/docs/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 20. Valid Parentheses\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Parentheses problem on LeetCode, click here!](https://leetcode.com/problems/valid-parentheses/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given a string s containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.\n",
-        "\n",
-        "An input string is valid if:\n",
-        "\n",
-        "1. Open brackets must be closed by the same type of brackets.\n",
-        "2. Open brackets must be closed in the correct order.\n",
-        "3. Every close bracket has a corresponding open bracket of the same type.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $10^4$\n",
-        "- `s` consists of parentheses only `'()[]{}'`."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 49,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isValid(s):\n",
-        "    # Create an empty stack to store opening brackets\n",
-        "    stack = []\n",
-        "    \n",
-        "    # Define a mapping for matching brackets\n",
-        "    bracket_mapping = {')': '(', '}': '{', ']': '['}\n",
-        "    \n",
-        "    # Iterate through the characters in the input string\n",
-        "    for char in s:\n",
-        "        # If the character is an opening bracket, push it onto the stack\n",
-        "        if char in bracket_mapping.values():\n",
-        "            stack.append(char)\n",
-        "        # If the character is a closing bracket\n",
-        "        elif char in bracket_mapping.keys():\n",
-        "            # Pop the top element from the stack if it exists, or use a dummy value '#'\n",
-        "            top_element = stack.pop() if stack else '#'\n",
-        "            # If the popped element does not match the corresponding opening bracket, return False\n",
-        "            if bracket_mapping[char] != top_element:\n",
-        "                return False\n",
-        "        # If the character is not a bracket, ignore it\n",
-        "        \n",
-        "    # After processing the entire string, if there are any unmatched opening brackets left in the stack, return False\n",
-        "    return len(stack) == 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining the `isValid` function that takes a single argument `s`, which is the input string containing parentheses and brackets.\n",
-        "\n",
-        "2. Inside the function, we create an empty stack, which is a list in Python, to store opening brackets as we encounter them in the input string. The stack will help us keep track of the brackets and their order.\n",
-        "\n",
-        "3. We define a `bracket_mapping` dictionary that maps each closing bracket to its corresponding opening bracket. This mapping will be used to check if a closing bracket matches the most recent opening bracket in the stack.\n",
-        "\n",
-        "4. We iterate through each character `char` in the input string `s` using a `for` loop.\n",
-        "\n",
-        "5. If the character `char` is an opening bracket (i.e., it exists in the `values()` of `bracket_mapping`), we push it onto the stack using the `append()` method.\n",
-        "\n",
-        "6. If the character `char` is a closing bracket (i.e., it exists in the `keys()` of `bracket_mapping`), we need to check if it matches the corresponding opening bracket. To do this, we pop the top element from the stack (if it exists) and store it in `top_element`. We use a dummy value `'#'` if the stack is empty to avoid errors.\n",
-        "\n",
-        "7. We then compare `top_element` with the corresponding opening bracket using `bracket_mapping[char]`. If they do not match, it means the string is not valid, and we return `False`.\n",
-        "\n",
-        "8. If the character `char` is not a bracket, we simply ignore it and continue the loop.\n",
-        "\n",
-        "9. After processing the entire string, we check if there are any unmatched opening brackets left in the stack. If the stack is empty, it means all opening brackets have been properly matched and closed, and we return `True`. Otherwise, we return `False`.\n",
-        "\n",
-        "10. Finally, we provide some test cases at the bottom of the code to demonstrate how the function works for different input strings."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 50,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(isValid(\"()\"))    "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 51,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(isValid(\"()[]{}\")) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 52,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(isValid(\"(]\"))     "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `isValid` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The function iterates through each character in the input string `s` once, performing constant-time operations for each character.\n",
-        "- Pushing and popping elements from the stack (list) also takes constant time in most cases.\n",
-        "- Therefore, the overall time complexity of the function is $O(n)$, where $n$ is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the function is determined by the space used by the stack and the `bracket_mapping` dictionary.\n",
-        "- In the worst case, if the input string `s` consists entirely of opening brackets, the stack can potentially contain all of these opening brackets, resulting in a space complexity of $O(n)$ in terms of the stack.\n",
-        "- The `bracket_mapping` dictionary has a constant number of key-value pairs (3 pairs in this case).\n",
-        "- Therefore, the overall space complexity of the function is $O(n)$, where $n$ is the length of the input string `s`, mainly due to the stack space.\n",
-        "\n",
-        "**In summary:**\n",
-        "- The time complexity is $O(n)$ because we iterate through the string once.\n",
-        "- The space complexity is $O(n)$ due to the stack used to keep track of opening brackets."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Valid Expressions:** Modify the problem to validate not only brackets but also arithmetic expressions containing parentheses, such as \"$2 * (3 + 5) / (4 - 2)$\".\n",
-        "2. **Valid Parentheses Combinations:**\n",
-        "Write a function to generate all valid combinations of n pairs of parentheses, where $n$ is a positive integer. For example, for $n = 3$, the valid combinations are [\"((()))\", \"(()())\", \"(())()\", \"()(())\", \"()()()\"]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/04. Stack/README.md b/docs/_build/html/_sources/04. Stack/README.md
deleted file mode 100644
index 2ed115b..0000000
--- a/docs/_build/html/_sources/04. Stack/README.md	
+++ /dev/null
@@ -1,16 +0,0 @@
-# Stack Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)                            | Easy       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb b/docs/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb
deleted file mode 100644
index 98f2331..0000000
--- a/docs/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,228 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 153. Find Minimum in Rotated Sorted Array\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Find Minimum in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:\n",
-        "\n",
-        "- `[4,5,6,7,0,1,2]` if it was rotated `4` times.\n",
-        "- `[0,1,2,4,5,6,7]` if it was rotated `7` times.\n",
-        "\n",
-        "Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.\n",
-        "\n",
-        "Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array.*\n",
-        "\n",
-        "You must write an algorithm that runs in $O(log\\ n)$ time.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `n == nums.length`\n",
-        "- `1 <= n <= 5000`\n",
-        "- `-5000 <= nums[i] <= 5000`\n",
-        "- All the integers of `nums` are **unique**.\n",
-        "- `nums` is sorted and rotated between `1` and `n` times."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 39,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def findMin(nums):\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "    \n",
-        "    while left < right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "        \n",
-        "        # If the mid element is greater than the rightmost element,\n",
-        "        # it means the minimum element is in the right half.\n",
-        "        if nums[mid] > nums[right]:\n",
-        "            left = mid + 1\n",
-        "        # If the mid element is less than or equal to the rightmost element,\n",
-        "        # it means the minimum element is in the left half or at the mid itself.\n",
-        "        else:\n",
-        "            right = mid\n",
-        "    \n",
-        "    # The loop will break when left and right converge to the minimum element.\n",
-        "    # At this point, left (or right) will be pointing to the minimum element.\n",
-        "    return nums[left]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The function `findMin(nums)` takes an input list `nums`, which represents the rotated sorted array.\n",
-        "\n",
-        "2. Initialize two pointers `left` and `right` to track the current search range. Initially, `left` is set to 0 (the beginning of the array), and `right` is set to `len(nums) - 1` (the end of the array).\n",
-        "\n",
-        "3. Enter a `while` loop with the condition `left < right`. This loop continues until `left` and `right` converge to a single element, which will be the minimum element in the rotated array.\n",
-        "\n",
-        "4. Inside the loop, calculate the middle index `mid` using integer division. This helps in finding the middle element of the current search range.\n",
-        "\n",
-        "5. Check if the element at the middle index (`nums[mid]`) is greater than the element at the rightmost index (`nums[right]`). If this condition is true, it means that the minimum element must be in the right half of the current search range. So, update `left` to `mid + 1`, effectively eliminating the left half of the search range.\n",
-        "\n",
-        "6. If the condition from step 5 is not met, it implies that the minimum element can be in the left half of the current search range or could be the element at the `mid` index itself. In this case, update `right` to `mid`, effectively eliminating the right half of the search range.\n",
-        "\n",
-        "7. Repeat steps 4-6 until `left` and `right` converge to the minimum element.\n",
-        "\n",
-        "8. When the loop exits, it means that `left` (or `right`) points to the minimum element in the array.\n",
-        "\n",
-        "9. Return `nums[left]` (or `nums[right]`) as the result, which is the minimum element in the rotated sorted array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 43,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1:\n",
-        "nums1 = [3, 4, 5, 1, 2]\n",
-        "print(findMin(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 41,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "0\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "print(findMin(nums2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 42,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "11\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "nums3 = [11, 13, 15, 17]\n",
-        "print(findMin(nums3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this code is $O(log\\ n)$, where '$n$' is the length of the input array `nums`. This is because the binary search algorithm is employed, which continually divides the search range in half with each iteration.\n",
-        "\n",
-        "In each iteration of the `while` loop, the search range is halved, and this process continues until the `left` and `right` pointers converge to the minimum element. In the worst case, it will take logarithmic time to reduce the search range to a single element.\n",
-        "\n",
-        "Therefore, the binary search used in this code runs in $O(log\\ n)$ time complexity.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of this code is $O(1)$, which means it uses a constant amount of additional space that does not depend on the size of the input array.\n",
-        "\n",
-        "The algorithm uses a fixed number of variables (`left`, `right`, `mid`) to keep track of the search range and indices, but the number of these variables does not grow with the size of the input array. Hence, the space complexity is constant.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: $O(log\\ n)$\n",
-        "- Space Complexity: $O(1)$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. Solve the problem with a variation where the input array may contain duplicates.\n",
-        "\n",
-        "2. Modify the problem to return the index of the minimum element in the rotated sorted array, rather than the element itself. The algorithm should still run in $O(log\\ n)$ time.\n",
-        "\n",
-        "3. Implement a function that finds the maximum element in a rotated sorted array. How would you adapt the binary search algorithm to solve this problem efficiently in $O(log\\ n)$ time?\n",
-        "\n",
-        "4. Implement a function that can find the kth smallest element in a rotated sorted array. This is an extension of the original problem, and the algorithm should still run in $O(log\\ n)$ time."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb b/docs/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb
deleted file mode 100644
index d8d4202..0000000
--- a/docs/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 33. Search in Rotated Sorted Array\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Search in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/search-in-rotated-sorted-array/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "There is an integer array `nums` sorted in ascending order (with **distinct** values).\n",
-        "\n",
-        "Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k (1 <= k < nums.length)` such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index 3 and become `[4,5,6,7,0,1,2]`.\n",
-        "\n",
-        "Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.*\n",
-        "\n",
-        "You must write an algorithm with $O(log\\ n)$ runtime complexity.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `1 <= nums.length <= 5000`\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- All values of `nums` are **unique**.\n",
-        "- `nums` is an ascending array that is possibly rotated.\n",
-        "- $-10^4$ <= `target` <= $10^4$"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 44,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def search(nums, target):\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "\n",
-        "    # Find the pivot index using binary search\n",
-        "    while left < right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "\n",
-        "        if nums[mid] > nums[right]:\n",
-        "            left = mid + 1\n",
-        "        else:\n",
-        "            right = mid\n",
-        "\n",
-        "    pivot = left\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "\n",
-        "    # Determine which part of the array to search in\n",
-        "    if target >= nums[pivot] and target <= nums[right]:\n",
-        "        left = pivot\n",
-        "    else:\n",
-        "        right = pivot\n",
-        "\n",
-        "    # Perform binary search to find the target\n",
-        "    while left <= right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "\n",
-        "        if nums[mid] == target:\n",
-        "            return mid\n",
-        "        elif nums[mid] < target:\n",
-        "            left = mid + 1\n",
-        "        else:\n",
-        "            right = mid - 1\n",
-        "\n",
-        "    return -1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `search` function takes two arguments: `nums`, which is the rotated sorted array, and `target`, which is the value we want to find in the array.\n",
-        "\n",
-        "2. We initialize two pointers, `left` and `right`, which represent the range in which we are going to perform binary search. Initially, `left` is set to 0, and `right` is set to the index of the last element in the array (`len(nums) - 1`).\n",
-        "\n",
-        "3. We start by finding the pivot index using binary search. The pivot index is the index at which the array is rotated. We use a `while` loop to continue the search until `left` is less than `right`.\n",
-        "\n",
-        "4. Inside the loop, we calculate the middle index `mid` using integer division. We then compare the value at `mid` with the value at `right`. If `nums[mid]` is greater than `nums[right]`, it means the pivot point lies to the right of `mid`, so we update `left` to `mid + 1`. Otherwise, the pivot point lies to the left of `mid`, so we update `right` to `mid`. This process continues until we find the pivot index.\n",
-        "\n",
-        "5. After finding the pivot index, we have divided the array into two parts: one part is sorted in ascending order, and the other part is also sorted in ascending order but rotated. We need to determine which part of the array contains the target value.\n",
-        "\n",
-        "6. We reset `left` and `right` pointers. If the target value is within the range `[nums[pivot], nums[right]]`, we set `left` to `pivot` (the start of the rotated part), indicating that we should search in the rotated part. Otherwise, we set `right` to `pivot` (the end of the sorted part), indicating that we should search in the sorted part.\n",
-        "\n",
-        "7. We then perform binary search again within the chosen range (`left` to `right`) to find the target element. The binary search continues until `left` is less than or equal to `right`.\n",
-        "\n",
-        "8. Inside the binary search loop, we calculate the middle index `mid` and compare `nums[mid]` with the target value. If they are equal, we return `mid` as the index of the target element. If `nums[mid]` is less than the target, we update `left` to `mid + 1` to search in the right half. If `nums[mid]` is greater than the target, we update `right` to `mid - 1` to search in the left half.\n",
-        "\n",
-        "9. If we exit the binary search loop without finding the target, we return -1 to indicate that the target is not present in the array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 45,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "nums1 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "target1 = 0\n",
-        "result1 = search(nums1, target1)\n",
-        "print(result1) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 46,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "-1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "target2 = 3\n",
-        "result2 = search(nums2, target2)\n",
-        "print(result2) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 47,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "-1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "nums3 = [1]\n",
-        "target3 = 0\n",
-        "result3 = search(nums3, target3)\n",
-        "print(result3) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. Finding the pivot index using binary search takes $O(log\\ n)$ time, where '$n$' is the number of elements in the array.\n",
-        "2. After finding the pivot index, performing binary search to find the target element also takes $O(log\\ n)$ time in the worst case.\n",
-        "\n",
-        "The dominant factor in the time complexity is the binary search, and since we perform two binary searches sequentially, the overall time complexity of the code is $O(log\\ n)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is very minimal:\n",
-        "\n",
-        "1. We use a constant amount of additional space for variables such as `left`, `right`, `pivot`, and `mid`. These variables do not depend on the input size, so they contribute $O(1)$ space complexity.\n",
-        "2. The function does not use any additional data structures that scale with the input size.\n",
-        "\n",
-        "Therefore, the space complexity of the code is $O(1)$, which means it has a constant space complexity and does not depend on the size of the input array.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: $O(log\\ n)$\n",
-        "- Space Complexity: $O(1)$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Find Minimum Element in Rotated Sorted Array**: Write an algorithm to find the minimum element in a rotated sorted array. This is a variation of the problem where you don't need to search for a target value but instead find the smallest element.\n",
-        "\n",
-        "2. **Search in a Circularly Sorted Array**: Consider an array that is sorted in a circular manner (e.g., [4, 5, 6, 7, 0, 1, 2, 3]). Adapt the search algorithm to work for circularly sorted arrays while maintaining $O(log\\ n)$ complexity."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/05. Binary Search/README.md b/docs/_build/html/_sources/05. Binary Search/README.md
deleted file mode 100644
index 6086b9b..0000000
--- a/docs/_build/html/_sources/05. Binary Search/README.md	
+++ /dev/null
@@ -1,17 +0,0 @@
-# Binary search Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [153. Find Minimum In Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) | Medium |
-| [33. Search In Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)   | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb b/docs/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb
deleted file mode 100644
index 0336bc8..0000000
--- a/docs/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb	
+++ /dev/null
@@ -1,271 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 141. Linked List Cycle\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Linked List Cycle problem on LeetCode, click here!](https://leetcode.com/problems/linked-list-cycle/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given `head`, the head of a linked list, determine if the linked list has a cycle in it.\n",
-        "\n",
-        "There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is connected to. **Note that `pos` is not passed as a parameter.**\n",
-        "\n",
-        "Return `true` if there is a cycle in the linked list. Otherwise, return `false`.\n",
-        "\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of the nodes in the list is in the range $[0, 10^4]$.\n",
-        "- $-10^5$ <= Node.val <= $10^5$\n",
-        "- `pos` is `-1` or a **valid index** in the linked-list.\n",
-        "\n",
-        "**Follow up:** Can you solve it using $O(1)$ (i.e. constant) memory?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "    def __init__(self, val=0):\n",
-        "        self.val = val\n",
-        "        self.next = None\n",
-        "\n",
-        "class Solution:\n",
-        "    def hasCycle(self, head: ListNode) -> bool:\n",
-        "        if not head:\n",
-        "            return False\n",
-        "        \n",
-        "        slow = head\n",
-        "        fast = head\n",
-        "        \n",
-        "        while fast and fast.next:\n",
-        "            slow = slow.next\n",
-        "            fast = fast.next.next\n",
-        "            \n",
-        "            if slow == fast:\n",
-        "                return True\n",
-        "        \n",
-        "        return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The given code defines a class `Solution` with a method `hasCycle` for determining if a linked list has a cycle.\n",
-        "\n",
-        "The `ListNode` class represents nodes in the linked list, where each node has a `val` (a value associated with the node) and a `next` attribute (a reference to the next node in the list).\n",
-        "\n",
-        "The `hasCycle` method takes the `head` of a linked list as input and returns `True` if there is a cycle in the linked list, and `False` otherwise. It uses two pointers, `slow` and `fast`, to traverse the linked list. If there is a cycle, the `fast` pointer will eventually catch up to the `slow` pointer.\n",
-        "\n",
-        "The algorithm works as follows:\n",
-        "1. It checks if the input `head` is `None` (an empty list). If it is, the method returns `False` because an empty list can't have a cycle.\n",
-        "2. Two pointers, `slow` and `fast`, initially point to the `head` of the list.\n",
-        "3. The code enters a `while` loop where `fast` moves two steps at a time, and `slow` moves one step at a time.\n",
-        "4. If there is a cycle, `fast` will eventually catch up to `slow`, and the method returns `True`.\n",
-        "5. If the loop completes without finding a cycle, the method returns `False`."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list with a cycle\n",
-        "def create_linked_list_with_cycle(values, pos):\n",
-        "    dummy = ListNode()\n",
-        "    current = dummy\n",
-        "    cycle_node = None\n",
-        "    \n",
-        "    for i, val in enumerate(values):\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "        \n",
-        "        if i == pos:\n",
-        "            cycle_node = current\n",
-        "    \n",
-        "    if cycle_node:\n",
-        "        current.next = cycle_node\n",
-        "    \n",
-        "    return dummy.next"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "This helper function is designed to create a linked list with a cycle for testing purposes. It takes two arguments:\n",
-        "\n",
-        "1. `values`: A list of values representing the nodes in the linked list.\n",
-        "2. `pos`: An integer representing the index of the node where the cycle begins.\n",
-        "\n",
-        "Here's how the function works:\n",
-        "\n",
-        "- It starts by creating an empty `ListNode` called `dummy`. This `dummy` node is used to simplify the creation of the linked list.\n",
-        "- It initializes a `current` pointer to the `dummy` node. This pointer will be used to traverse the linked list.\n",
-        "- It initializes a `cycle_node` variable to `None`. This variable will hold the reference to the node where the cycle begins, if any.\n",
-        "- It then iterates through the `values` list, creating a new `ListNode` for each value and appending it to the linked list.\n",
-        "- Inside the loop, it checks if the current index `i` is equal to the specified `pos`. If they match, it sets `cycle_node` to the current node. This simulates the creation of a cycle in the linked list.\n",
-        "- After the loop, if `cycle_node` is not `None` (indicating a cycle should be created), it connects the last node in the linked list to `cycle_node`, effectively creating a cycle.\n",
-        "- Finally, it returns the reference to the first node of the linked list (not the `dummy` node).\n",
-        "\n",
-        "This helper function allows you to easily create test cases where you can specify the values for the linked list and the position where the cycle starts. It's particularly useful for testing the code's ability to detect cycles in linked lists."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "head = [3,2,0,-4]\n",
-        "pos = 1\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "head = [1,2]\n",
-        "pos = 0\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 22,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "head = [1]\n",
-        "pos = -1\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's discuss the time and space complexity of the code for detecting a cycle in a linked list:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this code is $O(n)$, where \"$n$\" is the number of nodes in the linked list.\n",
-        "\n",
-        "The reason for this is that both the `slow` and `fast` pointers traverse the linked list, and they move at different speeds. In the worst case, when there is no cycle, the `fast` pointer will reach the end of the list after going through approximately $n/2$ nodes. In the case of a cycle, it may take some extra iterations for the `fast` pointer to catch up to the `slow` pointer. However, the total number of iterations is still proportional to the number of nodes in the linked list, making it $O(n)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this code is $O(1)$, which means it uses constant space.\n",
-        "\n",
-        "The reason for this is that regardless of the size of the linked list, the code only uses two additional pointers (`slow` and `fast`) to traverse the list. These pointers do not depend on the size of the input linked list, so the space complexity remains constant.\n",
-        "\n",
-        "**In summary**, this code efficiently detects cycles in a linked list with a time complexity of $O(n)$ and a space complexity of $O(1)$. It is an example of an algorithm that solves a complex problem with minimal memory usage."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/06. Linked List/143. Reorder List.ipynb b/docs/_build/html/_sources/06. Linked List/143. Reorder List.ipynb
deleted file mode 100644
index a83545b..0000000
--- a/docs/_build/html/_sources/06. Linked List/143. Reorder List.ipynb	
+++ /dev/null
@@ -1,285 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 143. Reorder List\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Reorder List problem on LeetCode, click here!](https://leetcode.com/problems/reorder-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given the head of a singly linked-list. The list can be represented as:\n",
-        "\n",
-        "$L_0 → L_1 → … → L_{n - 1} → L_n$\n",
-        "\n",
-        "Reorder the list to be on the following form:\n",
-        "\n",
-        "$L_0 → L_n → L_1 → L_{n - 1} → L_2 → L_{n - 2} → …$\n",
-        "\n",
-        "You may not modify the values in the list's nodes. Only nodes themselves may be changed.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is in the range $[1, 5 * 10^4]$.\n",
-        "- `1 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 28,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def reorderList(head):\n",
-        "    if not head or not head.next:\n",
-        "        return head\n",
-        "\n",
-        "    # Step 1: Find the middle of the linked list\n",
-        "    slow, fast = head, head\n",
-        "    while fast.next and fast.next.next:\n",
-        "        slow = slow.next\n",
-        "        fast = fast.next.next\n",
-        "\n",
-        "    # Split the list into two halves\n",
-        "    first_half = head\n",
-        "    second_half = slow.next\n",
-        "    slow.next = None\n",
-        "\n",
-        "    # Step 2: Reverse the second half of the linked list\n",
-        "    prev = None\n",
-        "    current = second_half\n",
-        "    while current:\n",
-        "        next_node = current.next\n",
-        "        current.next = prev\n",
-        "        prev = current\n",
-        "        current = next_node\n",
-        "    second_half = prev\n",
-        "\n",
-        "    # Step 3: Merge the first and reversed second halves alternately\n",
-        "    p1, p2 = first_half, second_half\n",
-        "    while p2:\n",
-        "        next_p1 = p1.next\n",
-        "        next_p2 = p2.next\n",
-        "        p1.next = p2\n",
-        "        p2.next = next_p1\n",
-        "        p1 = next_p1\n",
-        "        p2 = next_p2\n",
-        "\n",
-        "    return head"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `reorderList` function is designed to reorder a singly linked list following the specific pattern described in the problem statement. Here's a detailed explanation of how the function works:\n",
-        "\n",
-        "1. **Base Cases Handling**:\n",
-        "   - The function first checks if the input linked list is empty (`not head`) or contains only one element (`not head.next`). If either of these conditions is met, the list cannot be reordered, so the function returns the original list as-is.\n",
-        "\n",
-        "2. **Finding the Middle of the Linked List**:\n",
-        "   - To reorder the list, we first need to find the middle point so that we can split the list into two halves. This is done using two pointers, `slow` and `fast`, initialized to the head of the list.\n",
-        "   - `slow` moves one step at a time while `fast` moves two steps at a time. When `fast` reaches the end of the list or the second-to-last node, `slow` will be at the middle node.\n",
-        "\n",
-        "3. **Splitting the List**:\n",
-        "   - After finding the middle node, we split the list into two halves:\n",
-        "     - The first half, `first_half`, contains nodes from the beginning up to the middle.\n",
-        "     - The second half, `second_half`, contains nodes from the middle to the end.\n",
-        "   - To split the list, we set the `next` pointer of the node before the middle node to `None`.\n",
-        "\n",
-        "4. **Reversing the Second Half**:\n",
-        "   - We reverse the second half of the linked list using the `prev`, `current`, and `next_node` pointers.\n",
-        "   - `prev` is initially `None`, and we iterate through the second half. For each node, we:\n",
-        "     - Set the `next` of the current node to `prev`, effectively reversing the next pointer direction.\n",
-        "     - Update `prev` to the current node.\n",
-        "     - Move to the next node using the `next_node`.\n",
-        "\n",
-        "5. **Merging the Two Halves Alternately**:\n",
-        "   - We now have two linked lists: `first_half` and the reversed `second_half`.\n",
-        "   - We merge these two lists alternately by adjusting the `next` pointers of the nodes.\n",
-        "   - `p1` and `p2` are pointers to the current nodes in `first_half` and `second_half`, respectively.\n",
-        "   - We iterate through both lists while reordering nodes:\n",
-        "     - Set the `next` of `p1` to `p2` to link a node from the first half to a node from the reversed second half.\n",
-        "     - Update `p1` and `p2` to their respective next nodes.\n",
-        "     - Repeat this process until we have processed all nodes in both halves.\n",
-        "\n",
-        "6. **Returning the Reordered List**:\n",
-        "   - After the merging process is complete, the linked list is reordered as specified.\n",
-        "   - The function returns the `head` of the reordered list.\n",
-        "\n",
-        "The overall result is a singly linked list that has been reordered according to the given pattern. The time complexity of this algorithm is O(N), where N is the number of nodes in the linked list, as we traverse the list once to find the middle, once to reverse the second half, and once to merge the two halves."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 29,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to print the linked list\n",
-        "def printLinkedList(head):\n",
-        "    result = []\n",
-        "    while head:\n",
-        "        result.append(head.val)\n",
-        "        head = head.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the helper functions:\n",
-        "**printLinkedList Function**:\n",
-        "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-        "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-        "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-        "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-        "\n",
-        "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 30,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 4, 2, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "head1 = ListNode(1, ListNode(2, ListNode(3, ListNode(4))))\n",
-        "reorderList(head1)\n",
-        "print(printLinkedList(head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 5, 2, 4, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "head2 = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))\n",
-        "reorderList(head2)\n",
-        "print(printLinkedList(head2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `reorderList` function:\n",
-        "\n",
-        "**Time Complexity**:\n",
-        "1. **Finding the Middle of the Linked List**: In this step, we use two pointers, one moving one step at a time (`slow`) and the other moving two steps at a time (`fast`). This process takes $O(N/2)$ time, where $N$ is the number of nodes in the linked list.\n",
-        "\n",
-        "2. **Splitting the List**: After finding the middle, we split the list into two halves by setting the `next` pointer of the node before the middle node to `None`. This step takes $O(1)$ time.\n",
-        "\n",
-        "3. **Reversing the Second Half**: We reverse the second half of the linked list using a while loop that iterates through the second half of the list once. Therefore, this step also takes $O(N/2)$ time.\n",
-        "\n",
-        "4. **Merging the Two Halves Alternately**: In this step, we merge the two halves alternately by adjusting the `next` pointers of the nodes. We iterate through both halves once, so this step takes $O(N/2)$ time.\n",
-        "\n",
-        "Overall, the time complexity of the `reorderList` function is dominated by the steps involving reversing and merging the two halves, both of which take $O(N/2)$ time. Therefore, the total time complexity is $O(N)$.\n",
-        "\n",
-        "**Space Complexity**:\n",
-        "The space complexity of the `reorderList` function is primarily determined by the variables and data structures used within the function. Let's break down the space complexity components:\n",
-        "\n",
-        "1. **Constant Space Variables**: The variables `slow`, `fast`, `prev`, `current`, `next_node`, `p1`, and `p2` are used to traverse and manipulate the linked list. These variables occupy constant space, regardless of the input size. Therefore, they contribute $O(1)$ to the space complexity.\n",
-        "\n",
-        "2. **Split Linked Lists**: In the splitting step, we create two new linked list segments (`first_half` and `second_half`) that store references to nodes from the original linked list. These segments occupy space proportional to half of the input list, $O(N/2)$. \n",
-        "\n",
-        "3. **Reversed Second Half**: During the reversal step, we reverse the second half of the linked list in-place without creating any additional data structures. Therefore, it doesn't contribute to additional space complexity.\n",
-        "\n",
-        "4. **Overall**: Combining the above components, the total space complexity is $O(N/2)$, which simplifies to $O(N)$ in terms of space complexity.\n",
-        "\n",
-        "**In summary**, the `reorderList` function has a time complexity of $O(N)$ and a space complexity of $O(N)$ due to the creation of two new linked list segments while splitting the list. The constant space variables used for traversal do not significantly impact the overall space complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Reorder by K Elements:** Generalize the reorderList function to reorder the list in a different pattern, where you reorder the list by $K$ elements at a time. For example, for $K=3$, you would reorder it as ($L_0 → L_1 → L_2 → L_n → L_{n-1} → L_{n-2} → ...$).\n",
-        "You may need to handle cases where the number of nodes is not a multiple of $K$.\n",
-        "\n",
-        "2. **Reorder by Odd and Even Nodes:** Modify the reorderList function to reorder the list in a pattern where odd-indexed nodes (1-based index) come before even-indexed nodes. For example, ($L_0 → L_n → L_1 → L_{n-1} → L_2 → L_{n-2} → ...$)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb b/docs/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb
deleted file mode 100644
index 263124e..0000000
--- a/docs/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb	
+++ /dev/null
@@ -1,278 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 19. Remove Nth Node From End of List\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Remove Nth Node From End of List problem on LeetCode, click here!](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the head of a linked list, remove the $n^{th}$ node from the end of the list and return its head.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is `sz`.\n",
-        "- `1 <= sz <= 30`\n",
-        "- `0 <= Node.val <= 100`\n",
-        "- `1 <= n <= sz`\n",
-        "\n",
-        "**Follow up:** Could you do this in one pass?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 23,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def removeNthFromEnd(head, n):\n",
-        "    # Create a dummy node to handle the case of removing the head node.\n",
-        "    dummy = ListNode(0)\n",
-        "    dummy.next = head\n",
-        "    first = dummy\n",
-        "    second = dummy\n",
-        "    \n",
-        "    # Advance the first pointer by n+1 nodes.\n",
-        "    for _ in range(n + 1):\n",
-        "        first = first.next\n",
-        "    \n",
-        "    # Move both pointers simultaneously until first reaches the end.\n",
-        "    while first is not None:\n",
-        "        first = first.next\n",
-        "        second = second.next\n",
-        "    \n",
-        "    # Remove the nth node by updating the next pointer of the previous node.\n",
-        "    second.next = second.next.next\n",
-        "    \n",
-        "    return dummy.next  # Return the modified head of the linked list"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a `ListNode` class to represent individual nodes in a linked list. Each node has two attributes: `val` (the value of the node) and `next` (a reference to the next node in the list).\n",
-        "\n",
-        "2. The `removeNthFromEnd` function takes the head of a linked list (`head`) and the value `n`, which represents the position from the end of the list of the node to be removed. It returns the modified head of the linked list after the removal.\n",
-        "\n",
-        "3. We create a dummy node (`dummy`) at the beginning of the list. This dummy node simplifies the code by handling cases where the head of the list needs to be removed.\n",
-        "\n",
-        "4. We initialize two pointers, `first` and `second`, both initially pointing to the dummy node.\n",
-        "\n",
-        "5. To advance the `first` pointer by `n+1` nodes, we use a `for` loop. This positions the `first` pointer `n` nodes ahead of the `second` pointer.\n",
-        "\n",
-        "6. We then move both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This ensures that the `second` pointer ends up pointing to the node that needs to be removed.\n",
-        "\n",
-        "7. To remove the nth node from the end, we update the `next` pointer of the node pointed to by the `second` pointer to skip over the node to be removed.\n",
-        "\n",
-        "8. Finally, we return `dummy.next`, which is the modified head of the linked list without the removed node.\n",
-        "\n",
-        "9. Additionally, there are two helper functions:\n",
-        "   - `createLinkedList(values)` creates a linked list from a list of values.\n",
-        "   - `convertToList(head)` converts a linked list back into a list of values."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list from a list of values.\n",
-        "def createLinkedList(values):\n",
-        "    if not values:\n",
-        "        return None\n",
-        "    head = ListNode(values[0])\n",
-        "    current = head\n",
-        "    for val in values[1:]:\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "    return head\n",
-        "\n",
-        "# Helper function to convert a linked list to a list of values.\n",
-        "def convertToList(head):\n",
-        "    result = []\n",
-        "    current = head\n",
-        "    while current:\n",
-        "        result.append(current.val)\n",
-        "        current = current.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the two helper functions:\n",
-        "\n",
-        "1. `createLinkedList(values)`: This function creates a linked list from a list of values.\n",
-        "\n",
-        "   - Input: `values` is a list containing values that you want to insert into the linked list.\n",
-        "   - Output: The function returns the head of the created linked list.\n",
-        "\n",
-        "   Explanation:\n",
-        "   - The function starts by checking if the `values` list is empty. If it is, it returns `None` to indicate an empty linked list.\n",
-        "   - If the `values` list is not empty, it creates the first node of the linked list with the value from the first element of `values`.\n",
-        "   - It then iterates through the remaining elements of `values` and creates new nodes for each value, connecting them together using the `next` attribute to form a linked list.\n",
-        "   - Finally, it returns the head of the linked list, which is the first node created.\n",
-        "   \n",
-        "\n",
-        "2. `convertToList(head)`: This function converts a linked list into a list of values.\n",
-        "\n",
-        "   - Input: `head` is the head node of the linked list that you want to convert.\n",
-        "   - Output: The function returns a list containing the values of the nodes in the linked list in order.\n",
-        "\n",
-        "   Explanation:\n",
-        "   - The function starts by initializing an empty list, `result`, to store the values of the linked list nodes.\n",
-        "   - It then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - The iteration continues until it reaches the end of the linked list (i.e., the `next` attribute of the current node becomes `None`).\n",
-        "   - After the iteration is complete, the function returns the `result` list, which now contains the values of all the nodes in the linked list in the same order as they appear in the linked list.\n",
-        "\n",
-        "These helper functions are useful for creating linked lists from lists of values and converting linked lists back into lists of values, making it easier to work with linked list examples and test cases in a more familiar list format."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 2, 3, 5]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "head1 = createLinkedList([1, 2, 3, 4, 5])\n",
-        "n1 = 2\n",
-        "new_head1 = removeNthFromEnd(head1, n1)\n",
-        "print(convertToList(new_head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 26,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "head2 = createLinkedList([1])\n",
-        "n2 = 1\n",
-        "new_head2 = removeNthFromEnd(head2, n2)\n",
-        "print(convertToList(new_head2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 27,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "head3 = createLinkedList([1, 2])\n",
-        "n3 = 1\n",
-        "new_head3 = removeNthFromEnd(head3, n3)\n",
-        "print(convertToList(new_head3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Certainly! Let's analyze the time and space complexity of the `removeNthFromEnd` function in isolation:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The `removeNthFromEnd` function performs two passes through the linked list. In the worst case, it needs to traverse the entire linked list of length $N$.\n",
-        "- The first pass positions the `first` pointer $n$ nodes ahead of the `second` pointer. This pass takes $O(N)$ time because it involves iterating through all N nodes.\n",
-        "- The second pass moves both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This also takes $O(N)$ time in the worst case.\n",
-        "- Therefore, the overall time complexity of the `removeNthFromEnd` function is $O(N)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The `removeNthFromEnd` function uses a constant amount of extra space for its variables (e.g., `first`, `second`, and `dummy`). This space usage does not depend on the size of the linked list.\n",
-        "- The space complexity is $O(1)$, indicating that the space used by the function is constant and independent of the size of the input linked list.\n",
-        "\n",
-        "**In summary**, the `removeNthFromEnd` function has a time complexity of $O(N)$ and a space complexity of $O(1)$. It efficiently removes the nth node from the end of the linked list with a single pass through the list and constant extra space usage."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb b/docs/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb
deleted file mode 100644
index c97de2d..0000000
--- a/docs/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb	
+++ /dev/null
@@ -1,316 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 206. Reverse Linked List\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Reverse Linked List problem on LeetCode, click here!](https://leetcode.com/problems/reverse-linked-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the `head` of a singly linked list, reverse the list, and return the reversed list.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is the range `[0, 5000]`.\n",
-        "- `-5000 <= Node.val <= 5000`\n",
-        "\n",
-        "**Follow up:** A linked list can be reversed either iteratively or recursively. Could you implement both?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 33,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "# Function to reverse a linked list iteratively\n",
-        "def reverseListIterative(head):\n",
-        "    prev = None\n",
-        "    current = head\n",
-        "    \n",
-        "    while current:\n",
-        "        next_node = current.next\n",
-        "        current.next = prev\n",
-        "        prev = current\n",
-        "        current = next_node\n",
-        "    \n",
-        "    return prev\n",
-        "\n",
-        "# Function to reverse a linked list recursively\n",
-        "def reverseListRecursive(head):\n",
-        "    if not head or not head.next:\n",
-        "        return head\n",
-        "\n",
-        "    new_head = reverseListRecursive(head.next)\n",
-        "    head.next.next = head\n",
-        "    head.next = None\n",
-        "\n",
-        "    return new_head"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `reverseListIterative(head)`: This function reverses a linked list iteratively using a loop. Here's a step-by-step explanation:\n",
-        "\n",
-        "   - It takes the `head` of the input linked list as a parameter.\n",
-        "   - It initializes two pointers, `prev` and `current`, initially set to `None` and the `head` of the list, respectively.\n",
-        "   - It enters a `while` loop that continues until `current` reaches the end of the list (i.e., becomes `None`).\n",
-        "   - Inside the loop:\n",
-        "     - It stores the `next_node` which is the next node after `current`.\n",
-        "     - It updates the `next` pointer of `current` to point to the `prev` node. This effectively reverses the link direction.\n",
-        "     - It moves `prev` to `current` and `current` to `next_node`, advancing the pointers one step further in the list.\n",
-        "   - Once the loop completes, `prev` will be pointing to the new head of the reversed list (which was the last node of the original list), so it returns `prev`.\n",
-        "\n",
-        "2. `reverseListRecursive(head)`: This function reverses a linked list recursively. Here's how it works:\n",
-        "\n",
-        "   - It takes the `head` of the input linked list as a parameter.\n",
-        "   - The base case is checked: if `head` is `None` or `head.next` is `None`, meaning the list is empty or has only one node, it returns `head` as there's no need to reverse a list with zero or one element.\n",
-        "   - In the recursive case:\n",
-        "     - It calls itself with `head.next`, effectively moving down the list until it reaches the end.\n",
-        "     - Once it reaches the end, it starts reversing the links:\n",
-        "       - `head.next.next` is set to `head`, reversing the link between `head` and the next node.\n",
-        "       - `head.next` is set to `None` to avoid cycles.\n",
-        "   - Finally, it returns the new head of the reversed list, which will be the last node of the original list.\n",
-        "\n",
-        "**In summary,** `reverseListIterative` reverses the linked list by iterating through it and changing the next pointers, while `reverseListRecursive` reverses the linked list by recursively reaching the end and then reversing the links on the way back up the recursion stack. Both functions achieve the same result: reversing the linked list."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 36,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list from a list of values\n",
-        "def createLinkedList(values):\n",
-        "    if not values:\n",
-        "        return None\n",
-        "    head = ListNode(values[0])\n",
-        "    current = head\n",
-        "    for val in values[1:]:\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "    return head\n",
-        "\n",
-        "# Helper function to convert a linked list to a list for testing\n",
-        "def linkedListToList(head):\n",
-        "    result = []\n",
-        "    current = head\n",
-        "    while current:\n",
-        "        result.append(current.val)\n",
-        "        current = current.next\n",
-        "    return result\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the two helper functions used in the provided code:\n",
-        "\n",
-        "1. `createLinkedList(values)`:\n",
-        "\n",
-        "   - Purpose: This function is used to create a linked list from a list of values. It takes a list of values as input and returns the head of the linked list.\n",
-        "\n",
-        "   - How it works:\n",
-        "     - It first checks if the input list `values` is empty. If it's empty, it returns `None` to indicate an empty linked list.\n",
-        "     - If `values` is not empty, it initializes a `head` node with the value of the first element in `values`.\n",
-        "     - It then iterates through the remaining values in `values`, creating new nodes for each value and linking them together to form a linked list.\n",
-        "     - Finally, it returns the `head` node of the newly created linked list.\n",
-        "\n",
-        "2. `linkedListToList(head)`:\n",
-        "\n",
-        "   - Purpose: This function is used to convert a linked list back into a Python list for testing and output purposes.\n",
-        "\n",
-        "   - How it works:\n",
-        "     - It takes the `head` of the linked list as input.\n",
-        "     - It initializes an empty list called `result`.\n",
-        "     - It then iterates through the linked list, starting from `head`, and appends each node's `val` (value) to the `result` list.\n",
-        "     - This process continues until the end of the linked list is reached.\n",
-        "     - Finally, it returns the `result` list, which contains the values from the linked list in the same order.\n",
-        "\n",
-        "These helper functions are used to simplify the process of creating linked lists from lists of values and converting linked lists back into lists, making it easier to work with linked lists in the provided examples and test cases."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 35,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[1, 1, 2, 3, 4, 4]"
-            ]
-          },
-          "execution_count": 35,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 1: Iterative\n",
-        "head1 = [1,2,3,4,5]\n",
-        "head1 = createLinkedList(head1)\n",
-        "reversed_head1 = reverseListIterative(head1)\n",
-        "print(linkedListToList(reversed_head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[]"
-            ]
-          },
-          "execution_count": 37,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 2: Recursive\n",
-        "head2 = [1,2]\n",
-        "head2 = createLinkedList(head2)\n",
-        "reversed_head2 = reverseListRecursive(head2)\n",
-        "print(linkedListToList(reversed_head2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 38,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[0]"
-            ]
-          },
-          "execution_count": 38,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 3: Empty list\n",
-        "head3 = []\n",
-        "head3 = createLinkedList(head3)\n",
-        "reversed_head3 = reverseListIterative(head3)\n",
-        "print(linkedListToList(reversed_head3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the two functions:\n",
-        "\n",
-        "1. `reverseListIterative(head)`:\n",
-        "\n",
-        "   - Time Complexity: $O(n)$\n",
-        "     - The iterative function visits each node in the linked list exactly once in a single pass, where \"$n$\" is the number of nodes in the list. Therefore, the time complexity is linear in the number of nodes.\n",
-        "\n",
-        "   - Space Complexity: $O(1)$\n",
-        "     - This function uses a constant amount of extra space for the `prev`, `current`, and `next_node` pointers. Regardless of the size of the input list, the amount of additional memory used remains the same, so the space complexity is constant.\n",
-        "\n",
-        "2. `reverseListRecursive(head)`:\n",
-        "\n",
-        "   - Time Complexity: $O(n)$\n",
-        "     - The recursive function also visits each node in the linked list exactly once. It recursively traverses the list from the head to the tail, reversing the links on the way back. Therefore, like the iterative approach, the time complexity is $O(n)$, where \"$n$\" is the number of nodes.\n",
-        "\n",
-        "   - Space Complexity: $O(n)$\n",
-        "     - The recursive function uses space on the call stack for each recursive call. In the worst case, when the list has \"$n$\" nodes, it will create \"$n$\" recursive function calls on the stack, resulting in a space complexity of $O(n)$. This is because each function call stores information about its state, including the `head` pointer, until it reaches the base case and starts returning.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "- The time complexity of both functions is $O(n)$ because they both visit each node once.\n",
-        "- The space complexity of `reverseListIterative` is $O(1)$ because it uses a constant amount of extra space.\n",
-        "- The space complexity of `reverseListRecursive` is $O(n)$ due to the recursive function calls and the associated call stack space.\n",
-        "\n",
-        "Both functions are efficient in terms of time complexity, but `reverseListIterative` is more memory-efficient because it uses a constant amount of additional space, while `reverseListRecursive` uses space on the call stack proportional to the number of nodes in the list."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K-Group Reverse:** Modify the reverseListIterative function to reverse the linked list in groups of $K$ elements. For example, if the input linked list is [1, 2, 3, 4, 5, 6], and $K$ is 3, the output should be [3, 2, 1, 6, 5, 4].\n",
-        "\n",
-        "2. **Rotate Linked List:** Write a function to rotate a linked list to the right by $K$ places. For example, if the input is [1, 2, 3, 4, 5] and $K$ is 2, the output should be [4, 5, 1, 2, 3]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb b/docs/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb
deleted file mode 100644
index 876d633..0000000
--- a/docs/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 21. Merge Two Sorted Lists\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Merge Two Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-two-sorted-lists/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given the heads of two sorted linked lists `list1` and `list2`.\n",
-        "\n",
-        "Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.\n",
-        "\n",
-        "*Return the head of the merged linked list.*\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in both lists is in the range `[0, 50]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "- Both `list1` and `list2` are sorted in non-decreasing order."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def mergeTwoLists(list1, list2):\n",
-        "    # Create a dummy node to simplify the code.\n",
-        "    dummy = ListNode()\n",
-        "    current = dummy\n",
-        "\n",
-        "    while list1 and list2:\n",
-        "        if list1.val < list2.val:\n",
-        "            current.next = list1\n",
-        "            list1 = list1.next\n",
-        "        else:\n",
-        "            current.next = list2\n",
-        "            list2 = list2.next\n",
-        "        current = current.next\n",
-        "\n",
-        "    # Append the remaining elements from either list.\n",
-        "    if list1:\n",
-        "        current.next = list1\n",
-        "    else:\n",
-        "        current.next = list2\n",
-        "\n",
-        "    return dummy.next"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a class `ListNode` to represent the nodes of a singly-linked list. Each node has a `val` attribute to store the value of the node and a `next` attribute to point to the next node in the list.\n",
-        "\n",
-        "2. We define the `mergeTwoLists` function that takes two linked list heads, `list1` and `list2`, as input.\n",
-        "\n",
-        "3. We create a `dummy` node at the beginning of the merged list. The `dummy` node simplifies the code by serving as a placeholder, and its `next` attribute will point to the actual merged list.\n",
-        "\n",
-        "4. We initialize a `current` pointer to point to the `dummy` node initially. This `current` pointer helps us traverse and build the merged list.\n",
-        "\n",
-        "5. We enter a `while` loop that continues until either `list1` or `list2` becomes empty. Inside the loop, we compare the values of the nodes at the current positions of `list1` and `list2`.\n",
-        "\n",
-        "6. If the `val` of the node in `list1` is smaller than the `val` of the node in `list2`, we attach the node from `list1` to the merged list by updating the `next` attribute of the `current` node to point to the node in `list1`. We then move the `list1` pointer to the next node.\n",
-        "\n",
-        "7. If the `val` of the node in `list2` is smaller than or equal to the `val` of the node in `list1`, we attach the node from `list2` to the merged list in a similar manner. We move the `list2` pointer to the next node.\n",
-        "\n",
-        "8. After attaching a node to the merged list, we move the `current` pointer to the newly added node. This step is essential for keeping track of the end of the merged list.\n",
-        "\n",
-        "9. The loop continues until either `list1` or `list2` becomes empty.\n",
-        "\n",
-        "10. Once the loop exits, we check if there are any remaining elements in `list1` or `list2`. If `list1` is not empty, we attach the remaining elements of `list1` to the merged list. If `list2` is not empty, we attach the remaining elements of `list2`.\n",
-        "\n",
-        "11. Finally, we return the `next` attribute of the `dummy` node as the head of the merged list, which represents the sorted merged list."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to print the linked list\n",
-        "def printLinkedList(head):\n",
-        "    result = []\n",
-        "    while head:\n",
-        "        result.append(head.val)\n",
-        "        head = head.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the helper functions:\n",
-        "**printLinkedList Function**:\n",
-        "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-        "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-        "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-        "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-        "\n",
-        "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 1:\n",
-        "# Input: list1 = [1,2,4], list2 = [1,3,4]\n",
-        "list1 = ListNode(1, ListNode(2, ListNode(4)))\n",
-        "list2 = ListNode(1, ListNode(3, ListNode(4)))\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 2:\n",
-        "# Input: list1 = [], list2 = []\n",
-        "list1 = None\n",
-        "list2 = None\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 3:\n",
-        "# list1 = [], list2 = [0]\n",
-        "list1 = None\n",
-        "list2 = ListNode(0)\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `mergeTwoLists` function.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this function is $O(N + M$), where $N$ and $M$ are the lengths of `list1` and `list2`, respectively. Here's why:\n",
-        "\n",
-        "1. In the worst case, we need to traverse both `list1` and `list2` completely. This involves iterating through all the nodes in both lists once.\n",
-        "\n",
-        "2. The while loop runs until either `list1` or `list2` becomes empty. The number of iterations depends on the total number of nodes in both lists, which is N + M in the worst case.\n",
-        "\n",
-        "3. Inside the loop, we perform constant-time operations for each iteration, such as comparisons and updating pointers.\n",
-        "\n",
-        "Therefore, the dominant factor in the time complexity is the combined length of both input lists, resulting in a linear time complexity of $O(N + M)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this function is $O(1)$, which means it uses a constant amount of extra space regardless of the input sizes. Here's why:\n",
-        "\n",
-        "1. We create a few extra pointers (`dummy`, `current`) and temporary variables (`val`) to manage the merging process. These consume a fixed amount of memory, regardless of the input sizes. The number of these extra variables is independent of the lengths of `list1` and `list2`.\n",
-        "\n",
-        "2. We do not create a new data structure or allocate memory for the merged list. Instead, we rearrange the existing nodes from `list1` and `list2` to form the merged list. This operation does not consume additional memory proportional to the input sizes.\n",
-        "\n",
-        "**In summary**, the `mergeTwoLists` function has a time complexity of $O(N + M)$ and a space complexity of $O(1)$, making it an efficient solution for merging two sorted linked lists."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Merge K Sorted Lists:** Extend the problem to merge $K$ sorted linked lists instead of just two. You need to efficiently merge $K$ lists into one sorted list.\n",
-        "\n",
-        "2. **Merge Lists in a Zigzag Pattern:** Merge two sorted lists in a zigzag pattern. For example, given [1, 2, 4] and [1, 3, 5], the merged list should be [1, 1, 2, 3, 4, 5]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": "sql",
-      "file_extension": "",
-      "mimetype": "",
-      "name": "sql",
-      "version": "3.32.3"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb b/docs/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb
deleted file mode 100644
index a850734..0000000
--- a/docs/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb	
+++ /dev/null
@@ -1,285 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 23. Merge k Sorted Lists\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Merge k Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-k-sorted-lists/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given an array of $k$ linked-lists lists, each linked-list is sorted in ascending order.\n",
-        "\n",
-        "Merge all the linked-lists into one sorted linked-list and return it.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `k == lists.length`\n",
-        "- 0 <= k <= $10^4$\n",
-        "- `0 <= lists[i].length <= 500`\n",
-        "- $-10^4$ <= lists[i][j] <= $10^4$\n",
-        "- `lists[i]` is sorted in ascending order.\n",
-        "- The sum of `lists[i].length` will not exceed $10^4$.\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "     def __init__(self, val=0, next=None):\n",
-        "         self.val = val\n",
-        "         self.next = next\n",
-        "\n",
-        "class Solution:\n",
-        "    def mergeKLists(self, lists: [[ListNode]]) -> ListNode:\n",
-        "        # Check if the input list of linked lists is empty\n",
-        "        if not lists or len(lists) == 0:\n",
-        "            return None\n",
-        "\n",
-        "        # Loop until there is only one merged list left\n",
-        "        while len(lists) > 1:\n",
-        "            mergedLists = []\n",
-        "\n",
-        "            # Merge pairs of lists\n",
-        "            for i in range(0, len(lists), 2):\n",
-        "                l1 = lists[i]\n",
-        "                l2 = lists[i + 1] if (i + 1) < len(lists) else None\n",
-        "                mergedLists.append(self.mergeList(l1, l2))\n",
-        "\n",
-        "            # Update the list of lists with merged results\n",
-        "            lists = mergedLists\n",
-        "        \n",
-        "        # Return the final merged list\n",
-        "        return lists[0]\n",
-        "\n",
-        "    def mergeList(self, l1, l2):\n",
-        "        # Create a dummy node to simplify merging\n",
-        "        dummy = ListNode()\n",
-        "        tail = dummy\n",
-        "\n",
-        "        # Merge the two sorted lists\n",
-        "        while l1 and l2:\n",
-        "            if l1.val < l2.val:\n",
-        "                tail.next = l1\n",
-        "                l1 = l1.next\n",
-        "            else:\n",
-        "                tail.next = l2\n",
-        "                l2 = l2.next\n",
-        "            tail = tail.next\n",
-        "\n",
-        "        # Append any remaining elements from l1 or l2 (if any)\n",
-        "        if l1:\n",
-        "            tail.next = l1\n",
-        "        if l2:\n",
-        "            tail.next = l2\n",
-        "\n",
-        "        # Return the merged result starting from the next of the dummy node\n",
-        "        return dummy.next\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided code defines a Python class `Solution` with two methods for merging k sorted linked lists:\n",
-        "\n",
-        "1. `mergeKLists(self, lists: List[ListNode]) -> ListNode`: This method takes a list of k sorted linked lists as input and returns a single merged sorted linked list. It uses a divide-and-conquer approach to repeatedly merge pairs of lists until only one merged list remains.\n",
-        "\n",
-        "2. `mergeList(self, l1, l2)`: This method takes two sorted linked lists, `l1` and `l2`, as input and merges them into a single sorted linked list. It uses a dummy node to simplify the merging process.\n",
-        "\n",
-        "Here's a high-level overview of how the code works:\n",
-        "\n",
-        "- The `mergeKLists` method checks if the input list of linked lists is empty or contains no lists. If there are no lists, it returns `None`.\n",
-        "\n",
-        "- Inside a `while` loop, the code repeatedly merges pairs of linked lists until only one merged list remains in the `lists` array. It does this by iterating through the input lists in pairs and calling the `mergeList` method to merge each pair.\n",
-        "\n",
-        "- The `mergeList` method takes two sorted linked lists, `l1` and `l2`, and merges them into a single sorted linked list. It uses a dummy node (`dummy`) and a `tail` pointer to keep track of the merged list while comparing and merging elements from `l1` and `l2`.\n",
-        "\n",
-        "- After merging all pairs of lists and updating the `lists` array with the merged results, the loop continues until only one merged list remains in the `lists` array.\n",
-        "\n",
-        "- Finally, the `mergeKLists` method returns the merged list.\n",
-        "\n",
-        "Overall, this code efficiently merges k sorted linked lists using a divide-and-conquer strategy, resulting in a single merged sorted linked list as the output."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> "
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: lists = [[1,4,5],[1,3,4],[2,6]]\n",
-        "\n",
-        "lists1 = [\n",
-        "    ListNode(1, ListNode(4, ListNode(5))),\n",
-        "    ListNode(1, ListNode(3, ListNode(4))),\n",
-        "    ListNode(2, ListNode(6))\n",
-        "]\n",
-        "solution = Solution()\n",
-        "result1 = solution.mergeKLists(lists1)\n",
-        "\n",
-        "# Print the result\n",
-        "if result1:\n",
-        "    current = result1\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: lists = []\n",
-        "\n",
-        "lists2 = []\n",
-        "solution = Solution()\n",
-        "result2 = solution.mergeKLists(lists2)\n",
-        "\n",
-        "# Print the result\n",
-        "if result2:\n",
-        "    current = result2\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "# Input: lists = [[]]\n",
-        "\n",
-        "lists3 = [None]\n",
-        "solution = Solution()\n",
-        "result3 = solution.mergeKLists(lists3)\n",
-        "\n",
-        "# Print the result\n",
-        "if result3:\n",
-        "    current = result3\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. **Heap Initialization:** The code does not use a heap. Instead, it uses a divide-and-conquer approach. The initial check for empty input lists takes $O(1)$ time.\n",
-        "\n",
-        "2. **Merging:** The merging operation is performed in a divide-and-conquer fashion. In each iteration of the `while` loop, we merge pairs of linked lists, and the number of comparisons made is proportional to the total number of nodes across all the linked lists ($n$). In each merge step, we effectively process each node once. The number of iterations required to reduce k lists to 1 is $O(log\\ k)$.\n",
-        "\n",
-        "   Therefore, the overall time complexity of the code is $O(n * log\\ k)$, where n is the total number of nodes across all lists, and $k$ is the number of linked lists.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. **Heap Space:** The code doesn't use a heap data structure, so there's no additional space complexity due to a heap.\n",
-        "\n",
-        "2. **Merged Lists:** In the `mergeList` method, we create a new merged list. However, this list is not stored in memory for all lists; it's replaced with each merged pair. The space used for these merged lists is proportional to the size of the largest merged list, which is $O(n)$ in the worst case.\n",
-        "\n",
-        "3. **Additional Variables:** The code uses a few additional variables, such as `dummy` and `tail`, but these occupy a constant amount of space and don't depend on the input size.\n",
-        "\n",
-        "   Therefore, the overall space complexity of the code is $O(n)$, where n is the total number of nodes across all lists.\n",
-        "\n",
-        "In summary, the code's time complexity is $O(n * log(k))$, and its space complexity is $O(n)$. This code efficiently merges $k$ sorted linked lists using a divide-and-conquer approach with a relatively low space overhead."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/06. Linked List/README.md b/docs/_build/html/_sources/06. Linked List/README.md
deleted file mode 100644
index fb7b3ee..0000000
--- a/docs/_build/html/_sources/06. Linked List/README.md	
+++ /dev/null
@@ -1,21 +0,0 @@
-# Linked List Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)                       | Easy       |
-| [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)                   | Easy       |
-| [143. Reorder List](https://leetcode.com/problems/reorder-list/)                                     | Medium     |
-| [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | Medium |
-| [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)                           | Medium     |
-| [23. Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)                       | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/_build/html/_sources/07. Trees/100. Same Tree.ipynb b/docs/_build/html/_sources/07. Trees/100. Same Tree.ipynb
deleted file mode 100644
index a270fab..0000000
--- a/docs/_build/html/_sources/07. Trees/100. Same Tree.ipynb	
+++ /dev/null
@@ -1,211 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 100: Same Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-        "\n",
-        "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-        "\n",
-        "**Constraints**\n",
-        "\n",
-        "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-        "2. $-10^4$ <= Node.val <= $10^4$\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def isSameTree(p, q):\n",
-        "    # Base case: If both p and q are None, the trees are the same.\n",
-        "    if not p and not q:\n",
-        "        return True\n",
-        "    \n",
-        "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-        "    if not p or not q:\n",
-        "        return False\n",
-        "    \n",
-        "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-        "    if p.val != q.val:\n",
-        "        return False\n",
-        "    \n",
-        "    # Recursively check the left and right subtrees of p and q.\n",
-        "    # If both subtrees are the same, the entire trees are the same.\n",
-        "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-        "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-        "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-        "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-        "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-        "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "\n",
-        "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-        "\n",
-        "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "print(isSameTree(p1, q1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "\n",
-        "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-        "\n",
-        "p2 = TreeNode(1, TreeNode(2), None)\n",
-        "q2 = TreeNode(1, None, TreeNode(2))\n",
-        "print(isSameTree(p2, q2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "\n",
-        "#Input: p = [1,2,1], q = [1,1,2]\n",
-        "\n",
-        "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-        "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-        "print(isSameTree(p3, q3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time Complexity\n",
-        "\n",
-        "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-        "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-        "\n",
-        "## Space Complexity\n",
-        "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-        "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-        "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-        "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-        "\n",
-        "## In summary:\n",
-        "\n",
-        "+ **Time Complexity:** $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-        "+ **Space Complexity:** $O(n)$ due to the recursive call stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb b/docs/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb
deleted file mode 100644
index 8cb08b7..0000000
--- a/docs/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb	
+++ /dev/null
@@ -1,271 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 102. Binary Tree Level Order Traversal\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Level Order Traversal problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-level-order-traversal/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, return *the level order traversal of its nodes' values*. (i.e., from left to right, level by level).\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-        "- `-1000 <= Node.val <= 1000`\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def levelOrder(root):\n",
-        "    # Check if the tree is empty, if so, return an empty list.\n",
-        "    if not root:\n",
-        "        return []\n",
-        "\n",
-        "    # Initialize an empty list to store the result.\n",
-        "    result = []\n",
-        "    \n",
-        "    # Initialize a queue with the root node to perform BFS.\n",
-        "    queue = [root]\n",
-        "\n",
-        "    while queue:\n",
-        "        # Initialize an empty list to store the values of nodes at the current level.\n",
-        "        level_values = []\n",
-        "        \n",
-        "        # Get the number of nodes at the current level.\n",
-        "        level_size = len(queue)\n",
-        "\n",
-        "        # Iterate through the nodes at the current level.\n",
-        "        for _ in range(level_size):\n",
-        "            # Dequeue the front node from the queue.\n",
-        "            node = queue.pop(0)\n",
-        "            \n",
-        "            # Append the value of the current node to the level_values list.\n",
-        "            level_values.append(node.val)\n",
-        "\n",
-        "            # Enqueue the left and right children of the current node if they exist.\n",
-        "            if node.left:\n",
-        "                queue.append(node.left)\n",
-        "            if node.right:\n",
-        "                queue.append(node.right)\n",
-        "\n",
-        "        # Append the level_values list (values at the current level) to the result list.\n",
-        "        result.append(level_values)\n",
-        "\n",
-        "    # Return the final result, which is a list of lists representing level order traversal.\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a `TreeNode` class that represents a node in the binary tree. Each `TreeNode` object has a value (`val`) and two child nodes: `left` and `right`.\n",
-        "\n",
-        "2. The `levelOrder` function takes the root of the binary tree as its input and returns the level order traversal of the tree as a list of lists.\n",
-        "\n",
-        "3. We start by checking if the input `root` is `None`, which indicates an empty tree. If the tree is empty, we return an empty list because there are no nodes to traverse.\n",
-        "\n",
-        "4. We initialize an empty list called `result` to store the final result, which will be a list of lists containing node values at each level.\n",
-        "\n",
-        "5. We initialize a queue called `queue` with the root node. This queue will be used for breadth-first traversal of the tree.\n",
-        "\n",
-        "6. We enter a while loop that continues until the `queue` is empty. Inside the loop, we perform the following steps:\n",
-        "\n",
-        "   - We initialize an empty list called `level_values` to store the values of nodes at the current level.\n",
-        "\n",
-        "   - We determine the number of nodes at the current level by getting the length of the `queue`. This is done to process nodes level by level.\n",
-        "\n",
-        "   - We iterate through the nodes at the current level using a for loop. For each node in the current level:\n",
-        "\n",
-        "     - We dequeue (remove) the front node from the `queue`.\n",
-        "\n",
-        "     - We append the value of the dequeued node to the `level_values` list, effectively collecting the values of nodes at the current level.\n",
-        "\n",
-        "     - If the dequeued node has a left child, we enqueue the left child to the `queue`.\n",
-        "\n",
-        "     - If the dequeued node has a right child, we enqueue the right child to the `queue`.\n",
-        "\n",
-        "   - After processing all nodes at the current level, we append the `level_values` list to the `result` list. This represents the values at the current level.\n",
-        "\n",
-        "7. The loop continues until all levels have been traversed, and the `queue` becomes empty.\n",
-        "\n",
-        "8. Finally, we return the `result` list, which contains lists of node values at each level, representing the level order traversal of the binary tree.\n",
-        "\n",
-        "The code effectively performs a breadth-first traversal of the binary tree, processing nodes level by level, and constructs the result list that represents the level order traversal."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[3], [9, 20], [15, 7]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(9)\n",
-        "root.right = TreeNode(20)\n",
-        "root.right.left = TreeNode(15)\n",
-        "root.right.right = TreeNode(7)\n",
-        "\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Using the same tree as before\n",
-        "root = TreeNode(1)\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root = None\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `levelOrder` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this function is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the breadth-first traversal.\n",
-        "\n",
-        "In the worst case, we have to enqueue and dequeue all nodes in the binary tree, which is proportional to the number of nodes (N). In each level, we process all nodes in that level, and since there are a total of log(N) levels in a balanced binary tree, the time complexity can also be approximated as O(N) for unbalanced trees.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this function is O(N), where N is the number of nodes in the binary tree. Here's how the space complexity breaks down:\n",
-        "\n",
-        "1. The `result` list stores the level order traversal, and in the worst case, it contains N/2 levels (for a completely unbalanced binary tree). So, the space used by `result` is O(N).\n",
-        "\n",
-        "2. The `queue` data structure is used for BFS traversal. In the worst case, it can store all nodes at the last level of the tree. In a balanced binary tree, the maximum number of nodes at any level is 2^(log(N)), which is still O(N). In the case of an unbalanced tree, it can be even worse. So, the space used by `queue` is O(N).\n",
-        "\n",
-        "Overall, the dominant factor in terms of space complexity is the `queue`, and the space complexity is O(N).\n",
-        "\n",
-        "In summary, the function's time complexity is O(N), and its space complexity is also O(N). It is an efficient and optimal solution for performing a level order traversal of a binary tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Reverse Level Order Traversal:** Modify the `levelOrder` function to return the level order traversal in reverse order (from bottom to top). For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[15, 7], [9, 20], [3]]`.\n",
-        "\n",
-        "2. **Zigzag Level Order Traversal:** Write a function that performs a level order traversal of a binary tree in a zigzag pattern. In a zigzag traversal, the nodes at even levels are traversed from left to right, and nodes at odd levels are traversed from right to left. For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[3], [20, 9], [15, 7]]`."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb b/docs/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb
deleted file mode 100644
index 213989d..0000000
--- a/docs/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 104. Maximum Depth of Binary Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the `root` of a binary tree, return its maximum depth.\n",
-        "\n",
-        "A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 19,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def maxDepth(root):\n",
-        "    # Base case: If the root is None, the depth is 0.\n",
-        "    if root is None:\n",
-        "        return 0\n",
-        "    \n",
-        "    # Recursively calculate the maximum depth of the left and right subtrees.\n",
-        "    left_depth = maxDepth(root.left)\n",
-        "    right_depth = maxDepth(root.right)\n",
-        "    \n",
-        "    # Return the maximum depth of the tree by adding 1 to the maximum depth of the subtrees.\n",
-        "    return max(left_depth, right_depth) + 1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's go through the code step by step to understand how it works:\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class to represent the nodes of the binary tree. Each node has three attributes:\n",
-        "   - `val`: the value stored in the node.\n",
-        "   - `left`: a reference to the left child node.\n",
-        "   - `right`: a reference to the right child node.\n",
-        "\n",
-        "2. The `maxDepth` function is the main function that calculates the maximum depth of the binary tree. It takes a single argument, `root`, which is the root node of the binary tree.\n",
-        "\n",
-        "3. In the `maxDepth` function, we have a base case:\n",
-        "   - If the `root` is `None`, it means we have reached the end of a branch of the tree (a leaf node or an empty subtree). In this case, the depth is 0 because there are no nodes to count.\n",
-        "\n",
-        "4. If the `root` is not `None`, we recursively calculate the maximum depth of the left and right subtrees:\n",
-        "   - We call `maxDepth` on the `root.left` to calculate the maximum depth of the left subtree and store it in the variable `left_depth`.\n",
-        "   - We call `maxDepth` on the `root.right` to calculate the maximum depth of the right subtree and store it in the variable `right_depth`.\n",
-        "\n",
-        "5. Finally, we return the maximum depth of the tree by taking the maximum of `left_depth` and `right_depth` and adding 1 to it. This is because we are counting the current level as well."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: root = [3,9,20,null,null,15,7]\n",
-        "\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(9)\n",
-        "root.right = TreeNode(20)\n",
-        "root.right.left = TreeNode(15)\n",
-        "root.right.right = TreeNode(7)\n",
-        "\n",
-        "print(maxDepth(root))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a230be4f-1f47-4919-87e3-62fc3f072d97",
-      "metadata": {},
-      "source": [
-        "In this example, we create a binary tree based on the given input `[3,9,20,null,null,15,7]` and calculate its maximum depth, which is `3`. The tree structure is as follows:\n",
-        "   ```\n",
-        "       3\n",
-        "      / \\\n",
-        "     9  20\n",
-        "       /  \\\n",
-        "      15   7\n",
-        "   ```\n",
-        "\n",
-        "So, the maximum depth is the length of the longest path from the root to a leaf node, which is 3 in this case."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: root = [1,null,2]\n",
-        "\n",
-        "root = TreeNode(1)\n",
-        "root.right = TreeNode(2)\n",
-        "\n",
-        "# Calculate the maximum depth of the tree and print the result\n",
-        "print(maxDepth(root))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The time complexity of the `maxDepth` function is $O(N)$, where $N$ is the number of nodes in the binary tree.\n",
-        "- This is because in the worst case, the function visits every node exactly once in a depth-first manner.\n",
-        "- The recursion explores all nodes of the tree, so the time complexity is linear with respect to the number of nodes.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the `maxDepth` function is $O(H)$, where H is the height of the binary tree.\n",
-        "- In the worst case, if the binary tree is completely unbalanced (skewed), the recursion stack can go as deep as the height of the tree.\n",
-        "- In the best case, if the binary tree is perfectly balanced, the height is $O(log\\ N)$, where $N$ is the number of nodes.\n",
-        "- Therefore, the space complexity can vary from $O(log\\ N)$ to $O(N)$ depending on the shape of the tree.\n",
-        "- In addition to the recursion stack, there is a small constant amount of space used for variables and function call overhead.\n",
-        "\n",
-        "**In summary:**\n",
-        "- The time complexity of the code is $O(N)$ as it visits each node once.\n",
-        "- The space complexity is $O(H)$, where H is the height of the tree, which can vary from $O(log\\ N)$ to $O(N)$ depending on the tree's shape."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb b/docs/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb
deleted file mode 100644
index 31c9e69..0000000
--- a/docs/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb	
+++ /dev/null
@@ -1,313 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 105. Construct Binary Tree from Preorder and Inorder Traversal\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Construct Binary Tree from Preorder and Inorder Traversal problem on LeetCode, click here!](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)\n",
-        "\n",
-        "---\n",
-        "Given two integer arrays `preorder` and `inorder` where `preorder` is the preorder traversal of a binary tree and `inorder` is the inorder traversal of the same tree, construct and return *the binary tree.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= preorder.length <= 3000`\n",
-        "- `inorder.length == preorder.length`\n",
-        "- `-3000 <= preorder[i], inorder[i] <= 3000`\n",
-        "- `preorder` and `inorder` consist of **unique** values.\n",
-        "- Each value of `inorder` also appears in `preorder`.\n",
-        "- `preorder` is **guaranteed** to be the preorder traversal of the tree.\n",
-        "- `inorder` is **guaranteed** to be the inorder traversal of the tree.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def buildTree(preorder, inorder):\n",
-        "    if not preorder:\n",
-        "        return None\n",
-        "\n",
-        "    # The first element in the preorder traversal is the root of the current subtree\n",
-        "    root_val = preorder[0]\n",
-        "    root = TreeNode(root_val)\n",
-        "\n",
-        "    # Find the index of the root value in the inorder traversal\n",
-        "    root_idx_inorder = inorder.index(root_val)\n",
-        "\n",
-        "    # Recursively build left and right subtrees\n",
-        "    root.left = buildTree(preorder[1:1 + root_idx_inorder], inorder[:root_idx_inorder])\n",
-        "    root.right = buildTree(preorder[1 + root_idx_inorder:], inorder[root_idx_inorder + 1:])\n",
-        "\n",
-        "    return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `class TreeNode:`: This is a class definition for a binary tree node. It has three attributes:\n",
-        "   - `val`: The value of the node.\n",
-        "   - `left`: A reference to the left child node.\n",
-        "   - `right`: A reference to the right child node.\n",
-        "\n",
-        "   This class is used to create instances of binary tree nodes, which will be used to build the binary tree.\n",
-        "\n",
-        "2. `def buildTree(preorder, inorder):`: This is the main function that constructs a binary tree from its preorder and inorder traversals.\n",
-        "\n",
-        "   - `preorder`: A list representing the preorder traversal of the binary tree.\n",
-        "   - `inorder`: A list representing the inorder traversal of the binary tree.\n",
-        "\n",
-        "3. `if not preorder: return None`: This line checks if the `preorder` list is empty. If it is, it means there are no nodes to construct, so the function returns `None`.\n",
-        "\n",
-        "4. `root_val = preorder[0]`: The value of the root node is extracted from the first element of the `preorder` list.\n",
-        "\n",
-        "5. `root = TreeNode(root_val)`: A new `TreeNode` object is created with the `root_val` as its value. This represents the root of the current subtree.\n",
-        "\n",
-        "6. `root_idx_inorder = inorder.index(root_val)`: The index of the `root_val` in the `inorder` list is found. This index indicates the position of the root node in the inorder traversal.\n",
-        "\n",
-        "7. `root.left`: The left subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The left subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists up to the `root_idx_inorder`.\n",
-        "\n",
-        "8. `root.right`: The right subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The right subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists starting from `root_idx_inorder + 1`.\n",
-        "\n",
-        "9. Finally, the function returns the `root` of the subtree it just constructed.\n",
-        "\n",
-        "The recursive approach used here divides the problem of constructing the entire binary tree into smaller subproblems, starting with the root node and then recursively building the left and right subtrees until the entire tree is constructed."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "2d82857b-4fbe-474f-bf33-17b55c20147b",
-      "metadata": {},
-      "source": [
-        "## Helper Function \n",
-        "\n",
-        "Here is a helper function to represent the binary tree elements."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "13e6f3e1-659d-479d-9521-75629b8c782f",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def treeToList(root):\n",
-        "    # Initialize an empty list to store the elements of the binary tree\n",
-        "    result = []\n",
-        "    \n",
-        "    # Initialize a queue for level-order traversal starting with the root\n",
-        "    queue = [root]\n",
-        "\n",
-        "    while queue:\n",
-        "        # Pop the front node from the queue\n",
-        "        node = queue.pop(0)\n",
-        "        \n",
-        "        # If the node is not None (i.e., a valid node), add its value to the result list\n",
-        "        if node:\n",
-        "            result.append(node.val)\n",
-        "            \n",
-        "            # Add the left and right children of the node to the queue\n",
-        "            queue.append(node.left)\n",
-        "            queue.append(node.right)\n",
-        "        else:\n",
-        "            # If the node is None, add None to the result to represent an empty node\n",
-        "            result.append(None)\n",
-        "\n",
-        "    # Remove any trailing None values from the result list\n",
-        "    while result and result[-1] is None:\n",
-        "        result.pop()\n",
-        "\n",
-        "    # Return the resulting list representing the binary tree elements\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "86cd4c9d-a120-4dd1-a1bd-530436012574",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's break down the `treeToList` helper function step by step:\n",
-        "\n",
-        "1. `def treeToList(root):`: This function takes the root of a binary tree as input and converts it into a list.\n",
-        "\n",
-        "2. `result = []`: Initialize an empty list called `result` to store the elements of the binary tree in list format.\n",
-        "\n",
-        "3. `queue = [root]`: Initialize a queue data structure for a level-order traversal, starting with the root node.\n",
-        "\n",
-        "4. `while queue:`: This starts a loop that continues until the queue is empty, indicating that all nodes have been processed.\n",
-        "\n",
-        "5. `node = queue.pop(0)`: Dequeue the first node from the queue, effectively processing it.\n",
-        "\n",
-        "6. `if node:`: Check if the `node` is not None, which means it's a valid node in the binary tree.\n",
-        "\n",
-        "    a. `result.append(node.val)`: If the node is valid, append its value (`node.val`) to the `result` list. This represents the value of the current node in the binary tree.\n",
-        "\n",
-        "    b. `queue.append(node.left)`: Enqueue the left child of the current node if it exists. This adds the left child to the queue for processing in the next iteration.\n",
-        "\n",
-        "    c. `queue.append(node.right)`: Enqueue the right child of the current node if it exists. This adds the right child to the queue for processing in the next iteration.\n",
-        "\n",
-        "7. `else:`: If the `node` is None, it represents an empty node in the binary tree.\n",
-        "\n",
-        "    a. `result.append(None)`: Append `None` to the `result` list to indicate an empty node.\n",
-        "\n",
-        "8. `while result and result[-1] is None:`: After the traversal is complete, there might be trailing `None` values in the `result` list. This loop removes any such trailing `None` values to ensure a clean representation of the tree's elements.\n",
-        "\n",
-        "9. `return result`: Return the `result` list, which now contains the elements of the binary tree in a format where `None` represents empty nodes and the order of elements reflects a level-order traversal of the tree.\n",
-        "\n",
-        "In summary, the `treeToList` function performs a level-order traversal of the binary tree using a queue, constructing a list where each element corresponds to a node's value or represents an empty node with `None`. This list represents the binary tree's elements in a structured format."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[3, 9, 20, None, None, 15, 7]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "preorder = [3, 9, 20, 15, 7]\n",
-        "inorder = [9, 3, 15, 20, 7]\n",
-        "result = buildTree(preorder, inorder)\n",
-        "\n",
-        "print(treeToList(result))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[-1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "preorder = [-1]\n",
-        "inorder = [-1]\n",
-        "result = buildTree(preorder, inorder)\n",
-        "\n",
-        "print(treeToList(result))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `buildTree` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of the `buildTree` function can be analyzed in terms of the number of nodes in the binary tree. Let's assume there are `n` nodes in the binary tree.\n",
-        "\n",
-        "1. Finding the root value in the `preorder` traversal takes O(1) time as it's just extracting the first element from the list.\n",
-        "2. Finding the index of the root value in the `inorder` traversal using `inorder.index(root_val)` takes O(n) time in the worst case because in the worst case, it might have to search through all `n` elements in the `inorder` list to find the index.\n",
-        "3. The recursive calls to `buildTree` for the left and right subtrees are made once for each node in the tree. Therefore, the recursive calls have a combined time complexity of O(n) as they process each node once.\n",
-        "\n",
-        "The total time complexity of the `buildTree` function is O(n) due to the recursive calls and finding the root's index in the `inorder` list.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity is determined by the space used by the function's call stack during recursion and any additional data structures used. \n",
-        "\n",
-        "1. The space used by the function call stack during recursion depends on the height of the binary tree. In the worst case, where the tree is highly unbalanced (e.g., a skewed tree), the space complexity for the call stack is O(n) as it can go as deep as the number of nodes in the tree.\n",
-        "\n",
-        "2. Additionally, the function creates TreeNode objects for each node in the binary tree. Therefore, the space complexity for these objects is also O(n).\n",
-        "\n",
-        "Overall, the space complexity of the `buildTree` function is O(n) due to the space used by the call stack and the TreeNode objects.\n",
-        "\n",
-        "In summary, the time complexity of `buildTree` is O(n), and the space complexity is O(n), where 'n' is the number of nodes in the binary tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Construct Binary Tree from Postorder and Inorder Traversal:**\n",
-        "   Modify the problem to construct a binary tree from its postorder and inorder traversals. Implement a function similar to `buildTree` but for postorder and inorder traversals.\n",
-        "\n",
-        "2. **Reconstruct Binary Tree with Duplicate Values:**\n",
-        "   Extend the problem to handle binary trees with duplicate values. Ensure that your solution correctly handles scenarios where there are duplicate values in the preorder and inorder traversals."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb b/docs/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb
deleted file mode 100644
index 59d8e7d..0000000
--- a/docs/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 124. Binary Tree Maximum Path Sum\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Maximum Path Sum problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-maximum-path-sum/)\n",
-        "\n",
-        "---\n",
-        "A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.\n",
-        "\n",
-        "The **path sum** of a path is the sum of the node's values in the path.\n",
-        "\n",
-        "Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[1, 3*10^4]$.\n",
-        "- `-1000 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to find the maximum path sum in a binary tree. A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. The path sum of a path is the sum of the node values in that path.\n",
-        "\n",
-        "In this problem, you are given the root of a binary tree, and you need to find the maximum path sum among all possible paths in the tree. Note that the path does not need to start at the root node or end at a leaf node; it can start and end at any nodes in the tree.\n",
-        "\n",
-        "To find the maximum path sum, you need to consider both left and right subtrees of each node while traversing the tree. At each node, you have two choices:\n",
-        "\n",
-        "1. Include the current node in the path: In this case, you add the value of the current node to the sum and continue exploring both the left and right subtrees for possible extensions of the path.\n",
-        "\n",
-        "2. Start a new path from the current node: In this case, you do not include the current node in the path sum, and you choose either the left subtree or the right subtree to start a new path.\n",
-        "\n",
-        "To solve this problem, you can use a recursive approach to traverse the binary tree. For each node, you calculate the maximum path sum that can pass through that node (option 1) and also return the maximum path sum that can be extended from that node (option 2). You keep track of the global maximum path sum encountered during the traversal.\n",
-        "\n",
-        "Ultimately, the maximum path sum among all paths in the tree will be the maximum of the global maximum path sum and the maximum path sum calculated for each node.\n",
-        "\n",
-        "The constraints for this problem include:\n",
-        "\n",
-        "- The number of nodes in the tree is in the range $[1, 3 * 10^4]$.\n",
-        "- Node values are in the range [-1000, 1000].\n",
-        "\n",
-        "By considering all possible paths through the tree, the algorithm aims to find the most significant sum of node values in any path in the binary tree.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "class Solution:\n",
-        "    def maxPathSum(self, root):\n",
-        "        def max_path_sum_helper(node):\n",
-        "            if not node:\n",
-        "                return 0\n",
-        "            \n",
-        "            # Recursively calculate the maximum path sum for left and right subtrees\n",
-        "            left_max = max(0, max_path_sum_helper(node.left))\n",
-        "            right_max = max(0, max_path_sum_helper(node.right))\n",
-        "            \n",
-        "            # Update the global maximum path sum\n",
-        "            self.max_sum = max(self.max_sum, left_max + right_max + node.val)\n",
-        "            \n",
-        "            # Return the maximum path sum starting from the current node\n",
-        "            return max(left_max, right_max) + node.val\n",
-        "        \n",
-        "        self.max_sum = float('-inf')  # Initialize the global maximum to negative infinity\n",
-        "        max_path_sum_helper(root)  # Start the recursive traversal from the root\n",
-        "        return self.max_sum"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The code starts by defining a `TreeNode` class to represent nodes in the binary tree. Each `TreeNode` has a `val` attribute representing its value, a `left` attribute pointing to the left child, and a `right` attribute pointing to the right child.\n",
-        "\n",
-        "2. Next, a `Solution` class is defined to contain the `maxPathSum` method, which will be used to find the maximum path sum in the binary tree.\n",
-        "\n",
-        "3. Within the `maxPathSum` method, a helper function called `max_path_sum_helper` is defined. This recursive function takes a single argument, `node`, which is the current node being considered during traversal.\n",
-        "\n",
-        "4. Inside the `max_path_sum_helper` function:\n",
-        "   - If `node` is `None` (i.e., the current node is a null node), it returns 0. This is the base case of the recursion.\n",
-        "   - It calculates the maximum path sum starting from the left subtree (`left_max`) and the right subtree (`right_max`). Importantly, it uses `max(0, ...)` to ensure that negative values (which would make the path sum smaller) are not considered. If a subtree's path sum is negative, it's better to not include it in the path.\n",
-        "   - The code then updates the global maximum path sum (`self.max_sum`) by checking if the sum of `left_max`, `right_max`, and the current node's value is greater than the current maximum.\n",
-        "   - Finally, it returns the maximum path sum starting from the current node, which is the maximum of `left_max` and `right_max` plus the current node's value.\n",
-        "\n",
-        "5. Before calling the `max_path_sum_helper` function, the `max_sum` attribute of the `Solution` class is initialized to negative infinity (`float('-inf')`). This is done to ensure that any valid path sum encountered during traversal will be greater than the initial value of `max_sum`.\n",
-        "\n",
-        "6. The `max_path_sum_helper` function is called with the root of the binary tree, effectively starting the traversal from the root node.\n",
-        "\n",
-        "7. Once the traversal is complete, the method returns the value of `self.max_sum`, which contains the maximum path sum found in the binary tree.\n",
-        "\n",
-        "8. Example usage at the bottom of the code demonstrates how to create binary trees and use the `maxPathSum` method to find the maximum path sum for two different tree structures.\n",
-        "\n",
-        "In summary, the code uses a recursive approach to explore all possible paths in the binary tree, ensuring that negative path sums are not considered. It keeps track of the maximum path sum encountered and returns that maximum value as the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "6\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "root1 = TreeNode(1)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(3)\n",
-        "solution = Solution()\n",
-        "print(solution.maxPathSum(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "42\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "root2 = TreeNode(-10)\n",
-        "root2.left = TreeNode(9)\n",
-        "root2.right = TreeNode(20)\n",
-        "root2.right.left = TreeNode(15)\n",
-        "root2.right.right = TreeNode(7)\n",
-        "print(solution.maxPathSum(root2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "Time Complexity:\n",
-        "- The code uses a recursive depth-first traversal to explore each node in the binary tree exactly once.\n",
-        "- During the traversal, for each node, we perform constant time operations, such as updating the `max_sum` variable and calculating the maximum path sum starting from that node.\n",
-        "- Therefore, the time complexity of the code is O(N), where N is the number of nodes in the binary tree.\n",
-        "\n",
-        "Space Complexity:\n",
-        "- The space complexity of the code is determined by the space used in the call stack during the recursive traversal.\n",
-        "- In the worst case, the depth of the call stack can be equal to the height of the binary tree. In an unbalanced tree, this could be O(N), but in a balanced binary tree, the height is O(log N).\n",
-        "- Additionally, the code uses a constant amount of space for variables like `left_max`, `right_max`, and `max_sum`.\n",
-        "- Therefore, the overall space complexity is O(H), where H is the height of the binary tree. In the worst case, it's O(N), and in the best case (a balanced binary tree), it's O(log N).\n",
-        "\n",
-        "In summary:\n",
-        "- Time Complexity: O(N)\n",
-        "- Space Complexity: O(H), where H is the height of the binary tree, ranging from O(log N) to O(N) depending on the tree's balance."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Print the Maximum Path:**\n",
-        "   Modify the code to not only find the maximum path sum but also print the nodes involved in the maximum path. You can print the nodes in the correct order from the root to the leaf.\n",
-        "\n",
-        "2. **Path Sum Count:**\n",
-        "   Instead of finding the maximum path sum, find the count of unique paths that sum to a given target value. Each path can start and end anywhere in the tree."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb b/docs/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb
deleted file mode 100644
index 55dc944..0000000
--- a/docs/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb	
+++ /dev/null
@@ -1,257 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 199. Binary Tree Right Side View\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Right Side View problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-right-side-view/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, imagine yourself standing on the **right side** of it, return *the values of the nodes you can see ordered from top to bottom*.\n",
-        "\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def rightSideView(root):\n",
-        "    if not root:\n",
-        "        return []  # If the tree is empty, return an empty list.\n",
-        "\n",
-        "    result = []  # Initialize a list to store the right side view values.\n",
-        "    queue = [root]  # Initialize a queue for level-order traversal with the root node.\n",
-        "\n",
-        "    while queue:\n",
-        "        # Get the number of nodes at the current level.\n",
-        "        level_size = len(queue)\n",
-        "\n",
-        "        # Traverse all nodes at the current level and add the rightmost node to the result.\n",
-        "        for i in range(level_size):\n",
-        "            node = queue.pop(0)  # Dequeue the first node from the queue.\n",
-        "\n",
-        "            # If it's the rightmost node at this level, add its value to the result.\n",
-        "            if i == level_size - 1:\n",
-        "                result.append(node.val)\n",
-        "\n",
-        "            # Add the children of the current node to the queue.\n",
-        "            if node.left:\n",
-        "                queue.append(node.left)\n",
-        "            if node.right:\n",
-        "                queue.append(node.right)\n",
-        "\n",
-        "    return result  # Return the list of right side view values.\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start with a binary tree represented using a TreeNode class. Each node has a value (`val`) and can have a left child (`left`) and a right child (`right`).\n",
-        "\n",
-        "2. The `rightSideView` function takes the root node of the binary tree as input and returns a list of values representing the nodes you can see when standing on the right side of the tree.\n",
-        "\n",
-        "3. We check if the `root` is `None`, which means the tree is empty. If it's empty, we return an empty list because there are no nodes to see.\n",
-        "\n",
-        "4. We initialize an empty list called `result` to store the right side view values.\n",
-        "\n",
-        "5. We initialize a queue called `queue` with the root node. This queue is used for a level-order traversal of the tree.\n",
-        "\n",
-        "6. We enter a while loop that continues until the `queue` is empty. Inside this loop, we perform the following steps for each level of the tree:\n",
-        "\n",
-        "   a. We determine the number of nodes at the current level by getting the length of the `queue`. This is important for processing nodes at the same level together.\n",
-        "\n",
-        "   b. We then iterate through the nodes at the current level using a for loop. For each node in the level, we dequeue it from the `queue` using `queue.pop(0)`.\n",
-        "\n",
-        "   c. If the node we dequeue is the rightmost node at the current level (determined by `i == level_size - 1`), we add its `val` to the `result` list. This is because, when standing on the right side of the tree, you can see the rightmost node at each level.\n",
-        "\n",
-        "   d. We enqueue the left and right children of the current node (if they exist) to the `queue`. This ensures that we process the next level in the subsequent iterations of the while loop.\n",
-        "\n",
-        "7. After processing all levels of the tree, the `result` list contains the values of the rightmost nodes in each level, ordered from top to bottom.\n",
-        "\n",
-        "8. Finally, we return the `result` list as the output of the function, which represents the right side view of the binary tree.\n",
-        "\n",
-        "The code effectively performs a level-order traversal of the binary tree while keeping track of the rightmost nodes at each level and adding them to the result list. This ensures that we obtain the correct order of nodes visible from the right side of the tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 3, 4]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root1 = TreeNode(1)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(3)\n",
-        "root1.left.right = TreeNode(5)\n",
-        "root1.right.right = TreeNode(4)\n",
-        "print(rightSideView(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "root2 = TreeNode(1)\n",
-        "root2.right = TreeNode(3)\n",
-        "print(rightSideView(root2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root = None\n",
-        "result = rightSideView(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of the code is O(N), where N is the number of nodes in the binary tree. Here's why:\n",
-        "\n",
-        "1. We perform a level-order traversal of the tree using a queue. In the worst case, we visit all nodes once, which takes O(N) time since we visit each node exactly once.\n",
-        "\n",
-        "2. For each node, we perform constant-time operations like dequeuing it from the queue, checking if it's the rightmost node at the current level, and enqueuing its children (if they exist). These operations do not depend on the size of the tree, so they do not contribute to the overall time complexity.\n",
-        "\n",
-        "Therefore, the dominant factor in the time complexity is the level-order traversal, making it O(N).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is also O(N). Here's why:\n",
-        "\n",
-        "1. We use a queue to perform level-order traversal. In the worst case, the queue can contain all nodes at the maximum width of the tree, which can be up to N/2 nodes for a completely unbalanced tree. Therefore, the space required for the queue is O(N).\n",
-        "\n",
-        "2. The `result` list stores the values of the rightmost nodes. In the worst case, when the binary tree is a complete binary tree, it can have roughly N/2 rightmost nodes. Therefore, the space required for the `result` list is also O(N).\n",
-        "\n",
-        "3. Other auxiliary variables like `level_size`, `i`, and `node` require only constant space and do not depend on the size of the tree.\n",
-        "\n",
-        "Combining the space used by the queue and the `result` list, we get a space complexity of O(N).\n",
-        "\n",
-        "In summary, the code has a time complexity of O(N) and a space complexity of O(N), where N is the number of nodes in the binary tree. These complexities are efficient and scale linearly with the size of the input tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Variant with Left Side View**: Modify the code to find and return the values of nodes visible from the left side of the binary tree instead of the right side.\n",
-        "\n",
-        "2. **Zigzag Right Side View**: Extend the code to return the right side view values in a zigzag order. In a zigzag order, you alternate between starting from the rightmost node at level 0, then the leftmost at level 1, rightmost at level 2, and so on."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb b/docs/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb
deleted file mode 100644
index 8ea4609..0000000
--- a/docs/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb	
+++ /dev/null
@@ -1,223 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 226: Invert Binary Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the root of a binary tree, invert the tree, and return its root.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the tree is in the range `[0, 100]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def invertTree(root):\n",
-        "    # Base case: If the root is None or the tree is empty, return None.\n",
-        "    if not root:\n",
-        "        return None\n",
-        "\n",
-        "    # Swap the left and right subtrees of the current node.\n",
-        "    root.left, root.right = root.right, root.left\n",
-        "\n",
-        "    # Recursively invert the left and right subtrees.\n",
-        "    invertTree(root.left)\n",
-        "    invertTree(root.right)\n",
-        "\n",
-        "    return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a TreeNode class, which represents a node in a binary tree. Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-        "\n",
-        "2. The `invertTree` function is defined to invert the binary tree. It takes the root node of the tree as an argument.\n",
-        "\n",
-        "3. In the `invertTree` function, we have a base case to handle the scenario when the root is `None` (empty tree). In such cases, we return `None` because there's nothing to invert.\n",
-        "\n",
-        "4. For non-empty trees, we swap the left and right subtrees of the current node. This effectively inverts the tree at the current node.\n",
-        "\n",
-        "5. We then recursively call `invertTree` on the left and right subtrees to invert them.\n",
-        "\n",
-        "6. Finally, we return the root of the inverted tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inverted Tree (Example 1): <__main__.TreeNode object at 0x270c9c0>\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: root = [4,2,7,1,3,6,9]\n",
-        "\n",
-        "root1 = TreeNode(4)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(7)\n",
-        "root1.left.left = TreeNode(1)\n",
-        "root1.left.right = TreeNode(3)\n",
-        "root1.right.left = TreeNode(6)\n",
-        "root1.right.right = TreeNode(9)\n",
-        "\n",
-        "inverted_root1 = invertTree(root1)\n",
-        "print(\"Inverted Tree (Example 1):\", inverted_root1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inorder Traversal of Inverted Tree (Example 2): <__main__.TreeNode object at 0x2255fc0>\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: root = [2,1,3]\n",
-        "\n",
-        "root2 = TreeNode(2)\n",
-        "root2.left = TreeNode(1)\n",
-        "root2.right = TreeNode(3)\n",
-        "\n",
-        "inverted_root2 = invertTree(root2)\n",
-        "print(\"Inverted Tree (Example 2):\", inverted_root2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inverted Tree (Example 3): None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "# Input: root = []\n",
-        "\n",
-        "root3 = None  # Empty tree\n",
-        "\n",
-        "inverted_root3 = invertTree(root3)\n",
-        "print(\"Inverted Tree (Example 3):\", inverted_root3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's discuss the time and space complexity of the provided Python code to invert a binary tree.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of the `invertTree` function is $O(n)$, where $n$ is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal of the tree. In the worst case, we have to visit all nodes in the tree.\n",
-        "\n",
-        "The reason for this time complexity is the depth-first traversal of the tree, where we recursively process the left and right subtrees of each node.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is determined by the function call stack during the recursion. In the worst case, the space complexity is $O(h)$, where $h$ is the height of the binary tree. \n",
-        "\n",
-        "In a completely unbalanced binary tree (essentially a linked list), the height $h$ is equal to the number of nodes $n$, resulting in a space complexity of $O(n)$. This occurs when the tree is skewed to one side.\n",
-        "\n",
-        "In a balanced binary tree, such as a full binary tree, the height $h$ is $O(log\\ n)$, and the space complexity is $O(log\\ n)$.\n",
-        "\n",
-        "The space complexity depends on how balanced the tree is. In practical scenarios, binary trees are often approximately balanced, so the space complexity is typically closer to $O(log\\ n)$.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "- Time Complexity: $O(n)$ where $n$ is the number of nodes.\n",
-        "- Space Complexity: $O(h)$, where $h$ is the height of the binary tree. In the worst case, it can be $O(n)$, and in a balanced tree, it is $O(log\\ n)$.\n",
-        "\n",
-        "Keep in mind that these complexities are based on the recursive implementation provided. Iterative solutions can achieve the same task with $O(1)$ space complexity, using auxiliary data structures like stacks or queues to mimic the recursion stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb b/docs/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb
deleted file mode 100644
index 673ced2..0000000
--- a/docs/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 230. Kth Smallest Element in a BST\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Kth Smallest Element in a BST problem on LeetCode, click here!](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary search tree, and an integer `k`, return *the $k^{th}$ smallest value (**1-indexed**) of all the values of the nodes in the tree.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is `n`.\n",
-        "- 1 <= `k <= n` <= $10^4$\n",
-        "- 0 <= `Node.val` <= $10^4$\n",
-        "\n",
-        "**Follow up**: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def kthSmallest(root, k):\n",
-        "    stack = []  # Initialize an empty stack to simulate the traversal\n",
-        "    while root or stack:\n",
-        "        while root:\n",
-        "            stack.append(root)  # Push the current node onto the stack\n",
-        "            root = root.left  # Move to the left child\n",
-        "        root = stack.pop()  # Pop a node from the stack\n",
-        "        k -= 1  # Decrement k since we've visited a node\n",
-        "        if k == 0:\n",
-        "            return root.val  # If k becomes 0, return the current node's value as the kth smallest\n",
-        "        root = root.right  # Move to the right child to continue the traversal"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in the binary search tree (BST). Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-        "\n",
-        "2. The `kthSmallest` function takes two parameters: `root`, which is the root of the BST, and `k`, which represents the kth smallest element we want to find.\n",
-        "\n",
-        "3. We initialize an empty stack (`stack`) to simulate the traversal of the BST. The stack will be used to keep track of nodes as we traverse the tree in an iterative manner.\n",
-        "\n",
-        "4. We enter a while loop that continues until either `root` becomes `None` (indicating we have traversed the entire tree) or the `stack` is empty.\n",
-        "\n",
-        "5. Within the loop, we start another while loop to traverse as far left as possible in the BST. We repeatedly push nodes onto the `stack` and move to their left children until we reach the leftmost leaf node. This is the smallest node in the BST.\n",
-        "\n",
-        "6. Once we have reached the leftmost leaf node (the smallest node), we pop nodes from the `stack` one by one. As we pop each node, we decrement `k` by 1 to keep track of the number of nodes we have visited.\n",
-        "\n",
-        "7. If `k` becomes 0 after decrementing, it means we have found the kth smallest element. In this case, we return the `val` of the current node as the result.\n",
-        "\n",
-        "8. If `k` is still greater than 0, it means we haven't found the kth smallest element yet. In this case, we move to the right child of the current node to continue the traversal, as the kth smallest element, if it exists, will be in the right subtree of the current node.\n",
-        "\n",
-        "9. The process continues until we find the kth smallest element or traverse the entire tree.\n",
-        "\n",
-        "10. Finally, we return the kth smallest element found.\n",
-        "\n",
-        "The key idea in this code is to perform an in-order traversal of the BST while keeping track of the kth smallest element. By visiting nodes in ascending order, we can efficiently find the kth smallest element in O(h + k) time, where h is the height of the BST and k is the desired kth element. This approach is both concise and efficient for this problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Create the tree for the first example: root = [3,1,4,null,2], k = 1\n",
-        "root1 = TreeNode(3)\n",
-        "root1.left = TreeNode(1)\n",
-        "root1.right = TreeNode(4)\n",
-        "root1.left.right = TreeNode(2)\n",
-        "k1 = 1\n",
-        "print(kthSmallest(root1, k1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Create the tree for the second example: root = [5,3,6,2,4,null,null,1], k = 3\n",
-        "root2 = TreeNode(5)\n",
-        "root2.left = TreeNode(3)\n",
-        "root2.right = TreeNode(6)\n",
-        "root2.left.left = TreeNode(2)\n",
-        "root2.left.right = TreeNode(4)\n",
-        "root2.left.left.left = TreeNode(1)\n",
-        "k2 = 3\n",
-        "print(kthSmallest(root2, k2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this code is O(h + k), where:\n",
-        "- h is the height of the binary search tree (BST).\n",
-        "- k is the desired kth smallest element we want to find.\n",
-        "\n",
-        "1. In the worst case, where the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the time complexity is O(n + k).\n",
-        "\n",
-        "2. In the best case, where the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the time complexity is O(log(n) + k).\n",
-        "\n",
-        "So, the time complexity can vary from O(log(n) + k) in the best-case scenario to O(n + k) in the worst-case scenario. Typically, for balanced BSTs, the time complexity is closer to O(log(n) + k).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this code is O(h) due to the stack used for the iterative in-order traversal, where h is the height of the BST.\n",
-        "\n",
-        "1. In the worst case, when the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the space complexity is O(n) because the stack can potentially store all n nodes.\n",
-        "\n",
-        "2. In the best case, when the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the space complexity is O(log(n)) because the stack will have at most log(n) nodes.\n",
-        "\n",
-        "So, the space complexity depends on the height of the BST and can vary from O(log(n)) in the best-case scenario to O(n) in the worst-case scenario.\n",
-        "\n",
-        "In practical terms, for balanced BSTs or moderately unbalanced BSTs, the space complexity is usually close to O(log(n)), and the code is efficient in terms of space usage."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Kth Largest Element**: Modify the code to find the kth largest element in the BST instead of the kth smallest element.\n",
-        "\n",
-        "2. **Kth Smallest Element in Two BSTs**: Given two BSTs, find the kth smallest element when considering both BSTs as a single sorted list. This involves merging the two BSTs while finding the kth element efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb b/docs/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb
deleted file mode 100644
index 666bcd9..0000000
--- a/docs/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 235. Lowest Common Ancestor of a Binary Search Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Lowest Common Ancestor of a Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)\n",
-        "\n",
-        "---\n",
-        "Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.\n",
-        "\n",
-        "According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[2, 10^5]$.\n",
-        "- $-10^9$ <= `Node.val` <= $10^9$\n",
-        "- All `Node.val` are **unique**.\n",
-        "- `p != q`\n",
-        "- `p` and `q` will exist in the BST.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def lowestCommonAncestor(root, p, q):\n",
-        "    # Ensure p is less than q\n",
-        "    if p.val > q.val:\n",
-        "        p, q = q, p\n",
-        "\n",
-        "    while root:\n",
-        "        # If the current node value is greater than both p and q, move left\n",
-        "        if root.val > q.val:\n",
-        "            root = root.left\n",
-        "        # If the current node value is less than both p and q, move right\n",
-        "        elif root.val < p.val:\n",
-        "            root = root.right\n",
-        "        # If the current node value is between p and q (inclusive), or it matches either p or q, it's the LCA\n",
-        "        else:\n",
-        "            return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `class TreeNode:` defines a simple class for representing nodes in a binary tree. Each `TreeNode` has three attributes:\n",
-        "   - `val`: The value of the node.\n",
-        "   - `left`: A reference to the left child node.\n",
-        "   - `right`: A reference to the right child node.\n",
-        "\n",
-        "2. `def lowestCommonAncestor(root, p, q):` is a function that takes three arguments:\n",
-        "   - `root`: The root node of the BST.\n",
-        "   - `p`: A TreeNode representing one of the nodes for which we want to find the LCA.\n",
-        "   - `q`: A TreeNode representing the other node for which we want to find the LCA.\n",
-        "\n",
-        "3. `if p.val > q.val:` checks if the value of `p` is greater than the value of `q`. In a BST, it's essential to ensure that `p` represents the smaller value, and `q` represents the larger value. If `p` is greater than `q`, the code swaps their values.\n",
-        "\n",
-        "4. The main logic is inside the `while` loop, which runs until `root` becomes `None`. It performs the following steps to find the LCA:\n",
-        "   - If the current `root` node's value is greater than the value of `q`, it means both `p` and `q` are on the left subtree of the current node. So, we move to the left child of the current node by setting `root = root.left`.\n",
-        "   - If the current `root` node's value is less than the value of `p`, it means both `p` and `q` are on the right subtree of the current node. So, we move to the right child of the current node by setting `root = root.right`.\n",
-        "   - If neither of the above conditions is met, it means the current `root` node's value is between `p` and `q`, or it matches either `p` or `q`. In this case, the current node is the lowest common ancestor (LCA), so we return `root`.\n",
-        "\n",
-        "The algorithm is efficient because it takes advantage of the properties of a BST. It eliminates subtrees that cannot contain the LCA by comparing the values of `p`, `q`, and the current `root` node. Eventually, it reaches the LCA node, and that node is returned as the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "6\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root = TreeNode(6)\n",
-        "root.left = TreeNode(2)\n",
-        "root.right = TreeNode(8)\n",
-        "root.left.left = TreeNode(0)\n",
-        "root.left.right = TreeNode(4)\n",
-        "root.right.left = TreeNode(7)\n",
-        "root.right.right = TreeNode(9)\n",
-        "root.left.right.left = TreeNode(3)\n",
-        "root.left.right.right = TreeNode(5)\n",
-        "\n",
-        "p = root.left  # Node 2\n",
-        "q = root.right  # Node 8\n",
-        "\n",
-        "result = lowestCommonAncestor(root, p, q)\n",
-        "print(result.val)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Using the same tree as before\n",
-        "p = root.left  # Node 2\n",
-        "q = root.left.right  # Node 4\n",
-        "\n",
-        "result = lowestCommonAncestor(root, p, q)\n",
-        "print(result.val) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root2 = TreeNode(2)\n",
-        "root2.left = TreeNode(1)\n",
-        "\n",
-        "p = root2  # Node 2\n",
-        "q = root2.left  # Node 1\n",
-        "\n",
-        "result = lowestCommonAncestor(root2, p, q)\n",
-        "print(result.val)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided algorithm for finding the lowest common ancestor (LCA) in a Binary Search Tree (BST).\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this algorithm is O(h), where \"h\" is the height of the BST. In the worst case, where the tree is completely unbalanced (essentially a linked list), the height of the tree is O(n), where \"n\" is the number of nodes in the tree. However, in a well-balanced BST, the height is logarithmic, which is O(log n).\n",
-        "\n",
-        "The reason for this time complexity is that the algorithm efficiently narrows down the search space by traversing either left or right subtrees based on the values of the target nodes `p` and `q`. It eliminates entire subtrees that cannot contain the LCA, leading to a relatively quick search.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of the algorithm is O(1). This is because it uses a constant amount of extra space, regardless of the size of the input BST. The only variables used are `root`, `p`, and `q`, and there are no data structures like stacks or queues used for additional space. The algorithm performs a simple traversal without recursion, so it does not consume extra memory as the tree depth increases.\n",
-        "\n",
-        "**In summary**, the provided algorithm for finding the LCA in a BST is both time and space-efficient. Its time complexity is O(h), where \"h\" is the height of the tree, and its space complexity is O(1), making it suitable for practical use even in large BSTs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "### Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple LCAs:** Extend the algorithm to find all the lowest common ancestors of two given nodes `p` and `q` in a BST. In some cases, there can be multiple LCAs.\n",
-        "\n",
-        "2. **LCA with k Nodes:** Given a BST and k nodes, find the lowest common ancestor of these k nodes. This is an extension of the problem where you must find the LCA of more than two nodes."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb b/docs/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb
deleted file mode 100644
index 75c48ea..0000000
--- a/docs/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb	
+++ /dev/null
@@ -1,297 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 297. Serialize and Deserialize Binary Tree\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Serialize and Deserialize Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.\n",
-        "\n",
-        "Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.\n",
-        "\n",
-        "**Clarification:** The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-        "- `-1000 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "TThe problem is to design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of characters or bytes so that it can be easily stored in a file, transmitted over a network, or otherwise persisted. Deserialization is the reverse process of converting the serialized data back into the original data structure.\n",
-        "\n",
-        "In the context of this problem:\n",
-        "\n",
-        "1. **Serialization**: You are given a binary tree, and your task is to convert it into a string representation such that you can later recreate the same binary tree from this string. The format of serialization is flexible, but it should allow you to reconstruct the original binary tree accurately.\n",
-        "\n",
-        "2. **Deserialization**: Given a serialized string, you need to reconstruct the binary tree it represents, ensuring that it is identical to the original tree.\n",
-        "\n",
-        "Here are some key points to consider in solving this problem:\n",
-        "\n",
-        "- The input can include any valid binary tree, including trees with nodes having integer values within the range [-1000, 1000].\n",
-        "- You don't necessarily need to follow a specific format for serialization, but it should be designed in a way that allows unambiguous deserialization.\n",
-        "- The goal is to serialize the tree structure and its values and then deserialize it back into the same structure and values.\n",
-        "\n",
-        "A common approach for serialization is to use a traversal method like preorder traversal, where you visit nodes in the order: root, left subtree, right subtree. This way, you can serialize the tree into a string, and during deserialization, you can reconstruct the tree by parsing the string in the same order.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode(object):\n",
-        "    def __init__(self, x):\n",
-        "        self.val = x\n",
-        "        self.left = None\n",
-        "        self.right = None\n",
-        "\n",
-        "class Codec:\n",
-        "    def serialize(self, root):\n",
-        "        serialized  = []\n",
-        "\n",
-        "        def dfs(node):\n",
-        "            if not node:\n",
-        "                # If the node is None, represent it as \"Null\" in the serialized string\n",
-        "                serialized .append(\"Null\")\n",
-        "                return\n",
-        "            # Convert the node's value to a string and add it to the serialized string\n",
-        "            serialized .append(str(node.val))\n",
-        "            dfs(node.left)\n",
-        "            dfs(node.right)\n",
-        "\n",
-        "        dfs(root)\n",
-        "        # Join the serialized values with \",\"\n",
-        "        return \",\".join(serialized )\n",
-        "\n",
-        "    def deserialize(self, data):\n",
-        "        vals = data.split(\",\")\n",
-        "        self.i = 0\n",
-        "\n",
-        "        def dfs():\n",
-        "            if vals[self.i] == \"Null\":\n",
-        "                self.i += 1\n",
-        "                return None\n",
-        "            # Convert the value to an integer to create a new node\n",
-        "            node = TreeNode(int(vals[self.i]))\n",
-        "            self.i += 1\n",
-        "            node.left = dfs()\n",
-        "            node.right = dfs()\n",
-        "            return node\n",
-        "\n",
-        "        return dfs()"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Here's a step-by-step explanation of the code:\n",
-        "\n",
-        "1. **Definition of TreeNode**:\n",
-        "   - The code defines a simple `TreeNode` class that represents a node in a binary tree. Each `TreeNode` has three attributes:\n",
-        "     - `val`: The integer value stored in the node.\n",
-        "     - `left`: A reference to the left child node (or None if there is no left child).\n",
-        "     - `right`: A reference to the right child node (or None if there is no right child).\n",
-        "   - This class allows you to create binary tree nodes with integer values and left and right child references.\n",
-        "\n",
-        "2. **Serialization (serialize method)**:\n",
-        "   - The `serialize` method takes a `root` node as input, which is the root of the binary tree that needs to be serialized.\n",
-        "   - It initializes an empty list `serialized` to store the serialized elements.\n",
-        "   - The `dfs` (depth-first search) function is defined within the `serialize` method to perform a preorder traversal of the binary tree.\n",
-        "   - In the `dfs` function:\n",
-        "     - If the current `node` is `None` (i.e., a leaf node or a child of a leaf node), it appends the string \"Null\" to the `serialized` list to represent the absence of a node.\n",
-        "     - If the current `node` is not `None`, it appends the string representation of the `node.val` to the `serialized` list and then recursively calls `dfs` on the left and right children.\n",
-        "   - After the `dfs` traversal, the `serialized` list contains the serialized binary tree elements.\n",
-        "   - The method returns a string obtained by joining the elements in the `serialized` list with commas.\n",
-        "\n",
-        "3. **Deserialization (deserialize method)**:\n",
-        "   - The `deserialize` method takes a serialized string `data` as input, which represents a binary tree in the custom format.\n",
-        "   - It splits the `data` string by commas to obtain a list of elements called `vals`.\n",
-        "   - The method initializes an index `self.i` to 0. This index keeps track of the current position in the `vals` list during deserialization.\n",
-        "   - The `dfs` function is defined within the `deserialize` method to perform the deserialization process using a recursive approach.\n",
-        "   - In the `dfs` function:\n",
-        "     - If the current element in `vals` is \"Null,\" it means there is no node at this position in the binary tree, so it returns `None`.\n",
-        "     - If the current element is not \"Null,\" it converts the element to an integer to create a new `TreeNode` with that value.\n",
-        "     - It then recursively calls `dfs` to set the left and right children of the current node.\n",
-        "   - The method returns the root node of the reconstructed binary tree.\n",
-        "\n",
-        "Overall, this code demonstrates a way to serialize a binary tree into a string format and then deserialize it back into the original tree structure, allowing for the representation of both the tree structure and the values stored in the nodes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1,2,Null,Null,3,4,Null,Null,5,Null,Null\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Serialize a binary tree\n",
-        "root = TreeNode(1)\n",
-        "root.left = TreeNode(2)\n",
-        "root.right = TreeNode(3)\n",
-        "root.right.left = TreeNode(4)\n",
-        "root.right.right = TreeNode(5)\n",
-        "\n",
-        "codec = Codec()\n",
-        "serialized_tree = codec.serialize(root)\n",
-        "print(serialized_tree)\n",
-        "\n",
-        "# Deserialize the serialized tree\n",
-        "new_root = codec.deserialize(serialized_tree)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Null\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Example usage:\n",
-        "# Serialize a binary tree\n",
-        "root = None\n",
-        "\n",
-        "codec = Codec()\n",
-        "serialized_tree = codec.serialize(root)\n",
-        "print(serialized_tree)\n",
-        "\n",
-        "# Deserialize the serialized tree\n",
-        "new_root = codec.deserialize(serialized_tree)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for serializing and deserializing a binary tree:\n",
-        "\n",
-        "1. **Serialization (serialize method)**:\n",
-        "\n",
-        "   - Time Complexity:\n",
-        "     - The `serialize` method uses a depth-first traversal (preorder) of the binary tree.\n",
-        "     - It visits each node exactly once.\n",
-        "     - At each node, it performs constant-time operations (string conversion and list append).\n",
-        "     - Therefore, the time complexity for serialization is O(N), where N is the number of nodes in the binary tree.\n",
-        "\n",
-        "   - Space Complexity:\n",
-        "     - The space complexity for serialization includes the space used by the `serialized` list and the recursive call stack.\n",
-        "     - The `serialized` list stores the serialized elements, and its space is proportional to the number of nodes.\n",
-        "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-        "     - Therefore, the space complexity for serialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-        "\n",
-        "2. **Deserialization (deserialize method)**:\n",
-        "\n",
-        "   - Time Complexity:\n",
-        "     - The `deserialize` method splits the serialized string into a list of elements, which takes O(N) time.\n",
-        "     - The `dfs` function is a recursive function that visits each element in the list exactly once.\n",
-        "     - At each element, it performs constant-time operations (string comparison, integer conversion, and recursive function calls).\n",
-        "     - Therefore, the time complexity for deserialization is O(N).\n",
-        "\n",
-        "   - Space Complexity:\n",
-        "     - The space complexity for deserialization includes the space used by the `vals` list and the recursive call stack.\n",
-        "     - The `vals` list stores the elements from the serialized string, and its space is proportional to the number of nodes.\n",
-        "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-        "     - Therefore, the space complexity for deserialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-        "\n",
-        "Overall, the provided code has a time complexity of O(N) for both serialization and deserialization and a space complexity of O(N) in the majority of cases (unless the tree is severely unbalanced, in which case the height H dominates the space complexity). It efficiently serializes and deserializes a binary tree while using a linear amount of memory."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Efficient Serialization**:\n",
-        "   Optimize the serialization process to minimize the length of the serialized string. Consider using techniques like binary encoding to represent node values more efficiently.\n",
-        "\n",
-        "2. **Custom Serialization Format**:\n",
-        "   Modify the serialization and deserialization methods to use a custom format different from the one provided. Ensure that the new format allows for accurate reconstruction of the original binary tree."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb b/docs/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb
deleted file mode 100644
index 8a41c3b..0000000
--- a/docs/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb	
+++ /dev/null
@@ -1,265 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 572: Subtree of Another Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Subtree of Another Tree problem on LeetCode, click here!](https://leetcode.com/problems/subtree-of-another-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "**Problem Description:**\n",
-        "\n",
-        "Given the roots of two binary trees `root` and `subRoot`, return `True` if there is a subtree of `root` with the same structure and node values as `subRoot`, and `False` otherwise.\n",
-        "\n",
-        "A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "1. The number of nodes in the `root` tree is in the range $[1, 2000]$.\n",
-        "2. The number of nodes in the `subRoot` tree is in the range $[1, 1000]$.\n",
-        "3. $-10^4$ <= root.val <= $10^4$\n",
-        "4. $-10^4$ <= subRoot.val <= $10^4$\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "class Solution:\n",
-        "    def isSubtree(self, root, subRoot):\n",
-        "        if not root:\n",
-        "            return False\n",
-        "        \n",
-        "        # Check if the current subtree is equal to the subRoot\n",
-        "        if self.isSameTree(root, subRoot):\n",
-        "            return True\n",
-        "        \n",
-        "        # Recursively check the left and right subtrees\n",
-        "        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)\n",
-        "    \n",
-        "    def isSameTree(self, tree1, tree2):\n",
-        "        if not tree1 and not tree2:\n",
-        "            return True\n",
-        "        if not tree1 or not tree2:\n",
-        "            return False\n",
-        "        \n",
-        "        return (\n",
-        "            tree1.val == tree2.val and\n",
-        "            self.isSameTree(tree1.left, tree2.left) and\n",
-        "            self.isSameTree(tree1.right, tree2.right)\n",
-        "        )"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "This code defines a TreeNode class for the binary tree nodes and a Solution class with two methods:\n",
-        "\n",
-        "1. `isSubtree`: This method checks if there is a subtree of the `root` tree with the same structure and node values as the `subRoot` tree. It uses a helper function `isSameTree` to compare two trees for equality.\n",
-        "\n",
-        "2. `isSameTree`: This helper method recursively compares two trees to check if they are the same in structure and node values.\n",
-        "\n",
-        "### Here's a detailed explanation of the code:\n",
-        "\n",
-        "1. **TreeNode Class:**\n",
-        "\n",
-        "The `TreeNode` class is defined to represent nodes in a binary tree. Each node has a `val` (the node's value) and may have a `left` and `right` child.\n",
-        "\n",
-        "2. **Solution Class:**\n",
-        "\n",
-        "    + The `Solution` class contains the solution for the problem and defines two important methods:\n",
-        "\n",
-        "    + `isSubtree(self, root, subRoot)`:\n",
-        "\n",
-        "        - This method checks whether `subRoot` is a subtree of `root`. It takes two tree nodes, `root` and `subRoot`, as input arguments.\n",
-        "        - If `root` is `None`, it returns `False` because there is no subtree to search.\n",
-        "        - It then checks if the current subtree with `root` as its root is equal to `subRoot` using the `isSameTree` method. If they are the same, it returns `True`.\n",
-        "        - If the current subtree is not the same as `subRoot`, it recursively checks the left and right subtrees of `root` to see if `subRoot` is a subtree of any of them.\n",
-        "        - It returns `True` if `subRoot` is found in either the left or right subtree; otherwise, it returns `False`.\n",
-        "\n",
-        "    + `isSameTree(self, tree1, tree2)`:\n",
-        "\n",
-        "        - This method checks whether two trees, `tree1` and `tree2`, are the same.\n",
-        "        - If both `tree1` and `tree2` are `None`, they are considered the same tree, so it returns `True`.\n",
-        "        - If only one of them is `None` (but not both), they are different trees, so it returns `False`.\n",
-        "        - If both `tree1` and `tree2` have values, it checks if their values are equal and recursively checks if their left and right subtrees are the same.\n",
-        "        - It returns `True` if the trees are the same; otherwise, it returns `False`.\n",
-        "\n",
-        "The code effectively uses recursion to traverse the binary trees and check for subtree equality. The `isSubtree` method starts the recursive search, and the `isSameTree` method is used to compare individual subtrees. The approach is efficient and avoids unnecessary checks when possible.\n",
-        "\n",
-        "This solution allows you to determine if there exists a subtree within the `root` tree that matches the structure and node values of `subRoot`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Is subRoot a subtree of root? True\n"
-          ]
-        }
-      ],
-      "source": [
-        "#root = [3,4,5,1,2]\n",
-        "#subRoot = [4,1,2]\n",
-        "\n",
-        "# Example input\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(4)\n",
-        "root.right = TreeNode(5)\n",
-        "root.left.left = TreeNode(1)\n",
-        "root.left.right = TreeNode(2)\n",
-        "\n",
-        "subRoot = TreeNode(4)\n",
-        "subRoot.left = TreeNode(1)\n",
-        "subRoot.right = TreeNode(2)\n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test the isSubtree method\n",
-        "result = solution.isSubtree(root, subRoot)\n",
-        "print(\"Is subRoot a subtree of root?\", result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Is subRoot a subtree of root? False\n"
-          ]
-        }
-      ],
-      "source": [
-        "#root = [3,4,5,1,2,null,null,null,null,0] \n",
-        "#subRoot = [4,1,2]\n",
-        "\n",
-        "\n",
-        "# Example input\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(4)\n",
-        "root.right = TreeNode(5)\n",
-        "root.left.left = TreeNode(1)\n",
-        "root.left.right = TreeNode(2)\n",
-        "root.left.right.left = TreeNode(0)\n",
-        "\n",
-        "subRoot = TreeNode(4)\n",
-        "subRoot.left = TreeNode(1)\n",
-        "subRoot.right = TreeNode(2)\n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test the isSubtree method\n",
-        "result = solution.isSubtree(root, subRoot)\n",
-        "print(\"Is subRoot a subtree of root?\", result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "86a34f0d-0789-498c-818c-2d30f84754d0",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code for the \"Subtree of Another Tree\" problem:\n",
-        "\n",
-        "**Time Complexity**\n",
-        "\n",
-        "The time complexity of the code primarily depends on the recursive traversal of the binary tree. In both the `isSubtree` and `isSameTree` functions, we visit each node in the binary trees once. Let's break down the time complexity:\n",
-        "\n",
-        "1. The `isSubtree` function:\n",
-        "   - In the worst case, we need to visit every node in the `root` tree.\n",
-        "   - For each node in the `root` tree, we call the `isSameTree` function, which has its own traversal.\n",
-        "   - So, the total time complexity is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree.\n",
-        "\n",
-        "2. The `isSameTree` function:\n",
-        "   - In the worst case, we visit every node in both `tree1` and `tree2`.\n",
-        "   - The number of recursive calls made is proportional to the number of nodes in the trees.\n",
-        "   - So, the time complexity of this function is $O(max(n, m))$, where $n$ and $m$ are the numbers of nodes in `tree1` and `tree2`, respectively.\n",
-        "\n",
-        "Overall, the time complexity of the entire code is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree. In practice, it may be less than $O(n * m)$ if a subtree mismatch is detected early during the traversal.\n",
-        "\n",
-        "**Space Complexity**\n",
-        "\n",
-        "The space complexity of the code is determined by the function call stack during recursion and the space used by the recursive functions. Let's analyze the space complexity:\n",
-        "\n",
-        "1. The `isSubtree` function:\n",
-        "   - It uses the call stack for recursion.\n",
-        "   - The maximum depth of the recursion is equal to the height of the `root` tree, which can be $O(n)$ in the worst case (unbalanced tree).\n",
-        "   - Additionally, the function doesn't use any significant extra space other than the recursion stack.\n",
-        "\n",
-        "2. The `isSameTree` function:\n",
-        "   - It also uses the call stack for recursion.\n",
-        "   - The maximum depth of the recursion is equal to the height of the `tree1` or `tree2`, whichever is greater.\n",
-        "   - So, the maximum space used for the call stack is $O(max(n, m))$.\n",
-        "\n",
-        "In summary, the space complexity of the code is $O(max(n, m))$ due to the function call stack. It scales with the maximum height of the trees being compared.\n",
-        "\n",
-        "Overall, the code is efficient and works well for trees with moderate sizes. However, it's important to keep in mind that the worst-case time complexity is $O(n * m)$, so for very large trees, the performance may degrade."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb b/docs/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb
deleted file mode 100644
index ad86882..0000000
--- a/docs/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 98. Validate Binary Search Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Validate Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/validate-binary-search-tree/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, *determine if it is a valid binary search tree (BST)*.\n",
-        "\n",
-        "A **valid BST** is defined as follows:\n",
-        "\n",
-        "- The left subtree of a node contains only nodes with keys **less than** the node's key.\n",
-        "- The right subtree of a node contains only nodes with keys **greater than** the node's key.\n",
-        "- Both the left and right subtrees must also be binary search trees. \n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[1, 10^4]$.\n",
-        "- $-2^{31}$ <= `Node.val` <= $2^{31} - 1$\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def is_valid_BST(root):\n",
-        "    def is_valid(node, min_val, max_val):\n",
-        "        # Base case: If the node is None, it's a valid BST.\n",
-        "        if node is None:\n",
-        "            return True\n",
-        "        \n",
-        "        # Check if the current node's value is within the valid range.\n",
-        "        if not (min_val < node.val < max_val):\n",
-        "            return False\n",
-        "        \n",
-        "        # Recursively check the left and right subtrees with updated ranges.\n",
-        "        return (is_valid(node.left, min_val, node.val) and\n",
-        "                is_valid(node.right, node.val, max_val))\n",
-        "    \n",
-        "    # Call the helper function starting with a wide range for root node.\n",
-        "    return is_valid(root, float('-inf'), float('inf'))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- It defines a `TreeNode` class to represent nodes in the binary tree.\n",
-        "\n",
-        "- The `is_valid_BST` function takes the root node of the binary tree as input and returns `True` if the tree is a valid BST, and `False` otherwise.\n",
-        "\n",
-        "- Inside the `is_valid_BST` function, there is a helper function called `is_valid` that performs the actual validation using a recursive approach.\n",
-        "\n",
-        "- The `is_valid` function checks each node in the tree to ensure that it satisfies the properties of a BST:\n",
-        "  - The value of the current node must be within a valid range defined by `min_val` and `max_val`.\n",
-        "  - The left subtree of the current node should contain values less than the current node's value.\n",
-        "  - The right subtree of the current node should contain values greater than the current node's value.\n",
-        "\n",
-        "- If any of these conditions are violated, the function returns `False`, indicating that the tree is not a valid BST.\n",
-        "\n",
-        "- If all nodes satisfy these conditions, the function returns `True`, indicating that the tree is a valid BST.\n",
-        "\n",
-        "- The code calls the `is_valid` function with the root node and initial range values of negative infinity to positive infinity to start the validation.\n",
-        "\n",
-        "- The time complexity of the code is O(N), where N is the number of nodes in the tree, as it traverses each node once.\n",
-        "\n",
-        "- The space complexity depends on the height of the tree. In the average case for a balanced BST, the space complexity is O(log N), but in the worst case (skewed tree), it can be O(N) due to the recursive call stack.\n",
-        "\n",
-        "In summary, this code checks whether a given binary tree is a valid BST by recursively validating each node's value and its left and right subtrees while maintaining valid value ranges. If all nodes satisfy the BST properties, the tree is considered a valid BST."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Create the tree for the first example: [2, 1, 3]\n",
-        "root1 = TreeNode(2)\n",
-        "root1.left = TreeNode(1)\n",
-        "root1.right = TreeNode(3)\n",
-        "print(is_valid_BST(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Create the tree for the second example: [5, 1, 4, None, None, 3, 6]\n",
-        "root2 = TreeNode(5)\n",
-        "root2.left = TreeNode(1)\n",
-        "root2.right = TreeNode(4)\n",
-        "root2.right.left = TreeNode(3)\n",
-        "root2.right.right = TreeNode(6)\n",
-        "print(is_valid_BST(root2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `is_valid_BST` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The time complexity of the function primarily depends on the number of nodes in the binary tree.\n",
-        "- In the worst case, we may have to visit every node in the tree once to validate whether it's part of a valid BST.\n",
-        "- Since we are performing a depth-first search (DFS) traversal of the tree, the time complexity is O(N), where N is the number of nodes in the tree.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity is determined by the space used by the recursive call stack during the DFS traversal.\n",
-        "- In the worst case, if the tree is completely unbalanced (a skewed tree), the maximum depth of the call stack would be equal to the height of the tree.\n",
-        "- In a balanced BST, the height is approximately log2(N), where N is the number of nodes.\n",
-        "- Therefore, the space complexity of the call stack is O(log N) for a balanced BST.\n",
-        "- In the worst case (skewed tree), the space complexity can be O(N) as the height of the tree can be equal to N.\n",
-        "\n",
-        "**Overall:**\n",
-        "- Time Complexity: O(N)\n",
-        "- Space Complexity: O(log N) on average for a balanced BST, and O(N) in the worst case for a skewed tree.\n",
-        "\n",
-        "The space complexity is typically dominated by the recursive call stack, and it varies depending on the shape of the binary tree. In practice, for balanced binary trees, the space complexity is often close to O(log N), which is quite efficient."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Complex Constraints**: Modify the problem to have more complex constraints on the tree. For example, consider a binary tree with constraints like \"The left subtree of a node contains only nodes with keys less than or equal to the node's key,\" and \"The right subtree of a node contains only nodes with keys greater than or equal to the node's key.\" How would you adapt the code to handle these constraints?\n",
-        "\n",
-        "2. **Handling Duplicates**: Modify the code to handle binary search trees that allow duplicate values. In a standard BST, each value is unique, but in this case, multiple nodes can have the same value. The tree should still be considered valid if it follows the BST property."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/07. Trees/README.md b/docs/_build/html/_sources/07. Trees/README.md
deleted file mode 100644
index 1eba5d4..0000000
--- a/docs/_build/html/_sources/07. Trees/README.md	
+++ /dev/null
@@ -1,26 +0,0 @@
-# Trees Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)                          | Easy       |
-| [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)     | Easy       |
-| [100. Same Tree](https://leetcode.com/problems/same-tree/)                                           | Easy       |
-| [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)               | Easy       |
-| [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/) | Easy |
-| [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/) | Medium |
-| [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)       | Medium     |
-| [230. Kth Smallest Element In a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)   | Medium     |
-| [105. Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) | Medium |
-| [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)     | Hard       |
-| [297. Serialize And Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/) | Hard |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb b/docs/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb
deleted file mode 100644
index 9da2a17..0000000
--- a/docs/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 208. Implement Trie (Prefix Tree)\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Implement Trie problem on LeetCode, click here!](https://leetcode.com/problems/implement-trie-prefix-tree/)\n",
-        "\n",
-        "---\n",
-        "A **trie** (pronounced as \"try\") or **prefix tree** is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.\n",
-        "\n",
-        "Implement the Trie class:\n",
-        "\n",
-        "- `Trie()` Initializes the trie object.\n",
-        "- `void insert(String word)` Inserts the string `word` into the trie.\n",
-        "- `boolean search(String word)` Returns `true` if the string `word` is in the trie (i.e., was inserted before), and `false` otherwise.\n",
-        "- `boolean startsWith(String prefix)` Returns `true` if there is a previously inserted string `word` that has the prefix `prefix`, and `false` otherwise.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= word.length, prefix.length <= 2000`\n",
-        "- `word` and `prefix` consist only of lowercase English letters.\n",
-        "- At most $3 * 10^4$ calls in total will be made to `insert`, `search`, and `startsWith`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to implement a data structure known as a Trie (pronounced \"try\") or Prefix Tree. A Trie is a tree-like data structure that is used to efficiently store and retrieve a set of strings or words. It's particularly useful for tasks involving string manipulation, such as autocomplete and spell-checking. The problem statement defines the following operations for implementing the Trie class:\n",
-        "\n",
-        "1. `Trie()`: This is the constructor that initializes the Trie object.\n",
-        "\n",
-        "2. `void insert(String word)`: This method inserts the given string `word` into the Trie. It effectively adds the characters of the word to the Trie's structure, creating nodes for each character if they don't already exist. At the end of the word, a special flag is set to indicate that this node represents the end of a valid word.\n",
-        "\n",
-        "3. `boolean search(String word)`: This method checks if the given string `word` is present in the Trie. It starts from the root of the Trie and traverses the Trie by following the characters in the word. If it successfully traverses the Trie and reaches the end of the word, it returns `true`, indicating that the word exists in the Trie; otherwise, it returns `false`.\n",
-        "\n",
-        "4. `boolean startsWith(String prefix)`: This method checks if there is any previously inserted word in the Trie that has the given `prefix`. It's similar to the `search` method but does not require that the prefix be a complete word. If there is any word in the Trie that starts with the given prefix, it returns `true`; otherwise, it returns `false`.\n",
-        "\n",
-        "The problem also provides an example scenario where these operations are called, demonstrating the expected output for each operation.\n",
-        "\n",
-        "In summary, the goal is to create a data structure (Trie) that efficiently stores a set of strings and provides methods to insert new strings, search for complete words, and check if a given prefix exists in the stored set of strings.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        self.children = {}           # A dictionary to store child nodes\n",
-        "        self.is_end_of_word = False  # Indicates if a word ends at this node\n",
-        "\n",
-        "class Trie:\n",
-        "    def __init__(self):\n",
-        "        self.root = TrieNode()       # Initialize the Trie with a root node\n",
-        "\n",
-        "    def insert(self, word):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                node.children[char] = TrieNode()  # Create a new node if the character is not present\n",
-        "            node = node.children[char]           # Move to the child node\n",
-        "        node.is_end_of_word = True                # Mark the end of the inserted word\n",
-        "\n",
-        "    def search(self, word):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                return False         # If the character is not found, the word is not in the Trie\n",
-        "            node = node.children[char]  # Move to the child node\n",
-        "        return node.is_end_of_word    # Check if the node represents the end of a valid word\n",
-        "\n",
-        "    def startsWith(self, prefix):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in prefix:\n",
-        "            if char not in node.children:\n",
-        "                return False         # If the character is not found, no word starts with the prefix\n",
-        "            node = node.children[char]  # Move to the child node\n",
-        "        return True                  # Prefix found in the Trie"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's explain how the code works step by step:\n",
-        "\n",
-        "1. `TrieNode` Class:\n",
-        "   - `TrieNode` is a class that represents nodes in the Trie. Each node has two attributes:\n",
-        "     - `children`: This is a dictionary that stores child nodes. Each key in the dictionary represents a character, and the corresponding value is the child node for that character.\n",
-        "     - `is_end_of_word`: This boolean flag indicates whether the node represents the end of a complete word. Initially, it's set to `False` for all nodes.\n",
-        "\n",
-        "2. `Trie` Class:\n",
-        "   - `Trie` is the main class that implements the Trie data structure. It has the following methods:\n",
-        "\n",
-        "   - `__init__(self)`: The constructor initializes the Trie object by creating a root node, which serves as the starting point for all Trie operations.\n",
-        "\n",
-        "   - `insert(self, word)`: This method inserts a string `word` into the Trie. It starts from the root node and iterates through each character in the word.\n",
-        "     - If a character is not present as a child node, it creates a new node for that character.\n",
-        "     - It then moves to the child node and continues the process until the entire word is inserted.\n",
-        "     - Finally, it sets the `is_end_of_word` flag to `True` for the last node to mark the end of the inserted word.\n",
-        "\n",
-        "   - `search(self, word)`: This method checks if a complete word (string) exists in the Trie. It starts from the root node and iterates through each character in the word.\n",
-        "     - If a character is not found as a child node, it immediately returns `False` because the word is not in the Trie.\n",
-        "     - It continues to move to the child node for each character.\n",
-        "     - After reaching the end of the word, it checks if the `is_end_of_word` flag is `True` for the last node to confirm the presence of the word.\n",
-        "\n",
-        "   - `startsWith(self, prefix)`: This method checks if there is any previously inserted word in the Trie that starts with a given `prefix`. It follows the same logic as the `search` method but does not require the entire word to be present.\n",
-        "     - If the prefix is found in the Trie, it returns `True`; otherwise, it returns `False`.\n",
-        "\n",
-        "3. Example Usage:\n",
-        "   - The code demonstrates how to create a Trie object, insert words (\"apple\" and \"app\"), and perform operations on the Trie:\n",
-        "     - It inserts the word \"apple\" into the Trie.\n",
-        "     - It checks if \"apple\" is in the Trie, which returns `True`.\n",
-        "     - It checks if \"app\" is in the Trie, which returns `False`.\n",
-        "     - It checks if there is any word in the Trie that starts with \"app,\" which returns `True`.\n",
-        "     - It inserts the word \"app\" into the Trie.\n",
-        "     - It checks if \"app\" is in the Trie again, which now returns `True`.\n",
-        "\n",
-        "In summary, the code efficiently implements a Trie data structure with the ability to insert words, search for complete words, and check for the existence of words with a given prefix. The Trie is organized as a tree of nodes, with each node representing a character in the words being stored."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n",
-            "False\n",
-            "True\n",
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "\n",
-        "trie = Trie()\n",
-        "trie.insert(\"apple\")\n",
-        "print(trie.search(\"apple\"))   # Output: True\n",
-        "print(trie.search(\"app\"))     # Output: False\n",
-        "print(trie.startsWith(\"app\")) # Output: True\n",
-        "trie.insert(\"app\")\n",
-        "print(trie.search(\"app\"))     # Output: True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the Trie implementation:\n",
-        "\n",
-        "1. Time Complexity:\n",
-        "   - Insertion (`insert` method): The time complexity of inserting a word into the Trie is O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to insert it into the Trie.\n",
-        "   - Search (`search` method): The time complexity of searching for a word in the Trie is also O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to determine if it exists in the Trie.\n",
-        "   - Starts with (`startsWith` method): The time complexity of checking if a prefix exists in the Trie is O(p), where p is the length of the prefix. In the worst case, you may need to traverse the entire prefix to determine its presence.\n",
-        "\n",
-        "   Therefore, for each operation (insertion, search, or starts with), the time complexity is O(m) or O(p), where m is the length of the word being operated on, and p is the length of the prefix.\n",
-        "\n",
-        "2. Space Complexity:\n",
-        "   - The space complexity of the Trie is determined by the number of nodes and characters stored in it. In the worst case, where none of the inserted words share common prefixes, the space complexity is O(N), where N is the total number of characters across all inserted words.\n",
-        "   - Each node in the Trie represents a character, and the number of nodes is directly related to the total number of characters.\n",
-        "\n",
-        "   In practice, the space complexity can be less than O(N) because common prefixes are shared among words in the Trie, leading to space optimization.\n",
-        "\n",
-        "In summary:\n",
-        "- Time complexity for each operation (insertion, search, starts with) is O(m) or O(p), where m is the length of the word and p is the length of the prefix.\n",
-        "- Space complexity is O(N) in the worst case, where N is the total number of characters across all inserted words. However, space optimization occurs due to common prefix sharing in practice, potentially reducing the actual space used."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Count Prefixes**:\n",
-        "   Implement a method to count the number of words in the Trie that have a specific prefix. This exercise requires you to maintain additional data in the Trie nodes to keep track of the count.\n",
-        "\n",
-        "2. **Auto-Complete Suggestions**:\n",
-        "   Implement an auto-complete feature using the Trie. Given a prefix, the program should return a list of suggested words based on the words previously inserted into the Trie."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb b/docs/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb
deleted file mode 100644
index 2de472c..0000000
--- a/docs/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb	
+++ /dev/null
@@ -1,275 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 211. Design Add and Search Words Data Structure\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Design Add and Search Words Data Structure problem on LeetCode, click here!](https://leetcode.com/problems/design-add-and-search-words-data-structure/)\n",
-        "\n",
-        "---\n",
-        "Design a data structure that supports adding new words and finding if a string matches any previously added string.\n",
-        "\n",
-        "Implement the `WordDictionary` class:\n",
-        "\n",
-        "- `WordDictionary()` Initializes the object.\n",
-        "- `void addWord(word)` Adds `word` to the data structure, it can be matched later.\n",
-        "- `bool search(word)` Returns `true` if there is any string in the data structure that matches `word` or `false` otherwise. `word` may contain dots `'.'` where dots can be matched with any letter.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= word.length <= 25`\n",
-        "- `word` in `addWord` consists of lowercase English letters.\n",
-        "- `word` in `search` consist of `'.'` or lowercase English letters.\n",
-        "- There will be at most `2` dots in `word` for `search` queries.\n",
-        "- At most $10^4$ calls will be made to `addWord` and `search`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to design a data structure called `WordDictionary` that allows you to add words and search for words efficiently. The key feature of this data structure is that it should support wildcard search, where a dot ('.') can match any letter.\n",
-        "\n",
-        "Here are the specific requirements and explanations for the problem:\n",
-        "\n",
-        "1. Initialize the `WordDictionary` object using the constructor `WordDictionary()`.\n",
-        "\n",
-        "2. The `addWord(word)` method allows you to add a word to the data structure. Once added, you should be able to search for this word later. The words consist of lowercase English letters, and each word has a length between 1 and 25 characters.\n",
-        "\n",
-        "3. The `search(word)` method allows you to search for a word in the data structure. This method returns `True` if there is any string in the data structure that matches the given word or `False` otherwise. The word you are searching for may contain dots ('.'), where each dot can match any letter.\n",
-        "\n",
-        "4. The wildcard character ('.') in the `search` method allows for partial or fuzzy searching. For example, if the word dictionary contains the words \"bad,\" \"dad,\" and \"mad,\" then:\n",
-        "   - Searching for \"pad\" should return `False` because there is no exact match.\n",
-        "   - Searching for \"bad\" should return `True` because \"bad\" is in the dictionary.\n",
-        "   - Searching for \".ad\" should return `True` because it can match \"bad,\" \"dad,\" or \"mad.\"\n",
-        "   - Searching for \"b..\" should return `True` because it can match \"bad,\" \"bed,\" \"bet,\" etc.\n",
-        "\n",
-        "5. The problem specifies that there will be at most two dots ('.') in search queries, and there will be at most 10^4 calls to both `addWord` and `search`.\n",
-        "\n",
-        "In summary, you need to implement the `WordDictionary` class that efficiently supports adding words and searching for words, where the search can include wildcard characters ('.') that match any letter.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        # Initialize a TrieNode with children and a flag to indicate the end of a word.\n",
-        "        self.children = {}\n",
-        "        self.is_end_of_word = False\n",
-        "\n",
-        "class WordDictionary:\n",
-        "\n",
-        "    def __init__(self):\n",
-        "        # Initialize the WordDictionary with a root node.\n",
-        "        self.root = TrieNode()\n",
-        "\n",
-        "    def addWord(self, word: str) -> None:\n",
-        "        node = self.root\n",
-        "        # Iterate through the characters in the word.\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                # Create a new node for the character if it doesn't exist.\n",
-        "                node.children[char] = TrieNode()\n",
-        "            # Move to the next node.\n",
-        "            node = node.children[char]\n",
-        "        # Mark the end of the word by setting the flag to True.\n",
-        "        node.is_end_of_word = True\n",
-        "\n",
-        "    def search_helper(self, node: TrieNode, word: str) -> bool:\n",
-        "        if not word:\n",
-        "            # If there are no characters left in the word, check if we reached the end of a word in the trie.\n",
-        "            return node.is_end_of_word\n",
-        "        \n",
-        "        char = word[0]\n",
-        "        if char == '.':\n",
-        "            # If the character is a dot, explore all possible child nodes.\n",
-        "            for child in node.children.values():\n",
-        "                if self.search_helper(child, word[1:]):\n",
-        "                    return True\n",
-        "        elif char in node.children:\n",
-        "            # If the character is in the children of the current node, move to the next node.\n",
-        "            return self.search_helper(node.children[char], word[1:])\n",
-        "        \n",
-        "        # If the character is not found and is not a dot, the word is not in the trie.\n",
-        "        return False\n",
-        "\n",
-        "    def search(self, word: str) -> bool:\n",
-        "        # Start the search from the root of the trie.\n",
-        "        return self.search_helper(self.root, word)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "This code defines a `WordDictionary` class that uses a Trie data structure to efficiently store and search for words. Here's a step-by-step explanation:\n",
-        "\n",
-        "1. `TrieNode` class: This class represents nodes in the Trie data structure. Each node has two attributes: `children` (a dictionary to store child nodes) and `is_end_of_word` (a flag to indicate whether the node marks the end of a word).\n",
-        "\n",
-        "2. `WordDictionary` class: This class initializes with an empty Trie by creating a root node.\n",
-        "\n",
-        "3. `addWord` method: This method adds a word to the Trie. It iterates through each character in the word and creates Trie nodes as necessary. It sets the `is_end_of_word` flag to `True` for the final character of the word to mark the end of the word.\n",
-        "\n",
-        "4. `search_helper` method: This is a recursive helper function for searching words in the Trie. It takes a Trie node (`node`) and a word to search (`word`) as input. If there are no characters left in the word (`not word`), it checks if the current node marks the end of a word. If it does, it returns `True`.\n",
-        "\n",
-        "5. If the first character of the word is a dot ('.'), the function explores all possible child nodes by iterating through `node.children.values()` and recursively calls `search_helper` for each child with the remaining part of the word (`word[1:]`).\n",
-        "\n",
-        "6. If the first character of the word is not a dot and is found in the children of the current node, the function recursively moves to the next node by calling `search_helper` on the child node and the remaining part of the word.\n",
-        "\n",
-        "7. If the character is not found in the children and is not a dot, the word is not in the Trie, so the function returns `False`.\n",
-        "\n",
-        "8. `search` method: This is the public method for searching words. It initiates the search process by calling `search_helper` starting from the root node of the Trie.\n",
-        "\n",
-        "Overall, this code efficiently implements a Trie-based data structure that allows adding and searching for words, including support for wildcard characters represented by dots ('.')."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n",
-            "True\n",
-            "True\n",
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "\n",
-        "wordDictionary = WordDictionary()\n",
-        "wordDictionary.addWord(\"bad\")\n",
-        "wordDictionary.addWord(\"dad\")\n",
-        "wordDictionary.addWord(\"mad\")\n",
-        "print(wordDictionary.search(\"pad\"))  # Output: False\n",
-        "print(wordDictionary.search(\"bad\"))  # Output: True\n",
-        "print(wordDictionary.search(\".ad\"))  # Output: True\n",
-        "print(wordDictionary.search(\"b..\"))  # Output: True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `WordDictionary` class methods:\n",
-        "\n",
-        "1. `addWord` Method:\n",
-        "\n",
-        "   - Time Complexity: O(L)\n",
-        "     - L is the length of the word being added.\n",
-        "     - In the worst case, we need to iterate through each character in the word and create nodes in the Trie. This takes O(L) time.\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for storing the word in the Trie is also O(L) because we create nodes for each character in the word.\n",
-        "\n",
-        "2. `search_helper` Method (called by `search`):\n",
-        "\n",
-        "   - Time Complexity: O(2^M * L)\n",
-        "     - M is the maximum number of dots ('.') in the search word.\n",
-        "     - In the worst case, when there are M dots in the search word, we may explore all possible child nodes at each dot. This results in a branching factor of 26 (for lowercase letters) at each dot.\n",
-        "     - So, the time complexity is exponential in the number of dots and linear in the length of the word being searched (L).\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for the recursive call stack is O(L) in the worst case because we may have to recurse to a depth equal to the length of the search word.\n",
-        "\n",
-        "3. `search` Method:\n",
-        "\n",
-        "   - Time Complexity: O(2^M * L)\n",
-        "     - The `search` method calls the `search_helper` method, which has the same time complexity explained above.\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for the recursive call stack is O(L) in the worst case, as explained earlier.\n",
-        "\n",
-        "Overall, the time complexity of the `search` method is mainly affected by the number of dots ('.') in the search word. In the worst case, when there are multiple dots and many possibilities to explore at each dot, the time complexity can be exponential. However, for most practical cases, it performs reasonably well.\n",
-        "\n",
-        "The space complexity primarily depends on the space required to store the words in the Trie. It is proportional to the total number of characters in all added words and is linear with respect to the length of the words added.\n",
-        "\n",
-        "In summary:\n",
-        "\n",
-        "- Time Complexity for `addWord`: O(L)\n",
-        "- Time Complexity for `search`: O(2^M * L)\n",
-        "- Space Complexity: O(L) for storing words in the Trie, O(L) for the recursive call stack during search.\n",
-        "\n",
-        "Note: The space complexity analysis assumes that the dictionary contains a reasonable number of words and does not account for the overhead of Python objects or system-level memory allocation."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Dictionary Auto-Completion**: Extend the `WordDictionary` class to provide auto-completion suggestions based on the prefix of a word. Implement a method that returns a list of words that match the given prefix.\n",
-        "\n",
-        "2. **Support Word Deletion**: Extend the `WordDictionary` class to support word deletion. Add a method `deleteWord(word)` that removes a word from the data structure. Ensure that the search operation still works correctly after deletion."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/08. Tries/212. Word Search II.ipynb b/docs/_build/html/_sources/08. Tries/212. Word Search II.ipynb
deleted file mode 100644
index 71597b8..0000000
--- a/docs/_build/html/_sources/08. Tries/212. Word Search II.ipynb	
+++ /dev/null
@@ -1,316 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 212. Word Search II\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Word Search II problem on LeetCode, click here!](https://leetcode.com/problems/word-search-ii/)\n",
-        "\n",
-        "---\n",
-        "Given an `m x n` `board` of characters and a list of strings `words`, the task is to return *all words on the board*.\n",
-        "\n",
-        "Each word must be constructed from letters of sequentially **adjacent cells**, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == board.length`\n",
-        "- `n == board[i].length`\n",
-        "- `1 <= m, n <= 12`\n",
-        "- `board[i][j]` is a lowercase English letter.\n",
-        "- 1 <= `words.length` <= $3 * 10^4$\n",
-        "- `1 <= words[i].length <= 10`\n",
-        "- `words[i]` consists of lowercase English letters.\n",
-        "- All the strings of `words` are unique."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "To solve the \"Word Search II\" problem, you can use a Trie data structure and perform a depth-first search (DFS) on the board. Here's a step-by-step approach to solve the problem:\n",
-        "\n",
-        "1. Build a Trie from the given list of words:\n",
-        "   - Create a Trie data structure to store all the words from the input list.\n",
-        "   - For each word in the input list, insert it into the Trie, character by character. Make sure to mark the end of each word in the Trie.\n",
-        "\n",
-        "2. Perform DFS on the board:\n",
-        "   - Iterate through each cell (i, j) on the board.\n",
-        "   - For each cell, start a DFS traversal from that cell to search for words.\n",
-        "   - During the DFS traversal, maintain a current path string.\n",
-        "   - At each cell, check if the current path string, concatenated with the character in the cell, matches any word prefix in the Trie. If it does, continue the DFS.\n",
-        "   - If the current path string matches a word in the Trie, add it to the result.\n",
-        "\n",
-        "3. Implement DFS with backtracking:\n",
-        "   - During the DFS traversal, you will explore neighboring cells (horizontally and vertically) if they are within bounds and haven't been visited before.\n",
-        "   - To prevent revisiting the same cell within the same word, mark the cell as visited (e.g., change its character to a special character like '#') before exploring it and restore its original character after the DFS traversal.\n",
-        "\n",
-        "4. Return the result:\n",
-        "   - After completing the DFS traversal for all cells on the board, you will have collected all the valid words in the result set.\n",
-        "   - Convert the result set to a list and return it as the final output.\n",
-        "\n",
-        "This approach efficiently finds all words on the board by utilizing the Trie data structure to prune unnecessary DFS paths and avoiding duplicate visits to the same cell within the same word. It ensures that each word is constructed from letters of sequentially adjacent cells without using the same letter cell more than once in a word.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Define a TrieNode class to represent nodes in the Trie.\n",
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        self.children = {}  # Dictionary to store child nodes.\n",
-        "        self.is_end_of_word = False  # Flag to mark the end of a word.\n",
-        "\n",
-        "# Main function to find words on the board.\n",
-        "def findWords(board, words):\n",
-        "    # DFS function to search for words starting from a given cell.\n",
-        "    def dfs(node, i, j, path):\n",
-        "        char = board[i][j]  # Get the character at the current cell.\n",
-        "        curr_node = node.children.get(char)  # Find the corresponding Trie node.\n",
-        "\n",
-        "        if not curr_node:\n",
-        "            return  # If no matching Trie node, stop the search.\n",
-        "\n",
-        "        path += char  # Add the character to the current path.\n",
-        "\n",
-        "        if curr_node.is_end_of_word:\n",
-        "            result.add(path)  # If the path matches a word, add it to the result.\n",
-        "\n",
-        "        temp, board[i][j] = board[i][j], \"#\"  # Mark the cell as visited.\n",
-        "\n",
-        "        # Explore neighboring cells (up, down, left, right).\n",
-        "        if i > 0:\n",
-        "            dfs(curr_node, i - 1, j, path)\n",
-        "        if i < m - 1:\n",
-        "            dfs(curr_node, i + 1, j, path)\n",
-        "        if j > 0:\n",
-        "            dfs(curr_node, i, j - 1, path)\n",
-        "        if j < n - 1:\n",
-        "            dfs(curr_node, i, j + 1, path)\n",
-        "\n",
-        "        board[i][j] = temp  # Restore the cell's original character.\n",
-        "\n",
-        "    # Function to build a Trie from the list of words.\n",
-        "    def buildTrie():\n",
-        "        root = TrieNode()  # Create a root node.\n",
-        "        for word in words:\n",
-        "            node = root\n",
-        "            for char in word:\n",
-        "                if char not in node.children:\n",
-        "                    node.children[char] = TrieNode()  # Create a new node if needed.\n",
-        "                node = node.children[char]\n",
-        "            node.is_end_of_word = True  # Mark the end of the word.\n",
-        "        return root\n",
-        "\n",
-        "    m, n = len(board), len(board[0])  # Get the dimensions of the board.\n",
-        "    root = buildTrie()  # Build the Trie from the list of words.\n",
-        "    result = set()  # Use a set to store unique words found on the board.\n",
-        "\n",
-        "    # Iterate through each cell on the board and start DFS from there.\n",
-        "    for i in range(m):\n",
-        "        for j in range(n):\n",
-        "            dfs(root, i, j, \"\")  # Start DFS from each cell on the board.\n",
-        "\n",
-        "    return list(result)  # Convert the set to a list and return the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code solves the \"Word Search II\" problem, where you have a grid of characters and a list of words. The goal is to find all words from the list that can be constructed by traversing adjacent cells (horizontally or vertically) in the grid, without using the same cell more than once for a single word.\n",
-        "\n",
-        "The code does the following:\n",
-        "\n",
-        "1. Defines a `TrieNode` class to represent nodes in a Trie data structure. Each node has children (other characters in the word) and a flag to mark the end of a word.\n",
-        "\n",
-        "2. Defines the `findWords` function, which takes the grid (`board`) and a list of words (`words`) as input.\n",
-        "\n",
-        "3. Inside the `findWords` function, there is a `dfs` (depth-first search) function. This function is used to explore the grid starting from a specific cell. It checks if the current path matches any word in the Trie and adds it to the result if found.\n",
-        "\n",
-        "4. The code also includes a `buildTrie` function that constructs a Trie data structure from the list of words. It iterates through each word, creating nodes for each character in the Trie and marking the end of words.\n",
-        "\n",
-        "5. The dimensions of the grid are obtained (m x n).\n",
-        "\n",
-        "6. A Trie is built using the `buildTrie` function.\n",
-        "\n",
-        "7. A set called `result` is used to store unique words found in the grid.\n",
-        "\n",
-        "8. The code iterates through each cell on the grid (using nested loops) and starts a DFS search from each cell.\n",
-        "\n",
-        "9. During the DFS, it explores neighboring cells (up, down, left, right) if they haven't been visited before and if the current path matches a prefix in the Trie.\n",
-        "\n",
-        "10. When a word is found during the DFS traversal, it is added to the `result` set.\n",
-        "\n",
-        "11. The grid cell is marked as visited during the DFS by changing its character to \"#\", and it is restored to its original character after the DFS traversal.\n",
-        "\n",
-        "12. Finally, the set of unique words found is converted to a list and returned as the output.\n",
-        "\n",
-        "The code effectively uses a Trie data structure to optimize the search and ensures that each word is constructed from adjacent cells without reusing the same cell for the same word."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['eat', 'oath']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "board1 = [[\"o\",\"a\",\"a\",\"n\"],[\"e\",\"t\",\"a\",\"e\"],[\"i\",\"h\",\"k\",\"r\"],[\"i\",\"f\",\"l\",\"v\"]]\n",
-        "words1 = [\"oath\",\"pea\",\"eat\",\"rain\"]\n",
-        "print(findWords(board1, words1))  # Output: [\"eat\", \"oath\"]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "c748b879-a2d8-4298-a42c-c7f61f9c7fc6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "\n",
-        "board2 = [[\"a\",\"b\"],[\"c\",\"d\"]]\n",
-        "words2 = [\"abcb\"]\n",
-        "print(findWords(board2, words2))  # Output: []"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. **Building the Trie (`buildTrie` function):** \n",
-        "   - Suppose there are `N` words in the input list, and the average length of a word is `L`.\n",
-        "   - Building the Trie takes O(N * L) time, as we iterate through each word and insert each character into the Trie.\n",
-        "\n",
-        "2. **DFS Traversal (`dfs` function):** \n",
-        "   - In the worst case, the DFS function explores all possible paths in the grid.\n",
-        "   - The maximum number of recursive calls for each cell is 4 (up, down, left, right).\n",
-        "   - The maximum depth of the DFS is limited by the length of the longest word in the Trie, which is at most `L`.\n",
-        "   - Therefore, the DFS traversal for each cell takes $O(4^L)$ time.\n",
-        "\n",
-        "3. **Iterating Through the Grid (`findWords` function):**\n",
-        "   - In the worst case, we iterate through all `m` rows and `n` columns in the grid.\n",
-        "   - For each cell, we start a DFS traversal.\n",
-        "   - Therefore, iterating through the entire grid takes $O(m * n * 4^L)$ time.\n",
-        "\n",
-        "Overall, the time complexity of the code is $O(N * L + m * n * 4^L)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. **Trie Data Structure:**\n",
-        "   - The space required to store the Trie depends on the number of unique characters in all the words.\n",
-        "   - In the worst case, where all words are unique and have no common prefixes, the space complexity of the Trie is O(N * L).\n",
-        "   - In practice, it may be less if there are common prefixes among words.\n",
-        "\n",
-        "2. **DFS Stack (Recursive Calls):**\n",
-        "   - The depth of the recursive call stack during DFS is at most `L`, where `L` is the length of the longest word in the Trie.\n",
-        "   - Therefore, the space complexity for the recursive call stack is O(L).\n",
-        "\n",
-        "3. **Result Set (`result` set):**\n",
-        "   - The space used to store the result set depends on the number of valid words found in the grid.\n",
-        "   - In the worst case, when all words are found, the space complexity of the result set is O(N * L).\n",
-        "\n",
-        "Overall, the space complexity of the code is O(N * L) for the Trie and O(L) for the recursive call stack.\n",
-        "\n",
-        "In summary, the time complexity is dominated by the DFS traversal and is influenced by the number of words, their lengths, and the size of the grid. The space complexity is mainly determined by the Trie structure and the recursive call stack depth during DFS."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Word Search Paths:**\n",
-        "   Extend the basic word search exercise to find all possible paths (sequences of cells) that spell out a given word on the board. Return a list of paths if the word can be formed, otherwise an empty list.\n",
-        "\n",
-        "\n",
-        "2. **Reverse Word Search:**\n",
-        "   Modify the code to search for words in reverse order on the board. Given a 2D grid of characters and a list of words, find all words from the list that can be formed by traversing the board in reverse (right to left, bottom to top, etc.)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/08. Tries/README.md b/docs/_build/html/_sources/08. Tries/README.md
deleted file mode 100644
index 22c3bcd..0000000
--- a/docs/_build/html/_sources/08. Tries/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Trie Problems - Blind 75 LeetCode
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [208. Implement Trie Prefix Tree](https://leetcode.com/problems/implement-trie-prefix-tree/)         | Medium     |
-| [211. Design Add And Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/) | Medium |
-| [212. Word Search II](https://leetcode.com/problems/word-search-ii/)                                 | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb b/docs/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb
deleted file mode 100644
index cb66b6a..0000000
--- a/docs/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb	
+++ /dev/null
@@ -1,254 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 1046. Last Stone Weight\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Last Stone Weight problem on LeetCode, click here!](https://leetcode.com/problems/last-stone-weight/)\n",
-        "\n",
-        "---\n",
-        "You are given an array of integers `stones` where `stones[i]` is the weight of the `i`th stone.\n",
-        "\n",
-        "We are playing a game with the stones. On each turn, you choose the **heaviest two stones** and smash them together. Suppose the heaviest two stones have weights `x` and `y` with `x <= y`. The result of this smash is:\n",
-        "\n",
-        "1. If `x == y`, both stones are destroyed.\n",
-        "2. If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - x`.\n",
-        "\n",
-        "At the end of the game, there is **at most one** stone left.\n",
-        "\n",
-        "Return *the weight of the last remaining stone*. If there are no stones left, return `0`.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= stones.length <= 30`\n",
-        "- `1 <= stones[i] <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The \"Last Stone Weight\" problem involves a game played with a collection of stones, each stone having a certain weight. The goal is to find the weight of the last remaining stone after applying a specific set of rules.\n",
-        "\n",
-        "Here's how the game works:\n",
-        "\n",
-        "1. You start with an array of stones, where each stone is represented by its weight. The weights of the stones are given in the input as an array.\n",
-        "\n",
-        "2. In each turn of the game, you select the two heaviest stones from the remaining stones. If there is only one stone left, you've found the answer.\n",
-        "\n",
-        "3. If there are two stones with weights `x` and `y`, where `x` is less than or equal to `y`, the following happens:\n",
-        "   - If `x` is equal to `y`, both stones are completely destroyed, and they are removed from the array of stones.\n",
-        "   - If `x` is not equal to `y`, the stone with weight `x` is destroyed, and the stone with weight `y` now has a new weight of `y - x`. The stone with weight `x` is removed from the array, and the modified stone with weight `y - x` remains in the array.\n",
-        "\n",
-        "4. The game continues with the modified array of stones until there is at most one stone left.\n",
-        "\n",
-        "5. The goal is to find the weight of the last remaining stone, or if there are no stones left, return `0`.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "\n",
-        "def lastStoneWeight(stones):\n",
-        "    # Convert the input list to a max heap (negate the values to simulate a max heap)\n",
-        "    max_heap = [-stone for stone in stones]\n",
-        "    heapq.heapify(max_heap)\n",
-        "\n",
-        "    # Continue as long as there are more than 1 stone in the heap\n",
-        "    while len(max_heap) > 1:\n",
-        "        # Get the two heaviest stones\n",
-        "        y = -heapq.heappop(max_heap)  # Get the heaviest stone and negate it back to positive\n",
-        "        x = -heapq.heappop(max_heap)  # Get the second heaviest stone and negate it back to positive\n",
-        "\n",
-        "        # Calculate the new stone's weight after smashing\n",
-        "        if x != y:\n",
-        "            new_stone = y - x\n",
-        "            heapq.heappush(max_heap, -new_stone)  # Negate the value and push it back to the heap\n",
-        "\n",
-        "    # If there is one stone left, return its weight (negate it back to positive)\n",
-        "    if max_heap:\n",
-        "        return -max_heap[0]\n",
-        "    else:\n",
-        "        return 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a Python function called `lastStoneWeight` that takes a list of stone weights as input. It aims to find the weight of the last remaining stone after playing the stone smashing game described in the problem statement.\n",
-        "\n",
-        "Here's a step-by-step explanation of the code:\n",
-        "\n",
-        "1. We import the `heapq` library, which allows us to create and manipulate a heap (priority queue).\n",
-        "\n",
-        "2. Inside the `lastStoneWeight` function:\n",
-        "   - We create a max heap (negating stone weights to simulate a max heap) using the `heapq.heapify` function. This heap will keep track of the heaviest stones.\n",
-        "   \n",
-        "   - We enter a loop that continues as long as there are more than 1 stone in the heap.\n",
-        "   \n",
-        "   - Inside the loop, we:\n",
-        "     - Retrieve and negate the heaviest stone (`y`) from the heap.\n",
-        "     - Retrieve and negate the second heaviest stone (`x`) from the heap.\n",
-        "   \n",
-        "     - Calculate the new stone's weight after smashing (`y - x`), and negate it back to simulate a max heap.\n",
-        "     - Add the new stone back to the heap using `heapq.heappush`.\n",
-        "   \n",
-        "3. After the loop, if there is one stone left in the heap, we retrieve and negate it to get its actual weight, and return it as the result.\n",
-        "   \n",
-        "4. If there are no stones left (the heap is empty), we return `0`, indicating that no stones remain.\n",
-        "\n",
-        "5. Finally, we provide two test cases to demonstrate the function's usage and correctness.\n",
-        "\n",
-        "In summary, the code efficiently implements the stone smashing game by maintaining a max heap, selecting the heaviest stones in each turn, smashing them, and updating the heap until there is at most one stone left. It then returns the weight of the last remaining stone or 0 if there are no stones left."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "stones1 = [2, 7, 4, 1, 8, 1]\n",
-        "print(lastStoneWeight(stones1))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "b7e79514-4821-48d0-bb9d-0f2de8150945",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2: \n",
-        "\n",
-        "stones2 = [1]\n",
-        "print(lastStoneWeight(stones2))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided Python code for the \"Last Stone Weight\" problem:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The `heapq.heapify` operation to create the max heap from the input stone weights has a time complexity of O(n), where n is the number of stones.\n",
-        "\n",
-        "2. The loop that continues until there are more than 1 stone in the heap performs the following operations in each iteration:\n",
-        "   - Retrieving and negating the heaviest stone (`y`) from the heap, which takes O(log n) time.\n",
-        "   - Retrieving and negating the second heaviest stone (`x`) from the heap, also taking O(log n) time.\n",
-        "   - Calculating the new stone's weight and pushing it back into the heap using `heapq.heappush`, which takes O(log n) time.\n",
-        "\n",
-        "3. In the worst case, the loop runs until there is only one stone left, which requires approximately (n-1) iterations.\n",
-        "\n",
-        "Overall, the dominant time complexity is O(n) for the initial heap creation, and within the loop, each iteration takes O(log n) time. Therefore, the overall time complexity of the algorithm is O(n log n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The space complexity is primarily determined by the space required for the max heap. In the worst case, this heap can contain all the stones, so the space complexity for the heap is O(n).\n",
-        "\n",
-        "2. The rest of the variables used in the function (e.g., `x`, `y`, `new_stone`) require only constant space, so they do not significantly contribute to the space complexity.\n",
-        "\n",
-        "Therefore, the overall space complexity of the algorithm is O(n) due to the space used for the max heap.\n",
-        "\n",
-        "In summary:\n",
-        "- Time Complexity: $O(n\\ log\\ n)$\n",
-        "- Space Complexity: $O(n)$\n",
-        "\n",
-        "The algorithm is efficient enough to handle the problem's constraints, as the worst-case time and space complexities are both linearithmic in terms of the number of stones."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Kth Last Stone Weight**: Modify the problem to find the weight of the Kth last remaining stone, where K is an integer input. Your function should work efficiently for large values of K.\n",
-        "\n",
-        "\n",
-        "2. **Multiple Stones Smash**: Modify the problem so that instead of smashing two stones at a time, you can smash up to K stones at each turn, where K is an integer input. Determine the weight of the last remaining stone in this variation."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb b/docs/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb
deleted file mode 100644
index 847dd60..0000000
--- a/docs/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 295. Find Median from Data Stream\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Find Median from Data Stream problem on LeetCode, click here!](https://leetcode.com/problems/find-median-from-data-stream/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.\n",
-        "\n",
-        "Implement the MedianFinder class:\n",
-        "\n",
-        "- `MedianFinder()` initializes the `MedianFinder` object.\n",
-        "- `void addNum(int num)` adds the integer `num` from the data stream to the data structure.\n",
-        "- `double findMedian()` returns the median of all elements so far. Answers within $10^{-5}$ of the actual answer will be accepted.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- $-10^{5}$ <= `num` <= $10^{5}$\n",
-        "- There will be at least one element in the data structure before calling `findMedian`.\n",
-        "- At most $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-        "\n",
-        "**Follow-up:**\n",
-        "\n",
-        "1. If all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?\n",
-        "2. If `99%` of all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem, \"Find Median from Data Stream,\" deals with efficiently computing the median of a sequence of numbers as they are incrementally provided. The median is the middle value in an ordered list of numbers. When the list has an even number of elements, the median is the average of the two middle values.\n",
-        "\n",
-        "Here's a breakdown of the problem:\n",
-        "\n",
-        "1. **Initialization**: You are asked to implement a `MedianFinder` class that supports the following operations:\n",
-        "\n",
-        "    - `MedianFinder()`: Initializes the MedianFinder object.\n",
-        "    - `addNum(int num)`: Adds an integer `num` from the data stream to the data structure.\n",
-        "    - `findMedian()`: Returns the median of all elements added to the data structure.\n",
-        "\n",
-        "2. **Median Calculation**: The `findMedian` operation should efficiently compute the median, and the result should have a precision of at least $10^{-5}$ (i.e., answers within this range will be considered correct).\n",
-        "\n",
-        "3. **Examples**:\n",
-        "   \n",
-        "   - If you add numbers [1, 2], the median should be 1.5 because (1 + 2) / 2 = 1.5.\n",
-        "   - If you add numbers [1, 2, 3], the median should be 2 because it's the middle value.\n",
-        "   \n",
-        "4. **Constraints**:\n",
-        "\n",
-        "   - The integers provided in the data stream are in the range from $-10^5$ to $10^5$.\n",
-        "   - There will be at least one element in the data structure before calling `findMedian`.\n",
-        "   - At most, $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-        "\n",
-        "The problem can be solved by using two priority queues (heaps): one max-heap to store the smaller half of the numbers and one min-heap to store the larger half of the numbers. The max-heap ensures that the largest number in the smaller half is at the top, and the min-heap ensures that the smallest number in the larger half is at the top. These heaps are balanced to efficiently find the median when requested.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "\n",
-        "class MedianFinder:\n",
-        "\n",
-        "    def __init__(self):\n",
-        "        # Initialize two heaps: a max-heap (small_half) for the smaller half of numbers,\n",
-        "        # and a min-heap (large_half) for the larger half of numbers.\n",
-        "        self.small_half = []  # Max-heap for the smaller half of the numbers\n",
-        "        self.large_half = []  # Min-heap for the larger half of the numbers\n",
-        "\n",
-        "    def addNum(self, num: int) -> None:\n",
-        "        # Add the number to the appropriate heap based on its value\n",
-        "        if not self.small_half or num <= -self.small_half[0]:\n",
-        "            # If the number is less than or equal to the current maximum in the smaller half,\n",
-        "            # add it to the smaller half (max-heap)\n",
-        "            heapq.heappush(self.small_half, -num)\n",
-        "        else:\n",
-        "            # Otherwise, add it to the larger half (min-heap)\n",
-        "            heapq.heappush(self.large_half, num)\n",
-        "\n",
-        "        # Balance the heaps if necessary to ensure the size difference is at most 1\n",
-        "        if len(self.small_half) > len(self.large_half) + 1:\n",
-        "            # If the size of the smaller half is more than one greater than the size of the larger half,\n",
-        "            # move the maximum from the smaller half to the larger half to balance them.\n",
-        "            heapq.heappush(self.large_half, -heapq.heappop(self.small_half))\n",
-        "        elif len(self.large_half) > len(self.small_half):\n",
-        "            # If the size of the larger half is greater than the size of the smaller half,\n",
-        "            # move the minimum from the larger half to the smaller half to balance them.\n",
-        "            heapq.heappush(self.small_half, -heapq.heappop(self.large_half))\n",
-        "\n",
-        "    def findMedian(self) -> float:\n",
-        "        if len(self.small_half) == len(self.large_half):\n",
-        "            # If both halves have the same size, there's an even number of elements,\n",
-        "            # so the median is the average of the top elements in both heaps.\n",
-        "            return (-self.small_half[0] + self.large_half[0]) / 2.0\n",
-        "        else:\n",
-        "            # If the smaller half has more elements, it contains the median (odd number of elements).\n",
-        "            return -self.small_half[0]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a `MedianFinder` class for efficiently finding the median of a stream of numbers. It uses two heaps: a max-heap (`small_half`) to store the smaller half of the numbers and a min-heap (`large_half`) to store the larger half. Here's how the code works:\n",
-        "\n",
-        "1. The `__init__` method initializes the two heaps.\n",
-        "\n",
-        "2. The `addNum` method adds a number to the appropriate heap, ensuring that the heaps remain balanced. It does this by comparing the number to the current maximum in the smaller half and adding it to the appropriate heap. Then, it checks the sizes of the two heaps and rebalances them if necessary.\n",
-        "\n",
-        "3. The `findMedian` method calculates and returns the median. If both halves have the same size (even number of elements), it computes the average of the tops of both heaps. If the smaller half has more elements (odd number of elements), it returns the top of the max-heap, which is the median.\n",
-        "\n",
-        "In summary, this code efficiently maintains two heaps to divide the elements into smaller and larger halves, allowing for quick median retrieval. It offers a time complexity of O(log N) for adding elements and O(1) for finding the median, where N is the total number of elements processed. The space complexity is O(1) as the space used by the two heaps is independent of the input size."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1.5\n",
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "medianFinder = MedianFinder()\n",
-        "medianFinder.addNum(1)\n",
-        "medianFinder.addNum(2)\n",
-        "print(medianFinder.findMedian())  # Output: 1.5\n",
-        "medianFinder.addNum(3)\n",
-        "print(medianFinder.findMedian())  # Output: 2.0\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `MedianFinder` class:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. `MedianFinder()` Initialization: This operation has a time complexity of O(1). It simply initializes the two heaps.\n",
-        "\n",
-        "2. `addNum(int num)`: The time complexity for adding a number is O(log N), where N is the total number of elements processed so far. This is because, in the worst case, you might need to perform logarithmic operations on the heaps to maintain their balance.\n",
-        "\n",
-        "3. `findMedian()`: Finding the median has a time complexity of O(1). It simply involves accessing the tops of the two heaps, which takes constant time.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. `MedianFinder()` Initialization: The space complexity for initialization is O(1). It involves creating two empty heaps, which do not depend on the size of the data stream.\n",
-        "\n",
-        "2. `addNum(int num)`: The space complexity for adding a number is also O(1). It involves adding the number to one of the two heaps, which do not significantly affect the overall space complexity.\n",
-        "\n",
-        "3. `findMedian()`: The space complexity for finding the median is O(1). It doesn't require any additional data structures that grow with the input size.\n",
-        "\n",
-        "In summary, the time complexity of the `addNum` operation is O(log N), and the time complexity of the `findMedian` operation is O(1). The space complexity is O(1) as well, as the space used by the two heaps is independent of the size of the data stream. This solution efficiently maintains and retrieves the median, making it suitable for the given problem constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Generalized kth Element**: Extend the `MedianFinder` class to support finding the kth element in the data stream efficiently. This means finding the kth smallest or kth largest element in the stream.\n",
-        "\n",
-        "2. **Sliding Window Median**: Given an array and a sliding window size, find the median within the window as it slides over the array. This is an extension of the problem to a moving window scenario."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb b/docs/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb
deleted file mode 100644
index 2e1522d..0000000
--- a/docs/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb	
+++ /dev/null
@@ -1,239 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 703. Kth Largest Element in a Stream\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Kth Largest Element in a Stream problem on LeetCode, click here!](https://leetcode.com/problems/kth-largest-element-in-a-stream/)\n",
-        "\n",
-        "---\n",
-        "Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.\n",
-        "\n",
-        "Implement the `KthLargest` class with the following methods:\n",
-        "\n",
-        "- `KthLargest(int k, int[] nums)`: Initializes the object with the integer `k` and the stream of integers `nums`.\n",
-        "- `int add(int val)`: Appends the integer `val` to the stream and returns the element representing the `kth` largest element in the stream.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `k` <= $10^4$\n",
-        "- 0 <= `nums.length` <= $10^4$\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- $-10^4$ <= `val` <= $10^4$\n",
-        "- At most $10^4$ calls will be made to `add`.\n",
-        "- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem you're trying to solve is to design a class called `KthLargest` that can efficiently find the kth largest element in a stream of integers. You need to support two main operations:\n",
-        "\n",
-        "To solve this problem efficiently, you can use a min-heap data structure. Here's how the approach works:\n",
-        "\n",
-        "1. Initialize a min-heap to store the k largest elements. Initially, it's empty.\n",
-        "\n",
-        "2. In the `__init__` method, populate the min-heap with the first `k` elements from `nums`. This ensures that you have the k largest elements seen so far.\n",
-        "\n",
-        "3. Whenever you add a new element to the stream using the `add` method, follow these steps:\n",
-        "   - Add the new element to the min-heap.\n",
-        "   - If the size of the min-heap exceeds `k`, remove the smallest element from the min-heap. This ensures that you always have the k largest elements in the min-heap.\n",
-        "   - The smallest element in the min-heap (the root) will always represent the kth largest element in the stream.\n",
-        "\n",
-        "Here's why this approach works efficiently:\n",
-        "\n",
-        "- By using a min-heap, you can quickly maintain the k largest elements, and finding the smallest element in the heap (the root) takes constant time.\n",
-        "\n",
-        "- When you add a new element, the min-heap's size is kept at most `k`, which ensures that you only track the k largest elements and discard the smaller ones.\n",
-        "\n",
-        "- The time complexity for adding an element is O(log k), which is very efficient compared to sorting the entire stream, which would be O(n log n).\n",
-        "\n",
-        "- This approach meets the constraints of the problem, including handling large streams and having a low time complexity for both initialization and adding elements.\n",
-        "\n",
-        "In summary, the min-heap approach efficiently tracks the kth largest element in the stream by maintaining a heap of the k largest elements seen so far, updating it as new elements are added. This approach provides a fast and scalable solution to the problem.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "from typing import List  # Import the List type from the typing module\n",
-        "\n",
-        "class KthLargest:\n",
-        "\n",
-        "    def __init__(self, k: int, nums: List[int]):\n",
-        "        # Initialize the KthLargest object with k and nums.\n",
-        "        self.k = k\n",
-        "        # Create a min-heap to store the k largest elements.\n",
-        "        self.min_heap = []\n",
-        "        # Populate the min-heap with the first k elements from nums.\n",
-        "        for num in nums:\n",
-        "            self.add(num)\n",
-        "\n",
-        "    def add(self, val: int) -> int:\n",
-        "        # Add val to the min-heap.\n",
-        "        heapq.heappush(self.min_heap, val)\n",
-        "        # If the size of the min-heap exceeds k, remove the smallest element.\n",
-        "        if len(self.min_heap) > self.k:\n",
-        "            heapq.heappop(self.min_heap)\n",
-        "        # The root of the min-heap is the kth largest element.\n",
-        "        return self.min_heap[0]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- We start by importing the necessary modules:\n",
-        "  - `heapq`: This module provides functions to create and manipulate heaps.\n",
-        "  - `List` from the `typing` module: This is used to specify the type of the `nums` parameter.\n",
-        "\n",
-        "- We define the `KthLargest` class:\n",
-        "  - The `__init__` method initializes the object with the integer `k` and the list of integers `nums`.\n",
-        "  - It also creates an empty min-heap called `self.min_heap` to store the k largest elements.\n",
-        "  - It populates the min-heap with the first `k` elements from `nums` by calling the `add` method.\n",
-        "\n",
-        "- The `add` method:\n",
-        "  - Adds the new integer `val` to the min-heap using `heapq.heappush`. This maintains the min-heap property.\n",
-        "  - Checks if the size of the min-heap exceeds `k`. If it does, it removes the smallest element (the k+1th largest) using `heapq.heappop`.\n",
-        "  - Finally, it returns the smallest element in the min-heap, which is always the kth largest element.\n",
-        "\n",
-        "Overall, this code implements a class that efficiently finds the kth largest element in a stream of integers by maintaining a min-heap of the k largest elements seen so far.\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n",
-            "5\n",
-            "5\n",
-            "8\n",
-            "8\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "kthLargest = KthLargest(3, [4, 5, 8, 2])\n",
-        "print(kthLargest.add(3))  # Output: 4\n",
-        "print(kthLargest.add(5))  # Output: 5\n",
-        "print(kthLargest.add(10)) # Output: 5\n",
-        "print(kthLargest.add(9))  # Output: 8\n",
-        "print(kthLargest.add(4))  # Output: 8"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `KthLargest` class using the min-heap approach:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. `__init__` method:\n",
-        "   - In the `__init__` method, you iterate over the first `k` elements in `nums` and add them to the min-heap. Each `heapq.heappush` operation takes O(log k) time.\n",
-        "   - Therefore, the time complexity of the `__init__` method is O(k * log k).\n",
-        "\n",
-        "2. `add` method:\n",
-        "   - In the `add` method, you perform the following operations:\n",
-        "     - `heapq.heappush`: O(log k) to add the new element to the min-heap.\n",
-        "     - If the size of the min-heap exceeds `k`, you perform `heapq.heappop`, which is also O(log k).\n",
-        "     - Finally, you return the smallest element from the min-heap, which is O(1) because it's always the root.\n",
-        "   - Overall, the time complexity of the `add` method is O(log k).\n",
-        "\n",
-        "3. Overall, if you make `n` calls to the `add` method, the total time complexity is O(n * log k), where `n` is the total number of elements added to the stream.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The space complexity is determined by the space used to store the min-heap and the instance variables.\n",
-        "2. The min-heap stores at most `k` elements at any given time, so its space complexity is O(k).\n",
-        "3. The instance variables (such as `self.k` and `self.min_heap`) have constant space requirements.\n",
-        "4. Therefore, the overall space complexity of the `KthLargest` class is O(k).\n",
-        "\n",
-        "In summary, the time complexity of the `KthLargest` class is O(n * log k) for `n` add operations, and the space complexity is O(k), where `k` is the parameter passed during initialization. This implementation efficiently maintains the kth largest element in the stream while meeting the problem's constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Dynamic k:** Modify the `KthLargest` class to support dynamic changes in the value of `k`. Implement a method `update_k` that allows changing the value of `k` during the lifetime of the object. Ensure that the object can still correctly find the kth largest element based on the updated value of `k`.\n",
-        "\n",
-        "\n",
-        "2. **Implement kth Smallest:** Create a new class called `KthSmallest` that finds the kth smallest element in a stream instead of the kth largest. You may need to modify the data structure used in the implementation."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/09. Heap - Priority Queue/README.md b/docs/_build/html/_sources/09. Heap - Priority Queue/README.md
deleted file mode 100644
index 2eac70c..0000000
--- a/docs/_build/html/_sources/09. Heap - Priority Queue/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Heap / Priority Queue Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [703. Kth Largest Element in a Stream](https://leetcode.com/problems/kth-largest-element-in-a-stream/)     | Easy       |
-| [1046. Last Stone Weight](https://leetcode.com/problems/last-stone-weight/)     | Easy       |
-| [295. Find Median From Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)     | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/docs/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb b/docs/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb
deleted file mode 100644
index 228a62e..0000000
--- a/docs/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb	
+++ /dev/null
@@ -1,282 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 39. Combination Sum\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Combination Sum problem on LeetCode, click here!](https://leetcode.com/problems/combination-sum/)\n",
-        "\n",
-        "---\n",
-        "Given an array of **distinct** integers `candidates` and a target integer `target`, return *a list of all **unique combinations** of `candidates` where the chosen numbers sum to the `target`*. You may return the combinations in **any order**. \n",
-        "\n",
-        "The **same** number may be chosen from `candidates` an **unlimited number** of times. Two combinations are unique if the **frequency** of at least one of the chosen numbers is different.\n",
-        "\n",
-        "The test cases are generated such that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= candidates.length <= 30`\n",
-        "- `2 <= candidates[i] <= 40`\n",
-        "- All elements of `candidates` are **distinct**.\n",
-        "- `1 <= target <= 40`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to find all unique combinations of numbers from a given array (candidates) such that their sum equals a target number. Here are the details of the problem:\n",
-        "\n",
-        "- You are given an array of distinct integers called \"candidates.\"\n",
-        "- You are also given a target integer called \"target.\"\n",
-        "\n",
-        "The goal is to find all unique combinations of numbers from the candidates array where the sum of the selected numbers is equal to the target. You can use the same number from the candidates array an unlimited number of times. A combination is considered unique if it has a different frequency (i.e., a different number of occurrences) of at least one chosen number compared to other combinations.\n",
-        "\n",
-        "For example:\n",
-        "\n",
-        "**Example 1:**\n",
-        "```python\n",
-        "Input: candidates = [2, 3, 6, 7], target = 7\n",
-        "Output: [[2, 2, 3], [7]]\n",
-        "```\n",
-        "In this example, there are two unique combinations that sum up to the target:\n",
-        "- 2 + 2 + 3 = 7\n",
-        "- 7 = 7\n",
-        "\n",
-        "**Example 2:**\n",
-        "```python\n",
-        "Input: candidates = [2, 3, 5], target = 8\n",
-        "Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n",
-        "```\n",
-        "Here, there are three unique combinations that sum up to the target.\n",
-        "\n",
-        "**Example 3:**\n",
-        "```python\n",
-        "Input: candidates = [2], target = 1\n",
-        "Output: []\n",
-        "```\n",
-        "In this case, there are no combinations that can be formed from the candidates to reach the target of 1, so the output is an empty list.\n",
-        "\n",
-        "The problem asks you to find and return these combinations in any order.\n",
-        "\n",
-        "The constraints for this problem include the length of the candidates array, the values of candidates, and the target value.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def combinationSum(candidates, target):\n",
-        "    # Define a recursive DFS function to find combinations\n",
-        "    def dfs(remaining, start, path):\n",
-        "        # If the remaining target is 0, we found a valid combination\n",
-        "        if remaining == 0:\n",
-        "            result.append(path)\n",
-        "            return\n",
-        "        # If the remaining target is negative, this path is invalid\n",
-        "        if remaining < 0:\n",
-        "            return\n",
-        "        # Iterate through candidates starting from 'start' to avoid duplicates\n",
-        "        for i in range(start, len(candidates)):\n",
-        "            # Explore the current candidate by subtracting it from the remaining target\n",
-        "            # Add the current candidate to the path\n",
-        "            dfs(remaining - candidates[i], i, path + [candidates[i]])\n",
-        "\n",
-        "    result = []  # Initialize an empty list to store the result\n",
-        "    candidates.sort()  # Sort the candidates for deduplication and early stopping\n",
-        "    dfs(target, 0, [])  # Start the DFS from the target value with an empty path\n",
-        "    return result  # Return the list of unique combinations"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `combinationSum` function takes an array of distinct integers `candidates` and a target integer `target`. It returns a list of all unique combinations of `candidates` where the chosen numbers sum to the `target`. Each number in the `candidates` array can be used an unlimited number of times.\n",
-        "\n",
-        "Here's an overview of how the code works:\n",
-        "\n",
-        "1. The `combinationSum` function is defined, and it takes two arguments: `candidates` and `target`.\n",
-        "\n",
-        "2. Inside the function, there is a nested helper function called `backtrack`. This function is responsible for the actual combination generation using a recursive approach.\n",
-        "\n",
-        "3. The `backtrack` function is called with three arguments: `start`, `target`, and `path`. The `start` variable helps keep track of the current index in the `candidates` array, `target` keeps track of the remaining sum to be achieved, and `path` is a list that stores the current combination.\n",
-        "\n",
-        "4. Within the `backtrack` function, there are three main conditional statements:\n",
-        "   - If `target` becomes zero, it means we have found a combination that sums up to the target, so we add `path` to the `result` list.\n",
-        "   - If `target` becomes negative, it means the current combination doesn't work, so we return without doing anything.\n",
-        "   - Otherwise, we enter a loop that iterates over the `candidates` array, starting from the current index `start`.\n",
-        "\n",
-        "5. In the loop, the `backtrack` function is called recursively with the updated `target` and `path` after including the current candidate. This process explores different combinations by considering the current candidate or moving to the next candidate.\n",
-        "\n",
-        "6. After the loop completes, the `result` list contains all unique combinations that sum to the target.\n",
-        "\n",
-        "7. The `candidates` array is sorted to optimize the search. Sorting helps in avoiding unnecessary recursive branches and reduces the number of explored combinations.\n",
-        "\n",
-        "8. Finally, the `backtrack` function is initially called from the `combinationSum` function with `start` set to 0, `target` set to the original target value, and an empty `path`. The result is returned as a list of unique combinations.\n",
-        "\n",
-        "The code includes a few example cases to demonstrate how the function works with different inputs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[2, 2, 3], [7]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example 1\n",
-        "candidates1 = [2, 3, 6, 7]\n",
-        "target1 = 7\n",
-        "print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "candidates2 = [2, 3, 5]\n",
-        "target2 = 8\n",
-        "print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "candidates3 = [2]\n",
-        "target3 = 1\n",
-        "print(combinationSum(candidates3, target3))  # Output: []"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `combinationSum` function.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this function is influenced by the number of recursive calls made by the `backtrack` function and the amount of work done within each call. In the worst case, the function will explore all possible combinations.\n",
-        "\n",
-        "The worst-case scenario occurs when we have many combinations to reach the target. The time complexity can be expressed as $O(2^n)$, where 'n' is the maximum number of recursive calls. In this context, 'n' corresponds to the number of elements in the `candidates` list, and it may be up to 30 (as per the constraints). However, the actual number of combinations explored is generally much smaller because we eliminate branches when the target becomes negative, and we skip over candidates that cannot lead to a solution. The sorting step also helps to optimize the search.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity is determined by the auxiliary space used for storing the results and the stack space for the recursive calls.\n",
-        "\n",
-        "1. The space required for the `result` list can be up to $O(2^n)$ in the worst case because each combination is stored.\n",
-        "\n",
-        "2. The stack space for the recursive calls can also be up to O(n) in the worst case, where 'n' is the number of elements in the `candidates` list.\n",
-        "\n",
-        "Therefore, the overall space complexity of the function is $O(2^n + n)$. However, in practice, the space used for the results list is often the dominant factor, and the actual space used may be much smaller than 2^n because not all combinations are explored.\n",
-        "\n",
-        "In summary, the time complexity is exponential but typically smaller in practice due to optimization, and the space complexity is influenced by the number of results and the depth of the recursion."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Find the Number of Unique Combinations:** Instead of listing all unique combinations, modify the function to return the count of unique combinations that sum to the target.\n",
-        "\n",
-        "2. **Combinations with a Maximum Value:** Add a maximum value constraint for each combination, so they cannot exceed a certain value. Modify the code to find combinations respecting this constraint."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/10. Backtracking/79. Word Search.ipynb b/docs/_build/html/_sources/10. Backtracking/79. Word Search.ipynb
deleted file mode 100644
index cee509b..0000000
--- a/docs/_build/html/_sources/10. Backtracking/79. Word Search.ipynb	
+++ /dev/null
@@ -1,298 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 79. Word Search\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Word Search problem on LeetCode, click here!](https://leetcode.com/problems/word-search/)\n",
-        "\n",
-        "---\n",
-        "Given an `m x n` grid of characters `board` and a string `word`, return *`true` if `word` exists in the grid.*\n",
-        "\n",
-        "The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == board.length`\n",
-        "- `n = board[i].length`\n",
-        "- `1 <= m, n <= 6`\n",
-        "- `1 <= word.length <= 15`\n",
-        "- `board` and `word` consists of only lowercase and uppercase English letters.\n",
-        "\n",
-        "**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether a given word can be constructed from characters in an `m x n` grid of characters. The word can be formed by moving sequentially through horizontally or vertically neighboring cells in the grid, and you cannot use the same cell more than once.\n",
-        "\n",
-        "Here are the specifics of the problem:\n",
-        "\n",
-        "- You are given an `m x n` grid, which is a 2D board of characters.\n",
-        "- You are also given a string `word` consisting of lowercase and uppercase English letters.\n",
-        "- Your task is to determine if the word can be found in the grid by moving from one cell to an adjacent cell (horizontally or vertically) and collecting the characters sequentially to form the word.\n",
-        "- You can't use the same cell more than once when forming the word.\n",
-        "- If it's possible to form the word in the grid, return `True`; otherwise, return `False`.\n",
-        "\n",
-        "For example, consider the following grid and word:\n",
-        "\n",
-        "```python\n",
-        "board = [[\"A\", \"B\", \"C\", \"E\"],\n",
-        "         [\"S\", \"F\", \"C\", \"S\"],\n",
-        "         [\"A\", \"D\", \"E\", \"E\"]]         \n",
-        "word = \"ABCCED\"\n",
-        "```\n",
-        "\n",
-        "In this case, the word \"ABCCED\" can be formed by moving through the cells (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) in the grid. Therefore, the function should return `True`.\n",
-        "\n",
-        "Conversely, if the word is impossible to construct, such as in the case of the word \"ABCB\" for the same grid, where there's no valid path that forms the word, the function should return `False`.\n",
-        "\n",
-        "The problem involves searching for a path through the grid that matches the characters in the word while respecting the constraints mentioned. It can be solved using depth-first search (DFS) or backtracking.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def exist(self, board: List[List[str]], word: str) -> bool:\n",
-        "        def dfs(i, j, k):\n",
-        "            # Check if the current position is out of bounds or doesn't match the character in the word.\n",
-        "            if not (0 <= i < len(board)) or not (0 <= j < len(board[0])) or board[i][j] != word[k]:\n",
-        "                return False\n",
-        "            \n",
-        "            # If we have matched all characters in the word, return True.\n",
-        "            if k == len(word) - 1:\n",
-        "                return True\n",
-        "\n",
-        "            # Temporarily mark the current cell to prevent revisiting.\n",
-        "            tmp, board[i][j] = board[i][j], \"/\"\n",
-        "\n",
-        "            # Explore adjacent cells by making recursive calls in all four directions.\n",
-        "            found = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)\n",
-        "\n",
-        "            # Restore the original character in the cell.\n",
-        "            board[i][j] = tmp\n",
-        "\n",
-        "            return found\n",
-        "\n",
-        "        # Iterate through all cells in the board.\n",
-        "        for i in range(len(board)):\n",
-        "            for j in range(len(board[0])):\n",
-        "                # Check if the 'dfs' function starting from this cell returns True.\n",
-        "                if dfs(i, j, 0):\n",
-        "                    return True\n",
-        "\n",
-        "        # If no match is found after exploring all cells, return False.\n",
-        "        return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. Import the necessary module:\n",
-        "   - `from typing import List`: This line imports the `List` type from the `typing` module, which is used to specify the type of the `board` parameter in the method signature.\n",
-        "\n",
-        "2. Define the `Solution` class:\n",
-        "   - `class Solution:`: This line defines a class called `Solution` to encapsulate the solution for the word search problem.\n",
-        "\n",
-        "3. Define the `exist` method:\n",
-        "   - `def exist(self, board: List[List[str]], word: str) -> bool:`: This method is a member of the `Solution` class and takes two parameters: `board`, which is a 2D list of characters, and `word`, which is a string. It returns a boolean value, indicating whether the word exists in the board.\n",
-        "\n",
-        "4. Implement the inner `dfs` function:\n",
-        "   - `def dfs(i, j, k):`: This is a helper function used for depth-first search (DFS). It is called recursively to explore the board and search for the word. It takes three parameters: `i` and `j` representing the current position on the board, and `k` representing the current index in the `word` string.\n",
-        "\n",
-        "   - Inside `dfs`, it checks if the current position is out of bounds or if the character at the current position on the board does not match the character at the current index in the `word`. If either of these conditions is met, it returns `False`.\n",
-        "\n",
-        "   - If `k` is equal to the length of the `word` minus one, it means we have successfully matched all characters in the `word`, so it returns `True`.\n",
-        "\n",
-        "   - The function then temporarily replaces the character at the current position with a marker, \"/\", to prevent revisiting the same cell. It then explores adjacent cells by making recursive calls in all four directions: up, down, left, and right. If any of these recursive calls return `True`, it propagates the `True` result up the call stack. After the exploration is complete, it restores the original character in the cell.\n",
-        "\n",
-        "5. Iterate through the board:\n",
-        "   - The outer loop iterates through all cells in the `board` using two nested loops. For each cell, it checks if the `dfs` function starting from that cell (with `k` initially set to 0) returns `True`. If it does, it means the word exists, so the method returns `True`.\n",
-        "\n",
-        "6. Return `False`:\n",
-        "   - If the loop completes without finding the word in the board, the method returns `False`.\n",
-        "\n",
-        "To use this code, you can create an instance of the `Solution` class and call the `exist` method on that instance, passing the `board` and the `word` as arguments to check if the word exists in the board."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test case 1: Word \"ABCCED\" can be found in the grid\n",
-        "board1 = [[\"A\",\"B\",\"C\",\"E\"],\n",
-        "          [\"S\",\"F\",\"C\",\"S\"],\n",
-        "          [\"A\",\"D\",\"E\",\"E\"]]\n",
-        "word1 = \"ABCCED\"\n",
-        "print(solution.exist(board1, word1))  # Expected output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "\n",
-        "# Test case 2: Word \"SEE\" can be found in the grid\n",
-        "word2 = \"SEE\"\n",
-        "print(solution.exist(board1, word2))  # Expected output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "\n",
-        "# Test case 3: Word \"ABCB\" cannot be found in the grid\n",
-        "word3 = \"ABCB\"\n",
-        "print(solution.exist(board1, word3))  # Expected output: False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `exist` method in the given code.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The primary work in this code is done by the `dfs` function, which performs a depth-first search through the grid. The time complexity of the `dfs` function is $O(4^(n))$, where \"n\" is the length of the `word`. This is because, in the worst case, the function explores all possible directions (up, down, left, right) for each character in the `word`.\n",
-        "\n",
-        "The outer loops iterate through all cells in the `board`, which has dimensions `m x n`. So, the total number of iterations of `dfs` is O(m * n). However, since the `dfs` function has a higher time complexity, the overall time complexity is still dominated by the `dfs` function.\n",
-        "\n",
-        "Therefore, the overall time complexity of the `exist` method is $O(m * n * 4^n)$, where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The primary space usage comes from the recursive calls in the `dfs` function. The depth of the recursion can be at most equal to the length of the `word`, which is O(n).\n",
-        "\n",
-        "2. Additionally, the `tmp` variable and the recursive calls on the call stack consume a small amount of additional memory.\n",
-        "\n",
-        "3. The marking of visited cells with the \"/\" character temporarily modifies the `board` in-place but doesn't significantly contribute to the space complexity.\n",
-        "\n",
-        "Therefore, the overall space complexity is O(n) due to the recursive call stack.\n",
-        "\n",
-        "In summary, the time complexity is $O(m * n * 4^n)$, and the space complexity is O(n), where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word. The time complexity is exponential in the worst case due to the depth-first search through all possible paths in the grid."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the code to find all unique words that can be constructed from the grid. Instead of returning `True` or `False`, return a list of words found. Ensure that the same cell is not used more than once in constructing each word.\n",
-        "\n",
-        "2. **Board Generation:** Create a function to generate random grids of characters and then use the `exist` method to solve them. Test the efficiency and correctness of your code by generating larger boards and words."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/10. Backtracking/README.md b/docs/_build/html/_sources/10. Backtracking/README.md
deleted file mode 100644
index 0d1ea13..0000000
--- a/docs/_build/html/_sources/10. Backtracking/README.md	
+++ /dev/null
@@ -1,17 +0,0 @@
-# Backtracking Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [39. Combination Sum](https://leetcode.com/problems/combination-sum/)        | Medium     |
-| [79. Word Search](https://leetcode.com/problems/word-search/)                | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb b/docs/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb
deleted file mode 100644
index 07351b6..0000000
--- a/docs/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb	
+++ /dev/null
@@ -1,317 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 133. Clone Graph\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Clone Graph problem on LeetCode, click here!](https://leetcode.com/problems/clone-graph/)\n",
-        "\n",
-        "---\n",
-        "Given a reference of a node in a **connected** undirected graph.\n",
-        "\n",
-        "Return a **deep copy** (clone) of the graph.\n",
-        "\n",
-        "Each node in the graph contains a value (`int`) and a list (`List[Node]`) of its neighbors.\n",
-        "\n",
-        "```\n",
-        "class Node {\n",
-        "    public int val;\n",
-        "    public List<Node> neighbors;\n",
-        "}\n",
-        "```\n",
-        "\n",
-        "**Test case format:**\n",
-        "\n",
-        "For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with `val == 1`, the second node with `val == 2`, and so on. The graph is represented in the test case using an adjacency list.\n",
-        "\n",
-        "**An adjacency list** is a collection of unordered **lists** used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.\n",
-        "\n",
-        "The given node will always be the first node with `val = 1`. You must return the **copy of the given node** as a reference to the cloned graph.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the graph is in the range `[0, 100]`.\n",
-        "- `1 <= Node.val <= 100`\n",
-        "- `Node.val` is unique for each node.\n",
-        "- There are no repeated edges and no self-loops in the graph.\n",
-        "- The Graph is connected and all nodes can be visited starting from the given node."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to clone a connected undirected graph, which means we need to create a deep copy of the original graph while preserving its structure and relationships between nodes. The graph is represented using adjacency lists, where each node has a value (an integer) and a list of its neighbors.\n",
-        "\n",
-        "Here are the key aspects of the problem:\n",
-        "\n",
-        "1. **Input**: You are given a reference to one of the nodes in the original graph. This reference node serves as the entry point to the graph. The entire graph can be explored starting from this node.\n",
-        "\n",
-        "2. **Output**: The goal is to create a deep copy of the graph and return a reference to the cloned graph. The cloned graph should have the same structure and relationships as the original one but should be a separate instance in memory.\n",
-        "\n",
-        "3. **Graph Structure**: The graph is composed of nodes (vertices) and edges. Nodes have values (integers) and are connected to other nodes through edges (undirected connections). Each node has a list of neighbors, which are other nodes it is connected to.\n",
-        "\n",
-        "4. **Connected Graph**: The problem assumes that the graph is connected, meaning that you can start from the reference node provided and traverse the entire graph by following the edges. This ensures that there are no isolated components in the graph.\n",
-        "\n",
-        "To solve this problem, you typically need to perform a graph traversal (e.g., depth-first search or breadth-first search) to visit all the nodes in the original graph, create corresponding nodes in the clone, and establish the same connections between nodes in the cloned graph as in the original one. To avoid duplication of nodes, you also need to keep track of visited nodes to ensure that the cloning process is efficient and that nodes are not duplicated in the clone.\n",
-        "\n",
-        "The problem can be challenging because you need to create a deep copy of the graph while handling cyclic dependencies between nodes and ensuring that the structure and relationships of the original graph are preserved in the clone. The solution requires recursion or a data structure to maintain the mapping between original nodes and their clones to achieve an efficient and accurate cloning process.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a Node.\n",
-        "class Node:\n",
-        "    def __init__(self, val=0, neighbors=None):\n",
-        "        self.val = val\n",
-        "        self.neighbors = neighbors if neighbors is not None else []\n",
-        "\n",
-        "class Solution:\n",
-        "    def cloneGraph(self, node: 'Node') -> 'Node':\n",
-        "        if not node:\n",
-        "            return None\n",
-        "\n",
-        "        visited = {}  # Dictionary to keep track of visited nodes\n",
-        "\n",
-        "        def dfs(original_node):\n",
-        "            if original_node in visited:\n",
-        "                return visited[original_node]\n",
-        "\n",
-        "            # Create a copy of the original node\n",
-        "            copy_node = Node(original_node.val)\n",
-        "            visited[original_node] = copy_node\n",
-        "\n",
-        "            # Recursively clone the neighbors of the original node\n",
-        "            for neighbor in original_node.neighbors:\n",
-        "                copy_node.neighbors.append(dfs(neighbor))\n",
-        "\n",
-        "            return copy_node\n",
-        "\n",
-        "        return dfs(node)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The given code defines a Python class, `Node`, and a solution class, `Solution`, to clone a connected undirected graph represented by nodes and their neighbors. Here's an explanation of the code:\n",
-        "\n",
-        "1. The `Node` class defines a node in the graph. Each node has a value (`val`) which is an integer and a list of neighbors (`neighbors`) that are other nodes in the graph.\n",
-        "\n",
-        "2. The `Solution` class contains the method `cloneGraph`, which takes a reference to the first node of the graph (`node`) as input and returns a deep copy of the entire graph, preserving the structure and relationships between nodes.\n",
-        "\n",
-        "3. Inside the `cloneGraph` method:\n",
-        "   - It first checks if the input `node` is `None`. If it's `None`, it means there is no graph to clone, and the method returns `None`.\n",
-        "\n",
-        "   - It initializes a `visited` dictionary to keep track of visited nodes. This dictionary is used to ensure that each node is cloned only once to avoid duplication.\n",
-        "\n",
-        "   - The `dfs` (depth-first search) function is defined within the `cloneGraph` method. It takes an original node as input and returns its corresponding clone. This function performs the actual cloning of the graph.\n",
-        "\n",
-        "   - Inside the `dfs` function:\n",
-        "     - It checks if the original node has been visited before by looking it up in the `visited` dictionary. If it has been visited, it returns the corresponding clone from the `visited` dictionary.\n",
-        "\n",
-        "     - If the original node is not visited, it creates a new node, `copy_node`, with the same value as the original node. This is the first step in creating the clone of the original node.\n",
-        "\n",
-        "     - It then recursively clones the neighbors of the original node by iterating through the `neighbors` list of the original node and appending the corresponding cloned neighbors to the `neighbors` list of the `copy_node`. This step ensures that the relationships between nodes are preserved in the clone.\n",
-        "\n",
-        "     - The `copy_node` is added to the `visited` dictionary with the original node as the key and the clone as the value.\n",
-        "\n",
-        "     - Finally, the `copy_node` is returned as the result of the DFS for the original node.\n",
-        "\n",
-        "4. The `cloneGraph` method returns the result of the DFS on the input `node`, which is the clone of the entire graph with its structure and relationships intact.\n",
-        "\n",
-        "In summary, the code uses depth-first search (DFS) to traverse the original graph, creating a clone for each node and its neighbors. It ensures that each node is cloned only once to avoid duplication and returns a reference to the cloned graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1, 2, 4]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "def get_adjacency_list(node):\n",
-        "    visited = {}  # Dictionary to keep track of visited nodes\n",
-        "\n",
-        "    def dfs(node):\n",
-        "        if not node:\n",
-        "            return []\n",
-        "\n",
-        "        if node in visited:\n",
-        "            return visited[node]\n",
-        "\n",
-        "        neighbors = [n.val for n in node.neighbors]\n",
-        "        visited[node] = neighbors\n",
-        "\n",
-        "        for neighbor in node.neighbors:\n",
-        "            dfs(neighbor)\n",
-        "\n",
-        "    dfs(node)\n",
-        "    result = []\n",
-        "\n",
-        "    def get_node_and_neighbors(node):\n",
-        "        result.append([node.val] + visited[node])\n",
-        "\n",
-        "    get_node_and_neighbors(node)\n",
-        "    return result\n",
-        "\n",
-        "\n",
-        "# Testing the code with the provided examples\n",
-        "adjList1 = [[2, 4], [1, 3], [2, 4], [1, 3]]\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Create the graph from adjList\n",
-        "nodes = [Node(i) for i in range(1, len(adjList1) + 1)]\n",
-        "for i, adj in enumerate(adjList1):\n",
-        "    nodes[i].neighbors = [nodes[j - 1] for j in adj]\n",
-        "\n",
-        "# Clone the graph and get its adjacency list\n",
-        "cloned_node1 = solution.cloneGraph(nodes[0])\n",
-        "cloned_adjList1 = get_adjacency_list(cloned_node1)\n",
-        "\n",
-        "# Print the cloned graph as an adjacency list\n",
-        "print(cloned_adjList1)\n",
-        "\n",
-        "# Output should match adjList1\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "# Input: adjList = [[]]\n",
-        "\n",
-        "adjList2 = [[]]\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Create the graph from adjList\n",
-        "node1 = Node(1)\n",
-        "\n",
-        "# Clone the graph and get its adjacency list\n",
-        "cloned_node1 = solution.cloneGraph(node1)\n",
-        "cloned_adjList2 = get_adjacency_list(cloned_node1)\n",
-        "\n",
-        "# Print the cloned graph as an adjacency list\n",
-        "print(cloned_adjList2)\n",
-        "\n",
-        "# Output should be [[]], which represents an empty graph"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for cloning a graph using depth-first search (DFS).\n",
-        "\n",
-        "Time Complexity:\n",
-        "1. Creating the graph: The time complexity to create the original graph from `adjList` is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. We iterate through each node and its neighbors in `adjList`.\n",
-        "\n",
-        "2. Cloning the graph (DFS): The depth-first search (DFS) is used to traverse the original graph and create a clone. The time complexity of DFS is O(V + E), as we visit each node and each edge once.\n",
-        "\n",
-        "Overall, the time complexity is O(V + E) for both creating the original graph and cloning it.\n",
-        "\n",
-        "Space Complexity:\n",
-        "1. Storage for the original graph: We need to store the original graph, which includes all nodes and their adjacency lists. In the worst case, this requires O(V + E) space, where V is the number of nodes, and E is the number of edges.\n",
-        "\n",
-        "2. Storage for the visited dictionary: The `visited` dictionary stores mappings from original nodes to their corresponding clones. In the worst case, this requires O(V) space because each node is visited once.\n",
-        "\n",
-        "3. Recursive stack for DFS: The space used for the recursive call stack during DFS can be up to O(V) in the worst case, where V is the number of nodes.\n",
-        "\n",
-        "Overall, the space complexity is O(V + E) for storing the original graph, and O(V) for auxiliary data structures, including the `visited` dictionary and the DFS call stack. Therefore, the overall space complexity is O(V + E).\n",
-        "\n",
-        "In summary, the time and space complexities of the provided code for cloning a graph are both O(V + E), where V is the number of nodes, and E is the number of edges in the graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Cyclic Graph**: Design a graph with cycles (nodes connected in a cycle). Test if the code can correctly clone graphs with cycles and doesn't fall into an infinite loop during traversal.\n",
-        "\n",
-        "\n",
-        "2. **Partial Graph Cloning**: Modify the code to clone only a subset of the graph starting from a given reference node, rather than the entire graph. Implement this as an extension to the problem."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb b/docs/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb
deleted file mode 100644
index e0235d6..0000000
--- a/docs/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb	
+++ /dev/null
@@ -1,248 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 178. Graph Valid Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Graph Valid Tree problem on LintCode, click here!](https://www.lintcode.com/problem/178/)\n",
-        "\n",
-        "---\n",
-        "Given `n` nodes labeled from `0` to `n - 1` and a list of `undirected` edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- You can assume that no duplicate edges will appear in edges. Since all edges are `undirected`, `[0, 1]` is the same as `[1, 0]` and thus will not appear together in edges."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether a given set of undirected edges can form a valid tree. In this context, a \"tree\" is a specific type of graph with the following properties:\n",
-        "\n",
-        "1. It is a connected graph, meaning there is a path between any pair of nodes in the graph.\n",
-        "2. It has no cycles, which means there are no closed loops or circuits in the graph.\n",
-        "3. It has exactly `n - 1` edges, where `n` is the number of nodes in the tree.\n",
-        "\n",
-        "The problem is typically given with two pieces of information:\n",
-        "\n",
-        "1. The number of nodes (`n`), which are usually labeled from 0 to `n-1`.\n",
-        "2. A list of undirected edges, where each edge is represented as a pair of nodes.\n",
-        "\n",
-        "The task is to write a function or algorithm to determine whether the provided set of edges can be arranged to form a valid tree based on the properties mentioned above.\n",
-        "\n",
-        "To solve this problem, you need to check the following conditions:\n",
-        "\n",
-        "1. There should be exactly `n - 1` edges to ensure that the graph is connected and tree-like.\n",
-        "2. There should be no cycles in the graph. If there are any cycles, the set of edges cannot form a tree.\n",
-        "3. The graph should be connected, which means you can reach any node from any other node through a series of edges.\n",
-        "\n",
-        "The solution typically involves using graph traversal algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the graph and check these conditions. If all conditions are met, the given edges form a valid tree; otherwise, they do not.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def validTree(n, edges):\n",
-        "    if len(edges) != n - 1:\n",
-        "        return False  # A tree with n nodes must have n-1 edges.\n",
-        "\n",
-        "    # Create an adjacency list to represent the graph.\n",
-        "    adj_list = {i: [] for i in range(n)}\n",
-        "    for u, v in edges:\n",
-        "        adj_list[u].append(v)\n",
-        "        adj_list[v].append(u)\n",
-        "\n",
-        "    visited = set()\n",
-        "\n",
-        "    # Define a recursive DFS function.\n",
-        "    def dfs(node, parent):\n",
-        "        if node in visited:\n",
-        "            return False  # If we encounter a visited node, there's a cycle.\n",
-        "        visited.add(node)\n",
-        "\n",
-        "        for neighbor in adj_list[node]:\n",
-        "            if neighbor != parent:\n",
-        "                if not dfs(neighbor, node):\n",
-        "                    return False\n",
-        "\n",
-        "        return True\n",
-        "\n",
-        "    # Start DFS from any node (e.g., node 0).\n",
-        "    if not dfs(0, -1):\n",
-        "        return False  # If there's a cycle, it's not a valid tree.\n",
-        "\n",
-        "    return len(visited) == n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `validTree` function checks if a given set of undirected edges forms a valid tree. A valid tree must satisfy these conditions:\n",
-        "\n",
-        "1. It has exactly `n - 1` edges, where `n` is the number of nodes.\n",
-        "2. There are no cycles in the graph.\n",
-        "3. The graph is connected, meaning all nodes are reachable from any starting node.\n",
-        "\n",
-        "The code accomplishes this as follows:\n",
-        "\n",
-        "- It first checks if the number of edges is equal to `n - 1`. If not, it immediately returns `False` since a tree with `n` nodes must have `n - 1` edges to be connected.\n",
-        "\n",
-        "- It then constructs an adjacency list (`adj_list`) to represent the graph efficiently. This data structure stores the neighbors of each node.\n",
-        "\n",
-        "- The code uses a `visited` set to keep track of nodes visited during depth-first search (DFS).\n",
-        "\n",
-        "- The `dfs` function is a recursive function that checks for cycles in the graph. It does this by visiting nodes and marking them as visited. If it encounters a node that has already been visited (indicating a cycle), it returns `False`.\n",
-        "\n",
-        "- The DFS function explores the neighbors of each node, avoiding revisiting the parent node to prevent cycles.\n",
-        "\n",
-        "- After the DFS is complete, the code checks if the number of visited nodes is equal to `n`, ensuring the graph is connected.\n",
-        "\n",
-        "- Finally, if all conditions are met, the code returns `True`, indicating that the given edges form a valid tree. Otherwise, it returns `False`.\n",
-        "\n",
-        "The code provides examples to demonstrate how to use the `validTree` function to check whether a set of edges forms a valid tree or not."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "n1 = 5\n",
-        "edges1 = [[0, 1], [0, 2], [0, 3], [1, 4]]\n",
-        "print(validTree(n1, edges1))  # Output: true"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "n2 = 5\n",
-        "edges2 = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]\n",
-        "print(validTree(n2, edges2))  # Output: false"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexities of the `validTree` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. Constructing the adjacency list (`adj_list`) takes O(E) time, where E is the number of edges in the input graph (|edges|).\n",
-        "\n",
-        "2. The DFS function, when considering all nodes, has a time complexity of O(V + E), where V is the number of nodes in the graph. This is because, in the worst case, the DFS function visits each node and each edge once.\n",
-        "\n",
-        "3. Checking if the number of visited nodes is equal to `n` takes O(1) time.\n",
-        "\n",
-        "Overall, the time complexity of the `validTree` function is O(E + V) in the worst case.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The adjacency list (`adj_list`) consumes additional space to store the graph structure, taking O(V + E) space. This is because it stores information about each node and its adjacent nodes.\n",
-        "\n",
-        "2. The `visited` set keeps track of visited nodes, and its space complexity is O(V) in the worst case, as there can be at most V unique nodes in the set.\n",
-        "\n",
-        "3. The depth of the function call stack during DFS can be at most V, which contributes O(V) space to the space complexity.\n",
-        "\n",
-        "Overall, the space complexity of the `validTree` function is O(V + E) due to the adjacency list and O(V) due to the `visited` set and DFS call stack.\n",
-        "\n",
-        "In summary, the time complexity is O(E + V), and the space complexity is O(V + E). These complexities are based on the worst-case scenario, where all nodes and edges are considered. In practical cases, the actual complexities may be smaller, depending on the specific structure of the graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Minimum Spanning Tree (MST)**: Implement an algorithm to find the minimum spanning tree of a connected graph using either Kruskal's or Prim's algorithm. Compare the MST to the original graph to determine if the original graph is a tree. This exercise reinforces the concept of tree properties.\n",
-        "\n",
-        "2. **Detecting Cycles**: Modify the code to not only check if the input forms a tree but also to identify and print out any cycles present in the graph if it's not a tree. This exercise enhances your ability to detect and visualize cycles in a graph."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb b/docs/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb
deleted file mode 100644
index a8a7727..0000000
--- a/docs/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 200. Number of Islands\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Number of Islands problem on LeetCode, click here!](https://leetcode.com/problems/number-of-islands/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an `m x n` 2D binary grid `grid` which represents a map of `'1'`s (land) and `'0'`s (water), return *the number of islands*. \n",
-        "\n",
-        "An **island** is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == grid.length`\n",
-        "- `n == grid[i].length`\n",
-        "- `1 <= m, n <= 300`\n",
-        "- `grid[i][j]` is `'0'` or `'1'`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is about counting the number of islands in a 2D binary grid. In this grid, '1' represents land, and '0' represents water. An island is defined as a group of '1' cells that are adjacent to each other either horizontally or vertically. It's important to note that diagonal connections do not count. Additionally, it's assumed that all four edges of the grid are surrounded by water, meaning the grid is finite and doesn't \"wrap around.\"\n",
-        "\n",
-        "The objective is to determine how many distinct islands exist in the grid. To clarify, an island consists of connected '1' cells, and these connections can only occur horizontally or vertically. If there are no '1' cells in the grid, the number of islands would be zero.\n",
-        "\n",
-        "The problem often involves using graph traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to identify and count these islands. You start at one '1' cell and explore its adjacent cells to determine the extent of the island. Once you've visited all the '1' cells in an island, you move on to the next unvisited '1' cell and repeat the process until you've counted all the islands in the grid.\n",
-        "\n",
-        "In essence, the problem asks you to analyze the grid to find groups of connected '1' cells and count them as individual islands. The goal is to return the total count of islands in the given grid.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def numIslands(self, grid: List[List[str]]) -> int:\n",
-        "        if not grid:\n",
-        "            return 0\n",
-        "\n",
-        "        m, n = len(grid), len(grid[0])  # Get the dimensions of the grid\n",
-        "        num_islands = 0  # Initialize the count of islands\n",
-        "\n",
-        "        def dfs(row, col):\n",
-        "            if row < 0 or row >= m or col < 0 or col >= n or grid[row][col] == '0':\n",
-        "                return\n",
-        "\n",
-        "            grid[row][col] = '0'  # Mark the current cell as visited by changing it to '0'\n",
-        "            directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Define four possible directions to move\n",
-        "\n",
-        "            for dr, dc in directions:\n",
-        "                # Explore adjacent cells\n",
-        "                dfs(row + dr, col + dc)\n",
-        "\n",
-        "        for i in range(m):\n",
-        "            for j in range(n):\n",
-        "                if grid[i][j] == '1':\n",
-        "                    num_islands += 1  # Increment the island count\n",
-        "                    dfs(i, j)  # Start a DFS from the current '1' cell to explore the entire island\n",
-        "\n",
-        "        return num_islands\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided code defines a Python class called `Solution` with a method called `numIslands`. This class is designed to count the number of islands in a 2D binary grid.\n",
-        "\n",
-        "The `numIslands` method takes a 2D grid as input, where '1' represents land, and '0' represents water. It uses a depth-first search (DFS) approach to identify and count islands. The key steps in the code include:\n",
-        "\n",
-        "1. Checking if the input grid is empty, and if so, returning 0 (indicating there are no islands).\n",
-        "\n",
-        "2. Determining the dimensions of the grid (number of rows and columns).\n",
-        "\n",
-        "3. Initializing a variable `num_islands` to keep track of the count of islands.\n",
-        "\n",
-        "4. Defining a nested function `dfs` to perform depth-first search. This function recursively explores adjacent land cells connected to the current cell and marks them as visited by changing '1' to '0'.\n",
-        "\n",
-        "5. In the `dfs` function, it checks for boundary conditions and whether the current cell is '0'. If these conditions are met, it returns without further exploration.\n",
-        "\n",
-        "6. The `dfs` function explores neighboring cells in all four directions (up, down, left, right) using a loop.\n",
-        "\n",
-        "7. The main loop iterates through each cell of the grid. When a '1' cell is encountered, it increments the `num_islands` count and starts a DFS from that cell to explore the entire island.\n",
-        "\n",
-        "8. Finally, the method returns the total count of islands found in the grid.\n",
-        "\n",
-        "This code encapsulates the solution in a class, allowing you to create an instance of the `Solution` class and call the `numIslands` method on it to count the number of islands in a given grid."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 1 Output: 1\n"
-          ]
-        }
-      ],
-      "source": [
-        "grid1 = [\n",
-        "  [\"1\", \"1\", \"1\", \"1\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"1\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"0\", \"0\", \"0\"]\n",
-        "]\n",
-        "\n",
-        "solution = Solution()\n",
-        "result1 = solution.numIslands(grid1)\n",
-        "print(\"Example 1 Output:\", result1)  # Expected output: 1\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 2 Output: 3\n"
-          ]
-        }
-      ],
-      "source": [
-        "grid2 = [\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"1\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"0\", \"1\", \"1\"]\n",
-        "]\n",
-        "\n",
-        "result2 = solution.numIslands(grid2)\n",
-        "print(\"Example 2 Output:\", result2)  # Expected output: 3\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "Time Complexity:\n",
-        "- The code uses a depth-first search (DFS) to explore the grid and count the number of islands.\n",
-        "- In the worst case, it visits every cell in the grid exactly once.\n",
-        "- The DFS function explores neighboring cells in four directions (up, down, left, right), so it has a time complexity of $O(4^{max(m, n)})$, where m is the number of rows and n is the number of columns.\n",
-        "- Therefore, the overall time complexity of the code is O(m * n), where m is the number of rows, and n is the number of columns.\n",
-        "\n",
-        "Space Complexity:\n",
-        "- The space complexity is determined by the recursive calls and the depth of the DFS stack.\n",
-        "- In the worst case, when the entire grid consists of '1's, the depth of the DFS recursion can be as deep as max(m, n).\n",
-        "- Therefore, the space complexity of the code is O(max(m, n)), representing the maximum depth of the DFS stack.\n",
-        "\n",
-        "In practical terms, for reasonably sized grids, the time and space complexity is linear, and the code should perform well. However, it's important to keep in mind that the worst-case time complexity is O(m * n), and the space complexity can be O(max(m, n))."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Identify Largest Island:** In addition to counting the number of islands, modify the code to identify the largest island in the grid and return its size.\n",
-        "\n",
-        "2. **Optimize for Space:** Modify the solution to reduce its space complexity to O(1) without modifying the input grid. This is a more memory-efficient version of the problem."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb b/docs/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb
deleted file mode 100644
index 37624f6..0000000
--- a/docs/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 207. Course Schedule\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Course Schedule problem on LeetCode, click here!](https://leetcode.com/problems/course-schedule/)\n",
-        "\n",
-        "---\n",
-        "There are a total of `numCourses` courses to take, labeled from 0 to `numCourses - 1`. You are given an array `prerequisites` where `prerequisites[i] = [ai, bi]` indicates that you **must** take course `bi` first if you want to take course `ai`. \n",
-        "\n",
-        "- For example, the pair `[0, 1]`, indicates that to take course `0` you have to first take course `1`.\n",
-        "\n",
-        "Return `true` if you can finish all courses. Otherwise, return `false`.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `numCourses` <= 2000\n",
-        "- 0 <= `prerequisites.length` <= 5000\n",
-        "- `prerequisites[i].length` == 2\n",
-        "- 0 <= `ai`, `bi` < `numCourses`\n",
-        "- All the pairs `prerequisites[i]` are **unique**."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether it's possible to complete a set of courses given a list of prerequisites. You are given the following:\n",
-        "\n",
-        "- `numCourses`: The total number of courses to be taken, labeled from 0 to `numCourses - 1`.\n",
-        "- `prerequisites`: A list of prerequisite courses, where each element `prerequisites[i]` is a pair `[ai, bi]`, indicating that you must take course `bi` before you can take course `ai`.\n",
-        "\n",
-        "The goal is to check whether it's possible to complete all the courses while respecting the prerequisite requirements. In other words, you need to determine if there are any circular dependencies in the course prerequisites that would prevent you from taking all the courses.\n",
-        "\n",
-        "For example:\n",
-        "\n",
-        "- If `numCourses` is 2, and `prerequisites` is `[[1, 0]]`, it means you have two courses, and you must finish course 0 before you can take course 1. In this case, it's possible to complete all courses, so the function should return `True`.\n",
-        "\n",
-        "- If `numCourses` is 2, and `prerequisites` is `[[1, 0], [0, 1]]`, it means you have two courses, but there's a circular dependency where course 1 requires course 0 and course 0 requires course 1. In this case, it's impossible to complete all courses, so the function should return `False`.\n",
-        "\n",
-        "The problem is essentially about checking for the presence of cycles in a directed graph, where the courses are nodes, and the prerequisites represent directed edges between them. If there are no cycles, it's possible to complete all courses; otherwise, it's not possible.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict, deque\n",
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:\n",
-        "        # Create a graph using an adjacency list to represent the courses and their prerequisites.\n",
-        "        graph = defaultdict(list)  # Initialize an empty adjacency list.\n",
-        "        in_degree = [0] * numCourses  # Initialize an array to store in-degrees of courses.\n",
-        "\n",
-        "        # Populate the graph and calculate in-degrees.\n",
-        "        for course, prerequisite in prerequisites:\n",
-        "            graph[prerequisite].append(course)  # Add the course as a neighbor to its prerequisite.\n",
-        "            in_degree[course] += 1  # Increment the in-degree of the course.\n",
-        "\n",
-        "        # Initialize a queue with courses that have no prerequisites.\n",
-        "        queue = deque()\n",
-        "        for course in range(numCourses):\n",
-        "            if in_degree[course] == 0:\n",
-        "                queue.append(course)\n",
-        "\n",
-        "        # Perform topological sorting.\n",
-        "        while queue:\n",
-        "            course = queue.popleft()  # Take a course with no prerequisites.\n",
-        "            numCourses -= 1  # Decrement the count of remaining courses.\n",
-        "\n",
-        "            for neighbor in graph[course]:\n",
-        "                in_degree[neighbor] -= 1  # Remove the prerequisite relationship.\n",
-        "                if in_degree[neighbor] == 0:\n",
-        "                    queue.append(neighbor)  # If no more prerequisites, add to the queue.\n",
-        "\n",
-        "        # If all courses were successfully taken (numCourses becomes 0), return True.\n",
-        "        return numCourses == 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "This Python code is designed to determine whether it is possible to complete all the required courses given a set of prerequisites. It uses a topological sorting algorithm to accomplish this. Here's a brief explanation of the code:\n",
-        "\n",
-        "1. The code defines a `Solution` class with a `canFinish` method that takes two parameters: `numCourses` (the total number of courses) and `prerequisites` (a list of pairs indicating which course must be taken before another).\n",
-        "\n",
-        "2. It initializes an empty graph as an adjacency list using a `defaultdict` to represent the courses and their prerequisites. It also initializes a list called `in_degree` to keep track of the in-degrees of courses, which is used in the topological sorting process.\n",
-        "\n",
-        "3. The code then populates the graph by iterating through the `prerequisites` list. For each pair `(ai, bi)`, it adds course `ai` as a neighbor to its prerequisite course `bi` in the graph and increments the in-degree of course `ai`.\n",
-        "\n",
-        "4. Next, it initializes a queue with courses that have no prerequisites. It does this by iterating through all the courses and adding those with an in-degree of 0 to the queue.\n",
-        "\n",
-        "5. The code performs a topological sorting of the courses using a while loop. In each iteration of the loop, it dequeues a course from the front of the queue (a course with no prerequisites).\n",
-        "\n",
-        "6. For each neighbor (course that depends on the dequeued course), it decrements the in-degree of the neighbor. If the neighbor's in-degree becomes 0, it means all of its prerequisites have been taken, so the neighbor is added to the queue.\n",
-        "\n",
-        "7. The loop continues until there are no more courses with no prerequisites to dequeue, and the in-degrees are updated accordingly.\n",
-        "\n",
-        "8. After the loop, if all courses were successfully taken (i.e., `numCourses` becomes 0), it means that there were no circular dependencies, and it's possible to finish all courses. In this case, the method returns `True`. Otherwise, if there are remaining courses (indicating a circular dependency), it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "numCourses1 = 2\n",
-        "prerequisites1 = [[1, 0]]\n",
-        "sol = Solution()\n",
-        "print(sol.canFinish(numCourses1, prerequisites1))  # Output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "numCourses2 = 2\n",
-        "prerequisites2 = [[1, 0], [0, 1]]\n",
-        "print(sol.canFinish(numCourses2, prerequisites2))  # Output: False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `canFinish` method:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. Constructing the graph: In the first loop where we iterate through the `prerequisites` list, we populate the graph with prerequisites. This loop has a time complexity of O(E), where E is the number of prerequisites (edges).\n",
-        "\n",
-        "2. Initializing in-degrees and finding courses with no prerequisites: In the worst case, we iterate through all `numCourses` courses to initialize in-degrees and find the courses with no prerequisites. This has a time complexity of O(V), where V is the number of courses (vertices).\n",
-        "\n",
-        "3. Topological Sorting: In the while loop, we perform topological sorting, processing each course once and its outgoing edges. In the worst case, each course is processed once. This has a time complexity of O(V + E).\n",
-        "\n",
-        "The overall time complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. Graph Representation: The space complexity for storing the graph as an adjacency list is O(E), where E is the number of prerequisites.\n",
-        "\n",
-        "2. In-Degree Array: The space complexity for storing the in-degrees of courses is O(V), where V is the number of courses.\n",
-        "\n",
-        "3. Queue: The space complexity for the queue is O(V), as it may contain all the courses.\n",
-        "\n",
-        "The overall space complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-        "\n",
-        "In most practical cases, the number of prerequisites (E) is much smaller than the number of courses (V), so the time and space complexity can often be considered as O(V)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimizing Course Scheduling**: Given a list of courses with their durations and prerequisites, find the most efficient way to schedule these courses to minimize the time it takes to complete all of them.\n",
-        "\n",
-        "2. **Detecting Cycles in Course Dependencies**: Modify the original problem to not just determine if you can finish all courses but also to identify the specific courses that form a cycle in the prerequisite dependencies."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb b/docs/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb
deleted file mode 100644
index fb186f2..0000000
--- a/docs/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb	
+++ /dev/null
@@ -1,236 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 3651. Number of Connected Components in an Undirected Graph\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Number of Connected Components in an Undirected Graph problem on LintCode, click here!](https://www.lintcode.com/problem/3651/)\n",
-        "\n",
-        "---\n",
-        "In this problem, there is an undirected graph with `n` nodes. There is also an `edges` array. Where `edges[i] = [a, b]` means that there is an edge between *node a* and *node b* in the graph.\n",
-        "\n",
-        "You need to return *the number of connected components in that graph*."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to determine the number of connected components in an undirected graph. Here's a more detailed explanation of the problem:\n",
-        "\n",
-        "- You are given an undirected graph, which consists of nodes and edges. The graph may have multiple distinct connected components.\n",
-        "\n",
-        "- A connected component is a group of nodes within the graph where each node can be reached from every other node in the same component. In other words, there is a path between any two nodes in the same connected component, but there are no paths connecting nodes from different components.\n",
-        "\n",
-        "- Your task is to write a function that takes as input the number of nodes `n` and a list of edges `edges`, where each edge is represented as a pair of nodes `[a, b]` indicating an edge between node `a` and node `b`.\n",
-        "\n",
-        "- The goal is to determine the number of distinct connected components in the graph.\n",
-        "\n",
-        "- You need to implement an algorithm that can identify these connected components and count how many there are in the given graph.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def countConnectedComponents(n, edges):\n",
-        "    # Define a helper function for depth-first search (DFS).\n",
-        "    def dfs(node):\n",
-        "        if not visited[node]:  # If the node has not been visited yet:\n",
-        "            visited[node] = True  # Mark it as visited.\n",
-        "            for neighbor in graph[node]:  # Explore all neighbors of the current node.\n",
-        "                dfs(neighbor)  # Recursively call DFS on the neighbor.\n",
-        "\n",
-        "    # Create an empty adjacency list to represent the graph.\n",
-        "    graph = [[] for _ in range(n)]\n",
-        "\n",
-        "    # Initialize a boolean list to keep track of visited nodes.\n",
-        "    visited = [False] * n\n",
-        "\n",
-        "    # Populate the adjacency list with edges from the input.\n",
-        "    for a, b in edges:\n",
-        "        graph[a].append(b)\n",
-        "        graph[b].append(a)  # Since the graph is undirected, we add edges in both directions.\n",
-        "\n",
-        "    # Initialize a counter to keep track of the number of connected components.\n",
-        "    count = 0\n",
-        "\n",
-        "    # Iterate through all nodes in the graph.\n",
-        "    for i in range(n):\n",
-        "        if not visited[i]:  # If the node has not been visited:\n",
-        "            count += 1  # Increment the count since we've found a new connected component.\n",
-        "            dfs(i)  # Start a depth-first search from the unvisited node to explore the component.\n",
-        "\n",
-        "    return count"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided Python code calculates the number of connected components in an undirected graph. Here's a high-level explanation of how the code works:\n",
-        "\n",
-        "1. The `countConnectedComponents` function takes two arguments: `n` (the number of nodes in the graph) and `edges` (a list of edge pairs representing connections between nodes).\n",
-        "\n",
-        "2. Inside the function, there is a helper function called `dfs` (depth-first search), which is used to explore and mark nodes as visited.\n",
-        "\n",
-        "3. An empty adjacency list called `graph` is created to represent the graph. The `graph` is a list of lists where each element represents a node, and the list associated with each node contains its neighboring nodes.\n",
-        "\n",
-        "4. Another list called `visited` is created to keep track of visited nodes. Initially, all nodes are marked as not visited (`False`).\n",
-        "\n",
-        "5. The code populates the `graph` by iterating through the `edges` list and adding each edge to the adjacency list. Since the graph is undirected, both directions of each edge are added.\n",
-        "\n",
-        "6. A counter variable called `count` is initialized to keep track of the number of connected components.\n",
-        "\n",
-        "7. The code iterates through all nodes from `0` to `n-1`. For each node, it checks whether it has been visited yet. If it hasn't been visited, it means a new connected component is found. The counter `count` is incremented, and a depth-first search (DFS) is initiated from this unvisited node to explore the connected component.\n",
-        "\n",
-        "8. The `dfs` function recursively visits all nodes in the connected component, marking them as visited. It does this by checking each neighbor of the current node and recursively calling `dfs` on unvisited neighbors.\n",
-        "\n",
-        "9. After the loop through all nodes, the `count` variable contains the total number of connected components in the graph.\n",
-        "\n",
-        "10. The function returns the value of `count`, which represents the number of connected components.\n",
-        "\n",
-        "11. Two examples are provided at the end to demonstrate how to use the function with different inputs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "n1 = 3\n",
-        "edges1 = [[0, 1], [0, 2]]\n",
-        "print(countConnectedComponents(n1, edges1))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "n2 = 6\n",
-        "edges2 = [[0, 1], [1, 2], [2, 3], [4, 5]]\n",
-        "print(countConnectedComponents(n2, edges2))  # Output: 2"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the code for counting the number of connected components in an undirected graph using depth-first search (DFS).\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code iterates through all nodes in the graph (from 0 to `n-1`).\n",
-        "- For each unvisited node, it initiates a depth-first search (DFS) from that node to explore the connected component.\n",
-        "- In the worst case, every node may belong to a separate connected component, and you perform a DFS for each node.\n",
-        "- The time complexity of the DFS for a single connected component is O(V + E), where V is the number of nodes in the component, and E is the number of edges.\n",
-        "- In the worst case, you perform O(n) DFS calls, each with its own O(V + E) complexity.\n",
-        "\n",
-        "So, the overall time complexity of the code is O(n * (V + E)). In the worst case, if the graph is connected (a single connected component), this simplifies to O(n * (n + m)), where m is the number of edges.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The main data structures used are the `graph` (adjacency list), `visited` (boolean array), and the recursive call stack for DFS.\n",
-        "- The `graph` has a space complexity of O(n + m), where n is the number of nodes, and m is the number of edges.\n",
-        "- The `visited` boolean array also has a space complexity of O(n) to keep track of whether nodes have been visited.\n",
-        "- The depth of the recursive call stack for DFS can go up to O(n) in the worst case, when all nodes are part of a single connected component.\n",
-        "\n",
-        "So, the overall space complexity of the code is O(n + m) for data structures and O(n) for the call stack. In the worst case, this simplifies to O(n + m).\n",
-        "\n",
-        "In summary, the time complexity is O(n * (V + E)), and the space complexity is O(n + m)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Disconnected Components with Multiple Queries:** You are given an undirected graph with `n` nodes and `m` edges. Initially, the graph is disconnected, and you need to process `q` queries. Each query consists of adding an edge to connect two nodes. After each query, you need to output the number of connected components in the updated graph.\n",
-        "\n",
-        "2. **Maximum Size of Connected Components:** Given an undirected graph with `n` nodes and `m` edges, find the size of the largest connected component in the graph. Additionally, identify the nodes that belong to this largest component. You can assume there is at least one connected component in the graph.\n"
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb b/docs/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb
deleted file mode 100644
index eb34497..0000000
--- a/docs/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb	
+++ /dev/null
@@ -1,261 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 417. Pacific Atlantic Water Flow\n",
-        "\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Pacific Atlantic Water Flow problem on LeetCode, click here!](https://leetcode.com/problems/pacific-atlantic-water-flow/)\n",
-        "\n",
-        "---\n",
-        "There is an `m x n` rectangular island that borders both the **Pacific Ocean** and **Atlantic Ocean**. The **Pacific Ocea**n touches the island's left and top edges, and the **Atlantic Ocean** touches the island's right and bottom edges.\n",
-        "\n",
-        "The island is partitioned into a grid of square cells. You are given an `m x n` integer matrix `heights` where `heights[r][c]` represents the **height above sea level** of the cell at coordinate `(r, c)`.\n",
-        "\n",
-        "The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is **less than or equal to** the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.\n",
-        "\n",
-        "Return *a **2D list** of grid coordinates `result` where $result[i] = [r_i, c_i]$ denotes that rain water can flow from cell $(r_i, c_i)$ to **both** the Pacific and Atlantic oceans*.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == heights.length`\n",
-        "- `n == heights[r].length`\n",
-        "- `1 <= m, n <= 200`\n",
-        "- 0 <= `heights[r][c]` <= $10^5$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand involves a scenario with a rectangular island surrounded by both the Pacific Ocean and the Atlantic Ocean. The Pacific Ocean is in contact with the island's left and top edges, while the Atlantic Ocean is in contact with the island's right and bottom edges. The goal is to determine which cells on the island can allow rainwater to flow to both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "Here are the key details and constraints of the problem:\n",
-        "\n",
-        "1. The island is divided into a grid of square cells.\n",
-        "\n",
-        "2. Each cell is represented by an integer value in a matrix `heights`. The value at `heights[r][c]` represents the height above sea level of the cell at coordinates `(r, c)` on the island.\n",
-        "\n",
-        "3. Rainwater can flow from a cell to neighboring cells directly north, south, east, or west only if the neighboring cell's height is less than or equal to the current cell's height.\n",
-        "\n",
-        "4. Water can flow from any cell adjacent to the ocean into the ocean. This means water can flow from cells on the edge of the island to the Pacific Ocean and the Atlantic Ocean.\n",
-        "\n",
-        "The task is to find and return a 2D list of grid coordinates where rainwater can flow from a cell to both the Pacific and Atlantic Oceans. In other words, you need to identify the cells that can send water to both the left and top edges (Pacific) and the right and bottom edges (Atlantic) of the island.\n",
-        "\n",
-        "The problem is solved by performing a depth-first search (DFS) starting from the ocean edges and marking the cells that can flow to each ocean. Then, the algorithm identifies cells that are reachable from both oceans and returns their coordinates as the result.\n",
-        "\n",
-        "The problem involves both traversal of the island and backtracking to find the solution, which makes it a classic graph traversal problem.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:\n",
-        "        # Check if the input matrix is empty\n",
-        "        if not heights:\n",
-        "            return []\n",
-        "        \n",
-        "        m, n = len(heights), len(heights[0])\n",
-        "        \n",
-        "        # Initialize boolean matrices to keep track of cells reachable from each ocean\n",
-        "        pacific_reachable = [[False] * n for _ in range(m)]\n",
-        "        atlantic_reachable = [[False] * n for _ in range(m)]\n",
-        "        \n",
-        "        # Depth-First Search (DFS) function to mark cells that can be reached from an ocean\n",
-        "        def dfs(x, y, reachable):\n",
-        "            if reachable[x][y]:\n",
-        "                return\n",
-        "            reachable[x][y] = True\n",
-        "            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:\n",
-        "                new_x, new_y = x + dx, y + dy\n",
-        "                if 0 <= new_x < m and 0 <= new_y < n and heights[new_x][new_y] >= heights[x][y]:\n",
-        "                    dfs(new_x, new_y, reachable)\n",
-        "        \n",
-        "        # Start DFS from the Pacific Ocean (left and top edges)\n",
-        "        for i in range(m):\n",
-        "            dfs(i, 0, pacific_reachable)\n",
-        "            dfs(i, n - 1, atlantic_reachable)\n",
-        "        \n",
-        "        for j in range(n):\n",
-        "            dfs(0, j, pacific_reachable)\n",
-        "            dfs(m - 1, j, atlantic_reachable)\n",
-        "        \n",
-        "        result = []\n",
-        "        \n",
-        "        # Find cells that can flow to both the Pacific and Atlantic oceans\n",
-        "        for i in range(m):\n",
-        "            for j in range(n):\n",
-        "                if pacific_reachable[i][j] and atlantic_reachable[i][j]:\n",
-        "                    result.append([i, j])\n",
-        "        \n",
-        "        return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a class called `Solution` with a method named `pacificAtlantic`. This method takes a 2D matrix `heights`, where each element represents the height above sea level of a cell on a rectangular island. The goal of the method is to find and return a list of grid coordinates (cells) from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "The code uses a Depth-First Search (DFS) algorithm to traverse the island, starting from the ocean edges, and marks cells that are reachable from either the Pacific or Atlantic Ocean. It does this by creating two boolean matrices, `pacific_reachable` and `atlantic_reachable`, to keep track of cells that can be reached from each ocean.\n",
-        "\n",
-        "The code iterates through the entire island, initiating DFS searches from the edges of the island. If a cell is reachable from an ocean (either Pacific or Atlantic), it is marked as reachable in the corresponding boolean matrix. The DFS function ensures that the traversal follows the rule that water can flow to a neighboring cell if the neighboring cell's height is less than or equal to the current cell's height.\n",
-        "\n",
-        "After marking all reachable cells from both oceans, the code then looks for cells that are reachable from both the Pacific and Atlantic Oceans. If a cell satisfies this condition, it is added to the `result` list.\n",
-        "\n",
-        "Finally, the code returns the `result` list, which contains the grid coordinates of cells from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "The code provides a solution to the problem of finding cells on the island where water can reach both oceans, demonstrating a traversal algorithm that explores the connectivity of cells on the island."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "heights1 = [\n",
-        "    [1,2,2,3,5],\n",
-        "    [3,2,3,4,4],\n",
-        "    [2,4,5,3,1],\n",
-        "    [6,7,1,4,5],\n",
-        "    [5,1,1,2,4]\n",
-        "]\n",
-        "\n",
-        "solution = Solution()\n",
-        "print(solution.pacificAtlantic(heights1)) # Expected Output: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 26,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "heights2 = [[1]]\n",
-        "print(solution.pacificAtlantic(heights2)) # Expected Output: [[0, 0]]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Time Complexity:\n",
-        "The time complexity of this solution is O(m * n), where 'm' is the number of rows in the input matrix and 'n' is the number of columns. Here's a breakdown of the time complexity:\n",
-        "\n",
-        "1. Creating the `pacific_reachable` and `atlantic_reachable` matrices with dimensions m x n takes O(m * n) time.\n",
-        "\n",
-        "2. Initiating Depth-First Search (DFS) from the edges of the island takes O(m * n) time because it processes each cell once.\n",
-        "\n",
-        "3. The DFS function visits each cell at most once, and its time complexity is O(1) for each cell. The number of cells visited by the DFS in the worst case is m * n.\n",
-        "\n",
-        "4. The final step of identifying and appending cells to the result list takes O(m * n) time in the worst case.\n",
-        "\n",
-        "Overall, the time complexity is dominated by the DFS and is O(m * n).\n",
-        "\n",
-        "Space Complexity:\n",
-        "The space complexity of this solution is also O(m * n). Here's how the space is used:\n",
-        "\n",
-        "1. Two boolean matrices, `pacific_reachable` and `atlantic_reachable`, are created with dimensions m x n. Each matrix takes O(m * n) space, resulting in O(2 * m * n) space usage.\n",
-        "\n",
-        "2. The depth-first search (DFS) stack space during the recursive calls can go as deep as the diagonal of the grid, which is at most min(m, n). This additional space for the call stack is relatively small compared to the boolean matrices and can be considered O(min(m, n)).\n",
-        "\n",
-        "3. The `result` list stores the grid coordinates, and its size is determined by the number of cells that can flow to both oceans. In the worst case, this list can contain all m * n cells, so it takes O(m * n) space.\n",
-        "\n",
-        "The overall space complexity is the sum of these components, which is O(m * n) + O(min(m, n)) + O(m * n). In big O notation, we can simplify this to O(m * n)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the algorithm to find the maximum flow of water from any cell to both the Pacific and Atlantic Oceans. In other words, find the cell(s) that can send the most water to both oceans.\n",
-        "\n",
-        "2. **Efficiency Challenge:** Optimize the time and space complexity of the algorithm while maintaining correctness. Aim to reduce both time and space complexity as much as possible."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/11. Graphs/README.md b/docs/_build/html/_sources/11. Graphs/README.md
deleted file mode 100644
index aad262c..0000000
--- a/docs/_build/html/_sources/11. Graphs/README.md	
+++ /dev/null
@@ -1,23 +0,0 @@
-# Graphs Problems - Blind 75 LeetCode
-
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)                           | Medium     |
-| [133. Clone Graph](https://leetcode.com/problems/clone-graph/)                                       | Medium     |
-| [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)       | Medium     |
-| [207. Course Schedule](https://leetcode.com/problems/course-schedule/)                                 | Medium     |
-| [323. Number of Connected Components In An Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/) | Medium |
-| [261. Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)                               | Medium     |
-
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb b/docs/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb
deleted file mode 100644
index 81f0e5b..0000000
--- a/docs/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb	
+++ /dev/null
@@ -1,256 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 892. Alien Dictionary\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Alien Dictionary problem on LintCode, click here!](https://www.lintcode.com/problem/892/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of **non-empty** words from the dictionary, where words are **sorted lexicographically by the rules of this new language**. Derive the order of letters in this language.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- You may assume all letters are in lowercase.\n",
-        "- The dictionary is invalid, if string a is prefix of string b and b is appear before a.\n",
-        "- If the order is invalid, return an empty string.\n",
-        "- There may be multiple valid order of letters, return the smallest in normal lexicographical order.\n",
-        "- The letters in **one** string are of the same rank by default and are sorted in Human dictionary order."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand involves determining the order of letters in a new alien language that uses the Latin alphabet. However, the order of letters in this alien language is unknown. To solve this problem, we are given a list of non-empty words from the dictionary, where the words are sorted lexicographically based on the rules of the alien language. The goal is to derive the correct order of letters in this new language.\n",
-        "\n",
-        "Here's a more detailed explanation of the problem:\n",
-        "\n",
-        "1. The alphabet: The new alien language uses the Latin alphabet, which consists of lowercase letters.\n",
-        "\n",
-        "2. Dictionary: We are provided with a list of non-empty words. These words are sorted in lexicographical order based on the rules of the alien language. The order of letters in the words reflects the correct order in the new language.\n",
-        "\n",
-        "3. Invalid order: The problem specifies that the dictionary is considered invalid if there exists a situation where one string 'a' is a prefix of another string 'b,' and 'b' appears before 'a' in the dictionary. This condition ensures that there are no inconsistencies in the order of letters.\n",
-        "\n",
-        "4. Multiple valid orders: It's possible that there are multiple valid orders of letters that satisfy the provided dictionary. However, we are instructed to return the smallest valid order in normal lexicographical order.\n",
-        "\n",
-        "To solve this problem, you need to analyze the given list of words and determine the correct order of letters in the alien language while adhering to the specified constraints. You should return the order as a string in lexicographical order if it's valid, and an empty string if the order is invalid.\n",
-        "\n",
-        "The key to solving this problem is to construct a directed graph of letter relationships based on the order information provided by the words in the dictionary. Then, perform a topological sort on this graph to derive the order of letters, ensuring that the order is consistent and valid according to the given constraints.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict, deque\n",
-        "\n",
-        "def alienOrder(words):\n",
-        "    # Create an adjacency list to represent the graph\n",
-        "    graph = defaultdict(list)\n",
-        "    \n",
-        "    # Create a dictionary to store the in-degrees of each letter\n",
-        "    in_degree = {}\n",
-        "    \n",
-        "    # Initialize in-degrees to 0 for all letters\n",
-        "    for word in words:\n",
-        "        for char in word:\n",
-        "            if char not in in_degree:\n",
-        "                in_degree[char] = 0\n",
-        "    \n",
-        "    # Build the graph and update in-degrees\n",
-        "    for i in range(1, len(words)):\n",
-        "        word1, word2 = words[i-1], words[i]\n",
-        "        min_length = min(len(word1), len(word2))\n",
-        "        \n",
-        "        # Compare characters in the two words\n",
-        "        j = 0\n",
-        "        while j < min_length and word1[j] == word2[j]:\n",
-        "            j += 1\n",
-        "        \n",
-        "        if j < min_length:\n",
-        "            # If word1[j] is lexicographically before word2[j], add an edge\n",
-        "            graph[word1[j]].append(word2[j])\n",
-        "            in_degree[word2[j]] += 1\n",
-        "            \n",
-        "            # Break the loop to avoid further comparisons\n",
-        "            break\n",
-        "    \n",
-        "    # Perform topological sorting using Kahn's algorithm\n",
-        "    result = []\n",
-        "    queue = deque([char for char in in_degree if in_degree[char] == 0])\n",
-        "    \n",
-        "    while queue:\n",
-        "        char = queue.popleft()\n",
-        "        result.append(char)\n",
-        "        \n",
-        "        for neighbor in graph[char]:\n",
-        "            in_degree[neighbor] -= 1\n",
-        "            if in_degree[neighbor] == 0:\n",
-        "                queue.append(neighbor)\n",
-        "    \n",
-        "    # Check if there is a valid order\n",
-        "    if len(result) < len(in_degree):\n",
-        "        return \"\"\n",
-        "    \n",
-        "    return \"\".join(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code is designed to determine the order of letters in an unknown alien language that uses the Latin alphabet. It takes a list of non-empty words from a dictionary as input, where the words are sorted lexicographically based on the rules of this alien language.\n",
-        "\n",
-        "The code follows these main steps:\n",
-        "\n",
-        "1. Create a data structure to represent the graph of letters and their relationships in the alien language. It uses a defaultdict to store a list of letters that come after each letter. Additionally, it uses a dictionary called `in_degree` to keep track of the in-degrees (the number of letters that precede a letter) for each letter.\n",
-        "\n",
-        "2. Initialize the in-degrees of all letters to 0. This is done by iterating through all the words and characters in the words, adding each character to the `in_degree` dictionary with an initial in-degree of 0.\n",
-        "\n",
-        "3. Build the graph by comparing adjacent words in the dictionary. For each pair of adjacent words, the code finds the first differing character between the two words. If such a character exists, it means there's an order relationship between the letters represented by these characters. The code updates the graph and in-degrees accordingly.\n",
-        "\n",
-        "4. After constructing the graph, the code performs a topological sorting of the letters using Kahn's algorithm. It starts by initializing an empty result list and a queue containing letters with in-degrees of 0. The algorithm repeatedly removes a letter with an in-degree of 0 from the queue, adds it to the result list, and updates the in-degrees of its neighbors. This process continues until the queue is empty.\n",
-        "\n",
-        "5. Finally, the code checks if the topological sorting was successful by comparing the length of the result list with the number of unique letters in the input. If the lengths differ, it means there's a cycle in the graph, and the order is invalid. In such a case, an empty string is returned. Otherwise, the result list is joined together to form the smallest order of letters in normal lexicographical order, and it is returned as the output.\n",
-        "\n",
-        "The code provides a way to determine the order of letters in the alien language while handling various edge cases, including checking for invalid order conditions."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "wrtef\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "words1 = [\"wrt\", \"wrf\", \"er\", \"ett\", \"rftt\"]\n",
-        "print(alienOrder(words1))  # Output: \"wertf\""
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "zx\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "words2 = [\"z\", \"x\"]\n",
-        "print(alienOrder(words2))  # Output: \"zx\""
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for determining the order of letters in an alien language.\n",
-        "\n",
-        "Time Complexity:\n",
-        "1. Building the graph: The code iterates through the list of words and compares adjacent words. In the worst case, where all words have the same prefix, it takes O(N) time to build the graph, where N is the total number of characters in all the words.\n",
-        "2. Topological sorting: Performing a topological sort on the graph takes O(V + E) time, where V is the number of unique letters (vertices) and E is the number of relationships between letters (edges) in the graph.\n",
-        "\n",
-        "Overall, the time complexity is O(N + V + E), where N is the total number of characters in the words, V is the number of unique letters, and E is the number of order relationships between letters.\n",
-        "\n",
-        "Space Complexity:\n",
-        "1. Graph and In-degrees: The code uses data structures to represent the graph and in-degrees. The space complexity for these data structures is O(V + E), where V is the number of unique letters, and E is the number of order relationships between letters in the words.\n",
-        "2. Result List: The result list stores the order of letters, which can have a maximum size of V, where V is the number of unique letters.\n",
-        "3. Queue: The space used by the queue during topological sorting is also O(V).\n",
-        "\n",
-        "Overall, the space complexity is O(V + E) due to the data structures used for graph representation and the space needed for storing the result and the queue.\n",
-        "\n",
-        "In the worst case, where there are many unique letters and many order relationships, the space complexity is dominated by the number of unique letters and their relationships.\n",
-        "\n",
-        "In summary, the time complexity of the code is O(N + V + E), and the space complexity is O(V + E). The actual performance will depend on the specific input data, such as the number of unique letters and the relationships between them in the alien language."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Valid Orders**: Extend the problem to handle cases where there are multiple valid orders of letters in the alien language. Modify the code to return all valid orders instead of just one. Be mindful of performance.\n",
-        "\n",
-        "2. **Detect Invalid Dictionary**: Given a dictionary of words, write a function to detect whether the dictionary is valid or not based on the constraints mentioned in the problem statement."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/12. Advanced Graphs/README.md b/docs/_build/html/_sources/12. Advanced Graphs/README.md
deleted file mode 100644
index f362de7..0000000
--- a/docs/_build/html/_sources/12. Advanced Graphs/README.md	
+++ /dev/null
@@ -1,16 +0,0 @@
-# Advanced Graphs Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [269. Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)                               | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb b/docs/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb
deleted file mode 100644
index 8d32e21..0000000
--- a/docs/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb	
+++ /dev/null
@@ -1,103 +0,0 @@
-{
-  "metadata": {
-    "kernelspec": {
-      "name": "python",
-      "display_name": "Python (Pyodide)",
-      "language": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat_minor": 5,
-  "nbformat": 4,
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "source": "# House Robber\n**Difficulty:** Medium\n\n---\nYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and **it will automatically contact the police if two adjacent houses were broken into on the same night.**\n\nGiven an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight **without alerting the police**.*\n\n**Constraints:**\n- `1 <= nums.length <= 100`\n- `0 <= nums[i] <= 400`",
-      "metadata": {},
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Probelm Explanation:\n\nThe problem is a classic dynamic programming challenge known as the \"House Robber\" problem. The scenario is that you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, but the constraint is that adjacent houses have security systems connected. If two adjacent houses are broken into on the same night, the security systems will automatically contact the police.\n\nThe goal is to determine the maximum amount of money you can rob tonight without alerting the police. You need to find a strategy to maximize your earnings while avoiding robbing two adjacent houses.\n\nThe input to the problem is an array `nums`, where `nums[i]` represents the amount of money in the `i`-th house. The task is to return the maximum amount of money that can be robbed without triggering the security system.\n\nFor example:\n```\nInput: nums = [1,2,3,1]\nOutput: 4\nExplanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).\nTotal amount you can rob = 1 + 3 = 4.\n\nInput: nums = [2,7,9,3,1]\nOutput: 12\nExplanation: Rob house 1 (money = 2), rob house 3 (money = 9), and rob house 5 (money = 1).\nTotal amount you can rob = 2 + 9 + 1 = 12.\n```\n\nThe challenge involves designing an algorithm to efficiently solve this problem and find the optimal strategy for robbing houses to maximize the stolen money while adhering to the constraint of not robbing adjacent houses. The provided solution utilizes dynamic programming to achieve this.\n\n## Solution:\nHere's a Python function to implement this algorithm:",
-      "metadata": {},
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76"
-    },
-    {
-      "cell_type": "code",
-      "source": "from typing import List\n\nclass Solution:\n    def rob(self, nums: List[int]) -> int:\n        # Initialize variables to keep track of the maximum amount of money robbed at the previous and current houses\n        prev, curr = 0, 0\n\n        # Iterate through the houses\n        for n in nums:\n            # Calculate the maximum amount of money that can be robbed at the current house\n            # It is the maximum of either robbing the current house and the money robbed at the house before the previous one (prev),\n            # or skipping the current house and taking the maximum amount from the previous house (curr)\n            temp = max(n + prev, curr)\n\n            # Update the variables for the next iteration\n            prev, curr = curr, temp\n\n        # The final result is the maximum amount of money that can be robbed at the last house\n        return curr",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 4,
-      "outputs": [],
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Explanation:\n\nThe provided code is a Python implementation of the \"House Robber\" problem using a dynamic programming approach. It's part of a class `Solution` and includes a method `rob` that takes a list of integers `nums` as input and returns the maximum amount of money that can be robbed without alerting the police.\n\nThe key idea of the algorithm is to maintain two variables, `prev` and `curr`, which represent the maximum amount of money that can be robbed up to the previous house and the current house, respectively. The algorithm iterates through the houses, updating these variables based on the maximum amount that can be robbed at the current house, considering the constraint of not robbing adjacent houses.\n\nThe loop calculates a temporary variable `temp` that represents the maximum amount of money that can be robbed at the current house. It considers two scenarios: either robbing the current house and adding the money robbed at the house before the previous one (`n + prev`), or skipping the current house and taking the maximum amount from the previous house (`curr`). The maximum of these two scenarios is assigned to `temp`.\n\nThe `prev` and `curr` variables are then updated for the next iteration. `prev` is set to the previous value of `curr`, and `curr` is set to the calculated `temp`.\n\nThe final result is the value stored in the `curr` variable, representing the maximum amount of money that can be robbed at the last house. This value is returned as the result of the `rob` method.\n\nThe provided test cases at the end demonstrate the usage of the `rob` method with different input arrays.",
-      "metadata": {},
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Test cases:\n\nHere's how you can use this solution:",
-      "metadata": {},
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 1:\nsol = Solution()\n\nnums1 = [1, 2, 3, 1]\nprint(sol.rob(nums1))  # Output: 4",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 5,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "4\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums2 = [2, 7, 9, 3, 1]\nprint(sol.rob(nums2))  # Output: 12",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 6,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "12\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Time and Space Complexity Analysis\n\nLet's analyze the time and space complexity of the provided code.\n\n### Time Complexity:\n\nThe time complexity of the code is O(n), where n is the number of houses. The algorithm iterates through the array once, and in each iteration, it performs constant time operations. The loop runs for each house, and the number of operations inside the loop is independent of the input size. Therefore, the overall time complexity is linear, O(n).\n\n### Space Complexity:\n\nThe space complexity is O(1), constant space. The algorithm uses only a constant amount of extra space regardless of the input size. The variables `prev`, `curr`, and `temp` are the only additional variables used, and they do not depend on the size of the input array. Thus, the space complexity is constant, O(1).\n\nIn summary:\n- Time Complexity: O(n)\n- Space Complexity: O(1)",
-      "metadata": {},
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Challenging Exercises:\n\n1. **Variation with Maximum Number of Houses to Rob:** Modify the problem to include an additional constraint: you are allowed to rob at most 'k' houses without triggering the alarm. Find the maximum amount you can rob under this new constraint.\n\n2. **Robbing in a Circular Street:** Extend the problem to a circular street where the first and last houses are adjacent. You cannot rob both the first and last houses simultaneously. Find the maximum amount you can rob in this circular setting.\n\n\n**Link to Problem:** [To see the House Robber problem on LeetCode, click here!](https://leetcode.com/problems/house-robber/description/)\n",
-      "metadata": {},
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6"
-    }
-  ]
-}
\ No newline at end of file
diff --git a/docs/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb b/docs/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb
deleted file mode 100644
index b1080e8..0000000
--- a/docs/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb	
+++ /dev/null
@@ -1,119 +0,0 @@
-{
-  "metadata": {
-    "kernelspec": {
-      "name": "python",
-      "display_name": "Python (Pyodide)",
-      "language": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat_minor": 5,
-  "nbformat": 4,
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "source": "# House Robber II\n**Difficulty:** Medium\n\n---\nYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle**. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.\n\nGiven an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight without alerting the police*.\n\n**Constraints:**\n- `1 <= nums.length <= 100`\n- `0 <= nums[i] <= 1000`",
-      "metadata": {},
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Probelm Explanation:\n\nCertainly! The problem statement describes a scenario where you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. However, there is a security system in place: adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.\n\nThe twist in this problem is that the houses are arranged in a circle. That means the first house is the neighbor of the last one. So, robbing the first house has implications for the last house, and vice versa.\n\nThe goal is to find the maximum amount of money you can rob tonight without alerting the police. You need to implement a function that takes an integer array `nums` representing the amount of money in each house and returns the maximum amount of money that can be robbed without triggering the security system.\n\nHere are a few examples to illustrate the problem:\n\n1. **Example 1:**\n   ```python\n   Input: nums = [2, 3, 2]\n   Output: 3\n   Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2) because they are adjacent houses. The optimal strategy is to rob house 2 (money = 3).\n   ```\n\n2. **Example 2:**\n   ```python\n   Input: nums = [1, 2, 3, 1]\n   Output: 4\n   Explanation: Robbing house 1 (money = 1) and then robbing house 3 (money = 3) results in a total amount of 4. Robbing house 2 is skipped to avoid alerting the police.\n   ```\n\n3. **Example 3:**\n   ```python\n   Input: nums = [1, 2, 3]\n   Output: 3\n   Explanation: Since the houses are in a circle, you can choose either house 1, 2, or 3. The optimal strategy is to rob house 3 for a total amount of 3.\n   ```\n\nThe constraints for the problem are that the length of the `nums` array is between 1 and 100, and the amount of money in each house (`nums[i]`) is between 0 and 1000.\n\n## Solution:\nHere's a Python function to implement this algorithm:",
-      "metadata": {},
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76"
-    },
-    {
-      "cell_type": "code",
-      "source": "from typing import List\n\nclass Solution:\n    def rob(self, nums: List[int]) -> int:\n        def simple_rob(nums):\n            # Helper function for basic house robbery problem\n            prev, curr = 0, 0\n            for num in nums:\n                # Calculate the maximum amount of money that can be robbed\n                # without alerting the police for each house\n                prev, curr = curr, max(prev + num, curr)\n            return curr\n\n        # Check the base cases where the length of nums is 1 or 2\n        if len(nums) == 1:\n            return nums[0]\n        elif len(nums) == 2:\n            return max(nums[0], nums[1])\n\n        # Case 1: First house is robbed, last house is not\n        case1 = simple_rob(nums[:-1])\n\n        # Case 2: First house is not robbed, last house is\n        case2 = simple_rob(nums[1:])\n\n        # Maximum of the two cases\n        return max(case1, case2)",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 7,
-      "outputs": [],
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Explanation:\n\nThe provided code defines a class `Solution` with a method `rob`, which aims to determine the maximum amount of money a professional robber can steal from houses arranged in a circle without alerting the police. Each house contains a certain amount of money, and adjacent houses have a security system that triggers if both are robbed on the same night.\n\nThe code uses dynamic programming to solve the problem efficiently. It contains a helper function `simple_rob` that computes the maximum amount of money that can be robbed from a linear arrangement of houses. The main method `rob` handles the circular arrangement by considering two cases: robbing the first house and not robbing the last, and not robbing the first house and robbing the last. The result is the maximum of these two cases.\n\nThe class is instantiated as `solution`, and the `rob` method is applied to three different scenarios (test cases) to demonstrate its functionality. The final output represents the maximum amount of money the robber can steal without triggering the security system.",
-      "metadata": {},
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Test cases:\n\nHere's how you can use this solution:",
-      "metadata": {},
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 1:\n\n# Create an instance of the Solution class\nsolution = Solution()\n\nnums1 = [2, 3, 2]\nprint(solution.rob(nums1))  # Output: 3",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 8,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "3\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums2 = [1, 2, 3, 1]\nprint(solution.rob(nums2))  # Output: 4",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 9,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "4\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums3 = [1, 2, 3]\nprint(solution.rob(nums3))  # Output: 3",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 10,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "3\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "e6e1554f-ae77-410c-8af2-e59e5428166c"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Time and Space Complexity Analysis\n\nLet's analyze the time and space complexity of the provided code:\n\n### Time Complexity:\nThe time complexity is primarily determined by the `simple_rob` helper function, which has a time complexity of O(N), where N is the number of houses in the input array. The main `rob` function calls `simple_rob` twice, once for the case of robbing the first house and not robbing the last, and once for the case of not robbing the first house and robbing the last. Therefore, the overall time complexity is O(N).\n\n### Space Complexity:\nThe space complexity is O(1) because the space used by the algorithm is constant, regardless of the size of the input array. The only variables used are `prev` and `curr` in the `simple_rob` function, which represent the previous and current maximum amounts of money that can be robbed without alerting the police. These variables are updated iteratively, and no additional data structures are used that scale with the input size. Therefore, the space complexity is constant, or O(1).\n\nIn summary:\n- Time Complexity: O(N)\n- Space Complexity: O(1)",
-      "metadata": {},
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Challenging Exercises:\n\n1. **Optimization Challenge:** Optimize the space complexity of the solution. Can you solve the problem with O(1) space complexity without sacrificing the time complexity?\n\n2. **Circular Robbery Variations:** Consider variations of the circular house robbery problem, such as allowing the robber to skip a certain number of houses or restricting the number of consecutive houses that can be robbed. Modify the solution to accommodate these variations.\n\n**Link to Problem:** [To see the House Robber II problem on LeetCode, click here!](https://leetcode.com/problems/house-robber-ii/)",
-      "metadata": {},
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6"
-    }
-  ]
-}
\ No newline at end of file
diff --git a/docs/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb b/docs/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb
deleted file mode 100644
index b727489..0000000
--- a/docs/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb	
+++ /dev/null
@@ -1,210 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 70. Climbing Stairs\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Climbing Stairs problem on LeetCode, click here!](https://leetcode.com/problems/climbing-stairs/)\n",
-        "\n",
-        "---\n",
-        "You are climbing a staircase. It takes `n` steps to reach the top.\n",
-        "\n",
-        "Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= n <= 45`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The \"Climbing Stairs\" problem is a classic algorithmic problem that asks you to find the number of distinct ways to climb a staircase with a given number of steps. Here's the problem statement:\n",
-        "\n",
-        "You are given a staircase with 'n' steps. Each time you can either climb 1 step or 2 steps. You need to find out how many distinct ways there are to reach the top of the staircase.\n",
-        "\n",
-        "For example, if 'n' is 2, there are two ways to climb to the top:\n",
-        "\n",
-        "1. Climbing two steps at once.\n",
-        "2. Climbing one step, and then another step.\n",
-        "\n",
-        "If 'n' is 3, there are three ways:\n",
-        "\n",
-        "1. Climbing one step, then one step, and finally one step.\n",
-        "2. Climbing one step, then two steps.\n",
-        "3. Climbing two steps, then one step.\n",
-        "\n",
-        "The problem essentially asks you to find a solution that counts the possible combinations of 1-step and 2-step climbs that can be taken to reach the top of the staircase. It's a classic example of a dynamic programming problem where you can break it down into smaller subproblems and build a solution incrementally, as shown in the code I provided earlier.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class Solution:\n",
-        "    def climbStairs(self, n: int) -> int:\n",
-        "        # If there are 1 or 2 steps, there are exactly n ways to reach the top.\n",
-        "        if n <= 2:\n",
-        "            return n\n",
-        "\n",
-        "        # Create an array to store the number of ways to reach each step.\n",
-        "        dp = [0] * (n + 1)\n",
-        "\n",
-        "        # There is one way to reach the first step (base case).\n",
-        "        dp[1] = 1\n",
-        "\n",
-        "        # There are two ways to reach the second step (base case).\n",
-        "        dp[2] = 2\n",
-        "\n",
-        "        # Calculate the number of ways to reach each step from the previous steps.\n",
-        "        for i in range(3, n + 1):\n",
-        "            # The number of ways to reach step 'i' is the sum of the ways to reach\n",
-        "            # the previous two steps, as you can take 1 or 2 steps at a time.\n",
-        "            dp[i] = dp[i - 1] + dp[i - 2]\n",
-        "\n",
-        "        # The result is the number of ways to reach the top, which is stored in dp[n].\n",
-        "        return dp[n]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a class `Solution` with a method `climbStairs` that takes an integer `n` as input and returns the number of distinct ways to climb a staircase with `n` steps.\n",
-        "\n",
-        "1. If `n` is 1 or 2, the function returns `n` directly because there are only 1 or 2 distinct ways to reach the top in those cases.\n",
-        "\n",
-        "2. For `n` greater than 2, the code uses dynamic programming to calculate the number of ways to reach each step. It creates an array `dp` of length `n + 1` to store these values.\n",
-        "\n",
-        "3. The `dp[1]` and `dp[2]` are initialized to 1 and 2, respectively. This is because there is only one way to reach the first step (by taking 1 step), and there are two ways to reach the second step (by taking 2 steps or two 1-step climbs).\n",
-        "\n",
-        "4. The code then uses a loop to fill in the `dp` array for steps 3 and above. For each step `i` from 3 to `n`, it calculates the number of ways to reach that step by adding the number of ways to reach the previous two steps, as you can either take 1 step or 2 steps at a time.\n",
-        "\n",
-        "5. Finally, the function returns the value stored in `dp[n]`, which represents the total number of distinct ways to climb the staircase with `n` steps."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 1 - Input: n = 2, Output: 2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "n = 2\n",
-        "solution = Solution()\n",
-        "print(f\"Example 1 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 2"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 2 - Input: n = 3, Output: 3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "n = 3\n",
-        "print(f\"Example 2 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 3\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Time Complexity:\n",
-        "The time complexity of the solution is O(n) because we use a single for loop that iterates from 3 to n to fill the `dp` array. In each iteration, we perform constant time operations (addition and assignment). Therefore, the time complexity is linear with respect to the input value `n`.\n",
-        "\n",
-        "Space Complexity:\n",
-        "The space complexity of the solution is O(n) as well. We use an array `dp` of length `n + 1` to store the number of ways to reach each step. The size of this array is directly proportional to the input value `n`, which makes the space complexity linear in terms of `n`. The rest of the variables used in the function (e.g., `i`) occupy constant space and do not contribute significantly to the overall space complexity.\n",
-        "\n",
-        "In summary, the time complexity is O(n), and the space complexity is O(n) for this dynamic programming solution to the \"Climbing Stairs\" problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Generalized Step Sizes**: Instead of being limited to 1 step or 2 steps, consider a scenario where you can take steps of sizes `[1, 2, 3]`. Implement a function that calculates the number of ways to climb the staircase with these different step sizes.\n",
-        "\n",
-        "2. **Minimize Steps**: Given an integer `n`, find the minimum number of steps required to reach the top of the staircase. Return both the minimum number of steps and the specific sequence of steps taken."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/docs/_build/html/_sources/13. One-D Dynamic Programming/README.md b/docs/_build/html/_sources/13. One-D Dynamic Programming/README.md
deleted file mode 100644
index 73f4de3..0000000
--- a/docs/_build/html/_sources/13. One-D Dynamic Programming/README.md	
+++ /dev/null
@@ -1,25 +0,0 @@
-# 1-D Dynamic Programming Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)                                   | Easy       |
-| [198. House Robber](https://leetcode.com/problems/house-robber/)                                       | Medium     |
-| [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)                                 | Medium     |
-| [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)       | Medium     |
-| [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)                   | Medium     |
-| [91. Decode Ways](https://leetcode.com/problems/decode-ways/)                                         | Medium     |
-| [322. Coin Change](https://leetcode.com/problems/coin-change/)                                         | Medium     |
-| [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)             | Medium     |
-| [139. Word Break](https://leetcode.com/problems/word-break/)                                           | Medium     |
-| [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/14. Two-D Dynamic Programming/README.md b/docs/_build/html/_sources/14. Two-D Dynamic Programming/README.md
deleted file mode 100644
index 8bab876..0000000
--- a/docs/_build/html/_sources/14. Two-D Dynamic Programming/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### 2-D Dynamic Programming
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [62. Unique Paths](https://leetcode.com/problems/unique-paths/)                                         | Medium     |
-| [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)       | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/15. Greedy/README.md b/docs/_build/html/_sources/15. Greedy/README.md
deleted file mode 100644
index 82b6543..0000000
--- a/docs/_build/html/_sources/15. Greedy/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Greedy
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)                                   | Easy       |
-| [55. Jump Game](https://leetcode.com/problems/jump-game/)                                                 | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/16. Intervals/README.md b/docs/_build/html/_sources/16. Intervals/README.md
deleted file mode 100644
index a57713e..0000000
--- a/docs/_build/html/_sources/16. Intervals/README.md	
+++ /dev/null
@@ -1,50 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Intervals
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [57. Insert Interval](https://leetcode.com/problems/insert-interval/)                                     | Hard       |
-| [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)                                   | Medium     |
-| [435. Non Overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)             | Medium     |
-| [252. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)                                       | Easy       |
-| [253. Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)                                 | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/17. Math Geometry/README.md b/docs/_build/html/_sources/17. Math Geometry/README.md
deleted file mode 100644
index 8b2f180..0000000
--- a/docs/_build/html/_sources/17. Math Geometry/README.md	
+++ /dev/null
@@ -1,48 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Math & Geometry
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [48. Rotate Image](https://leetcode.com/problems/rotate-image/)                                           | Medium     |
-| [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)                                         | Medium     |
-| [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)                               | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/18. Bit Manipulation/README.md b/docs/_build/html/_sources/18. Bit Manipulation/README.md
deleted file mode 100644
index 05b2fef..0000000
--- a/docs/_build/html/_sources/18. Bit Manipulation/README.md	
+++ /dev/null
@@ -1,51 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Bit Manupulation
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)                                   | Easy       |
-| [338. Counting Bits](https://leetcode.com/problems/counting-bits/)                                         | Medium     |
-| [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)                                           | Easy       |
-| [268. Missing Number](https://leetcode.com/problems/missing-number/)                                       | Easy       |
-| [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)                             | Medium     |
-
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/docs/_build/html/_sources/Book.ipynb b/docs/_build/html/_sources/Book.ipynb
deleted file mode 100644
index d400192..0000000
--- a/docs/_build/html/_sources/Book.ipynb
+++ /dev/null
@@ -1,373 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!python -m pip install --upgrade pip\n",
-    "%pip install jupyter-book"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions\n"
-     ]
-    }
-   ],
-   "source": [
-    "%cd /workspaces/LeetCode-Solutions"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book toc migrate /workspaces/LeetCode-Solutions/_toc.yml -o /workspaces/LeetCode-Solutions/_toc.yml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "\u001b[01mmaking output directory... \u001b[39;49;00mdone\n",
-      "[etoc] Changing master_doc to 'README'\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 77 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 77 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]  \n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.26 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.00 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]     \n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.92 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 1.28 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executed notebook in 0.91 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executing notebook using local CWD [mystnb]      \n",
-      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executed notebook in 0.91 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executed notebook in 1.17 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executed notebook in 1.16 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executing notebook using local CWD [mystnb] \n",
-      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executed notebook in 1.27 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executed notebook in 4.03 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executing notebook using local CWD [mystnb]   \n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/Book.ipynb: Executing notebook using local CWD [mystnb]                          \n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mSECURITY\u001b[39;49;00m                                              \n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb:30008: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Executing notebook failed: CellTimeoutError [mystnb.exec]\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Notebook exception traceback saved in: /workspaces/LeetCode-Solutions/_build/html/reports/Book.err.log [mystnb.exec]\u001b[39;49;00m\n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CODE_OF_CONDUCT.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/SECURITY.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[ 94%] \u001b[32mBook\u001b[39;49;00m                                                   raph\u001b[39;49;00m0m\r"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build ."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book build /workspaces/LeetCode-Solutions/ --builder pdfhtml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/html": [
-       "<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>"
-      ],
-      "text/plain": [
-       "<IPython.core.display.HTML object>"
-      ]
-     },
-     "execution_count": 5,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "from IPython.display import HTML\n",
-    "HTML('<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>')\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Your book template can be found at\n",
-      "\n",
-      "    mybook/\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book create mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions/mybook\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_build/html\n",
-      "[sphinxcontrib-bibtex] Beware that docutils versions 0.18 and 0.19 (you are running 0.18.1) are known to generate invalid html for citations. If this issue affects you, please use docutils<0.18 (or >=0.20 once released) instead. For more details, see https://sourceforge.net/p/docutils/patches/195/\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "[etoc] Changing master_doc to 'intro'\n",
-      "checking bibtex cache... out of date\n",
-      "parsing bibtex file /workspaces/LeetCode-Solutions/mybook/references.bib... parsed 5 entries\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='force', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/mybook/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 14 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 14 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.35 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.97 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 0.83 seconds [mystnb]\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mintro\u001b[39;49;00m                                                 \n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mintro\u001b[39;49;00m                                                  \n",
-      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
-      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
-      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
-      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
-      "[etoc] missing index.html written as redirect to 'intro.html'\n",
-      "\u001b[01mbuild succeeded, 2 warnings.\u001b[39;49;00m\n",
-      "\n",
-      "The HTML pages are in mybook/_build/html.\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Finished generating HTML for book.\n",
-      "Your book's HTML pages are here:\n",
-      "    mybook/_build/html/\n",
-      "You can look at your book by opening this file in a browser:\n",
-      "    mybook/_build/html/index.html\n",
-      "Or paste this line directly into your browser bar:\n",
-      "    file:///workspaces/LeetCode-Solutions/mybook/_build/html/index.html            \n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "format: jb-book\n",
-      "root: intro\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book toc from-project /workspaces/LeetCode-Solutions/mybook/ -f jb-book"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 2
-}
diff --git a/docs/_build/html/_sources/CODE_OF_CONDUCT.md b/docs/_build/html/_sources/CODE_OF_CONDUCT.md
deleted file mode 100644
index 5fe5041..0000000
--- a/docs/_build/html/_sources/CODE_OF_CONDUCT.md
+++ /dev/null
@@ -1,128 +0,0 @@
-# Contributor Covenant Code of Conduct
-
-## Our Pledge
-
-We as members, contributors, and leaders pledge to make participation in our
-community a harassment-free experience for everyone, regardless of age, body
-size, visible or invisible disability, ethnicity, sex characteristics, gender
-identity and expression, level of experience, education, socio-economic status,
-nationality, personal appearance, race, religion, or sexual identity
-and orientation.
-
-We pledge to act and interact in ways that contribute to an open, welcoming,
-diverse, inclusive, and healthy community.
-
-## Our Standards
-
-Examples of behavior that contributes to a positive environment for our
-community include:
-
-* Demonstrating empathy and kindness toward other people
-* Being respectful of differing opinions, viewpoints, and experiences
-* Giving and gracefully accepting constructive feedback
-* Accepting responsibility and apologizing to those affected by our mistakes,
-  and learning from the experience
-* Focusing on what is best not just for us as individuals, but for the
-  overall community
-
-Examples of unacceptable behavior include:
-
-* The use of sexualized language or imagery, and sexual attention or
-  advances of any kind
-* Trolling, insulting or derogatory comments, and personal or political attacks
-* Public or private harassment
-* Publishing others' private information, such as a physical or email
-  address, without their explicit permission
-* Other conduct which could reasonably be considered inappropriate in a
-  professional setting
-
-## Enforcement Responsibilities
-
-Community leaders are responsible for clarifying and enforcing our standards of
-acceptable behavior and will take appropriate and fair corrective action in
-response to any behavior that they deem inappropriate, threatening, offensive,
-or harmful.
-
-Community leaders have the right and responsibility to remove, edit, or reject
-comments, commits, code, wiki edits, issues, and other contributions that are
-not aligned to this Code of Conduct, and will communicate reasons for moderation
-decisions when appropriate.
-
-## Scope
-
-This Code of Conduct applies within all community spaces, and also applies when
-an individual is officially representing the community in public spaces.
-Examples of representing our community include using an official e-mail address,
-posting via an official social media account, or acting as an appointed
-representative at an online or offline event.
-
-## Enforcement
-
-Instances of abusive, harassing, or otherwise unacceptable behavior may be
-reported to the community leaders responsible for enforcement at
-Email.
-All complaints will be reviewed and investigated promptly and fairly.
-
-All community leaders are obligated to respect the privacy and security of the
-reporter of any incident.
-
-## Enforcement Guidelines
-
-Community leaders will follow these Community Impact Guidelines in determining
-the consequences for any action they deem in violation of this Code of Conduct:
-
-### 1. Correction
-
-**Community Impact**: Use of inappropriate language or other behavior deemed
-unprofessional or unwelcome in the community.
-
-**Consequence**: A private, written warning from community leaders, providing
-clarity around the nature of the violation and an explanation of why the
-behavior was inappropriate. A public apology may be requested.
-
-### 2. Warning
-
-**Community Impact**: A violation through a single incident or series
-of actions.
-
-**Consequence**: A warning with consequences for continued behavior. No
-interaction with the people involved, including unsolicited interaction with
-those enforcing the Code of Conduct, for a specified period of time. This
-includes avoiding interactions in community spaces as well as external channels
-like social media. Violating these terms may lead to a temporary or
-permanent ban.
-
-### 3. Temporary Ban
-
-**Community Impact**: A serious violation of community standards, including
-sustained inappropriate behavior.
-
-**Consequence**: A temporary ban from any sort of interaction or public
-communication with the community for a specified period of time. No public or
-private interaction with the people involved, including unsolicited interaction
-with those enforcing the Code of Conduct, is allowed during this period.
-Violating these terms may lead to a permanent ban.
-
-### 4. Permanent Ban
-
-**Community Impact**: Demonstrating a pattern of violation of community
-standards, including sustained inappropriate behavior,  harassment of an
-individual, or aggression toward or disparagement of classes of individuals.
-
-**Consequence**: A permanent ban from any sort of public interaction within
-the community.
-
-## Attribution
-
-This Code of Conduct is adapted from the [Contributor Covenant][homepage],
-version 2.0, available at
-https://www.contributor-covenant.org/version/2/0/code_of_conduct.html.
-
-Community Impact Guidelines were inspired by [Mozilla's code of conduct
-enforcement ladder](https://github.com/mozilla/diversity).
-
-[homepage]: https://www.contributor-covenant.org
-
-For answers to common questions about this code of conduct, see the FAQ at
-https://www.contributor-covenant.org/faq. Translations are available at
-https://www.contributor-covenant.org/translations.
diff --git a/docs/_build/html/_sources/CONTRIBUTING.md b/docs/_build/html/_sources/CONTRIBUTING.md
deleted file mode 100644
index ca794ef..0000000
--- a/docs/_build/html/_sources/CONTRIBUTING.md
+++ /dev/null
@@ -1,79 +0,0 @@
-# Contributing Guidelines
-
-Thank you for considering contributing to the Blind 75 LeetCode repository! We welcome contributions from the community to help improve this resource and make it even more valuable.
-
-Please take a moment to review this document to understand how you can contribute.
-
-## Table of Contents
-- [Ways to Contribute](#ways-to-contribute)
-- [Getting Started](#getting-started)
-- [Contributing Process](#contributing-process)
-- [Code of Conduct](#code-of-conduct)
-
-## Ways to Contribute
-
-There are several ways you can contribute to this repository:
-
-1. **Suggest Improvements**: If you have ideas for making the Blind 75 LeetCode repository better, please open an issue to discuss your suggestions.
-
-2. **Report Issues**: If you encounter any issues with the existing content or code, please open an issue and provide as much detail as possible.
-
-3. **Contribute Solutions**: You can contribute solutions to the problems listed in the repository. Please follow the guidelines mentioned below when submitting your solutions.
-
-4. **Enhance Documentation**: Improve documentation, including explanations and time/space complexity analysis, for existing solutions.
-
-5. **Fix Bugs**: If you find and fix any bugs in the existing code or documentation, you are welcome to submit a pull request.
-
-## Getting Started
-
-Before you start contributing, please ensure you have the following prerequisites:
-
-- A GitHub account.
-- Familiarity with Git and GitHub.
-- Code editor or IDE for making changes.
-
-## Contributing Process
-
-Here's the general process for contributing:
-
-1. **Fork the Repository**: Click the "Fork" button at the top of this repository to create your copy.
-
-2. **Clone Your Fork**: Clone your fork to your local machine.
-
-    ```
-    git clone https://github.com/your-username/Blind-75-LeetCode.git
-    ```
-
-3. **Create a Branch**: Create a new branch for your contribution.
-
-    ```
-    git checkout -b feature/your-feature-name
-    ```
-
-4. **Make Changes**: Implement your changes, following the guidelines provided.
-
-5. **Test Your Changes**: Ensure that your changes do not introduce new issues and are consistent with the repository's style and structure.
-
-6. **Commit Changes**: Commit your changes with a meaningful commit message.
-
-    ```
-    git commit -m "Add solution to problem X"
-    ```
-
-7. **Push Changes**: Push your changes to your fork on GitHub.
-
-    ```
-    git push origin feature/your-feature-name
-    ```
-
-8. **Create a Pull Request**: Go to the original repository and create a pull request. Please provide a detailed description of your changes.
-
-9. **Review and Collaboration**: Your pull request will be reviewed, and you may need to make further changes. Be responsive to comments and feedback.
-
-10. **Merge**: Once your pull request is approved, it will be merged into the main branch.
-
-## Code of Conduct
-
-Please note that we have a [Code of Conduct](CODE_OF_CONDUCT.md) that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.
-
-We appreciate your contributions to the Blind 75 LeetCode repository and look forward to working together to make it an even better resource for the community!
diff --git a/docs/_build/html/_sources/README.md b/docs/_build/html/_sources/README.md
deleted file mode 100644
index a9b8b2e..0000000
--- a/docs/_build/html/_sources/README.md
+++ /dev/null
@@ -1,216 +0,0 @@
-# Blind 75 LeetCode Solutions
-
-![Blind 75 LeetCode](https://img.shields.io/badge/Blind_75_LeetCode-Solutions-blue?labelColor=red)
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Array & Hashing
-![Array & Hashing](https://img.shields.io/badge/Array_&_Hashing-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [217. Contains Duplicate](https://leetcode.com/problems/contains-duplicate/)                        | Easy       |
-| [242. Valid Anagram](https://leetcode.com/problems/valid-anagram/)                                  | Easy       |
-| [1. Two Sum](https://leetcode.com/problems/two-sum/)                                              | Easy       |
-| [49. Group Anagrams](https://leetcode.com/problems/group-anagrams/)                                | Medium     |
-| [347. Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/)              | Medium     |
-| [238. Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/)    | Medium     |
-| [659. Encode and Decode Strings](https://leetcode.com/problems/encode-and-decode-strings/)          | Medium     |
-| [128. Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/)    | Hard       |
-
-### Two Pointers
-![Array & Hashing](https://img.shields.io/badge/Two_Pointers-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)                            | Easy       |
-| [15. 3Sum](https://leetcode.com/problems/3sum/)                                                       | Medium     |
-| [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)           | Medium     |
-
-### Sliding Window
-![Array & Hashing](https://img.shields.io/badge/Sliding_Window-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [121. Best Time to Buy And Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | Easy       |
-| [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) | Medium |
-| [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) | Medium |
-| [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)             | Hard       |
-
-### Stack
-![Array & Hashing](https://img.shields.io/badge/Stack-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)                            | Easy       |
-
-### Binary Search
-![Array & Hashing](https://img.shields.io/badge/Binary_Search-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [153. Find Minimum In Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) | Medium |
-| [33. Search In Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)   | Medium     |
-
-### Linked List
-![Array & Hashing](https://img.shields.io/badge/Linked_List-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)                       | Easy       |
-| [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)                   | Easy       |
-| [143. Reorder List](https://leetcode.com/problems/reorder-list/)                                     | Medium     |
-| [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | Medium |
-| [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)                           | Medium     |
-| [23. Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)                       | Hard       |
-
-### Trees
-![Array & Hashing](https://img.shields.io/badge/Trees-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)                          | Easy       |
-| [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)     | Easy       |
-| [100. Same Tree](https://leetcode.com/problems/same-tree/)                                           | Easy       |
-| [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)               | Easy       |
-| [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/) | Easy |
-| [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/) | Medium |
-| [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)       | Medium     |
-| [230. Kth Smallest Element In a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)   | Medium     |
-| [105. Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) | Medium |
-| [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)     | Hard       |
-| [297. Serialize And Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/) | Hard |
-
-### Tries
-![Array & Hashing](https://img.shields.io/badge/Tries-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [208. Implement Trie Prefix Tree](https://leetcode.com/problems/implement-trie-prefix-tree/)         | Medium     |
-| [211. Design Add And Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/) | Medium |
-| [212. Word Search II](https://leetcode.com/problems/word-search-ii/)                                 | Hard       |
-
-### Heap / Priority Queue
-![Array & Hashing](https://img.shields.io/badge/Heap_/_Priority_Queue-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [295. Find Median From Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)     | Hard       |
-
-### Backtracking
-![Array & Hashing](https://img.shields.io/badge/Backtracking-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [39. Combination Sum](https://leetcode.com/problems/combination-sum/)                                 | Medium     |
-| [79. Word Search](https://leetcode.com/problems/word-search/)                                         | Medium     |
-
-### Graphs
-![Array & Hashing](https://img.shields.io/badge/Graphs-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)                           | Medium     |
-| [133. Clone Graph](https://leetcode.com/problems/clone-graph/)                                       | Medium     |
-| [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)       | Medium     |
-| [207. Course Schedule](https://leetcode.com/problems/course-schedule/)                                 | Medium     |
-| [323. Number of Connected Components In An Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/) | Medium |
-| [261. Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)                               | Medium     |
-
-### Advanced Graphs
-![Array & Hashing](https://img.shields.io/badge/Advanced_Graphs-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [269. Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)                               | Hard       |
-
-### 1-D Dynamic Programming
-![Array & Hashing](https://img.shields.io/badge/1_D_Dynamic_Programming-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)                                   | Easy       |
-| [198. House Robber](https://leetcode.com/problems/house-robber/)                                       | Easy       |
-| [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)                                 | Medium     |
-| [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)       | Medium     |
-| [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)                   | Medium     |
-| [91. Decode Ways](https://leetcode.com/problems/decode-ways/)                                         | Medium     |
-| [322. Coin Change](https://leetcode.com/problems/coin-change/)                                         | Medium     |
-| [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)             | Medium     |
-| [139. Word Break](https://leetcode.com/problems/word-break/)                                           | Medium     |
-| [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | Medium     |
-
-### 2-D Dynamic Programming
-![Array & Hashing](https://img.shields.io/badge/2_D_Dynamic_Programming-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [62. Unique Paths](https://leetcode.com/problems/unique-paths/)                                         | Medium     |
-| [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)       | Medium     |
-
-### Greedy
-![Array & Hashing](https://img.shields.io/badge/Greedy-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)                                   | Easy       |
-| [55. Jump Game](https://leetcode.com/problems/jump-game/)                                                 | Medium     |
-
-### Intervals
-![Array & Hashing](https://img.shields.io/badge/Intervals-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [57. Insert Interval](https://leetcode.com/problems/insert-interval/)                                     | Hard       |
-| [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)                                   | Medium     |
-| [435. Non Overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)             | Medium     |
-| [252. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)                                       | Easy       |
-| [253. Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)                                 | Medium     |
-
-### Math & Geometry
-![Array & Hashing](https://img.shields.io/badge/Math_&_Geometry-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [48. Rotate Image](https://leetcode.com/problems/rotate-image/)                                           | Medium     |
-| [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)                                         | Medium     |
-| [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)                               | Medium     |
-
-### Bit Manipulation
-![Array & Hashing](https://img.shields.io/badge/Bit_Manipulation-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)                                   | Easy       |
-| [338. Counting Bits](https://leetcode.com/problems/counting-bits/)                                         | Medium     |
-| [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)                                           | Easy       |
-| [268. Missing Number](https://leetcode.com/problems/missing-number/)                                       | Easy       |
-| [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)                             | Medium     |
-
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-## Code of Conduct
-
-We have a [Code of Conduct](CODE_OF_CONDUCT.md) that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.
-
-## Security Policy
-
-We take security seriously. Please review our [Security Policy](SECURITY.md) for information on reporting security vulnerabilities.
-
-Happy coding!
diff --git a/docs/_build/html/_sources/SECURITY.md b/docs/_build/html/_sources/SECURITY.md
deleted file mode 100644
index 7c19993..0000000
--- a/docs/_build/html/_sources/SECURITY.md
+++ /dev/null
@@ -1,42 +0,0 @@
-# Security Policy
-
-## Reporting a Vulnerability
-
-We take the security of the Blind 75 LeetCode repository seriously. If you believe you have found a security vulnerability, please help us by following these steps:
-
-1. **Do Not Disclose Publicly**: Please do not disclose the issue publicly until it has been addressed by our team.
-
-2. **Report Privately**: Create a GitHub issue with the title "Security Vulnerability Report" and provide a detailed description of the vulnerability. You can use a sample report template as follows:
-
-    ```
-    **Description**
-    A clear and concise description of the security vulnerability.
-
-    **Steps to Reproduce**
-    Provide detailed steps on how to reproduce the vulnerability.
-
-    **Expected Behavior**
-    Describe what you expected to happen.
-
-    **Actual Behavior**
-    Describe what actually happened.
-
-    **Additional Information**
-    Any additional information that can help us understand and reproduce the issue, such as environment details, configuration, or screenshots.
-    ```
-
-3. **Secure Communication**: If the vulnerability involves sensitive data, please encrypt your communication with our PGP key. Contact us for our PGP key information.
-
-4. **Response**: We will acknowledge your report within 5 business days and provide an estimated timeline for the resolution.
-
-5. **Fix and Disclosure**: We will work to fix the issue promptly and privately. Once resolved, we will provide a security advisory and give credit to the reporter if desired.
-
-6. **Responsible Disclosure**: We encourage responsible disclosure. Please do not take advantage of the vulnerability or attempt to exploit it for malicious purposes.
-
-## Security Updates
-
-We are committed to promptly addressing and resolving any security issues. We recommend that you stay up-to-date with the latest releases of this repository, as they may include security updates and fixes.
-
-## Contact
-
-If you have any questions or need further assistance regarding security, you can contact us at mohsentabibian@gmail.com.
diff --git a/docs/_build/html/_sphinx_design_static/design-style.4045f2051d55cab465a707391d5b2007.min.css b/docs/_build/html/_sphinx_design_static/design-style.4045f2051d55cab465a707391d5b2007.min.css
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diff --git a/docs/_build/html/_sphinx_design_static/design-tabs.js b/docs/_build/html/_sphinx_design_static/design-tabs.js
deleted file mode 100644
index 36b38cf..0000000
--- a/docs/_build/html/_sphinx_design_static/design-tabs.js
+++ /dev/null
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diff --git a/docs/_build/html/_static/_sphinx_javascript_frameworks_compat.js b/docs/_build/html/_static/_sphinx_javascript_frameworks_compat.js
deleted file mode 100644
index 8549469..0000000
--- a/docs/_build/html/_static/_sphinx_javascript_frameworks_compat.js
+++ /dev/null
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-/*
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diff --git a/docs/_build/html/_static/basic.css b/docs/_build/html/_static/basic.css
deleted file mode 100644
index 9e364ed..0000000
--- a/docs/_build/html/_static/basic.css
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t=o?(t=m(this).constructor,Reflect.construct(e,arguments,t)):e.apply(this,arguments),e=this,!(t=t)||"object"!==d(t)&&"function"!=typeof t?function(t){if(void 0!==t)return t;throw new ReferenceError("this hasn't been initialised - super() hasn't been called")}(e):t}}function m(t){return(m=Object.setPrototypeOf?Object.getPrototypeOf:function(t){return t.__proto__||Object.getPrototypeOf(t)})(t)}function v(t,e){t="data-clipboard-".concat(t);if(e.hasAttribute(t))return e.getAttribute(t)}var o=function(){!function(t,e){if("function"!=typeof e&&null!==e)throw new TypeError("Super expression must either be null or a function");t.prototype=Object.create(e&&e.prototype,{constructor:{value:t,writable:!0,configurable:!0}}),e&&y(t,e)}(r,i());var t,e,n,o=h(r);function r(t,e){var n;return function(t){if(!(t instanceof r))throw new TypeError("Cannot call a class as a function")}(this),(n=o.call(this)).resolveOptions(e),n.listenClick(t),n}return t=r,n=[{key:"copy",value:function(t){var 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\ No newline at end of file
diff --git a/docs/_build/html/_static/copy-button.svg b/docs/_build/html/_static/copy-button.svg
deleted file mode 100644
index 9c074da..0000000
--- a/docs/_build/html/_static/copy-button.svg
+++ /dev/null
@@ -1,5 +0,0 @@
-<svg xmlns="http://www.w3.org/2000/svg" class="icon icon-tabler icon-tabler-copy" width="44" height="44" viewBox="0 0 24 24" stroke-width="1.5" stroke="#000000" fill="none" stroke-linecap="round" stroke-linejoin="round">
-  <path stroke="none" d="M0 0h24v24H0z" fill="none"/>
-  <rect x="8" y="8" width="12" height="12" rx="2" />
-  <path d="M16 8v-2a2 2 0 0 0 -2 -2h-8a2 2 0 0 0 -2 2v8a2 2 0 0 0 2 2h2" />
-</svg>
diff --git a/docs/_build/html/_static/copybutton.css b/docs/_build/html/_static/copybutton.css
deleted file mode 100644
index f1916ec..0000000
--- a/docs/_build/html/_static/copybutton.css
+++ /dev/null
@@ -1,94 +0,0 @@
-/* Copy buttons */
-button.copybtn {
-    position: absolute;
-    display: flex;
-    top: .3em;
-    right: .3em;
-    width: 1.7em;
-    height: 1.7em;
-	opacity: 0;
-    transition: opacity 0.3s, border .3s, background-color .3s;
-    user-select: none;
-    padding: 0;
-    border: none;
-    outline: none;
-    border-radius: 0.4em;
-    /* The colors that GitHub uses */
-    border: #1b1f2426 1px solid;
-    background-color: #f6f8fa;
-    color: #57606a;
-}
-
-button.copybtn.success {
-    border-color: #22863a;
-    color: #22863a;
-}
-
-button.copybtn svg {
-    stroke: currentColor;
-    width: 1.5em;
-    height: 1.5em;
-    padding: 0.1em;
-}
-
-div.highlight  {
-    position: relative;
-}
-
-/* Show the copybutton */
-.highlight:hover button.copybtn, button.copybtn.success {
-	opacity: 1;
-}
-
-.highlight button.copybtn:hover {
-    background-color: rgb(235, 235, 235);
-}
-
-.highlight button.copybtn:active {
-    background-color: rgb(187, 187, 187);
-}
-
-/**
- * A minimal CSS-only tooltip copied from:
- *   https://codepen.io/mildrenben/pen/rVBrpK
- *
- * To use, write HTML like the following:
- *
- * <p class="o-tooltip--left" data-tooltip="Hey">Short</p>
- */
- .o-tooltip--left {
-  position: relative;
- }
-
- .o-tooltip--left:after {
-    opacity: 0;
-    visibility: hidden;
-    position: absolute;
-    content: attr(data-tooltip);
-    padding: .2em;
-    font-size: .8em;
-    left: -.2em;
-    background: grey;
-    color: white;
-    white-space: nowrap;
-    z-index: 2;
-    border-radius: 2px;
-    transform: translateX(-102%) translateY(0);
-    transition: opacity 0.2s cubic-bezier(0.64, 0.09, 0.08, 1), transform 0.2s cubic-bezier(0.64, 0.09, 0.08, 1);
-}
-
-.o-tooltip--left:hover:after {
-    display: block;
-    opacity: 1;
-    visibility: visible;
-    transform: translateX(-100%) translateY(0);
-    transition: opacity 0.2s cubic-bezier(0.64, 0.09, 0.08, 1), transform 0.2s cubic-bezier(0.64, 0.09, 0.08, 1);
-    transition-delay: .5s;
-}
-
-/* By default the copy button shouldn't show up when printing a page */
-@media print {
-    button.copybtn {
-        display: none;
-    }
-}
diff --git a/docs/_build/html/_static/copybutton.js b/docs/_build/html/_static/copybutton.js
deleted file mode 100644
index 2ea7ff3..0000000
--- a/docs/_build/html/_static/copybutton.js
+++ /dev/null
@@ -1,248 +0,0 @@
-// Localization support
-const messages = {
-  'en': {
-    'copy': 'Copy',
-    'copy_to_clipboard': 'Copy to clipboard',
-    'copy_success': 'Copied!',
-    'copy_failure': 'Failed to copy',
-  },
-  'es' : {
-    'copy': 'Copiar',
-    'copy_to_clipboard': 'Copiar al portapapeles',
-    'copy_success': '¡Copiado!',
-    'copy_failure': 'Error al copiar',
-  },
-  'de' : {
-    'copy': 'Kopieren',
-    'copy_to_clipboard': 'In die Zwischenablage kopieren',
-    'copy_success': 'Kopiert!',
-    'copy_failure': 'Fehler beim Kopieren',
-  },
-  'fr' : {
-    'copy': 'Copier',
-    'copy_to_clipboard': 'Copier dans le presse-papier',
-    'copy_success': 'Copié !',
-    'copy_failure': 'Échec de la copie',
-  },
-  'ru': {
-    'copy': 'Скопировать',
-    'copy_to_clipboard': 'Скопировать в буфер',
-    'copy_success': 'Скопировано!',
-    'copy_failure': 'Не удалось скопировать',
-  },
-  'zh-CN': {
-    'copy': '复制',
-    'copy_to_clipboard': '复制到剪贴板',
-    'copy_success': '复制成功!',
-    'copy_failure': '复制失败',
-  },
-  'it' : {
-    'copy': 'Copiare',
-    'copy_to_clipboard': 'Copiato negli appunti',
-    'copy_success': 'Copiato!',
-    'copy_failure': 'Errore durante la copia',
-  }
-}
-
-let locale = 'en'
-if( document.documentElement.lang !== undefined
-    && messages[document.documentElement.lang] !== undefined ) {
-  locale = document.documentElement.lang
-}
-
-let doc_url_root = DOCUMENTATION_OPTIONS.URL_ROOT;
-if (doc_url_root == '#') {
-    doc_url_root = '';
-}
-
-/**
- * SVG files for our copy buttons
- */
-let iconCheck = `<svg xmlns="http://www.w3.org/2000/svg" class="icon icon-tabler icon-tabler-check" width="44" height="44" viewBox="0 0 24 24" stroke-width="2" stroke="#22863a" fill="none" stroke-linecap="round" stroke-linejoin="round">
-  <title>${messages[locale]['copy_success']}</title>
-  <path stroke="none" d="M0 0h24v24H0z" fill="none"/>
-  <path d="M5 12l5 5l10 -10" />
-</svg>`
-
-// If the user specified their own SVG use that, otherwise use the default
-let iconCopy = ``;
-if (!iconCopy) {
-  iconCopy = `<svg xmlns="http://www.w3.org/2000/svg" class="icon icon-tabler icon-tabler-copy" width="44" height="44" viewBox="0 0 24 24" stroke-width="1.5" stroke="#000000" fill="none" stroke-linecap="round" stroke-linejoin="round">
-  <title>${messages[locale]['copy_to_clipboard']}</title>
-  <path stroke="none" d="M0 0h24v24H0z" fill="none"/>
-  <rect x="8" y="8" width="12" height="12" rx="2" />
-  <path d="M16 8v-2a2 2 0 0 0 -2 -2h-8a2 2 0 0 0 -2 2v8a2 2 0 0 0 2 2h2" />
-</svg>`
-}
-
-/**
- * Set up copy/paste for code blocks
- */
-
-const runWhenDOMLoaded = cb => {
-  if (document.readyState != 'loading') {
-    cb()
-  } else if (document.addEventListener) {
-    document.addEventListener('DOMContentLoaded', cb)
-  } else {
-    document.attachEvent('onreadystatechange', function() {
-      if (document.readyState == 'complete') cb()
-    })
-  }
-}
-
-const codeCellId = index => `codecell${index}`
-
-// Clears selected text since ClipboardJS will select the text when copying
-const clearSelection = () => {
-  if (window.getSelection) {
-    window.getSelection().removeAllRanges()
-  } else if (document.selection) {
-    document.selection.empty()
-  }
-}
-
-// Changes tooltip text for a moment, then changes it back
-// We want the timeout of our `success` class to be a bit shorter than the
-// tooltip and icon change, so that we can hide the icon before changing back.
-var timeoutIcon = 2000;
-var timeoutSuccessClass = 1500;
-
-const temporarilyChangeTooltip = (el, oldText, newText) => {
-  el.setAttribute('data-tooltip', newText)
-  el.classList.add('success')
-  // Remove success a little bit sooner than we change the tooltip
-  // So that we can use CSS to hide the copybutton first
-  setTimeout(() => el.classList.remove('success'), timeoutSuccessClass)
-  setTimeout(() => el.setAttribute('data-tooltip', oldText), timeoutIcon)
-}
-
-// Changes the copy button icon for two seconds, then changes it back
-const temporarilyChangeIcon = (el) => {
-  el.innerHTML = iconCheck;
-  setTimeout(() => {el.innerHTML = iconCopy}, timeoutIcon)
-}
-
-const addCopyButtonToCodeCells = () => {
-  // If ClipboardJS hasn't loaded, wait a bit and try again. This
-  // happens because we load ClipboardJS asynchronously.
-  if (window.ClipboardJS === undefined) {
-    setTimeout(addCopyButtonToCodeCells, 250)
-    return
-  }
-
-  // Add copybuttons to all of our code cells
-  const COPYBUTTON_SELECTOR = 'div.highlight pre';
-  const codeCells = document.querySelectorAll(COPYBUTTON_SELECTOR)
-  codeCells.forEach((codeCell, index) => {
-    const id = codeCellId(index)
-    codeCell.setAttribute('id', id)
-
-    const clipboardButton = id =>
-    `<button class="copybtn o-tooltip--left" data-tooltip="${messages[locale]['copy']}" data-clipboard-target="#${id}">
-      ${iconCopy}
-    </button>`
-    codeCell.insertAdjacentHTML('afterend', clipboardButton(id))
-  })
-
-function escapeRegExp(string) {
-    return string.replace(/[.*+?^${}()|[\]\\]/g, '\\$&'); // $& means the whole matched string
-}
-
-/**
- * Removes excluded text from a Node.
- *
- * @param {Node} target Node to filter.
- * @param {string} exclude CSS selector of nodes to exclude.
- * @returns {DOMString} Text from `target` with text removed.
- */
-function filterText(target, exclude) {
-    const clone = target.cloneNode(true);  // clone as to not modify the live DOM
-    if (exclude) {
-        // remove excluded nodes
-        clone.querySelectorAll(exclude).forEach(node => node.remove());
-    }
-    return clone.innerText;
-}
-
-// Callback when a copy button is clicked. Will be passed the node that was clicked
-// should then grab the text and replace pieces of text that shouldn't be used in output
-function formatCopyText(textContent, copybuttonPromptText, isRegexp = false, onlyCopyPromptLines = true, removePrompts = true, copyEmptyLines = true, lineContinuationChar = "", hereDocDelim = "") {
-    var regexp;
-    var match;
-
-    // Do we check for line continuation characters and "HERE-documents"?
-    var useLineCont = !!lineContinuationChar
-    var useHereDoc = !!hereDocDelim
-
-    // create regexp to capture prompt and remaining line
-    if (isRegexp) {
-        regexp = new RegExp('^(' + copybuttonPromptText + ')(.*)')
-    } else {
-        regexp = new RegExp('^(' + escapeRegExp(copybuttonPromptText) + ')(.*)')
-    }
-
-    const outputLines = [];
-    var promptFound = false;
-    var gotLineCont = false;
-    var gotHereDoc = false;
-    const lineGotPrompt = [];
-    for (const line of textContent.split('\n')) {
-        match = line.match(regexp)
-        if (match || gotLineCont || gotHereDoc) {
-            promptFound = regexp.test(line)
-            lineGotPrompt.push(promptFound)
-            if (removePrompts && promptFound) {
-                outputLines.push(match[2])
-            } else {
-                outputLines.push(line)
-            }
-            gotLineCont = line.endsWith(lineContinuationChar) & useLineCont
-            if (line.includes(hereDocDelim) & useHereDoc)
-                gotHereDoc = !gotHereDoc
-        } else if (!onlyCopyPromptLines) {
-            outputLines.push(line)
-        } else if (copyEmptyLines && line.trim() === '') {
-            outputLines.push(line)
-        }
-    }
-
-    // If no lines with the prompt were found then just use original lines
-    if (lineGotPrompt.some(v => v === true)) {
-        textContent = outputLines.join('\n');
-    }
-
-    // Remove a trailing newline to avoid auto-running when pasting
-    if (textContent.endsWith("\n")) {
-        textContent = textContent.slice(0, -1)
-    }
-    return textContent
-}
-
-
-var copyTargetText = (trigger) => {
-  var target = document.querySelector(trigger.attributes['data-clipboard-target'].value);
-
-  // get filtered text
-  let exclude = '.linenos';
-
-  let text = filterText(target, exclude);
-  return formatCopyText(text, '', false, true, true, true, '', '')
-}
-
-  // Initialize with a callback so we can modify the text before copy
-  const clipboard = new ClipboardJS('.copybtn', {text: copyTargetText})
-
-  // Update UI with error/success messages
-  clipboard.on('success', event => {
-    clearSelection()
-    temporarilyChangeTooltip(event.trigger, messages[locale]['copy'], messages[locale]['copy_success'])
-    temporarilyChangeIcon(event.trigger)
-  })
-
-  clipboard.on('error', event => {
-    temporarilyChangeTooltip(event.trigger, messages[locale]['copy'], messages[locale]['copy_failure'])
-  })
-}
-
-runWhenDOMLoaded(addCopyButtonToCodeCells)
\ No newline at end of file
diff --git a/docs/_build/html/_static/copybutton_funcs.js b/docs/_build/html/_static/copybutton_funcs.js
deleted file mode 100644
index dbe1aaa..0000000
--- a/docs/_build/html/_static/copybutton_funcs.js
+++ /dev/null
@@ -1,73 +0,0 @@
-function escapeRegExp(string) {
-    return string.replace(/[.*+?^${}()|[\]\\]/g, '\\$&'); // $& means the whole matched string
-}
-
-/**
- * Removes excluded text from a Node.
- *
- * @param {Node} target Node to filter.
- * @param {string} exclude CSS selector of nodes to exclude.
- * @returns {DOMString} Text from `target` with text removed.
- */
-export function filterText(target, exclude) {
-    const clone = target.cloneNode(true);  // clone as to not modify the live DOM
-    if (exclude) {
-        // remove excluded nodes
-        clone.querySelectorAll(exclude).forEach(node => node.remove());
-    }
-    return clone.innerText;
-}
-
-// Callback when a copy button is clicked. Will be passed the node that was clicked
-// should then grab the text and replace pieces of text that shouldn't be used in output
-export function formatCopyText(textContent, copybuttonPromptText, isRegexp = false, onlyCopyPromptLines = true, removePrompts = true, copyEmptyLines = true, lineContinuationChar = "", hereDocDelim = "") {
-    var regexp;
-    var match;
-
-    // Do we check for line continuation characters and "HERE-documents"?
-    var useLineCont = !!lineContinuationChar
-    var useHereDoc = !!hereDocDelim
-
-    // create regexp to capture prompt and remaining line
-    if (isRegexp) {
-        regexp = new RegExp('^(' + copybuttonPromptText + ')(.*)')
-    } else {
-        regexp = new RegExp('^(' + escapeRegExp(copybuttonPromptText) + ')(.*)')
-    }
-
-    const outputLines = [];
-    var promptFound = false;
-    var gotLineCont = false;
-    var gotHereDoc = false;
-    const lineGotPrompt = [];
-    for (const line of textContent.split('\n')) {
-        match = line.match(regexp)
-        if (match || gotLineCont || gotHereDoc) {
-            promptFound = regexp.test(line)
-            lineGotPrompt.push(promptFound)
-            if (removePrompts && promptFound) {
-                outputLines.push(match[2])
-            } else {
-                outputLines.push(line)
-            }
-            gotLineCont = line.endsWith(lineContinuationChar) & useLineCont
-            if (line.includes(hereDocDelim) & useHereDoc)
-                gotHereDoc = !gotHereDoc
-        } else if (!onlyCopyPromptLines) {
-            outputLines.push(line)
-        } else if (copyEmptyLines && line.trim() === '') {
-            outputLines.push(line)
-        }
-    }
-
-    // If no lines with the prompt were found then just use original lines
-    if (lineGotPrompt.some(v => v === true)) {
-        textContent = outputLines.join('\n');
-    }
-
-    // Remove a trailing newline to avoid auto-running when pasting
-    if (textContent.endsWith("\n")) {
-        textContent = textContent.slice(0, -1)
-    }
-    return textContent
-}
diff --git a/docs/_build/html/_static/design-style.4045f2051d55cab465a707391d5b2007.min.css b/docs/_build/html/_static/design-style.4045f2051d55cab465a707391d5b2007.min.css
deleted file mode 100644
index 3225661..0000000
--- a/docs/_build/html/_static/design-style.4045f2051d55cab465a707391d5b2007.min.css
+++ /dev/null
@@ -1 +0,0 @@
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diff --git a/docs/_build/html/_static/design-tabs.js b/docs/_build/html/_static/design-tabs.js
deleted file mode 100644
index 36b38cf..0000000
--- a/docs/_build/html/_static/design-tabs.js
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diff --git a/docs/_build/html/_static/doctools.js b/docs/_build/html/_static/doctools.js
deleted file mode 100644
index c3db08d..0000000
--- a/docs/_build/html/_static/doctools.js
+++ /dev/null
@@ -1,264 +0,0 @@
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-
-  /**
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-  highlightSearchWords: () => {
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-            '<a href="javascript:Documentation.hideSearchWords()">' +
-            Documentation.gettext("Hide Search Matches") +
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-        )
-    );
-  },
-
-  /**
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-  },
-
-  /**
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-      return;
-
-    const blacklistedElements = new Set([
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-      "INPUT",
-      "SELECT",
-      "BUTTON",
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-      if (event.altKey || event.ctrlKey || event.metaKey) return; // bail with special keys
-
-      if (!event.shiftKey) {
-        switch (event.key) {
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-
-            const prevLink = document.querySelector('link[rel="prev"]');
-            if (prevLink && prevLink.href) {
-              window.location.href = prevLink.href;
-              event.preventDefault();
-            }
-            break;
-          case "ArrowRight":
-            if (!DOCUMENTATION_OPTIONS.NAVIGATION_WITH_KEYS) break;
-
-            const nextLink = document.querySelector('link[rel="next"]');
-            if (nextLink && nextLink.href) {
-              window.location.href = nextLink.href;
-              event.preventDefault();
-            }
-            break;
-          case "Escape":
-            if (!DOCUMENTATION_OPTIONS.ENABLE_SEARCH_SHORTCUTS) break;
-            Documentation.hideSearchWords();
-            event.preventDefault();
-        }
-      }
-
-      // some keyboard layouts may need Shift to get /
-      switch (event.key) {
-        case "/":
-          if (!DOCUMENTATION_OPTIONS.ENABLE_SEARCH_SHORTCUTS) break;
-          Documentation.focusSearchBar();
-          event.preventDefault();
-      }
-    });
-  },
-};
-
-// quick alias for translations
-const _ = Documentation.gettext;
-
-_ready(Documentation.init);
diff --git a/docs/_build/html/_static/documentation_options.js b/docs/_build/html/_static/documentation_options.js
deleted file mode 100644
index 162a6ba..0000000
--- a/docs/_build/html/_static/documentation_options.js
+++ /dev/null
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diff --git a/docs/_build/html/_static/images/logo_deepnote.svg b/docs/_build/html/_static/images/logo_deepnote.svg
deleted file mode 100644
index fa77ebf..0000000
--- a/docs/_build/html/_static/images/logo_deepnote.svg
+++ /dev/null
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-<svg viewBox="0 0 128 128" fill="none" xmlns="http://www.w3.org/2000/svg"><path fill-rule="evenodd" clip-rule="evenodd" d="M0 128h52.512l29.539-11.077-11.077-43.487-34.051 3.693L0 128Z" fill="#0076D4"/><path fill-rule="evenodd" clip-rule="evenodd" d="M52.513 128s16.6-8.759 19.673-24.277c3.072-15.517-12.091-26.594-35.263-26.594 0-.41 20.343-28.718 20.343-28.718l49.4 1.435L95.71 107.7l-20.452 15.978L52.513 128Z" fill="#002868"/><path fill-rule="evenodd" clip-rule="evenodd" d="M0 60.718 41.025.001s1.006.01 3.282 0c16.082-.068 81.23 3.12 81.23 60.368 0 65.352-73.025 67.631-73.025 67.631s30.495-5.839 30.495-34.816c0-28.978-27.541-32.466-45.264-32.466H0Z" fill="#00A9FF"/></svg>
diff --git a/docs/_build/html/_static/images/logo_jupyterhub.svg b/docs/_build/html/_static/images/logo_jupyterhub.svg
deleted file mode 100644
index 60cfe9f..0000000
--- a/docs/_build/html/_static/images/logo_jupyterhub.svg
+++ /dev/null
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diff --git a/docs/_build/html/_static/jquery-3.6.0.js b/docs/_build/html/_static/jquery-3.6.0.js
deleted file mode 100644
index fc6c299..0000000
--- a/docs/_build/html/_static/jquery-3.6.0.js
+++ /dev/null
@@ -1,10881 +0,0 @@
-/*!
- * jQuery JavaScript Library v3.6.0
- * https://jquery.com/
- *
- * Includes Sizzle.js
- * https://sizzlejs.com/
- *
- * Copyright OpenJS Foundation and other contributors
- * Released under the MIT license
- * https://jquery.org/license
- *
- * Date: 2021-03-02T17:08Z
- */
-( function( global, factory ) {
-
-	"use strict";
-
-	if ( typeof module === "object" && typeof module.exports === "object" ) {
-
-		// For CommonJS and CommonJS-like environments where a proper `window`
-		// is present, execute the factory and get jQuery.
-		// For environments that do not have a `window` with a `document`
-		// (such as Node.js), expose a factory as module.exports.
-		// This accentuates the need for the creation of a real `window`.
-		// e.g. var jQuery = require("jquery")(window);
-		// See ticket #14549 for more info.
-		module.exports = global.document ?
-			factory( global, true ) :
-			function( w ) {
-				if ( !w.document ) {
-					throw new Error( "jQuery requires a window with a document" );
-				}
-				return factory( w );
-			};
-	} else {
-		factory( global );
-	}
-
-// Pass this if window is not defined yet
-} )( typeof window !== "undefined" ? window : this, function( window, noGlobal ) {
-
-// Edge <= 12 - 13+, Firefox <=18 - 45+, IE 10 - 11, Safari 5.1 - 9+, iOS 6 - 9.1
-// throw exceptions when non-strict code (e.g., ASP.NET 4.5) accesses strict mode
-// arguments.callee.caller (trac-13335). But as of jQuery 3.0 (2016), strict mode should be common
-// enough that all such attempts are guarded in a try block.
-"use strict";
-
-var arr = [];
-
-var getProto = Object.getPrototypeOf;
-
-var slice = arr.slice;
-
-var flat = arr.flat ? function( array ) {
-	return arr.flat.call( array );
-} : function( array ) {
-	return arr.concat.apply( [], array );
-};
-
-
-var push = arr.push;
-
-var indexOf = arr.indexOf;
-
-var class2type = {};
-
-var toString = class2type.toString;
-
-var hasOwn = class2type.hasOwnProperty;
-
-var fnToString = hasOwn.toString;
-
-var ObjectFunctionString = fnToString.call( Object );
-
-var support = {};
-
-var isFunction = function isFunction( obj ) {
-
-		// Support: Chrome <=57, Firefox <=52
-		// In some browsers, typeof returns "function" for HTML <object> elements
-		// (i.e., `typeof document.createElement( "object" ) === "function"`).
-		// We don't want to classify *any* DOM node as a function.
-		// Support: QtWeb <=3.8.5, WebKit <=534.34, wkhtmltopdf tool <=0.12.5
-		// Plus for old WebKit, typeof returns "function" for HTML collections
-		// (e.g., `typeof document.getElementsByTagName("div") === "function"`). (gh-4756)
-		return typeof obj === "function" && typeof obj.nodeType !== "number" &&
-			typeof obj.item !== "function";
-	};
-
-
-var isWindow = function isWindow( obj ) {
-		return obj != null && obj === obj.window;
-	};
-
-
-var document = window.document;
-
-
-
-	var preservedScriptAttributes = {
-		type: true,
-		src: true,
-		nonce: true,
-		noModule: true
-	};
-
-	function DOMEval( code, node, doc ) {
-		doc = doc || document;
-
-		var i, val,
-			script = doc.createElement( "script" );
-
-		script.text = code;
-		if ( node ) {
-			for ( i in preservedScriptAttributes ) {
-
-				// Support: Firefox 64+, Edge 18+
-				// Some browsers don't support the "nonce" property on scripts.
-				// On the other hand, just using `getAttribute` is not enough as
-				// the `nonce` attribute is reset to an empty string whenever it
-				// becomes browsing-context connected.
-				// See https://github.com/whatwg/html/issues/2369
-				// See https://html.spec.whatwg.org/#nonce-attributes
-				// The `node.getAttribute` check was added for the sake of
-				// `jQuery.globalEval` so that it can fake a nonce-containing node
-				// via an object.
-				val = node[ i ] || node.getAttribute && node.getAttribute( i );
-				if ( val ) {
-					script.setAttribute( i, val );
-				}
-			}
-		}
-		doc.head.appendChild( script ).parentNode.removeChild( script );
-	}
-
-
-function toType( obj ) {
-	if ( obj == null ) {
-		return obj + "";
-	}
-
-	// Support: Android <=2.3 only (functionish RegExp)
-	return typeof obj === "object" || typeof obj === "function" ?
-		class2type[ toString.call( obj ) ] || "object" :
-		typeof obj;
-}
-/* global Symbol */
-// Defining this global in .eslintrc.json would create a danger of using the global
-// unguarded in another place, it seems safer to define global only for this module
-
-
-
-var
-	version = "3.6.0",
-
-	// Define a local copy of jQuery
-	jQuery = function( selector, context ) {
-
-		// The jQuery object is actually just the init constructor 'enhanced'
-		// Need init if jQuery is called (just allow error to be thrown if not included)
-		return new jQuery.fn.init( selector, context );
-	};
-
-jQuery.fn = jQuery.prototype = {
-
-	// The current version of jQuery being used
-	jquery: version,
-
-	constructor: jQuery,
-
-	// The default length of a jQuery object is 0
-	length: 0,
-
-	toArray: function() {
-		return slice.call( this );
-	},
-
-	// Get the Nth element in the matched element set OR
-	// Get the whole matched element set as a clean array
-	get: function( num ) {
-
-		// Return all the elements in a clean array
-		if ( num == null ) {
-			return slice.call( this );
-		}
-
-		// Return just the one element from the set
-		return num < 0 ? this[ num + this.length ] : this[ num ];
-	},
-
-	// Take an array of elements and push it onto the stack
-	// (returning the new matched element set)
-	pushStack: function( elems ) {
-
-		// Build a new jQuery matched element set
-		var ret = jQuery.merge( this.constructor(), elems );
-
-		// Add the old object onto the stack (as a reference)
-		ret.prevObject = this;
-
-		// Return the newly-formed element set
-		return ret;
-	},
-
-	// Execute a callback for every element in the matched set.
-	each: function( callback ) {
-		return jQuery.each( this, callback );
-	},
-
-	map: function( callback ) {
-		return this.pushStack( jQuery.map( this, function( elem, i ) {
-			return callback.call( elem, i, elem );
-		} ) );
-	},
-
-	slice: function() {
-		return this.pushStack( slice.apply( this, arguments ) );
-	},
-
-	first: function() {
-		return this.eq( 0 );
-	},
-
-	last: function() {
-		return this.eq( -1 );
-	},
-
-	even: function() {
-		return this.pushStack( jQuery.grep( this, function( _elem, i ) {
-			return ( i + 1 ) % 2;
-		} ) );
-	},
-
-	odd: function() {
-		return this.pushStack( jQuery.grep( this, function( _elem, i ) {
-			return i % 2;
-		} ) );
-	},
-
-	eq: function( i ) {
-		var len = this.length,
-			j = +i + ( i < 0 ? len : 0 );
-		return this.pushStack( j >= 0 && j < len ? [ this[ j ] ] : [] );
-	},
-
-	end: function() {
-		return this.prevObject || this.constructor();
-	},
-
-	// For internal use only.
-	// Behaves like an Array's method, not like a jQuery method.
-	push: push,
-	sort: arr.sort,
-	splice: arr.splice
-};
-
-jQuery.extend = jQuery.fn.extend = function() {
-	var options, name, src, copy, copyIsArray, clone,
-		target = arguments[ 0 ] || {},
-		i = 1,
-		length = arguments.length,
-		deep = false;
-
-	// Handle a deep copy situation
-	if ( typeof target === "boolean" ) {
-		deep = target;
-
-		// Skip the boolean and the target
-		target = arguments[ i ] || {};
-		i++;
-	}
-
-	// Handle case when target is a string or something (possible in deep copy)
-	if ( typeof target !== "object" && !isFunction( target ) ) {
-		target = {};
-	}
-
-	// Extend jQuery itself if only one argument is passed
-	if ( i === length ) {
-		target = this;
-		i--;
-	}
-
-	for ( ; i < length; i++ ) {
-
-		// Only deal with non-null/undefined values
-		if ( ( options = arguments[ i ] ) != null ) {
-
-			// Extend the base object
-			for ( name in options ) {
-				copy = options[ name ];
-
-				// Prevent Object.prototype pollution
-				// Prevent never-ending loop
-				if ( name === "__proto__" || target === copy ) {
-					continue;
-				}
-
-				// Recurse if we're merging plain objects or arrays
-				if ( deep && copy && ( jQuery.isPlainObject( copy ) ||
-					( copyIsArray = Array.isArray( copy ) ) ) ) {
-					src = target[ name ];
-
-					// Ensure proper type for the source value
-					if ( copyIsArray && !Array.isArray( src ) ) {
-						clone = [];
-					} else if ( !copyIsArray && !jQuery.isPlainObject( src ) ) {
-						clone = {};
-					} else {
-						clone = src;
-					}
-					copyIsArray = false;
-
-					// Never move original objects, clone them
-					target[ name ] = jQuery.extend( deep, clone, copy );
-
-				// Don't bring in undefined values
-				} else if ( copy !== undefined ) {
-					target[ name ] = copy;
-				}
-			}
-		}
-	}
-
-	// Return the modified object
-	return target;
-};
-
-jQuery.extend( {
-
-	// Unique for each copy of jQuery on the page
-	expando: "jQuery" + ( version + Math.random() ).replace( /\D/g, "" ),
-
-	// Assume jQuery is ready without the ready module
-	isReady: true,
-
-	error: function( msg ) {
-		throw new Error( msg );
-	},
-
-	noop: function() {},
-
-	isPlainObject: function( obj ) {
-		var proto, Ctor;
-
-		// Detect obvious negatives
-		// Use toString instead of jQuery.type to catch host objects
-		if ( !obj || toString.call( obj ) !== "[object Object]" ) {
-			return false;
-		}
-
-		proto = getProto( obj );
-
-		// Objects with no prototype (e.g., `Object.create( null )`) are plain
-		if ( !proto ) {
-			return true;
-		}
-
-		// Objects with prototype are plain iff they were constructed by a global Object function
-		Ctor = hasOwn.call( proto, "constructor" ) && proto.constructor;
-		return typeof Ctor === "function" && fnToString.call( Ctor ) === ObjectFunctionString;
-	},
-
-	isEmptyObject: function( obj ) {
-		var name;
-
-		for ( name in obj ) {
-			return false;
-		}
-		return true;
-	},
-
-	// Evaluates a script in a provided context; falls back to the global one
-	// if not specified.
-	globalEval: function( code, options, doc ) {
-		DOMEval( code, { nonce: options && options.nonce }, doc );
-	},
-
-	each: function( obj, callback ) {
-		var length, i = 0;
-
-		if ( isArrayLike( obj ) ) {
-			length = obj.length;
-			for ( ; i < length; i++ ) {
-				if ( callback.call( obj[ i ], i, obj[ i ] ) === false ) {
-					break;
-				}
-			}
-		} else {
-			for ( i in obj ) {
-				if ( callback.call( obj[ i ], i, obj[ i ] ) === false ) {
-					break;
-				}
-			}
-		}
-
-		return obj;
-	},
-
-	// results is for internal usage only
-	makeArray: function( arr, results ) {
-		var ret = results || [];
-
-		if ( arr != null ) {
-			if ( isArrayLike( Object( arr ) ) ) {
-				jQuery.merge( ret,
-					typeof arr === "string" ?
-						[ arr ] : arr
-				);
-			} else {
-				push.call( ret, arr );
-			}
-		}
-
-		return ret;
-	},
-
-	inArray: function( elem, arr, i ) {
-		return arr == null ? -1 : indexOf.call( arr, elem, i );
-	},
-
-	// Support: Android <=4.0 only, PhantomJS 1 only
-	// push.apply(_, arraylike) throws on ancient WebKit
-	merge: function( first, second ) {
-		var len = +second.length,
-			j = 0,
-			i = first.length;
-
-		for ( ; j < len; j++ ) {
-			first[ i++ ] = second[ j ];
-		}
-
-		first.length = i;
-
-		return first;
-	},
-
-	grep: function( elems, callback, invert ) {
-		var callbackInverse,
-			matches = [],
-			i = 0,
-			length = elems.length,
-			callbackExpect = !invert;
-
-		// Go through the array, only saving the items
-		// that pass the validator function
-		for ( ; i < length; i++ ) {
-			callbackInverse = !callback( elems[ i ], i );
-			if ( callbackInverse !== callbackExpect ) {
-				matches.push( elems[ i ] );
-			}
-		}
-
-		return matches;
-	},
-
-	// arg is for internal usage only
-	map: function( elems, callback, arg ) {
-		var length, value,
-			i = 0,
-			ret = [];
-
-		// Go through the array, translating each of the items to their new values
-		if ( isArrayLike( elems ) ) {
-			length = elems.length;
-			for ( ; i < length; i++ ) {
-				value = callback( elems[ i ], i, arg );
-
-				if ( value != null ) {
-					ret.push( value );
-				}
-			}
-
-		// Go through every key on the object,
-		} else {
-			for ( i in elems ) {
-				value = callback( elems[ i ], i, arg );
-
-				if ( value != null ) {
-					ret.push( value );
-				}
-			}
-		}
-
-		// Flatten any nested arrays
-		return flat( ret );
-	},
-
-	// A global GUID counter for objects
-	guid: 1,
-
-	// jQuery.support is not used in Core but other projects attach their
-	// properties to it so it needs to exist.
-	support: support
-} );
-
-if ( typeof Symbol === "function" ) {
-	jQuery.fn[ Symbol.iterator ] = arr[ Symbol.iterator ];
-}
-
-// Populate the class2type map
-jQuery.each( "Boolean Number String Function Array Date RegExp Object Error Symbol".split( " " ),
-	function( _i, name ) {
-		class2type[ "[object " + name + "]" ] = name.toLowerCase();
-	} );
-
-function isArrayLike( obj ) {
-
-	// Support: real iOS 8.2 only (not reproducible in simulator)
-	// `in` check used to prevent JIT error (gh-2145)
-	// hasOwn isn't used here due to false negatives
-	// regarding Nodelist length in IE
-	var length = !!obj && "length" in obj && obj.length,
-		type = toType( obj );
-
-	if ( isFunction( obj ) || isWindow( obj ) ) {
-		return false;
-	}
-
-	return type === "array" || length === 0 ||
-		typeof length === "number" && length > 0 && ( length - 1 ) in obj;
-}
-var Sizzle =
-/*!
- * Sizzle CSS Selector Engine v2.3.6
- * https://sizzlejs.com/
- *
- * Copyright JS Foundation and other contributors
- * Released under the MIT license
- * https://js.foundation/
- *
- * Date: 2021-02-16
- */
-( function( window ) {
-var i,
-	support,
-	Expr,
-	getText,
-	isXML,
-	tokenize,
-	compile,
-	select,
-	outermostContext,
-	sortInput,
-	hasDuplicate,
-
-	// Local document vars
-	setDocument,
-	document,
-	docElem,
-	documentIsHTML,
-	rbuggyQSA,
-	rbuggyMatches,
-	matches,
-	contains,
-
-	// Instance-specific data
-	expando = "sizzle" + 1 * new Date(),
-	preferredDoc = window.document,
-	dirruns = 0,
-	done = 0,
-	classCache = createCache(),
-	tokenCache = createCache(),
-	compilerCache = createCache(),
-	nonnativeSelectorCache = createCache(),
-	sortOrder = function( a, b ) {
-		if ( a === b ) {
-			hasDuplicate = true;
-		}
-		return 0;
-	},
-
-	// Instance methods
-	hasOwn = ( {} ).hasOwnProperty,
-	arr = [],
-	pop = arr.pop,
-	pushNative = arr.push,
-	push = arr.push,
-	slice = arr.slice,
-
-	// Use a stripped-down indexOf as it's faster than native
-	// https://jsperf.com/thor-indexof-vs-for/5
-	indexOf = function( list, elem ) {
-		var i = 0,
-			len = list.length;
-		for ( ; i < len; i++ ) {
-			if ( list[ i ] === elem ) {
-				return i;
-			}
-		}
-		return -1;
-	},
-
-	booleans = "checked|selected|async|autofocus|autoplay|controls|defer|disabled|hidden|" +
-		"ismap|loop|multiple|open|readonly|required|scoped",
-
-	// Regular expressions
-
-	// http://www.w3.org/TR/css3-selectors/#whitespace
-	whitespace = "[\\x20\\t\\r\\n\\f]",
-
-	// https://www.w3.org/TR/css-syntax-3/#ident-token-diagram
-	identifier = "(?:\\\\[\\da-fA-F]{1,6}" + whitespace +
-		"?|\\\\[^\\r\\n\\f]|[\\w-]|[^\0-\\x7f])+",
-
-	// Attribute selectors: http://www.w3.org/TR/selectors/#attribute-selectors
-	attributes = "\\[" + whitespace + "*(" + identifier + ")(?:" + whitespace +
-
-		// Operator (capture 2)
-		"*([*^$|!~]?=)" + whitespace +
-
-		// "Attribute values must be CSS identifiers [capture 5]
-		// or strings [capture 3 or capture 4]"
-		"*(?:'((?:\\\\.|[^\\\\'])*)'|\"((?:\\\\.|[^\\\\\"])*)\"|(" + identifier + "))|)" +
-		whitespace + "*\\]",
-
-	pseudos = ":(" + identifier + ")(?:\\((" +
-
-		// To reduce the number of selectors needing tokenize in the preFilter, prefer arguments:
-		// 1. quoted (capture 3; capture 4 or capture 5)
-		"('((?:\\\\.|[^\\\\'])*)'|\"((?:\\\\.|[^\\\\\"])*)\")|" +
-
-		// 2. simple (capture 6)
-		"((?:\\\\.|[^\\\\()[\\]]|" + attributes + ")*)|" +
-
-		// 3. anything else (capture 2)
-		".*" +
-		")\\)|)",
-
-	// Leading and non-escaped trailing whitespace, capturing some non-whitespace characters preceding the latter
-	rwhitespace = new RegExp( whitespace + "+", "g" ),
-	rtrim = new RegExp( "^" + whitespace + "+|((?:^|[^\\\\])(?:\\\\.)*)" +
-		whitespace + "+$", "g" ),
-
-	rcomma = new RegExp( "^" + whitespace + "*," + whitespace + "*" ),
-	rcombinators = new RegExp( "^" + whitespace + "*([>+~]|" + whitespace + ")" + whitespace +
-		"*" ),
-	rdescend = new RegExp( whitespace + "|>" ),
-
-	rpseudo = new RegExp( pseudos ),
-	ridentifier = new RegExp( "^" + identifier + "$" ),
-
-	matchExpr = {
-		"ID": new RegExp( "^#(" + identifier + ")" ),
-		"CLASS": new RegExp( "^\\.(" + identifier + ")" ),
-		"TAG": new RegExp( "^(" + identifier + "|[*])" ),
-		"ATTR": new RegExp( "^" + attributes ),
-		"PSEUDO": new RegExp( "^" + pseudos ),
-		"CHILD": new RegExp( "^:(only|first|last|nth|nth-last)-(child|of-type)(?:\\(" +
-			whitespace + "*(even|odd|(([+-]|)(\\d*)n|)" + whitespace + "*(?:([+-]|)" +
-			whitespace + "*(\\d+)|))" + whitespace + "*\\)|)", "i" ),
-		"bool": new RegExp( "^(?:" + booleans + ")$", "i" ),
-
-		// For use in libraries implementing .is()
-		// We use this for POS matching in `select`
-		"needsContext": new RegExp( "^" + whitespace +
-			"*[>+~]|:(even|odd|eq|gt|lt|nth|first|last)(?:\\(" + whitespace +
-			"*((?:-\\d)?\\d*)" + whitespace + "*\\)|)(?=[^-]|$)", "i" )
-	},
-
-	rhtml = /HTML$/i,
-	rinputs = /^(?:input|select|textarea|button)$/i,
-	rheader = /^h\d$/i,
-
-	rnative = /^[^{]+\{\s*\[native \w/,
-
-	// Easily-parseable/retrievable ID or TAG or CLASS selectors
-	rquickExpr = /^(?:#([\w-]+)|(\w+)|\.([\w-]+))$/,
-
-	rsibling = /[+~]/,
-
-	// CSS escapes
-	// http://www.w3.org/TR/CSS21/syndata.html#escaped-characters
-	runescape = new RegExp( "\\\\[\\da-fA-F]{1,6}" + whitespace + "?|\\\\([^\\r\\n\\f])", "g" ),
-	funescape = function( escape, nonHex ) {
-		var high = "0x" + escape.slice( 1 ) - 0x10000;
-
-		return nonHex ?
-
-			// Strip the backslash prefix from a non-hex escape sequence
-			nonHex :
-
-			// Replace a hexadecimal escape sequence with the encoded Unicode code point
-			// Support: IE <=11+
-			// For values outside the Basic Multilingual Plane (BMP), manually construct a
-			// surrogate pair
-			high < 0 ?
-				String.fromCharCode( high + 0x10000 ) :
-				String.fromCharCode( high >> 10 | 0xD800, high & 0x3FF | 0xDC00 );
-	},
-
-	// CSS string/identifier serialization
-	// https://drafts.csswg.org/cssom/#common-serializing-idioms
-	rcssescape = /([\0-\x1f\x7f]|^-?\d)|^-$|[^\0-\x1f\x7f-\uFFFF\w-]/g,
-	fcssescape = function( ch, asCodePoint ) {
-		if ( asCodePoint ) {
-
-			// U+0000 NULL becomes U+FFFD REPLACEMENT CHARACTER
-			if ( ch === "\0" ) {
-				return "\uFFFD";
-			}
-
-			// Control characters and (dependent upon position) numbers get escaped as code points
-			return ch.slice( 0, -1 ) + "\\" +
-				ch.charCodeAt( ch.length - 1 ).toString( 16 ) + " ";
-		}
-
-		// Other potentially-special ASCII characters get backslash-escaped
-		return "\\" + ch;
-	},
-
-	// Used for iframes
-	// See setDocument()
-	// Removing the function wrapper causes a "Permission Denied"
-	// error in IE
-	unloadHandler = function() {
-		setDocument();
-	},
-
-	inDisabledFieldset = addCombinator(
-		function( elem ) {
-			return elem.disabled === true && elem.nodeName.toLowerCase() === "fieldset";
-		},
-		{ dir: "parentNode", next: "legend" }
-	);
-
-// Optimize for push.apply( _, NodeList )
-try {
-	push.apply(
-		( arr = slice.call( preferredDoc.childNodes ) ),
-		preferredDoc.childNodes
-	);
-
-	// Support: Android<4.0
-	// Detect silently failing push.apply
-	// eslint-disable-next-line no-unused-expressions
-	arr[ preferredDoc.childNodes.length ].nodeType;
-} catch ( e ) {
-	push = { apply: arr.length ?
-
-		// Leverage slice if possible
-		function( target, els ) {
-			pushNative.apply( target, slice.call( els ) );
-		} :
-
-		// Support: IE<9
-		// Otherwise append directly
-		function( target, els ) {
-			var j = target.length,
-				i = 0;
-
-			// Can't trust NodeList.length
-			while ( ( target[ j++ ] = els[ i++ ] ) ) {}
-			target.length = j - 1;
-		}
-	};
-}
-
-function Sizzle( selector, context, results, seed ) {
-	var m, i, elem, nid, match, groups, newSelector,
-		newContext = context && context.ownerDocument,
-
-		// nodeType defaults to 9, since context defaults to document
-		nodeType = context ? context.nodeType : 9;
-
-	results = results || [];
-
-	// Return early from calls with invalid selector or context
-	if ( typeof selector !== "string" || !selector ||
-		nodeType !== 1 && nodeType !== 9 && nodeType !== 11 ) {
-
-		return results;
-	}
-
-	// Try to shortcut find operations (as opposed to filters) in HTML documents
-	if ( !seed ) {
-		setDocument( context );
-		context = context || document;
-
-		if ( documentIsHTML ) {
-
-			// If the selector is sufficiently simple, try using a "get*By*" DOM method
-			// (excepting DocumentFragment context, where the methods don't exist)
-			if ( nodeType !== 11 && ( match = rquickExpr.exec( selector ) ) ) {
-
-				// ID selector
-				if ( ( m = match[ 1 ] ) ) {
-
-					// Document context
-					if ( nodeType === 9 ) {
-						if ( ( elem = context.getElementById( m ) ) ) {
-
-							// Support: IE, Opera, Webkit
-							// TODO: identify versions
-							// getElementById can match elements by name instead of ID
-							if ( elem.id === m ) {
-								results.push( elem );
-								return results;
-							}
-						} else {
-							return results;
-						}
-
-					// Element context
-					} else {
-
-						// Support: IE, Opera, Webkit
-						// TODO: identify versions
-						// getElementById can match elements by name instead of ID
-						if ( newContext && ( elem = newContext.getElementById( m ) ) &&
-							contains( context, elem ) &&
-							elem.id === m ) {
-
-							results.push( elem );
-							return results;
-						}
-					}
-
-				// Type selector
-				} else if ( match[ 2 ] ) {
-					push.apply( results, context.getElementsByTagName( selector ) );
-					return results;
-
-				// Class selector
-				} else if ( ( m = match[ 3 ] ) && support.getElementsByClassName &&
-					context.getElementsByClassName ) {
-
-					push.apply( results, context.getElementsByClassName( m ) );
-					return results;
-				}
-			}
-
-			// Take advantage of querySelectorAll
-			if ( support.qsa &&
-				!nonnativeSelectorCache[ selector + " " ] &&
-				( !rbuggyQSA || !rbuggyQSA.test( selector ) ) &&
-
-				// Support: IE 8 only
-				// Exclude object elements
-				( nodeType !== 1 || context.nodeName.toLowerCase() !== "object" ) ) {
-
-				newSelector = selector;
-				newContext = context;
-
-				// qSA considers elements outside a scoping root when evaluating child or
-				// descendant combinators, which is not what we want.
-				// In such cases, we work around the behavior by prefixing every selector in the
-				// list with an ID selector referencing the scope context.
-				// The technique has to be used as well when a leading combinator is used
-				// as such selectors are not recognized by querySelectorAll.
-				// Thanks to Andrew Dupont for this technique.
-				if ( nodeType === 1 &&
-					( rdescend.test( selector ) || rcombinators.test( selector ) ) ) {
-
-					// Expand context for sibling selectors
-					newContext = rsibling.test( selector ) && testContext( context.parentNode ) ||
-						context;
-
-					// We can use :scope instead of the ID hack if the browser
-					// supports it & if we're not changing the context.
-					if ( newContext !== context || !support.scope ) {
-
-						// Capture the context ID, setting it first if necessary
-						if ( ( nid = context.getAttribute( "id" ) ) ) {
-							nid = nid.replace( rcssescape, fcssescape );
-						} else {
-							context.setAttribute( "id", ( nid = expando ) );
-						}
-					}
-
-					// Prefix every selector in the list
-					groups = tokenize( selector );
-					i = groups.length;
-					while ( i-- ) {
-						groups[ i ] = ( nid ? "#" + nid : ":scope" ) + " " +
-							toSelector( groups[ i ] );
-					}
-					newSelector = groups.join( "," );
-				}
-
-				try {
-					push.apply( results,
-						newContext.querySelectorAll( newSelector )
-					);
-					return results;
-				} catch ( qsaError ) {
-					nonnativeSelectorCache( selector, true );
-				} finally {
-					if ( nid === expando ) {
-						context.removeAttribute( "id" );
-					}
-				}
-			}
-		}
-	}
-
-	// All others
-	return select( selector.replace( rtrim, "$1" ), context, results, seed );
-}
-
-/**
- * Create key-value caches of limited size
- * @returns {function(string, object)} Returns the Object data after storing it on itself with
- *	property name the (space-suffixed) string and (if the cache is larger than Expr.cacheLength)
- *	deleting the oldest entry
- */
-function createCache() {
-	var keys = [];
-
-	function cache( key, value ) {
-
-		// Use (key + " ") to avoid collision with native prototype properties (see Issue #157)
-		if ( keys.push( key + " " ) > Expr.cacheLength ) {
-
-			// Only keep the most recent entries
-			delete cache[ keys.shift() ];
-		}
-		return ( cache[ key + " " ] = value );
-	}
-	return cache;
-}
-
-/**
- * Mark a function for special use by Sizzle
- * @param {Function} fn The function to mark
- */
-function markFunction( fn ) {
-	fn[ expando ] = true;
-	return fn;
-}
-
-/**
- * Support testing using an element
- * @param {Function} fn Passed the created element and returns a boolean result
- */
-function assert( fn ) {
-	var el = document.createElement( "fieldset" );
-
-	try {
-		return !!fn( el );
-	} catch ( e ) {
-		return false;
-	} finally {
-
-		// Remove from its parent by default
-		if ( el.parentNode ) {
-			el.parentNode.removeChild( el );
-		}
-
-		// release memory in IE
-		el = null;
-	}
-}
-
-/**
- * Adds the same handler for all of the specified attrs
- * @param {String} attrs Pipe-separated list of attributes
- * @param {Function} handler The method that will be applied
- */
-function addHandle( attrs, handler ) {
-	var arr = attrs.split( "|" ),
-		i = arr.length;
-
-	while ( i-- ) {
-		Expr.attrHandle[ arr[ i ] ] = handler;
-	}
-}
-
-/**
- * Checks document order of two siblings
- * @param {Element} a
- * @param {Element} b
- * @returns {Number} Returns less than 0 if a precedes b, greater than 0 if a follows b
- */
-function siblingCheck( a, b ) {
-	var cur = b && a,
-		diff = cur && a.nodeType === 1 && b.nodeType === 1 &&
-			a.sourceIndex - b.sourceIndex;
-
-	// Use IE sourceIndex if available on both nodes
-	if ( diff ) {
-		return diff;
-	}
-
-	// Check if b follows a
-	if ( cur ) {
-		while ( ( cur = cur.nextSibling ) ) {
-			if ( cur === b ) {
-				return -1;
-			}
-		}
-	}
-
-	return a ? 1 : -1;
-}
-
-/**
- * Returns a function to use in pseudos for input types
- * @param {String} type
- */
-function createInputPseudo( type ) {
-	return function( elem ) {
-		var name = elem.nodeName.toLowerCase();
-		return name === "input" && elem.type === type;
-	};
-}
-
-/**
- * Returns a function to use in pseudos for buttons
- * @param {String} type
- */
-function createButtonPseudo( type ) {
-	return function( elem ) {
-		var name = elem.nodeName.toLowerCase();
-		return ( name === "input" || name === "button" ) && elem.type === type;
-	};
-}
-
-/**
- * Returns a function to use in pseudos for :enabled/:disabled
- * @param {Boolean} disabled true for :disabled; false for :enabled
- */
-function createDisabledPseudo( disabled ) {
-
-	// Known :disabled false positives: fieldset[disabled] > legend:nth-of-type(n+2) :can-disable
-	return function( elem ) {
-
-		// Only certain elements can match :enabled or :disabled
-		// https://html.spec.whatwg.org/multipage/scripting.html#selector-enabled
-		// https://html.spec.whatwg.org/multipage/scripting.html#selector-disabled
-		if ( "form" in elem ) {
-
-			// Check for inherited disabledness on relevant non-disabled elements:
-			// * listed form-associated elements in a disabled fieldset
-			//   https://html.spec.whatwg.org/multipage/forms.html#category-listed
-			//   https://html.spec.whatwg.org/multipage/forms.html#concept-fe-disabled
-			// * option elements in a disabled optgroup
-			//   https://html.spec.whatwg.org/multipage/forms.html#concept-option-disabled
-			// All such elements have a "form" property.
-			if ( elem.parentNode && elem.disabled === false ) {
-
-				// Option elements defer to a parent optgroup if present
-				if ( "label" in elem ) {
-					if ( "label" in elem.parentNode ) {
-						return elem.parentNode.disabled === disabled;
-					} else {
-						return elem.disabled === disabled;
-					}
-				}
-
-				// Support: IE 6 - 11
-				// Use the isDisabled shortcut property to check for disabled fieldset ancestors
-				return elem.isDisabled === disabled ||
-
-					// Where there is no isDisabled, check manually
-					/* jshint -W018 */
-					elem.isDisabled !== !disabled &&
-					inDisabledFieldset( elem ) === disabled;
-			}
-
-			return elem.disabled === disabled;
-
-		// Try to winnow out elements that can't be disabled before trusting the disabled property.
-		// Some victims get caught in our net (label, legend, menu, track), but it shouldn't
-		// even exist on them, let alone have a boolean value.
-		} else if ( "label" in elem ) {
-			return elem.disabled === disabled;
-		}
-
-		// Remaining elements are neither :enabled nor :disabled
-		return false;
-	};
-}
-
-/**
- * Returns a function to use in pseudos for positionals
- * @param {Function} fn
- */
-function createPositionalPseudo( fn ) {
-	return markFunction( function( argument ) {
-		argument = +argument;
-		return markFunction( function( seed, matches ) {
-			var j,
-				matchIndexes = fn( [], seed.length, argument ),
-				i = matchIndexes.length;
-
-			// Match elements found at the specified indexes
-			while ( i-- ) {
-				if ( seed[ ( j = matchIndexes[ i ] ) ] ) {
-					seed[ j ] = !( matches[ j ] = seed[ j ] );
-				}
-			}
-		} );
-	} );
-}
-
-/**
- * Checks a node for validity as a Sizzle context
- * @param {Element|Object=} context
- * @returns {Element|Object|Boolean} The input node if acceptable, otherwise a falsy value
- */
-function testContext( context ) {
-	return context && typeof context.getElementsByTagName !== "undefined" && context;
-}
-
-// Expose support vars for convenience
-support = Sizzle.support = {};
-
-/**
- * Detects XML nodes
- * @param {Element|Object} elem An element or a document
- * @returns {Boolean} True iff elem is a non-HTML XML node
- */
-isXML = Sizzle.isXML = function( elem ) {
-	var namespace = elem && elem.namespaceURI,
-		docElem = elem && ( elem.ownerDocument || elem ).documentElement;
-
-	// Support: IE <=8
-	// Assume HTML when documentElement doesn't yet exist, such as inside loading iframes
-	// https://bugs.jquery.com/ticket/4833
-	return !rhtml.test( namespace || docElem && docElem.nodeName || "HTML" );
-};
-
-/**
- * Sets document-related variables once based on the current document
- * @param {Element|Object} [doc] An element or document object to use to set the document
- * @returns {Object} Returns the current document
- */
-setDocument = Sizzle.setDocument = function( node ) {
-	var hasCompare, subWindow,
-		doc = node ? node.ownerDocument || node : preferredDoc;
-
-	// Return early if doc is invalid or already selected
-	// Support: IE 11+, Edge 17 - 18+
-	// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-	// two documents; shallow comparisons work.
-	// eslint-disable-next-line eqeqeq
-	if ( doc == document || doc.nodeType !== 9 || !doc.documentElement ) {
-		return document;
-	}
-
-	// Update global variables
-	document = doc;
-	docElem = document.documentElement;
-	documentIsHTML = !isXML( document );
-
-	// Support: IE 9 - 11+, Edge 12 - 18+
-	// Accessing iframe documents after unload throws "permission denied" errors (jQuery #13936)
-	// Support: IE 11+, Edge 17 - 18+
-	// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-	// two documents; shallow comparisons work.
-	// eslint-disable-next-line eqeqeq
-	if ( preferredDoc != document &&
-		( subWindow = document.defaultView ) && subWindow.top !== subWindow ) {
-
-		// Support: IE 11, Edge
-		if ( subWindow.addEventListener ) {
-			subWindow.addEventListener( "unload", unloadHandler, false );
-
-		// Support: IE 9 - 10 only
-		} else if ( subWindow.attachEvent ) {
-			subWindow.attachEvent( "onunload", unloadHandler );
-		}
-	}
-
-	// Support: IE 8 - 11+, Edge 12 - 18+, Chrome <=16 - 25 only, Firefox <=3.6 - 31 only,
-	// Safari 4 - 5 only, Opera <=11.6 - 12.x only
-	// IE/Edge & older browsers don't support the :scope pseudo-class.
-	// Support: Safari 6.0 only
-	// Safari 6.0 supports :scope but it's an alias of :root there.
-	support.scope = assert( function( el ) {
-		docElem.appendChild( el ).appendChild( document.createElement( "div" ) );
-		return typeof el.querySelectorAll !== "undefined" &&
-			!el.querySelectorAll( ":scope fieldset div" ).length;
-	} );
-
-	/* Attributes
-	---------------------------------------------------------------------- */
-
-	// Support: IE<8
-	// Verify that getAttribute really returns attributes and not properties
-	// (excepting IE8 booleans)
-	support.attributes = assert( function( el ) {
-		el.className = "i";
-		return !el.getAttribute( "className" );
-	} );
-
-	/* getElement(s)By*
-	---------------------------------------------------------------------- */
-
-	// Check if getElementsByTagName("*") returns only elements
-	support.getElementsByTagName = assert( function( el ) {
-		el.appendChild( document.createComment( "" ) );
-		return !el.getElementsByTagName( "*" ).length;
-	} );
-
-	// Support: IE<9
-	support.getElementsByClassName = rnative.test( document.getElementsByClassName );
-
-	// Support: IE<10
-	// Check if getElementById returns elements by name
-	// The broken getElementById methods don't pick up programmatically-set names,
-	// so use a roundabout getElementsByName test
-	support.getById = assert( function( el ) {
-		docElem.appendChild( el ).id = expando;
-		return !document.getElementsByName || !document.getElementsByName( expando ).length;
-	} );
-
-	// ID filter and find
-	if ( support.getById ) {
-		Expr.filter[ "ID" ] = function( id ) {
-			var attrId = id.replace( runescape, funescape );
-			return function( elem ) {
-				return elem.getAttribute( "id" ) === attrId;
-			};
-		};
-		Expr.find[ "ID" ] = function( id, context ) {
-			if ( typeof context.getElementById !== "undefined" && documentIsHTML ) {
-				var elem = context.getElementById( id );
-				return elem ? [ elem ] : [];
-			}
-		};
-	} else {
-		Expr.filter[ "ID" ] =  function( id ) {
-			var attrId = id.replace( runescape, funescape );
-			return function( elem ) {
-				var node = typeof elem.getAttributeNode !== "undefined" &&
-					elem.getAttributeNode( "id" );
-				return node && node.value === attrId;
-			};
-		};
-
-		// Support: IE 6 - 7 only
-		// getElementById is not reliable as a find shortcut
-		Expr.find[ "ID" ] = function( id, context ) {
-			if ( typeof context.getElementById !== "undefined" && documentIsHTML ) {
-				var node, i, elems,
-					elem = context.getElementById( id );
-
-				if ( elem ) {
-
-					// Verify the id attribute
-					node = elem.getAttributeNode( "id" );
-					if ( node && node.value === id ) {
-						return [ elem ];
-					}
-
-					// Fall back on getElementsByName
-					elems = context.getElementsByName( id );
-					i = 0;
-					while ( ( elem = elems[ i++ ] ) ) {
-						node = elem.getAttributeNode( "id" );
-						if ( node && node.value === id ) {
-							return [ elem ];
-						}
-					}
-				}
-
-				return [];
-			}
-		};
-	}
-
-	// Tag
-	Expr.find[ "TAG" ] = support.getElementsByTagName ?
-		function( tag, context ) {
-			if ( typeof context.getElementsByTagName !== "undefined" ) {
-				return context.getElementsByTagName( tag );
-
-			// DocumentFragment nodes don't have gEBTN
-			} else if ( support.qsa ) {
-				return context.querySelectorAll( tag );
-			}
-		} :
-
-		function( tag, context ) {
-			var elem,
-				tmp = [],
-				i = 0,
-
-				// By happy coincidence, a (broken) gEBTN appears on DocumentFragment nodes too
-				results = context.getElementsByTagName( tag );
-
-			// Filter out possible comments
-			if ( tag === "*" ) {
-				while ( ( elem = results[ i++ ] ) ) {
-					if ( elem.nodeType === 1 ) {
-						tmp.push( elem );
-					}
-				}
-
-				return tmp;
-			}
-			return results;
-		};
-
-	// Class
-	Expr.find[ "CLASS" ] = support.getElementsByClassName && function( className, context ) {
-		if ( typeof context.getElementsByClassName !== "undefined" && documentIsHTML ) {
-			return context.getElementsByClassName( className );
-		}
-	};
-
-	/* QSA/matchesSelector
-	---------------------------------------------------------------------- */
-
-	// QSA and matchesSelector support
-
-	// matchesSelector(:active) reports false when true (IE9/Opera 11.5)
-	rbuggyMatches = [];
-
-	// qSa(:focus) reports false when true (Chrome 21)
-	// We allow this because of a bug in IE8/9 that throws an error
-	// whenever `document.activeElement` is accessed on an iframe
-	// So, we allow :focus to pass through QSA all the time to avoid the IE error
-	// See https://bugs.jquery.com/ticket/13378
-	rbuggyQSA = [];
-
-	if ( ( support.qsa = rnative.test( document.querySelectorAll ) ) ) {
-
-		// Build QSA regex
-		// Regex strategy adopted from Diego Perini
-		assert( function( el ) {
-
-			var input;
-
-			// Select is set to empty string on purpose
-			// This is to test IE's treatment of not explicitly
-			// setting a boolean content attribute,
-			// since its presence should be enough
-			// https://bugs.jquery.com/ticket/12359
-			docElem.appendChild( el ).innerHTML = "<a id='" + expando + "'></a>" +
-				"<select id='" + expando + "-\r\\' msallowcapture=''>" +
-				"<option selected=''></option></select>";
-
-			// Support: IE8, Opera 11-12.16
-			// Nothing should be selected when empty strings follow ^= or $= or *=
-			// The test attribute must be unknown in Opera but "safe" for WinRT
-			// https://msdn.microsoft.com/en-us/library/ie/hh465388.aspx#attribute_section
-			if ( el.querySelectorAll( "[msallowcapture^='']" ).length ) {
-				rbuggyQSA.push( "[*^$]=" + whitespace + "*(?:''|\"\")" );
-			}
-
-			// Support: IE8
-			// Boolean attributes and "value" are not treated correctly
-			if ( !el.querySelectorAll( "[selected]" ).length ) {
-				rbuggyQSA.push( "\\[" + whitespace + "*(?:value|" + booleans + ")" );
-			}
-
-			// Support: Chrome<29, Android<4.4, Safari<7.0+, iOS<7.0+, PhantomJS<1.9.8+
-			if ( !el.querySelectorAll( "[id~=" + expando + "-]" ).length ) {
-				rbuggyQSA.push( "~=" );
-			}
-
-			// Support: IE 11+, Edge 15 - 18+
-			// IE 11/Edge don't find elements on a `[name='']` query in some cases.
-			// Adding a temporary attribute to the document before the selection works
-			// around the issue.
-			// Interestingly, IE 10 & older don't seem to have the issue.
-			input = document.createElement( "input" );
-			input.setAttribute( "name", "" );
-			el.appendChild( input );
-			if ( !el.querySelectorAll( "[name='']" ).length ) {
-				rbuggyQSA.push( "\\[" + whitespace + "*name" + whitespace + "*=" +
-					whitespace + "*(?:''|\"\")" );
-			}
-
-			// Webkit/Opera - :checked should return selected option elements
-			// http://www.w3.org/TR/2011/REC-css3-selectors-20110929/#checked
-			// IE8 throws error here and will not see later tests
-			if ( !el.querySelectorAll( ":checked" ).length ) {
-				rbuggyQSA.push( ":checked" );
-			}
-
-			// Support: Safari 8+, iOS 8+
-			// https://bugs.webkit.org/show_bug.cgi?id=136851
-			// In-page `selector#id sibling-combinator selector` fails
-			if ( !el.querySelectorAll( "a#" + expando + "+*" ).length ) {
-				rbuggyQSA.push( ".#.+[+~]" );
-			}
-
-			// Support: Firefox <=3.6 - 5 only
-			// Old Firefox doesn't throw on a badly-escaped identifier.
-			el.querySelectorAll( "\\\f" );
-			rbuggyQSA.push( "[\\r\\n\\f]" );
-		} );
-
-		assert( function( el ) {
-			el.innerHTML = "<a href='' disabled='disabled'></a>" +
-				"<select disabled='disabled'><option/></select>";
-
-			// Support: Windows 8 Native Apps
-			// The type and name attributes are restricted during .innerHTML assignment
-			var input = document.createElement( "input" );
-			input.setAttribute( "type", "hidden" );
-			el.appendChild( input ).setAttribute( "name", "D" );
-
-			// Support: IE8
-			// Enforce case-sensitivity of name attribute
-			if ( el.querySelectorAll( "[name=d]" ).length ) {
-				rbuggyQSA.push( "name" + whitespace + "*[*^$|!~]?=" );
-			}
-
-			// FF 3.5 - :enabled/:disabled and hidden elements (hidden elements are still enabled)
-			// IE8 throws error here and will not see later tests
-			if ( el.querySelectorAll( ":enabled" ).length !== 2 ) {
-				rbuggyQSA.push( ":enabled", ":disabled" );
-			}
-
-			// Support: IE9-11+
-			// IE's :disabled selector does not pick up the children of disabled fieldsets
-			docElem.appendChild( el ).disabled = true;
-			if ( el.querySelectorAll( ":disabled" ).length !== 2 ) {
-				rbuggyQSA.push( ":enabled", ":disabled" );
-			}
-
-			// Support: Opera 10 - 11 only
-			// Opera 10-11 does not throw on post-comma invalid pseudos
-			el.querySelectorAll( "*,:x" );
-			rbuggyQSA.push( ",.*:" );
-		} );
-	}
-
-	if ( ( support.matchesSelector = rnative.test( ( matches = docElem.matches ||
-		docElem.webkitMatchesSelector ||
-		docElem.mozMatchesSelector ||
-		docElem.oMatchesSelector ||
-		docElem.msMatchesSelector ) ) ) ) {
-
-		assert( function( el ) {
-
-			// Check to see if it's possible to do matchesSelector
-			// on a disconnected node (IE 9)
-			support.disconnectedMatch = matches.call( el, "*" );
-
-			// This should fail with an exception
-			// Gecko does not error, returns false instead
-			matches.call( el, "[s!='']:x" );
-			rbuggyMatches.push( "!=", pseudos );
-		} );
-	}
-
-	rbuggyQSA = rbuggyQSA.length && new RegExp( rbuggyQSA.join( "|" ) );
-	rbuggyMatches = rbuggyMatches.length && new RegExp( rbuggyMatches.join( "|" ) );
-
-	/* Contains
-	---------------------------------------------------------------------- */
-	hasCompare = rnative.test( docElem.compareDocumentPosition );
-
-	// Element contains another
-	// Purposefully self-exclusive
-	// As in, an element does not contain itself
-	contains = hasCompare || rnative.test( docElem.contains ) ?
-		function( a, b ) {
-			var adown = a.nodeType === 9 ? a.documentElement : a,
-				bup = b && b.parentNode;
-			return a === bup || !!( bup && bup.nodeType === 1 && (
-				adown.contains ?
-					adown.contains( bup ) :
-					a.compareDocumentPosition && a.compareDocumentPosition( bup ) & 16
-			) );
-		} :
-		function( a, b ) {
-			if ( b ) {
-				while ( ( b = b.parentNode ) ) {
-					if ( b === a ) {
-						return true;
-					}
-				}
-			}
-			return false;
-		};
-
-	/* Sorting
-	---------------------------------------------------------------------- */
-
-	// Document order sorting
-	sortOrder = hasCompare ?
-	function( a, b ) {
-
-		// Flag for duplicate removal
-		if ( a === b ) {
-			hasDuplicate = true;
-			return 0;
-		}
-
-		// Sort on method existence if only one input has compareDocumentPosition
-		var compare = !a.compareDocumentPosition - !b.compareDocumentPosition;
-		if ( compare ) {
-			return compare;
-		}
-
-		// Calculate position if both inputs belong to the same document
-		// Support: IE 11+, Edge 17 - 18+
-		// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-		// two documents; shallow comparisons work.
-		// eslint-disable-next-line eqeqeq
-		compare = ( a.ownerDocument || a ) == ( b.ownerDocument || b ) ?
-			a.compareDocumentPosition( b ) :
-
-			// Otherwise we know they are disconnected
-			1;
-
-		// Disconnected nodes
-		if ( compare & 1 ||
-			( !support.sortDetached && b.compareDocumentPosition( a ) === compare ) ) {
-
-			// Choose the first element that is related to our preferred document
-			// Support: IE 11+, Edge 17 - 18+
-			// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-			// two documents; shallow comparisons work.
-			// eslint-disable-next-line eqeqeq
-			if ( a == document || a.ownerDocument == preferredDoc &&
-				contains( preferredDoc, a ) ) {
-				return -1;
-			}
-
-			// Support: IE 11+, Edge 17 - 18+
-			// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-			// two documents; shallow comparisons work.
-			// eslint-disable-next-line eqeqeq
-			if ( b == document || b.ownerDocument == preferredDoc &&
-				contains( preferredDoc, b ) ) {
-				return 1;
-			}
-
-			// Maintain original order
-			return sortInput ?
-				( indexOf( sortInput, a ) - indexOf( sortInput, b ) ) :
-				0;
-		}
-
-		return compare & 4 ? -1 : 1;
-	} :
-	function( a, b ) {
-
-		// Exit early if the nodes are identical
-		if ( a === b ) {
-			hasDuplicate = true;
-			return 0;
-		}
-
-		var cur,
-			i = 0,
-			aup = a.parentNode,
-			bup = b.parentNode,
-			ap = [ a ],
-			bp = [ b ];
-
-		// Parentless nodes are either documents or disconnected
-		if ( !aup || !bup ) {
-
-			// Support: IE 11+, Edge 17 - 18+
-			// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-			// two documents; shallow comparisons work.
-			/* eslint-disable eqeqeq */
-			return a == document ? -1 :
-				b == document ? 1 :
-				/* eslint-enable eqeqeq */
-				aup ? -1 :
-				bup ? 1 :
-				sortInput ?
-				( indexOf( sortInput, a ) - indexOf( sortInput, b ) ) :
-				0;
-
-		// If the nodes are siblings, we can do a quick check
-		} else if ( aup === bup ) {
-			return siblingCheck( a, b );
-		}
-
-		// Otherwise we need full lists of their ancestors for comparison
-		cur = a;
-		while ( ( cur = cur.parentNode ) ) {
-			ap.unshift( cur );
-		}
-		cur = b;
-		while ( ( cur = cur.parentNode ) ) {
-			bp.unshift( cur );
-		}
-
-		// Walk down the tree looking for a discrepancy
-		while ( ap[ i ] === bp[ i ] ) {
-			i++;
-		}
-
-		return i ?
-
-			// Do a sibling check if the nodes have a common ancestor
-			siblingCheck( ap[ i ], bp[ i ] ) :
-
-			// Otherwise nodes in our document sort first
-			// Support: IE 11+, Edge 17 - 18+
-			// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-			// two documents; shallow comparisons work.
-			/* eslint-disable eqeqeq */
-			ap[ i ] == preferredDoc ? -1 :
-			bp[ i ] == preferredDoc ? 1 :
-			/* eslint-enable eqeqeq */
-			0;
-	};
-
-	return document;
-};
-
-Sizzle.matches = function( expr, elements ) {
-	return Sizzle( expr, null, null, elements );
-};
-
-Sizzle.matchesSelector = function( elem, expr ) {
-	setDocument( elem );
-
-	if ( support.matchesSelector && documentIsHTML &&
-		!nonnativeSelectorCache[ expr + " " ] &&
-		( !rbuggyMatches || !rbuggyMatches.test( expr ) ) &&
-		( !rbuggyQSA     || !rbuggyQSA.test( expr ) ) ) {
-
-		try {
-			var ret = matches.call( elem, expr );
-
-			// IE 9's matchesSelector returns false on disconnected nodes
-			if ( ret || support.disconnectedMatch ||
-
-				// As well, disconnected nodes are said to be in a document
-				// fragment in IE 9
-				elem.document && elem.document.nodeType !== 11 ) {
-				return ret;
-			}
-		} catch ( e ) {
-			nonnativeSelectorCache( expr, true );
-		}
-	}
-
-	return Sizzle( expr, document, null, [ elem ] ).length > 0;
-};
-
-Sizzle.contains = function( context, elem ) {
-
-	// Set document vars if needed
-	// Support: IE 11+, Edge 17 - 18+
-	// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-	// two documents; shallow comparisons work.
-	// eslint-disable-next-line eqeqeq
-	if ( ( context.ownerDocument || context ) != document ) {
-		setDocument( context );
-	}
-	return contains( context, elem );
-};
-
-Sizzle.attr = function( elem, name ) {
-
-	// Set document vars if needed
-	// Support: IE 11+, Edge 17 - 18+
-	// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-	// two documents; shallow comparisons work.
-	// eslint-disable-next-line eqeqeq
-	if ( ( elem.ownerDocument || elem ) != document ) {
-		setDocument( elem );
-	}
-
-	var fn = Expr.attrHandle[ name.toLowerCase() ],
-
-		// Don't get fooled by Object.prototype properties (jQuery #13807)
-		val = fn && hasOwn.call( Expr.attrHandle, name.toLowerCase() ) ?
-			fn( elem, name, !documentIsHTML ) :
-			undefined;
-
-	return val !== undefined ?
-		val :
-		support.attributes || !documentIsHTML ?
-			elem.getAttribute( name ) :
-			( val = elem.getAttributeNode( name ) ) && val.specified ?
-				val.value :
-				null;
-};
-
-Sizzle.escape = function( sel ) {
-	return ( sel + "" ).replace( rcssescape, fcssescape );
-};
-
-Sizzle.error = function( msg ) {
-	throw new Error( "Syntax error, unrecognized expression: " + msg );
-};
-
-/**
- * Document sorting and removing duplicates
- * @param {ArrayLike} results
- */
-Sizzle.uniqueSort = function( results ) {
-	var elem,
-		duplicates = [],
-		j = 0,
-		i = 0;
-
-	// Unless we *know* we can detect duplicates, assume their presence
-	hasDuplicate = !support.detectDuplicates;
-	sortInput = !support.sortStable && results.slice( 0 );
-	results.sort( sortOrder );
-
-	if ( hasDuplicate ) {
-		while ( ( elem = results[ i++ ] ) ) {
-			if ( elem === results[ i ] ) {
-				j = duplicates.push( i );
-			}
-		}
-		while ( j-- ) {
-			results.splice( duplicates[ j ], 1 );
-		}
-	}
-
-	// Clear input after sorting to release objects
-	// See https://github.com/jquery/sizzle/pull/225
-	sortInput = null;
-
-	return results;
-};
-
-/**
- * Utility function for retrieving the text value of an array of DOM nodes
- * @param {Array|Element} elem
- */
-getText = Sizzle.getText = function( elem ) {
-	var node,
-		ret = "",
-		i = 0,
-		nodeType = elem.nodeType;
-
-	if ( !nodeType ) {
-
-		// If no nodeType, this is expected to be an array
-		while ( ( node = elem[ i++ ] ) ) {
-
-			// Do not traverse comment nodes
-			ret += getText( node );
-		}
-	} else if ( nodeType === 1 || nodeType === 9 || nodeType === 11 ) {
-
-		// Use textContent for elements
-		// innerText usage removed for consistency of new lines (jQuery #11153)
-		if ( typeof elem.textContent === "string" ) {
-			return elem.textContent;
-		} else {
-
-			// Traverse its children
-			for ( elem = elem.firstChild; elem; elem = elem.nextSibling ) {
-				ret += getText( elem );
-			}
-		}
-	} else if ( nodeType === 3 || nodeType === 4 ) {
-		return elem.nodeValue;
-	}
-
-	// Do not include comment or processing instruction nodes
-
-	return ret;
-};
-
-Expr = Sizzle.selectors = {
-
-	// Can be adjusted by the user
-	cacheLength: 50,
-
-	createPseudo: markFunction,
-
-	match: matchExpr,
-
-	attrHandle: {},
-
-	find: {},
-
-	relative: {
-		">": { dir: "parentNode", first: true },
-		" ": { dir: "parentNode" },
-		"+": { dir: "previousSibling", first: true },
-		"~": { dir: "previousSibling" }
-	},
-
-	preFilter: {
-		"ATTR": function( match ) {
-			match[ 1 ] = match[ 1 ].replace( runescape, funescape );
-
-			// Move the given value to match[3] whether quoted or unquoted
-			match[ 3 ] = ( match[ 3 ] || match[ 4 ] ||
-				match[ 5 ] || "" ).replace( runescape, funescape );
-
-			if ( match[ 2 ] === "~=" ) {
-				match[ 3 ] = " " + match[ 3 ] + " ";
-			}
-
-			return match.slice( 0, 4 );
-		},
-
-		"CHILD": function( match ) {
-
-			/* matches from matchExpr["CHILD"]
-				1 type (only|nth|...)
-				2 what (child|of-type)
-				3 argument (even|odd|\d*|\d*n([+-]\d+)?|...)
-				4 xn-component of xn+y argument ([+-]?\d*n|)
-				5 sign of xn-component
-				6 x of xn-component
-				7 sign of y-component
-				8 y of y-component
-			*/
-			match[ 1 ] = match[ 1 ].toLowerCase();
-
-			if ( match[ 1 ].slice( 0, 3 ) === "nth" ) {
-
-				// nth-* requires argument
-				if ( !match[ 3 ] ) {
-					Sizzle.error( match[ 0 ] );
-				}
-
-				// numeric x and y parameters for Expr.filter.CHILD
-				// remember that false/true cast respectively to 0/1
-				match[ 4 ] = +( match[ 4 ] ?
-					match[ 5 ] + ( match[ 6 ] || 1 ) :
-					2 * ( match[ 3 ] === "even" || match[ 3 ] === "odd" ) );
-				match[ 5 ] = +( ( match[ 7 ] + match[ 8 ] ) || match[ 3 ] === "odd" );
-
-				// other types prohibit arguments
-			} else if ( match[ 3 ] ) {
-				Sizzle.error( match[ 0 ] );
-			}
-
-			return match;
-		},
-
-		"PSEUDO": function( match ) {
-			var excess,
-				unquoted = !match[ 6 ] && match[ 2 ];
-
-			if ( matchExpr[ "CHILD" ].test( match[ 0 ] ) ) {
-				return null;
-			}
-
-			// Accept quoted arguments as-is
-			if ( match[ 3 ] ) {
-				match[ 2 ] = match[ 4 ] || match[ 5 ] || "";
-
-			// Strip excess characters from unquoted arguments
-			} else if ( unquoted && rpseudo.test( unquoted ) &&
-
-				// Get excess from tokenize (recursively)
-				( excess = tokenize( unquoted, true ) ) &&
-
-				// advance to the next closing parenthesis
-				( excess = unquoted.indexOf( ")", unquoted.length - excess ) - unquoted.length ) ) {
-
-				// excess is a negative index
-				match[ 0 ] = match[ 0 ].slice( 0, excess );
-				match[ 2 ] = unquoted.slice( 0, excess );
-			}
-
-			// Return only captures needed by the pseudo filter method (type and argument)
-			return match.slice( 0, 3 );
-		}
-	},
-
-	filter: {
-
-		"TAG": function( nodeNameSelector ) {
-			var nodeName = nodeNameSelector.replace( runescape, funescape ).toLowerCase();
-			return nodeNameSelector === "*" ?
-				function() {
-					return true;
-				} :
-				function( elem ) {
-					return elem.nodeName && elem.nodeName.toLowerCase() === nodeName;
-				};
-		},
-
-		"CLASS": function( className ) {
-			var pattern = classCache[ className + " " ];
-
-			return pattern ||
-				( pattern = new RegExp( "(^|" + whitespace +
-					")" + className + "(" + whitespace + "|$)" ) ) && classCache(
-						className, function( elem ) {
-							return pattern.test(
-								typeof elem.className === "string" && elem.className ||
-								typeof elem.getAttribute !== "undefined" &&
-									elem.getAttribute( "class" ) ||
-								""
-							);
-				} );
-		},
-
-		"ATTR": function( name, operator, check ) {
-			return function( elem ) {
-				var result = Sizzle.attr( elem, name );
-
-				if ( result == null ) {
-					return operator === "!=";
-				}
-				if ( !operator ) {
-					return true;
-				}
-
-				result += "";
-
-				/* eslint-disable max-len */
-
-				return operator === "=" ? result === check :
-					operator === "!=" ? result !== check :
-					operator === "^=" ? check && result.indexOf( check ) === 0 :
-					operator === "*=" ? check && result.indexOf( check ) > -1 :
-					operator === "$=" ? check && result.slice( -check.length ) === check :
-					operator === "~=" ? ( " " + result.replace( rwhitespace, " " ) + " " ).indexOf( check ) > -1 :
-					operator === "|=" ? result === check || result.slice( 0, check.length + 1 ) === check + "-" :
-					false;
-				/* eslint-enable max-len */
-
-			};
-		},
-
-		"CHILD": function( type, what, _argument, first, last ) {
-			var simple = type.slice( 0, 3 ) !== "nth",
-				forward = type.slice( -4 ) !== "last",
-				ofType = what === "of-type";
-
-			return first === 1 && last === 0 ?
-
-				// Shortcut for :nth-*(n)
-				function( elem ) {
-					return !!elem.parentNode;
-				} :
-
-				function( elem, _context, xml ) {
-					var cache, uniqueCache, outerCache, node, nodeIndex, start,
-						dir = simple !== forward ? "nextSibling" : "previousSibling",
-						parent = elem.parentNode,
-						name = ofType && elem.nodeName.toLowerCase(),
-						useCache = !xml && !ofType,
-						diff = false;
-
-					if ( parent ) {
-
-						// :(first|last|only)-(child|of-type)
-						if ( simple ) {
-							while ( dir ) {
-								node = elem;
-								while ( ( node = node[ dir ] ) ) {
-									if ( ofType ?
-										node.nodeName.toLowerCase() === name :
-										node.nodeType === 1 ) {
-
-										return false;
-									}
-								}
-
-								// Reverse direction for :only-* (if we haven't yet done so)
-								start = dir = type === "only" && !start && "nextSibling";
-							}
-							return true;
-						}
-
-						start = [ forward ? parent.firstChild : parent.lastChild ];
-
-						// non-xml :nth-child(...) stores cache data on `parent`
-						if ( forward && useCache ) {
-
-							// Seek `elem` from a previously-cached index
-
-							// ...in a gzip-friendly way
-							node = parent;
-							outerCache = node[ expando ] || ( node[ expando ] = {} );
-
-							// Support: IE <9 only
-							// Defend against cloned attroperties (jQuery gh-1709)
-							uniqueCache = outerCache[ node.uniqueID ] ||
-								( outerCache[ node.uniqueID ] = {} );
-
-							cache = uniqueCache[ type ] || [];
-							nodeIndex = cache[ 0 ] === dirruns && cache[ 1 ];
-							diff = nodeIndex && cache[ 2 ];
-							node = nodeIndex && parent.childNodes[ nodeIndex ];
-
-							while ( ( node = ++nodeIndex && node && node[ dir ] ||
-
-								// Fallback to seeking `elem` from the start
-								( diff = nodeIndex = 0 ) || start.pop() ) ) {
-
-								// When found, cache indexes on `parent` and break
-								if ( node.nodeType === 1 && ++diff && node === elem ) {
-									uniqueCache[ type ] = [ dirruns, nodeIndex, diff ];
-									break;
-								}
-							}
-
-						} else {
-
-							// Use previously-cached element index if available
-							if ( useCache ) {
-
-								// ...in a gzip-friendly way
-								node = elem;
-								outerCache = node[ expando ] || ( node[ expando ] = {} );
-
-								// Support: IE <9 only
-								// Defend against cloned attroperties (jQuery gh-1709)
-								uniqueCache = outerCache[ node.uniqueID ] ||
-									( outerCache[ node.uniqueID ] = {} );
-
-								cache = uniqueCache[ type ] || [];
-								nodeIndex = cache[ 0 ] === dirruns && cache[ 1 ];
-								diff = nodeIndex;
-							}
-
-							// xml :nth-child(...)
-							// or :nth-last-child(...) or :nth(-last)?-of-type(...)
-							if ( diff === false ) {
-
-								// Use the same loop as above to seek `elem` from the start
-								while ( ( node = ++nodeIndex && node && node[ dir ] ||
-									( diff = nodeIndex = 0 ) || start.pop() ) ) {
-
-									if ( ( ofType ?
-										node.nodeName.toLowerCase() === name :
-										node.nodeType === 1 ) &&
-										++diff ) {
-
-										// Cache the index of each encountered element
-										if ( useCache ) {
-											outerCache = node[ expando ] ||
-												( node[ expando ] = {} );
-
-											// Support: IE <9 only
-											// Defend against cloned attroperties (jQuery gh-1709)
-											uniqueCache = outerCache[ node.uniqueID ] ||
-												( outerCache[ node.uniqueID ] = {} );
-
-											uniqueCache[ type ] = [ dirruns, diff ];
-										}
-
-										if ( node === elem ) {
-											break;
-										}
-									}
-								}
-							}
-						}
-
-						// Incorporate the offset, then check against cycle size
-						diff -= last;
-						return diff === first || ( diff % first === 0 && diff / first >= 0 );
-					}
-				};
-		},
-
-		"PSEUDO": function( pseudo, argument ) {
-
-			// pseudo-class names are case-insensitive
-			// http://www.w3.org/TR/selectors/#pseudo-classes
-			// Prioritize by case sensitivity in case custom pseudos are added with uppercase letters
-			// Remember that setFilters inherits from pseudos
-			var args,
-				fn = Expr.pseudos[ pseudo ] || Expr.setFilters[ pseudo.toLowerCase() ] ||
-					Sizzle.error( "unsupported pseudo: " + pseudo );
-
-			// The user may use createPseudo to indicate that
-			// arguments are needed to create the filter function
-			// just as Sizzle does
-			if ( fn[ expando ] ) {
-				return fn( argument );
-			}
-
-			// But maintain support for old signatures
-			if ( fn.length > 1 ) {
-				args = [ pseudo, pseudo, "", argument ];
-				return Expr.setFilters.hasOwnProperty( pseudo.toLowerCase() ) ?
-					markFunction( function( seed, matches ) {
-						var idx,
-							matched = fn( seed, argument ),
-							i = matched.length;
-						while ( i-- ) {
-							idx = indexOf( seed, matched[ i ] );
-							seed[ idx ] = !( matches[ idx ] = matched[ i ] );
-						}
-					} ) :
-					function( elem ) {
-						return fn( elem, 0, args );
-					};
-			}
-
-			return fn;
-		}
-	},
-
-	pseudos: {
-
-		// Potentially complex pseudos
-		"not": markFunction( function( selector ) {
-
-			// Trim the selector passed to compile
-			// to avoid treating leading and trailing
-			// spaces as combinators
-			var input = [],
-				results = [],
-				matcher = compile( selector.replace( rtrim, "$1" ) );
-
-			return matcher[ expando ] ?
-				markFunction( function( seed, matches, _context, xml ) {
-					var elem,
-						unmatched = matcher( seed, null, xml, [] ),
-						i = seed.length;
-
-					// Match elements unmatched by `matcher`
-					while ( i-- ) {
-						if ( ( elem = unmatched[ i ] ) ) {
-							seed[ i ] = !( matches[ i ] = elem );
-						}
-					}
-				} ) :
-				function( elem, _context, xml ) {
-					input[ 0 ] = elem;
-					matcher( input, null, xml, results );
-
-					// Don't keep the element (issue #299)
-					input[ 0 ] = null;
-					return !results.pop();
-				};
-		} ),
-
-		"has": markFunction( function( selector ) {
-			return function( elem ) {
-				return Sizzle( selector, elem ).length > 0;
-			};
-		} ),
-
-		"contains": markFunction( function( text ) {
-			text = text.replace( runescape, funescape );
-			return function( elem ) {
-				return ( elem.textContent || getText( elem ) ).indexOf( text ) > -1;
-			};
-		} ),
-
-		// "Whether an element is represented by a :lang() selector
-		// is based solely on the element's language value
-		// being equal to the identifier C,
-		// or beginning with the identifier C immediately followed by "-".
-		// The matching of C against the element's language value is performed case-insensitively.
-		// The identifier C does not have to be a valid language name."
-		// http://www.w3.org/TR/selectors/#lang-pseudo
-		"lang": markFunction( function( lang ) {
-
-			// lang value must be a valid identifier
-			if ( !ridentifier.test( lang || "" ) ) {
-				Sizzle.error( "unsupported lang: " + lang );
-			}
-			lang = lang.replace( runescape, funescape ).toLowerCase();
-			return function( elem ) {
-				var elemLang;
-				do {
-					if ( ( elemLang = documentIsHTML ?
-						elem.lang :
-						elem.getAttribute( "xml:lang" ) || elem.getAttribute( "lang" ) ) ) {
-
-						elemLang = elemLang.toLowerCase();
-						return elemLang === lang || elemLang.indexOf( lang + "-" ) === 0;
-					}
-				} while ( ( elem = elem.parentNode ) && elem.nodeType === 1 );
-				return false;
-			};
-		} ),
-
-		// Miscellaneous
-		"target": function( elem ) {
-			var hash = window.location && window.location.hash;
-			return hash && hash.slice( 1 ) === elem.id;
-		},
-
-		"root": function( elem ) {
-			return elem === docElem;
-		},
-
-		"focus": function( elem ) {
-			return elem === document.activeElement &&
-				( !document.hasFocus || document.hasFocus() ) &&
-				!!( elem.type || elem.href || ~elem.tabIndex );
-		},
-
-		// Boolean properties
-		"enabled": createDisabledPseudo( false ),
-		"disabled": createDisabledPseudo( true ),
-
-		"checked": function( elem ) {
-
-			// In CSS3, :checked should return both checked and selected elements
-			// http://www.w3.org/TR/2011/REC-css3-selectors-20110929/#checked
-			var nodeName = elem.nodeName.toLowerCase();
-			return ( nodeName === "input" && !!elem.checked ) ||
-				( nodeName === "option" && !!elem.selected );
-		},
-
-		"selected": function( elem ) {
-
-			// Accessing this property makes selected-by-default
-			// options in Safari work properly
-			if ( elem.parentNode ) {
-				// eslint-disable-next-line no-unused-expressions
-				elem.parentNode.selectedIndex;
-			}
-
-			return elem.selected === true;
-		},
-
-		// Contents
-		"empty": function( elem ) {
-
-			// http://www.w3.org/TR/selectors/#empty-pseudo
-			// :empty is negated by element (1) or content nodes (text: 3; cdata: 4; entity ref: 5),
-			//   but not by others (comment: 8; processing instruction: 7; etc.)
-			// nodeType < 6 works because attributes (2) do not appear as children
-			for ( elem = elem.firstChild; elem; elem = elem.nextSibling ) {
-				if ( elem.nodeType < 6 ) {
-					return false;
-				}
-			}
-			return true;
-		},
-
-		"parent": function( elem ) {
-			return !Expr.pseudos[ "empty" ]( elem );
-		},
-
-		// Element/input types
-		"header": function( elem ) {
-			return rheader.test( elem.nodeName );
-		},
-
-		"input": function( elem ) {
-			return rinputs.test( elem.nodeName );
-		},
-
-		"button": function( elem ) {
-			var name = elem.nodeName.toLowerCase();
-			return name === "input" && elem.type === "button" || name === "button";
-		},
-
-		"text": function( elem ) {
-			var attr;
-			return elem.nodeName.toLowerCase() === "input" &&
-				elem.type === "text" &&
-
-				// Support: IE<8
-				// New HTML5 attribute values (e.g., "search") appear with elem.type === "text"
-				( ( attr = elem.getAttribute( "type" ) ) == null ||
-					attr.toLowerCase() === "text" );
-		},
-
-		// Position-in-collection
-		"first": createPositionalPseudo( function() {
-			return [ 0 ];
-		} ),
-
-		"last": createPositionalPseudo( function( _matchIndexes, length ) {
-			return [ length - 1 ];
-		} ),
-
-		"eq": createPositionalPseudo( function( _matchIndexes, length, argument ) {
-			return [ argument < 0 ? argument + length : argument ];
-		} ),
-
-		"even": createPositionalPseudo( function( matchIndexes, length ) {
-			var i = 0;
-			for ( ; i < length; i += 2 ) {
-				matchIndexes.push( i );
-			}
-			return matchIndexes;
-		} ),
-
-		"odd": createPositionalPseudo( function( matchIndexes, length ) {
-			var i = 1;
-			for ( ; i < length; i += 2 ) {
-				matchIndexes.push( i );
-			}
-			return matchIndexes;
-		} ),
-
-		"lt": createPositionalPseudo( function( matchIndexes, length, argument ) {
-			var i = argument < 0 ?
-				argument + length :
-				argument > length ?
-					length :
-					argument;
-			for ( ; --i >= 0; ) {
-				matchIndexes.push( i );
-			}
-			return matchIndexes;
-		} ),
-
-		"gt": createPositionalPseudo( function( matchIndexes, length, argument ) {
-			var i = argument < 0 ? argument + length : argument;
-			for ( ; ++i < length; ) {
-				matchIndexes.push( i );
-			}
-			return matchIndexes;
-		} )
-	}
-};
-
-Expr.pseudos[ "nth" ] = Expr.pseudos[ "eq" ];
-
-// Add button/input type pseudos
-for ( i in { radio: true, checkbox: true, file: true, password: true, image: true } ) {
-	Expr.pseudos[ i ] = createInputPseudo( i );
-}
-for ( i in { submit: true, reset: true } ) {
-	Expr.pseudos[ i ] = createButtonPseudo( i );
-}
-
-// Easy API for creating new setFilters
-function setFilters() {}
-setFilters.prototype = Expr.filters = Expr.pseudos;
-Expr.setFilters = new setFilters();
-
-tokenize = Sizzle.tokenize = function( selector, parseOnly ) {
-	var matched, match, tokens, type,
-		soFar, groups, preFilters,
-		cached = tokenCache[ selector + " " ];
-
-	if ( cached ) {
-		return parseOnly ? 0 : cached.slice( 0 );
-	}
-
-	soFar = selector;
-	groups = [];
-	preFilters = Expr.preFilter;
-
-	while ( soFar ) {
-
-		// Comma and first run
-		if ( !matched || ( match = rcomma.exec( soFar ) ) ) {
-			if ( match ) {
-
-				// Don't consume trailing commas as valid
-				soFar = soFar.slice( match[ 0 ].length ) || soFar;
-			}
-			groups.push( ( tokens = [] ) );
-		}
-
-		matched = false;
-
-		// Combinators
-		if ( ( match = rcombinators.exec( soFar ) ) ) {
-			matched = match.shift();
-			tokens.push( {
-				value: matched,
-
-				// Cast descendant combinators to space
-				type: match[ 0 ].replace( rtrim, " " )
-			} );
-			soFar = soFar.slice( matched.length );
-		}
-
-		// Filters
-		for ( type in Expr.filter ) {
-			if ( ( match = matchExpr[ type ].exec( soFar ) ) && ( !preFilters[ type ] ||
-				( match = preFilters[ type ]( match ) ) ) ) {
-				matched = match.shift();
-				tokens.push( {
-					value: matched,
-					type: type,
-					matches: match
-				} );
-				soFar = soFar.slice( matched.length );
-			}
-		}
-
-		if ( !matched ) {
-			break;
-		}
-	}
-
-	// Return the length of the invalid excess
-	// if we're just parsing
-	// Otherwise, throw an error or return tokens
-	return parseOnly ?
-		soFar.length :
-		soFar ?
-			Sizzle.error( selector ) :
-
-			// Cache the tokens
-			tokenCache( selector, groups ).slice( 0 );
-};
-
-function toSelector( tokens ) {
-	var i = 0,
-		len = tokens.length,
-		selector = "";
-	for ( ; i < len; i++ ) {
-		selector += tokens[ i ].value;
-	}
-	return selector;
-}
-
-function addCombinator( matcher, combinator, base ) {
-	var dir = combinator.dir,
-		skip = combinator.next,
-		key = skip || dir,
-		checkNonElements = base && key === "parentNode",
-		doneName = done++;
-
-	return combinator.first ?
-
-		// Check against closest ancestor/preceding element
-		function( elem, context, xml ) {
-			while ( ( elem = elem[ dir ] ) ) {
-				if ( elem.nodeType === 1 || checkNonElements ) {
-					return matcher( elem, context, xml );
-				}
-			}
-			return false;
-		} :
-
-		// Check against all ancestor/preceding elements
-		function( elem, context, xml ) {
-			var oldCache, uniqueCache, outerCache,
-				newCache = [ dirruns, doneName ];
-
-			// We can't set arbitrary data on XML nodes, so they don't benefit from combinator caching
-			if ( xml ) {
-				while ( ( elem = elem[ dir ] ) ) {
-					if ( elem.nodeType === 1 || checkNonElements ) {
-						if ( matcher( elem, context, xml ) ) {
-							return true;
-						}
-					}
-				}
-			} else {
-				while ( ( elem = elem[ dir ] ) ) {
-					if ( elem.nodeType === 1 || checkNonElements ) {
-						outerCache = elem[ expando ] || ( elem[ expando ] = {} );
-
-						// Support: IE <9 only
-						// Defend against cloned attroperties (jQuery gh-1709)
-						uniqueCache = outerCache[ elem.uniqueID ] ||
-							( outerCache[ elem.uniqueID ] = {} );
-
-						if ( skip && skip === elem.nodeName.toLowerCase() ) {
-							elem = elem[ dir ] || elem;
-						} else if ( ( oldCache = uniqueCache[ key ] ) &&
-							oldCache[ 0 ] === dirruns && oldCache[ 1 ] === doneName ) {
-
-							// Assign to newCache so results back-propagate to previous elements
-							return ( newCache[ 2 ] = oldCache[ 2 ] );
-						} else {
-
-							// Reuse newcache so results back-propagate to previous elements
-							uniqueCache[ key ] = newCache;
-
-							// A match means we're done; a fail means we have to keep checking
-							if ( ( newCache[ 2 ] = matcher( elem, context, xml ) ) ) {
-								return true;
-							}
-						}
-					}
-				}
-			}
-			return false;
-		};
-}
-
-function elementMatcher( matchers ) {
-	return matchers.length > 1 ?
-		function( elem, context, xml ) {
-			var i = matchers.length;
-			while ( i-- ) {
-				if ( !matchers[ i ]( elem, context, xml ) ) {
-					return false;
-				}
-			}
-			return true;
-		} :
-		matchers[ 0 ];
-}
-
-function multipleContexts( selector, contexts, results ) {
-	var i = 0,
-		len = contexts.length;
-	for ( ; i < len; i++ ) {
-		Sizzle( selector, contexts[ i ], results );
-	}
-	return results;
-}
-
-function condense( unmatched, map, filter, context, xml ) {
-	var elem,
-		newUnmatched = [],
-		i = 0,
-		len = unmatched.length,
-		mapped = map != null;
-
-	for ( ; i < len; i++ ) {
-		if ( ( elem = unmatched[ i ] ) ) {
-			if ( !filter || filter( elem, context, xml ) ) {
-				newUnmatched.push( elem );
-				if ( mapped ) {
-					map.push( i );
-				}
-			}
-		}
-	}
-
-	return newUnmatched;
-}
-
-function setMatcher( preFilter, selector, matcher, postFilter, postFinder, postSelector ) {
-	if ( postFilter && !postFilter[ expando ] ) {
-		postFilter = setMatcher( postFilter );
-	}
-	if ( postFinder && !postFinder[ expando ] ) {
-		postFinder = setMatcher( postFinder, postSelector );
-	}
-	return markFunction( function( seed, results, context, xml ) {
-		var temp, i, elem,
-			preMap = [],
-			postMap = [],
-			preexisting = results.length,
-
-			// Get initial elements from seed or context
-			elems = seed || multipleContexts(
-				selector || "*",
-				context.nodeType ? [ context ] : context,
-				[]
-			),
-
-			// Prefilter to get matcher input, preserving a map for seed-results synchronization
-			matcherIn = preFilter && ( seed || !selector ) ?
-				condense( elems, preMap, preFilter, context, xml ) :
-				elems,
-
-			matcherOut = matcher ?
-
-				// If we have a postFinder, or filtered seed, or non-seed postFilter or preexisting results,
-				postFinder || ( seed ? preFilter : preexisting || postFilter ) ?
-
-					// ...intermediate processing is necessary
-					[] :
-
-					// ...otherwise use results directly
-					results :
-				matcherIn;
-
-		// Find primary matches
-		if ( matcher ) {
-			matcher( matcherIn, matcherOut, context, xml );
-		}
-
-		// Apply postFilter
-		if ( postFilter ) {
-			temp = condense( matcherOut, postMap );
-			postFilter( temp, [], context, xml );
-
-			// Un-match failing elements by moving them back to matcherIn
-			i = temp.length;
-			while ( i-- ) {
-				if ( ( elem = temp[ i ] ) ) {
-					matcherOut[ postMap[ i ] ] = !( matcherIn[ postMap[ i ] ] = elem );
-				}
-			}
-		}
-
-		if ( seed ) {
-			if ( postFinder || preFilter ) {
-				if ( postFinder ) {
-
-					// Get the final matcherOut by condensing this intermediate into postFinder contexts
-					temp = [];
-					i = matcherOut.length;
-					while ( i-- ) {
-						if ( ( elem = matcherOut[ i ] ) ) {
-
-							// Restore matcherIn since elem is not yet a final match
-							temp.push( ( matcherIn[ i ] = elem ) );
-						}
-					}
-					postFinder( null, ( matcherOut = [] ), temp, xml );
-				}
-
-				// Move matched elements from seed to results to keep them synchronized
-				i = matcherOut.length;
-				while ( i-- ) {
-					if ( ( elem = matcherOut[ i ] ) &&
-						( temp = postFinder ? indexOf( seed, elem ) : preMap[ i ] ) > -1 ) {
-
-						seed[ temp ] = !( results[ temp ] = elem );
-					}
-				}
-			}
-
-		// Add elements to results, through postFinder if defined
-		} else {
-			matcherOut = condense(
-				matcherOut === results ?
-					matcherOut.splice( preexisting, matcherOut.length ) :
-					matcherOut
-			);
-			if ( postFinder ) {
-				postFinder( null, results, matcherOut, xml );
-			} else {
-				push.apply( results, matcherOut );
-			}
-		}
-	} );
-}
-
-function matcherFromTokens( tokens ) {
-	var checkContext, matcher, j,
-		len = tokens.length,
-		leadingRelative = Expr.relative[ tokens[ 0 ].type ],
-		implicitRelative = leadingRelative || Expr.relative[ " " ],
-		i = leadingRelative ? 1 : 0,
-
-		// The foundational matcher ensures that elements are reachable from top-level context(s)
-		matchContext = addCombinator( function( elem ) {
-			return elem === checkContext;
-		}, implicitRelative, true ),
-		matchAnyContext = addCombinator( function( elem ) {
-			return indexOf( checkContext, elem ) > -1;
-		}, implicitRelative, true ),
-		matchers = [ function( elem, context, xml ) {
-			var ret = ( !leadingRelative && ( xml || context !== outermostContext ) ) || (
-				( checkContext = context ).nodeType ?
-					matchContext( elem, context, xml ) :
-					matchAnyContext( elem, context, xml ) );
-
-			// Avoid hanging onto element (issue #299)
-			checkContext = null;
-			return ret;
-		} ];
-
-	for ( ; i < len; i++ ) {
-		if ( ( matcher = Expr.relative[ tokens[ i ].type ] ) ) {
-			matchers = [ addCombinator( elementMatcher( matchers ), matcher ) ];
-		} else {
-			matcher = Expr.filter[ tokens[ i ].type ].apply( null, tokens[ i ].matches );
-
-			// Return special upon seeing a positional matcher
-			if ( matcher[ expando ] ) {
-
-				// Find the next relative operator (if any) for proper handling
-				j = ++i;
-				for ( ; j < len; j++ ) {
-					if ( Expr.relative[ tokens[ j ].type ] ) {
-						break;
-					}
-				}
-				return setMatcher(
-					i > 1 && elementMatcher( matchers ),
-					i > 1 && toSelector(
-
-					// If the preceding token was a descendant combinator, insert an implicit any-element `*`
-					tokens
-						.slice( 0, i - 1 )
-						.concat( { value: tokens[ i - 2 ].type === " " ? "*" : "" } )
-					).replace( rtrim, "$1" ),
-					matcher,
-					i < j && matcherFromTokens( tokens.slice( i, j ) ),
-					j < len && matcherFromTokens( ( tokens = tokens.slice( j ) ) ),
-					j < len && toSelector( tokens )
-				);
-			}
-			matchers.push( matcher );
-		}
-	}
-
-	return elementMatcher( matchers );
-}
-
-function matcherFromGroupMatchers( elementMatchers, setMatchers ) {
-	var bySet = setMatchers.length > 0,
-		byElement = elementMatchers.length > 0,
-		superMatcher = function( seed, context, xml, results, outermost ) {
-			var elem, j, matcher,
-				matchedCount = 0,
-				i = "0",
-				unmatched = seed && [],
-				setMatched = [],
-				contextBackup = outermostContext,
-
-				// We must always have either seed elements or outermost context
-				elems = seed || byElement && Expr.find[ "TAG" ]( "*", outermost ),
-
-				// Use integer dirruns iff this is the outermost matcher
-				dirrunsUnique = ( dirruns += contextBackup == null ? 1 : Math.random() || 0.1 ),
-				len = elems.length;
-
-			if ( outermost ) {
-
-				// Support: IE 11+, Edge 17 - 18+
-				// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-				// two documents; shallow comparisons work.
-				// eslint-disable-next-line eqeqeq
-				outermostContext = context == document || context || outermost;
-			}
-
-			// Add elements passing elementMatchers directly to results
-			// Support: IE<9, Safari
-			// Tolerate NodeList properties (IE: "length"; Safari: <number>) matching elements by id
-			for ( ; i !== len && ( elem = elems[ i ] ) != null; i++ ) {
-				if ( byElement && elem ) {
-					j = 0;
-
-					// Support: IE 11+, Edge 17 - 18+
-					// IE/Edge sometimes throw a "Permission denied" error when strict-comparing
-					// two documents; shallow comparisons work.
-					// eslint-disable-next-line eqeqeq
-					if ( !context && elem.ownerDocument != document ) {
-						setDocument( elem );
-						xml = !documentIsHTML;
-					}
-					while ( ( matcher = elementMatchers[ j++ ] ) ) {
-						if ( matcher( elem, context || document, xml ) ) {
-							results.push( elem );
-							break;
-						}
-					}
-					if ( outermost ) {
-						dirruns = dirrunsUnique;
-					}
-				}
-
-				// Track unmatched elements for set filters
-				if ( bySet ) {
-
-					// They will have gone through all possible matchers
-					if ( ( elem = !matcher && elem ) ) {
-						matchedCount--;
-					}
-
-					// Lengthen the array for every element, matched or not
-					if ( seed ) {
-						unmatched.push( elem );
-					}
-				}
-			}
-
-			// `i` is now the count of elements visited above, and adding it to `matchedCount`
-			// makes the latter nonnegative.
-			matchedCount += i;
-
-			// Apply set filters to unmatched elements
-			// NOTE: This can be skipped if there are no unmatched elements (i.e., `matchedCount`
-			// equals `i`), unless we didn't visit _any_ elements in the above loop because we have
-			// no element matchers and no seed.
-			// Incrementing an initially-string "0" `i` allows `i` to remain a string only in that
-			// case, which will result in a "00" `matchedCount` that differs from `i` but is also
-			// numerically zero.
-			if ( bySet && i !== matchedCount ) {
-				j = 0;
-				while ( ( matcher = setMatchers[ j++ ] ) ) {
-					matcher( unmatched, setMatched, context, xml );
-				}
-
-				if ( seed ) {
-
-					// Reintegrate element matches to eliminate the need for sorting
-					if ( matchedCount > 0 ) {
-						while ( i-- ) {
-							if ( !( unmatched[ i ] || setMatched[ i ] ) ) {
-								setMatched[ i ] = pop.call( results );
-							}
-						}
-					}
-
-					// Discard index placeholder values to get only actual matches
-					setMatched = condense( setMatched );
-				}
-
-				// Add matches to results
-				push.apply( results, setMatched );
-
-				// Seedless set matches succeeding multiple successful matchers stipulate sorting
-				if ( outermost && !seed && setMatched.length > 0 &&
-					( matchedCount + setMatchers.length ) > 1 ) {
-
-					Sizzle.uniqueSort( results );
-				}
-			}
-
-			// Override manipulation of globals by nested matchers
-			if ( outermost ) {
-				dirruns = dirrunsUnique;
-				outermostContext = contextBackup;
-			}
-
-			return unmatched;
-		};
-
-	return bySet ?
-		markFunction( superMatcher ) :
-		superMatcher;
-}
-
-compile = Sizzle.compile = function( selector, match /* Internal Use Only */ ) {
-	var i,
-		setMatchers = [],
-		elementMatchers = [],
-		cached = compilerCache[ selector + " " ];
-
-	if ( !cached ) {
-
-		// Generate a function of recursive functions that can be used to check each element
-		if ( !match ) {
-			match = tokenize( selector );
-		}
-		i = match.length;
-		while ( i-- ) {
-			cached = matcherFromTokens( match[ i ] );
-			if ( cached[ expando ] ) {
-				setMatchers.push( cached );
-			} else {
-				elementMatchers.push( cached );
-			}
-		}
-
-		// Cache the compiled function
-		cached = compilerCache(
-			selector,
-			matcherFromGroupMatchers( elementMatchers, setMatchers )
-		);
-
-		// Save selector and tokenization
-		cached.selector = selector;
-	}
-	return cached;
-};
-
-/**
- * A low-level selection function that works with Sizzle's compiled
- *  selector functions
- * @param {String|Function} selector A selector or a pre-compiled
- *  selector function built with Sizzle.compile
- * @param {Element} context
- * @param {Array} [results]
- * @param {Array} [seed] A set of elements to match against
- */
-select = Sizzle.select = function( selector, context, results, seed ) {
-	var i, tokens, token, type, find,
-		compiled = typeof selector === "function" && selector,
-		match = !seed && tokenize( ( selector = compiled.selector || selector ) );
-
-	results = results || [];
-
-	// Try to minimize operations if there is only one selector in the list and no seed
-	// (the latter of which guarantees us context)
-	if ( match.length === 1 ) {
-
-		// Reduce context if the leading compound selector is an ID
-		tokens = match[ 0 ] = match[ 0 ].slice( 0 );
-		if ( tokens.length > 2 && ( token = tokens[ 0 ] ).type === "ID" &&
-			context.nodeType === 9 && documentIsHTML && Expr.relative[ tokens[ 1 ].type ] ) {
-
-			context = ( Expr.find[ "ID" ]( token.matches[ 0 ]
-				.replace( runescape, funescape ), context ) || [] )[ 0 ];
-			if ( !context ) {
-				return results;
-
-			// Precompiled matchers will still verify ancestry, so step up a level
-			} else if ( compiled ) {
-				context = context.parentNode;
-			}
-
-			selector = selector.slice( tokens.shift().value.length );
-		}
-
-		// Fetch a seed set for right-to-left matching
-		i = matchExpr[ "needsContext" ].test( selector ) ? 0 : tokens.length;
-		while ( i-- ) {
-			token = tokens[ i ];
-
-			// Abort if we hit a combinator
-			if ( Expr.relative[ ( type = token.type ) ] ) {
-				break;
-			}
-			if ( ( find = Expr.find[ type ] ) ) {
-
-				// Search, expanding context for leading sibling combinators
-				if ( ( seed = find(
-					token.matches[ 0 ].replace( runescape, funescape ),
-					rsibling.test( tokens[ 0 ].type ) && testContext( context.parentNode ) ||
-						context
-				) ) ) {
-
-					// If seed is empty or no tokens remain, we can return early
-					tokens.splice( i, 1 );
-					selector = seed.length && toSelector( tokens );
-					if ( !selector ) {
-						push.apply( results, seed );
-						return results;
-					}
-
-					break;
-				}
-			}
-		}
-	}
-
-	// Compile and execute a filtering function if one is not provided
-	// Provide `match` to avoid retokenization if we modified the selector above
-	( compiled || compile( selector, match ) )(
-		seed,
-		context,
-		!documentIsHTML,
-		results,
-		!context || rsibling.test( selector ) && testContext( context.parentNode ) || context
-	);
-	return results;
-};
-
-// One-time assignments
-
-// Sort stability
-support.sortStable = expando.split( "" ).sort( sortOrder ).join( "" ) === expando;
-
-// Support: Chrome 14-35+
-// Always assume duplicates if they aren't passed to the comparison function
-support.detectDuplicates = !!hasDuplicate;
-
-// Initialize against the default document
-setDocument();
-
-// Support: Webkit<537.32 - Safari 6.0.3/Chrome 25 (fixed in Chrome 27)
-// Detached nodes confoundingly follow *each other*
-support.sortDetached = assert( function( el ) {
-
-	// Should return 1, but returns 4 (following)
-	return el.compareDocumentPosition( document.createElement( "fieldset" ) ) & 1;
-} );
-
-// Support: IE<8
-// Prevent attribute/property "interpolation"
-// https://msdn.microsoft.com/en-us/library/ms536429%28VS.85%29.aspx
-if ( !assert( function( el ) {
-	el.innerHTML = "<a href='#'></a>";
-	return el.firstChild.getAttribute( "href" ) === "#";
-} ) ) {
-	addHandle( "type|href|height|width", function( elem, name, isXML ) {
-		if ( !isXML ) {
-			return elem.getAttribute( name, name.toLowerCase() === "type" ? 1 : 2 );
-		}
-	} );
-}
-
-// Support: IE<9
-// Use defaultValue in place of getAttribute("value")
-if ( !support.attributes || !assert( function( el ) {
-	el.innerHTML = "<input/>";
-	el.firstChild.setAttribute( "value", "" );
-	return el.firstChild.getAttribute( "value" ) === "";
-} ) ) {
-	addHandle( "value", function( elem, _name, isXML ) {
-		if ( !isXML && elem.nodeName.toLowerCase() === "input" ) {
-			return elem.defaultValue;
-		}
-	} );
-}
-
-// Support: IE<9
-// Use getAttributeNode to fetch booleans when getAttribute lies
-if ( !assert( function( el ) {
-	return el.getAttribute( "disabled" ) == null;
-} ) ) {
-	addHandle( booleans, function( elem, name, isXML ) {
-		var val;
-		if ( !isXML ) {
-			return elem[ name ] === true ? name.toLowerCase() :
-				( val = elem.getAttributeNode( name ) ) && val.specified ?
-					val.value :
-					null;
-		}
-	} );
-}
-
-return Sizzle;
-
-} )( window );
-
-
-
-jQuery.find = Sizzle;
-jQuery.expr = Sizzle.selectors;
-
-// Deprecated
-jQuery.expr[ ":" ] = jQuery.expr.pseudos;
-jQuery.uniqueSort = jQuery.unique = Sizzle.uniqueSort;
-jQuery.text = Sizzle.getText;
-jQuery.isXMLDoc = Sizzle.isXML;
-jQuery.contains = Sizzle.contains;
-jQuery.escapeSelector = Sizzle.escape;
-
-
-
-
-var dir = function( elem, dir, until ) {
-	var matched = [],
-		truncate = until !== undefined;
-
-	while ( ( elem = elem[ dir ] ) && elem.nodeType !== 9 ) {
-		if ( elem.nodeType === 1 ) {
-			if ( truncate && jQuery( elem ).is( until ) ) {
-				break;
-			}
-			matched.push( elem );
-		}
-	}
-	return matched;
-};
-
-
-var siblings = function( n, elem ) {
-	var matched = [];
-
-	for ( ; n; n = n.nextSibling ) {
-		if ( n.nodeType === 1 && n !== elem ) {
-			matched.push( n );
-		}
-	}
-
-	return matched;
-};
-
-
-var rneedsContext = jQuery.expr.match.needsContext;
-
-
-
-function nodeName( elem, name ) {
-
-	return elem.nodeName && elem.nodeName.toLowerCase() === name.toLowerCase();
-
-}
-var rsingleTag = ( /^<([a-z][^\/\0>:\x20\t\r\n\f]*)[\x20\t\r\n\f]*\/?>(?:<\/\1>|)$/i );
-
-
-
-// Implement the identical functionality for filter and not
-function winnow( elements, qualifier, not ) {
-	if ( isFunction( qualifier ) ) {
-		return jQuery.grep( elements, function( elem, i ) {
-			return !!qualifier.call( elem, i, elem ) !== not;
-		} );
-	}
-
-	// Single element
-	if ( qualifier.nodeType ) {
-		return jQuery.grep( elements, function( elem ) {
-			return ( elem === qualifier ) !== not;
-		} );
-	}
-
-	// Arraylike of elements (jQuery, arguments, Array)
-	if ( typeof qualifier !== "string" ) {
-		return jQuery.grep( elements, function( elem ) {
-			return ( indexOf.call( qualifier, elem ) > -1 ) !== not;
-		} );
-	}
-
-	// Filtered directly for both simple and complex selectors
-	return jQuery.filter( qualifier, elements, not );
-}
-
-jQuery.filter = function( expr, elems, not ) {
-	var elem = elems[ 0 ];
-
-	if ( not ) {
-		expr = ":not(" + expr + ")";
-	}
-
-	if ( elems.length === 1 && elem.nodeType === 1 ) {
-		return jQuery.find.matchesSelector( elem, expr ) ? [ elem ] : [];
-	}
-
-	return jQuery.find.matches( expr, jQuery.grep( elems, function( elem ) {
-		return elem.nodeType === 1;
-	} ) );
-};
-
-jQuery.fn.extend( {
-	find: function( selector ) {
-		var i, ret,
-			len = this.length,
-			self = this;
-
-		if ( typeof selector !== "string" ) {
-			return this.pushStack( jQuery( selector ).filter( function() {
-				for ( i = 0; i < len; i++ ) {
-					if ( jQuery.contains( self[ i ], this ) ) {
-						return true;
-					}
-				}
-			} ) );
-		}
-
-		ret = this.pushStack( [] );
-
-		for ( i = 0; i < len; i++ ) {
-			jQuery.find( selector, self[ i ], ret );
-		}
-
-		return len > 1 ? jQuery.uniqueSort( ret ) : ret;
-	},
-	filter: function( selector ) {
-		return this.pushStack( winnow( this, selector || [], false ) );
-	},
-	not: function( selector ) {
-		return this.pushStack( winnow( this, selector || [], true ) );
-	},
-	is: function( selector ) {
-		return !!winnow(
-			this,
-
-			// If this is a positional/relative selector, check membership in the returned set
-			// so $("p:first").is("p:last") won't return true for a doc with two "p".
-			typeof selector === "string" && rneedsContext.test( selector ) ?
-				jQuery( selector ) :
-				selector || [],
-			false
-		).length;
-	}
-} );
-
-
-// Initialize a jQuery object
-
-
-// A central reference to the root jQuery(document)
-var rootjQuery,
-
-	// A simple way to check for HTML strings
-	// Prioritize #id over <tag> to avoid XSS via location.hash (#9521)
-	// Strict HTML recognition (#11290: must start with <)
-	// Shortcut simple #id case for speed
-	rquickExpr = /^(?:\s*(<[\w\W]+>)[^>]*|#([\w-]+))$/,
-
-	init = jQuery.fn.init = function( selector, context, root ) {
-		var match, elem;
-
-		// HANDLE: $(""), $(null), $(undefined), $(false)
-		if ( !selector ) {
-			return this;
-		}
-
-		// Method init() accepts an alternate rootjQuery
-		// so migrate can support jQuery.sub (gh-2101)
-		root = root || rootjQuery;
-
-		// Handle HTML strings
-		if ( typeof selector === "string" ) {
-			if ( selector[ 0 ] === "<" &&
-				selector[ selector.length - 1 ] === ">" &&
-				selector.length >= 3 ) {
-
-				// Assume that strings that start and end with <> are HTML and skip the regex check
-				match = [ null, selector, null ];
-
-			} else {
-				match = rquickExpr.exec( selector );
-			}
-
-			// Match html or make sure no context is specified for #id
-			if ( match && ( match[ 1 ] || !context ) ) {
-
-				// HANDLE: $(html) -> $(array)
-				if ( match[ 1 ] ) {
-					context = context instanceof jQuery ? context[ 0 ] : context;
-
-					// Option to run scripts is true for back-compat
-					// Intentionally let the error be thrown if parseHTML is not present
-					jQuery.merge( this, jQuery.parseHTML(
-						match[ 1 ],
-						context && context.nodeType ? context.ownerDocument || context : document,
-						true
-					) );
-
-					// HANDLE: $(html, props)
-					if ( rsingleTag.test( match[ 1 ] ) && jQuery.isPlainObject( context ) ) {
-						for ( match in context ) {
-
-							// Properties of context are called as methods if possible
-							if ( isFunction( this[ match ] ) ) {
-								this[ match ]( context[ match ] );
-
-							// ...and otherwise set as attributes
-							} else {
-								this.attr( match, context[ match ] );
-							}
-						}
-					}
-
-					return this;
-
-				// HANDLE: $(#id)
-				} else {
-					elem = document.getElementById( match[ 2 ] );
-
-					if ( elem ) {
-
-						// Inject the element directly into the jQuery object
-						this[ 0 ] = elem;
-						this.length = 1;
-					}
-					return this;
-				}
-
-			// HANDLE: $(expr, $(...))
-			} else if ( !context || context.jquery ) {
-				return ( context || root ).find( selector );
-
-			// HANDLE: $(expr, context)
-			// (which is just equivalent to: $(context).find(expr)
-			} else {
-				return this.constructor( context ).find( selector );
-			}
-
-		// HANDLE: $(DOMElement)
-		} else if ( selector.nodeType ) {
-			this[ 0 ] = selector;
-			this.length = 1;
-			return this;
-
-		// HANDLE: $(function)
-		// Shortcut for document ready
-		} else if ( isFunction( selector ) ) {
-			return root.ready !== undefined ?
-				root.ready( selector ) :
-
-				// Execute immediately if ready is not present
-				selector( jQuery );
-		}
-
-		return jQuery.makeArray( selector, this );
-	};
-
-// Give the init function the jQuery prototype for later instantiation
-init.prototype = jQuery.fn;
-
-// Initialize central reference
-rootjQuery = jQuery( document );
-
-
-var rparentsprev = /^(?:parents|prev(?:Until|All))/,
-
-	// Methods guaranteed to produce a unique set when starting from a unique set
-	guaranteedUnique = {
-		children: true,
-		contents: true,
-		next: true,
-		prev: true
-	};
-
-jQuery.fn.extend( {
-	has: function( target ) {
-		var targets = jQuery( target, this ),
-			l = targets.length;
-
-		return this.filter( function() {
-			var i = 0;
-			for ( ; i < l; i++ ) {
-				if ( jQuery.contains( this, targets[ i ] ) ) {
-					return true;
-				}
-			}
-		} );
-	},
-
-	closest: function( selectors, context ) {
-		var cur,
-			i = 0,
-			l = this.length,
-			matched = [],
-			targets = typeof selectors !== "string" && jQuery( selectors );
-
-		// Positional selectors never match, since there's no _selection_ context
-		if ( !rneedsContext.test( selectors ) ) {
-			for ( ; i < l; i++ ) {
-				for ( cur = this[ i ]; cur && cur !== context; cur = cur.parentNode ) {
-
-					// Always skip document fragments
-					if ( cur.nodeType < 11 && ( targets ?
-						targets.index( cur ) > -1 :
-
-						// Don't pass non-elements to Sizzle
-						cur.nodeType === 1 &&
-							jQuery.find.matchesSelector( cur, selectors ) ) ) {
-
-						matched.push( cur );
-						break;
-					}
-				}
-			}
-		}
-
-		return this.pushStack( matched.length > 1 ? jQuery.uniqueSort( matched ) : matched );
-	},
-
-	// Determine the position of an element within the set
-	index: function( elem ) {
-
-		// No argument, return index in parent
-		if ( !elem ) {
-			return ( this[ 0 ] && this[ 0 ].parentNode ) ? this.first().prevAll().length : -1;
-		}
-
-		// Index in selector
-		if ( typeof elem === "string" ) {
-			return indexOf.call( jQuery( elem ), this[ 0 ] );
-		}
-
-		// Locate the position of the desired element
-		return indexOf.call( this,
-
-			// If it receives a jQuery object, the first element is used
-			elem.jquery ? elem[ 0 ] : elem
-		);
-	},
-
-	add: function( selector, context ) {
-		return this.pushStack(
-			jQuery.uniqueSort(
-				jQuery.merge( this.get(), jQuery( selector, context ) )
-			)
-		);
-	},
-
-	addBack: function( selector ) {
-		return this.add( selector == null ?
-			this.prevObject : this.prevObject.filter( selector )
-		);
-	}
-} );
-
-function sibling( cur, dir ) {
-	while ( ( cur = cur[ dir ] ) && cur.nodeType !== 1 ) {}
-	return cur;
-}
-
-jQuery.each( {
-	parent: function( elem ) {
-		var parent = elem.parentNode;
-		return parent && parent.nodeType !== 11 ? parent : null;
-	},
-	parents: function( elem ) {
-		return dir( elem, "parentNode" );
-	},
-	parentsUntil: function( elem, _i, until ) {
-		return dir( elem, "parentNode", until );
-	},
-	next: function( elem ) {
-		return sibling( elem, "nextSibling" );
-	},
-	prev: function( elem ) {
-		return sibling( elem, "previousSibling" );
-	},
-	nextAll: function( elem ) {
-		return dir( elem, "nextSibling" );
-	},
-	prevAll: function( elem ) {
-		return dir( elem, "previousSibling" );
-	},
-	nextUntil: function( elem, _i, until ) {
-		return dir( elem, "nextSibling", until );
-	},
-	prevUntil: function( elem, _i, until ) {
-		return dir( elem, "previousSibling", until );
-	},
-	siblings: function( elem ) {
-		return siblings( ( elem.parentNode || {} ).firstChild, elem );
-	},
-	children: function( elem ) {
-		return siblings( elem.firstChild );
-	},
-	contents: function( elem ) {
-		if ( elem.contentDocument != null &&
-
-			// Support: IE 11+
-			// <object> elements with no `data` attribute has an object
-			// `contentDocument` with a `null` prototype.
-			getProto( elem.contentDocument ) ) {
-
-			return elem.contentDocument;
-		}
-
-		// Support: IE 9 - 11 only, iOS 7 only, Android Browser <=4.3 only
-		// Treat the template element as a regular one in browsers that
-		// don't support it.
-		if ( nodeName( elem, "template" ) ) {
-			elem = elem.content || elem;
-		}
-
-		return jQuery.merge( [], elem.childNodes );
-	}
-}, function( name, fn ) {
-	jQuery.fn[ name ] = function( until, selector ) {
-		var matched = jQuery.map( this, fn, until );
-
-		if ( name.slice( -5 ) !== "Until" ) {
-			selector = until;
-		}
-
-		if ( selector && typeof selector === "string" ) {
-			matched = jQuery.filter( selector, matched );
-		}
-
-		if ( this.length > 1 ) {
-
-			// Remove duplicates
-			if ( !guaranteedUnique[ name ] ) {
-				jQuery.uniqueSort( matched );
-			}
-
-			// Reverse order for parents* and prev-derivatives
-			if ( rparentsprev.test( name ) ) {
-				matched.reverse();
-			}
-		}
-
-		return this.pushStack( matched );
-	};
-} );
-var rnothtmlwhite = ( /[^\x20\t\r\n\f]+/g );
-
-
-
-// Convert String-formatted options into Object-formatted ones
-function createOptions( options ) {
-	var object = {};
-	jQuery.each( options.match( rnothtmlwhite ) || [], function( _, flag ) {
-		object[ flag ] = true;
-	} );
-	return object;
-}
-
-/*
- * Create a callback list using the following parameters:
- *
- *	options: an optional list of space-separated options that will change how
- *			the callback list behaves or a more traditional option object
- *
- * By default a callback list will act like an event callback list and can be
- * "fired" multiple times.
- *
- * Possible options:
- *
- *	once:			will ensure the callback list can only be fired once (like a Deferred)
- *
- *	memory:			will keep track of previous values and will call any callback added
- *					after the list has been fired right away with the latest "memorized"
- *					values (like a Deferred)
- *
- *	unique:			will ensure a callback can only be added once (no duplicate in the list)
- *
- *	stopOnFalse:	interrupt callings when a callback returns false
- *
- */
-jQuery.Callbacks = function( options ) {
-
-	// Convert options from String-formatted to Object-formatted if needed
-	// (we check in cache first)
-	options = typeof options === "string" ?
-		createOptions( options ) :
-		jQuery.extend( {}, options );
-
-	var // Flag to know if list is currently firing
-		firing,
-
-		// Last fire value for non-forgettable lists
-		memory,
-
-		// Flag to know if list was already fired
-		fired,
-
-		// Flag to prevent firing
-		locked,
-
-		// Actual callback list
-		list = [],
-
-		// Queue of execution data for repeatable lists
-		queue = [],
-
-		// Index of currently firing callback (modified by add/remove as needed)
-		firingIndex = -1,
-
-		// Fire callbacks
-		fire = function() {
-
-			// Enforce single-firing
-			locked = locked || options.once;
-
-			// Execute callbacks for all pending executions,
-			// respecting firingIndex overrides and runtime changes
-			fired = firing = true;
-			for ( ; queue.length; firingIndex = -1 ) {
-				memory = queue.shift();
-				while ( ++firingIndex < list.length ) {
-
-					// Run callback and check for early termination
-					if ( list[ firingIndex ].apply( memory[ 0 ], memory[ 1 ] ) === false &&
-						options.stopOnFalse ) {
-
-						// Jump to end and forget the data so .add doesn't re-fire
-						firingIndex = list.length;
-						memory = false;
-					}
-				}
-			}
-
-			// Forget the data if we're done with it
-			if ( !options.memory ) {
-				memory = false;
-			}
-
-			firing = false;
-
-			// Clean up if we're done firing for good
-			if ( locked ) {
-
-				// Keep an empty list if we have data for future add calls
-				if ( memory ) {
-					list = [];
-
-				// Otherwise, this object is spent
-				} else {
-					list = "";
-				}
-			}
-		},
-
-		// Actual Callbacks object
-		self = {
-
-			// Add a callback or a collection of callbacks to the list
-			add: function() {
-				if ( list ) {
-
-					// If we have memory from a past run, we should fire after adding
-					if ( memory && !firing ) {
-						firingIndex = list.length - 1;
-						queue.push( memory );
-					}
-
-					( function add( args ) {
-						jQuery.each( args, function( _, arg ) {
-							if ( isFunction( arg ) ) {
-								if ( !options.unique || !self.has( arg ) ) {
-									list.push( arg );
-								}
-							} else if ( arg && arg.length && toType( arg ) !== "string" ) {
-
-								// Inspect recursively
-								add( arg );
-							}
-						} );
-					} )( arguments );
-
-					if ( memory && !firing ) {
-						fire();
-					}
-				}
-				return this;
-			},
-
-			// Remove a callback from the list
-			remove: function() {
-				jQuery.each( arguments, function( _, arg ) {
-					var index;
-					while ( ( index = jQuery.inArray( arg, list, index ) ) > -1 ) {
-						list.splice( index, 1 );
-
-						// Handle firing indexes
-						if ( index <= firingIndex ) {
-							firingIndex--;
-						}
-					}
-				} );
-				return this;
-			},
-
-			// Check if a given callback is in the list.
-			// If no argument is given, return whether or not list has callbacks attached.
-			has: function( fn ) {
-				return fn ?
-					jQuery.inArray( fn, list ) > -1 :
-					list.length > 0;
-			},
-
-			// Remove all callbacks from the list
-			empty: function() {
-				if ( list ) {
-					list = [];
-				}
-				return this;
-			},
-
-			// Disable .fire and .add
-			// Abort any current/pending executions
-			// Clear all callbacks and values
-			disable: function() {
-				locked = queue = [];
-				list = memory = "";
-				return this;
-			},
-			disabled: function() {
-				return !list;
-			},
-
-			// Disable .fire
-			// Also disable .add unless we have memory (since it would have no effect)
-			// Abort any pending executions
-			lock: function() {
-				locked = queue = [];
-				if ( !memory && !firing ) {
-					list = memory = "";
-				}
-				return this;
-			},
-			locked: function() {
-				return !!locked;
-			},
-
-			// Call all callbacks with the given context and arguments
-			fireWith: function( context, args ) {
-				if ( !locked ) {
-					args = args || [];
-					args = [ context, args.slice ? args.slice() : args ];
-					queue.push( args );
-					if ( !firing ) {
-						fire();
-					}
-				}
-				return this;
-			},
-
-			// Call all the callbacks with the given arguments
-			fire: function() {
-				self.fireWith( this, arguments );
-				return this;
-			},
-
-			// To know if the callbacks have already been called at least once
-			fired: function() {
-				return !!fired;
-			}
-		};
-
-	return self;
-};
-
-
-function Identity( v ) {
-	return v;
-}
-function Thrower( ex ) {
-	throw ex;
-}
-
-function adoptValue( value, resolve, reject, noValue ) {
-	var method;
-
-	try {
-
-		// Check for promise aspect first to privilege synchronous behavior
-		if ( value && isFunction( ( method = value.promise ) ) ) {
-			method.call( value ).done( resolve ).fail( reject );
-
-		// Other thenables
-		} else if ( value && isFunction( ( method = value.then ) ) ) {
-			method.call( value, resolve, reject );
-
-		// Other non-thenables
-		} else {
-
-			// Control `resolve` arguments by letting Array#slice cast boolean `noValue` to integer:
-			// * false: [ value ].slice( 0 ) => resolve( value )
-			// * true: [ value ].slice( 1 ) => resolve()
-			resolve.apply( undefined, [ value ].slice( noValue ) );
-		}
-
-	// For Promises/A+, convert exceptions into rejections
-	// Since jQuery.when doesn't unwrap thenables, we can skip the extra checks appearing in
-	// Deferred#then to conditionally suppress rejection.
-	} catch ( value ) {
-
-		// Support: Android 4.0 only
-		// Strict mode functions invoked without .call/.apply get global-object context
-		reject.apply( undefined, [ value ] );
-	}
-}
-
-jQuery.extend( {
-
-	Deferred: function( func ) {
-		var tuples = [
-
-				// action, add listener, callbacks,
-				// ... .then handlers, argument index, [final state]
-				[ "notify", "progress", jQuery.Callbacks( "memory" ),
-					jQuery.Callbacks( "memory" ), 2 ],
-				[ "resolve", "done", jQuery.Callbacks( "once memory" ),
-					jQuery.Callbacks( "once memory" ), 0, "resolved" ],
-				[ "reject", "fail", jQuery.Callbacks( "once memory" ),
-					jQuery.Callbacks( "once memory" ), 1, "rejected" ]
-			],
-			state = "pending",
-			promise = {
-				state: function() {
-					return state;
-				},
-				always: function() {
-					deferred.done( arguments ).fail( arguments );
-					return this;
-				},
-				"catch": function( fn ) {
-					return promise.then( null, fn );
-				},
-
-				// Keep pipe for back-compat
-				pipe: function( /* fnDone, fnFail, fnProgress */ ) {
-					var fns = arguments;
-
-					return jQuery.Deferred( function( newDefer ) {
-						jQuery.each( tuples, function( _i, tuple ) {
-
-							// Map tuples (progress, done, fail) to arguments (done, fail, progress)
-							var fn = isFunction( fns[ tuple[ 4 ] ] ) && fns[ tuple[ 4 ] ];
-
-							// deferred.progress(function() { bind to newDefer or newDefer.notify })
-							// deferred.done(function() { bind to newDefer or newDefer.resolve })
-							// deferred.fail(function() { bind to newDefer or newDefer.reject })
-							deferred[ tuple[ 1 ] ]( function() {
-								var returned = fn && fn.apply( this, arguments );
-								if ( returned && isFunction( returned.promise ) ) {
-									returned.promise()
-										.progress( newDefer.notify )
-										.done( newDefer.resolve )
-										.fail( newDefer.reject );
-								} else {
-									newDefer[ tuple[ 0 ] + "With" ](
-										this,
-										fn ? [ returned ] : arguments
-									);
-								}
-							} );
-						} );
-						fns = null;
-					} ).promise();
-				},
-				then: function( onFulfilled, onRejected, onProgress ) {
-					var maxDepth = 0;
-					function resolve( depth, deferred, handler, special ) {
-						return function() {
-							var that = this,
-								args = arguments,
-								mightThrow = function() {
-									var returned, then;
-
-									// Support: Promises/A+ section 2.3.3.3.3
-									// https://promisesaplus.com/#point-59
-									// Ignore double-resolution attempts
-									if ( depth < maxDepth ) {
-										return;
-									}
-
-									returned = handler.apply( that, args );
-
-									// Support: Promises/A+ section 2.3.1
-									// https://promisesaplus.com/#point-48
-									if ( returned === deferred.promise() ) {
-										throw new TypeError( "Thenable self-resolution" );
-									}
-
-									// Support: Promises/A+ sections 2.3.3.1, 3.5
-									// https://promisesaplus.com/#point-54
-									// https://promisesaplus.com/#point-75
-									// Retrieve `then` only once
-									then = returned &&
-
-										// Support: Promises/A+ section 2.3.4
-										// https://promisesaplus.com/#point-64
-										// Only check objects and functions for thenability
-										( typeof returned === "object" ||
-											typeof returned === "function" ) &&
-										returned.then;
-
-									// Handle a returned thenable
-									if ( isFunction( then ) ) {
-
-										// Special processors (notify) just wait for resolution
-										if ( special ) {
-											then.call(
-												returned,
-												resolve( maxDepth, deferred, Identity, special ),
-												resolve( maxDepth, deferred, Thrower, special )
-											);
-
-										// Normal processors (resolve) also hook into progress
-										} else {
-
-											// ...and disregard older resolution values
-											maxDepth++;
-
-											then.call(
-												returned,
-												resolve( maxDepth, deferred, Identity, special ),
-												resolve( maxDepth, deferred, Thrower, special ),
-												resolve( maxDepth, deferred, Identity,
-													deferred.notifyWith )
-											);
-										}
-
-									// Handle all other returned values
-									} else {
-
-										// Only substitute handlers pass on context
-										// and multiple values (non-spec behavior)
-										if ( handler !== Identity ) {
-											that = undefined;
-											args = [ returned ];
-										}
-
-										// Process the value(s)
-										// Default process is resolve
-										( special || deferred.resolveWith )( that, args );
-									}
-								},
-
-								// Only normal processors (resolve) catch and reject exceptions
-								process = special ?
-									mightThrow :
-									function() {
-										try {
-											mightThrow();
-										} catch ( e ) {
-
-											if ( jQuery.Deferred.exceptionHook ) {
-												jQuery.Deferred.exceptionHook( e,
-													process.stackTrace );
-											}
-
-											// Support: Promises/A+ section 2.3.3.3.4.1
-											// https://promisesaplus.com/#point-61
-											// Ignore post-resolution exceptions
-											if ( depth + 1 >= maxDepth ) {
-
-												// Only substitute handlers pass on context
-												// and multiple values (non-spec behavior)
-												if ( handler !== Thrower ) {
-													that = undefined;
-													args = [ e ];
-												}
-
-												deferred.rejectWith( that, args );
-											}
-										}
-									};
-
-							// Support: Promises/A+ section 2.3.3.3.1
-							// https://promisesaplus.com/#point-57
-							// Re-resolve promises immediately to dodge false rejection from
-							// subsequent errors
-							if ( depth ) {
-								process();
-							} else {
-
-								// Call an optional hook to record the stack, in case of exception
-								// since it's otherwise lost when execution goes async
-								if ( jQuery.Deferred.getStackHook ) {
-									process.stackTrace = jQuery.Deferred.getStackHook();
-								}
-								window.setTimeout( process );
-							}
-						};
-					}
-
-					return jQuery.Deferred( function( newDefer ) {
-
-						// progress_handlers.add( ... )
-						tuples[ 0 ][ 3 ].add(
-							resolve(
-								0,
-								newDefer,
-								isFunction( onProgress ) ?
-									onProgress :
-									Identity,
-								newDefer.notifyWith
-							)
-						);
-
-						// fulfilled_handlers.add( ... )
-						tuples[ 1 ][ 3 ].add(
-							resolve(
-								0,
-								newDefer,
-								isFunction( onFulfilled ) ?
-									onFulfilled :
-									Identity
-							)
-						);
-
-						// rejected_handlers.add( ... )
-						tuples[ 2 ][ 3 ].add(
-							resolve(
-								0,
-								newDefer,
-								isFunction( onRejected ) ?
-									onRejected :
-									Thrower
-							)
-						);
-					} ).promise();
-				},
-
-				// Get a promise for this deferred
-				// If obj is provided, the promise aspect is added to the object
-				promise: function( obj ) {
-					return obj != null ? jQuery.extend( obj, promise ) : promise;
-				}
-			},
-			deferred = {};
-
-		// Add list-specific methods
-		jQuery.each( tuples, function( i, tuple ) {
-			var list = tuple[ 2 ],
-				stateString = tuple[ 5 ];
-
-			// promise.progress = list.add
-			// promise.done = list.add
-			// promise.fail = list.add
-			promise[ tuple[ 1 ] ] = list.add;
-
-			// Handle state
-			if ( stateString ) {
-				list.add(
-					function() {
-
-						// state = "resolved" (i.e., fulfilled)
-						// state = "rejected"
-						state = stateString;
-					},
-
-					// rejected_callbacks.disable
-					// fulfilled_callbacks.disable
-					tuples[ 3 - i ][ 2 ].disable,
-
-					// rejected_handlers.disable
-					// fulfilled_handlers.disable
-					tuples[ 3 - i ][ 3 ].disable,
-
-					// progress_callbacks.lock
-					tuples[ 0 ][ 2 ].lock,
-
-					// progress_handlers.lock
-					tuples[ 0 ][ 3 ].lock
-				);
-			}
-
-			// progress_handlers.fire
-			// fulfilled_handlers.fire
-			// rejected_handlers.fire
-			list.add( tuple[ 3 ].fire );
-
-			// deferred.notify = function() { deferred.notifyWith(...) }
-			// deferred.resolve = function() { deferred.resolveWith(...) }
-			// deferred.reject = function() { deferred.rejectWith(...) }
-			deferred[ tuple[ 0 ] ] = function() {
-				deferred[ tuple[ 0 ] + "With" ]( this === deferred ? undefined : this, arguments );
-				return this;
-			};
-
-			// deferred.notifyWith = list.fireWith
-			// deferred.resolveWith = list.fireWith
-			// deferred.rejectWith = list.fireWith
-			deferred[ tuple[ 0 ] + "With" ] = list.fireWith;
-		} );
-
-		// Make the deferred a promise
-		promise.promise( deferred );
-
-		// Call given func if any
-		if ( func ) {
-			func.call( deferred, deferred );
-		}
-
-		// All done!
-		return deferred;
-	},
-
-	// Deferred helper
-	when: function( singleValue ) {
-		var
-
-			// count of uncompleted subordinates
-			remaining = arguments.length,
-
-			// count of unprocessed arguments
-			i = remaining,
-
-			// subordinate fulfillment data
-			resolveContexts = Array( i ),
-			resolveValues = slice.call( arguments ),
-
-			// the primary Deferred
-			primary = jQuery.Deferred(),
-
-			// subordinate callback factory
-			updateFunc = function( i ) {
-				return function( value ) {
-					resolveContexts[ i ] = this;
-					resolveValues[ i ] = arguments.length > 1 ? slice.call( arguments ) : value;
-					if ( !( --remaining ) ) {
-						primary.resolveWith( resolveContexts, resolveValues );
-					}
-				};
-			};
-
-		// Single- and empty arguments are adopted like Promise.resolve
-		if ( remaining <= 1 ) {
-			adoptValue( singleValue, primary.done( updateFunc( i ) ).resolve, primary.reject,
-				!remaining );
-
-			// Use .then() to unwrap secondary thenables (cf. gh-3000)
-			if ( primary.state() === "pending" ||
-				isFunction( resolveValues[ i ] && resolveValues[ i ].then ) ) {
-
-				return primary.then();
-			}
-		}
-
-		// Multiple arguments are aggregated like Promise.all array elements
-		while ( i-- ) {
-			adoptValue( resolveValues[ i ], updateFunc( i ), primary.reject );
-		}
-
-		return primary.promise();
-	}
-} );
-
-
-// These usually indicate a programmer mistake during development,
-// warn about them ASAP rather than swallowing them by default.
-var rerrorNames = /^(Eval|Internal|Range|Reference|Syntax|Type|URI)Error$/;
-
-jQuery.Deferred.exceptionHook = function( error, stack ) {
-
-	// Support: IE 8 - 9 only
-	// Console exists when dev tools are open, which can happen at any time
-	if ( window.console && window.console.warn && error && rerrorNames.test( error.name ) ) {
-		window.console.warn( "jQuery.Deferred exception: " + error.message, error.stack, stack );
-	}
-};
-
-
-
-
-jQuery.readyException = function( error ) {
-	window.setTimeout( function() {
-		throw error;
-	} );
-};
-
-
-
-
-// The deferred used on DOM ready
-var readyList = jQuery.Deferred();
-
-jQuery.fn.ready = function( fn ) {
-
-	readyList
-		.then( fn )
-
-		// Wrap jQuery.readyException in a function so that the lookup
-		// happens at the time of error handling instead of callback
-		// registration.
-		.catch( function( error ) {
-			jQuery.readyException( error );
-		} );
-
-	return this;
-};
-
-jQuery.extend( {
-
-	// Is the DOM ready to be used? Set to true once it occurs.
-	isReady: false,
-
-	// A counter to track how many items to wait for before
-	// the ready event fires. See #6781
-	readyWait: 1,
-
-	// Handle when the DOM is ready
-	ready: function( wait ) {
-
-		// Abort if there are pending holds or we're already ready
-		if ( wait === true ? --jQuery.readyWait : jQuery.isReady ) {
-			return;
-		}
-
-		// Remember that the DOM is ready
-		jQuery.isReady = true;
-
-		// If a normal DOM Ready event fired, decrement, and wait if need be
-		if ( wait !== true && --jQuery.readyWait > 0 ) {
-			return;
-		}
-
-		// If there are functions bound, to execute
-		readyList.resolveWith( document, [ jQuery ] );
-	}
-} );
-
-jQuery.ready.then = readyList.then;
-
-// The ready event handler and self cleanup method
-function completed() {
-	document.removeEventListener( "DOMContentLoaded", completed );
-	window.removeEventListener( "load", completed );
-	jQuery.ready();
-}
-
-// Catch cases where $(document).ready() is called
-// after the browser event has already occurred.
-// Support: IE <=9 - 10 only
-// Older IE sometimes signals "interactive" too soon
-if ( document.readyState === "complete" ||
-	( document.readyState !== "loading" && !document.documentElement.doScroll ) ) {
-
-	// Handle it asynchronously to allow scripts the opportunity to delay ready
-	window.setTimeout( jQuery.ready );
-
-} else {
-
-	// Use the handy event callback
-	document.addEventListener( "DOMContentLoaded", completed );
-
-	// A fallback to window.onload, that will always work
-	window.addEventListener( "load", completed );
-}
-
-
-
-
-// Multifunctional method to get and set values of a collection
-// The value/s can optionally be executed if it's a function
-var access = function( elems, fn, key, value, chainable, emptyGet, raw ) {
-	var i = 0,
-		len = elems.length,
-		bulk = key == null;
-
-	// Sets many values
-	if ( toType( key ) === "object" ) {
-		chainable = true;
-		for ( i in key ) {
-			access( elems, fn, i, key[ i ], true, emptyGet, raw );
-		}
-
-	// Sets one value
-	} else if ( value !== undefined ) {
-		chainable = true;
-
-		if ( !isFunction( value ) ) {
-			raw = true;
-		}
-
-		if ( bulk ) {
-
-			// Bulk operations run against the entire set
-			if ( raw ) {
-				fn.call( elems, value );
-				fn = null;
-
-			// ...except when executing function values
-			} else {
-				bulk = fn;
-				fn = function( elem, _key, value ) {
-					return bulk.call( jQuery( elem ), value );
-				};
-			}
-		}
-
-		if ( fn ) {
-			for ( ; i < len; i++ ) {
-				fn(
-					elems[ i ], key, raw ?
-						value :
-						value.call( elems[ i ], i, fn( elems[ i ], key ) )
-				);
-			}
-		}
-	}
-
-	if ( chainable ) {
-		return elems;
-	}
-
-	// Gets
-	if ( bulk ) {
-		return fn.call( elems );
-	}
-
-	return len ? fn( elems[ 0 ], key ) : emptyGet;
-};
-
-
-// Matches dashed string for camelizing
-var rmsPrefix = /^-ms-/,
-	rdashAlpha = /-([a-z])/g;
-
-// Used by camelCase as callback to replace()
-function fcamelCase( _all, letter ) {
-	return letter.toUpperCase();
-}
-
-// Convert dashed to camelCase; used by the css and data modules
-// Support: IE <=9 - 11, Edge 12 - 15
-// Microsoft forgot to hump their vendor prefix (#9572)
-function camelCase( string ) {
-	return string.replace( rmsPrefix, "ms-" ).replace( rdashAlpha, fcamelCase );
-}
-var acceptData = function( owner ) {
-
-	// Accepts only:
-	//  - Node
-	//    - Node.ELEMENT_NODE
-	//    - Node.DOCUMENT_NODE
-	//  - Object
-	//    - Any
-	return owner.nodeType === 1 || owner.nodeType === 9 || !( +owner.nodeType );
-};
-
-
-
-
-function Data() {
-	this.expando = jQuery.expando + Data.uid++;
-}
-
-Data.uid = 1;
-
-Data.prototype = {
-
-	cache: function( owner ) {
-
-		// Check if the owner object already has a cache
-		var value = owner[ this.expando ];
-
-		// If not, create one
-		if ( !value ) {
-			value = {};
-
-			// We can accept data for non-element nodes in modern browsers,
-			// but we should not, see #8335.
-			// Always return an empty object.
-			if ( acceptData( owner ) ) {
-
-				// If it is a node unlikely to be stringify-ed or looped over
-				// use plain assignment
-				if ( owner.nodeType ) {
-					owner[ this.expando ] = value;
-
-				// Otherwise secure it in a non-enumerable property
-				// configurable must be true to allow the property to be
-				// deleted when data is removed
-				} else {
-					Object.defineProperty( owner, this.expando, {
-						value: value,
-						configurable: true
-					} );
-				}
-			}
-		}
-
-		return value;
-	},
-	set: function( owner, data, value ) {
-		var prop,
-			cache = this.cache( owner );
-
-		// Handle: [ owner, key, value ] args
-		// Always use camelCase key (gh-2257)
-		if ( typeof data === "string" ) {
-			cache[ camelCase( data ) ] = value;
-
-		// Handle: [ owner, { properties } ] args
-		} else {
-
-			// Copy the properties one-by-one to the cache object
-			for ( prop in data ) {
-				cache[ camelCase( prop ) ] = data[ prop ];
-			}
-		}
-		return cache;
-	},
-	get: function( owner, key ) {
-		return key === undefined ?
-			this.cache( owner ) :
-
-			// Always use camelCase key (gh-2257)
-			owner[ this.expando ] && owner[ this.expando ][ camelCase( key ) ];
-	},
-	access: function( owner, key, value ) {
-
-		// In cases where either:
-		//
-		//   1. No key was specified
-		//   2. A string key was specified, but no value provided
-		//
-		// Take the "read" path and allow the get method to determine
-		// which value to return, respectively either:
-		//
-		//   1. The entire cache object
-		//   2. The data stored at the key
-		//
-		if ( key === undefined ||
-				( ( key && typeof key === "string" ) && value === undefined ) ) {
-
-			return this.get( owner, key );
-		}
-
-		// When the key is not a string, or both a key and value
-		// are specified, set or extend (existing objects) with either:
-		//
-		//   1. An object of properties
-		//   2. A key and value
-		//
-		this.set( owner, key, value );
-
-		// Since the "set" path can have two possible entry points
-		// return the expected data based on which path was taken[*]
-		return value !== undefined ? value : key;
-	},
-	remove: function( owner, key ) {
-		var i,
-			cache = owner[ this.expando ];
-
-		if ( cache === undefined ) {
-			return;
-		}
-
-		if ( key !== undefined ) {
-
-			// Support array or space separated string of keys
-			if ( Array.isArray( key ) ) {
-
-				// If key is an array of keys...
-				// We always set camelCase keys, so remove that.
-				key = key.map( camelCase );
-			} else {
-				key = camelCase( key );
-
-				// If a key with the spaces exists, use it.
-				// Otherwise, create an array by matching non-whitespace
-				key = key in cache ?
-					[ key ] :
-					( key.match( rnothtmlwhite ) || [] );
-			}
-
-			i = key.length;
-
-			while ( i-- ) {
-				delete cache[ key[ i ] ];
-			}
-		}
-
-		// Remove the expando if there's no more data
-		if ( key === undefined || jQuery.isEmptyObject( cache ) ) {
-
-			// Support: Chrome <=35 - 45
-			// Webkit & Blink performance suffers when deleting properties
-			// from DOM nodes, so set to undefined instead
-			// https://bugs.chromium.org/p/chromium/issues/detail?id=378607 (bug restricted)
-			if ( owner.nodeType ) {
-				owner[ this.expando ] = undefined;
-			} else {
-				delete owner[ this.expando ];
-			}
-		}
-	},
-	hasData: function( owner ) {
-		var cache = owner[ this.expando ];
-		return cache !== undefined && !jQuery.isEmptyObject( cache );
-	}
-};
-var dataPriv = new Data();
-
-var dataUser = new Data();
-
-
-
-//	Implementation Summary
-//
-//	1. Enforce API surface and semantic compatibility with 1.9.x branch
-//	2. Improve the module's maintainability by reducing the storage
-//		paths to a single mechanism.
-//	3. Use the same single mechanism to support "private" and "user" data.
-//	4. _Never_ expose "private" data to user code (TODO: Drop _data, _removeData)
-//	5. Avoid exposing implementation details on user objects (eg. expando properties)
-//	6. Provide a clear path for implementation upgrade to WeakMap in 2014
-
-var rbrace = /^(?:\{[\w\W]*\}|\[[\w\W]*\])$/,
-	rmultiDash = /[A-Z]/g;
-
-function getData( data ) {
-	if ( data === "true" ) {
-		return true;
-	}
-
-	if ( data === "false" ) {
-		return false;
-	}
-
-	if ( data === "null" ) {
-		return null;
-	}
-
-	// Only convert to a number if it doesn't change the string
-	if ( data === +data + "" ) {
-		return +data;
-	}
-
-	if ( rbrace.test( data ) ) {
-		return JSON.parse( data );
-	}
-
-	return data;
-}
-
-function dataAttr( elem, key, data ) {
-	var name;
-
-	// If nothing was found internally, try to fetch any
-	// data from the HTML5 data-* attribute
-	if ( data === undefined && elem.nodeType === 1 ) {
-		name = "data-" + key.replace( rmultiDash, "-$&" ).toLowerCase();
-		data = elem.getAttribute( name );
-
-		if ( typeof data === "string" ) {
-			try {
-				data = getData( data );
-			} catch ( e ) {}
-
-			// Make sure we set the data so it isn't changed later
-			dataUser.set( elem, key, data );
-		} else {
-			data = undefined;
-		}
-	}
-	return data;
-}
-
-jQuery.extend( {
-	hasData: function( elem ) {
-		return dataUser.hasData( elem ) || dataPriv.hasData( elem );
-	},
-
-	data: function( elem, name, data ) {
-		return dataUser.access( elem, name, data );
-	},
-
-	removeData: function( elem, name ) {
-		dataUser.remove( elem, name );
-	},
-
-	// TODO: Now that all calls to _data and _removeData have been replaced
-	// with direct calls to dataPriv methods, these can be deprecated.
-	_data: function( elem, name, data ) {
-		return dataPriv.access( elem, name, data );
-	},
-
-	_removeData: function( elem, name ) {
-		dataPriv.remove( elem, name );
-	}
-} );
-
-jQuery.fn.extend( {
-	data: function( key, value ) {
-		var i, name, data,
-			elem = this[ 0 ],
-			attrs = elem && elem.attributes;
-
-		// Gets all values
-		if ( key === undefined ) {
-			if ( this.length ) {
-				data = dataUser.get( elem );
-
-				if ( elem.nodeType === 1 && !dataPriv.get( elem, "hasDataAttrs" ) ) {
-					i = attrs.length;
-					while ( i-- ) {
-
-						// Support: IE 11 only
-						// The attrs elements can be null (#14894)
-						if ( attrs[ i ] ) {
-							name = attrs[ i ].name;
-							if ( name.indexOf( "data-" ) === 0 ) {
-								name = camelCase( name.slice( 5 ) );
-								dataAttr( elem, name, data[ name ] );
-							}
-						}
-					}
-					dataPriv.set( elem, "hasDataAttrs", true );
-				}
-			}
-
-			return data;
-		}
-
-		// Sets multiple values
-		if ( typeof key === "object" ) {
-			return this.each( function() {
-				dataUser.set( this, key );
-			} );
-		}
-
-		return access( this, function( value ) {
-			var data;
-
-			// The calling jQuery object (element matches) is not empty
-			// (and therefore has an element appears at this[ 0 ]) and the
-			// `value` parameter was not undefined. An empty jQuery object
-			// will result in `undefined` for elem = this[ 0 ] which will
-			// throw an exception if an attempt to read a data cache is made.
-			if ( elem && value === undefined ) {
-
-				// Attempt to get data from the cache
-				// The key will always be camelCased in Data
-				data = dataUser.get( elem, key );
-				if ( data !== undefined ) {
-					return data;
-				}
-
-				// Attempt to "discover" the data in
-				// HTML5 custom data-* attrs
-				data = dataAttr( elem, key );
-				if ( data !== undefined ) {
-					return data;
-				}
-
-				// We tried really hard, but the data doesn't exist.
-				return;
-			}
-
-			// Set the data...
-			this.each( function() {
-
-				// We always store the camelCased key
-				dataUser.set( this, key, value );
-			} );
-		}, null, value, arguments.length > 1, null, true );
-	},
-
-	removeData: function( key ) {
-		return this.each( function() {
-			dataUser.remove( this, key );
-		} );
-	}
-} );
-
-
-jQuery.extend( {
-	queue: function( elem, type, data ) {
-		var queue;
-
-		if ( elem ) {
-			type = ( type || "fx" ) + "queue";
-			queue = dataPriv.get( elem, type );
-
-			// Speed up dequeue by getting out quickly if this is just a lookup
-			if ( data ) {
-				if ( !queue || Array.isArray( data ) ) {
-					queue = dataPriv.access( elem, type, jQuery.makeArray( data ) );
-				} else {
-					queue.push( data );
-				}
-			}
-			return queue || [];
-		}
-	},
-
-	dequeue: function( elem, type ) {
-		type = type || "fx";
-
-		var queue = jQuery.queue( elem, type ),
-			startLength = queue.length,
-			fn = queue.shift(),
-			hooks = jQuery._queueHooks( elem, type ),
-			next = function() {
-				jQuery.dequeue( elem, type );
-			};
-
-		// If the fx queue is dequeued, always remove the progress sentinel
-		if ( fn === "inprogress" ) {
-			fn = queue.shift();
-			startLength--;
-		}
-
-		if ( fn ) {
-
-			// Add a progress sentinel to prevent the fx queue from being
-			// automatically dequeued
-			if ( type === "fx" ) {
-				queue.unshift( "inprogress" );
-			}
-
-			// Clear up the last queue stop function
-			delete hooks.stop;
-			fn.call( elem, next, hooks );
-		}
-
-		if ( !startLength && hooks ) {
-			hooks.empty.fire();
-		}
-	},
-
-	// Not public - generate a queueHooks object, or return the current one
-	_queueHooks: function( elem, type ) {
-		var key = type + "queueHooks";
-		return dataPriv.get( elem, key ) || dataPriv.access( elem, key, {
-			empty: jQuery.Callbacks( "once memory" ).add( function() {
-				dataPriv.remove( elem, [ type + "queue", key ] );
-			} )
-		} );
-	}
-} );
-
-jQuery.fn.extend( {
-	queue: function( type, data ) {
-		var setter = 2;
-
-		if ( typeof type !== "string" ) {
-			data = type;
-			type = "fx";
-			setter--;
-		}
-
-		if ( arguments.length < setter ) {
-			return jQuery.queue( this[ 0 ], type );
-		}
-
-		return data === undefined ?
-			this :
-			this.each( function() {
-				var queue = jQuery.queue( this, type, data );
-
-				// Ensure a hooks for this queue
-				jQuery._queueHooks( this, type );
-
-				if ( type === "fx" && queue[ 0 ] !== "inprogress" ) {
-					jQuery.dequeue( this, type );
-				}
-			} );
-	},
-	dequeue: function( type ) {
-		return this.each( function() {
-			jQuery.dequeue( this, type );
-		} );
-	},
-	clearQueue: function( type ) {
-		return this.queue( type || "fx", [] );
-	},
-
-	// Get a promise resolved when queues of a certain type
-	// are emptied (fx is the type by default)
-	promise: function( type, obj ) {
-		var tmp,
-			count = 1,
-			defer = jQuery.Deferred(),
-			elements = this,
-			i = this.length,
-			resolve = function() {
-				if ( !( --count ) ) {
-					defer.resolveWith( elements, [ elements ] );
-				}
-			};
-
-		if ( typeof type !== "string" ) {
-			obj = type;
-			type = undefined;
-		}
-		type = type || "fx";
-
-		while ( i-- ) {
-			tmp = dataPriv.get( elements[ i ], type + "queueHooks" );
-			if ( tmp && tmp.empty ) {
-				count++;
-				tmp.empty.add( resolve );
-			}
-		}
-		resolve();
-		return defer.promise( obj );
-	}
-} );
-var pnum = ( /[+-]?(?:\d*\.|)\d+(?:[eE][+-]?\d+|)/ ).source;
-
-var rcssNum = new RegExp( "^(?:([+-])=|)(" + pnum + ")([a-z%]*)$", "i" );
-
-
-var cssExpand = [ "Top", "Right", "Bottom", "Left" ];
-
-var documentElement = document.documentElement;
-
-
-
-	var isAttached = function( elem ) {
-			return jQuery.contains( elem.ownerDocument, elem );
-		},
-		composed = { composed: true };
-
-	// Support: IE 9 - 11+, Edge 12 - 18+, iOS 10.0 - 10.2 only
-	// Check attachment across shadow DOM boundaries when possible (gh-3504)
-	// Support: iOS 10.0-10.2 only
-	// Early iOS 10 versions support `attachShadow` but not `getRootNode`,
-	// leading to errors. We need to check for `getRootNode`.
-	if ( documentElement.getRootNode ) {
-		isAttached = function( elem ) {
-			return jQuery.contains( elem.ownerDocument, elem ) ||
-				elem.getRootNode( composed ) === elem.ownerDocument;
-		};
-	}
-var isHiddenWithinTree = function( elem, el ) {
-
-		// isHiddenWithinTree might be called from jQuery#filter function;
-		// in that case, element will be second argument
-		elem = el || elem;
-
-		// Inline style trumps all
-		return elem.style.display === "none" ||
-			elem.style.display === "" &&
-
-			// Otherwise, check computed style
-			// Support: Firefox <=43 - 45
-			// Disconnected elements can have computed display: none, so first confirm that elem is
-			// in the document.
-			isAttached( elem ) &&
-
-			jQuery.css( elem, "display" ) === "none";
-	};
-
-
-
-function adjustCSS( elem, prop, valueParts, tween ) {
-	var adjusted, scale,
-		maxIterations = 20,
-		currentValue = tween ?
-			function() {
-				return tween.cur();
-			} :
-			function() {
-				return jQuery.css( elem, prop, "" );
-			},
-		initial = currentValue(),
-		unit = valueParts && valueParts[ 3 ] || ( jQuery.cssNumber[ prop ] ? "" : "px" ),
-
-		// Starting value computation is required for potential unit mismatches
-		initialInUnit = elem.nodeType &&
-			( jQuery.cssNumber[ prop ] || unit !== "px" && +initial ) &&
-			rcssNum.exec( jQuery.css( elem, prop ) );
-
-	if ( initialInUnit && initialInUnit[ 3 ] !== unit ) {
-
-		// Support: Firefox <=54
-		// Halve the iteration target value to prevent interference from CSS upper bounds (gh-2144)
-		initial = initial / 2;
-
-		// Trust units reported by jQuery.css
-		unit = unit || initialInUnit[ 3 ];
-
-		// Iteratively approximate from a nonzero starting point
-		initialInUnit = +initial || 1;
-
-		while ( maxIterations-- ) {
-
-			// Evaluate and update our best guess (doubling guesses that zero out).
-			// Finish if the scale equals or crosses 1 (making the old*new product non-positive).
-			jQuery.style( elem, prop, initialInUnit + unit );
-			if ( ( 1 - scale ) * ( 1 - ( scale = currentValue() / initial || 0.5 ) ) <= 0 ) {
-				maxIterations = 0;
-			}
-			initialInUnit = initialInUnit / scale;
-
-		}
-
-		initialInUnit = initialInUnit * 2;
-		jQuery.style( elem, prop, initialInUnit + unit );
-
-		// Make sure we update the tween properties later on
-		valueParts = valueParts || [];
-	}
-
-	if ( valueParts ) {
-		initialInUnit = +initialInUnit || +initial || 0;
-
-		// Apply relative offset (+=/-=) if specified
-		adjusted = valueParts[ 1 ] ?
-			initialInUnit + ( valueParts[ 1 ] + 1 ) * valueParts[ 2 ] :
-			+valueParts[ 2 ];
-		if ( tween ) {
-			tween.unit = unit;
-			tween.start = initialInUnit;
-			tween.end = adjusted;
-		}
-	}
-	return adjusted;
-}
-
-
-var defaultDisplayMap = {};
-
-function getDefaultDisplay( elem ) {
-	var temp,
-		doc = elem.ownerDocument,
-		nodeName = elem.nodeName,
-		display = defaultDisplayMap[ nodeName ];
-
-	if ( display ) {
-		return display;
-	}
-
-	temp = doc.body.appendChild( doc.createElement( nodeName ) );
-	display = jQuery.css( temp, "display" );
-
-	temp.parentNode.removeChild( temp );
-
-	if ( display === "none" ) {
-		display = "block";
-	}
-	defaultDisplayMap[ nodeName ] = display;
-
-	return display;
-}
-
-function showHide( elements, show ) {
-	var display, elem,
-		values = [],
-		index = 0,
-		length = elements.length;
-
-	// Determine new display value for elements that need to change
-	for ( ; index < length; index++ ) {
-		elem = elements[ index ];
-		if ( !elem.style ) {
-			continue;
-		}
-
-		display = elem.style.display;
-		if ( show ) {
-
-			// Since we force visibility upon cascade-hidden elements, an immediate (and slow)
-			// check is required in this first loop unless we have a nonempty display value (either
-			// inline or about-to-be-restored)
-			if ( display === "none" ) {
-				values[ index ] = dataPriv.get( elem, "display" ) || null;
-				if ( !values[ index ] ) {
-					elem.style.display = "";
-				}
-			}
-			if ( elem.style.display === "" && isHiddenWithinTree( elem ) ) {
-				values[ index ] = getDefaultDisplay( elem );
-			}
-		} else {
-			if ( display !== "none" ) {
-				values[ index ] = "none";
-
-				// Remember what we're overwriting
-				dataPriv.set( elem, "display", display );
-			}
-		}
-	}
-
-	// Set the display of the elements in a second loop to avoid constant reflow
-	for ( index = 0; index < length; index++ ) {
-		if ( values[ index ] != null ) {
-			elements[ index ].style.display = values[ index ];
-		}
-	}
-
-	return elements;
-}
-
-jQuery.fn.extend( {
-	show: function() {
-		return showHide( this, true );
-	},
-	hide: function() {
-		return showHide( this );
-	},
-	toggle: function( state ) {
-		if ( typeof state === "boolean" ) {
-			return state ? this.show() : this.hide();
-		}
-
-		return this.each( function() {
-			if ( isHiddenWithinTree( this ) ) {
-				jQuery( this ).show();
-			} else {
-				jQuery( this ).hide();
-			}
-		} );
-	}
-} );
-var rcheckableType = ( /^(?:checkbox|radio)$/i );
-
-var rtagName = ( /<([a-z][^\/\0>\x20\t\r\n\f]*)/i );
-
-var rscriptType = ( /^$|^module$|\/(?:java|ecma)script/i );
-
-
-
-( function() {
-	var fragment = document.createDocumentFragment(),
-		div = fragment.appendChild( document.createElement( "div" ) ),
-		input = document.createElement( "input" );
-
-	// Support: Android 4.0 - 4.3 only
-	// Check state lost if the name is set (#11217)
-	// Support: Windows Web Apps (WWA)
-	// `name` and `type` must use .setAttribute for WWA (#14901)
-	input.setAttribute( "type", "radio" );
-	input.setAttribute( "checked", "checked" );
-	input.setAttribute( "name", "t" );
-
-	div.appendChild( input );
-
-	// Support: Android <=4.1 only
-	// Older WebKit doesn't clone checked state correctly in fragments
-	support.checkClone = div.cloneNode( true ).cloneNode( true ).lastChild.checked;
-
-	// Support: IE <=11 only
-	// Make sure textarea (and checkbox) defaultValue is properly cloned
-	div.innerHTML = "<textarea>x</textarea>";
-	support.noCloneChecked = !!div.cloneNode( true ).lastChild.defaultValue;
-
-	// Support: IE <=9 only
-	// IE <=9 replaces <option> tags with their contents when inserted outside of
-	// the select element.
-	div.innerHTML = "<option></option>";
-	support.option = !!div.lastChild;
-} )();
-
-
-// We have to close these tags to support XHTML (#13200)
-var wrapMap = {
-
-	// XHTML parsers do not magically insert elements in the
-	// same way that tag soup parsers do. So we cannot shorten
-	// this by omitting <tbody> or other required elements.
-	thead: [ 1, "<table>", "</table>" ],
-	col: [ 2, "<table><colgroup>", "</colgroup></table>" ],
-	tr: [ 2, "<table><tbody>", "</tbody></table>" ],
-	td: [ 3, "<table><tbody><tr>", "</tr></tbody></table>" ],
-
-	_default: [ 0, "", "" ]
-};
-
-wrapMap.tbody = wrapMap.tfoot = wrapMap.colgroup = wrapMap.caption = wrapMap.thead;
-wrapMap.th = wrapMap.td;
-
-// Support: IE <=9 only
-if ( !support.option ) {
-	wrapMap.optgroup = wrapMap.option = [ 1, "<select multiple='multiple'>", "</select>" ];
-}
-
-
-function getAll( context, tag ) {
-
-	// Support: IE <=9 - 11 only
-	// Use typeof to avoid zero-argument method invocation on host objects (#15151)
-	var ret;
-
-	if ( typeof context.getElementsByTagName !== "undefined" ) {
-		ret = context.getElementsByTagName( tag || "*" );
-
-	} else if ( typeof context.querySelectorAll !== "undefined" ) {
-		ret = context.querySelectorAll( tag || "*" );
-
-	} else {
-		ret = [];
-	}
-
-	if ( tag === undefined || tag && nodeName( context, tag ) ) {
-		return jQuery.merge( [ context ], ret );
-	}
-
-	return ret;
-}
-
-
-// Mark scripts as having already been evaluated
-function setGlobalEval( elems, refElements ) {
-	var i = 0,
-		l = elems.length;
-
-	for ( ; i < l; i++ ) {
-		dataPriv.set(
-			elems[ i ],
-			"globalEval",
-			!refElements || dataPriv.get( refElements[ i ], "globalEval" )
-		);
-	}
-}
-
-
-var rhtml = /<|&#?\w+;/;
-
-function buildFragment( elems, context, scripts, selection, ignored ) {
-	var elem, tmp, tag, wrap, attached, j,
-		fragment = context.createDocumentFragment(),
-		nodes = [],
-		i = 0,
-		l = elems.length;
-
-	for ( ; i < l; i++ ) {
-		elem = elems[ i ];
-
-		if ( elem || elem === 0 ) {
-
-			// Add nodes directly
-			if ( toType( elem ) === "object" ) {
-
-				// Support: Android <=4.0 only, PhantomJS 1 only
-				// push.apply(_, arraylike) throws on ancient WebKit
-				jQuery.merge( nodes, elem.nodeType ? [ elem ] : elem );
-
-			// Convert non-html into a text node
-			} else if ( !rhtml.test( elem ) ) {
-				nodes.push( context.createTextNode( elem ) );
-
-			// Convert html into DOM nodes
-			} else {
-				tmp = tmp || fragment.appendChild( context.createElement( "div" ) );
-
-				// Deserialize a standard representation
-				tag = ( rtagName.exec( elem ) || [ "", "" ] )[ 1 ].toLowerCase();
-				wrap = wrapMap[ tag ] || wrapMap._default;
-				tmp.innerHTML = wrap[ 1 ] + jQuery.htmlPrefilter( elem ) + wrap[ 2 ];
-
-				// Descend through wrappers to the right content
-				j = wrap[ 0 ];
-				while ( j-- ) {
-					tmp = tmp.lastChild;
-				}
-
-				// Support: Android <=4.0 only, PhantomJS 1 only
-				// push.apply(_, arraylike) throws on ancient WebKit
-				jQuery.merge( nodes, tmp.childNodes );
-
-				// Remember the top-level container
-				tmp = fragment.firstChild;
-
-				// Ensure the created nodes are orphaned (#12392)
-				tmp.textContent = "";
-			}
-		}
-	}
-
-	// Remove wrapper from fragment
-	fragment.textContent = "";
-
-	i = 0;
-	while ( ( elem = nodes[ i++ ] ) ) {
-
-		// Skip elements already in the context collection (trac-4087)
-		if ( selection && jQuery.inArray( elem, selection ) > -1 ) {
-			if ( ignored ) {
-				ignored.push( elem );
-			}
-			continue;
-		}
-
-		attached = isAttached( elem );
-
-		// Append to fragment
-		tmp = getAll( fragment.appendChild( elem ), "script" );
-
-		// Preserve script evaluation history
-		if ( attached ) {
-			setGlobalEval( tmp );
-		}
-
-		// Capture executables
-		if ( scripts ) {
-			j = 0;
-			while ( ( elem = tmp[ j++ ] ) ) {
-				if ( rscriptType.test( elem.type || "" ) ) {
-					scripts.push( elem );
-				}
-			}
-		}
-	}
-
-	return fragment;
-}
-
-
-var rtypenamespace = /^([^.]*)(?:\.(.+)|)/;
-
-function returnTrue() {
-	return true;
-}
-
-function returnFalse() {
-	return false;
-}
-
-// Support: IE <=9 - 11+
-// focus() and blur() are asynchronous, except when they are no-op.
-// So expect focus to be synchronous when the element is already active,
-// and blur to be synchronous when the element is not already active.
-// (focus and blur are always synchronous in other supported browsers,
-// this just defines when we can count on it).
-function expectSync( elem, type ) {
-	return ( elem === safeActiveElement() ) === ( type === "focus" );
-}
-
-// Support: IE <=9 only
-// Accessing document.activeElement can throw unexpectedly
-// https://bugs.jquery.com/ticket/13393
-function safeActiveElement() {
-	try {
-		return document.activeElement;
-	} catch ( err ) { }
-}
-
-function on( elem, types, selector, data, fn, one ) {
-	var origFn, type;
-
-	// Types can be a map of types/handlers
-	if ( typeof types === "object" ) {
-
-		// ( types-Object, selector, data )
-		if ( typeof selector !== "string" ) {
-
-			// ( types-Object, data )
-			data = data || selector;
-			selector = undefined;
-		}
-		for ( type in types ) {
-			on( elem, type, selector, data, types[ type ], one );
-		}
-		return elem;
-	}
-
-	if ( data == null && fn == null ) {
-
-		// ( types, fn )
-		fn = selector;
-		data = selector = undefined;
-	} else if ( fn == null ) {
-		if ( typeof selector === "string" ) {
-
-			// ( types, selector, fn )
-			fn = data;
-			data = undefined;
-		} else {
-
-			// ( types, data, fn )
-			fn = data;
-			data = selector;
-			selector = undefined;
-		}
-	}
-	if ( fn === false ) {
-		fn = returnFalse;
-	} else if ( !fn ) {
-		return elem;
-	}
-
-	if ( one === 1 ) {
-		origFn = fn;
-		fn = function( event ) {
-
-			// Can use an empty set, since event contains the info
-			jQuery().off( event );
-			return origFn.apply( this, arguments );
-		};
-
-		// Use same guid so caller can remove using origFn
-		fn.guid = origFn.guid || ( origFn.guid = jQuery.guid++ );
-	}
-	return elem.each( function() {
-		jQuery.event.add( this, types, fn, data, selector );
-	} );
-}
-
-/*
- * Helper functions for managing events -- not part of the public interface.
- * Props to Dean Edwards' addEvent library for many of the ideas.
- */
-jQuery.event = {
-
-	global: {},
-
-	add: function( elem, types, handler, data, selector ) {
-
-		var handleObjIn, eventHandle, tmp,
-			events, t, handleObj,
-			special, handlers, type, namespaces, origType,
-			elemData = dataPriv.get( elem );
-
-		// Only attach events to objects that accept data
-		if ( !acceptData( elem ) ) {
-			return;
-		}
-
-		// Caller can pass in an object of custom data in lieu of the handler
-		if ( handler.handler ) {
-			handleObjIn = handler;
-			handler = handleObjIn.handler;
-			selector = handleObjIn.selector;
-		}
-
-		// Ensure that invalid selectors throw exceptions at attach time
-		// Evaluate against documentElement in case elem is a non-element node (e.g., document)
-		if ( selector ) {
-			jQuery.find.matchesSelector( documentElement, selector );
-		}
-
-		// Make sure that the handler has a unique ID, used to find/remove it later
-		if ( !handler.guid ) {
-			handler.guid = jQuery.guid++;
-		}
-
-		// Init the element's event structure and main handler, if this is the first
-		if ( !( events = elemData.events ) ) {
-			events = elemData.events = Object.create( null );
-		}
-		if ( !( eventHandle = elemData.handle ) ) {
-			eventHandle = elemData.handle = function( e ) {
-
-				// Discard the second event of a jQuery.event.trigger() and
-				// when an event is called after a page has unloaded
-				return typeof jQuery !== "undefined" && jQuery.event.triggered !== e.type ?
-					jQuery.event.dispatch.apply( elem, arguments ) : undefined;
-			};
-		}
-
-		// Handle multiple events separated by a space
-		types = ( types || "" ).match( rnothtmlwhite ) || [ "" ];
-		t = types.length;
-		while ( t-- ) {
-			tmp = rtypenamespace.exec( types[ t ] ) || [];
-			type = origType = tmp[ 1 ];
-			namespaces = ( tmp[ 2 ] || "" ).split( "." ).sort();
-
-			// There *must* be a type, no attaching namespace-only handlers
-			if ( !type ) {
-				continue;
-			}
-
-			// If event changes its type, use the special event handlers for the changed type
-			special = jQuery.event.special[ type ] || {};
-
-			// If selector defined, determine special event api type, otherwise given type
-			type = ( selector ? special.delegateType : special.bindType ) || type;
-
-			// Update special based on newly reset type
-			special = jQuery.event.special[ type ] || {};
-
-			// handleObj is passed to all event handlers
-			handleObj = jQuery.extend( {
-				type: type,
-				origType: origType,
-				data: data,
-				handler: handler,
-				guid: handler.guid,
-				selector: selector,
-				needsContext: selector && jQuery.expr.match.needsContext.test( selector ),
-				namespace: namespaces.join( "." )
-			}, handleObjIn );
-
-			// Init the event handler queue if we're the first
-			if ( !( handlers = events[ type ] ) ) {
-				handlers = events[ type ] = [];
-				handlers.delegateCount = 0;
-
-				// Only use addEventListener if the special events handler returns false
-				if ( !special.setup ||
-					special.setup.call( elem, data, namespaces, eventHandle ) === false ) {
-
-					if ( elem.addEventListener ) {
-						elem.addEventListener( type, eventHandle );
-					}
-				}
-			}
-
-			if ( special.add ) {
-				special.add.call( elem, handleObj );
-
-				if ( !handleObj.handler.guid ) {
-					handleObj.handler.guid = handler.guid;
-				}
-			}
-
-			// Add to the element's handler list, delegates in front
-			if ( selector ) {
-				handlers.splice( handlers.delegateCount++, 0, handleObj );
-			} else {
-				handlers.push( handleObj );
-			}
-
-			// Keep track of which events have ever been used, for event optimization
-			jQuery.event.global[ type ] = true;
-		}
-
-	},
-
-	// Detach an event or set of events from an element
-	remove: function( elem, types, handler, selector, mappedTypes ) {
-
-		var j, origCount, tmp,
-			events, t, handleObj,
-			special, handlers, type, namespaces, origType,
-			elemData = dataPriv.hasData( elem ) && dataPriv.get( elem );
-
-		if ( !elemData || !( events = elemData.events ) ) {
-			return;
-		}
-
-		// Once for each type.namespace in types; type may be omitted
-		types = ( types || "" ).match( rnothtmlwhite ) || [ "" ];
-		t = types.length;
-		while ( t-- ) {
-			tmp = rtypenamespace.exec( types[ t ] ) || [];
-			type = origType = tmp[ 1 ];
-			namespaces = ( tmp[ 2 ] || "" ).split( "." ).sort();
-
-			// Unbind all events (on this namespace, if provided) for the element
-			if ( !type ) {
-				for ( type in events ) {
-					jQuery.event.remove( elem, type + types[ t ], handler, selector, true );
-				}
-				continue;
-			}
-
-			special = jQuery.event.special[ type ] || {};
-			type = ( selector ? special.delegateType : special.bindType ) || type;
-			handlers = events[ type ] || [];
-			tmp = tmp[ 2 ] &&
-				new RegExp( "(^|\\.)" + namespaces.join( "\\.(?:.*\\.|)" ) + "(\\.|$)" );
-
-			// Remove matching events
-			origCount = j = handlers.length;
-			while ( j-- ) {
-				handleObj = handlers[ j ];
-
-				if ( ( mappedTypes || origType === handleObj.origType ) &&
-					( !handler || handler.guid === handleObj.guid ) &&
-					( !tmp || tmp.test( handleObj.namespace ) ) &&
-					( !selector || selector === handleObj.selector ||
-						selector === "**" && handleObj.selector ) ) {
-					handlers.splice( j, 1 );
-
-					if ( handleObj.selector ) {
-						handlers.delegateCount--;
-					}
-					if ( special.remove ) {
-						special.remove.call( elem, handleObj );
-					}
-				}
-			}
-
-			// Remove generic event handler if we removed something and no more handlers exist
-			// (avoids potential for endless recursion during removal of special event handlers)
-			if ( origCount && !handlers.length ) {
-				if ( !special.teardown ||
-					special.teardown.call( elem, namespaces, elemData.handle ) === false ) {
-
-					jQuery.removeEvent( elem, type, elemData.handle );
-				}
-
-				delete events[ type ];
-			}
-		}
-
-		// Remove data and the expando if it's no longer used
-		if ( jQuery.isEmptyObject( events ) ) {
-			dataPriv.remove( elem, "handle events" );
-		}
-	},
-
-	dispatch: function( nativeEvent ) {
-
-		var i, j, ret, matched, handleObj, handlerQueue,
-			args = new Array( arguments.length ),
-
-			// Make a writable jQuery.Event from the native event object
-			event = jQuery.event.fix( nativeEvent ),
-
-			handlers = (
-				dataPriv.get( this, "events" ) || Object.create( null )
-			)[ event.type ] || [],
-			special = jQuery.event.special[ event.type ] || {};
-
-		// Use the fix-ed jQuery.Event rather than the (read-only) native event
-		args[ 0 ] = event;
-
-		for ( i = 1; i < arguments.length; i++ ) {
-			args[ i ] = arguments[ i ];
-		}
-
-		event.delegateTarget = this;
-
-		// Call the preDispatch hook for the mapped type, and let it bail if desired
-		if ( special.preDispatch && special.preDispatch.call( this, event ) === false ) {
-			return;
-		}
-
-		// Determine handlers
-		handlerQueue = jQuery.event.handlers.call( this, event, handlers );
-
-		// Run delegates first; they may want to stop propagation beneath us
-		i = 0;
-		while ( ( matched = handlerQueue[ i++ ] ) && !event.isPropagationStopped() ) {
-			event.currentTarget = matched.elem;
-
-			j = 0;
-			while ( ( handleObj = matched.handlers[ j++ ] ) &&
-				!event.isImmediatePropagationStopped() ) {
-
-				// If the event is namespaced, then each handler is only invoked if it is
-				// specially universal or its namespaces are a superset of the event's.
-				if ( !event.rnamespace || handleObj.namespace === false ||
-					event.rnamespace.test( handleObj.namespace ) ) {
-
-					event.handleObj = handleObj;
-					event.data = handleObj.data;
-
-					ret = ( ( jQuery.event.special[ handleObj.origType ] || {} ).handle ||
-						handleObj.handler ).apply( matched.elem, args );
-
-					if ( ret !== undefined ) {
-						if ( ( event.result = ret ) === false ) {
-							event.preventDefault();
-							event.stopPropagation();
-						}
-					}
-				}
-			}
-		}
-
-		// Call the postDispatch hook for the mapped type
-		if ( special.postDispatch ) {
-			special.postDispatch.call( this, event );
-		}
-
-		return event.result;
-	},
-
-	handlers: function( event, handlers ) {
-		var i, handleObj, sel, matchedHandlers, matchedSelectors,
-			handlerQueue = [],
-			delegateCount = handlers.delegateCount,
-			cur = event.target;
-
-		// Find delegate handlers
-		if ( delegateCount &&
-
-			// Support: IE <=9
-			// Black-hole SVG <use> instance trees (trac-13180)
-			cur.nodeType &&
-
-			// Support: Firefox <=42
-			// Suppress spec-violating clicks indicating a non-primary pointer button (trac-3861)
-			// https://www.w3.org/TR/DOM-Level-3-Events/#event-type-click
-			// Support: IE 11 only
-			// ...but not arrow key "clicks" of radio inputs, which can have `button` -1 (gh-2343)
-			!( event.type === "click" && event.button >= 1 ) ) {
-
-			for ( ; cur !== this; cur = cur.parentNode || this ) {
-
-				// Don't check non-elements (#13208)
-				// Don't process clicks on disabled elements (#6911, #8165, #11382, #11764)
-				if ( cur.nodeType === 1 && !( event.type === "click" && cur.disabled === true ) ) {
-					matchedHandlers = [];
-					matchedSelectors = {};
-					for ( i = 0; i < delegateCount; i++ ) {
-						handleObj = handlers[ i ];
-
-						// Don't conflict with Object.prototype properties (#13203)
-						sel = handleObj.selector + " ";
-
-						if ( matchedSelectors[ sel ] === undefined ) {
-							matchedSelectors[ sel ] = handleObj.needsContext ?
-								jQuery( sel, this ).index( cur ) > -1 :
-								jQuery.find( sel, this, null, [ cur ] ).length;
-						}
-						if ( matchedSelectors[ sel ] ) {
-							matchedHandlers.push( handleObj );
-						}
-					}
-					if ( matchedHandlers.length ) {
-						handlerQueue.push( { elem: cur, handlers: matchedHandlers } );
-					}
-				}
-			}
-		}
-
-		// Add the remaining (directly-bound) handlers
-		cur = this;
-		if ( delegateCount < handlers.length ) {
-			handlerQueue.push( { elem: cur, handlers: handlers.slice( delegateCount ) } );
-		}
-
-		return handlerQueue;
-	},
-
-	addProp: function( name, hook ) {
-		Object.defineProperty( jQuery.Event.prototype, name, {
-			enumerable: true,
-			configurable: true,
-
-			get: isFunction( hook ) ?
-				function() {
-					if ( this.originalEvent ) {
-						return hook( this.originalEvent );
-					}
-				} :
-				function() {
-					if ( this.originalEvent ) {
-						return this.originalEvent[ name ];
-					}
-				},
-
-			set: function( value ) {
-				Object.defineProperty( this, name, {
-					enumerable: true,
-					configurable: true,
-					writable: true,
-					value: value
-				} );
-			}
-		} );
-	},
-
-	fix: function( originalEvent ) {
-		return originalEvent[ jQuery.expando ] ?
-			originalEvent :
-			new jQuery.Event( originalEvent );
-	},
-
-	special: {
-		load: {
-
-			// Prevent triggered image.load events from bubbling to window.load
-			noBubble: true
-		},
-		click: {
-
-			// Utilize native event to ensure correct state for checkable inputs
-			setup: function( data ) {
-
-				// For mutual compressibility with _default, replace `this` access with a local var.
-				// `|| data` is dead code meant only to preserve the variable through minification.
-				var el = this || data;
-
-				// Claim the first handler
-				if ( rcheckableType.test( el.type ) &&
-					el.click && nodeName( el, "input" ) ) {
-
-					// dataPriv.set( el, "click", ... )
-					leverageNative( el, "click", returnTrue );
-				}
-
-				// Return false to allow normal processing in the caller
-				return false;
-			},
-			trigger: function( data ) {
-
-				// For mutual compressibility with _default, replace `this` access with a local var.
-				// `|| data` is dead code meant only to preserve the variable through minification.
-				var el = this || data;
-
-				// Force setup before triggering a click
-				if ( rcheckableType.test( el.type ) &&
-					el.click && nodeName( el, "input" ) ) {
-
-					leverageNative( el, "click" );
-				}
-
-				// Return non-false to allow normal event-path propagation
-				return true;
-			},
-
-			// For cross-browser consistency, suppress native .click() on links
-			// Also prevent it if we're currently inside a leveraged native-event stack
-			_default: function( event ) {
-				var target = event.target;
-				return rcheckableType.test( target.type ) &&
-					target.click && nodeName( target, "input" ) &&
-					dataPriv.get( target, "click" ) ||
-					nodeName( target, "a" );
-			}
-		},
-
-		beforeunload: {
-			postDispatch: function( event ) {
-
-				// Support: Firefox 20+
-				// Firefox doesn't alert if the returnValue field is not set.
-				if ( event.result !== undefined && event.originalEvent ) {
-					event.originalEvent.returnValue = event.result;
-				}
-			}
-		}
-	}
-};
-
-// Ensure the presence of an event listener that handles manually-triggered
-// synthetic events by interrupting progress until reinvoked in response to
-// *native* events that it fires directly, ensuring that state changes have
-// already occurred before other listeners are invoked.
-function leverageNative( el, type, expectSync ) {
-
-	// Missing expectSync indicates a trigger call, which must force setup through jQuery.event.add
-	if ( !expectSync ) {
-		if ( dataPriv.get( el, type ) === undefined ) {
-			jQuery.event.add( el, type, returnTrue );
-		}
-		return;
-	}
-
-	// Register the controller as a special universal handler for all event namespaces
-	dataPriv.set( el, type, false );
-	jQuery.event.add( el, type, {
-		namespace: false,
-		handler: function( event ) {
-			var notAsync, result,
-				saved = dataPriv.get( this, type );
-
-			if ( ( event.isTrigger & 1 ) && this[ type ] ) {
-
-				// Interrupt processing of the outer synthetic .trigger()ed event
-				// Saved data should be false in such cases, but might be a leftover capture object
-				// from an async native handler (gh-4350)
-				if ( !saved.length ) {
-
-					// Store arguments for use when handling the inner native event
-					// There will always be at least one argument (an event object), so this array
-					// will not be confused with a leftover capture object.
-					saved = slice.call( arguments );
-					dataPriv.set( this, type, saved );
-
-					// Trigger the native event and capture its result
-					// Support: IE <=9 - 11+
-					// focus() and blur() are asynchronous
-					notAsync = expectSync( this, type );
-					this[ type ]();
-					result = dataPriv.get( this, type );
-					if ( saved !== result || notAsync ) {
-						dataPriv.set( this, type, false );
-					} else {
-						result = {};
-					}
-					if ( saved !== result ) {
-
-						// Cancel the outer synthetic event
-						event.stopImmediatePropagation();
-						event.preventDefault();
-
-						// Support: Chrome 86+
-						// In Chrome, if an element having a focusout handler is blurred by
-						// clicking outside of it, it invokes the handler synchronously. If
-						// that handler calls `.remove()` on the element, the data is cleared,
-						// leaving `result` undefined. We need to guard against this.
-						return result && result.value;
-					}
-
-				// If this is an inner synthetic event for an event with a bubbling surrogate
-				// (focus or blur), assume that the surrogate already propagated from triggering the
-				// native event and prevent that from happening again here.
-				// This technically gets the ordering wrong w.r.t. to `.trigger()` (in which the
-				// bubbling surrogate propagates *after* the non-bubbling base), but that seems
-				// less bad than duplication.
-				} else if ( ( jQuery.event.special[ type ] || {} ).delegateType ) {
-					event.stopPropagation();
-				}
-
-			// If this is a native event triggered above, everything is now in order
-			// Fire an inner synthetic event with the original arguments
-			} else if ( saved.length ) {
-
-				// ...and capture the result
-				dataPriv.set( this, type, {
-					value: jQuery.event.trigger(
-
-						// Support: IE <=9 - 11+
-						// Extend with the prototype to reset the above stopImmediatePropagation()
-						jQuery.extend( saved[ 0 ], jQuery.Event.prototype ),
-						saved.slice( 1 ),
-						this
-					)
-				} );
-
-				// Abort handling of the native event
-				event.stopImmediatePropagation();
-			}
-		}
-	} );
-}
-
-jQuery.removeEvent = function( elem, type, handle ) {
-
-	// This "if" is needed for plain objects
-	if ( elem.removeEventListener ) {
-		elem.removeEventListener( type, handle );
-	}
-};
-
-jQuery.Event = function( src, props ) {
-
-	// Allow instantiation without the 'new' keyword
-	if ( !( this instanceof jQuery.Event ) ) {
-		return new jQuery.Event( src, props );
-	}
-
-	// Event object
-	if ( src && src.type ) {
-		this.originalEvent = src;
-		this.type = src.type;
-
-		// Events bubbling up the document may have been marked as prevented
-		// by a handler lower down the tree; reflect the correct value.
-		this.isDefaultPrevented = src.defaultPrevented ||
-				src.defaultPrevented === undefined &&
-
-				// Support: Android <=2.3 only
-				src.returnValue === false ?
-			returnTrue :
-			returnFalse;
-
-		// Create target properties
-		// Support: Safari <=6 - 7 only
-		// Target should not be a text node (#504, #13143)
-		this.target = ( src.target && src.target.nodeType === 3 ) ?
-			src.target.parentNode :
-			src.target;
-
-		this.currentTarget = src.currentTarget;
-		this.relatedTarget = src.relatedTarget;
-
-	// Event type
-	} else {
-		this.type = src;
-	}
-
-	// Put explicitly provided properties onto the event object
-	if ( props ) {
-		jQuery.extend( this, props );
-	}
-
-	// Create a timestamp if incoming event doesn't have one
-	this.timeStamp = src && src.timeStamp || Date.now();
-
-	// Mark it as fixed
-	this[ jQuery.expando ] = true;
-};
-
-// jQuery.Event is based on DOM3 Events as specified by the ECMAScript Language Binding
-// https://www.w3.org/TR/2003/WD-DOM-Level-3-Events-20030331/ecma-script-binding.html
-jQuery.Event.prototype = {
-	constructor: jQuery.Event,
-	isDefaultPrevented: returnFalse,
-	isPropagationStopped: returnFalse,
-	isImmediatePropagationStopped: returnFalse,
-	isSimulated: false,
-
-	preventDefault: function() {
-		var e = this.originalEvent;
-
-		this.isDefaultPrevented = returnTrue;
-
-		if ( e && !this.isSimulated ) {
-			e.preventDefault();
-		}
-	},
-	stopPropagation: function() {
-		var e = this.originalEvent;
-
-		this.isPropagationStopped = returnTrue;
-
-		if ( e && !this.isSimulated ) {
-			e.stopPropagation();
-		}
-	},
-	stopImmediatePropagation: function() {
-		var e = this.originalEvent;
-
-		this.isImmediatePropagationStopped = returnTrue;
-
-		if ( e && !this.isSimulated ) {
-			e.stopImmediatePropagation();
-		}
-
-		this.stopPropagation();
-	}
-};
-
-// Includes all common event props including KeyEvent and MouseEvent specific props
-jQuery.each( {
-	altKey: true,
-	bubbles: true,
-	cancelable: true,
-	changedTouches: true,
-	ctrlKey: true,
-	detail: true,
-	eventPhase: true,
-	metaKey: true,
-	pageX: true,
-	pageY: true,
-	shiftKey: true,
-	view: true,
-	"char": true,
-	code: true,
-	charCode: true,
-	key: true,
-	keyCode: true,
-	button: true,
-	buttons: true,
-	clientX: true,
-	clientY: true,
-	offsetX: true,
-	offsetY: true,
-	pointerId: true,
-	pointerType: true,
-	screenX: true,
-	screenY: true,
-	targetTouches: true,
-	toElement: true,
-	touches: true,
-	which: true
-}, jQuery.event.addProp );
-
-jQuery.each( { focus: "focusin", blur: "focusout" }, function( type, delegateType ) {
-	jQuery.event.special[ type ] = {
-
-		// Utilize native event if possible so blur/focus sequence is correct
-		setup: function() {
-
-			// Claim the first handler
-			// dataPriv.set( this, "focus", ... )
-			// dataPriv.set( this, "blur", ... )
-			leverageNative( this, type, expectSync );
-
-			// Return false to allow normal processing in the caller
-			return false;
-		},
-		trigger: function() {
-
-			// Force setup before trigger
-			leverageNative( this, type );
-
-			// Return non-false to allow normal event-path propagation
-			return true;
-		},
-
-		// Suppress native focus or blur as it's already being fired
-		// in leverageNative.
-		_default: function() {
-			return true;
-		},
-
-		delegateType: delegateType
-	};
-} );
-
-// Create mouseenter/leave events using mouseover/out and event-time checks
-// so that event delegation works in jQuery.
-// Do the same for pointerenter/pointerleave and pointerover/pointerout
-//
-// Support: Safari 7 only
-// Safari sends mouseenter too often; see:
-// https://bugs.chromium.org/p/chromium/issues/detail?id=470258
-// for the description of the bug (it existed in older Chrome versions as well).
-jQuery.each( {
-	mouseenter: "mouseover",
-	mouseleave: "mouseout",
-	pointerenter: "pointerover",
-	pointerleave: "pointerout"
-}, function( orig, fix ) {
-	jQuery.event.special[ orig ] = {
-		delegateType: fix,
-		bindType: fix,
-
-		handle: function( event ) {
-			var ret,
-				target = this,
-				related = event.relatedTarget,
-				handleObj = event.handleObj;
-
-			// For mouseenter/leave call the handler if related is outside the target.
-			// NB: No relatedTarget if the mouse left/entered the browser window
-			if ( !related || ( related !== target && !jQuery.contains( target, related ) ) ) {
-				event.type = handleObj.origType;
-				ret = handleObj.handler.apply( this, arguments );
-				event.type = fix;
-			}
-			return ret;
-		}
-	};
-} );
-
-jQuery.fn.extend( {
-
-	on: function( types, selector, data, fn ) {
-		return on( this, types, selector, data, fn );
-	},
-	one: function( types, selector, data, fn ) {
-		return on( this, types, selector, data, fn, 1 );
-	},
-	off: function( types, selector, fn ) {
-		var handleObj, type;
-		if ( types && types.preventDefault && types.handleObj ) {
-
-			// ( event )  dispatched jQuery.Event
-			handleObj = types.handleObj;
-			jQuery( types.delegateTarget ).off(
-				handleObj.namespace ?
-					handleObj.origType + "." + handleObj.namespace :
-					handleObj.origType,
-				handleObj.selector,
-				handleObj.handler
-			);
-			return this;
-		}
-		if ( typeof types === "object" ) {
-
-			// ( types-object [, selector] )
-			for ( type in types ) {
-				this.off( type, selector, types[ type ] );
-			}
-			return this;
-		}
-		if ( selector === false || typeof selector === "function" ) {
-
-			// ( types [, fn] )
-			fn = selector;
-			selector = undefined;
-		}
-		if ( fn === false ) {
-			fn = returnFalse;
-		}
-		return this.each( function() {
-			jQuery.event.remove( this, types, fn, selector );
-		} );
-	}
-} );
-
-
-var
-
-	// Support: IE <=10 - 11, Edge 12 - 13 only
-	// In IE/Edge using regex groups here causes severe slowdowns.
-	// See https://connect.microsoft.com/IE/feedback/details/1736512/
-	rnoInnerhtml = /<script|<style|<link/i,
-
-	// checked="checked" or checked
-	rchecked = /checked\s*(?:[^=]|=\s*.checked.)/i,
-	rcleanScript = /^\s*<!(?:\[CDATA\[|--)|(?:\]\]|--)>\s*$/g;
-
-// Prefer a tbody over its parent table for containing new rows
-function manipulationTarget( elem, content ) {
-	if ( nodeName( elem, "table" ) &&
-		nodeName( content.nodeType !== 11 ? content : content.firstChild, "tr" ) ) {
-
-		return jQuery( elem ).children( "tbody" )[ 0 ] || elem;
-	}
-
-	return elem;
-}
-
-// Replace/restore the type attribute of script elements for safe DOM manipulation
-function disableScript( elem ) {
-	elem.type = ( elem.getAttribute( "type" ) !== null ) + "/" + elem.type;
-	return elem;
-}
-function restoreScript( elem ) {
-	if ( ( elem.type || "" ).slice( 0, 5 ) === "true/" ) {
-		elem.type = elem.type.slice( 5 );
-	} else {
-		elem.removeAttribute( "type" );
-	}
-
-	return elem;
-}
-
-function cloneCopyEvent( src, dest ) {
-	var i, l, type, pdataOld, udataOld, udataCur, events;
-
-	if ( dest.nodeType !== 1 ) {
-		return;
-	}
-
-	// 1. Copy private data: events, handlers, etc.
-	if ( dataPriv.hasData( src ) ) {
-		pdataOld = dataPriv.get( src );
-		events = pdataOld.events;
-
-		if ( events ) {
-			dataPriv.remove( dest, "handle events" );
-
-			for ( type in events ) {
-				for ( i = 0, l = events[ type ].length; i < l; i++ ) {
-					jQuery.event.add( dest, type, events[ type ][ i ] );
-				}
-			}
-		}
-	}
-
-	// 2. Copy user data
-	if ( dataUser.hasData( src ) ) {
-		udataOld = dataUser.access( src );
-		udataCur = jQuery.extend( {}, udataOld );
-
-		dataUser.set( dest, udataCur );
-	}
-}
-
-// Fix IE bugs, see support tests
-function fixInput( src, dest ) {
-	var nodeName = dest.nodeName.toLowerCase();
-
-	// Fails to persist the checked state of a cloned checkbox or radio button.
-	if ( nodeName === "input" && rcheckableType.test( src.type ) ) {
-		dest.checked = src.checked;
-
-	// Fails to return the selected option to the default selected state when cloning options
-	} else if ( nodeName === "input" || nodeName === "textarea" ) {
-		dest.defaultValue = src.defaultValue;
-	}
-}
-
-function domManip( collection, args, callback, ignored ) {
-
-	// Flatten any nested arrays
-	args = flat( args );
-
-	var fragment, first, scripts, hasScripts, node, doc,
-		i = 0,
-		l = collection.length,
-		iNoClone = l - 1,
-		value = args[ 0 ],
-		valueIsFunction = isFunction( value );
-
-	// We can't cloneNode fragments that contain checked, in WebKit
-	if ( valueIsFunction ||
-			( l > 1 && typeof value === "string" &&
-				!support.checkClone && rchecked.test( value ) ) ) {
-		return collection.each( function( index ) {
-			var self = collection.eq( index );
-			if ( valueIsFunction ) {
-				args[ 0 ] = value.call( this, index, self.html() );
-			}
-			domManip( self, args, callback, ignored );
-		} );
-	}
-
-	if ( l ) {
-		fragment = buildFragment( args, collection[ 0 ].ownerDocument, false, collection, ignored );
-		first = fragment.firstChild;
-
-		if ( fragment.childNodes.length === 1 ) {
-			fragment = first;
-		}
-
-		// Require either new content or an interest in ignored elements to invoke the callback
-		if ( first || ignored ) {
-			scripts = jQuery.map( getAll( fragment, "script" ), disableScript );
-			hasScripts = scripts.length;
-
-			// Use the original fragment for the last item
-			// instead of the first because it can end up
-			// being emptied incorrectly in certain situations (#8070).
-			for ( ; i < l; i++ ) {
-				node = fragment;
-
-				if ( i !== iNoClone ) {
-					node = jQuery.clone( node, true, true );
-
-					// Keep references to cloned scripts for later restoration
-					if ( hasScripts ) {
-
-						// Support: Android <=4.0 only, PhantomJS 1 only
-						// push.apply(_, arraylike) throws on ancient WebKit
-						jQuery.merge( scripts, getAll( node, "script" ) );
-					}
-				}
-
-				callback.call( collection[ i ], node, i );
-			}
-
-			if ( hasScripts ) {
-				doc = scripts[ scripts.length - 1 ].ownerDocument;
-
-				// Reenable scripts
-				jQuery.map( scripts, restoreScript );
-
-				// Evaluate executable scripts on first document insertion
-				for ( i = 0; i < hasScripts; i++ ) {
-					node = scripts[ i ];
-					if ( rscriptType.test( node.type || "" ) &&
-						!dataPriv.access( node, "globalEval" ) &&
-						jQuery.contains( doc, node ) ) {
-
-						if ( node.src && ( node.type || "" ).toLowerCase()  !== "module" ) {
-
-							// Optional AJAX dependency, but won't run scripts if not present
-							if ( jQuery._evalUrl && !node.noModule ) {
-								jQuery._evalUrl( node.src, {
-									nonce: node.nonce || node.getAttribute( "nonce" )
-								}, doc );
-							}
-						} else {
-							DOMEval( node.textContent.replace( rcleanScript, "" ), node, doc );
-						}
-					}
-				}
-			}
-		}
-	}
-
-	return collection;
-}
-
-function remove( elem, selector, keepData ) {
-	var node,
-		nodes = selector ? jQuery.filter( selector, elem ) : elem,
-		i = 0;
-
-	for ( ; ( node = nodes[ i ] ) != null; i++ ) {
-		if ( !keepData && node.nodeType === 1 ) {
-			jQuery.cleanData( getAll( node ) );
-		}
-
-		if ( node.parentNode ) {
-			if ( keepData && isAttached( node ) ) {
-				setGlobalEval( getAll( node, "script" ) );
-			}
-			node.parentNode.removeChild( node );
-		}
-	}
-
-	return elem;
-}
-
-jQuery.extend( {
-	htmlPrefilter: function( html ) {
-		return html;
-	},
-
-	clone: function( elem, dataAndEvents, deepDataAndEvents ) {
-		var i, l, srcElements, destElements,
-			clone = elem.cloneNode( true ),
-			inPage = isAttached( elem );
-
-		// Fix IE cloning issues
-		if ( !support.noCloneChecked && ( elem.nodeType === 1 || elem.nodeType === 11 ) &&
-				!jQuery.isXMLDoc( elem ) ) {
-
-			// We eschew Sizzle here for performance reasons: https://jsperf.com/getall-vs-sizzle/2
-			destElements = getAll( clone );
-			srcElements = getAll( elem );
-
-			for ( i = 0, l = srcElements.length; i < l; i++ ) {
-				fixInput( srcElements[ i ], destElements[ i ] );
-			}
-		}
-
-		// Copy the events from the original to the clone
-		if ( dataAndEvents ) {
-			if ( deepDataAndEvents ) {
-				srcElements = srcElements || getAll( elem );
-				destElements = destElements || getAll( clone );
-
-				for ( i = 0, l = srcElements.length; i < l; i++ ) {
-					cloneCopyEvent( srcElements[ i ], destElements[ i ] );
-				}
-			} else {
-				cloneCopyEvent( elem, clone );
-			}
-		}
-
-		// Preserve script evaluation history
-		destElements = getAll( clone, "script" );
-		if ( destElements.length > 0 ) {
-			setGlobalEval( destElements, !inPage && getAll( elem, "script" ) );
-		}
-
-		// Return the cloned set
-		return clone;
-	},
-
-	cleanData: function( elems ) {
-		var data, elem, type,
-			special = jQuery.event.special,
-			i = 0;
-
-		for ( ; ( elem = elems[ i ] ) !== undefined; i++ ) {
-			if ( acceptData( elem ) ) {
-				if ( ( data = elem[ dataPriv.expando ] ) ) {
-					if ( data.events ) {
-						for ( type in data.events ) {
-							if ( special[ type ] ) {
-								jQuery.event.remove( elem, type );
-
-							// This is a shortcut to avoid jQuery.event.remove's overhead
-							} else {
-								jQuery.removeEvent( elem, type, data.handle );
-							}
-						}
-					}
-
-					// Support: Chrome <=35 - 45+
-					// Assign undefined instead of using delete, see Data#remove
-					elem[ dataPriv.expando ] = undefined;
-				}
-				if ( elem[ dataUser.expando ] ) {
-
-					// Support: Chrome <=35 - 45+
-					// Assign undefined instead of using delete, see Data#remove
-					elem[ dataUser.expando ] = undefined;
-				}
-			}
-		}
-	}
-} );
-
-jQuery.fn.extend( {
-	detach: function( selector ) {
-		return remove( this, selector, true );
-	},
-
-	remove: function( selector ) {
-		return remove( this, selector );
-	},
-
-	text: function( value ) {
-		return access( this, function( value ) {
-			return value === undefined ?
-				jQuery.text( this ) :
-				this.empty().each( function() {
-					if ( this.nodeType === 1 || this.nodeType === 11 || this.nodeType === 9 ) {
-						this.textContent = value;
-					}
-				} );
-		}, null, value, arguments.length );
-	},
-
-	append: function() {
-		return domManip( this, arguments, function( elem ) {
-			if ( this.nodeType === 1 || this.nodeType === 11 || this.nodeType === 9 ) {
-				var target = manipulationTarget( this, elem );
-				target.appendChild( elem );
-			}
-		} );
-	},
-
-	prepend: function() {
-		return domManip( this, arguments, function( elem ) {
-			if ( this.nodeType === 1 || this.nodeType === 11 || this.nodeType === 9 ) {
-				var target = manipulationTarget( this, elem );
-				target.insertBefore( elem, target.firstChild );
-			}
-		} );
-	},
-
-	before: function() {
-		return domManip( this, arguments, function( elem ) {
-			if ( this.parentNode ) {
-				this.parentNode.insertBefore( elem, this );
-			}
-		} );
-	},
-
-	after: function() {
-		return domManip( this, arguments, function( elem ) {
-			if ( this.parentNode ) {
-				this.parentNode.insertBefore( elem, this.nextSibling );
-			}
-		} );
-	},
-
-	empty: function() {
-		var elem,
-			i = 0;
-
-		for ( ; ( elem = this[ i ] ) != null; i++ ) {
-			if ( elem.nodeType === 1 ) {
-
-				// Prevent memory leaks
-				jQuery.cleanData( getAll( elem, false ) );
-
-				// Remove any remaining nodes
-				elem.textContent = "";
-			}
-		}
-
-		return this;
-	},
-
-	clone: function( dataAndEvents, deepDataAndEvents ) {
-		dataAndEvents = dataAndEvents == null ? false : dataAndEvents;
-		deepDataAndEvents = deepDataAndEvents == null ? dataAndEvents : deepDataAndEvents;
-
-		return this.map( function() {
-			return jQuery.clone( this, dataAndEvents, deepDataAndEvents );
-		} );
-	},
-
-	html: function( value ) {
-		return access( this, function( value ) {
-			var elem = this[ 0 ] || {},
-				i = 0,
-				l = this.length;
-
-			if ( value === undefined && elem.nodeType === 1 ) {
-				return elem.innerHTML;
-			}
-
-			// See if we can take a shortcut and just use innerHTML
-			if ( typeof value === "string" && !rnoInnerhtml.test( value ) &&
-				!wrapMap[ ( rtagName.exec( value ) || [ "", "" ] )[ 1 ].toLowerCase() ] ) {
-
-				value = jQuery.htmlPrefilter( value );
-
-				try {
-					for ( ; i < l; i++ ) {
-						elem = this[ i ] || {};
-
-						// Remove element nodes and prevent memory leaks
-						if ( elem.nodeType === 1 ) {
-							jQuery.cleanData( getAll( elem, false ) );
-							elem.innerHTML = value;
-						}
-					}
-
-					elem = 0;
-
-				// If using innerHTML throws an exception, use the fallback method
-				} catch ( e ) {}
-			}
-
-			if ( elem ) {
-				this.empty().append( value );
-			}
-		}, null, value, arguments.length );
-	},
-
-	replaceWith: function() {
-		var ignored = [];
-
-		// Make the changes, replacing each non-ignored context element with the new content
-		return domManip( this, arguments, function( elem ) {
-			var parent = this.parentNode;
-
-			if ( jQuery.inArray( this, ignored ) < 0 ) {
-				jQuery.cleanData( getAll( this ) );
-				if ( parent ) {
-					parent.replaceChild( elem, this );
-				}
-			}
-
-		// Force callback invocation
-		}, ignored );
-	}
-} );
-
-jQuery.each( {
-	appendTo: "append",
-	prependTo: "prepend",
-	insertBefore: "before",
-	insertAfter: "after",
-	replaceAll: "replaceWith"
-}, function( name, original ) {
-	jQuery.fn[ name ] = function( selector ) {
-		var elems,
-			ret = [],
-			insert = jQuery( selector ),
-			last = insert.length - 1,
-			i = 0;
-
-		for ( ; i <= last; i++ ) {
-			elems = i === last ? this : this.clone( true );
-			jQuery( insert[ i ] )[ original ]( elems );
-
-			// Support: Android <=4.0 only, PhantomJS 1 only
-			// .get() because push.apply(_, arraylike) throws on ancient WebKit
-			push.apply( ret, elems.get() );
-		}
-
-		return this.pushStack( ret );
-	};
-} );
-var rnumnonpx = new RegExp( "^(" + pnum + ")(?!px)[a-z%]+$", "i" );
-
-var getStyles = function( elem ) {
-
-		// Support: IE <=11 only, Firefox <=30 (#15098, #14150)
-		// IE throws on elements created in popups
-		// FF meanwhile throws on frame elements through "defaultView.getComputedStyle"
-		var view = elem.ownerDocument.defaultView;
-
-		if ( !view || !view.opener ) {
-			view = window;
-		}
-
-		return view.getComputedStyle( elem );
-	};
-
-var swap = function( elem, options, callback ) {
-	var ret, name,
-		old = {};
-
-	// Remember the old values, and insert the new ones
-	for ( name in options ) {
-		old[ name ] = elem.style[ name ];
-		elem.style[ name ] = options[ name ];
-	}
-
-	ret = callback.call( elem );
-
-	// Revert the old values
-	for ( name in options ) {
-		elem.style[ name ] = old[ name ];
-	}
-
-	return ret;
-};
-
-
-var rboxStyle = new RegExp( cssExpand.join( "|" ), "i" );
-
-
-
-( function() {
-
-	// Executing both pixelPosition & boxSizingReliable tests require only one layout
-	// so they're executed at the same time to save the second computation.
-	function computeStyleTests() {
-
-		// This is a singleton, we need to execute it only once
-		if ( !div ) {
-			return;
-		}
-
-		container.style.cssText = "position:absolute;left:-11111px;width:60px;" +
-			"margin-top:1px;padding:0;border:0";
-		div.style.cssText =
-			"position:relative;display:block;box-sizing:border-box;overflow:scroll;" +
-			"margin:auto;border:1px;padding:1px;" +
-			"width:60%;top:1%";
-		documentElement.appendChild( container ).appendChild( div );
-
-		var divStyle = window.getComputedStyle( div );
-		pixelPositionVal = divStyle.top !== "1%";
-
-		// Support: Android 4.0 - 4.3 only, Firefox <=3 - 44
-		reliableMarginLeftVal = roundPixelMeasures( divStyle.marginLeft ) === 12;
-
-		// Support: Android 4.0 - 4.3 only, Safari <=9.1 - 10.1, iOS <=7.0 - 9.3
-		// Some styles come back with percentage values, even though they shouldn't
-		div.style.right = "60%";
-		pixelBoxStylesVal = roundPixelMeasures( divStyle.right ) === 36;
-
-		// Support: IE 9 - 11 only
-		// Detect misreporting of content dimensions for box-sizing:border-box elements
-		boxSizingReliableVal = roundPixelMeasures( divStyle.width ) === 36;
-
-		// Support: IE 9 only
-		// Detect overflow:scroll screwiness (gh-3699)
-		// Support: Chrome <=64
-		// Don't get tricked when zoom affects offsetWidth (gh-4029)
-		div.style.position = "absolute";
-		scrollboxSizeVal = roundPixelMeasures( div.offsetWidth / 3 ) === 12;
-
-		documentElement.removeChild( container );
-
-		// Nullify the div so it wouldn't be stored in the memory and
-		// it will also be a sign that checks already performed
-		div = null;
-	}
-
-	function roundPixelMeasures( measure ) {
-		return Math.round( parseFloat( measure ) );
-	}
-
-	var pixelPositionVal, boxSizingReliableVal, scrollboxSizeVal, pixelBoxStylesVal,
-		reliableTrDimensionsVal, reliableMarginLeftVal,
-		container = document.createElement( "div" ),
-		div = document.createElement( "div" );
-
-	// Finish early in limited (non-browser) environments
-	if ( !div.style ) {
-		return;
-	}
-
-	// Support: IE <=9 - 11 only
-	// Style of cloned element affects source element cloned (#8908)
-	div.style.backgroundClip = "content-box";
-	div.cloneNode( true ).style.backgroundClip = "";
-	support.clearCloneStyle = div.style.backgroundClip === "content-box";
-
-	jQuery.extend( support, {
-		boxSizingReliable: function() {
-			computeStyleTests();
-			return boxSizingReliableVal;
-		},
-		pixelBoxStyles: function() {
-			computeStyleTests();
-			return pixelBoxStylesVal;
-		},
-		pixelPosition: function() {
-			computeStyleTests();
-			return pixelPositionVal;
-		},
-		reliableMarginLeft: function() {
-			computeStyleTests();
-			return reliableMarginLeftVal;
-		},
-		scrollboxSize: function() {
-			computeStyleTests();
-			return scrollboxSizeVal;
-		},
-
-		// Support: IE 9 - 11+, Edge 15 - 18+
-		// IE/Edge misreport `getComputedStyle` of table rows with width/height
-		// set in CSS while `offset*` properties report correct values.
-		// Behavior in IE 9 is more subtle than in newer versions & it passes
-		// some versions of this test; make sure not to make it pass there!
-		//
-		// Support: Firefox 70+
-		// Only Firefox includes border widths
-		// in computed dimensions. (gh-4529)
-		reliableTrDimensions: function() {
-			var table, tr, trChild, trStyle;
-			if ( reliableTrDimensionsVal == null ) {
-				table = document.createElement( "table" );
-				tr = document.createElement( "tr" );
-				trChild = document.createElement( "div" );
-
-				table.style.cssText = "position:absolute;left:-11111px;border-collapse:separate";
-				tr.style.cssText = "border:1px solid";
-
-				// Support: Chrome 86+
-				// Height set through cssText does not get applied.
-				// Computed height then comes back as 0.
-				tr.style.height = "1px";
-				trChild.style.height = "9px";
-
-				// Support: Android 8 Chrome 86+
-				// In our bodyBackground.html iframe,
-				// display for all div elements is set to "inline",
-				// which causes a problem only in Android 8 Chrome 86.
-				// Ensuring the div is display: block
-				// gets around this issue.
-				trChild.style.display = "block";
-
-				documentElement
-					.appendChild( table )
-					.appendChild( tr )
-					.appendChild( trChild );
-
-				trStyle = window.getComputedStyle( tr );
-				reliableTrDimensionsVal = ( parseInt( trStyle.height, 10 ) +
-					parseInt( trStyle.borderTopWidth, 10 ) +
-					parseInt( trStyle.borderBottomWidth, 10 ) ) === tr.offsetHeight;
-
-				documentElement.removeChild( table );
-			}
-			return reliableTrDimensionsVal;
-		}
-	} );
-} )();
-
-
-function curCSS( elem, name, computed ) {
-	var width, minWidth, maxWidth, ret,
-
-		// Support: Firefox 51+
-		// Retrieving style before computed somehow
-		// fixes an issue with getting wrong values
-		// on detached elements
-		style = elem.style;
-
-	computed = computed || getStyles( elem );
-
-	// getPropertyValue is needed for:
-	//   .css('filter') (IE 9 only, #12537)
-	//   .css('--customProperty) (#3144)
-	if ( computed ) {
-		ret = computed.getPropertyValue( name ) || computed[ name ];
-
-		if ( ret === "" && !isAttached( elem ) ) {
-			ret = jQuery.style( elem, name );
-		}
-
-		// A tribute to the "awesome hack by Dean Edwards"
-		// Android Browser returns percentage for some values,
-		// but width seems to be reliably pixels.
-		// This is against the CSSOM draft spec:
-		// https://drafts.csswg.org/cssom/#resolved-values
-		if ( !support.pixelBoxStyles() && rnumnonpx.test( ret ) && rboxStyle.test( name ) ) {
-
-			// Remember the original values
-			width = style.width;
-			minWidth = style.minWidth;
-			maxWidth = style.maxWidth;
-
-			// Put in the new values to get a computed value out
-			style.minWidth = style.maxWidth = style.width = ret;
-			ret = computed.width;
-
-			// Revert the changed values
-			style.width = width;
-			style.minWidth = minWidth;
-			style.maxWidth = maxWidth;
-		}
-	}
-
-	return ret !== undefined ?
-
-		// Support: IE <=9 - 11 only
-		// IE returns zIndex value as an integer.
-		ret + "" :
-		ret;
-}
-
-
-function addGetHookIf( conditionFn, hookFn ) {
-
-	// Define the hook, we'll check on the first run if it's really needed.
-	return {
-		get: function() {
-			if ( conditionFn() ) {
-
-				// Hook not needed (or it's not possible to use it due
-				// to missing dependency), remove it.
-				delete this.get;
-				return;
-			}
-
-			// Hook needed; redefine it so that the support test is not executed again.
-			return ( this.get = hookFn ).apply( this, arguments );
-		}
-	};
-}
-
-
-var cssPrefixes = [ "Webkit", "Moz", "ms" ],
-	emptyStyle = document.createElement( "div" ).style,
-	vendorProps = {};
-
-// Return a vendor-prefixed property or undefined
-function vendorPropName( name ) {
-
-	// Check for vendor prefixed names
-	var capName = name[ 0 ].toUpperCase() + name.slice( 1 ),
-		i = cssPrefixes.length;
-
-	while ( i-- ) {
-		name = cssPrefixes[ i ] + capName;
-		if ( name in emptyStyle ) {
-			return name;
-		}
-	}
-}
-
-// Return a potentially-mapped jQuery.cssProps or vendor prefixed property
-function finalPropName( name ) {
-	var final = jQuery.cssProps[ name ] || vendorProps[ name ];
-
-	if ( final ) {
-		return final;
-	}
-	if ( name in emptyStyle ) {
-		return name;
-	}
-	return vendorProps[ name ] = vendorPropName( name ) || name;
-}
-
-
-var
-
-	// Swappable if display is none or starts with table
-	// except "table", "table-cell", or "table-caption"
-	// See here for display values: https://developer.mozilla.org/en-US/docs/CSS/display
-	rdisplayswap = /^(none|table(?!-c[ea]).+)/,
-	rcustomProp = /^--/,
-	cssShow = { position: "absolute", visibility: "hidden", display: "block" },
-	cssNormalTransform = {
-		letterSpacing: "0",
-		fontWeight: "400"
-	};
-
-function setPositiveNumber( _elem, value, subtract ) {
-
-	// Any relative (+/-) values have already been
-	// normalized at this point
-	var matches = rcssNum.exec( value );
-	return matches ?
-
-		// Guard against undefined "subtract", e.g., when used as in cssHooks
-		Math.max( 0, matches[ 2 ] - ( subtract || 0 ) ) + ( matches[ 3 ] || "px" ) :
-		value;
-}
-
-function boxModelAdjustment( elem, dimension, box, isBorderBox, styles, computedVal ) {
-	var i = dimension === "width" ? 1 : 0,
-		extra = 0,
-		delta = 0;
-
-	// Adjustment may not be necessary
-	if ( box === ( isBorderBox ? "border" : "content" ) ) {
-		return 0;
-	}
-
-	for ( ; i < 4; i += 2 ) {
-
-		// Both box models exclude margin
-		if ( box === "margin" ) {
-			delta += jQuery.css( elem, box + cssExpand[ i ], true, styles );
-		}
-
-		// If we get here with a content-box, we're seeking "padding" or "border" or "margin"
-		if ( !isBorderBox ) {
-
-			// Add padding
-			delta += jQuery.css( elem, "padding" + cssExpand[ i ], true, styles );
-
-			// For "border" or "margin", add border
-			if ( box !== "padding" ) {
-				delta += jQuery.css( elem, "border" + cssExpand[ i ] + "Width", true, styles );
-
-			// But still keep track of it otherwise
-			} else {
-				extra += jQuery.css( elem, "border" + cssExpand[ i ] + "Width", true, styles );
-			}
-
-		// If we get here with a border-box (content + padding + border), we're seeking "content" or
-		// "padding" or "margin"
-		} else {
-
-			// For "content", subtract padding
-			if ( box === "content" ) {
-				delta -= jQuery.css( elem, "padding" + cssExpand[ i ], true, styles );
-			}
-
-			// For "content" or "padding", subtract border
-			if ( box !== "margin" ) {
-				delta -= jQuery.css( elem, "border" + cssExpand[ i ] + "Width", true, styles );
-			}
-		}
-	}
-
-	// Account for positive content-box scroll gutter when requested by providing computedVal
-	if ( !isBorderBox && computedVal >= 0 ) {
-
-		// offsetWidth/offsetHeight is a rounded sum of content, padding, scroll gutter, and border
-		// Assuming integer scroll gutter, subtract the rest and round down
-		delta += Math.max( 0, Math.ceil(
-			elem[ "offset" + dimension[ 0 ].toUpperCase() + dimension.slice( 1 ) ] -
-			computedVal -
-			delta -
-			extra -
-			0.5
-
-		// If offsetWidth/offsetHeight is unknown, then we can't determine content-box scroll gutter
-		// Use an explicit zero to avoid NaN (gh-3964)
-		) ) || 0;
-	}
-
-	return delta;
-}
-
-function getWidthOrHeight( elem, dimension, extra ) {
-
-	// Start with computed style
-	var styles = getStyles( elem ),
-
-		// To avoid forcing a reflow, only fetch boxSizing if we need it (gh-4322).
-		// Fake content-box until we know it's needed to know the true value.
-		boxSizingNeeded = !support.boxSizingReliable() || extra,
-		isBorderBox = boxSizingNeeded &&
-			jQuery.css( elem, "boxSizing", false, styles ) === "border-box",
-		valueIsBorderBox = isBorderBox,
-
-		val = curCSS( elem, dimension, styles ),
-		offsetProp = "offset" + dimension[ 0 ].toUpperCase() + dimension.slice( 1 );
-
-	// Support: Firefox <=54
-	// Return a confounding non-pixel value or feign ignorance, as appropriate.
-	if ( rnumnonpx.test( val ) ) {
-		if ( !extra ) {
-			return val;
-		}
-		val = "auto";
-	}
-
-
-	// Support: IE 9 - 11 only
-	// Use offsetWidth/offsetHeight for when box sizing is unreliable.
-	// In those cases, the computed value can be trusted to be border-box.
-	if ( ( !support.boxSizingReliable() && isBorderBox ||
-
-		// Support: IE 10 - 11+, Edge 15 - 18+
-		// IE/Edge misreport `getComputedStyle` of table rows with width/height
-		// set in CSS while `offset*` properties report correct values.
-		// Interestingly, in some cases IE 9 doesn't suffer from this issue.
-		!support.reliableTrDimensions() && nodeName( elem, "tr" ) ||
-
-		// Fall back to offsetWidth/offsetHeight when value is "auto"
-		// This happens for inline elements with no explicit setting (gh-3571)
-		val === "auto" ||
-
-		// Support: Android <=4.1 - 4.3 only
-		// Also use offsetWidth/offsetHeight for misreported inline dimensions (gh-3602)
-		!parseFloat( val ) && jQuery.css( elem, "display", false, styles ) === "inline" ) &&
-
-		// Make sure the element is visible & connected
-		elem.getClientRects().length ) {
-
-		isBorderBox = jQuery.css( elem, "boxSizing", false, styles ) === "border-box";
-
-		// Where available, offsetWidth/offsetHeight approximate border box dimensions.
-		// Where not available (e.g., SVG), assume unreliable box-sizing and interpret the
-		// retrieved value as a content box dimension.
-		valueIsBorderBox = offsetProp in elem;
-		if ( valueIsBorderBox ) {
-			val = elem[ offsetProp ];
-		}
-	}
-
-	// Normalize "" and auto
-	val = parseFloat( val ) || 0;
-
-	// Adjust for the element's box model
-	return ( val +
-		boxModelAdjustment(
-			elem,
-			dimension,
-			extra || ( isBorderBox ? "border" : "content" ),
-			valueIsBorderBox,
-			styles,
-
-			// Provide the current computed size to request scroll gutter calculation (gh-3589)
-			val
-		)
-	) + "px";
-}
-
-jQuery.extend( {
-
-	// Add in style property hooks for overriding the default
-	// behavior of getting and setting a style property
-	cssHooks: {
-		opacity: {
-			get: function( elem, computed ) {
-				if ( computed ) {
-
-					// We should always get a number back from opacity
-					var ret = curCSS( elem, "opacity" );
-					return ret === "" ? "1" : ret;
-				}
-			}
-		}
-	},
-
-	// Don't automatically add "px" to these possibly-unitless properties
-	cssNumber: {
-		"animationIterationCount": true,
-		"columnCount": true,
-		"fillOpacity": true,
-		"flexGrow": true,
-		"flexShrink": true,
-		"fontWeight": true,
-		"gridArea": true,
-		"gridColumn": true,
-		"gridColumnEnd": true,
-		"gridColumnStart": true,
-		"gridRow": true,
-		"gridRowEnd": true,
-		"gridRowStart": true,
-		"lineHeight": true,
-		"opacity": true,
-		"order": true,
-		"orphans": true,
-		"widows": true,
-		"zIndex": true,
-		"zoom": true
-	},
-
-	// Add in properties whose names you wish to fix before
-	// setting or getting the value
-	cssProps: {},
-
-	// Get and set the style property on a DOM Node
-	style: function( elem, name, value, extra ) {
-
-		// Don't set styles on text and comment nodes
-		if ( !elem || elem.nodeType === 3 || elem.nodeType === 8 || !elem.style ) {
-			return;
-		}
-
-		// Make sure that we're working with the right name
-		var ret, type, hooks,
-			origName = camelCase( name ),
-			isCustomProp = rcustomProp.test( name ),
-			style = elem.style;
-
-		// Make sure that we're working with the right name. We don't
-		// want to query the value if it is a CSS custom property
-		// since they are user-defined.
-		if ( !isCustomProp ) {
-			name = finalPropName( origName );
-		}
-
-		// Gets hook for the prefixed version, then unprefixed version
-		hooks = jQuery.cssHooks[ name ] || jQuery.cssHooks[ origName ];
-
-		// Check if we're setting a value
-		if ( value !== undefined ) {
-			type = typeof value;
-
-			// Convert "+=" or "-=" to relative numbers (#7345)
-			if ( type === "string" && ( ret = rcssNum.exec( value ) ) && ret[ 1 ] ) {
-				value = adjustCSS( elem, name, ret );
-
-				// Fixes bug #9237
-				type = "number";
-			}
-
-			// Make sure that null and NaN values aren't set (#7116)
-			if ( value == null || value !== value ) {
-				return;
-			}
-
-			// If a number was passed in, add the unit (except for certain CSS properties)
-			// The isCustomProp check can be removed in jQuery 4.0 when we only auto-append
-			// "px" to a few hardcoded values.
-			if ( type === "number" && !isCustomProp ) {
-				value += ret && ret[ 3 ] || ( jQuery.cssNumber[ origName ] ? "" : "px" );
-			}
-
-			// background-* props affect original clone's values
-			if ( !support.clearCloneStyle && value === "" && name.indexOf( "background" ) === 0 ) {
-				style[ name ] = "inherit";
-			}
-
-			// If a hook was provided, use that value, otherwise just set the specified value
-			if ( !hooks || !( "set" in hooks ) ||
-				( value = hooks.set( elem, value, extra ) ) !== undefined ) {
-
-				if ( isCustomProp ) {
-					style.setProperty( name, value );
-				} else {
-					style[ name ] = value;
-				}
-			}
-
-		} else {
-
-			// If a hook was provided get the non-computed value from there
-			if ( hooks && "get" in hooks &&
-				( ret = hooks.get( elem, false, extra ) ) !== undefined ) {
-
-				return ret;
-			}
-
-			// Otherwise just get the value from the style object
-			return style[ name ];
-		}
-	},
-
-	css: function( elem, name, extra, styles ) {
-		var val, num, hooks,
-			origName = camelCase( name ),
-			isCustomProp = rcustomProp.test( name );
-
-		// Make sure that we're working with the right name. We don't
-		// want to modify the value if it is a CSS custom property
-		// since they are user-defined.
-		if ( !isCustomProp ) {
-			name = finalPropName( origName );
-		}
-
-		// Try prefixed name followed by the unprefixed name
-		hooks = jQuery.cssHooks[ name ] || jQuery.cssHooks[ origName ];
-
-		// If a hook was provided get the computed value from there
-		if ( hooks && "get" in hooks ) {
-			val = hooks.get( elem, true, extra );
-		}
-
-		// Otherwise, if a way to get the computed value exists, use that
-		if ( val === undefined ) {
-			val = curCSS( elem, name, styles );
-		}
-
-		// Convert "normal" to computed value
-		if ( val === "normal" && name in cssNormalTransform ) {
-			val = cssNormalTransform[ name ];
-		}
-
-		// Make numeric if forced or a qualifier was provided and val looks numeric
-		if ( extra === "" || extra ) {
-			num = parseFloat( val );
-			return extra === true || isFinite( num ) ? num || 0 : val;
-		}
-
-		return val;
-	}
-} );
-
-jQuery.each( [ "height", "width" ], function( _i, dimension ) {
-	jQuery.cssHooks[ dimension ] = {
-		get: function( elem, computed, extra ) {
-			if ( computed ) {
-
-				// Certain elements can have dimension info if we invisibly show them
-				// but it must have a current display style that would benefit
-				return rdisplayswap.test( jQuery.css( elem, "display" ) ) &&
-
-					// Support: Safari 8+
-					// Table columns in Safari have non-zero offsetWidth & zero
-					// getBoundingClientRect().width unless display is changed.
-					// Support: IE <=11 only
-					// Running getBoundingClientRect on a disconnected node
-					// in IE throws an error.
-					( !elem.getClientRects().length || !elem.getBoundingClientRect().width ) ?
-					swap( elem, cssShow, function() {
-						return getWidthOrHeight( elem, dimension, extra );
-					} ) :
-					getWidthOrHeight( elem, dimension, extra );
-			}
-		},
-
-		set: function( elem, value, extra ) {
-			var matches,
-				styles = getStyles( elem ),
-
-				// Only read styles.position if the test has a chance to fail
-				// to avoid forcing a reflow.
-				scrollboxSizeBuggy = !support.scrollboxSize() &&
-					styles.position === "absolute",
-
-				// To avoid forcing a reflow, only fetch boxSizing if we need it (gh-3991)
-				boxSizingNeeded = scrollboxSizeBuggy || extra,
-				isBorderBox = boxSizingNeeded &&
-					jQuery.css( elem, "boxSizing", false, styles ) === "border-box",
-				subtract = extra ?
-					boxModelAdjustment(
-						elem,
-						dimension,
-						extra,
-						isBorderBox,
-						styles
-					) :
-					0;
-
-			// Account for unreliable border-box dimensions by comparing offset* to computed and
-			// faking a content-box to get border and padding (gh-3699)
-			if ( isBorderBox && scrollboxSizeBuggy ) {
-				subtract -= Math.ceil(
-					elem[ "offset" + dimension[ 0 ].toUpperCase() + dimension.slice( 1 ) ] -
-					parseFloat( styles[ dimension ] ) -
-					boxModelAdjustment( elem, dimension, "border", false, styles ) -
-					0.5
-				);
-			}
-
-			// Convert to pixels if value adjustment is needed
-			if ( subtract && ( matches = rcssNum.exec( value ) ) &&
-				( matches[ 3 ] || "px" ) !== "px" ) {
-
-				elem.style[ dimension ] = value;
-				value = jQuery.css( elem, dimension );
-			}
-
-			return setPositiveNumber( elem, value, subtract );
-		}
-	};
-} );
-
-jQuery.cssHooks.marginLeft = addGetHookIf( support.reliableMarginLeft,
-	function( elem, computed ) {
-		if ( computed ) {
-			return ( parseFloat( curCSS( elem, "marginLeft" ) ) ||
-				elem.getBoundingClientRect().left -
-					swap( elem, { marginLeft: 0 }, function() {
-						return elem.getBoundingClientRect().left;
-					} )
-			) + "px";
-		}
-	}
-);
-
-// These hooks are used by animate to expand properties
-jQuery.each( {
-	margin: "",
-	padding: "",
-	border: "Width"
-}, function( prefix, suffix ) {
-	jQuery.cssHooks[ prefix + suffix ] = {
-		expand: function( value ) {
-			var i = 0,
-				expanded = {},
-
-				// Assumes a single number if not a string
-				parts = typeof value === "string" ? value.split( " " ) : [ value ];
-
-			for ( ; i < 4; i++ ) {
-				expanded[ prefix + cssExpand[ i ] + suffix ] =
-					parts[ i ] || parts[ i - 2 ] || parts[ 0 ];
-			}
-
-			return expanded;
-		}
-	};
-
-	if ( prefix !== "margin" ) {
-		jQuery.cssHooks[ prefix + suffix ].set = setPositiveNumber;
-	}
-} );
-
-jQuery.fn.extend( {
-	css: function( name, value ) {
-		return access( this, function( elem, name, value ) {
-			var styles, len,
-				map = {},
-				i = 0;
-
-			if ( Array.isArray( name ) ) {
-				styles = getStyles( elem );
-				len = name.length;
-
-				for ( ; i < len; i++ ) {
-					map[ name[ i ] ] = jQuery.css( elem, name[ i ], false, styles );
-				}
-
-				return map;
-			}
-
-			return value !== undefined ?
-				jQuery.style( elem, name, value ) :
-				jQuery.css( elem, name );
-		}, name, value, arguments.length > 1 );
-	}
-} );
-
-
-function Tween( elem, options, prop, end, easing ) {
-	return new Tween.prototype.init( elem, options, prop, end, easing );
-}
-jQuery.Tween = Tween;
-
-Tween.prototype = {
-	constructor: Tween,
-	init: function( elem, options, prop, end, easing, unit ) {
-		this.elem = elem;
-		this.prop = prop;
-		this.easing = easing || jQuery.easing._default;
-		this.options = options;
-		this.start = this.now = this.cur();
-		this.end = end;
-		this.unit = unit || ( jQuery.cssNumber[ prop ] ? "" : "px" );
-	},
-	cur: function() {
-		var hooks = Tween.propHooks[ this.prop ];
-
-		return hooks && hooks.get ?
-			hooks.get( this ) :
-			Tween.propHooks._default.get( this );
-	},
-	run: function( percent ) {
-		var eased,
-			hooks = Tween.propHooks[ this.prop ];
-
-		if ( this.options.duration ) {
-			this.pos = eased = jQuery.easing[ this.easing ](
-				percent, this.options.duration * percent, 0, 1, this.options.duration
-			);
-		} else {
-			this.pos = eased = percent;
-		}
-		this.now = ( this.end - this.start ) * eased + this.start;
-
-		if ( this.options.step ) {
-			this.options.step.call( this.elem, this.now, this );
-		}
-
-		if ( hooks && hooks.set ) {
-			hooks.set( this );
-		} else {
-			Tween.propHooks._default.set( this );
-		}
-		return this;
-	}
-};
-
-Tween.prototype.init.prototype = Tween.prototype;
-
-Tween.propHooks = {
-	_default: {
-		get: function( tween ) {
-			var result;
-
-			// Use a property on the element directly when it is not a DOM element,
-			// or when there is no matching style property that exists.
-			if ( tween.elem.nodeType !== 1 ||
-				tween.elem[ tween.prop ] != null && tween.elem.style[ tween.prop ] == null ) {
-				return tween.elem[ tween.prop ];
-			}
-
-			// Passing an empty string as a 3rd parameter to .css will automatically
-			// attempt a parseFloat and fallback to a string if the parse fails.
-			// Simple values such as "10px" are parsed to Float;
-			// complex values such as "rotate(1rad)" are returned as-is.
-			result = jQuery.css( tween.elem, tween.prop, "" );
-
-			// Empty strings, null, undefined and "auto" are converted to 0.
-			return !result || result === "auto" ? 0 : result;
-		},
-		set: function( tween ) {
-
-			// Use step hook for back compat.
-			// Use cssHook if its there.
-			// Use .style if available and use plain properties where available.
-			if ( jQuery.fx.step[ tween.prop ] ) {
-				jQuery.fx.step[ tween.prop ]( tween );
-			} else if ( tween.elem.nodeType === 1 && (
-				jQuery.cssHooks[ tween.prop ] ||
-					tween.elem.style[ finalPropName( tween.prop ) ] != null ) ) {
-				jQuery.style( tween.elem, tween.prop, tween.now + tween.unit );
-			} else {
-				tween.elem[ tween.prop ] = tween.now;
-			}
-		}
-	}
-};
-
-// Support: IE <=9 only
-// Panic based approach to setting things on disconnected nodes
-Tween.propHooks.scrollTop = Tween.propHooks.scrollLeft = {
-	set: function( tween ) {
-		if ( tween.elem.nodeType && tween.elem.parentNode ) {
-			tween.elem[ tween.prop ] = tween.now;
-		}
-	}
-};
-
-jQuery.easing = {
-	linear: function( p ) {
-		return p;
-	},
-	swing: function( p ) {
-		return 0.5 - Math.cos( p * Math.PI ) / 2;
-	},
-	_default: "swing"
-};
-
-jQuery.fx = Tween.prototype.init;
-
-// Back compat <1.8 extension point
-jQuery.fx.step = {};
-
-
-
-
-var
-	fxNow, inProgress,
-	rfxtypes = /^(?:toggle|show|hide)$/,
-	rrun = /queueHooks$/;
-
-function schedule() {
-	if ( inProgress ) {
-		if ( document.hidden === false && window.requestAnimationFrame ) {
-			window.requestAnimationFrame( schedule );
-		} else {
-			window.setTimeout( schedule, jQuery.fx.interval );
-		}
-
-		jQuery.fx.tick();
-	}
-}
-
-// Animations created synchronously will run synchronously
-function createFxNow() {
-	window.setTimeout( function() {
-		fxNow = undefined;
-	} );
-	return ( fxNow = Date.now() );
-}
-
-// Generate parameters to create a standard animation
-function genFx( type, includeWidth ) {
-	var which,
-		i = 0,
-		attrs = { height: type };
-
-	// If we include width, step value is 1 to do all cssExpand values,
-	// otherwise step value is 2 to skip over Left and Right
-	includeWidth = includeWidth ? 1 : 0;
-	for ( ; i < 4; i += 2 - includeWidth ) {
-		which = cssExpand[ i ];
-		attrs[ "margin" + which ] = attrs[ "padding" + which ] = type;
-	}
-
-	if ( includeWidth ) {
-		attrs.opacity = attrs.width = type;
-	}
-
-	return attrs;
-}
-
-function createTween( value, prop, animation ) {
-	var tween,
-		collection = ( Animation.tweeners[ prop ] || [] ).concat( Animation.tweeners[ "*" ] ),
-		index = 0,
-		length = collection.length;
-	for ( ; index < length; index++ ) {
-		if ( ( tween = collection[ index ].call( animation, prop, value ) ) ) {
-
-			// We're done with this property
-			return tween;
-		}
-	}
-}
-
-function defaultPrefilter( elem, props, opts ) {
-	var prop, value, toggle, hooks, oldfire, propTween, restoreDisplay, display,
-		isBox = "width" in props || "height" in props,
-		anim = this,
-		orig = {},
-		style = elem.style,
-		hidden = elem.nodeType && isHiddenWithinTree( elem ),
-		dataShow = dataPriv.get( elem, "fxshow" );
-
-	// Queue-skipping animations hijack the fx hooks
-	if ( !opts.queue ) {
-		hooks = jQuery._queueHooks( elem, "fx" );
-		if ( hooks.unqueued == null ) {
-			hooks.unqueued = 0;
-			oldfire = hooks.empty.fire;
-			hooks.empty.fire = function() {
-				if ( !hooks.unqueued ) {
-					oldfire();
-				}
-			};
-		}
-		hooks.unqueued++;
-
-		anim.always( function() {
-
-			// Ensure the complete handler is called before this completes
-			anim.always( function() {
-				hooks.unqueued--;
-				if ( !jQuery.queue( elem, "fx" ).length ) {
-					hooks.empty.fire();
-				}
-			} );
-		} );
-	}
-
-	// Detect show/hide animations
-	for ( prop in props ) {
-		value = props[ prop ];
-		if ( rfxtypes.test( value ) ) {
-			delete props[ prop ];
-			toggle = toggle || value === "toggle";
-			if ( value === ( hidden ? "hide" : "show" ) ) {
-
-				// Pretend to be hidden if this is a "show" and
-				// there is still data from a stopped show/hide
-				if ( value === "show" && dataShow && dataShow[ prop ] !== undefined ) {
-					hidden = true;
-
-				// Ignore all other no-op show/hide data
-				} else {
-					continue;
-				}
-			}
-			orig[ prop ] = dataShow && dataShow[ prop ] || jQuery.style( elem, prop );
-		}
-	}
-
-	// Bail out if this is a no-op like .hide().hide()
-	propTween = !jQuery.isEmptyObject( props );
-	if ( !propTween && jQuery.isEmptyObject( orig ) ) {
-		return;
-	}
-
-	// Restrict "overflow" and "display" styles during box animations
-	if ( isBox && elem.nodeType === 1 ) {
-
-		// Support: IE <=9 - 11, Edge 12 - 15
-		// Record all 3 overflow attributes because IE does not infer the shorthand
-		// from identically-valued overflowX and overflowY and Edge just mirrors
-		// the overflowX value there.
-		opts.overflow = [ style.overflow, style.overflowX, style.overflowY ];
-
-		// Identify a display type, preferring old show/hide data over the CSS cascade
-		restoreDisplay = dataShow && dataShow.display;
-		if ( restoreDisplay == null ) {
-			restoreDisplay = dataPriv.get( elem, "display" );
-		}
-		display = jQuery.css( elem, "display" );
-		if ( display === "none" ) {
-			if ( restoreDisplay ) {
-				display = restoreDisplay;
-			} else {
-
-				// Get nonempty value(s) by temporarily forcing visibility
-				showHide( [ elem ], true );
-				restoreDisplay = elem.style.display || restoreDisplay;
-				display = jQuery.css( elem, "display" );
-				showHide( [ elem ] );
-			}
-		}
-
-		// Animate inline elements as inline-block
-		if ( display === "inline" || display === "inline-block" && restoreDisplay != null ) {
-			if ( jQuery.css( elem, "float" ) === "none" ) {
-
-				// Restore the original display value at the end of pure show/hide animations
-				if ( !propTween ) {
-					anim.done( function() {
-						style.display = restoreDisplay;
-					} );
-					if ( restoreDisplay == null ) {
-						display = style.display;
-						restoreDisplay = display === "none" ? "" : display;
-					}
-				}
-				style.display = "inline-block";
-			}
-		}
-	}
-
-	if ( opts.overflow ) {
-		style.overflow = "hidden";
-		anim.always( function() {
-			style.overflow = opts.overflow[ 0 ];
-			style.overflowX = opts.overflow[ 1 ];
-			style.overflowY = opts.overflow[ 2 ];
-		} );
-	}
-
-	// Implement show/hide animations
-	propTween = false;
-	for ( prop in orig ) {
-
-		// General show/hide setup for this element animation
-		if ( !propTween ) {
-			if ( dataShow ) {
-				if ( "hidden" in dataShow ) {
-					hidden = dataShow.hidden;
-				}
-			} else {
-				dataShow = dataPriv.access( elem, "fxshow", { display: restoreDisplay } );
-			}
-
-			// Store hidden/visible for toggle so `.stop().toggle()` "reverses"
-			if ( toggle ) {
-				dataShow.hidden = !hidden;
-			}
-
-			// Show elements before animating them
-			if ( hidden ) {
-				showHide( [ elem ], true );
-			}
-
-			/* eslint-disable no-loop-func */
-
-			anim.done( function() {
-
-				/* eslint-enable no-loop-func */
-
-				// The final step of a "hide" animation is actually hiding the element
-				if ( !hidden ) {
-					showHide( [ elem ] );
-				}
-				dataPriv.remove( elem, "fxshow" );
-				for ( prop in orig ) {
-					jQuery.style( elem, prop, orig[ prop ] );
-				}
-			} );
-		}
-
-		// Per-property setup
-		propTween = createTween( hidden ? dataShow[ prop ] : 0, prop, anim );
-		if ( !( prop in dataShow ) ) {
-			dataShow[ prop ] = propTween.start;
-			if ( hidden ) {
-				propTween.end = propTween.start;
-				propTween.start = 0;
-			}
-		}
-	}
-}
-
-function propFilter( props, specialEasing ) {
-	var index, name, easing, value, hooks;
-
-	// camelCase, specialEasing and expand cssHook pass
-	for ( index in props ) {
-		name = camelCase( index );
-		easing = specialEasing[ name ];
-		value = props[ index ];
-		if ( Array.isArray( value ) ) {
-			easing = value[ 1 ];
-			value = props[ index ] = value[ 0 ];
-		}
-
-		if ( index !== name ) {
-			props[ name ] = value;
-			delete props[ index ];
-		}
-
-		hooks = jQuery.cssHooks[ name ];
-		if ( hooks && "expand" in hooks ) {
-			value = hooks.expand( value );
-			delete props[ name ];
-
-			// Not quite $.extend, this won't overwrite existing keys.
-			// Reusing 'index' because we have the correct "name"
-			for ( index in value ) {
-				if ( !( index in props ) ) {
-					props[ index ] = value[ index ];
-					specialEasing[ index ] = easing;
-				}
-			}
-		} else {
-			specialEasing[ name ] = easing;
-		}
-	}
-}
-
-function Animation( elem, properties, options ) {
-	var result,
-		stopped,
-		index = 0,
-		length = Animation.prefilters.length,
-		deferred = jQuery.Deferred().always( function() {
-
-			// Don't match elem in the :animated selector
-			delete tick.elem;
-		} ),
-		tick = function() {
-			if ( stopped ) {
-				return false;
-			}
-			var currentTime = fxNow || createFxNow(),
-				remaining = Math.max( 0, animation.startTime + animation.duration - currentTime ),
-
-				// Support: Android 2.3 only
-				// Archaic crash bug won't allow us to use `1 - ( 0.5 || 0 )` (#12497)
-				temp = remaining / animation.duration || 0,
-				percent = 1 - temp,
-				index = 0,
-				length = animation.tweens.length;
-
-			for ( ; index < length; index++ ) {
-				animation.tweens[ index ].run( percent );
-			}
-
-			deferred.notifyWith( elem, [ animation, percent, remaining ] );
-
-			// If there's more to do, yield
-			if ( percent < 1 && length ) {
-				return remaining;
-			}
-
-			// If this was an empty animation, synthesize a final progress notification
-			if ( !length ) {
-				deferred.notifyWith( elem, [ animation, 1, 0 ] );
-			}
-
-			// Resolve the animation and report its conclusion
-			deferred.resolveWith( elem, [ animation ] );
-			return false;
-		},
-		animation = deferred.promise( {
-			elem: elem,
-			props: jQuery.extend( {}, properties ),
-			opts: jQuery.extend( true, {
-				specialEasing: {},
-				easing: jQuery.easing._default
-			}, options ),
-			originalProperties: properties,
-			originalOptions: options,
-			startTime: fxNow || createFxNow(),
-			duration: options.duration,
-			tweens: [],
-			createTween: function( prop, end ) {
-				var tween = jQuery.Tween( elem, animation.opts, prop, end,
-					animation.opts.specialEasing[ prop ] || animation.opts.easing );
-				animation.tweens.push( tween );
-				return tween;
-			},
-			stop: function( gotoEnd ) {
-				var index = 0,
-
-					// If we are going to the end, we want to run all the tweens
-					// otherwise we skip this part
-					length = gotoEnd ? animation.tweens.length : 0;
-				if ( stopped ) {
-					return this;
-				}
-				stopped = true;
-				for ( ; index < length; index++ ) {
-					animation.tweens[ index ].run( 1 );
-				}
-
-				// Resolve when we played the last frame; otherwise, reject
-				if ( gotoEnd ) {
-					deferred.notifyWith( elem, [ animation, 1, 0 ] );
-					deferred.resolveWith( elem, [ animation, gotoEnd ] );
-				} else {
-					deferred.rejectWith( elem, [ animation, gotoEnd ] );
-				}
-				return this;
-			}
-		} ),
-		props = animation.props;
-
-	propFilter( props, animation.opts.specialEasing );
-
-	for ( ; index < length; index++ ) {
-		result = Animation.prefilters[ index ].call( animation, elem, props, animation.opts );
-		if ( result ) {
-			if ( isFunction( result.stop ) ) {
-				jQuery._queueHooks( animation.elem, animation.opts.queue ).stop =
-					result.stop.bind( result );
-			}
-			return result;
-		}
-	}
-
-	jQuery.map( props, createTween, animation );
-
-	if ( isFunction( animation.opts.start ) ) {
-		animation.opts.start.call( elem, animation );
-	}
-
-	// Attach callbacks from options
-	animation
-		.progress( animation.opts.progress )
-		.done( animation.opts.done, animation.opts.complete )
-		.fail( animation.opts.fail )
-		.always( animation.opts.always );
-
-	jQuery.fx.timer(
-		jQuery.extend( tick, {
-			elem: elem,
-			anim: animation,
-			queue: animation.opts.queue
-		} )
-	);
-
-	return animation;
-}
-
-jQuery.Animation = jQuery.extend( Animation, {
-
-	tweeners: {
-		"*": [ function( prop, value ) {
-			var tween = this.createTween( prop, value );
-			adjustCSS( tween.elem, prop, rcssNum.exec( value ), tween );
-			return tween;
-		} ]
-	},
-
-	tweener: function( props, callback ) {
-		if ( isFunction( props ) ) {
-			callback = props;
-			props = [ "*" ];
-		} else {
-			props = props.match( rnothtmlwhite );
-		}
-
-		var prop,
-			index = 0,
-			length = props.length;
-
-		for ( ; index < length; index++ ) {
-			prop = props[ index ];
-			Animation.tweeners[ prop ] = Animation.tweeners[ prop ] || [];
-			Animation.tweeners[ prop ].unshift( callback );
-		}
-	},
-
-	prefilters: [ defaultPrefilter ],
-
-	prefilter: function( callback, prepend ) {
-		if ( prepend ) {
-			Animation.prefilters.unshift( callback );
-		} else {
-			Animation.prefilters.push( callback );
-		}
-	}
-} );
-
-jQuery.speed = function( speed, easing, fn ) {
-	var opt = speed && typeof speed === "object" ? jQuery.extend( {}, speed ) : {
-		complete: fn || !fn && easing ||
-			isFunction( speed ) && speed,
-		duration: speed,
-		easing: fn && easing || easing && !isFunction( easing ) && easing
-	};
-
-	// Go to the end state if fx are off
-	if ( jQuery.fx.off ) {
-		opt.duration = 0;
-
-	} else {
-		if ( typeof opt.duration !== "number" ) {
-			if ( opt.duration in jQuery.fx.speeds ) {
-				opt.duration = jQuery.fx.speeds[ opt.duration ];
-
-			} else {
-				opt.duration = jQuery.fx.speeds._default;
-			}
-		}
-	}
-
-	// Normalize opt.queue - true/undefined/null -> "fx"
-	if ( opt.queue == null || opt.queue === true ) {
-		opt.queue = "fx";
-	}
-
-	// Queueing
-	opt.old = opt.complete;
-
-	opt.complete = function() {
-		if ( isFunction( opt.old ) ) {
-			opt.old.call( this );
-		}
-
-		if ( opt.queue ) {
-			jQuery.dequeue( this, opt.queue );
-		}
-	};
-
-	return opt;
-};
-
-jQuery.fn.extend( {
-	fadeTo: function( speed, to, easing, callback ) {
-
-		// Show any hidden elements after setting opacity to 0
-		return this.filter( isHiddenWithinTree ).css( "opacity", 0 ).show()
-
-			// Animate to the value specified
-			.end().animate( { opacity: to }, speed, easing, callback );
-	},
-	animate: function( prop, speed, easing, callback ) {
-		var empty = jQuery.isEmptyObject( prop ),
-			optall = jQuery.speed( speed, easing, callback ),
-			doAnimation = function() {
-
-				// Operate on a copy of prop so per-property easing won't be lost
-				var anim = Animation( this, jQuery.extend( {}, prop ), optall );
-
-				// Empty animations, or finishing resolves immediately
-				if ( empty || dataPriv.get( this, "finish" ) ) {
-					anim.stop( true );
-				}
-			};
-
-		doAnimation.finish = doAnimation;
-
-		return empty || optall.queue === false ?
-			this.each( doAnimation ) :
-			this.queue( optall.queue, doAnimation );
-	},
-	stop: function( type, clearQueue, gotoEnd ) {
-		var stopQueue = function( hooks ) {
-			var stop = hooks.stop;
-			delete hooks.stop;
-			stop( gotoEnd );
-		};
-
-		if ( typeof type !== "string" ) {
-			gotoEnd = clearQueue;
-			clearQueue = type;
-			type = undefined;
-		}
-		if ( clearQueue ) {
-			this.queue( type || "fx", [] );
-		}
-
-		return this.each( function() {
-			var dequeue = true,
-				index = type != null && type + "queueHooks",
-				timers = jQuery.timers,
-				data = dataPriv.get( this );
-
-			if ( index ) {
-				if ( data[ index ] && data[ index ].stop ) {
-					stopQueue( data[ index ] );
-				}
-			} else {
-				for ( index in data ) {
-					if ( data[ index ] && data[ index ].stop && rrun.test( index ) ) {
-						stopQueue( data[ index ] );
-					}
-				}
-			}
-
-			for ( index = timers.length; index--; ) {
-				if ( timers[ index ].elem === this &&
-					( type == null || timers[ index ].queue === type ) ) {
-
-					timers[ index ].anim.stop( gotoEnd );
-					dequeue = false;
-					timers.splice( index, 1 );
-				}
-			}
-
-			// Start the next in the queue if the last step wasn't forced.
-			// Timers currently will call their complete callbacks, which
-			// will dequeue but only if they were gotoEnd.
-			if ( dequeue || !gotoEnd ) {
-				jQuery.dequeue( this, type );
-			}
-		} );
-	},
-	finish: function( type ) {
-		if ( type !== false ) {
-			type = type || "fx";
-		}
-		return this.each( function() {
-			var index,
-				data = dataPriv.get( this ),
-				queue = data[ type + "queue" ],
-				hooks = data[ type + "queueHooks" ],
-				timers = jQuery.timers,
-				length = queue ? queue.length : 0;
-
-			// Enable finishing flag on private data
-			data.finish = true;
-
-			// Empty the queue first
-			jQuery.queue( this, type, [] );
-
-			if ( hooks && hooks.stop ) {
-				hooks.stop.call( this, true );
-			}
-
-			// Look for any active animations, and finish them
-			for ( index = timers.length; index--; ) {
-				if ( timers[ index ].elem === this && timers[ index ].queue === type ) {
-					timers[ index ].anim.stop( true );
-					timers.splice( index, 1 );
-				}
-			}
-
-			// Look for any animations in the old queue and finish them
-			for ( index = 0; index < length; index++ ) {
-				if ( queue[ index ] && queue[ index ].finish ) {
-					queue[ index ].finish.call( this );
-				}
-			}
-
-			// Turn off finishing flag
-			delete data.finish;
-		} );
-	}
-} );
-
-jQuery.each( [ "toggle", "show", "hide" ], function( _i, name ) {
-	var cssFn = jQuery.fn[ name ];
-	jQuery.fn[ name ] = function( speed, easing, callback ) {
-		return speed == null || typeof speed === "boolean" ?
-			cssFn.apply( this, arguments ) :
-			this.animate( genFx( name, true ), speed, easing, callback );
-	};
-} );
-
-// Generate shortcuts for custom animations
-jQuery.each( {
-	slideDown: genFx( "show" ),
-	slideUp: genFx( "hide" ),
-	slideToggle: genFx( "toggle" ),
-	fadeIn: { opacity: "show" },
-	fadeOut: { opacity: "hide" },
-	fadeToggle: { opacity: "toggle" }
-}, function( name, props ) {
-	jQuery.fn[ name ] = function( speed, easing, callback ) {
-		return this.animate( props, speed, easing, callback );
-	};
-} );
-
-jQuery.timers = [];
-jQuery.fx.tick = function() {
-	var timer,
-		i = 0,
-		timers = jQuery.timers;
-
-	fxNow = Date.now();
-
-	for ( ; i < timers.length; i++ ) {
-		timer = timers[ i ];
-
-		// Run the timer and safely remove it when done (allowing for external removal)
-		if ( !timer() && timers[ i ] === timer ) {
-			timers.splice( i--, 1 );
-		}
-	}
-
-	if ( !timers.length ) {
-		jQuery.fx.stop();
-	}
-	fxNow = undefined;
-};
-
-jQuery.fx.timer = function( timer ) {
-	jQuery.timers.push( timer );
-	jQuery.fx.start();
-};
-
-jQuery.fx.interval = 13;
-jQuery.fx.start = function() {
-	if ( inProgress ) {
-		return;
-	}
-
-	inProgress = true;
-	schedule();
-};
-
-jQuery.fx.stop = function() {
-	inProgress = null;
-};
-
-jQuery.fx.speeds = {
-	slow: 600,
-	fast: 200,
-
-	// Default speed
-	_default: 400
-};
-
-
-// Based off of the plugin by Clint Helfers, with permission.
-// https://web.archive.org/web/20100324014747/http://blindsignals.com/index.php/2009/07/jquery-delay/
-jQuery.fn.delay = function( time, type ) {
-	time = jQuery.fx ? jQuery.fx.speeds[ time ] || time : time;
-	type = type || "fx";
-
-	return this.queue( type, function( next, hooks ) {
-		var timeout = window.setTimeout( next, time );
-		hooks.stop = function() {
-			window.clearTimeout( timeout );
-		};
-	} );
-};
-
-
-( function() {
-	var input = document.createElement( "input" ),
-		select = document.createElement( "select" ),
-		opt = select.appendChild( document.createElement( "option" ) );
-
-	input.type = "checkbox";
-
-	// Support: Android <=4.3 only
-	// Default value for a checkbox should be "on"
-	support.checkOn = input.value !== "";
-
-	// Support: IE <=11 only
-	// Must access selectedIndex to make default options select
-	support.optSelected = opt.selected;
-
-	// Support: IE <=11 only
-	// An input loses its value after becoming a radio
-	input = document.createElement( "input" );
-	input.value = "t";
-	input.type = "radio";
-	support.radioValue = input.value === "t";
-} )();
-
-
-var boolHook,
-	attrHandle = jQuery.expr.attrHandle;
-
-jQuery.fn.extend( {
-	attr: function( name, value ) {
-		return access( this, jQuery.attr, name, value, arguments.length > 1 );
-	},
-
-	removeAttr: function( name ) {
-		return this.each( function() {
-			jQuery.removeAttr( this, name );
-		} );
-	}
-} );
-
-jQuery.extend( {
-	attr: function( elem, name, value ) {
-		var ret, hooks,
-			nType = elem.nodeType;
-
-		// Don't get/set attributes on text, comment and attribute nodes
-		if ( nType === 3 || nType === 8 || nType === 2 ) {
-			return;
-		}
-
-		// Fallback to prop when attributes are not supported
-		if ( typeof elem.getAttribute === "undefined" ) {
-			return jQuery.prop( elem, name, value );
-		}
-
-		// Attribute hooks are determined by the lowercase version
-		// Grab necessary hook if one is defined
-		if ( nType !== 1 || !jQuery.isXMLDoc( elem ) ) {
-			hooks = jQuery.attrHooks[ name.toLowerCase() ] ||
-				( jQuery.expr.match.bool.test( name ) ? boolHook : undefined );
-		}
-
-		if ( value !== undefined ) {
-			if ( value === null ) {
-				jQuery.removeAttr( elem, name );
-				return;
-			}
-
-			if ( hooks && "set" in hooks &&
-				( ret = hooks.set( elem, value, name ) ) !== undefined ) {
-				return ret;
-			}
-
-			elem.setAttribute( name, value + "" );
-			return value;
-		}
-
-		if ( hooks && "get" in hooks && ( ret = hooks.get( elem, name ) ) !== null ) {
-			return ret;
-		}
-
-		ret = jQuery.find.attr( elem, name );
-
-		// Non-existent attributes return null, we normalize to undefined
-		return ret == null ? undefined : ret;
-	},
-
-	attrHooks: {
-		type: {
-			set: function( elem, value ) {
-				if ( !support.radioValue && value === "radio" &&
-					nodeName( elem, "input" ) ) {
-					var val = elem.value;
-					elem.setAttribute( "type", value );
-					if ( val ) {
-						elem.value = val;
-					}
-					return value;
-				}
-			}
-		}
-	},
-
-	removeAttr: function( elem, value ) {
-		var name,
-			i = 0,
-
-			// Attribute names can contain non-HTML whitespace characters
-			// https://html.spec.whatwg.org/multipage/syntax.html#attributes-2
-			attrNames = value && value.match( rnothtmlwhite );
-
-		if ( attrNames && elem.nodeType === 1 ) {
-			while ( ( name = attrNames[ i++ ] ) ) {
-				elem.removeAttribute( name );
-			}
-		}
-	}
-} );
-
-// Hooks for boolean attributes
-boolHook = {
-	set: function( elem, value, name ) {
-		if ( value === false ) {
-
-			// Remove boolean attributes when set to false
-			jQuery.removeAttr( elem, name );
-		} else {
-			elem.setAttribute( name, name );
-		}
-		return name;
-	}
-};
-
-jQuery.each( jQuery.expr.match.bool.source.match( /\w+/g ), function( _i, name ) {
-	var getter = attrHandle[ name ] || jQuery.find.attr;
-
-	attrHandle[ name ] = function( elem, name, isXML ) {
-		var ret, handle,
-			lowercaseName = name.toLowerCase();
-
-		if ( !isXML ) {
-
-			// Avoid an infinite loop by temporarily removing this function from the getter
-			handle = attrHandle[ lowercaseName ];
-			attrHandle[ lowercaseName ] = ret;
-			ret = getter( elem, name, isXML ) != null ?
-				lowercaseName :
-				null;
-			attrHandle[ lowercaseName ] = handle;
-		}
-		return ret;
-	};
-} );
-
-
-
-
-var rfocusable = /^(?:input|select|textarea|button)$/i,
-	rclickable = /^(?:a|area)$/i;
-
-jQuery.fn.extend( {
-	prop: function( name, value ) {
-		return access( this, jQuery.prop, name, value, arguments.length > 1 );
-	},
-
-	removeProp: function( name ) {
-		return this.each( function() {
-			delete this[ jQuery.propFix[ name ] || name ];
-		} );
-	}
-} );
-
-jQuery.extend( {
-	prop: function( elem, name, value ) {
-		var ret, hooks,
-			nType = elem.nodeType;
-
-		// Don't get/set properties on text, comment and attribute nodes
-		if ( nType === 3 || nType === 8 || nType === 2 ) {
-			return;
-		}
-
-		if ( nType !== 1 || !jQuery.isXMLDoc( elem ) ) {
-
-			// Fix name and attach hooks
-			name = jQuery.propFix[ name ] || name;
-			hooks = jQuery.propHooks[ name ];
-		}
-
-		if ( value !== undefined ) {
-			if ( hooks && "set" in hooks &&
-				( ret = hooks.set( elem, value, name ) ) !== undefined ) {
-				return ret;
-			}
-
-			return ( elem[ name ] = value );
-		}
-
-		if ( hooks && "get" in hooks && ( ret = hooks.get( elem, name ) ) !== null ) {
-			return ret;
-		}
-
-		return elem[ name ];
-	},
-
-	propHooks: {
-		tabIndex: {
-			get: function( elem ) {
-
-				// Support: IE <=9 - 11 only
-				// elem.tabIndex doesn't always return the
-				// correct value when it hasn't been explicitly set
-				// https://web.archive.org/web/20141116233347/http://fluidproject.org/blog/2008/01/09/getting-setting-and-removing-tabindex-values-with-javascript/
-				// Use proper attribute retrieval(#12072)
-				var tabindex = jQuery.find.attr( elem, "tabindex" );
-
-				if ( tabindex ) {
-					return parseInt( tabindex, 10 );
-				}
-
-				if (
-					rfocusable.test( elem.nodeName ) ||
-					rclickable.test( elem.nodeName ) &&
-					elem.href
-				) {
-					return 0;
-				}
-
-				return -1;
-			}
-		}
-	},
-
-	propFix: {
-		"for": "htmlFor",
-		"class": "className"
-	}
-} );
-
-// Support: IE <=11 only
-// Accessing the selectedIndex property
-// forces the browser to respect setting selected
-// on the option
-// The getter ensures a default option is selected
-// when in an optgroup
-// eslint rule "no-unused-expressions" is disabled for this code
-// since it considers such accessions noop
-if ( !support.optSelected ) {
-	jQuery.propHooks.selected = {
-		get: function( elem ) {
-
-			/* eslint no-unused-expressions: "off" */
-
-			var parent = elem.parentNode;
-			if ( parent && parent.parentNode ) {
-				parent.parentNode.selectedIndex;
-			}
-			return null;
-		},
-		set: function( elem ) {
-
-			/* eslint no-unused-expressions: "off" */
-
-			var parent = elem.parentNode;
-			if ( parent ) {
-				parent.selectedIndex;
-
-				if ( parent.parentNode ) {
-					parent.parentNode.selectedIndex;
-				}
-			}
-		}
-	};
-}
-
-jQuery.each( [
-	"tabIndex",
-	"readOnly",
-	"maxLength",
-	"cellSpacing",
-	"cellPadding",
-	"rowSpan",
-	"colSpan",
-	"useMap",
-	"frameBorder",
-	"contentEditable"
-], function() {
-	jQuery.propFix[ this.toLowerCase() ] = this;
-} );
-
-
-
-
-	// Strip and collapse whitespace according to HTML spec
-	// https://infra.spec.whatwg.org/#strip-and-collapse-ascii-whitespace
-	function stripAndCollapse( value ) {
-		var tokens = value.match( rnothtmlwhite ) || [];
-		return tokens.join( " " );
-	}
-
-
-function getClass( elem ) {
-	return elem.getAttribute && elem.getAttribute( "class" ) || "";
-}
-
-function classesToArray( value ) {
-	if ( Array.isArray( value ) ) {
-		return value;
-	}
-	if ( typeof value === "string" ) {
-		return value.match( rnothtmlwhite ) || [];
-	}
-	return [];
-}
-
-jQuery.fn.extend( {
-	addClass: function( value ) {
-		var classes, elem, cur, curValue, clazz, j, finalValue,
-			i = 0;
-
-		if ( isFunction( value ) ) {
-			return this.each( function( j ) {
-				jQuery( this ).addClass( value.call( this, j, getClass( this ) ) );
-			} );
-		}
-
-		classes = classesToArray( value );
-
-		if ( classes.length ) {
-			while ( ( elem = this[ i++ ] ) ) {
-				curValue = getClass( elem );
-				cur = elem.nodeType === 1 && ( " " + stripAndCollapse( curValue ) + " " );
-
-				if ( cur ) {
-					j = 0;
-					while ( ( clazz = classes[ j++ ] ) ) {
-						if ( cur.indexOf( " " + clazz + " " ) < 0 ) {
-							cur += clazz + " ";
-						}
-					}
-
-					// Only assign if different to avoid unneeded rendering.
-					finalValue = stripAndCollapse( cur );
-					if ( curValue !== finalValue ) {
-						elem.setAttribute( "class", finalValue );
-					}
-				}
-			}
-		}
-
-		return this;
-	},
-
-	removeClass: function( value ) {
-		var classes, elem, cur, curValue, clazz, j, finalValue,
-			i = 0;
-
-		if ( isFunction( value ) ) {
-			return this.each( function( j ) {
-				jQuery( this ).removeClass( value.call( this, j, getClass( this ) ) );
-			} );
-		}
-
-		if ( !arguments.length ) {
-			return this.attr( "class", "" );
-		}
-
-		classes = classesToArray( value );
-
-		if ( classes.length ) {
-			while ( ( elem = this[ i++ ] ) ) {
-				curValue = getClass( elem );
-
-				// This expression is here for better compressibility (see addClass)
-				cur = elem.nodeType === 1 && ( " " + stripAndCollapse( curValue ) + " " );
-
-				if ( cur ) {
-					j = 0;
-					while ( ( clazz = classes[ j++ ] ) ) {
-
-						// Remove *all* instances
-						while ( cur.indexOf( " " + clazz + " " ) > -1 ) {
-							cur = cur.replace( " " + clazz + " ", " " );
-						}
-					}
-
-					// Only assign if different to avoid unneeded rendering.
-					finalValue = stripAndCollapse( cur );
-					if ( curValue !== finalValue ) {
-						elem.setAttribute( "class", finalValue );
-					}
-				}
-			}
-		}
-
-		return this;
-	},
-
-	toggleClass: function( value, stateVal ) {
-		var type = typeof value,
-			isValidValue = type === "string" || Array.isArray( value );
-
-		if ( typeof stateVal === "boolean" && isValidValue ) {
-			return stateVal ? this.addClass( value ) : this.removeClass( value );
-		}
-
-		if ( isFunction( value ) ) {
-			return this.each( function( i ) {
-				jQuery( this ).toggleClass(
-					value.call( this, i, getClass( this ), stateVal ),
-					stateVal
-				);
-			} );
-		}
-
-		return this.each( function() {
-			var className, i, self, classNames;
-
-			if ( isValidValue ) {
-
-				// Toggle individual class names
-				i = 0;
-				self = jQuery( this );
-				classNames = classesToArray( value );
-
-				while ( ( className = classNames[ i++ ] ) ) {
-
-					// Check each className given, space separated list
-					if ( self.hasClass( className ) ) {
-						self.removeClass( className );
-					} else {
-						self.addClass( className );
-					}
-				}
-
-			// Toggle whole class name
-			} else if ( value === undefined || type === "boolean" ) {
-				className = getClass( this );
-				if ( className ) {
-
-					// Store className if set
-					dataPriv.set( this, "__className__", className );
-				}
-
-				// If the element has a class name or if we're passed `false`,
-				// then remove the whole classname (if there was one, the above saved it).
-				// Otherwise bring back whatever was previously saved (if anything),
-				// falling back to the empty string if nothing was stored.
-				if ( this.setAttribute ) {
-					this.setAttribute( "class",
-						className || value === false ?
-							"" :
-							dataPriv.get( this, "__className__" ) || ""
-					);
-				}
-			}
-		} );
-	},
-
-	hasClass: function( selector ) {
-		var className, elem,
-			i = 0;
-
-		className = " " + selector + " ";
-		while ( ( elem = this[ i++ ] ) ) {
-			if ( elem.nodeType === 1 &&
-				( " " + stripAndCollapse( getClass( elem ) ) + " " ).indexOf( className ) > -1 ) {
-				return true;
-			}
-		}
-
-		return false;
-	}
-} );
-
-
-
-
-var rreturn = /\r/g;
-
-jQuery.fn.extend( {
-	val: function( value ) {
-		var hooks, ret, valueIsFunction,
-			elem = this[ 0 ];
-
-		if ( !arguments.length ) {
-			if ( elem ) {
-				hooks = jQuery.valHooks[ elem.type ] ||
-					jQuery.valHooks[ elem.nodeName.toLowerCase() ];
-
-				if ( hooks &&
-					"get" in hooks &&
-					( ret = hooks.get( elem, "value" ) ) !== undefined
-				) {
-					return ret;
-				}
-
-				ret = elem.value;
-
-				// Handle most common string cases
-				if ( typeof ret === "string" ) {
-					return ret.replace( rreturn, "" );
-				}
-
-				// Handle cases where value is null/undef or number
-				return ret == null ? "" : ret;
-			}
-
-			return;
-		}
-
-		valueIsFunction = isFunction( value );
-
-		return this.each( function( i ) {
-			var val;
-
-			if ( this.nodeType !== 1 ) {
-				return;
-			}
-
-			if ( valueIsFunction ) {
-				val = value.call( this, i, jQuery( this ).val() );
-			} else {
-				val = value;
-			}
-
-			// Treat null/undefined as ""; convert numbers to string
-			if ( val == null ) {
-				val = "";
-
-			} else if ( typeof val === "number" ) {
-				val += "";
-
-			} else if ( Array.isArray( val ) ) {
-				val = jQuery.map( val, function( value ) {
-					return value == null ? "" : value + "";
-				} );
-			}
-
-			hooks = jQuery.valHooks[ this.type ] || jQuery.valHooks[ this.nodeName.toLowerCase() ];
-
-			// If set returns undefined, fall back to normal setting
-			if ( !hooks || !( "set" in hooks ) || hooks.set( this, val, "value" ) === undefined ) {
-				this.value = val;
-			}
-		} );
-	}
-} );
-
-jQuery.extend( {
-	valHooks: {
-		option: {
-			get: function( elem ) {
-
-				var val = jQuery.find.attr( elem, "value" );
-				return val != null ?
-					val :
-
-					// Support: IE <=10 - 11 only
-					// option.text throws exceptions (#14686, #14858)
-					// Strip and collapse whitespace
-					// https://html.spec.whatwg.org/#strip-and-collapse-whitespace
-					stripAndCollapse( jQuery.text( elem ) );
-			}
-		},
-		select: {
-			get: function( elem ) {
-				var value, option, i,
-					options = elem.options,
-					index = elem.selectedIndex,
-					one = elem.type === "select-one",
-					values = one ? null : [],
-					max = one ? index + 1 : options.length;
-
-				if ( index < 0 ) {
-					i = max;
-
-				} else {
-					i = one ? index : 0;
-				}
-
-				// Loop through all the selected options
-				for ( ; i < max; i++ ) {
-					option = options[ i ];
-
-					// Support: IE <=9 only
-					// IE8-9 doesn't update selected after form reset (#2551)
-					if ( ( option.selected || i === index ) &&
-
-							// Don't return options that are disabled or in a disabled optgroup
-							!option.disabled &&
-							( !option.parentNode.disabled ||
-								!nodeName( option.parentNode, "optgroup" ) ) ) {
-
-						// Get the specific value for the option
-						value = jQuery( option ).val();
-
-						// We don't need an array for one selects
-						if ( one ) {
-							return value;
-						}
-
-						// Multi-Selects return an array
-						values.push( value );
-					}
-				}
-
-				return values;
-			},
-
-			set: function( elem, value ) {
-				var optionSet, option,
-					options = elem.options,
-					values = jQuery.makeArray( value ),
-					i = options.length;
-
-				while ( i-- ) {
-					option = options[ i ];
-
-					/* eslint-disable no-cond-assign */
-
-					if ( option.selected =
-						jQuery.inArray( jQuery.valHooks.option.get( option ), values ) > -1
-					) {
-						optionSet = true;
-					}
-
-					/* eslint-enable no-cond-assign */
-				}
-
-				// Force browsers to behave consistently when non-matching value is set
-				if ( !optionSet ) {
-					elem.selectedIndex = -1;
-				}
-				return values;
-			}
-		}
-	}
-} );
-
-// Radios and checkboxes getter/setter
-jQuery.each( [ "radio", "checkbox" ], function() {
-	jQuery.valHooks[ this ] = {
-		set: function( elem, value ) {
-			if ( Array.isArray( value ) ) {
-				return ( elem.checked = jQuery.inArray( jQuery( elem ).val(), value ) > -1 );
-			}
-		}
-	};
-	if ( !support.checkOn ) {
-		jQuery.valHooks[ this ].get = function( elem ) {
-			return elem.getAttribute( "value" ) === null ? "on" : elem.value;
-		};
-	}
-} );
-
-
-
-
-// Return jQuery for attributes-only inclusion
-
-
-support.focusin = "onfocusin" in window;
-
-
-var rfocusMorph = /^(?:focusinfocus|focusoutblur)$/,
-	stopPropagationCallback = function( e ) {
-		e.stopPropagation();
-	};
-
-jQuery.extend( jQuery.event, {
-
-	trigger: function( event, data, elem, onlyHandlers ) {
-
-		var i, cur, tmp, bubbleType, ontype, handle, special, lastElement,
-			eventPath = [ elem || document ],
-			type = hasOwn.call( event, "type" ) ? event.type : event,
-			namespaces = hasOwn.call( event, "namespace" ) ? event.namespace.split( "." ) : [];
-
-		cur = lastElement = tmp = elem = elem || document;
-
-		// Don't do events on text and comment nodes
-		if ( elem.nodeType === 3 || elem.nodeType === 8 ) {
-			return;
-		}
-
-		// focus/blur morphs to focusin/out; ensure we're not firing them right now
-		if ( rfocusMorph.test( type + jQuery.event.triggered ) ) {
-			return;
-		}
-
-		if ( type.indexOf( "." ) > -1 ) {
-
-			// Namespaced trigger; create a regexp to match event type in handle()
-			namespaces = type.split( "." );
-			type = namespaces.shift();
-			namespaces.sort();
-		}
-		ontype = type.indexOf( ":" ) < 0 && "on" + type;
-
-		// Caller can pass in a jQuery.Event object, Object, or just an event type string
-		event = event[ jQuery.expando ] ?
-			event :
-			new jQuery.Event( type, typeof event === "object" && event );
-
-		// Trigger bitmask: & 1 for native handlers; & 2 for jQuery (always true)
-		event.isTrigger = onlyHandlers ? 2 : 3;
-		event.namespace = namespaces.join( "." );
-		event.rnamespace = event.namespace ?
-			new RegExp( "(^|\\.)" + namespaces.join( "\\.(?:.*\\.|)" ) + "(\\.|$)" ) :
-			null;
-
-		// Clean up the event in case it is being reused
-		event.result = undefined;
-		if ( !event.target ) {
-			event.target = elem;
-		}
-
-		// Clone any incoming data and prepend the event, creating the handler arg list
-		data = data == null ?
-			[ event ] :
-			jQuery.makeArray( data, [ event ] );
-
-		// Allow special events to draw outside the lines
-		special = jQuery.event.special[ type ] || {};
-		if ( !onlyHandlers && special.trigger && special.trigger.apply( elem, data ) === false ) {
-			return;
-		}
-
-		// Determine event propagation path in advance, per W3C events spec (#9951)
-		// Bubble up to document, then to window; watch for a global ownerDocument var (#9724)
-		if ( !onlyHandlers && !special.noBubble && !isWindow( elem ) ) {
-
-			bubbleType = special.delegateType || type;
-			if ( !rfocusMorph.test( bubbleType + type ) ) {
-				cur = cur.parentNode;
-			}
-			for ( ; cur; cur = cur.parentNode ) {
-				eventPath.push( cur );
-				tmp = cur;
-			}
-
-			// Only add window if we got to document (e.g., not plain obj or detached DOM)
-			if ( tmp === ( elem.ownerDocument || document ) ) {
-				eventPath.push( tmp.defaultView || tmp.parentWindow || window );
-			}
-		}
-
-		// Fire handlers on the event path
-		i = 0;
-		while ( ( cur = eventPath[ i++ ] ) && !event.isPropagationStopped() ) {
-			lastElement = cur;
-			event.type = i > 1 ?
-				bubbleType :
-				special.bindType || type;
-
-			// jQuery handler
-			handle = ( dataPriv.get( cur, "events" ) || Object.create( null ) )[ event.type ] &&
-				dataPriv.get( cur, "handle" );
-			if ( handle ) {
-				handle.apply( cur, data );
-			}
-
-			// Native handler
-			handle = ontype && cur[ ontype ];
-			if ( handle && handle.apply && acceptData( cur ) ) {
-				event.result = handle.apply( cur, data );
-				if ( event.result === false ) {
-					event.preventDefault();
-				}
-			}
-		}
-		event.type = type;
-
-		// If nobody prevented the default action, do it now
-		if ( !onlyHandlers && !event.isDefaultPrevented() ) {
-
-			if ( ( !special._default ||
-				special._default.apply( eventPath.pop(), data ) === false ) &&
-				acceptData( elem ) ) {
-
-				// Call a native DOM method on the target with the same name as the event.
-				// Don't do default actions on window, that's where global variables be (#6170)
-				if ( ontype && isFunction( elem[ type ] ) && !isWindow( elem ) ) {
-
-					// Don't re-trigger an onFOO event when we call its FOO() method
-					tmp = elem[ ontype ];
-
-					if ( tmp ) {
-						elem[ ontype ] = null;
-					}
-
-					// Prevent re-triggering of the same event, since we already bubbled it above
-					jQuery.event.triggered = type;
-
-					if ( event.isPropagationStopped() ) {
-						lastElement.addEventListener( type, stopPropagationCallback );
-					}
-
-					elem[ type ]();
-
-					if ( event.isPropagationStopped() ) {
-						lastElement.removeEventListener( type, stopPropagationCallback );
-					}
-
-					jQuery.event.triggered = undefined;
-
-					if ( tmp ) {
-						elem[ ontype ] = tmp;
-					}
-				}
-			}
-		}
-
-		return event.result;
-	},
-
-	// Piggyback on a donor event to simulate a different one
-	// Used only for `focus(in | out)` events
-	simulate: function( type, elem, event ) {
-		var e = jQuery.extend(
-			new jQuery.Event(),
-			event,
-			{
-				type: type,
-				isSimulated: true
-			}
-		);
-
-		jQuery.event.trigger( e, null, elem );
-	}
-
-} );
-
-jQuery.fn.extend( {
-
-	trigger: function( type, data ) {
-		return this.each( function() {
-			jQuery.event.trigger( type, data, this );
-		} );
-	},
-	triggerHandler: function( type, data ) {
-		var elem = this[ 0 ];
-		if ( elem ) {
-			return jQuery.event.trigger( type, data, elem, true );
-		}
-	}
-} );
-
-
-// Support: Firefox <=44
-// Firefox doesn't have focus(in | out) events
-// Related ticket - https://bugzilla.mozilla.org/show_bug.cgi?id=687787
-//
-// Support: Chrome <=48 - 49, Safari <=9.0 - 9.1
-// focus(in | out) events fire after focus & blur events,
-// which is spec violation - http://www.w3.org/TR/DOM-Level-3-Events/#events-focusevent-event-order
-// Related ticket - https://bugs.chromium.org/p/chromium/issues/detail?id=449857
-if ( !support.focusin ) {
-	jQuery.each( { focus: "focusin", blur: "focusout" }, function( orig, fix ) {
-
-		// Attach a single capturing handler on the document while someone wants focusin/focusout
-		var handler = function( event ) {
-			jQuery.event.simulate( fix, event.target, jQuery.event.fix( event ) );
-		};
-
-		jQuery.event.special[ fix ] = {
-			setup: function() {
-
-				// Handle: regular nodes (via `this.ownerDocument`), window
-				// (via `this.document`) & document (via `this`).
-				var doc = this.ownerDocument || this.document || this,
-					attaches = dataPriv.access( doc, fix );
-
-				if ( !attaches ) {
-					doc.addEventListener( orig, handler, true );
-				}
-				dataPriv.access( doc, fix, ( attaches || 0 ) + 1 );
-			},
-			teardown: function() {
-				var doc = this.ownerDocument || this.document || this,
-					attaches = dataPriv.access( doc, fix ) - 1;
-
-				if ( !attaches ) {
-					doc.removeEventListener( orig, handler, true );
-					dataPriv.remove( doc, fix );
-
-				} else {
-					dataPriv.access( doc, fix, attaches );
-				}
-			}
-		};
-	} );
-}
-var location = window.location;
-
-var nonce = { guid: Date.now() };
-
-var rquery = ( /\?/ );
-
-
-
-// Cross-browser xml parsing
-jQuery.parseXML = function( data ) {
-	var xml, parserErrorElem;
-	if ( !data || typeof data !== "string" ) {
-		return null;
-	}
-
-	// Support: IE 9 - 11 only
-	// IE throws on parseFromString with invalid input.
-	try {
-		xml = ( new window.DOMParser() ).parseFromString( data, "text/xml" );
-	} catch ( e ) {}
-
-	parserErrorElem = xml && xml.getElementsByTagName( "parsererror" )[ 0 ];
-	if ( !xml || parserErrorElem ) {
-		jQuery.error( "Invalid XML: " + (
-			parserErrorElem ?
-				jQuery.map( parserErrorElem.childNodes, function( el ) {
-					return el.textContent;
-				} ).join( "\n" ) :
-				data
-		) );
-	}
-	return xml;
-};
-
-
-var
-	rbracket = /\[\]$/,
-	rCRLF = /\r?\n/g,
-	rsubmitterTypes = /^(?:submit|button|image|reset|file)$/i,
-	rsubmittable = /^(?:input|select|textarea|keygen)/i;
-
-function buildParams( prefix, obj, traditional, add ) {
-	var name;
-
-	if ( Array.isArray( obj ) ) {
-
-		// Serialize array item.
-		jQuery.each( obj, function( i, v ) {
-			if ( traditional || rbracket.test( prefix ) ) {
-
-				// Treat each array item as a scalar.
-				add( prefix, v );
-
-			} else {
-
-				// Item is non-scalar (array or object), encode its numeric index.
-				buildParams(
-					prefix + "[" + ( typeof v === "object" && v != null ? i : "" ) + "]",
-					v,
-					traditional,
-					add
-				);
-			}
-		} );
-
-	} else if ( !traditional && toType( obj ) === "object" ) {
-
-		// Serialize object item.
-		for ( name in obj ) {
-			buildParams( prefix + "[" + name + "]", obj[ name ], traditional, add );
-		}
-
-	} else {
-
-		// Serialize scalar item.
-		add( prefix, obj );
-	}
-}
-
-// Serialize an array of form elements or a set of
-// key/values into a query string
-jQuery.param = function( a, traditional ) {
-	var prefix,
-		s = [],
-		add = function( key, valueOrFunction ) {
-
-			// If value is a function, invoke it and use its return value
-			var value = isFunction( valueOrFunction ) ?
-				valueOrFunction() :
-				valueOrFunction;
-
-			s[ s.length ] = encodeURIComponent( key ) + "=" +
-				encodeURIComponent( value == null ? "" : value );
-		};
-
-	if ( a == null ) {
-		return "";
-	}
-
-	// If an array was passed in, assume that it is an array of form elements.
-	if ( Array.isArray( a ) || ( a.jquery && !jQuery.isPlainObject( a ) ) ) {
-
-		// Serialize the form elements
-		jQuery.each( a, function() {
-			add( this.name, this.value );
-		} );
-
-	} else {
-
-		// If traditional, encode the "old" way (the way 1.3.2 or older
-		// did it), otherwise encode params recursively.
-		for ( prefix in a ) {
-			buildParams( prefix, a[ prefix ], traditional, add );
-		}
-	}
-
-	// Return the resulting serialization
-	return s.join( "&" );
-};
-
-jQuery.fn.extend( {
-	serialize: function() {
-		return jQuery.param( this.serializeArray() );
-	},
-	serializeArray: function() {
-		return this.map( function() {
-
-			// Can add propHook for "elements" to filter or add form elements
-			var elements = jQuery.prop( this, "elements" );
-			return elements ? jQuery.makeArray( elements ) : this;
-		} ).filter( function() {
-			var type = this.type;
-
-			// Use .is( ":disabled" ) so that fieldset[disabled] works
-			return this.name && !jQuery( this ).is( ":disabled" ) &&
-				rsubmittable.test( this.nodeName ) && !rsubmitterTypes.test( type ) &&
-				( this.checked || !rcheckableType.test( type ) );
-		} ).map( function( _i, elem ) {
-			var val = jQuery( this ).val();
-
-			if ( val == null ) {
-				return null;
-			}
-
-			if ( Array.isArray( val ) ) {
-				return jQuery.map( val, function( val ) {
-					return { name: elem.name, value: val.replace( rCRLF, "\r\n" ) };
-				} );
-			}
-
-			return { name: elem.name, value: val.replace( rCRLF, "\r\n" ) };
-		} ).get();
-	}
-} );
-
-
-var
-	r20 = /%20/g,
-	rhash = /#.*$/,
-	rantiCache = /([?&])_=[^&]*/,
-	rheaders = /^(.*?):[ \t]*([^\r\n]*)$/mg,
-
-	// #7653, #8125, #8152: local protocol detection
-	rlocalProtocol = /^(?:about|app|app-storage|.+-extension|file|res|widget):$/,
-	rnoContent = /^(?:GET|HEAD)$/,
-	rprotocol = /^\/\//,
-
-	/* Prefilters
-	 * 1) They are useful to introduce custom dataTypes (see ajax/jsonp.js for an example)
-	 * 2) These are called:
-	 *    - BEFORE asking for a transport
-	 *    - AFTER param serialization (s.data is a string if s.processData is true)
-	 * 3) key is the dataType
-	 * 4) the catchall symbol "*" can be used
-	 * 5) execution will start with transport dataType and THEN continue down to "*" if needed
-	 */
-	prefilters = {},
-
-	/* Transports bindings
-	 * 1) key is the dataType
-	 * 2) the catchall symbol "*" can be used
-	 * 3) selection will start with transport dataType and THEN go to "*" if needed
-	 */
-	transports = {},
-
-	// Avoid comment-prolog char sequence (#10098); must appease lint and evade compression
-	allTypes = "*/".concat( "*" ),
-
-	// Anchor tag for parsing the document origin
-	originAnchor = document.createElement( "a" );
-
-originAnchor.href = location.href;
-
-// Base "constructor" for jQuery.ajaxPrefilter and jQuery.ajaxTransport
-function addToPrefiltersOrTransports( structure ) {
-
-	// dataTypeExpression is optional and defaults to "*"
-	return function( dataTypeExpression, func ) {
-
-		if ( typeof dataTypeExpression !== "string" ) {
-			func = dataTypeExpression;
-			dataTypeExpression = "*";
-		}
-
-		var dataType,
-			i = 0,
-			dataTypes = dataTypeExpression.toLowerCase().match( rnothtmlwhite ) || [];
-
-		if ( isFunction( func ) ) {
-
-			// For each dataType in the dataTypeExpression
-			while ( ( dataType = dataTypes[ i++ ] ) ) {
-
-				// Prepend if requested
-				if ( dataType[ 0 ] === "+" ) {
-					dataType = dataType.slice( 1 ) || "*";
-					( structure[ dataType ] = structure[ dataType ] || [] ).unshift( func );
-
-				// Otherwise append
-				} else {
-					( structure[ dataType ] = structure[ dataType ] || [] ).push( func );
-				}
-			}
-		}
-	};
-}
-
-// Base inspection function for prefilters and transports
-function inspectPrefiltersOrTransports( structure, options, originalOptions, jqXHR ) {
-
-	var inspected = {},
-		seekingTransport = ( structure === transports );
-
-	function inspect( dataType ) {
-		var selected;
-		inspected[ dataType ] = true;
-		jQuery.each( structure[ dataType ] || [], function( _, prefilterOrFactory ) {
-			var dataTypeOrTransport = prefilterOrFactory( options, originalOptions, jqXHR );
-			if ( typeof dataTypeOrTransport === "string" &&
-				!seekingTransport && !inspected[ dataTypeOrTransport ] ) {
-
-				options.dataTypes.unshift( dataTypeOrTransport );
-				inspect( dataTypeOrTransport );
-				return false;
-			} else if ( seekingTransport ) {
-				return !( selected = dataTypeOrTransport );
-			}
-		} );
-		return selected;
-	}
-
-	return inspect( options.dataTypes[ 0 ] ) || !inspected[ "*" ] && inspect( "*" );
-}
-
-// A special extend for ajax options
-// that takes "flat" options (not to be deep extended)
-// Fixes #9887
-function ajaxExtend( target, src ) {
-	var key, deep,
-		flatOptions = jQuery.ajaxSettings.flatOptions || {};
-
-	for ( key in src ) {
-		if ( src[ key ] !== undefined ) {
-			( flatOptions[ key ] ? target : ( deep || ( deep = {} ) ) )[ key ] = src[ key ];
-		}
-	}
-	if ( deep ) {
-		jQuery.extend( true, target, deep );
-	}
-
-	return target;
-}
-
-/* Handles responses to an ajax request:
- * - finds the right dataType (mediates between content-type and expected dataType)
- * - returns the corresponding response
- */
-function ajaxHandleResponses( s, jqXHR, responses ) {
-
-	var ct, type, finalDataType, firstDataType,
-		contents = s.contents,
-		dataTypes = s.dataTypes;
-
-	// Remove auto dataType and get content-type in the process
-	while ( dataTypes[ 0 ] === "*" ) {
-		dataTypes.shift();
-		if ( ct === undefined ) {
-			ct = s.mimeType || jqXHR.getResponseHeader( "Content-Type" );
-		}
-	}
-
-	// Check if we're dealing with a known content-type
-	if ( ct ) {
-		for ( type in contents ) {
-			if ( contents[ type ] && contents[ type ].test( ct ) ) {
-				dataTypes.unshift( type );
-				break;
-			}
-		}
-	}
-
-	// Check to see if we have a response for the expected dataType
-	if ( dataTypes[ 0 ] in responses ) {
-		finalDataType = dataTypes[ 0 ];
-	} else {
-
-		// Try convertible dataTypes
-		for ( type in responses ) {
-			if ( !dataTypes[ 0 ] || s.converters[ type + " " + dataTypes[ 0 ] ] ) {
-				finalDataType = type;
-				break;
-			}
-			if ( !firstDataType ) {
-				firstDataType = type;
-			}
-		}
-
-		// Or just use first one
-		finalDataType = finalDataType || firstDataType;
-	}
-
-	// If we found a dataType
-	// We add the dataType to the list if needed
-	// and return the corresponding response
-	if ( finalDataType ) {
-		if ( finalDataType !== dataTypes[ 0 ] ) {
-			dataTypes.unshift( finalDataType );
-		}
-		return responses[ finalDataType ];
-	}
-}
-
-/* Chain conversions given the request and the original response
- * Also sets the responseXXX fields on the jqXHR instance
- */
-function ajaxConvert( s, response, jqXHR, isSuccess ) {
-	var conv2, current, conv, tmp, prev,
-		converters = {},
-
-		// Work with a copy of dataTypes in case we need to modify it for conversion
-		dataTypes = s.dataTypes.slice();
-
-	// Create converters map with lowercased keys
-	if ( dataTypes[ 1 ] ) {
-		for ( conv in s.converters ) {
-			converters[ conv.toLowerCase() ] = s.converters[ conv ];
-		}
-	}
-
-	current = dataTypes.shift();
-
-	// Convert to each sequential dataType
-	while ( current ) {
-
-		if ( s.responseFields[ current ] ) {
-			jqXHR[ s.responseFields[ current ] ] = response;
-		}
-
-		// Apply the dataFilter if provided
-		if ( !prev && isSuccess && s.dataFilter ) {
-			response = s.dataFilter( response, s.dataType );
-		}
-
-		prev = current;
-		current = dataTypes.shift();
-
-		if ( current ) {
-
-			// There's only work to do if current dataType is non-auto
-			if ( current === "*" ) {
-
-				current = prev;
-
-			// Convert response if prev dataType is non-auto and differs from current
-			} else if ( prev !== "*" && prev !== current ) {
-
-				// Seek a direct converter
-				conv = converters[ prev + " " + current ] || converters[ "* " + current ];
-
-				// If none found, seek a pair
-				if ( !conv ) {
-					for ( conv2 in converters ) {
-
-						// If conv2 outputs current
-						tmp = conv2.split( " " );
-						if ( tmp[ 1 ] === current ) {
-
-							// If prev can be converted to accepted input
-							conv = converters[ prev + " " + tmp[ 0 ] ] ||
-								converters[ "* " + tmp[ 0 ] ];
-							if ( conv ) {
-
-								// Condense equivalence converters
-								if ( conv === true ) {
-									conv = converters[ conv2 ];
-
-								// Otherwise, insert the intermediate dataType
-								} else if ( converters[ conv2 ] !== true ) {
-									current = tmp[ 0 ];
-									dataTypes.unshift( tmp[ 1 ] );
-								}
-								break;
-							}
-						}
-					}
-				}
-
-				// Apply converter (if not an equivalence)
-				if ( conv !== true ) {
-
-					// Unless errors are allowed to bubble, catch and return them
-					if ( conv && s.throws ) {
-						response = conv( response );
-					} else {
-						try {
-							response = conv( response );
-						} catch ( e ) {
-							return {
-								state: "parsererror",
-								error: conv ? e : "No conversion from " + prev + " to " + current
-							};
-						}
-					}
-				}
-			}
-		}
-	}
-
-	return { state: "success", data: response };
-}
-
-jQuery.extend( {
-
-	// Counter for holding the number of active queries
-	active: 0,
-
-	// Last-Modified header cache for next request
-	lastModified: {},
-	etag: {},
-
-	ajaxSettings: {
-		url: location.href,
-		type: "GET",
-		isLocal: rlocalProtocol.test( location.protocol ),
-		global: true,
-		processData: true,
-		async: true,
-		contentType: "application/x-www-form-urlencoded; charset=UTF-8",
-
-		/*
-		timeout: 0,
-		data: null,
-		dataType: null,
-		username: null,
-		password: null,
-		cache: null,
-		throws: false,
-		traditional: false,
-		headers: {},
-		*/
-
-		accepts: {
-			"*": allTypes,
-			text: "text/plain",
-			html: "text/html",
-			xml: "application/xml, text/xml",
-			json: "application/json, text/javascript"
-		},
-
-		contents: {
-			xml: /\bxml\b/,
-			html: /\bhtml/,
-			json: /\bjson\b/
-		},
-
-		responseFields: {
-			xml: "responseXML",
-			text: "responseText",
-			json: "responseJSON"
-		},
-
-		// Data converters
-		// Keys separate source (or catchall "*") and destination types with a single space
-		converters: {
-
-			// Convert anything to text
-			"* text": String,
-
-			// Text to html (true = no transformation)
-			"text html": true,
-
-			// Evaluate text as a json expression
-			"text json": JSON.parse,
-
-			// Parse text as xml
-			"text xml": jQuery.parseXML
-		},
-
-		// For options that shouldn't be deep extended:
-		// you can add your own custom options here if
-		// and when you create one that shouldn't be
-		// deep extended (see ajaxExtend)
-		flatOptions: {
-			url: true,
-			context: true
-		}
-	},
-
-	// Creates a full fledged settings object into target
-	// with both ajaxSettings and settings fields.
-	// If target is omitted, writes into ajaxSettings.
-	ajaxSetup: function( target, settings ) {
-		return settings ?
-
-			// Building a settings object
-			ajaxExtend( ajaxExtend( target, jQuery.ajaxSettings ), settings ) :
-
-			// Extending ajaxSettings
-			ajaxExtend( jQuery.ajaxSettings, target );
-	},
-
-	ajaxPrefilter: addToPrefiltersOrTransports( prefilters ),
-	ajaxTransport: addToPrefiltersOrTransports( transports ),
-
-	// Main method
-	ajax: function( url, options ) {
-
-		// If url is an object, simulate pre-1.5 signature
-		if ( typeof url === "object" ) {
-			options = url;
-			url = undefined;
-		}
-
-		// Force options to be an object
-		options = options || {};
-
-		var transport,
-
-			// URL without anti-cache param
-			cacheURL,
-
-			// Response headers
-			responseHeadersString,
-			responseHeaders,
-
-			// timeout handle
-			timeoutTimer,
-
-			// Url cleanup var
-			urlAnchor,
-
-			// Request state (becomes false upon send and true upon completion)
-			completed,
-
-			// To know if global events are to be dispatched
-			fireGlobals,
-
-			// Loop variable
-			i,
-
-			// uncached part of the url
-			uncached,
-
-			// Create the final options object
-			s = jQuery.ajaxSetup( {}, options ),
-
-			// Callbacks context
-			callbackContext = s.context || s,
-
-			// Context for global events is callbackContext if it is a DOM node or jQuery collection
-			globalEventContext = s.context &&
-				( callbackContext.nodeType || callbackContext.jquery ) ?
-				jQuery( callbackContext ) :
-				jQuery.event,
-
-			// Deferreds
-			deferred = jQuery.Deferred(),
-			completeDeferred = jQuery.Callbacks( "once memory" ),
-
-			// Status-dependent callbacks
-			statusCode = s.statusCode || {},
-
-			// Headers (they are sent all at once)
-			requestHeaders = {},
-			requestHeadersNames = {},
-
-			// Default abort message
-			strAbort = "canceled",
-
-			// Fake xhr
-			jqXHR = {
-				readyState: 0,
-
-				// Builds headers hashtable if needed
-				getResponseHeader: function( key ) {
-					var match;
-					if ( completed ) {
-						if ( !responseHeaders ) {
-							responseHeaders = {};
-							while ( ( match = rheaders.exec( responseHeadersString ) ) ) {
-								responseHeaders[ match[ 1 ].toLowerCase() + " " ] =
-									( responseHeaders[ match[ 1 ].toLowerCase() + " " ] || [] )
-										.concat( match[ 2 ] );
-							}
-						}
-						match = responseHeaders[ key.toLowerCase() + " " ];
-					}
-					return match == null ? null : match.join( ", " );
-				},
-
-				// Raw string
-				getAllResponseHeaders: function() {
-					return completed ? responseHeadersString : null;
-				},
-
-				// Caches the header
-				setRequestHeader: function( name, value ) {
-					if ( completed == null ) {
-						name = requestHeadersNames[ name.toLowerCase() ] =
-							requestHeadersNames[ name.toLowerCase() ] || name;
-						requestHeaders[ name ] = value;
-					}
-					return this;
-				},
-
-				// Overrides response content-type header
-				overrideMimeType: function( type ) {
-					if ( completed == null ) {
-						s.mimeType = type;
-					}
-					return this;
-				},
-
-				// Status-dependent callbacks
-				statusCode: function( map ) {
-					var code;
-					if ( map ) {
-						if ( completed ) {
-
-							// Execute the appropriate callbacks
-							jqXHR.always( map[ jqXHR.status ] );
-						} else {
-
-							// Lazy-add the new callbacks in a way that preserves old ones
-							for ( code in map ) {
-								statusCode[ code ] = [ statusCode[ code ], map[ code ] ];
-							}
-						}
-					}
-					return this;
-				},
-
-				// Cancel the request
-				abort: function( statusText ) {
-					var finalText = statusText || strAbort;
-					if ( transport ) {
-						transport.abort( finalText );
-					}
-					done( 0, finalText );
-					return this;
-				}
-			};
-
-		// Attach deferreds
-		deferred.promise( jqXHR );
-
-		// Add protocol if not provided (prefilters might expect it)
-		// Handle falsy url in the settings object (#10093: consistency with old signature)
-		// We also use the url parameter if available
-		s.url = ( ( url || s.url || location.href ) + "" )
-			.replace( rprotocol, location.protocol + "//" );
-
-		// Alias method option to type as per ticket #12004
-		s.type = options.method || options.type || s.method || s.type;
-
-		// Extract dataTypes list
-		s.dataTypes = ( s.dataType || "*" ).toLowerCase().match( rnothtmlwhite ) || [ "" ];
-
-		// A cross-domain request is in order when the origin doesn't match the current origin.
-		if ( s.crossDomain == null ) {
-			urlAnchor = document.createElement( "a" );
-
-			// Support: IE <=8 - 11, Edge 12 - 15
-			// IE throws exception on accessing the href property if url is malformed,
-			// e.g. http://example.com:80x/
-			try {
-				urlAnchor.href = s.url;
-
-				// Support: IE <=8 - 11 only
-				// Anchor's host property isn't correctly set when s.url is relative
-				urlAnchor.href = urlAnchor.href;
-				s.crossDomain = originAnchor.protocol + "//" + originAnchor.host !==
-					urlAnchor.protocol + "//" + urlAnchor.host;
-			} catch ( e ) {
-
-				// If there is an error parsing the URL, assume it is crossDomain,
-				// it can be rejected by the transport if it is invalid
-				s.crossDomain = true;
-			}
-		}
-
-		// Convert data if not already a string
-		if ( s.data && s.processData && typeof s.data !== "string" ) {
-			s.data = jQuery.param( s.data, s.traditional );
-		}
-
-		// Apply prefilters
-		inspectPrefiltersOrTransports( prefilters, s, options, jqXHR );
-
-		// If request was aborted inside a prefilter, stop there
-		if ( completed ) {
-			return jqXHR;
-		}
-
-		// We can fire global events as of now if asked to
-		// Don't fire events if jQuery.event is undefined in an AMD-usage scenario (#15118)
-		fireGlobals = jQuery.event && s.global;
-
-		// Watch for a new set of requests
-		if ( fireGlobals && jQuery.active++ === 0 ) {
-			jQuery.event.trigger( "ajaxStart" );
-		}
-
-		// Uppercase the type
-		s.type = s.type.toUpperCase();
-
-		// Determine if request has content
-		s.hasContent = !rnoContent.test( s.type );
-
-		// Save the URL in case we're toying with the If-Modified-Since
-		// and/or If-None-Match header later on
-		// Remove hash to simplify url manipulation
-		cacheURL = s.url.replace( rhash, "" );
-
-		// More options handling for requests with no content
-		if ( !s.hasContent ) {
-
-			// Remember the hash so we can put it back
-			uncached = s.url.slice( cacheURL.length );
-
-			// If data is available and should be processed, append data to url
-			if ( s.data && ( s.processData || typeof s.data === "string" ) ) {
-				cacheURL += ( rquery.test( cacheURL ) ? "&" : "?" ) + s.data;
-
-				// #9682: remove data so that it's not used in an eventual retry
-				delete s.data;
-			}
-
-			// Add or update anti-cache param if needed
-			if ( s.cache === false ) {
-				cacheURL = cacheURL.replace( rantiCache, "$1" );
-				uncached = ( rquery.test( cacheURL ) ? "&" : "?" ) + "_=" + ( nonce.guid++ ) +
-					uncached;
-			}
-
-			// Put hash and anti-cache on the URL that will be requested (gh-1732)
-			s.url = cacheURL + uncached;
-
-		// Change '%20' to '+' if this is encoded form body content (gh-2658)
-		} else if ( s.data && s.processData &&
-			( s.contentType || "" ).indexOf( "application/x-www-form-urlencoded" ) === 0 ) {
-			s.data = s.data.replace( r20, "+" );
-		}
-
-		// Set the If-Modified-Since and/or If-None-Match header, if in ifModified mode.
-		if ( s.ifModified ) {
-			if ( jQuery.lastModified[ cacheURL ] ) {
-				jqXHR.setRequestHeader( "If-Modified-Since", jQuery.lastModified[ cacheURL ] );
-			}
-			if ( jQuery.etag[ cacheURL ] ) {
-				jqXHR.setRequestHeader( "If-None-Match", jQuery.etag[ cacheURL ] );
-			}
-		}
-
-		// Set the correct header, if data is being sent
-		if ( s.data && s.hasContent && s.contentType !== false || options.contentType ) {
-			jqXHR.setRequestHeader( "Content-Type", s.contentType );
-		}
-
-		// Set the Accepts header for the server, depending on the dataType
-		jqXHR.setRequestHeader(
-			"Accept",
-			s.dataTypes[ 0 ] && s.accepts[ s.dataTypes[ 0 ] ] ?
-				s.accepts[ s.dataTypes[ 0 ] ] +
-					( s.dataTypes[ 0 ] !== "*" ? ", " + allTypes + "; q=0.01" : "" ) :
-				s.accepts[ "*" ]
-		);
-
-		// Check for headers option
-		for ( i in s.headers ) {
-			jqXHR.setRequestHeader( i, s.headers[ i ] );
-		}
-
-		// Allow custom headers/mimetypes and early abort
-		if ( s.beforeSend &&
-			( s.beforeSend.call( callbackContext, jqXHR, s ) === false || completed ) ) {
-
-			// Abort if not done already and return
-			return jqXHR.abort();
-		}
-
-		// Aborting is no longer a cancellation
-		strAbort = "abort";
-
-		// Install callbacks on deferreds
-		completeDeferred.add( s.complete );
-		jqXHR.done( s.success );
-		jqXHR.fail( s.error );
-
-		// Get transport
-		transport = inspectPrefiltersOrTransports( transports, s, options, jqXHR );
-
-		// If no transport, we auto-abort
-		if ( !transport ) {
-			done( -1, "No Transport" );
-		} else {
-			jqXHR.readyState = 1;
-
-			// Send global event
-			if ( fireGlobals ) {
-				globalEventContext.trigger( "ajaxSend", [ jqXHR, s ] );
-			}
-
-			// If request was aborted inside ajaxSend, stop there
-			if ( completed ) {
-				return jqXHR;
-			}
-
-			// Timeout
-			if ( s.async && s.timeout > 0 ) {
-				timeoutTimer = window.setTimeout( function() {
-					jqXHR.abort( "timeout" );
-				}, s.timeout );
-			}
-
-			try {
-				completed = false;
-				transport.send( requestHeaders, done );
-			} catch ( e ) {
-
-				// Rethrow post-completion exceptions
-				if ( completed ) {
-					throw e;
-				}
-
-				// Propagate others as results
-				done( -1, e );
-			}
-		}
-
-		// Callback for when everything is done
-		function done( status, nativeStatusText, responses, headers ) {
-			var isSuccess, success, error, response, modified,
-				statusText = nativeStatusText;
-
-			// Ignore repeat invocations
-			if ( completed ) {
-				return;
-			}
-
-			completed = true;
-
-			// Clear timeout if it exists
-			if ( timeoutTimer ) {
-				window.clearTimeout( timeoutTimer );
-			}
-
-			// Dereference transport for early garbage collection
-			// (no matter how long the jqXHR object will be used)
-			transport = undefined;
-
-			// Cache response headers
-			responseHeadersString = headers || "";
-
-			// Set readyState
-			jqXHR.readyState = status > 0 ? 4 : 0;
-
-			// Determine if successful
-			isSuccess = status >= 200 && status < 300 || status === 304;
-
-			// Get response data
-			if ( responses ) {
-				response = ajaxHandleResponses( s, jqXHR, responses );
-			}
-
-			// Use a noop converter for missing script but not if jsonp
-			if ( !isSuccess &&
-				jQuery.inArray( "script", s.dataTypes ) > -1 &&
-				jQuery.inArray( "json", s.dataTypes ) < 0 ) {
-				s.converters[ "text script" ] = function() {};
-			}
-
-			// Convert no matter what (that way responseXXX fields are always set)
-			response = ajaxConvert( s, response, jqXHR, isSuccess );
-
-			// If successful, handle type chaining
-			if ( isSuccess ) {
-
-				// Set the If-Modified-Since and/or If-None-Match header, if in ifModified mode.
-				if ( s.ifModified ) {
-					modified = jqXHR.getResponseHeader( "Last-Modified" );
-					if ( modified ) {
-						jQuery.lastModified[ cacheURL ] = modified;
-					}
-					modified = jqXHR.getResponseHeader( "etag" );
-					if ( modified ) {
-						jQuery.etag[ cacheURL ] = modified;
-					}
-				}
-
-				// if no content
-				if ( status === 204 || s.type === "HEAD" ) {
-					statusText = "nocontent";
-
-				// if not modified
-				} else if ( status === 304 ) {
-					statusText = "notmodified";
-
-				// If we have data, let's convert it
-				} else {
-					statusText = response.state;
-					success = response.data;
-					error = response.error;
-					isSuccess = !error;
-				}
-			} else {
-
-				// Extract error from statusText and normalize for non-aborts
-				error = statusText;
-				if ( status || !statusText ) {
-					statusText = "error";
-					if ( status < 0 ) {
-						status = 0;
-					}
-				}
-			}
-
-			// Set data for the fake xhr object
-			jqXHR.status = status;
-			jqXHR.statusText = ( nativeStatusText || statusText ) + "";
-
-			// Success/Error
-			if ( isSuccess ) {
-				deferred.resolveWith( callbackContext, [ success, statusText, jqXHR ] );
-			} else {
-				deferred.rejectWith( callbackContext, [ jqXHR, statusText, error ] );
-			}
-
-			// Status-dependent callbacks
-			jqXHR.statusCode( statusCode );
-			statusCode = undefined;
-
-			if ( fireGlobals ) {
-				globalEventContext.trigger( isSuccess ? "ajaxSuccess" : "ajaxError",
-					[ jqXHR, s, isSuccess ? success : error ] );
-			}
-
-			// Complete
-			completeDeferred.fireWith( callbackContext, [ jqXHR, statusText ] );
-
-			if ( fireGlobals ) {
-				globalEventContext.trigger( "ajaxComplete", [ jqXHR, s ] );
-
-				// Handle the global AJAX counter
-				if ( !( --jQuery.active ) ) {
-					jQuery.event.trigger( "ajaxStop" );
-				}
-			}
-		}
-
-		return jqXHR;
-	},
-
-	getJSON: function( url, data, callback ) {
-		return jQuery.get( url, data, callback, "json" );
-	},
-
-	getScript: function( url, callback ) {
-		return jQuery.get( url, undefined, callback, "script" );
-	}
-} );
-
-jQuery.each( [ "get", "post" ], function( _i, method ) {
-	jQuery[ method ] = function( url, data, callback, type ) {
-
-		// Shift arguments if data argument was omitted
-		if ( isFunction( data ) ) {
-			type = type || callback;
-			callback = data;
-			data = undefined;
-		}
-
-		// The url can be an options object (which then must have .url)
-		return jQuery.ajax( jQuery.extend( {
-			url: url,
-			type: method,
-			dataType: type,
-			data: data,
-			success: callback
-		}, jQuery.isPlainObject( url ) && url ) );
-	};
-} );
-
-jQuery.ajaxPrefilter( function( s ) {
-	var i;
-	for ( i in s.headers ) {
-		if ( i.toLowerCase() === "content-type" ) {
-			s.contentType = s.headers[ i ] || "";
-		}
-	}
-} );
-
-
-jQuery._evalUrl = function( url, options, doc ) {
-	return jQuery.ajax( {
-		url: url,
-
-		// Make this explicit, since user can override this through ajaxSetup (#11264)
-		type: "GET",
-		dataType: "script",
-		cache: true,
-		async: false,
-		global: false,
-
-		// Only evaluate the response if it is successful (gh-4126)
-		// dataFilter is not invoked for failure responses, so using it instead
-		// of the default converter is kludgy but it works.
-		converters: {
-			"text script": function() {}
-		},
-		dataFilter: function( response ) {
-			jQuery.globalEval( response, options, doc );
-		}
-	} );
-};
-
-
-jQuery.fn.extend( {
-	wrapAll: function( html ) {
-		var wrap;
-
-		if ( this[ 0 ] ) {
-			if ( isFunction( html ) ) {
-				html = html.call( this[ 0 ] );
-			}
-
-			// The elements to wrap the target around
-			wrap = jQuery( html, this[ 0 ].ownerDocument ).eq( 0 ).clone( true );
-
-			if ( this[ 0 ].parentNode ) {
-				wrap.insertBefore( this[ 0 ] );
-			}
-
-			wrap.map( function() {
-				var elem = this;
-
-				while ( elem.firstElementChild ) {
-					elem = elem.firstElementChild;
-				}
-
-				return elem;
-			} ).append( this );
-		}
-
-		return this;
-	},
-
-	wrapInner: function( html ) {
-		if ( isFunction( html ) ) {
-			return this.each( function( i ) {
-				jQuery( this ).wrapInner( html.call( this, i ) );
-			} );
-		}
-
-		return this.each( function() {
-			var self = jQuery( this ),
-				contents = self.contents();
-
-			if ( contents.length ) {
-				contents.wrapAll( html );
-
-			} else {
-				self.append( html );
-			}
-		} );
-	},
-
-	wrap: function( html ) {
-		var htmlIsFunction = isFunction( html );
-
-		return this.each( function( i ) {
-			jQuery( this ).wrapAll( htmlIsFunction ? html.call( this, i ) : html );
-		} );
-	},
-
-	unwrap: function( selector ) {
-		this.parent( selector ).not( "body" ).each( function() {
-			jQuery( this ).replaceWith( this.childNodes );
-		} );
-		return this;
-	}
-} );
-
-
-jQuery.expr.pseudos.hidden = function( elem ) {
-	return !jQuery.expr.pseudos.visible( elem );
-};
-jQuery.expr.pseudos.visible = function( elem ) {
-	return !!( elem.offsetWidth || elem.offsetHeight || elem.getClientRects().length );
-};
-
-
-
-
-jQuery.ajaxSettings.xhr = function() {
-	try {
-		return new window.XMLHttpRequest();
-	} catch ( e ) {}
-};
-
-var xhrSuccessStatus = {
-
-		// File protocol always yields status code 0, assume 200
-		0: 200,
-
-		// Support: IE <=9 only
-		// #1450: sometimes IE returns 1223 when it should be 204
-		1223: 204
-	},
-	xhrSupported = jQuery.ajaxSettings.xhr();
-
-support.cors = !!xhrSupported && ( "withCredentials" in xhrSupported );
-support.ajax = xhrSupported = !!xhrSupported;
-
-jQuery.ajaxTransport( function( options ) {
-	var callback, errorCallback;
-
-	// Cross domain only allowed if supported through XMLHttpRequest
-	if ( support.cors || xhrSupported && !options.crossDomain ) {
-		return {
-			send: function( headers, complete ) {
-				var i,
-					xhr = options.xhr();
-
-				xhr.open(
-					options.type,
-					options.url,
-					options.async,
-					options.username,
-					options.password
-				);
-
-				// Apply custom fields if provided
-				if ( options.xhrFields ) {
-					for ( i in options.xhrFields ) {
-						xhr[ i ] = options.xhrFields[ i ];
-					}
-				}
-
-				// Override mime type if needed
-				if ( options.mimeType && xhr.overrideMimeType ) {
-					xhr.overrideMimeType( options.mimeType );
-				}
-
-				// X-Requested-With header
-				// For cross-domain requests, seeing as conditions for a preflight are
-				// akin to a jigsaw puzzle, we simply never set it to be sure.
-				// (it can always be set on a per-request basis or even using ajaxSetup)
-				// For same-domain requests, won't change header if already provided.
-				if ( !options.crossDomain && !headers[ "X-Requested-With" ] ) {
-					headers[ "X-Requested-With" ] = "XMLHttpRequest";
-				}
-
-				// Set headers
-				for ( i in headers ) {
-					xhr.setRequestHeader( i, headers[ i ] );
-				}
-
-				// Callback
-				callback = function( type ) {
-					return function() {
-						if ( callback ) {
-							callback = errorCallback = xhr.onload =
-								xhr.onerror = xhr.onabort = xhr.ontimeout =
-									xhr.onreadystatechange = null;
-
-							if ( type === "abort" ) {
-								xhr.abort();
-							} else if ( type === "error" ) {
-
-								// Support: IE <=9 only
-								// On a manual native abort, IE9 throws
-								// errors on any property access that is not readyState
-								if ( typeof xhr.status !== "number" ) {
-									complete( 0, "error" );
-								} else {
-									complete(
-
-										// File: protocol always yields status 0; see #8605, #14207
-										xhr.status,
-										xhr.statusText
-									);
-								}
-							} else {
-								complete(
-									xhrSuccessStatus[ xhr.status ] || xhr.status,
-									xhr.statusText,
-
-									// Support: IE <=9 only
-									// IE9 has no XHR2 but throws on binary (trac-11426)
-									// For XHR2 non-text, let the caller handle it (gh-2498)
-									( xhr.responseType || "text" ) !== "text"  ||
-									typeof xhr.responseText !== "string" ?
-										{ binary: xhr.response } :
-										{ text: xhr.responseText },
-									xhr.getAllResponseHeaders()
-								);
-							}
-						}
-					};
-				};
-
-				// Listen to events
-				xhr.onload = callback();
-				errorCallback = xhr.onerror = xhr.ontimeout = callback( "error" );
-
-				// Support: IE 9 only
-				// Use onreadystatechange to replace onabort
-				// to handle uncaught aborts
-				if ( xhr.onabort !== undefined ) {
-					xhr.onabort = errorCallback;
-				} else {
-					xhr.onreadystatechange = function() {
-
-						// Check readyState before timeout as it changes
-						if ( xhr.readyState === 4 ) {
-
-							// Allow onerror to be called first,
-							// but that will not handle a native abort
-							// Also, save errorCallback to a variable
-							// as xhr.onerror cannot be accessed
-							window.setTimeout( function() {
-								if ( callback ) {
-									errorCallback();
-								}
-							} );
-						}
-					};
-				}
-
-				// Create the abort callback
-				callback = callback( "abort" );
-
-				try {
-
-					// Do send the request (this may raise an exception)
-					xhr.send( options.hasContent && options.data || null );
-				} catch ( e ) {
-
-					// #14683: Only rethrow if this hasn't been notified as an error yet
-					if ( callback ) {
-						throw e;
-					}
-				}
-			},
-
-			abort: function() {
-				if ( callback ) {
-					callback();
-				}
-			}
-		};
-	}
-} );
-
-
-
-
-// Prevent auto-execution of scripts when no explicit dataType was provided (See gh-2432)
-jQuery.ajaxPrefilter( function( s ) {
-	if ( s.crossDomain ) {
-		s.contents.script = false;
-	}
-} );
-
-// Install script dataType
-jQuery.ajaxSetup( {
-	accepts: {
-		script: "text/javascript, application/javascript, " +
-			"application/ecmascript, application/x-ecmascript"
-	},
-	contents: {
-		script: /\b(?:java|ecma)script\b/
-	},
-	converters: {
-		"text script": function( text ) {
-			jQuery.globalEval( text );
-			return text;
-		}
-	}
-} );
-
-// Handle cache's special case and crossDomain
-jQuery.ajaxPrefilter( "script", function( s ) {
-	if ( s.cache === undefined ) {
-		s.cache = false;
-	}
-	if ( s.crossDomain ) {
-		s.type = "GET";
-	}
-} );
-
-// Bind script tag hack transport
-jQuery.ajaxTransport( "script", function( s ) {
-
-	// This transport only deals with cross domain or forced-by-attrs requests
-	if ( s.crossDomain || s.scriptAttrs ) {
-		var script, callback;
-		return {
-			send: function( _, complete ) {
-				script = jQuery( "<script>" )
-					.attr( s.scriptAttrs || {} )
-					.prop( { charset: s.scriptCharset, src: s.url } )
-					.on( "load error", callback = function( evt ) {
-						script.remove();
-						callback = null;
-						if ( evt ) {
-							complete( evt.type === "error" ? 404 : 200, evt.type );
-						}
-					} );
-
-				// Use native DOM manipulation to avoid our domManip AJAX trickery
-				document.head.appendChild( script[ 0 ] );
-			},
-			abort: function() {
-				if ( callback ) {
-					callback();
-				}
-			}
-		};
-	}
-} );
-
-
-
-
-var oldCallbacks = [],
-	rjsonp = /(=)\?(?=&|$)|\?\?/;
-
-// Default jsonp settings
-jQuery.ajaxSetup( {
-	jsonp: "callback",
-	jsonpCallback: function() {
-		var callback = oldCallbacks.pop() || ( jQuery.expando + "_" + ( nonce.guid++ ) );
-		this[ callback ] = true;
-		return callback;
-	}
-} );
-
-// Detect, normalize options and install callbacks for jsonp requests
-jQuery.ajaxPrefilter( "json jsonp", function( s, originalSettings, jqXHR ) {
-
-	var callbackName, overwritten, responseContainer,
-		jsonProp = s.jsonp !== false && ( rjsonp.test( s.url ) ?
-			"url" :
-			typeof s.data === "string" &&
-				( s.contentType || "" )
-					.indexOf( "application/x-www-form-urlencoded" ) === 0 &&
-				rjsonp.test( s.data ) && "data"
-		);
-
-	// Handle iff the expected data type is "jsonp" or we have a parameter to set
-	if ( jsonProp || s.dataTypes[ 0 ] === "jsonp" ) {
-
-		// Get callback name, remembering preexisting value associated with it
-		callbackName = s.jsonpCallback = isFunction( s.jsonpCallback ) ?
-			s.jsonpCallback() :
-			s.jsonpCallback;
-
-		// Insert callback into url or form data
-		if ( jsonProp ) {
-			s[ jsonProp ] = s[ jsonProp ].replace( rjsonp, "$1" + callbackName );
-		} else if ( s.jsonp !== false ) {
-			s.url += ( rquery.test( s.url ) ? "&" : "?" ) + s.jsonp + "=" + callbackName;
-		}
-
-		// Use data converter to retrieve json after script execution
-		s.converters[ "script json" ] = function() {
-			if ( !responseContainer ) {
-				jQuery.error( callbackName + " was not called" );
-			}
-			return responseContainer[ 0 ];
-		};
-
-		// Force json dataType
-		s.dataTypes[ 0 ] = "json";
-
-		// Install callback
-		overwritten = window[ callbackName ];
-		window[ callbackName ] = function() {
-			responseContainer = arguments;
-		};
-
-		// Clean-up function (fires after converters)
-		jqXHR.always( function() {
-
-			// If previous value didn't exist - remove it
-			if ( overwritten === undefined ) {
-				jQuery( window ).removeProp( callbackName );
-
-			// Otherwise restore preexisting value
-			} else {
-				window[ callbackName ] = overwritten;
-			}
-
-			// Save back as free
-			if ( s[ callbackName ] ) {
-
-				// Make sure that re-using the options doesn't screw things around
-				s.jsonpCallback = originalSettings.jsonpCallback;
-
-				// Save the callback name for future use
-				oldCallbacks.push( callbackName );
-			}
-
-			// Call if it was a function and we have a response
-			if ( responseContainer && isFunction( overwritten ) ) {
-				overwritten( responseContainer[ 0 ] );
-			}
-
-			responseContainer = overwritten = undefined;
-		} );
-
-		// Delegate to script
-		return "script";
-	}
-} );
-
-
-
-
-// Support: Safari 8 only
-// In Safari 8 documents created via document.implementation.createHTMLDocument
-// collapse sibling forms: the second one becomes a child of the first one.
-// Because of that, this security measure has to be disabled in Safari 8.
-// https://bugs.webkit.org/show_bug.cgi?id=137337
-support.createHTMLDocument = ( function() {
-	var body = document.implementation.createHTMLDocument( "" ).body;
-	body.innerHTML = "<form></form><form></form>";
-	return body.childNodes.length === 2;
-} )();
-
-
-// Argument "data" should be string of html
-// context (optional): If specified, the fragment will be created in this context,
-// defaults to document
-// keepScripts (optional): If true, will include scripts passed in the html string
-jQuery.parseHTML = function( data, context, keepScripts ) {
-	if ( typeof data !== "string" ) {
-		return [];
-	}
-	if ( typeof context === "boolean" ) {
-		keepScripts = context;
-		context = false;
-	}
-
-	var base, parsed, scripts;
-
-	if ( !context ) {
-
-		// Stop scripts or inline event handlers from being executed immediately
-		// by using document.implementation
-		if ( support.createHTMLDocument ) {
-			context = document.implementation.createHTMLDocument( "" );
-
-			// Set the base href for the created document
-			// so any parsed elements with URLs
-			// are based on the document's URL (gh-2965)
-			base = context.createElement( "base" );
-			base.href = document.location.href;
-			context.head.appendChild( base );
-		} else {
-			context = document;
-		}
-	}
-
-	parsed = rsingleTag.exec( data );
-	scripts = !keepScripts && [];
-
-	// Single tag
-	if ( parsed ) {
-		return [ context.createElement( parsed[ 1 ] ) ];
-	}
-
-	parsed = buildFragment( [ data ], context, scripts );
-
-	if ( scripts && scripts.length ) {
-		jQuery( scripts ).remove();
-	}
-
-	return jQuery.merge( [], parsed.childNodes );
-};
-
-
-/**
- * Load a url into a page
- */
-jQuery.fn.load = function( url, params, callback ) {
-	var selector, type, response,
-		self = this,
-		off = url.indexOf( " " );
-
-	if ( off > -1 ) {
-		selector = stripAndCollapse( url.slice( off ) );
-		url = url.slice( 0, off );
-	}
-
-	// If it's a function
-	if ( isFunction( params ) ) {
-
-		// We assume that it's the callback
-		callback = params;
-		params = undefined;
-
-	// Otherwise, build a param string
-	} else if ( params && typeof params === "object" ) {
-		type = "POST";
-	}
-
-	// If we have elements to modify, make the request
-	if ( self.length > 0 ) {
-		jQuery.ajax( {
-			url: url,
-
-			// If "type" variable is undefined, then "GET" method will be used.
-			// Make value of this field explicit since
-			// user can override it through ajaxSetup method
-			type: type || "GET",
-			dataType: "html",
-			data: params
-		} ).done( function( responseText ) {
-
-			// Save response for use in complete callback
-			response = arguments;
-
-			self.html( selector ?
-
-				// If a selector was specified, locate the right elements in a dummy div
-				// Exclude scripts to avoid IE 'Permission Denied' errors
-				jQuery( "<div>" ).append( jQuery.parseHTML( responseText ) ).find( selector ) :
-
-				// Otherwise use the full result
-				responseText );
-
-		// If the request succeeds, this function gets "data", "status", "jqXHR"
-		// but they are ignored because response was set above.
-		// If it fails, this function gets "jqXHR", "status", "error"
-		} ).always( callback && function( jqXHR, status ) {
-			self.each( function() {
-				callback.apply( this, response || [ jqXHR.responseText, status, jqXHR ] );
-			} );
-		} );
-	}
-
-	return this;
-};
-
-
-
-
-jQuery.expr.pseudos.animated = function( elem ) {
-	return jQuery.grep( jQuery.timers, function( fn ) {
-		return elem === fn.elem;
-	} ).length;
-};
-
-
-
-
-jQuery.offset = {
-	setOffset: function( elem, options, i ) {
-		var curPosition, curLeft, curCSSTop, curTop, curOffset, curCSSLeft, calculatePosition,
-			position = jQuery.css( elem, "position" ),
-			curElem = jQuery( elem ),
-			props = {};
-
-		// Set position first, in-case top/left are set even on static elem
-		if ( position === "static" ) {
-			elem.style.position = "relative";
-		}
-
-		curOffset = curElem.offset();
-		curCSSTop = jQuery.css( elem, "top" );
-		curCSSLeft = jQuery.css( elem, "left" );
-		calculatePosition = ( position === "absolute" || position === "fixed" ) &&
-			( curCSSTop + curCSSLeft ).indexOf( "auto" ) > -1;
-
-		// Need to be able to calculate position if either
-		// top or left is auto and position is either absolute or fixed
-		if ( calculatePosition ) {
-			curPosition = curElem.position();
-			curTop = curPosition.top;
-			curLeft = curPosition.left;
-
-		} else {
-			curTop = parseFloat( curCSSTop ) || 0;
-			curLeft = parseFloat( curCSSLeft ) || 0;
-		}
-
-		if ( isFunction( options ) ) {
-
-			// Use jQuery.extend here to allow modification of coordinates argument (gh-1848)
-			options = options.call( elem, i, jQuery.extend( {}, curOffset ) );
-		}
-
-		if ( options.top != null ) {
-			props.top = ( options.top - curOffset.top ) + curTop;
-		}
-		if ( options.left != null ) {
-			props.left = ( options.left - curOffset.left ) + curLeft;
-		}
-
-		if ( "using" in options ) {
-			options.using.call( elem, props );
-
-		} else {
-			curElem.css( props );
-		}
-	}
-};
-
-jQuery.fn.extend( {
-
-	// offset() relates an element's border box to the document origin
-	offset: function( options ) {
-
-		// Preserve chaining for setter
-		if ( arguments.length ) {
-			return options === undefined ?
-				this :
-				this.each( function( i ) {
-					jQuery.offset.setOffset( this, options, i );
-				} );
-		}
-
-		var rect, win,
-			elem = this[ 0 ];
-
-		if ( !elem ) {
-			return;
-		}
-
-		// Return zeros for disconnected and hidden (display: none) elements (gh-2310)
-		// Support: IE <=11 only
-		// Running getBoundingClientRect on a
-		// disconnected node in IE throws an error
-		if ( !elem.getClientRects().length ) {
-			return { top: 0, left: 0 };
-		}
-
-		// Get document-relative position by adding viewport scroll to viewport-relative gBCR
-		rect = elem.getBoundingClientRect();
-		win = elem.ownerDocument.defaultView;
-		return {
-			top: rect.top + win.pageYOffset,
-			left: rect.left + win.pageXOffset
-		};
-	},
-
-	// position() relates an element's margin box to its offset parent's padding box
-	// This corresponds to the behavior of CSS absolute positioning
-	position: function() {
-		if ( !this[ 0 ] ) {
-			return;
-		}
-
-		var offsetParent, offset, doc,
-			elem = this[ 0 ],
-			parentOffset = { top: 0, left: 0 };
-
-		// position:fixed elements are offset from the viewport, which itself always has zero offset
-		if ( jQuery.css( elem, "position" ) === "fixed" ) {
-
-			// Assume position:fixed implies availability of getBoundingClientRect
-			offset = elem.getBoundingClientRect();
-
-		} else {
-			offset = this.offset();
-
-			// Account for the *real* offset parent, which can be the document or its root element
-			// when a statically positioned element is identified
-			doc = elem.ownerDocument;
-			offsetParent = elem.offsetParent || doc.documentElement;
-			while ( offsetParent &&
-				( offsetParent === doc.body || offsetParent === doc.documentElement ) &&
-				jQuery.css( offsetParent, "position" ) === "static" ) {
-
-				offsetParent = offsetParent.parentNode;
-			}
-			if ( offsetParent && offsetParent !== elem && offsetParent.nodeType === 1 ) {
-
-				// Incorporate borders into its offset, since they are outside its content origin
-				parentOffset = jQuery( offsetParent ).offset();
-				parentOffset.top += jQuery.css( offsetParent, "borderTopWidth", true );
-				parentOffset.left += jQuery.css( offsetParent, "borderLeftWidth", true );
-			}
-		}
-
-		// Subtract parent offsets and element margins
-		return {
-			top: offset.top - parentOffset.top - jQuery.css( elem, "marginTop", true ),
-			left: offset.left - parentOffset.left - jQuery.css( elem, "marginLeft", true )
-		};
-	},
-
-	// This method will return documentElement in the following cases:
-	// 1) For the element inside the iframe without offsetParent, this method will return
-	//    documentElement of the parent window
-	// 2) For the hidden or detached element
-	// 3) For body or html element, i.e. in case of the html node - it will return itself
-	//
-	// but those exceptions were never presented as a real life use-cases
-	// and might be considered as more preferable results.
-	//
-	// This logic, however, is not guaranteed and can change at any point in the future
-	offsetParent: function() {
-		return this.map( function() {
-			var offsetParent = this.offsetParent;
-
-			while ( offsetParent && jQuery.css( offsetParent, "position" ) === "static" ) {
-				offsetParent = offsetParent.offsetParent;
-			}
-
-			return offsetParent || documentElement;
-		} );
-	}
-} );
-
-// Create scrollLeft and scrollTop methods
-jQuery.each( { scrollLeft: "pageXOffset", scrollTop: "pageYOffset" }, function( method, prop ) {
-	var top = "pageYOffset" === prop;
-
-	jQuery.fn[ method ] = function( val ) {
-		return access( this, function( elem, method, val ) {
-
-			// Coalesce documents and windows
-			var win;
-			if ( isWindow( elem ) ) {
-				win = elem;
-			} else if ( elem.nodeType === 9 ) {
-				win = elem.defaultView;
-			}
-
-			if ( val === undefined ) {
-				return win ? win[ prop ] : elem[ method ];
-			}
-
-			if ( win ) {
-				win.scrollTo(
-					!top ? val : win.pageXOffset,
-					top ? val : win.pageYOffset
-				);
-
-			} else {
-				elem[ method ] = val;
-			}
-		}, method, val, arguments.length );
-	};
-} );
-
-// Support: Safari <=7 - 9.1, Chrome <=37 - 49
-// Add the top/left cssHooks using jQuery.fn.position
-// Webkit bug: https://bugs.webkit.org/show_bug.cgi?id=29084
-// Blink bug: https://bugs.chromium.org/p/chromium/issues/detail?id=589347
-// getComputedStyle returns percent when specified for top/left/bottom/right;
-// rather than make the css module depend on the offset module, just check for it here
-jQuery.each( [ "top", "left" ], function( _i, prop ) {
-	jQuery.cssHooks[ prop ] = addGetHookIf( support.pixelPosition,
-		function( elem, computed ) {
-			if ( computed ) {
-				computed = curCSS( elem, prop );
-
-				// If curCSS returns percentage, fallback to offset
-				return rnumnonpx.test( computed ) ?
-					jQuery( elem ).position()[ prop ] + "px" :
-					computed;
-			}
-		}
-	);
-} );
-
-
-// Create innerHeight, innerWidth, height, width, outerHeight and outerWidth methods
-jQuery.each( { Height: "height", Width: "width" }, function( name, type ) {
-	jQuery.each( {
-		padding: "inner" + name,
-		content: type,
-		"": "outer" + name
-	}, function( defaultExtra, funcName ) {
-
-		// Margin is only for outerHeight, outerWidth
-		jQuery.fn[ funcName ] = function( margin, value ) {
-			var chainable = arguments.length && ( defaultExtra || typeof margin !== "boolean" ),
-				extra = defaultExtra || ( margin === true || value === true ? "margin" : "border" );
-
-			return access( this, function( elem, type, value ) {
-				var doc;
-
-				if ( isWindow( elem ) ) {
-
-					// $( window ).outerWidth/Height return w/h including scrollbars (gh-1729)
-					return funcName.indexOf( "outer" ) === 0 ?
-						elem[ "inner" + name ] :
-						elem.document.documentElement[ "client" + name ];
-				}
-
-				// Get document width or height
-				if ( elem.nodeType === 9 ) {
-					doc = elem.documentElement;
-
-					// Either scroll[Width/Height] or offset[Width/Height] or client[Width/Height],
-					// whichever is greatest
-					return Math.max(
-						elem.body[ "scroll" + name ], doc[ "scroll" + name ],
-						elem.body[ "offset" + name ], doc[ "offset" + name ],
-						doc[ "client" + name ]
-					);
-				}
-
-				return value === undefined ?
-
-					// Get width or height on the element, requesting but not forcing parseFloat
-					jQuery.css( elem, type, extra ) :
-
-					// Set width or height on the element
-					jQuery.style( elem, type, value, extra );
-			}, type, chainable ? margin : undefined, chainable );
-		};
-	} );
-} );
-
-
-jQuery.each( [
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-	"ajaxStop",
-	"ajaxComplete",
-	"ajaxError",
-	"ajaxSuccess",
-	"ajaxSend"
-], function( _i, type ) {
-	jQuery.fn[ type ] = function( fn ) {
-		return this.on( type, fn );
-	};
-} );
-
-
-
-
-jQuery.fn.extend( {
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-	bind: function( types, data, fn ) {
-		return this.on( types, null, data, fn );
-	},
-	unbind: function( types, fn ) {
-		return this.off( types, null, fn );
-	},
-
-	delegate: function( selector, types, data, fn ) {
-		return this.on( types, selector, data, fn );
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-		// ( namespace ) or ( selector, types [, fn] )
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-	},
-
-	hover: function( fnOver, fnOut ) {
-		return this.mouseenter( fnOver ).mouseleave( fnOut || fnOver );
-	}
-} );
-
-jQuery.each(
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-	"mousedown mouseup mousemove mouseover mouseout mouseenter mouseleave " +
-	"change select submit keydown keypress keyup contextmenu" ).split( " " ),
-	function( _i, name ) {
-
-		// Handle event binding
-		jQuery.fn[ name ] = function( data, fn ) {
-			return arguments.length > 0 ?
-				this.on( name, null, data, fn ) :
-				this.trigger( name );
-		};
-	}
-);
-
-
-
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-var rtrim = /^[\s\uFEFF\xA0]+|[\s\uFEFF\xA0]+$/g;
-
-// Bind a function to a context, optionally partially applying any
-// arguments.
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-// However, it is not slated for removal any time soon
-jQuery.proxy = function( fn, context ) {
-	var tmp, args, proxy;
-
-	if ( typeof context === "string" ) {
-		tmp = fn[ context ];
-		context = fn;
-		fn = tmp;
-	}
-
-	// Quick check to determine if target is callable, in the spec
-	// this throws a TypeError, but we will just return undefined.
-	if ( !isFunction( fn ) ) {
-		return undefined;
-	}
-
-	// Simulated bind
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-	proxy = function() {
-		return fn.apply( context || this, args.concat( slice.call( arguments ) ) );
-	};
-
-	// Set the guid of unique handler to the same of original handler, so it can be removed
-	proxy.guid = fn.guid = fn.guid || jQuery.guid++;
-
-	return proxy;
-};
-
-jQuery.holdReady = function( hold ) {
-	if ( hold ) {
-		jQuery.readyWait++;
-	} else {
-		jQuery.ready( true );
-	}
-};
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-jQuery.isWindow = isWindow;
-jQuery.camelCase = camelCase;
-jQuery.type = toType;
-
-jQuery.now = Date.now;
-
-jQuery.isNumeric = function( obj ) {
-
-	// As of jQuery 3.0, isNumeric is limited to
-	// strings and numbers (primitives or objects)
-	// that can be coerced to finite numbers (gh-2662)
-	var type = jQuery.type( obj );
-	return ( type === "number" || type === "string" ) &&
-
-		// parseFloat NaNs numeric-cast false positives ("")
-		// ...but misinterprets leading-number strings, particularly hex literals ("0x...")
-		// subtraction forces infinities to NaN
-		!isNaN( obj - parseFloat( obj ) );
-};
-
-jQuery.trim = function( text ) {
-	return text == null ?
-		"" :
-		( text + "" ).replace( rtrim, "" );
-};
-
-
-
-// Register as a named AMD module, since jQuery can be concatenated with other
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-
-if ( typeof define === "function" && define.amd ) {
-	define( "jquery", [], function() {
-		return jQuery;
-	} );
-}
-
-
-
-
-var
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-	// Map over jQuery in case of overwrite
-	_jQuery = window.jQuery,
-
-	// Map over the $ in case of overwrite
-	_$ = window.$;
-
-jQuery.noConflict = function( deep ) {
-	if ( window.$ === jQuery ) {
-		window.$ = _$;
-	}
-
-	if ( deep && window.jQuery === jQuery ) {
-		window.jQuery = _jQuery;
-	}
-
-	return jQuery;
-};
-
-// Expose jQuery and $ identifiers, even in AMD
-// (#7102#comment:10, https://github.com/jquery/jquery/pull/557)
-// and CommonJS for browser emulators (#13566)
-if ( typeof noGlobal === "undefined" ) {
-	window.jQuery = window.$ = jQuery;
-}
-
-
-
-
-return jQuery;
-} );
diff --git a/docs/_build/html/_static/jquery.js b/docs/_build/html/_static/jquery.js
deleted file mode 100644
index c4c6022..0000000
--- a/docs/_build/html/_static/jquery.js
+++ /dev/null
@@ -1,2 +0,0 @@
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diff --git a/docs/_build/html/_static/language_data.js b/docs/_build/html/_static/language_data.js
deleted file mode 100644
index 2e22b06..0000000
--- a/docs/_build/html/_static/language_data.js
+++ /dev/null
@@ -1,199 +0,0 @@
-/*
- * language_data.js
- * ~~~~~~~~~~~~~~~~
- *
- * This script contains the language-specific data used by searchtools.js,
- * namely the list of stopwords, stemmer, scorer and splitter.
- *
- * :copyright: Copyright 2007-2022 by the Sphinx team, see AUTHORS.
- * :license: BSD, see LICENSE for details.
- *
- */
-
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-
-
-/* Non-minified version is copied as a separate JS file, is available */
-
-/**
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- */
-var Stemmer = function() {
-
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-    tional: 'tion',
-    enci: 'ence',
-    anci: 'ance',
-    izer: 'ize',
-    bli: 'ble',
-    alli: 'al',
-    entli: 'ent',
-    eli: 'e',
-    ousli: 'ous',
-    ization: 'ize',
-    ation: 'ate',
-    ator: 'ate',
-    alism: 'al',
-    iveness: 'ive',
-    fulness: 'ful',
-    ousness: 'ous',
-    aliti: 'al',
-    iviti: 'ive',
-    biliti: 'ble',
-    logi: 'log'
-  };
-
-  var step3list = {
-    icate: 'ic',
-    ative: '',
-    alize: 'al',
-    iciti: 'ic',
-    ical: 'ic',
-    ful: '',
-    ness: ''
-  };
-
-  var c = "[^aeiou]";          // consonant
-  var v = "[aeiouy]";          // vowel
-  var C = c + "[^aeiouy]*";    // consonant sequence
-  var V = v + "[aeiou]*";      // vowel sequence
-
-  var mgr0 = "^(" + C + ")?" + V + C;                      // [C]VC... is m>0
-  var meq1 = "^(" + C + ")?" + V + C + "(" + V + ")?$";    // [C]VC[V] is m=1
-  var mgr1 = "^(" + C + ")?" + V + C + V + C;              // [C]VCVC... is m>1
-  var s_v   = "^(" + C + ")?" + v;                         // vowel in stem
-
-  this.stemWord = function (w) {
-    var stem;
-    var suffix;
-    var firstch;
-    var origword = w;
-
-    if (w.length < 3)
-      return w;
-
-    var re;
-    var re2;
-    var re3;
-    var re4;
-
-    firstch = w.substr(0,1);
-    if (firstch == "y")
-      w = firstch.toUpperCase() + w.substr(1);
-
-    // Step 1a
-    re = /^(.+?)(ss|i)es$/;
-    re2 = /^(.+?)([^s])s$/;
-
-    if (re.test(w))
-      w = w.replace(re,"$1$2");
-    else if (re2.test(w))
-      w = w.replace(re2,"$1$2");
-
-    // Step 1b
-    re = /^(.+?)eed$/;
-    re2 = /^(.+?)(ed|ing)$/;
-    if (re.test(w)) {
-      var fp = re.exec(w);
-      re = new RegExp(mgr0);
-      if (re.test(fp[1])) {
-        re = /.$/;
-        w = w.replace(re,"");
-      }
-    }
-    else if (re2.test(w)) {
-      var fp = re2.exec(w);
-      stem = fp[1];
-      re2 = new RegExp(s_v);
-      if (re2.test(stem)) {
-        w = stem;
-        re2 = /(at|bl|iz)$/;
-        re3 = new RegExp("([^aeiouylsz])\\1$");
-        re4 = new RegExp("^" + C + v + "[^aeiouwxy]$");
-        if (re2.test(w))
-          w = w + "e";
-        else if (re3.test(w)) {
-          re = /.$/;
-          w = w.replace(re,"");
-        }
-        else if (re4.test(w))
-          w = w + "e";
-      }
-    }
-
-    // Step 1c
-    re = /^(.+?)y$/;
-    if (re.test(w)) {
-      var fp = re.exec(w);
-      stem = fp[1];
-      re = new RegExp(s_v);
-      if (re.test(stem))
-        w = stem + "i";
-    }
-
-    // Step 2
-    re = /^(.+?)(ational|tional|enci|anci|izer|bli|alli|entli|eli|ousli|ization|ation|ator|alism|iveness|fulness|ousness|aliti|iviti|biliti|logi)$/;
-    if (re.test(w)) {
-      var fp = re.exec(w);
-      stem = fp[1];
-      suffix = fp[2];
-      re = new RegExp(mgr0);
-      if (re.test(stem))
-        w = stem + step2list[suffix];
-    }
-
-    // Step 3
-    re = /^(.+?)(icate|ative|alize|iciti|ical|ful|ness)$/;
-    if (re.test(w)) {
-      var fp = re.exec(w);
-      stem = fp[1];
-      suffix = fp[2];
-      re = new RegExp(mgr0);
-      if (re.test(stem))
-        w = stem + step3list[suffix];
-    }
-
-    // Step 4
-    re = /^(.+?)(al|ance|ence|er|ic|able|ible|ant|ement|ment|ent|ou|ism|ate|iti|ous|ive|ize)$/;
-    re2 = /^(.+?)(s|t)(ion)$/;
-    if (re.test(w)) {
-      var fp = re.exec(w);
-      stem = fp[1];
-      re = new RegExp(mgr1);
-      if (re.test(stem))
-        w = stem;
-    }
-    else if (re2.test(w)) {
-      var fp = re2.exec(w);
-      stem = fp[1] + fp[2];
-      re2 = new RegExp(mgr1);
-      if (re2.test(stem))
-        w = stem;
-    }
-
-    // Step 5
-    re = /^(.+?)e$/;
-    if (re.test(w)) {
-      var fp = re.exec(w);
-      stem = fp[1];
-      re = new RegExp(mgr1);
-      re2 = new RegExp(meq1);
-      re3 = new RegExp("^" + C + v + "[^aeiouwxy]$");
-      if (re.test(stem) || (re2.test(stem) && !(re3.test(stem))))
-        w = stem;
-    }
-    re = /ll$/;
-    re2 = new RegExp(mgr1);
-    if (re.test(w) && re2.test(w)) {
-      re = /.$/;
-      w = w.replace(re,"");
-    }
-
-    // and turn initial Y back to y
-    if (firstch == "y")
-      w = firstch.toLowerCase() + w.substr(1);
-    return w;
-  }
-}
-
diff --git a/docs/_build/html/_static/locales/ar/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/ar/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 15541a6a375f93efed3f410c2f892174ba91aaa0..0000000000000000000000000000000000000000
GIT binary patch
literal 0
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diff --git a/docs/_build/html/_static/locales/ar/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/ar/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 2e8d682..0000000
--- a/docs/_build/html/_static/locales/ar/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: ar\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "أقترح تحرير"
-
-msgid "Last updated on"
-msgstr "آخر تحديث في"
-
-msgid "Edit this page"
-msgstr "قم بتحرير هذه الصفحة"
-
-msgid "Launch"
-msgstr "إطلاق"
-
-msgid "Print to PDF"
-msgstr "طباعة إلى PDF"
-
-msgid "open issue"
-msgstr "قضية مفتوحة"
-
-msgid "Download notebook file"
-msgstr "تنزيل ملف دفتر الملاحظات"
-
-msgid "Toggle navigation"
-msgstr "تبديل التنقل"
-
-msgid "Source repository"
-msgstr "مستودع المصدر"
-
-msgid "By the"
-msgstr "بواسطة"
-
-msgid "next page"
-msgstr "الصفحة التالية"
-
-msgid "repository"
-msgstr "مخزن"
-
-msgid "Sphinx Book Theme"
-msgstr "موضوع كتاب أبو الهول"
-
-msgid "Download source file"
-msgstr "تنزيل ملف المصدر"
-
-msgid "Contents"
-msgstr "محتويات"
-
-msgid "By"
-msgstr "بواسطة"
-
-msgid "Copyright"
-msgstr "حقوق النشر"
-
-msgid "Fullscreen mode"
-msgstr "وضع ملء الشاشة"
-
-msgid "Open an issue"
-msgstr "افتح قضية"
-
-msgid "previous page"
-msgstr "الصفحة السابقة"
-
-msgid "Download this page"
-msgstr "قم بتنزيل هذه الصفحة"
-
-msgid "Theme by the"
-msgstr "موضوع بواسطة"
diff --git a/docs/_build/html/_static/locales/bg/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/bg/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index da95120037f8d7c52ad1f166a7682ea55ca47a01..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/bg/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/bg/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 56ef0eb..0000000
--- a/docs/_build/html/_static/locales/bg/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: bg\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "предложи редактиране"
-
-msgid "Last updated on"
-msgstr "Последна актуализация на"
-
-msgid "Edit this page"
-msgstr "Редактирайте тази страница"
-
-msgid "Launch"
-msgstr "Стартиране"
-
-msgid "Print to PDF"
-msgstr "Печат в PDF"
-
-msgid "open issue"
-msgstr "отворен брой"
-
-msgid "Download notebook file"
-msgstr "Изтеглете файла на бележника"
-
-msgid "Toggle navigation"
-msgstr "Превключване на навигацията"
-
-msgid "Source repository"
-msgstr "Хранилище на източника"
-
-msgid "By the"
-msgstr "По"
-
-msgid "next page"
-msgstr "Следваща страница"
-
-msgid "repository"
-msgstr "хранилище"
-
-msgid "Sphinx Book Theme"
-msgstr "Тема на книгата Sphinx"
-
-msgid "Download source file"
-msgstr "Изтеглете изходния файл"
-
-msgid "Contents"
-msgstr "Съдържание"
-
-msgid "By"
-msgstr "От"
-
-msgid "Copyright"
-msgstr "Авторско право"
-
-msgid "Fullscreen mode"
-msgstr "Режим на цял екран"
-
-msgid "Open an issue"
-msgstr "Отворете проблем"
-
-msgid "previous page"
-msgstr "предишна страница"
-
-msgid "Download this page"
-msgstr "Изтеглете тази страница"
-
-msgid "Theme by the"
-msgstr "Тема от"
diff --git a/docs/_build/html/_static/locales/bn/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/bn/LC_MESSAGES/booktheme.mo
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GIT binary patch
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diff --git a/docs/_build/html/_static/locales/bn/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/bn/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 243ca31..0000000
--- a/docs/_build/html/_static/locales/bn/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,63 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: bn\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "Last updated on"
-msgstr "সর্বশেষ আপডেট"
-
-msgid "Edit this page"
-msgstr "এই পৃষ্ঠাটি সম্পাদনা করুন"
-
-msgid "Launch"
-msgstr "শুরু করা"
-
-msgid "Print to PDF"
-msgstr "পিডিএফ প্রিন্ট করুন"
-
-msgid "open issue"
-msgstr "খোলা সমস্যা"
-
-msgid "Download notebook file"
-msgstr "নোটবুক ফাইল ডাউনলোড করুন"
-
-msgid "Toggle navigation"
-msgstr "নেভিগেশন টগল করুন"
-
-msgid "Source repository"
-msgstr "উত্স সংগ্রহস্থল"
-
-msgid "By the"
-msgstr "দ্বারা"
-
-msgid "next page"
-msgstr "পরবর্তী পৃষ্ঠা"
-
-msgid "Sphinx Book Theme"
-msgstr "স্পিনিক্স বুক থিম"
-
-msgid "Download source file"
-msgstr "উত্স ফাইল ডাউনলোড করুন"
-
-msgid "By"
-msgstr "দ্বারা"
-
-msgid "Copyright"
-msgstr "কপিরাইট"
-
-msgid "Open an issue"
-msgstr "একটি সমস্যা খুলুন"
-
-msgid "previous page"
-msgstr "আগের পৃষ্ঠা"
-
-msgid "Download this page"
-msgstr "এই পৃষ্ঠাটি ডাউনলোড করুন"
-
-msgid "Theme by the"
-msgstr "থিম দ্বারা"
diff --git a/docs/_build/html/_static/locales/ca/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/ca/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index a4dd30e9bd762578b119c4401d082ea4fa430613..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/ca/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/ca/LC_MESSAGES/booktheme.po
deleted file mode 100644
index b27a13d..0000000
--- a/docs/_build/html/_static/locales/ca/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,66 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: ca\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "suggerir edició"
-
-msgid "Last updated on"
-msgstr "Darrera actualització el"
-
-msgid "Edit this page"
-msgstr "Editeu aquesta pàgina"
-
-msgid "Launch"
-msgstr "Llançament"
-
-msgid "Print to PDF"
-msgstr "Imprimeix a PDF"
-
-msgid "open issue"
-msgstr "número obert"
-
-msgid "Download notebook file"
-msgstr "Descarregar fitxer de quadern"
-
-msgid "Toggle navigation"
-msgstr "Commuta la navegació"
-
-msgid "Source repository"
-msgstr "Dipòsit de fonts"
-
-msgid "By the"
-msgstr "Per la"
-
-msgid "next page"
-msgstr "pàgina següent"
-
-msgid "Sphinx Book Theme"
-msgstr "Tema del llibre Esfinx"
-
-msgid "Download source file"
-msgstr "Baixeu el fitxer font"
-
-msgid "By"
-msgstr "Per"
-
-msgid "Copyright"
-msgstr "Copyright"
-
-msgid "Open an issue"
-msgstr "Obriu un número"
-
-msgid "previous page"
-msgstr "Pàgina anterior"
-
-msgid "Download this page"
-msgstr "Descarregueu aquesta pàgina"
-
-msgid "Theme by the"
-msgstr "Tema del"
diff --git a/docs/_build/html/_static/locales/cs/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/cs/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index c39e01a6ae9ae5242f04566ffa2cda96b328ca18..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/cs/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/cs/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 3818df9..0000000
--- a/docs/_build/html/_static/locales/cs/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: cs\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "navrhnout úpravy"
-
-msgid "Last updated on"
-msgstr "Naposledy aktualizováno"
-
-msgid "Edit this page"
-msgstr "Upravit tuto stránku"
-
-msgid "Launch"
-msgstr "Zahájení"
-
-msgid "Print to PDF"
-msgstr "Tisk do PDF"
-
-msgid "open issue"
-msgstr "otevřené číslo"
-
-msgid "Download notebook file"
-msgstr "Stáhnout soubor poznámkového bloku"
-
-msgid "Toggle navigation"
-msgstr "Přepnout navigaci"
-
-msgid "Source repository"
-msgstr "Zdrojové úložiště"
-
-msgid "By the"
-msgstr "Podle"
-
-msgid "next page"
-msgstr "další strana"
-
-msgid "repository"
-msgstr "úložiště"
-
-msgid "Sphinx Book Theme"
-msgstr "Téma knihy Sfinga"
-
-msgid "Download source file"
-msgstr "Stáhněte si zdrojový soubor"
-
-msgid "Contents"
-msgstr "Obsah"
-
-msgid "By"
-msgstr "Podle"
-
-msgid "Copyright"
-msgstr "autorská práva"
-
-msgid "Fullscreen mode"
-msgstr "Režim celé obrazovky"
-
-msgid "Open an issue"
-msgstr "Otevřete problém"
-
-msgid "previous page"
-msgstr "předchozí stránka"
-
-msgid "Download this page"
-msgstr "Stáhněte si tuto stránku"
-
-msgid "Theme by the"
-msgstr "Téma od"
diff --git a/docs/_build/html/_static/locales/da/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/da/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index f43157d70c63ff21b4385dc36cb73f5b14eb6b01..0000000000000000000000000000000000000000
GIT binary patch
literal 0
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diff --git a/docs/_build/html/_static/locales/da/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/da/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 7f20a3b..0000000
--- a/docs/_build/html/_static/locales/da/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: da\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "foreslå redigering"
-
-msgid "Last updated on"
-msgstr "Sidst opdateret den"
-
-msgid "Edit this page"
-msgstr "Rediger denne side"
-
-msgid "Launch"
-msgstr "Start"
-
-msgid "Print to PDF"
-msgstr "Udskriv til PDF"
-
-msgid "open issue"
-msgstr "åbent nummer"
-
-msgid "Download notebook file"
-msgstr "Download notesbog-fil"
-
-msgid "Toggle navigation"
-msgstr "Skift navigation"
-
-msgid "Source repository"
-msgstr "Kildelager"
-
-msgid "By the"
-msgstr "Ved"
-
-msgid "next page"
-msgstr "Næste side"
-
-msgid "repository"
-msgstr "lager"
-
-msgid "Sphinx Book Theme"
-msgstr "Sphinx bogtema"
-
-msgid "Download source file"
-msgstr "Download kildefil"
-
-msgid "Contents"
-msgstr "Indhold"
-
-msgid "By"
-msgstr "Ved"
-
-msgid "Copyright"
-msgstr "ophavsret"
-
-msgid "Fullscreen mode"
-msgstr "Fuldskærmstilstand"
-
-msgid "Open an issue"
-msgstr "Åbn et problem"
-
-msgid "previous page"
-msgstr "forrige side"
-
-msgid "Download this page"
-msgstr "Download denne side"
-
-msgid "Theme by the"
-msgstr "Tema af"
diff --git a/docs/_build/html/_static/locales/de/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/de/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 648b565c7829535bc5f26f86909f458d0cd0fe1b..0000000000000000000000000000000000000000
GIT binary patch
literal 0
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diff --git a/docs/_build/html/_static/locales/de/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/de/LC_MESSAGES/booktheme.po
deleted file mode 100644
index c0027d3..0000000
--- a/docs/_build/html/_static/locales/de/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: de\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "vorschlagen zu bearbeiten"
-
-msgid "Last updated on"
-msgstr "Zuletzt aktualisiert am"
-
-msgid "Edit this page"
-msgstr "Bearbeite diese Seite"
-
-msgid "Launch"
-msgstr "Starten"
-
-msgid "Print to PDF"
-msgstr "In PDF drucken"
-
-msgid "open issue"
-msgstr "offenes Thema"
-
-msgid "Download notebook file"
-msgstr "Notebook-Datei herunterladen"
-
-msgid "Toggle navigation"
-msgstr "Navigation umschalten"
-
-msgid "Source repository"
-msgstr "Quell-Repository"
-
-msgid "By the"
-msgstr "Bis zum"
-
-msgid "next page"
-msgstr "Nächste Seite"
-
-msgid "repository"
-msgstr "Repository"
-
-msgid "Sphinx Book Theme"
-msgstr "Sphinx-Buch-Thema"
-
-msgid "Download source file"
-msgstr "Quelldatei herunterladen"
-
-msgid "Contents"
-msgstr "Inhalt"
-
-msgid "By"
-msgstr "Durch"
-
-msgid "Copyright"
-msgstr "Urheberrechte ©"
-
-msgid "Fullscreen mode"
-msgstr "Vollbildmodus"
-
-msgid "Open an issue"
-msgstr "Öffnen Sie ein Problem"
-
-msgid "previous page"
-msgstr "vorherige Seite"
-
-msgid "Download this page"
-msgstr "Laden Sie diese Seite herunter"
-
-msgid "Theme by the"
-msgstr "Thema von der"
diff --git a/docs/_build/html/_static/locales/el/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/el/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index fca6e9355f314677f2890f5679fd79dd67a7a5a2..0000000000000000000000000000000000000000
GIT binary patch
literal 0
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diff --git a/docs/_build/html/_static/locales/el/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/el/LC_MESSAGES/booktheme.po
deleted file mode 100644
index bdeb327..0000000
--- a/docs/_build/html/_static/locales/el/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: el\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "προτείνω επεξεργασία"
-
-msgid "Last updated on"
-msgstr "Τελευταία ενημέρωση στις"
-
-msgid "Edit this page"
-msgstr "Επεξεργαστείτε αυτήν τη σελίδα"
-
-msgid "Launch"
-msgstr "Εκτόξευση"
-
-msgid "Print to PDF"
-msgstr "Εκτύπωση σε PDF"
-
-msgid "open issue"
-msgstr "ανοιχτό ζήτημα"
-
-msgid "Download notebook file"
-msgstr "Λήψη αρχείου σημειωματάριου"
-
-msgid "Toggle navigation"
-msgstr "Εναλλαγή πλοήγησης"
-
-msgid "Source repository"
-msgstr "Αποθήκη πηγής"
-
-msgid "By the"
-msgstr "Από το"
-
-msgid "next page"
-msgstr "επόμενη σελίδα"
-
-msgid "repository"
-msgstr "αποθήκη"
-
-msgid "Sphinx Book Theme"
-msgstr "Θέμα βιβλίου Sphinx"
-
-msgid "Download source file"
-msgstr "Λήψη αρχείου προέλευσης"
-
-msgid "Contents"
-msgstr "Περιεχόμενα"
-
-msgid "By"
-msgstr "Με"
-
-msgid "Copyright"
-msgstr "Πνευματική ιδιοκτησία"
-
-msgid "Fullscreen mode"
-msgstr "ΛΕΙΤΟΥΡΓΙΑ ΠΛΗΡΟΥΣ ΟΘΟΝΗΣ"
-
-msgid "Open an issue"
-msgstr "Ανοίξτε ένα ζήτημα"
-
-msgid "previous page"
-msgstr "προηγούμενη σελίδα"
-
-msgid "Download this page"
-msgstr "Λήψη αυτής της σελίδας"
-
-msgid "Theme by the"
-msgstr "Θέμα από το"
diff --git a/docs/_build/html/_static/locales/eo/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/eo/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index d1072bbec62959524a87953d3f41b7c272ecbe37..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1337
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diff --git a/docs/_build/html/_static/locales/eo/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/eo/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 6749f3a..0000000
--- a/docs/_build/html/_static/locales/eo/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: eo\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "sugesti redaktadon"
-
-msgid "Last updated on"
-msgstr "Laste ĝisdatigita la"
-
-msgid "Edit this page"
-msgstr "Redaktu ĉi tiun paĝon"
-
-msgid "Launch"
-msgstr "Lanĉo"
-
-msgid "Print to PDF"
-msgstr "Presi al PDF"
-
-msgid "open issue"
-msgstr "malferma numero"
-
-msgid "Download notebook file"
-msgstr "Elŝutu kajeran dosieron"
-
-msgid "Toggle navigation"
-msgstr "Ŝalti navigadon"
-
-msgid "Source repository"
-msgstr "Fonto-deponejo"
-
-msgid "By the"
-msgstr "Per la"
-
-msgid "next page"
-msgstr "sekva paĝo"
-
-msgid "repository"
-msgstr "deponejo"
-
-msgid "Sphinx Book Theme"
-msgstr "Sfinksa Libro-Temo"
-
-msgid "Download source file"
-msgstr "Elŝutu fontodosieron"
-
-msgid "Contents"
-msgstr "Enhavo"
-
-msgid "By"
-msgstr "De"
-
-msgid "Copyright"
-msgstr "Kopirajto"
-
-msgid "Fullscreen mode"
-msgstr "Plenekrana reĝimo"
-
-msgid "Open an issue"
-msgstr "Malfermu numeron"
-
-msgid "previous page"
-msgstr "antaŭa paĝo"
-
-msgid "Download this page"
-msgstr "Elŝutu ĉi tiun paĝon"
-
-msgid "Theme by the"
-msgstr "Temo de la"
diff --git a/docs/_build/html/_static/locales/es/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/es/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index ba2ee4dc22148ed53f2aeba32c5965654d4a5a2f..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1396
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diff --git a/docs/_build/html/_static/locales/es/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/es/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 71dde37..0000000
--- a/docs/_build/html/_static/locales/es/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: es\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "sugerir editar"
-
-msgid "Last updated on"
-msgstr "Ultima actualización en"
-
-msgid "Edit this page"
-msgstr "Edita esta página"
-
-msgid "Launch"
-msgstr "Lanzamiento"
-
-msgid "Print to PDF"
-msgstr "Imprimir en PDF"
-
-msgid "open issue"
-msgstr "Tema abierto"
-
-msgid "Download notebook file"
-msgstr "Descargar archivo de cuaderno"
-
-msgid "Toggle navigation"
-msgstr "Navegación de palanca"
-
-msgid "Source repository"
-msgstr "Repositorio de origen"
-
-msgid "By the"
-msgstr "Por el"
-
-msgid "next page"
-msgstr "siguiente página"
-
-msgid "repository"
-msgstr "repositorio"
-
-msgid "Sphinx Book Theme"
-msgstr "Tema del libro de la esfinge"
-
-msgid "Download source file"
-msgstr "Descargar archivo fuente"
-
-msgid "Contents"
-msgstr "Contenido"
-
-msgid "By"
-msgstr "Por"
-
-msgid "Copyright"
-msgstr "Derechos de autor"
-
-msgid "Fullscreen mode"
-msgstr "Modo de pantalla completa"
-
-msgid "Open an issue"
-msgstr "Abrir un problema"
-
-msgid "previous page"
-msgstr "pagina anterior"
-
-msgid "Download this page"
-msgstr "Descarga esta pagina"
-
-msgid "Theme by the"
-msgstr "Tema por el"
diff --git a/docs/_build/html/_static/locales/et/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/et/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 983b82391f499b67a9046c15d0dd8744650ad925..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1341
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diff --git a/docs/_build/html/_static/locales/et/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/et/LC_MESSAGES/booktheme.po
deleted file mode 100644
index cdcd07c..0000000
--- a/docs/_build/html/_static/locales/et/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: et\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "soovita muuta"
-
-msgid "Last updated on"
-msgstr "Viimati uuendatud"
-
-msgid "Edit this page"
-msgstr "Muutke seda lehte"
-
-msgid "Launch"
-msgstr "Käivitage"
-
-msgid "Print to PDF"
-msgstr "Prindi PDF-i"
-
-msgid "open issue"
-msgstr "avatud küsimus"
-
-msgid "Download notebook file"
-msgstr "Laadige sülearvuti fail alla"
-
-msgid "Toggle navigation"
-msgstr "Lülita navigeerimine sisse"
-
-msgid "Source repository"
-msgstr "Allikahoidla"
-
-msgid "By the"
-msgstr "Autor"
-
-msgid "next page"
-msgstr "järgmine leht"
-
-msgid "repository"
-msgstr "hoidla"
-
-msgid "Sphinx Book Theme"
-msgstr "Sfinksiraamatu teema"
-
-msgid "Download source file"
-msgstr "Laadige alla lähtefail"
-
-msgid "Contents"
-msgstr "Sisu"
-
-msgid "By"
-msgstr "Kõrval"
-
-msgid "Copyright"
-msgstr "Autoriõigus"
-
-msgid "Fullscreen mode"
-msgstr "Täisekraanirežiim"
-
-msgid "Open an issue"
-msgstr "Avage probleem"
-
-msgid "previous page"
-msgstr "eelmine leht"
-
-msgid "Download this page"
-msgstr "Laadige see leht alla"
-
-msgid "Theme by the"
-msgstr "Teema"
diff --git a/docs/_build/html/_static/locales/fi/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/fi/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index d8ac054597c924e3010f629caeac1c748b7211cd..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1368
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diff --git a/docs/_build/html/_static/locales/fi/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/fi/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 3c3dd08..0000000
--- a/docs/_build/html/_static/locales/fi/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: fi\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "ehdottaa muokkausta"
-
-msgid "Last updated on"
-msgstr "Viimeksi päivitetty"
-
-msgid "Edit this page"
-msgstr "Muokkaa tätä sivua"
-
-msgid "Launch"
-msgstr "Tuoda markkinoille"
-
-msgid "Print to PDF"
-msgstr "Tulosta PDF-tiedostoon"
-
-msgid "open issue"
-msgstr "avoin ongelma"
-
-msgid "Download notebook file"
-msgstr "Lataa muistikirjatiedosto"
-
-msgid "Toggle navigation"
-msgstr "Vaihda navigointia"
-
-msgid "Source repository"
-msgstr "Lähteen arkisto"
-
-msgid "By the"
-msgstr "Mukaan"
-
-msgid "next page"
-msgstr "seuraava sivu"
-
-msgid "repository"
-msgstr "arkisto"
-
-msgid "Sphinx Book Theme"
-msgstr "Sphinx-kirjan teema"
-
-msgid "Download source file"
-msgstr "Lataa lähdetiedosto"
-
-msgid "Contents"
-msgstr "Sisällys"
-
-msgid "By"
-msgstr "Tekijä"
-
-msgid "Copyright"
-msgstr "Tekijänoikeus"
-
-msgid "Fullscreen mode"
-msgstr "Koko näytön tila"
-
-msgid "Open an issue"
-msgstr "Avaa ongelma"
-
-msgid "previous page"
-msgstr "Edellinen sivu"
-
-msgid "Download this page"
-msgstr "Lataa tämä sivu"
-
-msgid "Theme by the"
-msgstr "Teeman tekijä"
diff --git a/docs/_build/html/_static/locales/fr/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/fr/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index f663d39f0faa76c5b9bd504c51252eef74cca5de..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/fr/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/fr/LC_MESSAGES/booktheme.po
deleted file mode 100644
index b57d2fe..0000000
--- a/docs/_build/html/_static/locales/fr/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: fr\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "suggestion de modification"
-
-msgid "Last updated on"
-msgstr "Dernière mise à jour le"
-
-msgid "Edit this page"
-msgstr "Modifier cette page"
-
-msgid "Launch"
-msgstr "lancement"
-
-msgid "Print to PDF"
-msgstr "Imprimer au format PDF"
-
-msgid "open issue"
-msgstr "signaler un problème"
-
-msgid "Download notebook file"
-msgstr "Télécharger le fichier notebook"
-
-msgid "Toggle navigation"
-msgstr "Basculer la navigation"
-
-msgid "Source repository"
-msgstr "Dépôt source"
-
-msgid "By the"
-msgstr "Par le"
-
-msgid "next page"
-msgstr "page suivante"
-
-msgid "repository"
-msgstr "dépôt"
-
-msgid "Sphinx Book Theme"
-msgstr "Thème du livre Sphinx"
-
-msgid "Download source file"
-msgstr "Télécharger le fichier source"
-
-msgid "Contents"
-msgstr "Contenu"
-
-msgid "By"
-msgstr "Par"
-
-msgid "Copyright"
-msgstr "droits d'auteur"
-
-msgid "Fullscreen mode"
-msgstr "Mode plein écran"
-
-msgid "Open an issue"
-msgstr "Ouvrez un problème"
-
-msgid "previous page"
-msgstr "page précédente"
-
-msgid "Download this page"
-msgstr "Téléchargez cette page"
-
-msgid "Theme by the"
-msgstr "Thème par le"
diff --git a/docs/_build/html/_static/locales/hr/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/hr/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index eca4a1a2842830f06bd5f6235bf01d07bdd313d2..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1402
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diff --git a/docs/_build/html/_static/locales/hr/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/hr/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 4c425e8..0000000
--- a/docs/_build/html/_static/locales/hr/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: hr\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "predloži uređivanje"
-
-msgid "Last updated on"
-msgstr "Posljednje ažuriranje:"
-
-msgid "Edit this page"
-msgstr "Uredite ovu stranicu"
-
-msgid "Launch"
-msgstr "Pokrenite"
-
-msgid "Print to PDF"
-msgstr "Ispis u PDF"
-
-msgid "open issue"
-msgstr "otvoreno izdanje"
-
-msgid "Download notebook file"
-msgstr "Preuzmi datoteku bilježnice"
-
-msgid "Toggle navigation"
-msgstr "Uključi / isključi navigaciju"
-
-msgid "Source repository"
-msgstr "Izvorno spremište"
-
-msgid "By the"
-msgstr "Od strane"
-
-msgid "next page"
-msgstr "sljedeća stranica"
-
-msgid "repository"
-msgstr "spremište"
-
-msgid "Sphinx Book Theme"
-msgstr "Tema knjige Sphinx"
-
-msgid "Download source file"
-msgstr "Preuzmi izvornu datoteku"
-
-msgid "Contents"
-msgstr "Sadržaj"
-
-msgid "By"
-msgstr "Po"
-
-msgid "Copyright"
-msgstr "Autorska prava"
-
-msgid "Fullscreen mode"
-msgstr "Način preko cijelog zaslona"
-
-msgid "Open an issue"
-msgstr "Otvorite izdanje"
-
-msgid "previous page"
-msgstr "Prethodna stranica"
-
-msgid "Download this page"
-msgstr "Preuzmite ovu stranicu"
-
-msgid "Theme by the"
-msgstr "Tema autora"
diff --git a/docs/_build/html/_static/locales/id/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/id/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index d07a06a9d27aaf6d67234c227c22d61d5a23203b..0000000000000000000000000000000000000000
GIT binary patch
literal 0
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diff --git a/docs/_build/html/_static/locales/id/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/id/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 5db2ae1..0000000
--- a/docs/_build/html/_static/locales/id/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: id\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "menyarankan edit"
-
-msgid "Last updated on"
-msgstr "Terakhir diperbarui saat"
-
-msgid "Edit this page"
-msgstr "Edit halaman ini"
-
-msgid "Launch"
-msgstr "Meluncurkan"
-
-msgid "Print to PDF"
-msgstr "Cetak ke PDF"
-
-msgid "open issue"
-msgstr "masalah terbuka"
-
-msgid "Download notebook file"
-msgstr "Unduh file notebook"
-
-msgid "Toggle navigation"
-msgstr "Alihkan navigasi"
-
-msgid "Source repository"
-msgstr "Repositori sumber"
-
-msgid "By the"
-msgstr "Oleh"
-
-msgid "next page"
-msgstr "halaman selanjutnya"
-
-msgid "repository"
-msgstr "gudang"
-
-msgid "Sphinx Book Theme"
-msgstr "Tema Buku Sphinx"
-
-msgid "Download source file"
-msgstr "Unduh file sumber"
-
-msgid "Contents"
-msgstr "Isi"
-
-msgid "By"
-msgstr "Oleh"
-
-msgid "Copyright"
-msgstr "hak cipta"
-
-msgid "Fullscreen mode"
-msgstr "Mode layar penuh"
-
-msgid "Open an issue"
-msgstr "Buka masalah"
-
-msgid "previous page"
-msgstr "halaman sebelumnya"
-
-msgid "Download this page"
-msgstr "Unduh halaman ini"
-
-msgid "Theme by the"
-msgstr "Tema oleh"
diff --git a/docs/_build/html/_static/locales/it/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/it/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 53ba476edd2df2a802917e9df402257ceca1a130..0000000000000000000000000000000000000000
GIT binary patch
literal 0
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diff --git a/docs/_build/html/_static/locales/it/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/it/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 7d54fde..0000000
--- a/docs/_build/html/_static/locales/it/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: it\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "suggerisci modifica"
-
-msgid "Last updated on"
-msgstr "Ultimo aggiornamento il"
-
-msgid "Edit this page"
-msgstr "Modifica questa pagina"
-
-msgid "Launch"
-msgstr "Lanciare"
-
-msgid "Print to PDF"
-msgstr "Stampa in PDF"
-
-msgid "open issue"
-msgstr "questione aperta"
-
-msgid "Download notebook file"
-msgstr "Scarica il file del taccuino"
-
-msgid "Toggle navigation"
-msgstr "Attiva / disattiva la navigazione"
-
-msgid "Source repository"
-msgstr "Repository di origine"
-
-msgid "By the"
-msgstr "Dal"
-
-msgid "next page"
-msgstr "pagina successiva"
-
-msgid "repository"
-msgstr "repository"
-
-msgid "Sphinx Book Theme"
-msgstr "Tema del libro della Sfinge"
-
-msgid "Download source file"
-msgstr "Scarica il file sorgente"
-
-msgid "Contents"
-msgstr "Contenuti"
-
-msgid "By"
-msgstr "Di"
-
-msgid "Copyright"
-msgstr "Diritto d'autore"
-
-msgid "Fullscreen mode"
-msgstr "Modalità schermo intero"
-
-msgid "Open an issue"
-msgstr "Apri un problema"
-
-msgid "previous page"
-msgstr "pagina precedente"
-
-msgid "Download this page"
-msgstr "Scarica questa pagina"
-
-msgid "Theme by the"
-msgstr "Tema di"
diff --git a/docs/_build/html/_static/locales/iw/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/iw/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index a45c6575e469914582f8223ac54793afc2f26575..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1445
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diff --git a/docs/_build/html/_static/locales/iw/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/iw/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 32b017c..0000000
--- a/docs/_build/html/_static/locales/iw/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: iw\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "מציע לערוך"
-
-msgid "Last updated on"
-msgstr "עודכן לאחרונה ב"
-
-msgid "Edit this page"
-msgstr "ערוך דף זה"
-
-msgid "Launch"
-msgstr "לְהַשִׁיק"
-
-msgid "Print to PDF"
-msgstr "הדפס לקובץ PDF"
-
-msgid "open issue"
-msgstr "בעיה פתוחה"
-
-msgid "Download notebook file"
-msgstr "הורד קובץ מחברת"
-
-msgid "Toggle navigation"
-msgstr "החלף ניווט"
-
-msgid "Source repository"
-msgstr "מאגר המקורות"
-
-msgid "By the"
-msgstr "דרך"
-
-msgid "next page"
-msgstr "עמוד הבא"
-
-msgid "repository"
-msgstr "מאגר"
-
-msgid "Sphinx Book Theme"
-msgstr "נושא ספר ספינקס"
-
-msgid "Download source file"
-msgstr "הורד את קובץ המקור"
-
-msgid "Contents"
-msgstr "תוכן"
-
-msgid "By"
-msgstr "על ידי"
-
-msgid "Copyright"
-msgstr "זכויות יוצרים"
-
-msgid "Fullscreen mode"
-msgstr "מצב מסך מלא"
-
-msgid "Open an issue"
-msgstr "פתח גיליון"
-
-msgid "previous page"
-msgstr "עמוד קודם"
-
-msgid "Download this page"
-msgstr "הורד דף זה"
-
-msgid "Theme by the"
-msgstr "נושא מאת"
diff --git a/docs/_build/html/_static/locales/ja/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/ja/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 1cefd29ce3cc08792667a82dc7ff47e9843107be..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/ja/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/ja/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 16924e1..0000000
--- a/docs/_build/html/_static/locales/ja/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: ja\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "編集を提案する"
-
-msgid "Last updated on"
-msgstr "最終更新日"
-
-msgid "Edit this page"
-msgstr "このページを編集"
-
-msgid "Launch"
-msgstr "起動"
-
-msgid "Print to PDF"
-msgstr "PDFに印刷"
-
-msgid "open issue"
-msgstr "未解決の問題"
-
-msgid "Download notebook file"
-msgstr "ノートブックファイルをダウンロード"
-
-msgid "Toggle navigation"
-msgstr "ナビゲーションを切り替え"
-
-msgid "Source repository"
-msgstr "ソースリポジトリ"
-
-msgid "By the"
-msgstr "によって"
-
-msgid "next page"
-msgstr "次のページ"
-
-msgid "repository"
-msgstr "リポジトリ"
-
-msgid "Sphinx Book Theme"
-msgstr "スフィンクスの本のテーマ"
-
-msgid "Download source file"
-msgstr "ソースファイルをダウンロード"
-
-msgid "Contents"
-msgstr "目次"
-
-msgid "By"
-msgstr "著者"
-
-msgid "Copyright"
-msgstr "Copyright"
-
-msgid "Fullscreen mode"
-msgstr "全画面モード"
-
-msgid "Open an issue"
-msgstr "問題を報告"
-
-msgid "previous page"
-msgstr "前のページ"
-
-msgid "Download this page"
-msgstr "このページをダウンロード"
-
-msgid "Theme by the"
-msgstr "のテーマ"
diff --git a/docs/_build/html/_static/locales/ko/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/ko/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 06c7ec938b4975c721caff60b956e343eef11570..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1375
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diff --git a/docs/_build/html/_static/locales/ko/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/ko/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 69dd18f..0000000
--- a/docs/_build/html/_static/locales/ko/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: ko\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "편집 제안"
-
-msgid "Last updated on"
-msgstr "마지막 업데이트"
-
-msgid "Edit this page"
-msgstr "이 페이지 편집"
-
-msgid "Launch"
-msgstr "시작하다"
-
-msgid "Print to PDF"
-msgstr "PDF로 인쇄"
-
-msgid "open issue"
-msgstr "열린 문제"
-
-msgid "Download notebook file"
-msgstr "노트북 파일 다운로드"
-
-msgid "Toggle navigation"
-msgstr "탐색 전환"
-
-msgid "Source repository"
-msgstr "소스 저장소"
-
-msgid "By the"
-msgstr "에 의해"
-
-msgid "next page"
-msgstr "다음 페이지"
-
-msgid "repository"
-msgstr "저장소"
-
-msgid "Sphinx Book Theme"
-msgstr "스핑크스 도서 테마"
-
-msgid "Download source file"
-msgstr "소스 파일 다운로드"
-
-msgid "Contents"
-msgstr "내용"
-
-msgid "By"
-msgstr "으로"
-
-msgid "Copyright"
-msgstr "저작권"
-
-msgid "Fullscreen mode"
-msgstr "전체 화면으로보기"
-
-msgid "Open an issue"
-msgstr "이슈 열기"
-
-msgid "previous page"
-msgstr "이전 페이지"
-
-msgid "Download this page"
-msgstr "이 페이지 다운로드"
-
-msgid "Theme by the"
-msgstr "테마별"
diff --git a/docs/_build/html/_static/locales/lt/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/lt/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 4468ba04bc134a84fea5e3c973461cf02c9c2da3..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1413
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diff --git a/docs/_build/html/_static/locales/lt/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/lt/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 9f03775..0000000
--- a/docs/_build/html/_static/locales/lt/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: lt\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "pasiūlyti redaguoti"
-
-msgid "Last updated on"
-msgstr "Paskutinį kartą atnaujinta"
-
-msgid "Edit this page"
-msgstr "Redaguoti šį puslapį"
-
-msgid "Launch"
-msgstr "Paleiskite"
-
-msgid "Print to PDF"
-msgstr "Spausdinti į PDF"
-
-msgid "open issue"
-msgstr "atviras klausimas"
-
-msgid "Download notebook file"
-msgstr "Atsisiųsti nešiojamojo kompiuterio failą"
-
-msgid "Toggle navigation"
-msgstr "Perjungti naršymą"
-
-msgid "Source repository"
-msgstr "Šaltinio saugykla"
-
-msgid "By the"
-msgstr "Prie"
-
-msgid "next page"
-msgstr "Kitas puslapis"
-
-msgid "repository"
-msgstr "saugykla"
-
-msgid "Sphinx Book Theme"
-msgstr "Sfinkso knygos tema"
-
-msgid "Download source file"
-msgstr "Atsisiųsti šaltinio failą"
-
-msgid "Contents"
-msgstr "Turinys"
-
-msgid "By"
-msgstr "Iki"
-
-msgid "Copyright"
-msgstr "Autorių teisės"
-
-msgid "Fullscreen mode"
-msgstr "Pilno ekrano režimas"
-
-msgid "Open an issue"
-msgstr "Atidarykite problemą"
-
-msgid "previous page"
-msgstr "Ankstesnis puslapis"
-
-msgid "Download this page"
-msgstr "Atsisiųskite šį puslapį"
-
-msgid "Theme by the"
-msgstr "Tema"
diff --git a/docs/_build/html/_static/locales/lv/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/lv/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 74aa4d8985d96a07c4c9be96f006f3b58d884342..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1404
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diff --git a/docs/_build/html/_static/locales/lv/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/lv/LC_MESSAGES/booktheme.po
deleted file mode 100644
index c9633b5..0000000
--- a/docs/_build/html/_static/locales/lv/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: lv\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "ieteikt rediģēt"
-
-msgid "Last updated on"
-msgstr "Pēdējoreiz atjaunināts"
-
-msgid "Edit this page"
-msgstr "Rediģēt šo lapu"
-
-msgid "Launch"
-msgstr "Uzsākt"
-
-msgid "Print to PDF"
-msgstr "Drukāt PDF formātā"
-
-msgid "open issue"
-msgstr "atklāts jautājums"
-
-msgid "Download notebook file"
-msgstr "Lejupielādēt piezīmju grāmatiņu"
-
-msgid "Toggle navigation"
-msgstr "Pārslēgt navigāciju"
-
-msgid "Source repository"
-msgstr "Avota krātuve"
-
-msgid "By the"
-msgstr "Ar"
-
-msgid "next page"
-msgstr "nākamā lapaspuse"
-
-msgid "repository"
-msgstr "krātuve"
-
-msgid "Sphinx Book Theme"
-msgstr "Sfinksa grāmatas tēma"
-
-msgid "Download source file"
-msgstr "Lejupielādēt avota failu"
-
-msgid "Contents"
-msgstr "Saturs"
-
-msgid "By"
-msgstr "Autors"
-
-msgid "Copyright"
-msgstr "Autortiesības"
-
-msgid "Fullscreen mode"
-msgstr "Pilnekrāna režīms"
-
-msgid "Open an issue"
-msgstr "Atveriet problēmu"
-
-msgid "previous page"
-msgstr "iepriekšējā lapa"
-
-msgid "Download this page"
-msgstr "Lejupielādējiet šo lapu"
-
-msgid "Theme by the"
-msgstr "Autora tēma"
diff --git a/docs/_build/html/_static/locales/ml/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/ml/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 2736e8fcf6f9e923c2403307f0b366086d37b335..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/ml/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/ml/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 9a6a41e..0000000
--- a/docs/_build/html/_static/locales/ml/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,66 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: ml\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "എഡിറ്റുചെയ്യാൻ നിർദ്ദേശിക്കുക"
-
-msgid "Last updated on"
-msgstr "അവസാനം അപ്‌ഡേറ്റുചെയ്‌തത്"
-
-msgid "Edit this page"
-msgstr "ഈ പേജ് എഡിറ്റുചെയ്യുക"
-
-msgid "Launch"
-msgstr "സമാരംഭിക്കുക"
-
-msgid "Print to PDF"
-msgstr "PDF- ലേക്ക് പ്രിന്റുചെയ്യുക"
-
-msgid "open issue"
-msgstr "തുറന്ന പ്രശ്നം"
-
-msgid "Download notebook file"
-msgstr "നോട്ട്ബുക്ക് ഫയൽ ഡൺലോഡ് ചെയ്യുക"
-
-msgid "Toggle navigation"
-msgstr "നാവിഗേഷൻ ടോഗിൾ ചെയ്യുക"
-
-msgid "Source repository"
-msgstr "ഉറവിട ശേഖരം"
-
-msgid "By the"
-msgstr "എഴുതിയത്"
-
-msgid "next page"
-msgstr "അടുത്ത പേജ്"
-
-msgid "Sphinx Book Theme"
-msgstr "സ്ഫിങ്ക്സ് പുസ്തക തീം"
-
-msgid "Download source file"
-msgstr "ഉറവിട ഫയൽ ഡൗൺലോഡുചെയ്യുക"
-
-msgid "By"
-msgstr "എഴുതിയത്"
-
-msgid "Copyright"
-msgstr "പകർപ്പവകാശം"
-
-msgid "Open an issue"
-msgstr "ഒരു പ്രശ്നം തുറക്കുക"
-
-msgid "previous page"
-msgstr "മുൻപത്തെ താൾ"
-
-msgid "Download this page"
-msgstr "ഈ പേജ് ഡൗൺലോഡുചെയ്യുക"
-
-msgid "Theme by the"
-msgstr "പ്രമേയം"
diff --git a/docs/_build/html/_static/locales/mr/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/mr/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index fe530100d7715cdc19a6f9db33a971665835f3c4..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/mr/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/mr/LC_MESSAGES/booktheme.po
deleted file mode 100644
index ef72d8c..0000000
--- a/docs/_build/html/_static/locales/mr/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,66 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: mr\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "संपादन सुचवा"
-
-msgid "Last updated on"
-msgstr "अखेरचे अद्यतनित"
-
-msgid "Edit this page"
-msgstr "हे पृष्ठ संपादित करा"
-
-msgid "Launch"
-msgstr "लाँच करा"
-
-msgid "Print to PDF"
-msgstr "पीडीएफवर मुद्रित करा"
-
-msgid "open issue"
-msgstr "खुला मुद्दा"
-
-msgid "Download notebook file"
-msgstr "नोटबुक फाईल डाउनलोड करा"
-
-msgid "Toggle navigation"
-msgstr "नेव्हिगेशन टॉगल करा"
-
-msgid "Source repository"
-msgstr "स्त्रोत भांडार"
-
-msgid "By the"
-msgstr "द्वारा"
-
-msgid "next page"
-msgstr "पुढील पृष्ठ"
-
-msgid "Sphinx Book Theme"
-msgstr "स्फिंक्स बुक थीम"
-
-msgid "Download source file"
-msgstr "स्त्रोत फाइल डाउनलोड करा"
-
-msgid "By"
-msgstr "द्वारा"
-
-msgid "Copyright"
-msgstr "कॉपीराइट"
-
-msgid "Open an issue"
-msgstr "एक मुद्दा उघडा"
-
-msgid "previous page"
-msgstr "मागील पान"
-
-msgid "Download this page"
-msgstr "हे पृष्ठ डाउनलोड करा"
-
-msgid "Theme by the"
-msgstr "द्वारा थीम"
diff --git a/docs/_build/html/_static/locales/ms/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/ms/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index f02603fa2522a40060bd3f1b5d65052c77530de6..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1213
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diff --git a/docs/_build/html/_static/locales/ms/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/ms/LC_MESSAGES/booktheme.po
deleted file mode 100644
index e29cbe2..0000000
--- a/docs/_build/html/_static/locales/ms/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,66 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: ms\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "cadangkan edit"
-
-msgid "Last updated on"
-msgstr "Terakhir dikemas kini pada"
-
-msgid "Edit this page"
-msgstr "Edit halaman ini"
-
-msgid "Launch"
-msgstr "Lancarkan"
-
-msgid "Print to PDF"
-msgstr "Cetak ke PDF"
-
-msgid "open issue"
-msgstr "isu terbuka"
-
-msgid "Download notebook file"
-msgstr "Muat turun fail buku nota"
-
-msgid "Toggle navigation"
-msgstr "Togol navigasi"
-
-msgid "Source repository"
-msgstr "Repositori sumber"
-
-msgid "By the"
-msgstr "Oleh"
-
-msgid "next page"
-msgstr "muka surat seterusnya"
-
-msgid "Sphinx Book Theme"
-msgstr "Tema Buku Sphinx"
-
-msgid "Download source file"
-msgstr "Muat turun fail sumber"
-
-msgid "By"
-msgstr "Oleh"
-
-msgid "Copyright"
-msgstr "hak cipta"
-
-msgid "Open an issue"
-msgstr "Buka masalah"
-
-msgid "previous page"
-msgstr "halaman sebelumnya"
-
-msgid "Download this page"
-msgstr "Muat turun halaman ini"
-
-msgid "Theme by the"
-msgstr "Tema oleh"
diff --git a/docs/_build/html/_static/locales/nl/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/nl/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index e59e7ecb308a7648cd23aa6342a9ad98a61d0009..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/nl/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/nl/LC_MESSAGES/booktheme.po
deleted file mode 100644
index e4844d7..0000000
--- a/docs/_build/html/_static/locales/nl/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: nl\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "suggereren bewerken"
-
-msgid "Last updated on"
-msgstr "Laatst geupdate op"
-
-msgid "Edit this page"
-msgstr "bewerk deze pagina"
-
-msgid "Launch"
-msgstr "Lancering"
-
-msgid "Print to PDF"
-msgstr "Afdrukken naar pdf"
-
-msgid "open issue"
-msgstr "open probleem"
-
-msgid "Download notebook file"
-msgstr "Download notebookbestand"
-
-msgid "Toggle navigation"
-msgstr "Schakel navigatie"
-
-msgid "Source repository"
-msgstr "Bronopslagplaats"
-
-msgid "By the"
-msgstr "Door de"
-
-msgid "next page"
-msgstr "volgende bladzijde"
-
-msgid "repository"
-msgstr "repository"
-
-msgid "Sphinx Book Theme"
-msgstr "Sphinx-boekthema"
-
-msgid "Download source file"
-msgstr "Download het bronbestand"
-
-msgid "Contents"
-msgstr "Inhoud"
-
-msgid "By"
-msgstr "Door"
-
-msgid "Copyright"
-msgstr "auteursrechten"
-
-msgid "Fullscreen mode"
-msgstr "Volledig scherm"
-
-msgid "Open an issue"
-msgstr "Open een probleem"
-
-msgid "previous page"
-msgstr "vorige pagina"
-
-msgid "Download this page"
-msgstr "Download deze pagina"
-
-msgid "Theme by the"
-msgstr "Thema door de"
diff --git a/docs/_build/html/_static/locales/no/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/no/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 6cd15c88de675226c00ca4d0430171075e5559ff..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1317
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diff --git a/docs/_build/html/_static/locales/no/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/no/LC_MESSAGES/booktheme.po
deleted file mode 100644
index d079dd9..0000000
--- a/docs/_build/html/_static/locales/no/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: no\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "foreslå redigering"
-
-msgid "Last updated on"
-msgstr "Sist oppdatert den"
-
-msgid "Edit this page"
-msgstr "Rediger denne siden"
-
-msgid "Launch"
-msgstr "Start"
-
-msgid "Print to PDF"
-msgstr "Skriv ut til PDF"
-
-msgid "open issue"
-msgstr "åpent nummer"
-
-msgid "Download notebook file"
-msgstr "Last ned notatbokfilen"
-
-msgid "Toggle navigation"
-msgstr "Bytt navigasjon"
-
-msgid "Source repository"
-msgstr "Kildedepot"
-
-msgid "By the"
-msgstr "Ved"
-
-msgid "next page"
-msgstr "neste side"
-
-msgid "repository"
-msgstr "oppbevaringssted"
-
-msgid "Sphinx Book Theme"
-msgstr "Sphinx boktema"
-
-msgid "Download source file"
-msgstr "Last ned kildefilen"
-
-msgid "Contents"
-msgstr "Innhold"
-
-msgid "By"
-msgstr "Av"
-
-msgid "Copyright"
-msgstr "opphavsrett"
-
-msgid "Fullscreen mode"
-msgstr "Fullskjerm-modus"
-
-msgid "Open an issue"
-msgstr "Åpne et problem"
-
-msgid "previous page"
-msgstr "forrige side"
-
-msgid "Download this page"
-msgstr "Last ned denne siden"
-
-msgid "Theme by the"
-msgstr "Tema av"
diff --git a/docs/_build/html/_static/locales/pl/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/pl/LC_MESSAGES/booktheme.mo
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diff --git a/docs/_build/html/_static/locales/pl/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/pl/LC_MESSAGES/booktheme.po
deleted file mode 100644
index fcac51d..0000000
--- a/docs/_build/html/_static/locales/pl/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
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-"Content-Transfer-Encoding: 8bit\n"
-"Language: pl\n"
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-msgid "suggest edit"
-msgstr "zaproponuj edycję"
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-msgid "Last updated on"
-msgstr "Ostatnia aktualizacja"
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-msgid "Edit this page"
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-msgid "Launch"
-msgstr "Uruchomić"
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-msgid "Print to PDF"
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-msgid "open issue"
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-msgid "Download notebook file"
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-msgid "Toggle navigation"
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-
-msgid "Source repository"
-msgstr "Repozytorium źródłowe"
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-msgid "By the"
-msgstr "Przez"
-
-msgid "next page"
-msgstr "Następna strona"
-
-msgid "repository"
-msgstr "magazyn"
-
-msgid "Sphinx Book Theme"
-msgstr "Motyw książki Sphinx"
-
-msgid "Download source file"
-msgstr "Pobierz plik źródłowy"
-
-msgid "Contents"
-msgstr "Zawartość"
-
-msgid "By"
-msgstr "Przez"
-
-msgid "Copyright"
-msgstr "prawa autorskie"
-
-msgid "Fullscreen mode"
-msgstr "Pełny ekran"
-
-msgid "Open an issue"
-msgstr "Otwórz problem"
-
-msgid "previous page"
-msgstr "Poprzednia strona"
-
-msgid "Download this page"
-msgstr "Pobierz tę stronę"
-
-msgid "Theme by the"
-msgstr "Motyw autorstwa"
diff --git a/docs/_build/html/_static/locales/pt/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/pt/LC_MESSAGES/booktheme.mo
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GIT binary patch
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diff --git a/docs/_build/html/_static/locales/pt/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/pt/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 1761db0..0000000
--- a/docs/_build/html/_static/locales/pt/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
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-"Content-Transfer-Encoding: 8bit\n"
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-
-msgid "suggest edit"
-msgstr "sugerir edição"
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-msgid "Last updated on"
-msgstr "Última atualização em"
-
-msgid "Edit this page"
-msgstr "Edite essa página"
-
-msgid "Launch"
-msgstr "Lançamento"
-
-msgid "Print to PDF"
-msgstr "Imprimir em PDF"
-
-msgid "open issue"
-msgstr "questão aberta"
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-msgid "Download notebook file"
-msgstr "Baixar arquivo de notebook"
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-msgid "Toggle navigation"
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-msgid "Source repository"
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-
-msgid "next page"
-msgstr "próxima página"
-
-msgid "repository"
-msgstr "repositório"
-
-msgid "Sphinx Book Theme"
-msgstr "Tema do livro Sphinx"
-
-msgid "Download source file"
-msgstr "Baixar arquivo fonte"
-
-msgid "Contents"
-msgstr "Conteúdo"
-
-msgid "By"
-msgstr "De"
-
-msgid "Copyright"
-msgstr "direito autoral"
-
-msgid "Fullscreen mode"
-msgstr "Modo tela cheia"
-
-msgid "Open an issue"
-msgstr "Abra um problema"
-
-msgid "previous page"
-msgstr "página anterior"
-
-msgid "Download this page"
-msgstr "Baixe esta página"
-
-msgid "Theme by the"
-msgstr "Tema por"
diff --git a/docs/_build/html/_static/locales/ro/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/ro/LC_MESSAGES/booktheme.mo
deleted file mode 100644
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diff --git a/docs/_build/html/_static/locales/ro/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/ro/LC_MESSAGES/booktheme.po
deleted file mode 100644
index db865c8..0000000
--- a/docs/_build/html/_static/locales/ro/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: ro\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "sugerează editare"
-
-msgid "Last updated on"
-msgstr "Ultima actualizare la"
-
-msgid "Edit this page"
-msgstr "Editați această pagină"
-
-msgid "Launch"
-msgstr "Lansa"
-
-msgid "Print to PDF"
-msgstr "Imprimați în PDF"
-
-msgid "open issue"
-msgstr "problema deschisă"
-
-msgid "Download notebook file"
-msgstr "Descărcați fișierul notebook"
-
-msgid "Toggle navigation"
-msgstr "Comutare navigare"
-
-msgid "Source repository"
-msgstr "Depozit sursă"
-
-msgid "By the"
-msgstr "Langa"
-
-msgid "next page"
-msgstr "pagina următoare"
-
-msgid "repository"
-msgstr "repertoriu"
-
-msgid "Sphinx Book Theme"
-msgstr "Tema Sphinx Book"
-
-msgid "Download source file"
-msgstr "Descărcați fișierul sursă"
-
-msgid "Contents"
-msgstr "Cuprins"
-
-msgid "By"
-msgstr "De"
-
-msgid "Copyright"
-msgstr "Drepturi de autor"
-
-msgid "Fullscreen mode"
-msgstr "Modul ecran întreg"
-
-msgid "Open an issue"
-msgstr "Deschideți o problemă"
-
-msgid "previous page"
-msgstr "pagina anterioară"
-
-msgid "Download this page"
-msgstr "Descarcă această pagină"
-
-msgid "Theme by the"
-msgstr "Tema de"
diff --git a/docs/_build/html/_static/locales/ru/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/ru/LC_MESSAGES/booktheme.mo
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diff --git a/docs/_build/html/_static/locales/ru/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/ru/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 84ab6eb..0000000
--- a/docs/_build/html/_static/locales/ru/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
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-
-msgid "suggest edit"
-msgstr "предложить редактировать"
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-msgid "Last updated on"
-msgstr "Последнее обновление"
-
-msgid "Edit this page"
-msgstr "Редактировать эту страницу"
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-msgid "Launch"
-msgstr "Запуск"
-
-msgid "Print to PDF"
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-
-msgid "open issue"
-msgstr "открытый вопрос"
-
-msgid "Download notebook file"
-msgstr "Скачать файл записной книжки"
-
-msgid "Toggle navigation"
-msgstr "Переключить навигацию"
-
-msgid "Source repository"
-msgstr "Исходный репозиторий"
-
-msgid "By the"
-msgstr "Посредством"
-
-msgid "next page"
-msgstr "Следующая страница"
-
-msgid "repository"
-msgstr "хранилище"
-
-msgid "Sphinx Book Theme"
-msgstr "Тема книги Сфинкс"
-
-msgid "Download source file"
-msgstr "Скачать исходный файл"
-
-msgid "Contents"
-msgstr "Содержание"
-
-msgid "By"
-msgstr "По"
-
-msgid "Copyright"
-msgstr "авторское право"
-
-msgid "Fullscreen mode"
-msgstr "Полноэкранный режим"
-
-msgid "Open an issue"
-msgstr "Открыть вопрос"
-
-msgid "previous page"
-msgstr "Предыдущая страница"
-
-msgid "Download this page"
-msgstr "Загрузите эту страницу"
-
-msgid "Theme by the"
-msgstr "Тема от"
diff --git a/docs/_build/html/_static/locales/sk/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/sk/LC_MESSAGES/booktheme.mo
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diff --git a/docs/_build/html/_static/locales/sk/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/sk/LC_MESSAGES/booktheme.po
deleted file mode 100644
index e44878b..0000000
--- a/docs/_build/html/_static/locales/sk/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: sk\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "navrhnúť úpravu"
-
-msgid "Last updated on"
-msgstr "Posledná aktualizácia dňa"
-
-msgid "Edit this page"
-msgstr "Upraviť túto stránku"
-
-msgid "Launch"
-msgstr "Spustiť"
-
-msgid "Print to PDF"
-msgstr "Tlač do PDF"
-
-msgid "open issue"
-msgstr "otvorené vydanie"
-
-msgid "Download notebook file"
-msgstr "Stiahnite si zošit"
-
-msgid "Toggle navigation"
-msgstr "Prepnúť navigáciu"
-
-msgid "Source repository"
-msgstr "Zdrojové úložisko"
-
-msgid "By the"
-msgstr "Podľa"
-
-msgid "next page"
-msgstr "ďalšia strana"
-
-msgid "repository"
-msgstr "Úložisko"
-
-msgid "Sphinx Book Theme"
-msgstr "Téma knihy Sfinga"
-
-msgid "Download source file"
-msgstr "Stiahnite si zdrojový súbor"
-
-msgid "Contents"
-msgstr "Obsah"
-
-msgid "By"
-msgstr "Autor:"
-
-msgid "Copyright"
-msgstr "Autorské práva"
-
-msgid "Fullscreen mode"
-msgstr "Režim celej obrazovky"
-
-msgid "Open an issue"
-msgstr "Otvorte problém"
-
-msgid "previous page"
-msgstr "predchádzajúca strana"
-
-msgid "Download this page"
-msgstr "Stiahnite si túto stránku"
-
-msgid "Theme by the"
-msgstr "Téma od"
diff --git a/docs/_build/html/_static/locales/sl/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/sl/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 87bf26de683cb18c73bd23c2b4c57ef6a02545ec..0000000000000000000000000000000000000000
GIT binary patch
literal 0
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diff --git a/docs/_build/html/_static/locales/sl/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/sl/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 228939b..0000000
--- a/docs/_build/html/_static/locales/sl/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: sl\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "predlagajte urejanje"
-
-msgid "Last updated on"
-msgstr "Nazadnje posodobljeno dne"
-
-msgid "Edit this page"
-msgstr "Uredite to stran"
-
-msgid "Launch"
-msgstr "Kosilo"
-
-msgid "Print to PDF"
-msgstr "Natisni v PDF"
-
-msgid "open issue"
-msgstr "odprto vprašanje"
-
-msgid "Download notebook file"
-msgstr "Prenesite datoteko zvezka"
-
-msgid "Toggle navigation"
-msgstr "Preklopi navigacijo"
-
-msgid "Source repository"
-msgstr "Izvorno skladišče"
-
-msgid "By the"
-msgstr "Avtor"
-
-msgid "next page"
-msgstr "Naslednja stran"
-
-msgid "repository"
-msgstr "odlagališče"
-
-msgid "Sphinx Book Theme"
-msgstr "Tema knjige Sphinx"
-
-msgid "Download source file"
-msgstr "Prenesite izvorno datoteko"
-
-msgid "Contents"
-msgstr "Vsebina"
-
-msgid "By"
-msgstr "Avtor"
-
-msgid "Copyright"
-msgstr "avtorske pravice"
-
-msgid "Fullscreen mode"
-msgstr "Celozaslonski način"
-
-msgid "Open an issue"
-msgstr "Odprite številko"
-
-msgid "previous page"
-msgstr "Prejšnja stran"
-
-msgid "Download this page"
-msgstr "Prenesite to stran"
-
-msgid "Theme by the"
-msgstr "Tema avtorja"
diff --git a/docs/_build/html/_static/locales/sr/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/sr/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index ec740f4852f3973fa72ed0c7f7cf59273b8dba41..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1679
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diff --git a/docs/_build/html/_static/locales/sr/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/sr/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 1a712a1..0000000
--- a/docs/_build/html/_static/locales/sr/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: sr\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "предложи уређивање"
-
-msgid "Last updated on"
-msgstr "Последње ажурирање"
-
-msgid "Edit this page"
-msgstr "Уредите ову страницу"
-
-msgid "Launch"
-msgstr "Лансирање"
-
-msgid "Print to PDF"
-msgstr "Испис у ПДФ"
-
-msgid "open issue"
-msgstr "отворено издање"
-
-msgid "Download notebook file"
-msgstr "Преузмите датотеку бележнице"
-
-msgid "Toggle navigation"
-msgstr "Укључи / искључи навигацију"
-
-msgid "Source repository"
-msgstr "Изворно спремиште"
-
-msgid "By the"
-msgstr "Од"
-
-msgid "next page"
-msgstr "Следећа страна"
-
-msgid "repository"
-msgstr "спремиште"
-
-msgid "Sphinx Book Theme"
-msgstr "Тема књиге Спхинк"
-
-msgid "Download source file"
-msgstr "Преузми изворну датотеку"
-
-msgid "Contents"
-msgstr "Садржај"
-
-msgid "By"
-msgstr "Од стране"
-
-msgid "Copyright"
-msgstr "Ауторско право"
-
-msgid "Fullscreen mode"
-msgstr "Режим целог екрана"
-
-msgid "Open an issue"
-msgstr "Отворите издање"
-
-msgid "previous page"
-msgstr "Претходна страница"
-
-msgid "Download this page"
-msgstr "Преузмите ову страницу"
-
-msgid "Theme by the"
-msgstr "Тхеме би"
diff --git a/docs/_build/html/_static/locales/sv/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/sv/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index be951bec201ccd86d534499749af1df8938a14e3..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/sv/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/sv/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 7d2b56d..0000000
--- a/docs/_build/html/_static/locales/sv/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: sv\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "föreslå redigering"
-
-msgid "Last updated on"
-msgstr "Senast uppdaterad den"
-
-msgid "Edit this page"
-msgstr "Redigera den här sidan"
-
-msgid "Launch"
-msgstr "Lansera"
-
-msgid "Print to PDF"
-msgstr "Skriv ut till PDF"
-
-msgid "open issue"
-msgstr "öppet problem"
-
-msgid "Download notebook file"
-msgstr "Ladda ner anteckningsbokfilen"
-
-msgid "Toggle navigation"
-msgstr "Växla navigering"
-
-msgid "Source repository"
-msgstr "Källförvar"
-
-msgid "By the"
-msgstr "Vid"
-
-msgid "next page"
-msgstr "nästa sida"
-
-msgid "repository"
-msgstr "förvar"
-
-msgid "Sphinx Book Theme"
-msgstr "Sphinx boktema"
-
-msgid "Download source file"
-msgstr "Ladda ner källfil"
-
-msgid "Contents"
-msgstr "Innehåll"
-
-msgid "By"
-msgstr "Förbi"
-
-msgid "Copyright"
-msgstr "upphovsrätt"
-
-msgid "Fullscreen mode"
-msgstr "Fullskärmsläge"
-
-msgid "Open an issue"
-msgstr "Öppna ett problem"
-
-msgid "previous page"
-msgstr "föregående sida"
-
-msgid "Download this page"
-msgstr "Ladda ner den här sidan"
-
-msgid "Theme by the"
-msgstr "Tema av"
diff --git a/docs/_build/html/_static/locales/ta/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/ta/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 29f52e1f6f1ec96aceb0f883d8ccc82c23eb35c2..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/ta/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/ta/LC_MESSAGES/booktheme.po
deleted file mode 100644
index c75ffe1..0000000
--- a/docs/_build/html/_static/locales/ta/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,66 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: ta\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "திருத்த பரிந்துரைக்கவும்"
-
-msgid "Last updated on"
-msgstr "கடைசியாக புதுப்பிக்கப்பட்டது"
-
-msgid "Edit this page"
-msgstr "இந்தப் பக்கத்தைத் திருத்தவும்"
-
-msgid "Launch"
-msgstr "தொடங்க"
-
-msgid "Print to PDF"
-msgstr "PDF இல் அச்சிடுக"
-
-msgid "open issue"
-msgstr "திறந்த பிரச்சினை"
-
-msgid "Download notebook file"
-msgstr "நோட்புக் கோப்பைப் பதிவிறக்கவும்"
-
-msgid "Toggle navigation"
-msgstr "வழிசெலுத்தலை நிலைமாற்று"
-
-msgid "Source repository"
-msgstr "மூல களஞ்சியம்"
-
-msgid "By the"
-msgstr "மூலம்"
-
-msgid "next page"
-msgstr "அடுத்த பக்கம்"
-
-msgid "Sphinx Book Theme"
-msgstr "ஸ்பிங்க்ஸ் புத்தக தீம்"
-
-msgid "Download source file"
-msgstr "மூல கோப்பைப் பதிவிறக்குக"
-
-msgid "By"
-msgstr "வழங்கியவர்"
-
-msgid "Copyright"
-msgstr "பதிப்புரிமை"
-
-msgid "Open an issue"
-msgstr "சிக்கலைத் திறக்கவும்"
-
-msgid "previous page"
-msgstr "முந்தைய பக்கம்"
-
-msgid "Download this page"
-msgstr "இந்தப் பக்கத்தைப் பதிவிறக்கவும்"
-
-msgid "Theme by the"
-msgstr "வழங்கிய தீம்"
diff --git a/docs/_build/html/_static/locales/te/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/te/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 0a5f4b46adfda1551ae6791c3c3782525b61be04..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/te/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/te/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 2595c03..0000000
--- a/docs/_build/html/_static/locales/te/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,66 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: te\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "సవరించమని సూచించండి"
-
-msgid "Last updated on"
-msgstr "చివరిగా నవీకరించబడింది"
-
-msgid "Edit this page"
-msgstr "ఈ పేజీని సవరించండి"
-
-msgid "Launch"
-msgstr "ప్రారంభించండి"
-
-msgid "Print to PDF"
-msgstr "PDF కి ముద్రించండి"
-
-msgid "open issue"
-msgstr "ఓపెన్ ఇష్యూ"
-
-msgid "Download notebook file"
-msgstr "నోట్బుక్ ఫైల్ను డౌన్లోడ్ చేయండి"
-
-msgid "Toggle navigation"
-msgstr "నావిగేషన్‌ను టోగుల్ చేయండి"
-
-msgid "Source repository"
-msgstr "మూల రిపోజిటరీ"
-
-msgid "By the"
-msgstr "ద్వారా"
-
-msgid "next page"
-msgstr "తరువాతి పేజీ"
-
-msgid "Sphinx Book Theme"
-msgstr "సింహిక పుస్తక థీమ్"
-
-msgid "Download source file"
-msgstr "మూల ఫైల్‌ను డౌన్‌లోడ్ చేయండి"
-
-msgid "By"
-msgstr "ద్వారా"
-
-msgid "Copyright"
-msgstr "కాపీరైట్"
-
-msgid "Open an issue"
-msgstr "సమస్యను తెరవండి"
-
-msgid "previous page"
-msgstr "ముందు పేజి"
-
-msgid "Download this page"
-msgstr "ఈ పేజీని డౌన్‌లోడ్ చేయండి"
-
-msgid "Theme by the"
-msgstr "ద్వారా థీమ్"
diff --git a/docs/_build/html/_static/locales/tg/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/tg/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index b21c6c6340194fdf35f7e5660deb42479c8dbff9..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1628
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diff --git a/docs/_build/html/_static/locales/tg/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/tg/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 73cd30e..0000000
--- a/docs/_build/html/_static/locales/tg/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: tg\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "пешниҳод вироиш"
-
-msgid "Last updated on"
-msgstr "Last навсозӣ дар"
-
-msgid "Edit this page"
-msgstr "Ин саҳифаро таҳрир кунед"
-
-msgid "Launch"
-msgstr "Оғоз"
-
-msgid "Print to PDF"
-msgstr "Чоп ба PDF"
-
-msgid "open issue"
-msgstr "барориши кушод"
-
-msgid "Download notebook file"
-msgstr "Файли дафтарро зеркашӣ кунед"
-
-msgid "Toggle navigation"
-msgstr "Гузаришро иваз кунед"
-
-msgid "Source repository"
-msgstr "Анбори манбаъ"
-
-msgid "By the"
-msgstr "Бо"
-
-msgid "next page"
-msgstr "саҳифаи оянда"
-
-msgid "repository"
-msgstr "анбор"
-
-msgid "Sphinx Book Theme"
-msgstr "Сфинкс Мавзӯи китоб"
-
-msgid "Download source file"
-msgstr "Файли манбаъро зеркашӣ кунед"
-
-msgid "Contents"
-msgstr "Мундариҷа"
-
-msgid "By"
-msgstr "Бо"
-
-msgid "Copyright"
-msgstr "Ҳуқуқи муаллиф"
-
-msgid "Fullscreen mode"
-msgstr "Ҳолати экрани пурра"
-
-msgid "Open an issue"
-msgstr "Масъаларо кушоед"
-
-msgid "previous page"
-msgstr "саҳифаи қаблӣ"
-
-msgid "Download this page"
-msgstr "Ин саҳифаро зеркашӣ кунед"
-
-msgid "Theme by the"
-msgstr "Мавзӯъи аз"
diff --git a/docs/_build/html/_static/locales/th/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/th/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index abede98aa11b163e580a26b545980ace31e61ccf..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

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diff --git a/docs/_build/html/_static/locales/th/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/th/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 0392b4a..0000000
--- a/docs/_build/html/_static/locales/th/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: th\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "แนะนำแก้ไข"
-
-msgid "Last updated on"
-msgstr "ปรับปรุงล่าสุดเมื่อ"
-
-msgid "Edit this page"
-msgstr "แก้ไขหน้านี้"
-
-msgid "Launch"
-msgstr "เปิด"
-
-msgid "Print to PDF"
-msgstr "พิมพ์เป็น PDF"
-
-msgid "open issue"
-msgstr "เปิดปัญหา"
-
-msgid "Download notebook file"
-msgstr "ดาวน์โหลดไฟล์สมุดบันทึก"
-
-msgid "Toggle navigation"
-msgstr "ไม่ต้องสลับช่องทาง"
-
-msgid "Source repository"
-msgstr "ที่เก็บซอร์ส"
-
-msgid "By the"
-msgstr "โดย"
-
-msgid "next page"
-msgstr "หน้าต่อไป"
-
-msgid "repository"
-msgstr "ที่เก็บ"
-
-msgid "Sphinx Book Theme"
-msgstr "ธีมหนังสือสฟิงซ์"
-
-msgid "Download source file"
-msgstr "ดาวน์โหลดไฟล์ต้นฉบับ"
-
-msgid "Contents"
-msgstr "สารบัญ"
-
-msgid "By"
-msgstr "โดย"
-
-msgid "Copyright"
-msgstr "ลิขสิทธิ์"
-
-msgid "Fullscreen mode"
-msgstr "โหมดเต็มหน้าจอ"
-
-msgid "Open an issue"
-msgstr "เปิดปัญหา"
-
-msgid "previous page"
-msgstr "หน้าที่แล้ว"
-
-msgid "Download this page"
-msgstr "ดาวน์โหลดหน้านี้"
-
-msgid "Theme by the"
-msgstr "ธีมโดย"
diff --git a/docs/_build/html/_static/locales/tl/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/tl/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 8df1b73310e0c606251d1abff7e6033d1f6b273f..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1273
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diff --git a/docs/_build/html/_static/locales/tl/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/tl/LC_MESSAGES/booktheme.po
deleted file mode 100644
index c8375b5..0000000
--- a/docs/_build/html/_static/locales/tl/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,66 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: tl\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "iminumungkahi i-edit"
-
-msgid "Last updated on"
-msgstr "Huling na-update noong"
-
-msgid "Edit this page"
-msgstr "I-edit ang pahinang ito"
-
-msgid "Launch"
-msgstr "Ilunsad"
-
-msgid "Print to PDF"
-msgstr "I-print sa PDF"
-
-msgid "open issue"
-msgstr "bukas na isyu"
-
-msgid "Download notebook file"
-msgstr "Mag-download ng file ng notebook"
-
-msgid "Toggle navigation"
-msgstr "I-toggle ang pag-navigate"
-
-msgid "Source repository"
-msgstr "Pinagmulan ng imbakan"
-
-msgid "By the"
-msgstr "Sa pamamagitan ng"
-
-msgid "next page"
-msgstr "Susunod na pahina"
-
-msgid "Sphinx Book Theme"
-msgstr "Tema ng Sphinx Book"
-
-msgid "Download source file"
-msgstr "Mag-download ng file ng pinagmulan"
-
-msgid "By"
-msgstr "Ni"
-
-msgid "Copyright"
-msgstr "Copyright"
-
-msgid "Open an issue"
-msgstr "Magbukas ng isyu"
-
-msgid "previous page"
-msgstr "Nakaraang pahina"
-
-msgid "Download this page"
-msgstr "I-download ang pahinang ito"
-
-msgid "Theme by the"
-msgstr "Tema ng"
diff --git a/docs/_build/html/_static/locales/tr/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/tr/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 029ae18afb5360a8c238a052e8d3971761759d3d..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1373
zcmZ{izi%8x6vqdWKynFz!~~?sQXUOIiX{o62)3@sVLOWA2n!$d<J)(;liizH&CG_Z
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z>s|syPtwI-fS96gfHHp*lr0wzl2>wx9;q$97n`h;HKI?N=#nP7WUs~N1$3mG*A#z?
zP4-{T@J)1V%lv%Rpus~iXZ5ACk!=k4s2XpwJgPPKoh_YC$-2lp&J{E^B_9n3SNjqF
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diff --git a/docs/_build/html/_static/locales/tr/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/tr/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 47d7bdf..0000000
--- a/docs/_build/html/_static/locales/tr/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: tr\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "düzenleme öner"
-
-msgid "Last updated on"
-msgstr "Son güncelleme tarihi"
-
-msgid "Edit this page"
-msgstr "Bu sayfayı düzenle"
-
-msgid "Launch"
-msgstr "Başlatmak"
-
-msgid "Print to PDF"
-msgstr "PDF olarak yazdır"
-
-msgid "open issue"
-msgstr "Açık konu"
-
-msgid "Download notebook file"
-msgstr "Defter dosyasını indirin"
-
-msgid "Toggle navigation"
-msgstr "Gezinmeyi değiştir"
-
-msgid "Source repository"
-msgstr "Kaynak kod deposu"
-
-msgid "By the"
-msgstr "Tarafından"
-
-msgid "next page"
-msgstr "sonraki Sayfa"
-
-msgid "repository"
-msgstr "depo"
-
-msgid "Sphinx Book Theme"
-msgstr "Sfenks Kitap Teması"
-
-msgid "Download source file"
-msgstr "Kaynak dosyayı indirin"
-
-msgid "Contents"
-msgstr "İçindekiler"
-
-msgid "By"
-msgstr "Tarafından"
-
-msgid "Copyright"
-msgstr "Telif hakkı"
-
-msgid "Fullscreen mode"
-msgstr "Tam ekran modu"
-
-msgid "Open an issue"
-msgstr "Bir sorunu açın"
-
-msgid "previous page"
-msgstr "önceki sayfa"
-
-msgid "Download this page"
-msgstr "Bu sayfayı indirin"
-
-msgid "Theme by the"
-msgstr "Tarafından tema"
diff --git a/docs/_build/html/_static/locales/uk/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/uk/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 16ab78909cfbaba7fa199f47fdfc2934426ac699..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1681
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zPJ@uD1fdWPh|e20*TKPc;zZ)a@r}d@A+B8DUJjM|f4dvvq>McK+j;xuef*#QJbLI0
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LPmM*`H%jps*u#;t

diff --git a/docs/_build/html/_static/locales/uk/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/uk/LC_MESSAGES/booktheme.po
deleted file mode 100644
index e85f6f1..0000000
--- a/docs/_build/html/_static/locales/uk/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: uk\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "запропонувати редагувати"
-
-msgid "Last updated on"
-msgstr "Останнє оновлення:"
-
-msgid "Edit this page"
-msgstr "Редагувати цю сторінку"
-
-msgid "Launch"
-msgstr "Запуск"
-
-msgid "Print to PDF"
-msgstr "Друк у форматі PDF"
-
-msgid "open issue"
-msgstr "відкритий випуск"
-
-msgid "Download notebook file"
-msgstr "Завантажте файл блокнота"
-
-msgid "Toggle navigation"
-msgstr "Переключити навігацію"
-
-msgid "Source repository"
-msgstr "Джерело сховища"
-
-msgid "By the"
-msgstr "По"
-
-msgid "next page"
-msgstr "Наступна сторінка"
-
-msgid "repository"
-msgstr "сховище"
-
-msgid "Sphinx Book Theme"
-msgstr "Тема книги \"Сфінкс\""
-
-msgid "Download source file"
-msgstr "Завантажити вихідний файл"
-
-msgid "Contents"
-msgstr "Зміст"
-
-msgid "By"
-msgstr "Автор"
-
-msgid "Copyright"
-msgstr "Авторське право"
-
-msgid "Fullscreen mode"
-msgstr "Повноекранний режим"
-
-msgid "Open an issue"
-msgstr "Відкрийте випуск"
-
-msgid "previous page"
-msgstr "Попередня сторінка"
-
-msgid "Download this page"
-msgstr "Завантажте цю сторінку"
-
-msgid "Theme by the"
-msgstr "Тема від"
diff --git a/docs/_build/html/_static/locales/ur/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/ur/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index de8c84b935ed5c3db23e304015d45b638a3d2db3..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1462
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z@uTBCHyGACc;3gei{}G8S6;vi>pu7*_yBwvd<eb@w!uE|FOYJdfXBgJ1iuNM0*`@b
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diff --git a/docs/_build/html/_static/locales/ur/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/ur/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 0f90726..0000000
--- a/docs/_build/html/_static/locales/ur/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,66 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: ur\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "ترمیم کی تجویز کریں"
-
-msgid "Last updated on"
-msgstr "آخری بار تازہ کاری ہوئی"
-
-msgid "Edit this page"
-msgstr "اس صفحے میں ترمیم کریں"
-
-msgid "Launch"
-msgstr "لانچ کریں"
-
-msgid "Print to PDF"
-msgstr "پی ڈی ایف پرنٹ کریں"
-
-msgid "open issue"
-msgstr "کھلا مسئلہ"
-
-msgid "Download notebook file"
-msgstr "نوٹ بک فائل ڈاؤن لوڈ کریں"
-
-msgid "Toggle navigation"
-msgstr "نیویگیشن ٹوگل کریں"
-
-msgid "Source repository"
-msgstr "ماخذ ذخیرہ"
-
-msgid "By the"
-msgstr "کی طرف"
-
-msgid "next page"
-msgstr "اگلا صفحہ"
-
-msgid "Sphinx Book Theme"
-msgstr "سپنکس بک تھیم"
-
-msgid "Download source file"
-msgstr "سورس فائل ڈاؤن لوڈ کریں"
-
-msgid "By"
-msgstr "بذریعہ"
-
-msgid "Copyright"
-msgstr "کاپی رائٹ"
-
-msgid "Open an issue"
-msgstr "ایک مسئلہ کھولیں"
-
-msgid "previous page"
-msgstr "سابقہ ​​صفحہ"
-
-msgid "Download this page"
-msgstr "اس صفحے کو ڈاؤن لوڈ کریں"
-
-msgid "Theme by the"
-msgstr "کے ذریعہ تھیم"
diff --git a/docs/_build/html/_static/locales/vi/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/vi/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 2bb32555c3ea0dab08ed37b71cb863fbe75698c6..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1431
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diff --git a/docs/_build/html/_static/locales/vi/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/vi/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 2cb5cf3..0000000
--- a/docs/_build/html/_static/locales/vi/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
-
-msgid ""
-msgstr ""
-"Project-Id-Version: Sphinx-Book-Theme\n"
-"MIME-Version: 1.0\n"
-"Content-Type: text/plain; charset=UTF-8\n"
-"Content-Transfer-Encoding: 8bit\n"
-"Language: vi\n"
-"Plural-Forms: nplurals=2; plural=(n != 1);\n"
-
-msgid "suggest edit"
-msgstr "đề nghị chỉnh sửa"
-
-msgid "Last updated on"
-msgstr "Cập nhật lần cuối vào"
-
-msgid "Edit this page"
-msgstr "chỉnh sửa trang này"
-
-msgid "Launch"
-msgstr "Phóng"
-
-msgid "Print to PDF"
-msgstr "In sang PDF"
-
-msgid "open issue"
-msgstr "vấn đề mở"
-
-msgid "Download notebook file"
-msgstr "Tải xuống tệp sổ tay"
-
-msgid "Toggle navigation"
-msgstr "Chuyển đổi điều hướng thành"
-
-msgid "Source repository"
-msgstr "Kho nguồn"
-
-msgid "By the"
-msgstr "Bằng"
-
-msgid "next page"
-msgstr "Trang tiếp theo"
-
-msgid "repository"
-msgstr "kho"
-
-msgid "Sphinx Book Theme"
-msgstr "Chủ đề sách nhân sư"
-
-msgid "Download source file"
-msgstr "Tải xuống tệp nguồn"
-
-msgid "Contents"
-msgstr "Nội dung"
-
-msgid "By"
-msgstr "Bởi"
-
-msgid "Copyright"
-msgstr "Bản quyền"
-
-msgid "Fullscreen mode"
-msgstr "Chế độ toàn màn hình"
-
-msgid "Open an issue"
-msgstr "Mở một vấn đề"
-
-msgid "previous page"
-msgstr "trang trước"
-
-msgid "Download this page"
-msgstr "Tải xuống trang này"
-
-msgid "Theme by the"
-msgstr "Chủ đề của"
diff --git a/docs/_build/html/_static/locales/zh_CN/LC_MESSAGES/booktheme.mo b/docs/_build/html/_static/locales/zh_CN/LC_MESSAGES/booktheme.mo
deleted file mode 100644
index 0e3235d090c2a05c0e278d38520f0437847dc79f..0000000000000000000000000000000000000000
GIT binary patch
literal 0
HcmV?d00001

literal 1310
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diff --git a/docs/_build/html/_static/locales/zh_CN/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/zh_CN/LC_MESSAGES/booktheme.po
deleted file mode 100644
index f91f3ba..0000000
--- a/docs/_build/html/_static/locales/zh_CN/LC_MESSAGES/booktheme.po
+++ /dev/null
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-msgid "repository"
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-msgid "Sphinx Book Theme"
-msgstr "Sphinx Book 主题"
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-msgid "Download source file"
-msgstr "下载源文件"
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-msgid "By"
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-msgid "Copyright"
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-msgid "previous page"
-msgstr "上一页"
-
-msgid "Download this page"
-msgstr "下载此页面"
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-msgid "Theme by the"
-msgstr "主题作者："
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diff --git a/docs/_build/html/_static/locales/zh_TW/LC_MESSAGES/booktheme.po b/docs/_build/html/_static/locales/zh_TW/LC_MESSAGES/booktheme.po
deleted file mode 100644
index 7833d90..0000000
--- a/docs/_build/html/_static/locales/zh_TW/LC_MESSAGES/booktheme.po
+++ /dev/null
@@ -1,75 +0,0 @@
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-msgid ""
-msgstr ""
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-"MIME-Version: 1.0\n"
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-"Content-Transfer-Encoding: 8bit\n"
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-msgid "suggest edit"
-msgstr "提出修改建議"
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-msgid "Last updated on"
-msgstr "最後更新時間："
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-msgid "Edit this page"
-msgstr "編輯此頁面"
-
-msgid "Launch"
-msgstr "啟動"
-
-msgid "Print to PDF"
-msgstr "列印成 PDF"
-
-msgid "open issue"
-msgstr "公開的問題"
-
-msgid "Download notebook file"
-msgstr "下載 Notebook 檔案"
-
-msgid "Toggle navigation"
-msgstr "顯示或隱藏導覽列"
-
-msgid "Source repository"
-msgstr "來源儲存庫"
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-msgid "By the"
-msgstr "作者："
-
-msgid "next page"
-msgstr "下一頁"
-
-msgid "repository"
-msgstr "儲存庫"
-
-msgid "Sphinx Book Theme"
-msgstr "Sphinx Book 佈景主題"
-
-msgid "Download source file"
-msgstr "下載原始檔"
-
-msgid "Contents"
-msgstr "目錄"
-
-msgid "By"
-msgstr "作者："
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-msgid "Copyright"
-msgstr "Copyright"
-
-msgid "Fullscreen mode"
-msgstr "全螢幕模式"
-
-msgid "Open an issue"
-msgstr "開啟議題"
-
-msgid "previous page"
-msgstr "上一頁"
-
-msgid "Download this page"
-msgstr "下載此頁面"
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-msgstr "佈景主題作者："
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-
-div.highlight .-Color[class*=-C26] {
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-
-div.highlight .-Color[class*=-BGC26] {
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-
-div.highlight .-Color[class*=-C27] {
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-
-div.highlight .-Color[class*=-BGC27] {
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-
-div.highlight .-Color[class*=-C28] {
-  color: #008700
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-
-div.highlight .-Color[class*=-BGC28] {
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-
-div.highlight .-Color[class*=-C29] {
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-
-div.highlight .-Color[class*=-BGC29] {
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-
-div.highlight .-Color[class*=-C30] {
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-
-div.highlight .-Color[class*=-BGC30] {
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-
-div.highlight .-Color[class*=-C31] {
-  color: #0087AF
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-
-div.highlight .-Color[class*=-BGC31] {
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-
-div.highlight .-Color[class*=-C32] {
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-
-div.highlight .-Color[class*=-BGC32] {
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-
-div.highlight .-Color[class*=-C33] {
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-
-div.highlight .-Color[class*=-BGC33] {
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-
-div.highlight .-Color[class*=-C34] {
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-
-div.highlight .-Color[class*=-BGC34] {
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-
-div.highlight .-Color[class*=-C35] {
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-
-div.highlight .-Color[class*=-BGC35] {
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-
-div.highlight .-Color[class*=-C36] {
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-
-div.highlight .-Color[class*=-BGC36] {
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-
-div.highlight .-Color[class*=-C37] {
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-
-div.highlight .-Color[class*=-BGC37] {
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-
-div.highlight .-Color[class*=-C38] {
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-
-div.highlight .-Color[class*=-BGC38] {
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-
-div.highlight .-Color[class*=-C39] {
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-
-div.highlight .-Color[class*=-BGC39] {
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-
-div.highlight .-Color[class*=-C40] {
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-
-div.highlight .-Color[class*=-BGC40] {
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-
-div.highlight .-Color[class*=-C41] {
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-div.highlight .-Color[class*=-BGC41] {
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-
-div.highlight .-Color[class*=-C42] {
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-div.highlight .-Color[class*=-BGC42] {
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-
-div.highlight .-Color[class*=-C43] {
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-div.highlight .-Color[class*=-BGC43] {
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-div.highlight .-Color[class*=-C44] {
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-div.highlight .-Color[class*=-BGC44] {
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-div.highlight .-Color[class*=-C45] {
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-div.highlight .-Color[class*=-BGC45] {
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-div.highlight .-Color[class*=-C46] {
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-div.highlight .-Color[class*=-BGC46] {
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-div.highlight .-Color[class*=-C47] {
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-div.highlight .-Color[class*=-BGC47] {
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-div.highlight .-Color[class*=-C48] {
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-div.highlight .-Color[class*=-BGC48] {
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-div.highlight .-Color[class*=-C49] {
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-div.highlight .-Color[class*=-BGC49] {
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-div.highlight .-Color[class*=-C50] {
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-div.highlight .-Color[class*=-BGC50] {
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-div.highlight .-Color[class*=-C51] {
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-div.highlight .-Color[class*=-BGC51] {
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-div.highlight .-Color[class*=-C52] {
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-div.highlight .-Color[class*=-BGC52] {
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-
-div.highlight .-Color[class*=-C53] {
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-div.highlight .-Color[class*=-BGC53] {
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-div.highlight .-Color[class*=-C54] {
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-div.highlight .-Color[class*=-BGC54] {
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-div.highlight .-Color[class*=-C55] {
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-div.highlight .-Color[class*=-BGC55] {
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-div.highlight .-Color[class*=-C56] {
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-div.highlight .-Color[class*=-BGC56] {
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-div.highlight .-Color[class*=-C57] {
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-div.highlight .-Color[class*=-BGC57] {
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-div.highlight .-Color[class*=-C58] {
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-div.highlight .-Color[class*=-BGC58] {
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-div.highlight .-Color[class*=-C59] {
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-div.highlight .-Color[class*=-BGC59] {
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-div.highlight .-Color[class*=-C60] {
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-div.highlight .-Color[class*=-BGC60] {
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-div.highlight .-Color[class*=-C61] {
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-div.highlight .-Color[class*=-BGC61] {
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-div.highlight .-Color[class*=-C62] {
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-div.highlight .-Color[class*=-BGC62] {
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-div.highlight .-Color[class*=-C63] {
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-div.highlight .-Color[class*=-BGC63] {
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-div.highlight .-Color[class*=-C64] {
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-div.highlight .-Color[class*=-BGC64] {
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-div.highlight .-Color[class*=-C65] {
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-div.highlight .-Color[class*=-BGC65] {
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-div.highlight .-Color[class*=-C66] {
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-div.highlight .-Color[class*=-BGC66] {
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-div.highlight .-Color[class*=-BGC68] {
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-div.highlight .-Color[class*=-C141] {
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-div.highlight .-Color[class*=-BGC141] {
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-div.highlight .-Color[class*=-C142] {
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-div.highlight .-Color[class*=-BGC142] {
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-div.highlight .-Color[class*=-C143] {
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-div.highlight .-Color[class*=-BGC143] {
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-
-div.highlight .-Color[class*=-C144] {
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-div.highlight .-Color[class*=-BGC144] {
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-div.highlight .-Color[class*=-C145] {
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-div.highlight .-Color[class*=-BGC145] {
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-
-div.highlight .-Color[class*=-C146] {
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-div.highlight .-Color[class*=-BGC146] {
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-div.highlight .-Color[class*=-C147] {
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-div.highlight .-Color[class*=-BGC147] {
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-div.highlight .-Color[class*=-C148] {
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-div.highlight .-Color[class*=-BGC148] {
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-div.highlight .-Color[class*=-C149] {
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-div.highlight .-Color[class*=-BGC149] {
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-div.highlight .-Color[class*=-C150] {
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-div.highlight .-Color[class*=-BGC150] {
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-div.highlight .-Color[class*=-C151] {
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-div.highlight .-Color[class*=-BGC151] {
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-
-div.highlight .-Color[class*=-C152] {
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-div.highlight .-Color[class*=-BGC152] {
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-div.highlight .-Color[class*=-C153] {
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-div.highlight .-Color[class*=-BGC153] {
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-div.highlight .-Color[class*=-C154] {
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-div.highlight .-Color[class*=-BGC154] {
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-div.highlight .-Color[class*=-C155] {
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-div.highlight .-Color[class*=-BGC155] {
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-div.highlight .-Color[class*=-C156] {
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-div.highlight .-Color[class*=-BGC156] {
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-div.highlight .-Color[class*=-C157] {
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-div.highlight .-Color[class*=-BGC157] {
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-div.highlight .-Color[class*=-C158] {
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-div.highlight .-Color[class*=-BGC158] {
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-div.highlight .-Color[class*=-C159] {
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-div.highlight .-Color[class*=-BGC159] {
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-div.highlight .-Color[class*=-C160] {
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-div.highlight .-Color[class*=-BGC160] {
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-div.highlight .-Color[class*=-C161] {
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-div.highlight .-Color[class*=-BGC161] {
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-div.highlight .-Color[class*=-C162] {
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-div.highlight .-Color[class*=-BGC162] {
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-div.highlight .-Color[class*=-C163] {
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-
-div.highlight .-Color[class*=-BGC163] {
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-div.highlight .-Color[class*=-C164] {
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-div.highlight .-Color[class*=-BGC164] {
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-div.highlight .-Color[class*=-C165] {
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-div.highlight .-Color[class*=-BGC165] {
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-div.highlight .-Color[class*=-C166] {
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-div.highlight .-Color[class*=-BGC166] {
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-div.highlight .-Color[class*=-C167] {
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-div.highlight .-Color[class*=-BGC167] {
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-div.highlight .-Color[class*=-C168] {
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-div.highlight .-Color[class*=-BGC168] {
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-div.highlight .-Color[class*=-C169] {
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-div.highlight .-Color[class*=-BGC169] {
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-div.highlight .-Color[class*=-BGC170] {
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-div.highlight .-Color[class*=-C171] {
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-div.highlight .-Color[class*=-BGC171] {
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-div.highlight .-Color[class*=-C172] {
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-div.highlight .-Color[class*=-BGC172] {
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-div.highlight .-Color[class*=-C173] {
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-div.highlight .-Color[class*=-BGC173] {
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-div.highlight .-Color[class*=-C174] {
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-div.highlight .-Color[class*=-BGC174] {
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-div.highlight .-Color[class*=-C175] {
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-div.highlight .-Color[class*=-C176] {
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index 6cbf559..0000000
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l=n?"carousel-item-start":"carousel-item-end",c=n?"carousel-item-next":"carousel-item-prev";s.classList.add(c),Rt(s),i.classList.add(l),s.classList.add(l),this._queueCallback((()=>{s.classList.remove(l,c),s.classList.add(Je),i.classList.remove(Je,c,l),this._isSliding=!1,r(qe)}),i,this._isAnimated()),a&&this.cycle()}_isAnimated(){return this._element.classList.contains("slide")}_getActive(){return ye.findOne(ei,this._element)}_getItems(){return ye.find(ti,this._element)}_clearInterval(){this._interval&&(clearInterval(this._interval),this._interval=null)}_directionToOrder(t){return Yt()?t===We?Be:He:t===We?He:Be}_orderToDirection(t){return Yt()?t===Be?We:ze:t===Be?ze:We}static jQueryInterface(t){return this.each((function(){const e=oi.getOrCreateInstance(this,t);if("number"!=typeof t){if("string"==typeof t){if(void 0===e[t]||t.startsWith("_")||"constructor"===t)throw new TypeError(`No method named "${t}"`);e[t]()}}else e.to(t)}))}}ue.on(document,Ue,"[data-bs-slide], [data-bs-slide-to]",(function(t){const e=ye.getElementFromSelector(this);if(!e||!e.classList.contains(Ge))return;t.preventDefault();const i=oi.getOrCreateInstance(e),n=this.getAttribute("data-bs-slide-to");return n?(i.to(n),void i._maybeEnableCycle()):"next"===ge.getDataAttribute(this,"slide")?(i.next(),void i._maybeEnableCycle()):(i.prev(),void i._maybeEnableCycle())})),ue.on(window,Xe,(()=>{const t=ye.find('[data-bs-ride="carousel"]');for(const e of t)oi.getOrCreateInstance(e)})),Kt(oi);const ri=".bs.collapse",ai=`show${ri}`,li=`shown${ri}`,ci=`hide${ri}`,hi=`hidden${ri}`,di=`click${ri}.data-api`,ui="show",fi="collapse",pi="collapsing",mi=`:scope .${fi} .${fi}`,gi='[data-bs-toggle="collapse"]',_i={parent:null,toggle:!0},bi={parent:"(null|element)",toggle:"boolean"};class vi extends be{constructor(t,e){super(t,e),this._isTransitioning=!1,this._triggerArray=[];const i=ye.find(gi);for(const t of i){const e=ye.getSelectorFromElement(t),i=ye.find(e).filter((t=>t===this._element));null!==e&&i.length&&this._triggerArray.push(t)}this._initializeChildren(),this._config.parent||this._addAriaAndCollapsedClass(this._triggerArray,this._isShown()),this._config.toggle&&this.toggle()}static get Default(){return _i}static get DefaultType(){return bi}static get NAME(){return"collapse"}toggle(){this._isShown()?this.hide():this.show()}show(){if(this._isTransitioning||this._isShown())return;let t=[];if(this._config.parent&&(t=this._getFirstLevelChildren(".collapse.show, .collapse.collapsing").filter((t=>t!==this._element)).map((t=>vi.getOrCreateInstance(t,{toggle:!1})))),t.length&&t[0]._isTransitioning)return;if(ue.trigger(this._element,ai).defaultPrevented)return;for(const e of t)e.hide();const e=this._getDimension();this._element.classList.remove(fi),this._element.classList.add(pi),this._element.style[e]=0,this._addAriaAndCollapsedClass(this._triggerArray,!0),this._isTransitioning=!0;const i=`scroll${e[0].toUpperCase()+e.slice(1)}`;this._queueCallback((()=>{this._isTransitioning=!1,this._element.classList.remove(pi),this._element.classList.add(fi,ui),this._element.style[e]="",ue.trigger(this._element,li)}),this._element,!0),this._element.style[e]=`${this._element[i]}px`}hide(){if(this._isTransitioning||!this._isShown())return;if(ue.trigger(this._element,ci).defaultPrevented)return;const t=this._getDimension();this._element.style[t]=`${this._element.getBoundingClientRect()[t]}px`,Rt(this._element),this._element.classList.add(pi),this._element.classList.remove(fi,ui);for(const t of this._triggerArray){const e=ye.getElementFromSelector(t);e&&!this._isShown(e)&&this._addAriaAndCollapsedClass([t],!1)}this._isTransitioning=!0,this._element.style[t]="",this._queueCallback((()=>{this._isTransitioning=!1,this._element.classList.remove(pi),this._element.classList.add(fi),ue.trigger(this._element,hi)}),this._element,!0)}_isShown(t=this._element){return t.classList.contains(ui)}_configAfterMerge(t){return t.toggle=Boolean(t.toggle),t.parent=Ft(t.parent),t}_getDimension(){return this._element.classList.contains("collapse-horizontal")?"width":"height"}_initializeChildren(){if(!this._config.parent)return;const t=this._getFirstLevelChildren(gi);for(const e of t){const t=ye.getElementFromSelector(e);t&&this._addAriaAndCollapsedClass([e],this._isShown(t))}}_getFirstLevelChildren(t){const e=ye.find(mi,this._config.parent);return ye.find(t,this._config.parent).filter((t=>!e.includes(t)))}_addAriaAndCollapsedClass(t,e){if(t.length)for(const i of t)i.classList.toggle("collapsed",!e),i.setAttribute("aria-expanded",e)}static jQueryInterface(t){const e={};return"string"==typeof t&&/show|hide/.test(t)&&(e.toggle=!1),this.each((function(){const i=vi.getOrCreateInstance(this,e);if("string"==typeof t){if(void 0===i[t])throw new TypeError(`No method named "${t}"`);i[t]()}}))}}ue.on(document,di,gi,(function(t){("A"===t.target.tagName||t.delegateTarget&&"A"===t.delegateTarget.tagName)&&t.preventDefault();for(const t of ye.getMultipleElementsFromSelector(this))vi.getOrCreateInstance(t,{toggle:!1}).toggle()})),Kt(vi);const yi="dropdown",wi=".bs.dropdown",Ei=".data-api",Ai="ArrowUp",Ti="ArrowDown",Ci=`hide${wi}`,Oi=`hidden${wi}`,xi=`show${wi}`,ki=`shown${wi}`,Li=`click${wi}${Ei}`,Si=`keydown${wi}${Ei}`,Di=`keyup${wi}${Ei}`,$i="show",Ii='[data-bs-toggle="dropdown"]:not(.disabled):not(:disabled)',Ni=`${Ii}.${$i}`,Pi=".dropdown-menu",Mi=Yt()?"top-end":"top-start",ji=Yt()?"top-start":"top-end",Fi=Yt()?"bottom-end":"bottom-start",Hi=Yt()?"bottom-start":"bottom-end",Bi=Yt()?"left-start":"right-start",Wi=Yt()?"right-start":"left-start",zi={autoClose:!0,boundary:"clippingParents",display:"dynamic",offset:[0,2],popperConfig:null,reference:"toggle"},Ri={autoClose:"(boolean|string)",boundary:"(string|element)",display:"string",offset:"(array|string|function)",popperConfig:"(null|object|function)",reference:"(string|element|object)"};class qi extends be{constructor(t,e){super(t,e),this._popper=null,this._parent=this._element.parentNode,this._menu=ye.next(this._element,Pi)[0]||ye.prev(this._element,Pi)[0]||ye.findOne(Pi,this._parent),this._inNavbar=this._detectNavbar()}static get Default(){return zi}static get DefaultType(){return Ri}static get NAME(){return yi}toggle(){return this._isShown()?this.hide():this.show()}show(){if(Bt(this._element)||this._isShown())return;const t={relatedTarget:this._element};if(!ue.trigger(this._element,xi,t).defaultPrevented){if(this._createPopper(),"ontouchstart"in document.documentElement&&!this._parent.closest(".navbar-nav"))for(const t of[].concat(...document.body.children))ue.on(t,"mouseover",zt);this._element.focus(),this._element.setAttribute("aria-expanded",!0),this._menu.classList.add($i),this._element.classList.add($i),ue.trigger(this._element,ki,t)}}hide(){if(Bt(this._element)||!this._isShown())return;const t={relatedTarget:this._element};this._completeHide(t)}dispose(){this._popper&&this._popper.destroy(),super.dispose()}update(){this._inNavbar=this._detectNavbar(),this._popper&&this._popper.update()}_completeHide(t){if(!ue.trigger(this._element,Ci,t).defaultPrevented){if("ontouchstart"in document.documentElement)for(const t of[].concat(...document.body.children))ue.off(t,"mouseover",zt);this._popper&&this._popper.destroy(),this._menu.classList.remove($i),this._element.classList.remove($i),this._element.setAttribute("aria-expanded","false"),ge.removeDataAttribute(this._menu,"popper"),ue.trigger(this._element,Oi,t)}}_getConfig(t){if("object"==typeof(t=super._getConfig(t)).reference&&!jt(t.reference)&&"function"!=typeof t.reference.getBoundingClientRect)throw new TypeError(`${yi.toUpperCase()}: Option "reference" provided type "object" without a required "getBoundingClientRect" method.`);return t}_createPopper(){if(void 0===e)throw new TypeError("Bootstrap's dropdowns require Popper (https://popper.js.org)");let t=this._element;"parent"===this._config.reference?t=this._parent:jt(this._config.reference)?t=Ft(this._config.reference):"object"==typeof this._config.reference&&(t=this._config.reference);const i=this._getPopperConfig();this._popper=St(t,this._menu,i)}_isShown(){return this._menu.classList.contains($i)}_getPlacement(){const t=this._parent;if(t.classList.contains("dropend"))return Bi;if(t.classList.contains("dropstart"))return Wi;if(t.classList.contains("dropup-center"))return"top";if(t.classList.contains("dropdown-center"))return"bottom";const e="end"===getComputedStyle(this._menu).getPropertyValue("--bs-position").trim();return t.classList.contains("dropup")?e?ji:Mi:e?Hi:Fi}_detectNavbar(){return null!==this._element.closest(".navbar")}_getOffset(){const{offset:t}=this._config;return"string"==typeof t?t.split(",").map((t=>Number.parseInt(t,10))):"function"==typeof t?e=>t(e,this._element):t}_getPopperConfig(){const t={placement:this._getPlacement(),modifiers:[{name:"preventOverflow",options:{boundary:this._config.boundary}},{name:"offset",options:{offset:this._getOffset()}}]};return(this._inNavbar||"static"===this._config.display)&&(ge.setDataAttribute(this._menu,"popper","static"),t.modifiers=[{name:"applyStyles",enabled:!1}]),{...t,...Qt(this._config.popperConfig,[t])}}_selectMenuItem({key:t,target:e}){const i=ye.find(".dropdown-menu .dropdown-item:not(.disabled):not(:disabled)",this._menu).filter((t=>Ht(t)));i.length&&Ut(i,e,t===Ti,!i.includes(e)).focus()}static jQueryInterface(t){return this.each((function(){const e=qi.getOrCreateInstance(this,t);if("string"==typeof t){if(void 0===e[t])throw new TypeError(`No method named "${t}"`);e[t]()}}))}static clearMenus(t){if(2===t.button||"keyup"===t.type&&"Tab"!==t.key)return;const e=ye.find(Ni);for(const i of e){const e=qi.getInstance(i);if(!e||!1===e._config.autoClose)continue;const n=t.composedPath(),s=n.includes(e._menu);if(n.includes(e._element)||"inside"===e._config.autoClose&&!s||"outside"===e._config.autoClose&&s)continue;if(e._menu.contains(t.target)&&("keyup"===t.type&&"Tab"===t.key||/input|select|option|textarea|form/i.test(t.target.tagName)))continue;const o={relatedTarget:e._element};"click"===t.type&&(o.clickEvent=t),e._completeHide(o)}}static dataApiKeydownHandler(t){const e=/input|textarea/i.test(t.target.tagName),i="Escape"===t.key,n=[Ai,Ti].includes(t.key);if(!n&&!i)return;if(e&&!i)return;t.preventDefault();const s=this.matches(Ii)?this:ye.prev(this,Ii)[0]||ye.next(this,Ii)[0]||ye.findOne(Ii,t.delegateTarget.parentNode),o=qi.getOrCreateInstance(s);if(n)return t.stopPropagation(),o.show(),void o._selectMenuItem(t);o._isShown()&&(t.stopPropagation(),o.hide(),s.focus())}}ue.on(document,Si,Ii,qi.dataApiKeydownHandler),ue.on(document,Si,Pi,qi.dataApiKeydownHandler),ue.on(document,Li,qi.clearMenus),ue.on(document,Di,qi.clearMenus),ue.on(document,Li,Ii,(function(t){t.preventDefault(),qi.getOrCreateInstance(this).toggle()})),Kt(qi);const Vi="backdrop",Yi="show",Ki=`mousedown.bs.${Vi}`,Qi={className:"modal-backdrop",clickCallback:null,isAnimated:!1,isVisible:!0,rootElement:"body"},Xi={className:"string",clickCallback:"(function|null)",isAnimated:"boolean",isVisible:"boolean",rootElement:"(element|string)"};class Ui extends _e{constructor(t){super(),this._config=this._getConfig(t),this._isAppended=!1,this._element=null}static get Default(){return Qi}static get DefaultType(){return Xi}static get NAME(){return Vi}show(t){if(!this._config.isVisible)return void Qt(t);this._append();const e=this._getElement();this._config.isAnimated&&Rt(e),e.classList.add(Yi),this._emulateAnimation((()=>{Qt(t)}))}hide(t){this._config.isVisible?(this._getElement().classList.remove(Yi),this._emulateAnimation((()=>{this.dispose(),Qt(t)}))):Qt(t)}dispose(){this._isAppended&&(ue.off(this._element,Ki),this._element.remove(),this._isAppended=!1)}_getElement(){if(!this._element){const t=document.createElement("div");t.className=this._config.className,this._config.isAnimated&&t.classList.add("fade"),this._element=t}return this._element}_configAfterMerge(t){return t.rootElement=Ft(t.rootElement),t}_append(){if(this._isAppended)return;const t=this._getElement();this._config.rootElement.append(t),ue.on(t,Ki,(()=>{Qt(this._config.clickCallback)})),this._isAppended=!0}_emulateAnimation(t){Xt(t,this._getElement(),this._config.isAnimated)}}const Gi=".bs.focustrap",Ji=`focusin${Gi}`,Zi=`keydown.tab${Gi}`,tn="backward",en={autofocus:!0,trapElement:null},nn={autofocus:"boolean",trapElement:"element"};class sn extends _e{constructor(t){super(),this._config=this._getConfig(t),this._isActive=!1,this._lastTabNavDirection=null}static get Default(){return en}static get DefaultType(){return nn}static get NAME(){return"focustrap"}activate(){this._isActive||(this._config.autofocus&&this._config.trapElement.focus(),ue.off(document,Gi),ue.on(document,Ji,(t=>this._handleFocusin(t))),ue.on(document,Zi,(t=>this._handleKeydown(t))),this._isActive=!0)}deactivate(){this._isActive&&(this._isActive=!1,ue.off(document,Gi))}_handleFocusin(t){const{trapElement:e}=this._config;if(t.target===document||t.target===e||e.contains(t.target))return;const i=ye.focusableChildren(e);0===i.length?e.focus():this._lastTabNavDirection===tn?i[i.length-1].focus():i[0].focus()}_handleKeydown(t){"Tab"===t.key&&(this._lastTabNavDirection=t.shiftKey?tn:"forward")}}const on=".fixed-top, .fixed-bottom, .is-fixed, .sticky-top",rn=".sticky-top",an="padding-right",ln="margin-right";class cn{constructor(){this._element=document.body}getWidth(){const t=document.documentElement.clientWidth;return Math.abs(window.innerWidth-t)}hide(){const t=this.getWidth();this._disableOverFlow(),this._setElementAttributes(this._element,an,(e=>e+t)),this._setElementAttributes(on,an,(e=>e+t)),this._setElementAttributes(rn,ln,(e=>e-t))}reset(){this._resetElementAttributes(this._element,"overflow"),this._resetElementAttributes(this._element,an),this._resetElementAttributes(on,an),this._resetElementAttributes(rn,ln)}isOverflowing(){return this.getWidth()>0}_disableOverFlow(){this._saveInitialAttribute(this._element,"overflow"),this._element.style.overflow="hidden"}_setElementAttributes(t,e,i){const n=this.getWidth();this._applyManipulationCallback(t,(t=>{if(t!==this._element&&window.innerWidth>t.clientWidth+n)return;this._saveInitialAttribute(t,e);const s=window.getComputedStyle(t).getPropertyValue(e);t.style.setProperty(e,`${i(Number.parseFloat(s))}px`)}))}_saveInitialAttribute(t,e){const i=t.style.getPropertyValue(e);i&&ge.setDataAttribute(t,e,i)}_resetElementAttributes(t,e){this._applyManipulationCallback(t,(t=>{const i=ge.getDataAttribute(t,e);null!==i?(ge.removeDataAttribute(t,e),t.style.setProperty(e,i)):t.style.removeProperty(e)}))}_applyManipulationCallback(t,e){if(jt(t))e(t);else for(const i of ye.find(t,this._element))e(i)}}const hn=".bs.modal",dn=`hide${hn}`,un=`hidePrevented${hn}`,fn=`hidden${hn}`,pn=`show${hn}`,mn=`shown${hn}`,gn=`resize${hn}`,_n=`click.dismiss${hn}`,bn=`mousedown.dismiss${hn}`,vn=`keydown.dismiss${hn}`,yn=`click${hn}.data-api`,wn="modal-open",En="show",An="modal-static",Tn={backdrop:!0,focus:!0,keyboard:!0},Cn={backdrop:"(boolean|string)",focus:"boolean",keyboard:"boolean"};class On extends be{constructor(t,e){super(t,e),this._dialog=ye.findOne(".modal-dialog",this._element),this._backdrop=this._initializeBackDrop(),this._focustrap=this._initializeFocusTrap(),this._isShown=!1,this._isTransitioning=!1,this._scrollBar=new cn,this._addEventListeners()}static get Default(){return Tn}static get DefaultType(){return Cn}static get NAME(){return"modal"}toggle(t){return this._isShown?this.hide():this.show(t)}show(t){this._isShown||this._isTransitioning||ue.trigger(this._element,pn,{relatedTarget:t}).defaultPrevented||(this._isShown=!0,this._isTransitioning=!0,this._scrollBar.hide(),document.body.classList.add(wn),this._adjustDialog(),this._backdrop.show((()=>this._showElement(t))))}hide(){this._isShown&&!this._isTransitioning&&(ue.trigger(this._element,dn).defaultPrevented||(this._isShown=!1,this._isTransitioning=!0,this._focustrap.deactivate(),this._element.classList.remove(En),this._queueCallback((()=>this._hideModal()),this._element,this._isAnimated())))}dispose(){ue.off(window,hn),ue.off(this._dialog,hn),this._backdrop.dispose(),this._focustrap.deactivate(),super.dispose()}handleUpdate(){this._adjustDialog()}_initializeBackDrop(){return new Ui({isVisible:Boolean(this._config.backdrop),isAnimated:this._isAnimated()})}_initializeFocusTrap(){return new sn({trapElement:this._element})}_showElement(t){document.body.contains(this._element)||document.body.append(this._element),this._element.style.display="block",this._element.removeAttribute("aria-hidden"),this._element.setAttribute("aria-modal",!0),this._element.setAttribute("role","dialog"),this._element.scrollTop=0;const e=ye.findOne(".modal-body",this._dialog);e&&(e.scrollTop=0),Rt(this._element),this._element.classList.add(En),this._queueCallback((()=>{this._config.focus&&this._focustrap.activate(),this._isTransitioning=!1,ue.trigger(this._element,mn,{relatedTarget:t})}),this._dialog,this._isAnimated())}_addEventListeners(){ue.on(this._element,vn,(t=>{"Escape"===t.key&&(this._config.keyboard?this.hide():this._triggerBackdropTransition())})),ue.on(window,gn,(()=>{this._isShown&&!this._isTransitioning&&this._adjustDialog()})),ue.on(this._element,bn,(t=>{ue.one(this._element,_n,(e=>{this._element===t.target&&this._element===e.target&&("static"!==this._config.backdrop?this._config.backdrop&&this.hide():this._triggerBackdropTransition())}))}))}_hideModal(){this._element.style.display="none",this._element.setAttribute("aria-hidden",!0),this._element.removeAttribute("aria-modal"),this._element.removeAttribute("role"),this._isTransitioning=!1,this._backdrop.hide((()=>{document.body.classList.remove(wn),this._resetAdjustments(),this._scrollBar.reset(),ue.trigger(this._element,fn)}))}_isAnimated(){return this._element.classList.contains("fade")}_triggerBackdropTransition(){if(ue.trigger(this._element,un).defaultPrevented)return;const t=this._element.scrollHeight>document.documentElement.clientHeight,e=this._element.style.overflowY;"hidden"===e||this._element.classList.contains(An)||(t||(this._element.style.overflowY="hidden"),this._element.classList.add(An),this._queueCallback((()=>{this._element.classList.remove(An),this._queueCallback((()=>{this._element.style.overflowY=e}),this._dialog)}),this._dialog),this._element.focus())}_adjustDialog(){const t=this._element.scrollHeight>document.documentElement.clientHeight,e=this._scrollBar.getWidth(),i=e>0;if(i&&!t){const t=Yt()?"paddingLeft":"paddingRight";this._element.style[t]=`${e}px`}if(!i&&t){const t=Yt()?"paddingRight":"paddingLeft";this._element.style[t]=`${e}px`}}_resetAdjustments(){this._element.style.paddingLeft="",this._element.style.paddingRight=""}static jQueryInterface(t,e){return this.each((function(){const i=On.getOrCreateInstance(this,t);if("string"==typeof t){if(void 0===i[t])throw new TypeError(`No method named "${t}"`);i[t](e)}}))}}ue.on(document,yn,'[data-bs-toggle="modal"]',(function(t){const e=ye.getElementFromSelector(this);["A","AREA"].includes(this.tagName)&&t.preventDefault(),ue.one(e,pn,(t=>{t.defaultPrevented||ue.one(e,fn,(()=>{Ht(this)&&this.focus()}))}));const i=ye.findOne(".modal.show");i&&On.getInstance(i).hide(),On.getOrCreateInstance(e).toggle(this)})),we(On),Kt(On);const xn=".bs.offcanvas",kn=".data-api",Ln=`load${xn}${kn}`,Sn="show",Dn="showing",$n="hiding",In=".offcanvas.show",Nn=`show${xn}`,Pn=`shown${xn}`,Mn=`hide${xn}`,jn=`hidePrevented${xn}`,Fn=`hidden${xn}`,Hn=`resize${xn}`,Bn=`click${xn}${kn}`,Wn=`keydown.dismiss${xn}`,zn={backdrop:!0,keyboard:!0,scroll:!1},Rn={backdrop:"(boolean|string)",keyboard:"boolean",scroll:"boolean"};class qn extends be{constructor(t,e){super(t,e),this._isShown=!1,this._backdrop=this._initializeBackDrop(),this._focustrap=this._initializeFocusTrap(),this._addEventListeners()}static get Default(){return zn}static get DefaultType(){return Rn}static get NAME(){return"offcanvas"}toggle(t){return this._isShown?this.hide():this.show(t)}show(t){this._isShown||ue.trigger(this._element,Nn,{relatedTarget:t}).defaultPrevented||(this._isShown=!0,this._backdrop.show(),this._config.scroll||(new cn).hide(),this._element.setAttribute("aria-modal",!0),this._element.setAttribute("role","dialog"),this._element.classList.add(Dn),this._queueCallback((()=>{this._config.scroll&&!this._config.backdrop||this._focustrap.activate(),this._element.classList.add(Sn),this._element.classList.remove(Dn),ue.trigger(this._element,Pn,{relatedTarget:t})}),this._element,!0))}hide(){this._isShown&&(ue.trigger(this._element,Mn).defaultPrevented||(this._focustrap.deactivate(),this._element.blur(),this._isShown=!1,this._element.classList.add($n),this._backdrop.hide(),this._queueCallback((()=>{this._element.classList.remove(Sn,$n),this._element.removeAttribute("aria-modal"),this._element.removeAttribute("role"),this._config.scroll||(new cn).reset(),ue.trigger(this._element,Fn)}),this._element,!0)))}dispose(){this._backdrop.dispose(),this._focustrap.deactivate(),super.dispose()}_initializeBackDrop(){const t=Boolean(this._config.backdrop);return new Ui({className:"offcanvas-backdrop",isVisible:t,isAnimated:!0,rootElement:this._element.parentNode,clickCallback:t?()=>{"static"!==this._config.backdrop?this.hide():ue.trigger(this._element,jn)}:null})}_initializeFocusTrap(){return new sn({trapElement:this._element})}_addEventListeners(){ue.on(this._element,Wn,(t=>{"Escape"===t.key&&(this._config.keyboard?this.hide():ue.trigger(this._element,jn))}))}static jQueryInterface(t){return this.each((function(){const e=qn.getOrCreateInstance(this,t);if("string"==typeof t){if(void 0===e[t]||t.startsWith("_")||"constructor"===t)throw new TypeError(`No method named "${t}"`);e[t](this)}}))}}ue.on(document,Bn,'[data-bs-toggle="offcanvas"]',(function(t){const e=ye.getElementFromSelector(this);if(["A","AREA"].includes(this.tagName)&&t.preventDefault(),Bt(this))return;ue.one(e,Fn,(()=>{Ht(this)&&this.focus()}));const i=ye.findOne(In);i&&i!==e&&qn.getInstance(i).hide(),qn.getOrCreateInstance(e).toggle(this)})),ue.on(window,Ln,(()=>{for(const t of ye.find(In))qn.getOrCreateInstance(t).show()})),ue.on(window,Hn,(()=>{for(const t of ye.find("[aria-modal][class*=show][class*=offcanvas-]"))"fixed"!==getComputedStyle(t).position&&qn.getOrCreateInstance(t).hide()})),we(qn),Kt(qn);const Vn={"*":["class","dir","id","lang","role",/^aria-[\w-]*$/i],a:["target","href","title","rel"],area:[],b:[],br:[],col:[],code:[],div:[],em:[],hr:[],h1:[],h2:[],h3:[],h4:[],h5:[],h6:[],i:[],img:["src","srcset","alt","title","width","height"],li:[],ol:[],p:[],pre:[],s:[],small:[],span:[],sub:[],sup:[],strong:[],u:[],ul:[]},Yn=new Set(["background","cite","href","itemtype","longdesc","poster","src","xlink:href"]),Kn=/^(?!javascript:)(?:[a-z0-9+.-]+:|[^&:/?#]*(?:[/?#]|$))/i,Qn=(t,e)=>{const i=t.nodeName.toLowerCase();return e.includes(i)?!Yn.has(i)||Boolean(Kn.test(t.nodeValue)):e.filter((t=>t instanceof RegExp)).some((t=>t.test(i)))},Xn={allowList:Vn,content:{},extraClass:"",html:!1,sanitize:!0,sanitizeFn:null,template:"<div></div>"},Un={allowList:"object",content:"object",extraClass:"(string|function)",html:"boolean",sanitize:"boolean",sanitizeFn:"(null|function)",template:"string"},Gn={entry:"(string|element|function|null)",selector:"(string|element)"};class Jn extends _e{constructor(t){super(),this._config=this._getConfig(t)}static get Default(){return Xn}static get DefaultType(){return Un}static get NAME(){return"TemplateFactory"}getContent(){return Object.values(this._config.content).map((t=>this._resolvePossibleFunction(t))).filter(Boolean)}hasContent(){return this.getContent().length>0}changeContent(t){return this._checkContent(t),this._config.content={...this._config.content,...t},this}toHtml(){const t=document.createElement("div");t.innerHTML=this._maybeSanitize(this._config.template);for(const[e,i]of Object.entries(this._config.content))this._setContent(t,i,e);const e=t.children[0],i=this._resolvePossibleFunction(this._config.extraClass);return i&&e.classList.add(...i.split(" ")),e}_typeCheckConfig(t){super._typeCheckConfig(t),this._checkContent(t.content)}_checkContent(t){for(const[e,i]of Object.entries(t))super._typeCheckConfig({selector:e,entry:i},Gn)}_setContent(t,e,i){const n=ye.findOne(i,t);n&&((e=this._resolvePossibleFunction(e))?jt(e)?this._putElementInTemplate(Ft(e),n):this._config.html?n.innerHTML=this._maybeSanitize(e):n.textContent=e:n.remove())}_maybeSanitize(t){return this._config.sanitize?function(t,e,i){if(!t.length)return t;if(i&&"function"==typeof i)return i(t);const n=(new window.DOMParser).parseFromString(t,"text/html"),s=[].concat(...n.body.querySelectorAll("*"));for(const t of s){const i=t.nodeName.toLowerCase();if(!Object.keys(e).includes(i)){t.remove();continue}const n=[].concat(...t.attributes),s=[].concat(e["*"]||[],e[i]||[]);for(const e of n)Qn(e,s)||t.removeAttribute(e.nodeName)}return n.body.innerHTML}(t,this._config.allowList,this._config.sanitizeFn):t}_resolvePossibleFunction(t){return Qt(t,[this])}_putElementInTemplate(t,e){if(this._config.html)return e.innerHTML="",void e.append(t);e.textContent=t.textContent}}const Zn=new Set(["sanitize","allowList","sanitizeFn"]),ts="fade",es="show",is=".modal",ns="hide.bs.modal",ss="hover",os="focus",rs={AUTO:"auto",TOP:"top",RIGHT:Yt()?"left":"right",BOTTOM:"bottom",LEFT:Yt()?"right":"left"},as={allowList:Vn,animation:!0,boundary:"clippingParents",container:!1,customClass:"",delay:0,fallbackPlacements:["top","right","bottom","left"],html:!1,offset:[0,6],placement:"top",popperConfig:null,sanitize:!0,sanitizeFn:null,selector:!1,template:'<div class="tooltip" role="tooltip"><div class="tooltip-arrow"></div><div class="tooltip-inner"></div></div>',title:"",trigger:"hover focus"},ls={allowList:"object",animation:"boolean",boundary:"(string|element)",container:"(string|element|boolean)",customClass:"(string|function)",delay:"(number|object)",fallbackPlacements:"array",html:"boolean",offset:"(array|string|function)",placement:"(string|function)",popperConfig:"(null|object|function)",sanitize:"boolean",sanitizeFn:"(null|function)",selector:"(string|boolean)",template:"string",title:"(string|element|function)",trigger:"string"};class cs extends be{constructor(t,i){if(void 0===e)throw new TypeError("Bootstrap's tooltips require Popper (https://popper.js.org)");super(t,i),this._isEnabled=!0,this._timeout=0,this._isHovered=null,this._activeTrigger={},this._popper=null,this._templateFactory=null,this._newContent=null,this.tip=null,this._setListeners(),this._config.selector||this._fixTitle()}static get Default(){return as}static get DefaultType(){return ls}static get NAME(){return"tooltip"}enable(){this._isEnabled=!0}disable(){this._isEnabled=!1}toggleEnabled(){this._isEnabled=!this._isEnabled}toggle(){this._isEnabled&&(this._activeTrigger.click=!this._activeTrigger.click,this._isShown()?this._leave():this._enter())}dispose(){clearTimeout(this._timeout),ue.off(this._element.closest(is),ns,this._hideModalHandler),this._element.getAttribute("data-bs-original-title")&&this._element.setAttribute("title",this._element.getAttribute("data-bs-original-title")),this._disposePopper(),super.dispose()}show(){if("none"===this._element.style.display)throw new Error("Please use show on visible elements");if(!this._isWithContent()||!this._isEnabled)return;const t=ue.trigger(this._element,this.constructor.eventName("show")),e=(Wt(this._element)||this._element.ownerDocument.documentElement).contains(this._element);if(t.defaultPrevented||!e)return;this._disposePopper();const i=this._getTipElement();this._element.setAttribute("aria-describedby",i.getAttribute("id"));const{container:n}=this._config;if(this._element.ownerDocument.documentElement.contains(this.tip)||(n.append(i),ue.trigger(this._element,this.constructor.eventName("inserted"))),this._popper=this._createPopper(i),i.classList.add(es),"ontouchstart"in document.documentElement)for(const t of[].concat(...document.body.children))ue.on(t,"mouseover",zt);this._queueCallback((()=>{ue.trigger(this._element,this.constructor.eventName("shown")),!1===this._isHovered&&this._leave(),this._isHovered=!1}),this.tip,this._isAnimated())}hide(){if(this._isShown()&&!ue.trigger(this._element,this.constructor.eventName("hide")).defaultPrevented){if(this._getTipElement().classList.remove(es),"ontouchstart"in document.documentElement)for(const t of[].concat(...document.body.children))ue.off(t,"mouseover",zt);this._activeTrigger.click=!1,this._activeTrigger[os]=!1,this._activeTrigger[ss]=!1,this._isHovered=null,this._queueCallback((()=>{this._isWithActiveTrigger()||(this._isHovered||this._disposePopper(),this._element.removeAttribute("aria-describedby"),ue.trigger(this._element,this.constructor.eventName("hidden")))}),this.tip,this._isAnimated())}}update(){this._popper&&this._popper.update()}_isWithContent(){return Boolean(this._getTitle())}_getTipElement(){return this.tip||(this.tip=this._createTipElement(this._newContent||this._getContentForTemplate())),this.tip}_createTipElement(t){const e=this._getTemplateFactory(t).toHtml();if(!e)return null;e.classList.remove(ts,es),e.classList.add(`bs-${this.constructor.NAME}-auto`);const i=(t=>{do{t+=Math.floor(1e6*Math.random())}while(document.getElementById(t));return t})(this.constructor.NAME).toString();return e.setAttribute("id",i),this._isAnimated()&&e.classList.add(ts),e}setContent(t){this._newContent=t,this._isShown()&&(this._disposePopper(),this.show())}_getTemplateFactory(t){return this._templateFactory?this._templateFactory.changeContent(t):this._templateFactory=new Jn({...this._config,content:t,extraClass:this._resolvePossibleFunction(this._config.customClass)}),this._templateFactory}_getContentForTemplate(){return{".tooltip-inner":this._getTitle()}}_getTitle(){return this._resolvePossibleFunction(this._config.title)||this._element.getAttribute("data-bs-original-title")}_initializeOnDelegatedTarget(t){return this.constructor.getOrCreateInstance(t.delegateTarget,this._getDelegateConfig())}_isAnimated(){return this._config.animation||this.tip&&this.tip.classList.contains(ts)}_isShown(){return this.tip&&this.tip.classList.contains(es)}_createPopper(t){const e=Qt(this._config.placement,[this,t,this._element]),i=rs[e.toUpperCase()];return St(this._element,t,this._getPopperConfig(i))}_getOffset(){const{offset:t}=this._config;return"string"==typeof t?t.split(",").map((t=>Number.parseInt(t,10))):"function"==typeof t?e=>t(e,this._element):t}_resolvePossibleFunction(t){return Qt(t,[this._element])}_getPopperConfig(t){const e={placement:t,modifiers:[{name:"flip",options:{fallbackPlacements:this._config.fallbackPlacements}},{name:"offset",options:{offset:this._getOffset()}},{name:"preventOverflow",options:{boundary:this._config.boundary}},{name:"arrow",options:{element:`.${this.constructor.NAME}-arrow`}},{name:"preSetPlacement",enabled:!0,phase:"beforeMain",fn:t=>{this._getTipElement().setAttribute("data-popper-placement",t.state.placement)}}]};return{...e,...Qt(this._config.popperConfig,[e])}}_setListeners(){const t=this._config.trigger.split(" ");for(const e of t)if("click"===e)ue.on(this._element,this.constructor.eventName("click"),this._config.selector,(t=>{this._initializeOnDelegatedTarget(t).toggle()}));else if("manual"!==e){const t=e===ss?this.constructor.eventName("mouseenter"):this.constructor.eventName("focusin"),i=e===ss?this.constructor.eventName("mouseleave"):this.constructor.eventName("focusout");ue.on(this._element,t,this._config.selector,(t=>{const e=this._initializeOnDelegatedTarget(t);e._activeTrigger["focusin"===t.type?os:ss]=!0,e._enter()})),ue.on(this._element,i,this._config.selector,(t=>{const e=this._initializeOnDelegatedTarget(t);e._activeTrigger["focusout"===t.type?os:ss]=e._element.contains(t.relatedTarget),e._leave()}))}this._hideModalHandler=()=>{this._element&&this.hide()},ue.on(this._element.closest(is),ns,this._hideModalHandler)}_fixTitle(){const t=this._element.getAttribute("title");t&&(this._element.getAttribute("aria-label")||this._element.textContent.trim()||this._element.setAttribute("aria-label",t),this._element.setAttribute("data-bs-original-title",t),this._element.removeAttribute("title"))}_enter(){this._isShown()||this._isHovered?this._isHovered=!0:(this._isHovered=!0,this._setTimeout((()=>{this._isHovered&&this.show()}),this._config.delay.show))}_leave(){this._isWithActiveTrigger()||(this._isHovered=!1,this._setTimeout((()=>{this._isHovered||this.hide()}),this._config.delay.hide))}_setTimeout(t,e){clearTimeout(this._timeout),this._timeout=setTimeout(t,e)}_isWithActiveTrigger(){return Object.values(this._activeTrigger).includes(!0)}_getConfig(t){const e=ge.getDataAttributes(this._element);for(const t of Object.keys(e))Zn.has(t)&&delete e[t];return t={...e,..."object"==typeof t&&t?t:{}},t=this._mergeConfigObj(t),t=this._configAfterMerge(t),this._typeCheckConfig(t),t}_configAfterMerge(t){return t.container=!1===t.container?document.body:Ft(t.container),"number"==typeof t.delay&&(t.delay={show:t.delay,hide:t.delay}),"number"==typeof t.title&&(t.title=t.title.toString()),"number"==typeof t.content&&(t.content=t.content.toString()),t}_getDelegateConfig(){const t={};for(const[e,i]of Object.entries(this._config))this.constructor.Default[e]!==i&&(t[e]=i);return t.selector=!1,t.trigger="manual",t}_disposePopper(){this._popper&&(this._popper.destroy(),this._popper=null),this.tip&&(this.tip.remove(),this.tip=null)}static jQueryInterface(t){return this.each((function(){const e=cs.getOrCreateInstance(this,t);if("string"==typeof t){if(void 0===e[t])throw new TypeError(`No method named "${t}"`);e[t]()}}))}}Kt(cs);const hs={...cs.Default,content:"",offset:[0,8],placement:"right",template:'<div class="popover" role="tooltip"><div class="popover-arrow"></div><h3 class="popover-header"></h3><div class="popover-body"></div></div>',trigger:"click"},ds={...cs.DefaultType,content:"(null|string|element|function)"};class us extends cs{static get Default(){return hs}static get DefaultType(){return ds}static get NAME(){return"popover"}_isWithContent(){return this._getTitle()||this._getContent()}_getContentForTemplate(){return{".popover-header":this._getTitle(),".popover-body":this._getContent()}}_getContent(){return this._resolvePossibleFunction(this._config.content)}static jQueryInterface(t){return this.each((function(){const e=us.getOrCreateInstance(this,t);if("string"==typeof t){if(void 0===e[t])throw new TypeError(`No method named "${t}"`);e[t]()}}))}}Kt(us);const fs=".bs.scrollspy",ps=`activate${fs}`,ms=`click${fs}`,gs=`load${fs}.data-api`,_s="active",bs="[href]",vs=".nav-link",ys=`${vs}, .nav-item > ${vs}, .list-group-item`,ws={offset:null,rootMargin:"0px 0px -25%",smoothScroll:!1,target:null,threshold:[.1,.5,1]},Es={offset:"(number|null)",rootMargin:"string",smoothScroll:"boolean",target:"element",threshold:"array"};class As extends be{constructor(t,e){super(t,e),this._targetLinks=new Map,this._observableSections=new Map,this._rootElement="visible"===getComputedStyle(this._element).overflowY?null:this._element,this._activeTarget=null,this._observer=null,this._previousScrollData={visibleEntryTop:0,parentScrollTop:0},this.refresh()}static get Default(){return ws}static get DefaultType(){return Es}static get NAME(){return"scrollspy"}refresh(){this._initializeTargetsAndObservables(),this._maybeEnableSmoothScroll(),this._observer?this._observer.disconnect():this._observer=this._getNewObserver();for(const t of this._observableSections.values())this._observer.observe(t)}dispose(){this._observer.disconnect(),super.dispose()}_configAfterMerge(t){return t.target=Ft(t.target)||document.body,t.rootMargin=t.offset?`${t.offset}px 0px -30%`:t.rootMargin,"string"==typeof t.threshold&&(t.threshold=t.threshold.split(",").map((t=>Number.parseFloat(t)))),t}_maybeEnableSmoothScroll(){this._config.smoothScroll&&(ue.off(this._config.target,ms),ue.on(this._config.target,ms,bs,(t=>{const e=this._observableSections.get(t.target.hash);if(e){t.preventDefault();const i=this._rootElement||window,n=e.offsetTop-this._element.offsetTop;if(i.scrollTo)return void i.scrollTo({top:n,behavior:"smooth"});i.scrollTop=n}})))}_getNewObserver(){const t={root:this._rootElement,threshold:this._config.threshold,rootMargin:this._config.rootMargin};return new IntersectionObserver((t=>this._observerCallback(t)),t)}_observerCallback(t){const e=t=>this._targetLinks.get(`#${t.target.id}`),i=t=>{this._previousScrollData.visibleEntryTop=t.target.offsetTop,this._process(e(t))},n=(this._rootElement||document.documentElement).scrollTop,s=n>=this._previousScrollData.parentScrollTop;this._previousScrollData.parentScrollTop=n;for(const o of t){if(!o.isIntersecting){this._activeTarget=null,this._clearActiveClass(e(o));continue}const t=o.target.offsetTop>=this._previousScrollData.visibleEntryTop;if(s&&t){if(i(o),!n)return}else s||t||i(o)}}_initializeTargetsAndObservables(){this._targetLinks=new Map,this._observableSections=new Map;const t=ye.find(bs,this._config.target);for(const e of t){if(!e.hash||Bt(e))continue;const t=ye.findOne(decodeURI(e.hash),this._element);Ht(t)&&(this._targetLinks.set(decodeURI(e.hash),e),this._observableSections.set(e.hash,t))}}_process(t){this._activeTarget!==t&&(this._clearActiveClass(this._config.target),this._activeTarget=t,t.classList.add(_s),this._activateParents(t),ue.trigger(this._element,ps,{relatedTarget:t}))}_activateParents(t){if(t.classList.contains("dropdown-item"))ye.findOne(".dropdown-toggle",t.closest(".dropdown")).classList.add(_s);else for(const e of ye.parents(t,".nav, .list-group"))for(const t of ye.prev(e,ys))t.classList.add(_s)}_clearActiveClass(t){t.classList.remove(_s);const e=ye.find(`${bs}.${_s}`,t);for(const t of e)t.classList.remove(_s)}static jQueryInterface(t){return this.each((function(){const e=As.getOrCreateInstance(this,t);if("string"==typeof t){if(void 0===e[t]||t.startsWith("_")||"constructor"===t)throw new TypeError(`No method named "${t}"`);e[t]()}}))}}ue.on(window,gs,(()=>{for(const t of ye.find('[data-bs-spy="scroll"]'))As.getOrCreateInstance(t)})),Kt(As);const Ts=".bs.tab",Cs=`hide${Ts}`,Os=`hidden${Ts}`,xs=`show${Ts}`,ks=`shown${Ts}`,Ls=`click${Ts}`,Ss=`keydown${Ts}`,Ds=`load${Ts}`,$s="ArrowLeft",Is="ArrowRight",Ns="ArrowUp",Ps="ArrowDown",Ms="Home",js="End",Fs="active",Hs="fade",Bs="show",Ws=".dropdown-toggle",zs=`:not(${Ws})`,Rs='[data-bs-toggle="tab"], [data-bs-toggle="pill"], [data-bs-toggle="list"]',qs=`.nav-link${zs}, .list-group-item${zs}, [role="tab"]${zs}, ${Rs}`,Vs=`.${Fs}[data-bs-toggle="tab"], .${Fs}[data-bs-toggle="pill"], .${Fs}[data-bs-toggle="list"]`;class Ys extends be{constructor(t){super(t),this._parent=this._element.closest('.list-group, .nav, [role="tablist"]'),this._parent&&(this._setInitialAttributes(this._parent,this._getChildren()),ue.on(this._element,Ss,(t=>this._keydown(t))))}static get NAME(){return"tab"}show(){const t=this._element;if(this._elemIsActive(t))return;const e=this._getActiveElem(),i=e?ue.trigger(e,Cs,{relatedTarget:t}):null;ue.trigger(t,xs,{relatedTarget:e}).defaultPrevented||i&&i.defaultPrevented||(this._deactivate(e,t),this._activate(t,e))}_activate(t,e){t&&(t.classList.add(Fs),this._activate(ye.getElementFromSelector(t)),this._queueCallback((()=>{"tab"===t.getAttribute("role")?(t.removeAttribute("tabindex"),t.setAttribute("aria-selected",!0),this._toggleDropDown(t,!0),ue.trigger(t,ks,{relatedTarget:e})):t.classList.add(Bs)}),t,t.classList.contains(Hs)))}_deactivate(t,e){t&&(t.classList.remove(Fs),t.blur(),this._deactivate(ye.getElementFromSelector(t)),this._queueCallback((()=>{"tab"===t.getAttribute("role")?(t.setAttribute("aria-selected",!1),t.setAttribute("tabindex","-1"),this._toggleDropDown(t,!1),ue.trigger(t,Os,{relatedTarget:e})):t.classList.remove(Bs)}),t,t.classList.contains(Hs)))}_keydown(t){if(![$s,Is,Ns,Ps,Ms,js].includes(t.key))return;t.stopPropagation(),t.preventDefault();const e=this._getChildren().filter((t=>!Bt(t)));let i;if([Ms,js].includes(t.key))i=e[t.key===Ms?0:e.length-1];else{const n=[Is,Ps].includes(t.key);i=Ut(e,t.target,n,!0)}i&&(i.focus({preventScroll:!0}),Ys.getOrCreateInstance(i).show())}_getChildren(){return ye.find(qs,this._parent)}_getActiveElem(){return this._getChildren().find((t=>this._elemIsActive(t)))||null}_setInitialAttributes(t,e){this._setAttributeIfNotExists(t,"role","tablist");for(const t of e)this._setInitialAttributesOnChild(t)}_setInitialAttributesOnChild(t){t=this._getInnerElement(t);const e=this._elemIsActive(t),i=this._getOuterElement(t);t.setAttribute("aria-selected",e),i!==t&&this._setAttributeIfNotExists(i,"role","presentation"),e||t.setAttribute("tabindex","-1"),this._setAttributeIfNotExists(t,"role","tab"),this._setInitialAttributesOnTargetPanel(t)}_setInitialAttributesOnTargetPanel(t){const e=ye.getElementFromSelector(t);e&&(this._setAttributeIfNotExists(e,"role","tabpanel"),t.id&&this._setAttributeIfNotExists(e,"aria-labelledby",`${t.id}`))}_toggleDropDown(t,e){const i=this._getOuterElement(t);if(!i.classList.contains("dropdown"))return;const n=(t,n)=>{const s=ye.findOne(t,i);s&&s.classList.toggle(n,e)};n(Ws,Fs),n(".dropdown-menu",Bs),i.setAttribute("aria-expanded",e)}_setAttributeIfNotExists(t,e,i){t.hasAttribute(e)||t.setAttribute(e,i)}_elemIsActive(t){return t.classList.contains(Fs)}_getInnerElement(t){return t.matches(qs)?t:ye.findOne(qs,t)}_getOuterElement(t){return t.closest(".nav-item, .list-group-item")||t}static jQueryInterface(t){return this.each((function(){const e=Ys.getOrCreateInstance(this);if("string"==typeof t){if(void 0===e[t]||t.startsWith("_")||"constructor"===t)throw new TypeError(`No method named "${t}"`);e[t]()}}))}}ue.on(document,Ls,Rs,(function(t){["A","AREA"].includes(this.tagName)&&t.preventDefault(),Bt(this)||Ys.getOrCreateInstance(this).show()})),ue.on(window,Ds,(()=>{for(const t of ye.find(Vs))Ys.getOrCreateInstance(t)})),Kt(Ys);const Ks=".bs.toast",Qs=`mouseover${Ks}`,Xs=`mouseout${Ks}`,Us=`focusin${Ks}`,Gs=`focusout${Ks}`,Js=`hide${Ks}`,Zs=`hidden${Ks}`,to=`show${Ks}`,eo=`shown${Ks}`,io="hide",no="show",so="showing",oo={animation:"boolean",autohide:"boolean",delay:"number"},ro={animation:!0,autohide:!0,delay:5e3};class ao extends be{constructor(t,e){super(t,e),this._timeout=null,this._hasMouseInteraction=!1,this._hasKeyboardInteraction=!1,this._setListeners()}static get Default(){return ro}static get DefaultType(){return oo}static get NAME(){return"toast"}show(){ue.trigger(this._element,to).defaultPrevented||(this._clearTimeout(),this._config.animation&&this._element.classList.add("fade"),this._element.classList.remove(io),Rt(this._element),this._element.classList.add(no,so),this._queueCallback((()=>{this._element.classList.remove(so),ue.trigger(this._element,eo),this._maybeScheduleHide()}),this._element,this._config.animation))}hide(){this.isShown()&&(ue.trigger(this._element,Js).defaultPrevented||(this._element.classList.add(so),this._queueCallback((()=>{this._element.classList.add(io),this._element.classList.remove(so,no),ue.trigger(this._element,Zs)}),this._element,this._config.animation)))}dispose(){this._clearTimeout(),this.isShown()&&this._element.classList.remove(no),super.dispose()}isShown(){return this._element.classList.contains(no)}_maybeScheduleHide(){this._config.autohide&&(this._hasMouseInteraction||this._hasKeyboardInteraction||(this._timeout=setTimeout((()=>{this.hide()}),this._config.delay)))}_onInteraction(t,e){switch(t.type){case"mouseover":case"mouseout":this._hasMouseInteraction=e;break;case"focusin":case"focusout":this._hasKeyboardInteraction=e}if(e)return void this._clearTimeout();const i=t.relatedTarget;this._element===i||this._element.contains(i)||this._maybeScheduleHide()}_setListeners(){ue.on(this._element,Qs,(t=>this._onInteraction(t,!0))),ue.on(this._element,Xs,(t=>this._onInteraction(t,!1))),ue.on(this._element,Us,(t=>this._onInteraction(t,!0))),ue.on(this._element,Gs,(t=>this._onInteraction(t,!1)))}_clearTimeout(){clearTimeout(this._timeout),this._timeout=null}static jQueryInterface(t){return this.each((function(){const e=ao.getOrCreateInstance(this,t);if("string"==typeof t){if(void 0===e[t])throw new TypeError(`No method named 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diff --git a/docs/_build/html/_static/scripts/bootstrap.js.LICENSE.txt b/docs/_build/html/_static/scripts/bootstrap.js.LICENSE.txt
deleted file mode 100644
index 10f979d..0000000
--- a/docs/_build/html/_static/scripts/bootstrap.js.LICENSE.txt
+++ /dev/null
@@ -1,5 +0,0 @@
-/*!
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-  * Copyright 2011-2023 The Bootstrap Authors (https://github.com/twbs/bootstrap/graphs/contributors)
-  * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)
-  */
diff --git a/docs/_build/html/_static/scripts/bootstrap.js.map b/docs/_build/html/_static/scripts/bootstrap.js.map
deleted file mode 100644
index e5bc157..0000000
--- a/docs/_build/html/_static/scripts/bootstrap.js.map
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(element.nodeName || '').toLowerCase() : null;\n}","export default function getWindow(node) {\n  if (node == null) {\n    return window;\n  }\n\n  if (node.toString() !== '[object Window]') {\n    var ownerDocument = node.ownerDocument;\n    return ownerDocument ? ownerDocument.defaultView || window : window;\n  }\n\n  return node;\n}","import getWindow from \"./getWindow.js\";\n\nfunction isElement(node) {\n  var OwnElement = getWindow(node).Element;\n  return node instanceof OwnElement || node instanceof Element;\n}\n\nfunction isHTMLElement(node) {\n  var OwnElement = getWindow(node).HTMLElement;\n  return node instanceof OwnElement || node instanceof HTMLElement;\n}\n\nfunction isShadowRoot(node) {\n  // IE 11 has no ShadowRoot\n  if (typeof ShadowRoot === 'undefined') {\n    return false;\n  }\n\n  var OwnElement = getWindow(node).ShadowRoot;\n  return node instanceof OwnElement || node instanceof ShadowRoot;\n}\n\nexport { isElement, isHTMLElement, isShadowRoot };","import getNodeName from \"../dom-utils/getNodeName.js\";\nimport { isHTMLElement } from \"../dom-utils/instanceOf.js\"; // This modifier takes the styles prepared by the `computeStyles` modifier\n// and applies them to the HTMLElements such as popper and arrow\n\nfunction applyStyles(_ref) {\n  var state = _ref.state;\n  Object.keys(state.elements).forEach(function (name) {\n    var style = state.styles[name] || {};\n    var attributes = state.attributes[name] || {};\n    var element = state.elements[name]; // arrow is optional + virtual elements\n\n    if (!isHTMLElement(element) || !getNodeName(element)) {\n      return;\n    } // Flow doesn't support to extend this property, but it's the most\n    // effective way to apply styles to an HTMLElement\n    // $FlowFixMe[cannot-write]\n\n\n    Object.assign(element.style, style);\n    Object.keys(attributes).forEach(function (name) {\n      var value = attributes[name];\n\n      if (value === false) {\n        element.removeAttribute(name);\n      } else {\n        element.setAttribute(name, value === true ? '' : value);\n      }\n    });\n  });\n}\n\nfunction effect(_ref2) {\n  var state = _ref2.state;\n  var initialStyles = {\n    popper: {\n      position: state.options.strategy,\n      left: '0',\n      top: '0',\n      margin: '0'\n    },\n    arrow: {\n      position: 'absolute'\n    },\n    reference: {}\n  };\n  Object.assign(state.elements.popper.style, initialStyles.popper);\n  state.styles = initialStyles;\n\n  if (state.elements.arrow) {\n    Object.assign(state.elements.arrow.style, initialStyles.arrow);\n  }\n\n  return function () {\n    Object.keys(state.elements).forEach(function (name) {\n      var element = state.elements[name];\n      var attributes = state.attributes[name] || {};\n      var styleProperties = Object.keys(state.styles.hasOwnProperty(name) ? state.styles[name] : initialStyles[name]); // Set all values to an empty string to unset them\n\n      var style = styleProperties.reduce(function (style, property) {\n        style[property] = '';\n        return style;\n      }, {}); // arrow is optional + virtual elements\n\n      if (!isHTMLElement(element) || !getNodeName(element)) {\n        return;\n      }\n\n      Object.assign(element.style, style);\n      Object.keys(attributes).forEach(function (attribute) {\n        element.removeAttribute(attribute);\n      });\n    });\n  };\n} // eslint-disable-next-line import/no-unused-modules\n\n\nexport default {\n  name: 'applyStyles',\n  enabled: true,\n  phase: 'write',\n  fn: applyStyles,\n  effect: effect,\n  requires: ['computeStyles']\n};","import { auto } from \"../enums.js\";\nexport default function getBasePlacement(placement) {\n  return placement.split('-')[0];\n}","export var max = Math.max;\nexport var min = Math.min;\nexport var round = Math.round;","export default function getUAString() {\n  var uaData = navigator.userAgentData;\n\n  if (uaData != null && uaData.brands && Array.isArray(uaData.brands)) {\n    return uaData.brands.map(function (item) {\n      return item.brand + \"/\" + item.version;\n    }).join(' ');\n  }\n\n  return navigator.userAgent;\n}","import getUAString from \"../utils/userAgent.js\";\nexport default function isLayoutViewport() {\n  return !/^((?!chrome|android).)*safari/i.test(getUAString());\n}","import { isElement, isHTMLElement } from \"./instanceOf.js\";\nimport { round } from \"../utils/math.js\";\nimport getWindow from \"./getWindow.js\";\nimport isLayoutViewport from \"./isLayoutViewport.js\";\nexport default function getBoundingClientRect(element, includeScale, isFixedStrategy) {\n  if (includeScale === void 0) {\n    includeScale = false;\n  }\n\n  if (isFixedStrategy === void 0) {\n    isFixedStrategy = false;\n  }\n\n  var clientRect = element.getBoundingClientRect();\n  var scaleX = 1;\n  var scaleY = 1;\n\n  if (includeScale && isHTMLElement(element)) {\n    scaleX = element.offsetWidth > 0 ? round(clientRect.width) / element.offsetWidth || 1 : 1;\n    scaleY = element.offsetHeight > 0 ? round(clientRect.height) / element.offsetHeight || 1 : 1;\n  }\n\n  var _ref = isElement(element) ? getWindow(element) : window,\n      visualViewport = _ref.visualViewport;\n\n  var addVisualOffsets = !isLayoutViewport() && isFixedStrategy;\n  var x = (clientRect.left + (addVisualOffsets && visualViewport ? visualViewport.offsetLeft : 0)) / scaleX;\n  var y = (clientRect.top + (addVisualOffsets && visualViewport ? visualViewport.offsetTop : 0)) / scaleY;\n  var width = clientRect.width / scaleX;\n  var height = clientRect.height / scaleY;\n  return {\n    width: width,\n    height: height,\n    top: y,\n    right: x + width,\n    bottom: y + height,\n    left: x,\n    x: x,\n    y: y\n  };\n}","import getBoundingClientRect from \"./getBoundingClientRect.js\"; // Returns the layout rect of an element relative to its offsetParent. Layout\n// means it doesn't take into account transforms.\n\nexport default function getLayoutRect(element) {\n  var clientRect = getBoundingClientRect(element); // Use the clientRect sizes if it's not been transformed.\n  // Fixes https://github.com/popperjs/popper-core/issues/1223\n\n  var width = element.offsetWidth;\n  var height = element.offsetHeight;\n\n  if (Math.abs(clientRect.width - width) <= 1) {\n    width = clientRect.width;\n  }\n\n  if (Math.abs(clientRect.height - height) <= 1) {\n    height = clientRect.height;\n  }\n\n  return {\n    x: element.offsetLeft,\n    y: element.offsetTop,\n    width: width,\n    height: height\n  };\n}","import { isShadowRoot } from \"./instanceOf.js\";\nexport default function contains(parent, child) {\n  var rootNode = child.getRootNode && child.getRootNode(); // First, attempt with faster native method\n\n  if (parent.contains(child)) {\n    return true;\n  } // then fallback to custom implementation with Shadow DOM support\n  else if (rootNode && isShadowRoot(rootNode)) {\n      var next = child;\n\n      do {\n        if (next && parent.isSameNode(next)) {\n          return true;\n        } // $FlowFixMe[prop-missing]: need a better way to handle this...\n\n\n        next = next.parentNode || next.host;\n      } while (next);\n    } // Give up, the result is false\n\n\n  return false;\n}","import getWindow from \"./getWindow.js\";\nexport default function getComputedStyle(element) {\n  return getWindow(element).getComputedStyle(element);\n}","import getNodeName from \"./getNodeName.js\";\nexport default function isTableElement(element) {\n  return ['table', 'td', 'th'].indexOf(getNodeName(element)) >= 0;\n}","import { isElement } from \"./instanceOf.js\";\nexport default function getDocumentElement(element) {\n  // $FlowFixMe[incompatible-return]: assume body is always available\n  return ((isElement(element) ? element.ownerDocument : // $FlowFixMe[prop-missing]\n  element.document) || window.document).documentElement;\n}","import getNodeName from \"./getNodeName.js\";\nimport getDocumentElement from \"./getDocumentElement.js\";\nimport { isShadowRoot } from \"./instanceOf.js\";\nexport default function getParentNode(element) {\n  if (getNodeName(element) === 'html') {\n    return element;\n  }\n\n  return (// this is a quicker (but less type safe) way to save quite some bytes from the bundle\n    // $FlowFixMe[incompatible-return]\n    // $FlowFixMe[prop-missing]\n    element.assignedSlot || // step into the shadow DOM of the parent of a slotted node\n    element.parentNode || ( // DOM Element detected\n    isShadowRoot(element) ? element.host : null) || // ShadowRoot detected\n    // $FlowFixMe[incompatible-call]: HTMLElement is a Node\n    getDocumentElement(element) // fallback\n\n  );\n}","import getWindow from \"./getWindow.js\";\nimport getNodeName from \"./getNodeName.js\";\nimport getComputedStyle from \"./getComputedStyle.js\";\nimport { isHTMLElement, isShadowRoot } from \"./instanceOf.js\";\nimport isTableElement from \"./isTableElement.js\";\nimport getParentNode from \"./getParentNode.js\";\nimport getUAString from \"../utils/userAgent.js\";\n\nfunction getTrueOffsetParent(element) {\n  if (!isHTMLElement(element) || // https://github.com/popperjs/popper-core/issues/837\n  getComputedStyle(element).position === 'fixed') {\n    return null;\n  }\n\n  return element.offsetParent;\n} // `.offsetParent` reports `null` for fixed elements, while absolute elements\n// return the containing block\n\n\nfunction getContainingBlock(element) {\n  var isFirefox = /firefox/i.test(getUAString());\n  var isIE = /Trident/i.test(getUAString());\n\n  if (isIE && isHTMLElement(element)) {\n    // In IE 9, 10 and 11 fixed elements containing block is always established by the viewport\n    var elementCss = getComputedStyle(element);\n\n    if (elementCss.position === 'fixed') {\n      return null;\n    }\n  }\n\n  var currentNode = getParentNode(element);\n\n  if (isShadowRoot(currentNode)) {\n    currentNode = currentNode.host;\n  }\n\n  while (isHTMLElement(currentNode) && ['html', 'body'].indexOf(getNodeName(currentNode)) < 0) {\n    var css = getComputedStyle(currentNode); // This is non-exhaustive but covers the most common CSS properties that\n    // create a containing block.\n    // https://developer.mozilla.org/en-US/docs/Web/CSS/Containing_block#identifying_the_containing_block\n\n    if (css.transform !== 'none' || css.perspective !== 'none' || css.contain === 'paint' || ['transform', 'perspective'].indexOf(css.willChange) !== -1 || isFirefox && css.willChange === 'filter' || isFirefox && css.filter && css.filter !== 'none') {\n      return currentNode;\n    } else {\n      currentNode = currentNode.parentNode;\n    }\n  }\n\n  return null;\n} // Gets the closest ancestor positioned element. Handles some edge cases,\n// such as table ancestors and cross browser bugs.\n\n\nexport default function getOffsetParent(element) {\n  var window = getWindow(element);\n  var offsetParent = getTrueOffsetParent(element);\n\n  while (offsetParent && isTableElement(offsetParent) && getComputedStyle(offsetParent).position === 'static') {\n    offsetParent = getTrueOffsetParent(offsetParent);\n  }\n\n  if (offsetParent && (getNodeName(offsetParent) === 'html' || getNodeName(offsetParent) === 'body' && getComputedStyle(offsetParent).position === 'static')) {\n    return window;\n  }\n\n  return offsetParent || getContainingBlock(element) || window;\n}","export default function getMainAxisFromPlacement(placement) {\n  return ['top', 'bottom'].indexOf(placement) >= 0 ? 'x' : 'y';\n}","import { max as mathMax, min as mathMin } from \"./math.js\";\nexport function within(min, value, max) {\n  return mathMax(min, mathMin(value, max));\n}\nexport function withinMaxClamp(min, value, max) {\n  var v = within(min, value, max);\n  return v > max ? max : v;\n}","import getFreshSideObject from \"./getFreshSideObject.js\";\nexport default function mergePaddingObject(paddingObject) {\n  return Object.assign({}, getFreshSideObject(), paddingObject);\n}","export default function getFreshSideObject() {\n  return {\n    top: 0,\n    right: 0,\n    bottom: 0,\n    left: 0\n  };\n}","export default function expandToHashMap(value, keys) {\n  return keys.reduce(function (hashMap, key) {\n    hashMap[key] = value;\n    return hashMap;\n  }, {});\n}","import getBasePlacement from \"../utils/getBasePlacement.js\";\nimport getLayoutRect from \"../dom-utils/getLayoutRect.js\";\nimport contains from \"../dom-utils/contains.js\";\nimport getOffsetParent from \"../dom-utils/getOffsetParent.js\";\nimport getMainAxisFromPlacement from \"../utils/getMainAxisFromPlacement.js\";\nimport { within } from \"../utils/within.js\";\nimport mergePaddingObject from \"../utils/mergePaddingObject.js\";\nimport expandToHashMap from \"../utils/expandToHashMap.js\";\nimport { left, right, basePlacements, top, bottom } from \"../enums.js\"; // eslint-disable-next-line import/no-unused-modules\n\nvar toPaddingObject = function toPaddingObject(padding, state) {\n  padding = typeof padding === 'function' ? padding(Object.assign({}, state.rects, {\n    placement: state.placement\n  })) : padding;\n  return mergePaddingObject(typeof padding !== 'number' ? padding : expandToHashMap(padding, basePlacements));\n};\n\nfunction arrow(_ref) {\n  var _state$modifiersData$;\n\n  var state = _ref.state,\n      name = _ref.name,\n      options = _ref.options;\n  var arrowElement = state.elements.arrow;\n  var popperOffsets = state.modifiersData.popperOffsets;\n  var basePlacement = getBasePlacement(state.placement);\n  var axis = getMainAxisFromPlacement(basePlacement);\n  var isVertical = [left, right].indexOf(basePlacement) >= 0;\n  var len = isVertical ? 'height' : 'width';\n\n  if (!arrowElement || !popperOffsets) {\n    return;\n  }\n\n  var paddingObject = toPaddingObject(options.padding, state);\n  var arrowRect = getLayoutRect(arrowElement);\n  var minProp = axis === 'y' ? top : left;\n  var maxProp = axis === 'y' ? bottom : right;\n  var endDiff = state.rects.reference[len] + state.rects.reference[axis] - popperOffsets[axis] - state.rects.popper[len];\n  var startDiff = popperOffsets[axis] - state.rects.reference[axis];\n  var arrowOffsetParent = getOffsetParent(arrowElement);\n  var clientSize = arrowOffsetParent ? axis === 'y' ? arrowOffsetParent.clientHeight || 0 : arrowOffsetParent.clientWidth || 0 : 0;\n  var centerToReference = endDiff / 2 - startDiff / 2; // Make sure the arrow doesn't overflow the popper if the center point is\n  // outside of the popper bounds\n\n  var min = paddingObject[minProp];\n  var max = clientSize - arrowRect[len] - paddingObject[maxProp];\n  var center = clientSize / 2 - arrowRect[len] / 2 + centerToReference;\n  var offset = within(min, center, max); // Prevents breaking syntax highlighting...\n\n  var axisProp = axis;\n  state.modifiersData[name] = (_state$modifiersData$ = {}, _state$modifiersData$[axisProp] = offset, _state$modifiersData$.centerOffset = offset - center, _state$modifiersData$);\n}\n\nfunction effect(_ref2) {\n  var state = _ref2.state,\n      options = _ref2.options;\n  var _options$element = options.element,\n      arrowElement = _options$element === void 0 ? '[data-popper-arrow]' : _options$element;\n\n  if (arrowElement == null) {\n    return;\n  } // CSS selector\n\n\n  if (typeof arrowElement === 'string') {\n    arrowElement = state.elements.popper.querySelector(arrowElement);\n\n    if (!arrowElement) {\n      return;\n    }\n  }\n\n  if (!contains(state.elements.popper, arrowElement)) {\n    return;\n  }\n\n  state.elements.arrow = arrowElement;\n} // eslint-disable-next-line import/no-unused-modules\n\n\nexport default {\n  name: 'arrow',\n  enabled: true,\n  phase: 'main',\n  fn: arrow,\n  effect: effect,\n  requires: ['popperOffsets'],\n  requiresIfExists: ['preventOverflow']\n};","export default function getVariation(placement) {\n  return placement.split('-')[1];\n}","import { top, left, right, bottom, end } from \"../enums.js\";\nimport getOffsetParent from \"../dom-utils/getOffsetParent.js\";\nimport getWindow from \"../dom-utils/getWindow.js\";\nimport getDocumentElement from \"../dom-utils/getDocumentElement.js\";\nimport getComputedStyle from \"../dom-utils/getComputedStyle.js\";\nimport getBasePlacement from \"../utils/getBasePlacement.js\";\nimport getVariation from \"../utils/getVariation.js\";\nimport { round } from \"../utils/math.js\"; // eslint-disable-next-line import/no-unused-modules\n\nvar unsetSides = {\n  top: 'auto',\n  right: 'auto',\n  bottom: 'auto',\n  left: 'auto'\n}; // Round the offsets to the nearest suitable subpixel based on the DPR.\n// Zooming can change the DPR, but it seems to report a value that will\n// cleanly divide the values into the appropriate subpixels.\n\nfunction roundOffsetsByDPR(_ref, win) {\n  var x = _ref.x,\n      y = _ref.y;\n  var dpr = win.devicePixelRatio || 1;\n  return {\n    x: round(x * dpr) / dpr || 0,\n    y: round(y * dpr) / dpr || 0\n  };\n}\n\nexport function mapToStyles(_ref2) {\n  var _Object$assign2;\n\n  var popper = _ref2.popper,\n      popperRect = _ref2.popperRect,\n      placement = _ref2.placement,\n      variation = _ref2.variation,\n      offsets = _ref2.offsets,\n      position = _ref2.position,\n      gpuAcceleration = _ref2.gpuAcceleration,\n      adaptive = _ref2.adaptive,\n      roundOffsets = _ref2.roundOffsets,\n      isFixed = _ref2.isFixed;\n  var _offsets$x = offsets.x,\n      x = _offsets$x === void 0 ? 0 : _offsets$x,\n      _offsets$y = offsets.y,\n      y = _offsets$y === void 0 ? 0 : _offsets$y;\n\n  var _ref3 = typeof roundOffsets === 'function' ? roundOffsets({\n    x: x,\n    y: y\n  }) : {\n    x: x,\n    y: y\n  };\n\n  x = _ref3.x;\n  y = _ref3.y;\n  var hasX = offsets.hasOwnProperty('x');\n  var hasY = offsets.hasOwnProperty('y');\n  var sideX = left;\n  var sideY = top;\n  var win = window;\n\n  if (adaptive) {\n    var offsetParent = getOffsetParent(popper);\n    var heightProp = 'clientHeight';\n    var widthProp = 'clientWidth';\n\n    if (offsetParent === getWindow(popper)) {\n      offsetParent = getDocumentElement(popper);\n\n      if (getComputedStyle(offsetParent).position !== 'static' && position === 'absolute') {\n        heightProp = 'scrollHeight';\n        widthProp = 'scrollWidth';\n      }\n    } // $FlowFixMe[incompatible-cast]: force type refinement, we compare offsetParent with window above, but Flow doesn't detect it\n\n\n    offsetParent = offsetParent;\n\n    if (placement === top || (placement === left || placement === right) && variation === end) {\n      sideY = bottom;\n      var offsetY = isFixed && offsetParent === win && win.visualViewport ? win.visualViewport.height : // $FlowFixMe[prop-missing]\n      offsetParent[heightProp];\n      y -= offsetY - popperRect.height;\n      y *= gpuAcceleration ? 1 : -1;\n    }\n\n    if (placement === left || (placement === top || placement === bottom) && variation === end) {\n      sideX = right;\n      var offsetX = isFixed && offsetParent === win && win.visualViewport ? win.visualViewport.width : // $FlowFixMe[prop-missing]\n      offsetParent[widthProp];\n      x -= offsetX - popperRect.width;\n      x *= gpuAcceleration ? 1 : -1;\n    }\n  }\n\n  var commonStyles = Object.assign({\n    position: position\n  }, adaptive && unsetSides);\n\n  var _ref4 = roundOffsets === true ? roundOffsetsByDPR({\n    x: x,\n    y: y\n  }, getWindow(popper)) : {\n    x: x,\n    y: y\n  };\n\n  x = _ref4.x;\n  y = _ref4.y;\n\n  if (gpuAcceleration) {\n    var _Object$assign;\n\n    return Object.assign({}, commonStyles, (_Object$assign = {}, _Object$assign[sideY] = hasY ? '0' : '', _Object$assign[sideX] = hasX ? '0' : '', _Object$assign.transform = (win.devicePixelRatio || 1) <= 1 ? \"translate(\" + x + \"px, \" + y + \"px)\" : \"translate3d(\" + x + \"px, \" + y + \"px, 0)\", _Object$assign));\n  }\n\n  return Object.assign({}, commonStyles, (_Object$assign2 = {}, _Object$assign2[sideY] = hasY ? y + \"px\" : '', _Object$assign2[sideX] = hasX ? x + \"px\" : '', _Object$assign2.transform = '', _Object$assign2));\n}\n\nfunction computeStyles(_ref5) {\n  var state = _ref5.state,\n      options = _ref5.options;\n  var _options$gpuAccelerat = options.gpuAcceleration,\n      gpuAcceleration = _options$gpuAccelerat === void 0 ? true : _options$gpuAccelerat,\n      _options$adaptive = options.adaptive,\n      adaptive = _options$adaptive === void 0 ? true : _options$adaptive,\n      _options$roundOffsets = options.roundOffsets,\n      roundOffsets = _options$roundOffsets === void 0 ? true : _options$roundOffsets;\n  var commonStyles = {\n    placement: getBasePlacement(state.placement),\n    variation: getVariation(state.placement),\n    popper: state.elements.popper,\n    popperRect: state.rects.popper,\n    gpuAcceleration: gpuAcceleration,\n    isFixed: state.options.strategy === 'fixed'\n  };\n\n  if (state.modifiersData.popperOffsets != null) {\n    state.styles.popper = Object.assign({}, state.styles.popper, mapToStyles(Object.assign({}, commonStyles, {\n      offsets: state.modifiersData.popperOffsets,\n      position: state.options.strategy,\n      adaptive: adaptive,\n      roundOffsets: roundOffsets\n    })));\n  }\n\n  if (state.modifiersData.arrow != null) {\n    state.styles.arrow = Object.assign({}, state.styles.arrow, mapToStyles(Object.assign({}, commonStyles, {\n      offsets: state.modifiersData.arrow,\n      position: 'absolute',\n      adaptive: false,\n      roundOffsets: roundOffsets\n    })));\n  }\n\n  state.attributes.popper = Object.assign({}, state.attributes.popper, {\n    'data-popper-placement': state.placement\n  });\n} // eslint-disable-next-line import/no-unused-modules\n\n\nexport default {\n  name: 'computeStyles',\n  enabled: true,\n  phase: 'beforeWrite',\n  fn: computeStyles,\n  data: {}\n};","import getWindow from \"../dom-utils/getWindow.js\"; // eslint-disable-next-line import/no-unused-modules\n\nvar passive = {\n  passive: true\n};\n\nfunction effect(_ref) {\n  var state = _ref.state,\n      instance = _ref.instance,\n      options = _ref.options;\n  var _options$scroll = options.scroll,\n      scroll = _options$scroll === void 0 ? true : _options$scroll,\n      _options$resize = options.resize,\n      resize = _options$resize === void 0 ? true : _options$resize;\n  var window = getWindow(state.elements.popper);\n  var scrollParents = [].concat(state.scrollParents.reference, state.scrollParents.popper);\n\n  if (scroll) {\n    scrollParents.forEach(function (scrollParent) {\n      scrollParent.addEventListener('scroll', instance.update, passive);\n    });\n  }\n\n  if (resize) {\n    window.addEventListener('resize', instance.update, passive);\n  }\n\n  return function () {\n    if (scroll) {\n      scrollParents.forEach(function (scrollParent) {\n        scrollParent.removeEventListener('scroll', instance.update, passive);\n      });\n    }\n\n    if (resize) {\n      window.removeEventListener('resize', instance.update, passive);\n    }\n  };\n} // eslint-disable-next-line import/no-unused-modules\n\n\nexport default {\n  name: 'eventListeners',\n  enabled: true,\n  phase: 'write',\n  fn: function fn() {},\n  effect: effect,\n  data: {}\n};","var hash = {\n  left: 'right',\n  right: 'left',\n  bottom: 'top',\n  top: 'bottom'\n};\nexport default function getOppositePlacement(placement) {\n  return placement.replace(/left|right|bottom|top/g, function (matched) {\n    return hash[matched];\n  });\n}","var hash = {\n  start: 'end',\n  end: 'start'\n};\nexport default function getOppositeVariationPlacement(placement) {\n  return placement.replace(/start|end/g, function (matched) {\n    return hash[matched];\n  });\n}","import getWindow from \"./getWindow.js\";\nexport default function getWindowScroll(node) {\n  var win = getWindow(node);\n  var scrollLeft = win.pageXOffset;\n  var scrollTop = win.pageYOffset;\n  return {\n    scrollLeft: scrollLeft,\n    scrollTop: scrollTop\n  };\n}","import getBoundingClientRect from \"./getBoundingClientRect.js\";\nimport getDocumentElement from \"./getDocumentElement.js\";\nimport getWindowScroll from \"./getWindowScroll.js\";\nexport default function getWindowScrollBarX(element) {\n  // If <html> has a CSS width greater than the viewport, then this will be\n  // incorrect for RTL.\n  // Popper 1 is broken in this case and never had a bug report so let's assume\n  // it's not an issue. I don't think anyone ever specifies width on <html>\n  // anyway.\n  // Browsers where the left scrollbar doesn't cause an issue report `0` for\n  // this (e.g. Edge 2019, IE11, Safari)\n  return getBoundingClientRect(getDocumentElement(element)).left + getWindowScroll(element).scrollLeft;\n}","import getComputedStyle from \"./getComputedStyle.js\";\nexport default function isScrollParent(element) {\n  // Firefox wants us to check `-x` and `-y` variations as well\n  var _getComputedStyle = getComputedStyle(element),\n      overflow = _getComputedStyle.overflow,\n      overflowX = _getComputedStyle.overflowX,\n      overflowY = _getComputedStyle.overflowY;\n\n  return /auto|scroll|overlay|hidden/.test(overflow + overflowY + overflowX);\n}","import getParentNode from \"./getParentNode.js\";\nimport isScrollParent from \"./isScrollParent.js\";\nimport getNodeName from \"./getNodeName.js\";\nimport { isHTMLElement } from \"./instanceOf.js\";\nexport default function getScrollParent(node) {\n  if (['html', 'body', '#document'].indexOf(getNodeName(node)) >= 0) {\n    // $FlowFixMe[incompatible-return]: assume body is always available\n    return node.ownerDocument.body;\n  }\n\n  if (isHTMLElement(node) && isScrollParent(node)) {\n    return node;\n  }\n\n  return getScrollParent(getParentNode(node));\n}","import getScrollParent from \"./getScrollParent.js\";\nimport getParentNode from \"./getParentNode.js\";\nimport getWindow from \"./getWindow.js\";\nimport isScrollParent from \"./isScrollParent.js\";\n/*\ngiven a DOM element, return the list of all scroll parents, up the list of ancesors\nuntil we get to the top window object. This list is what we attach scroll listeners\nto, because if any of these parent elements scroll, we'll need to re-calculate the\nreference element's position.\n*/\n\nexport default function listScrollParents(element, list) {\n  var _element$ownerDocumen;\n\n  if (list === void 0) {\n    list = [];\n  }\n\n  var scrollParent = getScrollParent(element);\n  var isBody = scrollParent === ((_element$ownerDocumen = element.ownerDocument) == null ? void 0 : _element$ownerDocumen.body);\n  var win = getWindow(scrollParent);\n  var target = isBody ? [win].concat(win.visualViewport || [], isScrollParent(scrollParent) ? scrollParent : []) : scrollParent;\n  var updatedList = list.concat(target);\n  return isBody ? updatedList : // $FlowFixMe[incompatible-call]: isBody tells us target will be an HTMLElement here\n  updatedList.concat(listScrollParents(getParentNode(target)));\n}","export default function rectToClientRect(rect) {\n  return Object.assign({}, rect, {\n    left: rect.x,\n    top: rect.y,\n    right: rect.x + rect.width,\n    bottom: rect.y + rect.height\n  });\n}","import { viewport } from \"../enums.js\";\nimport getViewportRect from \"./getViewportRect.js\";\nimport getDocumentRect from \"./getDocumentRect.js\";\nimport listScrollParents from \"./listScrollParents.js\";\nimport getOffsetParent from \"./getOffsetParent.js\";\nimport getDocumentElement from \"./getDocumentElement.js\";\nimport getComputedStyle from \"./getComputedStyle.js\";\nimport { isElement, isHTMLElement } from \"./instanceOf.js\";\nimport getBoundingClientRect from \"./getBoundingClientRect.js\";\nimport getParentNode from \"./getParentNode.js\";\nimport contains from \"./contains.js\";\nimport getNodeName from \"./getNodeName.js\";\nimport rectToClientRect from \"../utils/rectToClientRect.js\";\nimport { max, min } from \"../utils/math.js\";\n\nfunction getInnerBoundingClientRect(element, strategy) {\n  var rect = getBoundingClientRect(element, false, strategy === 'fixed');\n  rect.top = rect.top + element.clientTop;\n  rect.left = rect.left + element.clientLeft;\n  rect.bottom = rect.top + element.clientHeight;\n  rect.right = rect.left + element.clientWidth;\n  rect.width = element.clientWidth;\n  rect.height = element.clientHeight;\n  rect.x = rect.left;\n  rect.y = rect.top;\n  return rect;\n}\n\nfunction getClientRectFromMixedType(element, clippingParent, strategy) {\n  return clippingParent === viewport ? rectToClientRect(getViewportRect(element, strategy)) : isElement(clippingParent) ? getInnerBoundingClientRect(clippingParent, strategy) : rectToClientRect(getDocumentRect(getDocumentElement(element)));\n} // A \"clipping parent\" is an overflowable container with the characteristic of\n// clipping (or hiding) overflowing elements with a position different from\n// `initial`\n\n\nfunction getClippingParents(element) {\n  var clippingParents = listScrollParents(getParentNode(element));\n  var canEscapeClipping = ['absolute', 'fixed'].indexOf(getComputedStyle(element).position) >= 0;\n  var clipperElement = canEscapeClipping && isHTMLElement(element) ? getOffsetParent(element) : element;\n\n  if (!isElement(clipperElement)) {\n    return [];\n  } // $FlowFixMe[incompatible-return]: https://github.com/facebook/flow/issues/1414\n\n\n  return clippingParents.filter(function (clippingParent) {\n    return isElement(clippingParent) && contains(clippingParent, clipperElement) && getNodeName(clippingParent) !== 'body';\n  });\n} // Gets the maximum area that the element is visible in due to any number of\n// clipping parents\n\n\nexport default function getClippingRect(element, boundary, rootBoundary, strategy) {\n  var mainClippingParents = boundary === 'clippingParents' ? getClippingParents(element) : [].concat(boundary);\n  var clippingParents = [].concat(mainClippingParents, [rootBoundary]);\n  var firstClippingParent = clippingParents[0];\n  var clippingRect = clippingParents.reduce(function (accRect, clippingParent) {\n    var rect = getClientRectFromMixedType(element, clippingParent, strategy);\n    accRect.top = max(rect.top, accRect.top);\n    accRect.right = min(rect.right, accRect.right);\n    accRect.bottom = min(rect.bottom, accRect.bottom);\n    accRect.left = max(rect.left, accRect.left);\n    return accRect;\n  }, getClientRectFromMixedType(element, firstClippingParent, strategy));\n  clippingRect.width = clippingRect.right - clippingRect.left;\n  clippingRect.height = clippingRect.bottom - clippingRect.top;\n  clippingRect.x = clippingRect.left;\n  clippingRect.y = clippingRect.top;\n  return clippingRect;\n}","import getWindow from \"./getWindow.js\";\nimport getDocumentElement from \"./getDocumentElement.js\";\nimport getWindowScrollBarX from \"./getWindowScrollBarX.js\";\nimport isLayoutViewport from \"./isLayoutViewport.js\";\nexport default function getViewportRect(element, strategy) {\n  var win = getWindow(element);\n  var html = getDocumentElement(element);\n  var visualViewport = win.visualViewport;\n  var width = html.clientWidth;\n  var height = html.clientHeight;\n  var x = 0;\n  var y = 0;\n\n  if (visualViewport) {\n    width = visualViewport.width;\n    height = visualViewport.height;\n    var layoutViewport = isLayoutViewport();\n\n    if (layoutViewport || !layoutViewport && strategy === 'fixed') {\n      x = visualViewport.offsetLeft;\n      y = visualViewport.offsetTop;\n    }\n  }\n\n  return {\n    width: width,\n    height: height,\n    x: x + getWindowScrollBarX(element),\n    y: y\n  };\n}","import getDocumentElement from \"./getDocumentElement.js\";\nimport getComputedStyle from \"./getComputedStyle.js\";\nimport getWindowScrollBarX from \"./getWindowScrollBarX.js\";\nimport getWindowScroll from \"./getWindowScroll.js\";\nimport { max } from \"../utils/math.js\"; // Gets the entire size of the scrollable document area, even extending outside\n// of the `<html>` and `<body>` rect bounds if horizontally scrollable\n\nexport default function getDocumentRect(element) {\n  var _element$ownerDocumen;\n\n  var html = getDocumentElement(element);\n  var winScroll = getWindowScroll(element);\n  var body = (_element$ownerDocumen = element.ownerDocument) == null ? void 0 : _element$ownerDocumen.body;\n  var width = max(html.scrollWidth, html.clientWidth, body ? body.scrollWidth : 0, body ? body.clientWidth : 0);\n  var height = max(html.scrollHeight, html.clientHeight, body ? body.scrollHeight : 0, body ? body.clientHeight : 0);\n  var x = -winScroll.scrollLeft + getWindowScrollBarX(element);\n  var y = -winScroll.scrollTop;\n\n  if (getComputedStyle(body || html).direction === 'rtl') {\n    x += max(html.clientWidth, body ? body.clientWidth : 0) - width;\n  }\n\n  return {\n    width: width,\n    height: height,\n    x: x,\n    y: y\n  };\n}","import getBasePlacement from \"./getBasePlacement.js\";\nimport getVariation from \"./getVariation.js\";\nimport getMainAxisFromPlacement from \"./getMainAxisFromPlacement.js\";\nimport { top, right, bottom, left, start, end } from \"../enums.js\";\nexport default function computeOffsets(_ref) {\n  var reference = _ref.reference,\n      element = _ref.element,\n      placement = _ref.placement;\n  var basePlacement = placement ? getBasePlacement(placement) : null;\n  var variation = placement ? getVariation(placement) : null;\n  var commonX = reference.x + reference.width / 2 - element.width / 2;\n  var commonY = reference.y + reference.height / 2 - element.height / 2;\n  var offsets;\n\n  switch (basePlacement) {\n    case top:\n      offsets = {\n        x: commonX,\n        y: reference.y - element.height\n      };\n      break;\n\n    case bottom:\n      offsets = {\n        x: commonX,\n        y: reference.y + reference.height\n      };\n      break;\n\n    case right:\n      offsets = {\n        x: reference.x + reference.width,\n        y: commonY\n      };\n      break;\n\n    case left:\n      offsets = {\n        x: reference.x - element.width,\n        y: commonY\n      };\n      break;\n\n    default:\n      offsets = {\n        x: reference.x,\n        y: reference.y\n      };\n  }\n\n  var mainAxis = basePlacement ? getMainAxisFromPlacement(basePlacement) : null;\n\n  if (mainAxis != null) {\n    var len = mainAxis === 'y' ? 'height' : 'width';\n\n    switch (variation) {\n      case start:\n        offsets[mainAxis] = offsets[mainAxis] - (reference[len] / 2 - element[len] / 2);\n        break;\n\n      case end:\n        offsets[mainAxis] = offsets[mainAxis] + (reference[len] / 2 - element[len] / 2);\n        break;\n\n      default:\n    }\n  }\n\n  return offsets;\n}","import getClippingRect from \"../dom-utils/getClippingRect.js\";\nimport getDocumentElement from \"../dom-utils/getDocumentElement.js\";\nimport getBoundingClientRect from \"../dom-utils/getBoundingClientRect.js\";\nimport computeOffsets from \"./computeOffsets.js\";\nimport rectToClientRect from \"./rectToClientRect.js\";\nimport { clippingParents, reference, popper, bottom, top, right, basePlacements, viewport } from \"../enums.js\";\nimport { isElement } from \"../dom-utils/instanceOf.js\";\nimport mergePaddingObject from \"./mergePaddingObject.js\";\nimport expandToHashMap from \"./expandToHashMap.js\"; // eslint-disable-next-line import/no-unused-modules\n\nexport default function detectOverflow(state, options) {\n  if (options === void 0) {\n    options = {};\n  }\n\n  var _options = options,\n      _options$placement = _options.placement,\n      placement = _options$placement === void 0 ? state.placement : _options$placement,\n      _options$strategy = _options.strategy,\n      strategy = _options$strategy === void 0 ? state.strategy : _options$strategy,\n      _options$boundary = _options.boundary,\n      boundary = _options$boundary === void 0 ? clippingParents : _options$boundary,\n      _options$rootBoundary = _options.rootBoundary,\n      rootBoundary = _options$rootBoundary === void 0 ? viewport : _options$rootBoundary,\n      _options$elementConte = _options.elementContext,\n      elementContext = _options$elementConte === void 0 ? popper : _options$elementConte,\n      _options$altBoundary = _options.altBoundary,\n      altBoundary = _options$altBoundary === void 0 ? false : _options$altBoundary,\n      _options$padding = _options.padding,\n      padding = _options$padding === void 0 ? 0 : _options$padding;\n  var paddingObject = mergePaddingObject(typeof padding !== 'number' ? padding : expandToHashMap(padding, basePlacements));\n  var altContext = elementContext === popper ? reference : popper;\n  var popperRect = state.rects.popper;\n  var element = state.elements[altBoundary ? altContext : elementContext];\n  var clippingClientRect = getClippingRect(isElement(element) ? element : element.contextElement || getDocumentElement(state.elements.popper), boundary, rootBoundary, strategy);\n  var referenceClientRect = getBoundingClientRect(state.elements.reference);\n  var popperOffsets = computeOffsets({\n    reference: referenceClientRect,\n    element: popperRect,\n    strategy: 'absolute',\n    placement: placement\n  });\n  var popperClientRect = rectToClientRect(Object.assign({}, popperRect, popperOffsets));\n  var elementClientRect = elementContext === popper ? popperClientRect : referenceClientRect; // positive = overflowing the clipping rect\n  // 0 or negative = within the clipping rect\n\n  var overflowOffsets = {\n    top: clippingClientRect.top - elementClientRect.top + paddingObject.top,\n    bottom: elementClientRect.bottom - clippingClientRect.bottom + paddingObject.bottom,\n    left: clippingClientRect.left - elementClientRect.left + paddingObject.left,\n    right: elementClientRect.right - clippingClientRect.right + paddingObject.right\n  };\n  var offsetData = state.modifiersData.offset; // Offsets can be applied only to the popper element\n\n  if (elementContext === popper && offsetData) {\n    var offset = offsetData[placement];\n    Object.keys(overflowOffsets).forEach(function (key) {\n      var multiply = [right, bottom].indexOf(key) >= 0 ? 1 : -1;\n      var axis = [top, bottom].indexOf(key) >= 0 ? 'y' : 'x';\n      overflowOffsets[key] += offset[axis] * multiply;\n    });\n  }\n\n  return overflowOffsets;\n}","import getOppositePlacement from \"../utils/getOppositePlacement.js\";\nimport getBasePlacement from \"../utils/getBasePlacement.js\";\nimport getOppositeVariationPlacement from \"../utils/getOppositeVariationPlacement.js\";\nimport detectOverflow from \"../utils/detectOverflow.js\";\nimport computeAutoPlacement from \"../utils/computeAutoPlacement.js\";\nimport { bottom, top, start, right, left, auto } from \"../enums.js\";\nimport getVariation from \"../utils/getVariation.js\"; // eslint-disable-next-line import/no-unused-modules\n\nfunction getExpandedFallbackPlacements(placement) {\n  if (getBasePlacement(placement) === auto) {\n    return [];\n  }\n\n  var oppositePlacement = getOppositePlacement(placement);\n  return [getOppositeVariationPlacement(placement), oppositePlacement, getOppositeVariationPlacement(oppositePlacement)];\n}\n\nfunction flip(_ref) {\n  var state = _ref.state,\n      options = _ref.options,\n      name = _ref.name;\n\n  if (state.modifiersData[name]._skip) {\n    return;\n  }\n\n  var _options$mainAxis = options.mainAxis,\n      checkMainAxis = _options$mainAxis === void 0 ? true : _options$mainAxis,\n      _options$altAxis = options.altAxis,\n      checkAltAxis = _options$altAxis === void 0 ? true : _options$altAxis,\n      specifiedFallbackPlacements = options.fallbackPlacements,\n      padding = options.padding,\n      boundary = options.boundary,\n      rootBoundary = options.rootBoundary,\n      altBoundary = options.altBoundary,\n      _options$flipVariatio = options.flipVariations,\n      flipVariations = _options$flipVariatio === void 0 ? true : _options$flipVariatio,\n      allowedAutoPlacements = options.allowedAutoPlacements;\n  var preferredPlacement = state.options.placement;\n  var basePlacement = getBasePlacement(preferredPlacement);\n  var isBasePlacement = basePlacement === preferredPlacement;\n  var fallbackPlacements = specifiedFallbackPlacements || (isBasePlacement || !flipVariations ? [getOppositePlacement(preferredPlacement)] : getExpandedFallbackPlacements(preferredPlacement));\n  var placements = [preferredPlacement].concat(fallbackPlacements).reduce(function (acc, placement) {\n    return acc.concat(getBasePlacement(placement) === auto ? computeAutoPlacement(state, {\n      placement: placement,\n      boundary: boundary,\n      rootBoundary: rootBoundary,\n      padding: padding,\n      flipVariations: flipVariations,\n      allowedAutoPlacements: allowedAutoPlacements\n    }) : placement);\n  }, []);\n  var referenceRect = state.rects.reference;\n  var popperRect = state.rects.popper;\n  var checksMap = new Map();\n  var makeFallbackChecks = true;\n  var firstFittingPlacement = placements[0];\n\n  for (var i = 0; i < placements.length; i++) {\n    var placement = placements[i];\n\n    var _basePlacement = getBasePlacement(placement);\n\n    var isStartVariation = getVariation(placement) === start;\n    var isVertical = [top, bottom].indexOf(_basePlacement) >= 0;\n    var len = isVertical ? 'width' : 'height';\n    var overflow = detectOverflow(state, {\n      placement: placement,\n      boundary: boundary,\n      rootBoundary: rootBoundary,\n      altBoundary: altBoundary,\n      padding: padding\n    });\n    var mainVariationSide = isVertical ? isStartVariation ? right : left : isStartVariation ? bottom : top;\n\n    if (referenceRect[len] > popperRect[len]) {\n      mainVariationSide = getOppositePlacement(mainVariationSide);\n    }\n\n    var altVariationSide = getOppositePlacement(mainVariationSide);\n    var checks = [];\n\n    if (checkMainAxis) {\n      checks.push(overflow[_basePlacement] <= 0);\n    }\n\n    if (checkAltAxis) {\n      checks.push(overflow[mainVariationSide] <= 0, overflow[altVariationSide] <= 0);\n    }\n\n    if (checks.every(function (check) {\n      return check;\n    })) {\n      firstFittingPlacement = placement;\n      makeFallbackChecks = false;\n      break;\n    }\n\n    checksMap.set(placement, checks);\n  }\n\n  if (makeFallbackChecks) {\n    // `2` may be desired in some cases – research later\n    var numberOfChecks = flipVariations ? 3 : 1;\n\n    var _loop = function _loop(_i) {\n      var fittingPlacement = placements.find(function (placement) {\n        var checks = checksMap.get(placement);\n\n        if (checks) {\n          return checks.slice(0, _i).every(function (check) {\n            return check;\n          });\n        }\n      });\n\n      if (fittingPlacement) {\n        firstFittingPlacement = fittingPlacement;\n        return \"break\";\n      }\n    };\n\n    for (var _i = numberOfChecks; _i > 0; _i--) {\n      var _ret = _loop(_i);\n\n      if (_ret === \"break\") break;\n    }\n  }\n\n  if (state.placement !== firstFittingPlacement) {\n    state.modifiersData[name]._skip = true;\n    state.placement = firstFittingPlacement;\n    state.reset = true;\n  }\n} // eslint-disable-next-line import/no-unused-modules\n\n\nexport default {\n  name: 'flip',\n  enabled: true,\n  phase: 'main',\n  fn: flip,\n  requiresIfExists: ['offset'],\n  data: {\n    _skip: false\n  }\n};","import getVariation from \"./getVariation.js\";\nimport { variationPlacements, basePlacements, placements as allPlacements } from \"../enums.js\";\nimport detectOverflow from \"./detectOverflow.js\";\nimport getBasePlacement from \"./getBasePlacement.js\";\nexport default function computeAutoPlacement(state, options) {\n  if (options === void 0) {\n    options = {};\n  }\n\n  var _options = options,\n      placement = _options.placement,\n      boundary = _options.boundary,\n      rootBoundary = _options.rootBoundary,\n      padding = _options.padding,\n      flipVariations = _options.flipVariations,\n      _options$allowedAutoP = _options.allowedAutoPlacements,\n      allowedAutoPlacements = _options$allowedAutoP === void 0 ? allPlacements : _options$allowedAutoP;\n  var variation = getVariation(placement);\n  var placements = variation ? flipVariations ? variationPlacements : variationPlacements.filter(function (placement) {\n    return getVariation(placement) === variation;\n  }) : basePlacements;\n  var allowedPlacements = placements.filter(function (placement) {\n    return allowedAutoPlacements.indexOf(placement) >= 0;\n  });\n\n  if (allowedPlacements.length === 0) {\n    allowedPlacements = placements;\n  } // $FlowFixMe[incompatible-type]: Flow seems to have problems with two array unions...\n\n\n  var overflows = allowedPlacements.reduce(function (acc, placement) {\n    acc[placement] = detectOverflow(state, {\n      placement: placement,\n      boundary: boundary,\n      rootBoundary: rootBoundary,\n      padding: padding\n    })[getBasePlacement(placement)];\n    return acc;\n  }, {});\n  return Object.keys(overflows).sort(function (a, b) {\n    return overflows[a] - overflows[b];\n  });\n}","import { top, bottom, left, right } from \"../enums.js\";\nimport detectOverflow from \"../utils/detectOverflow.js\";\n\nfunction getSideOffsets(overflow, rect, preventedOffsets) {\n  if (preventedOffsets === void 0) {\n    preventedOffsets = {\n      x: 0,\n      y: 0\n    };\n  }\n\n  return {\n    top: overflow.top - rect.height - preventedOffsets.y,\n    right: overflow.right - rect.width + preventedOffsets.x,\n    bottom: overflow.bottom - rect.height + preventedOffsets.y,\n    left: overflow.left - rect.width - preventedOffsets.x\n  };\n}\n\nfunction isAnySideFullyClipped(overflow) {\n  return [top, right, bottom, left].some(function (side) {\n    return overflow[side] >= 0;\n  });\n}\n\nfunction hide(_ref) {\n  var state = _ref.state,\n      name = _ref.name;\n  var referenceRect = state.rects.reference;\n  var popperRect = state.rects.popper;\n  var preventedOffsets = state.modifiersData.preventOverflow;\n  var referenceOverflow = detectOverflow(state, {\n    elementContext: 'reference'\n  });\n  var popperAltOverflow = detectOverflow(state, {\n    altBoundary: true\n  });\n  var referenceClippingOffsets = getSideOffsets(referenceOverflow, referenceRect);\n  var popperEscapeOffsets = getSideOffsets(popperAltOverflow, popperRect, preventedOffsets);\n  var isReferenceHidden = isAnySideFullyClipped(referenceClippingOffsets);\n  var hasPopperEscaped = isAnySideFullyClipped(popperEscapeOffsets);\n  state.modifiersData[name] = {\n    referenceClippingOffsets: referenceClippingOffsets,\n    popperEscapeOffsets: popperEscapeOffsets,\n    isReferenceHidden: isReferenceHidden,\n    hasPopperEscaped: hasPopperEscaped\n  };\n  state.attributes.popper = Object.assign({}, state.attributes.popper, {\n    'data-popper-reference-hidden': isReferenceHidden,\n    'data-popper-escaped': hasPopperEscaped\n  });\n} // eslint-disable-next-line import/no-unused-modules\n\n\nexport default {\n  name: 'hide',\n  enabled: true,\n  phase: 'main',\n  requiresIfExists: ['preventOverflow'],\n  fn: hide\n};","import getBasePlacement from \"../utils/getBasePlacement.js\";\nimport { top, left, right, placements } from \"../enums.js\"; // eslint-disable-next-line import/no-unused-modules\n\nexport function distanceAndSkiddingToXY(placement, rects, offset) {\n  var basePlacement = getBasePlacement(placement);\n  var invertDistance = [left, top].indexOf(basePlacement) >= 0 ? -1 : 1;\n\n  var _ref = typeof offset === 'function' ? offset(Object.assign({}, rects, {\n    placement: placement\n  })) : offset,\n      skidding = _ref[0],\n      distance = _ref[1];\n\n  skidding = skidding || 0;\n  distance = (distance || 0) * invertDistance;\n  return [left, right].indexOf(basePlacement) >= 0 ? {\n    x: distance,\n    y: skidding\n  } : {\n    x: skidding,\n    y: distance\n  };\n}\n\nfunction offset(_ref2) {\n  var state = _ref2.state,\n      options = _ref2.options,\n      name = _ref2.name;\n  var _options$offset = options.offset,\n      offset = _options$offset === void 0 ? [0, 0] : _options$offset;\n  var data = placements.reduce(function (acc, placement) {\n    acc[placement] = distanceAndSkiddingToXY(placement, state.rects, offset);\n    return acc;\n  }, {});\n  var _data$state$placement = data[state.placement],\n      x = _data$state$placement.x,\n      y = _data$state$placement.y;\n\n  if (state.modifiersData.popperOffsets != null) {\n    state.modifiersData.popperOffsets.x += x;\n    state.modifiersData.popperOffsets.y += y;\n  }\n\n  state.modifiersData[name] = data;\n} // eslint-disable-next-line import/no-unused-modules\n\n\nexport default {\n  name: 'offset',\n  enabled: true,\n  phase: 'main',\n  requires: ['popperOffsets'],\n  fn: offset\n};","import computeOffsets from \"../utils/computeOffsets.js\";\n\nfunction popperOffsets(_ref) {\n  var state = _ref.state,\n      name = _ref.name;\n  // Offsets are the actual position the popper needs to have to be\n  // properly positioned near its reference element\n  // This is the most basic placement, and will be adjusted by\n  // the modifiers in the next step\n  state.modifiersData[name] = computeOffsets({\n    reference: state.rects.reference,\n    element: state.rects.popper,\n    strategy: 'absolute',\n    placement: state.placement\n  });\n} // eslint-disable-next-line import/no-unused-modules\n\n\nexport default {\n  name: 'popperOffsets',\n  enabled: true,\n  phase: 'read',\n  fn: popperOffsets,\n  data: {}\n};","import { top, left, right, bottom, start } from \"../enums.js\";\nimport getBasePlacement from \"../utils/getBasePlacement.js\";\nimport getMainAxisFromPlacement from \"../utils/getMainAxisFromPlacement.js\";\nimport getAltAxis from \"../utils/getAltAxis.js\";\nimport { within, withinMaxClamp } from \"../utils/within.js\";\nimport getLayoutRect from \"../dom-utils/getLayoutRect.js\";\nimport getOffsetParent from \"../dom-utils/getOffsetParent.js\";\nimport detectOverflow from \"../utils/detectOverflow.js\";\nimport getVariation from \"../utils/getVariation.js\";\nimport getFreshSideObject from \"../utils/getFreshSideObject.js\";\nimport { min as mathMin, max as mathMax } from \"../utils/math.js\";\n\nfunction preventOverflow(_ref) {\n  var state = _ref.state,\n      options = _ref.options,\n      name = _ref.name;\n  var _options$mainAxis = options.mainAxis,\n      checkMainAxis = _options$mainAxis === void 0 ? true : _options$mainAxis,\n      _options$altAxis = options.altAxis,\n      checkAltAxis = _options$altAxis === void 0 ? false : _options$altAxis,\n      boundary = options.boundary,\n      rootBoundary = options.rootBoundary,\n      altBoundary = options.altBoundary,\n      padding = options.padding,\n      _options$tether = options.tether,\n      tether = _options$tether === void 0 ? true : _options$tether,\n      _options$tetherOffset = options.tetherOffset,\n      tetherOffset = _options$tetherOffset === void 0 ? 0 : _options$tetherOffset;\n  var overflow = detectOverflow(state, {\n    boundary: boundary,\n    rootBoundary: rootBoundary,\n    padding: padding,\n    altBoundary: altBoundary\n  });\n  var basePlacement = getBasePlacement(state.placement);\n  var variation = getVariation(state.placement);\n  var isBasePlacement = !variation;\n  var mainAxis = getMainAxisFromPlacement(basePlacement);\n  var altAxis = getAltAxis(mainAxis);\n  var popperOffsets = state.modifiersData.popperOffsets;\n  var referenceRect = state.rects.reference;\n  var popperRect = state.rects.popper;\n  var tetherOffsetValue = typeof tetherOffset === 'function' ? tetherOffset(Object.assign({}, state.rects, {\n    placement: state.placement\n  })) : tetherOffset;\n  var normalizedTetherOffsetValue = typeof tetherOffsetValue === 'number' ? {\n    mainAxis: tetherOffsetValue,\n    altAxis: tetherOffsetValue\n  } : Object.assign({\n    mainAxis: 0,\n    altAxis: 0\n  }, tetherOffsetValue);\n  var offsetModifierState = state.modifiersData.offset ? state.modifiersData.offset[state.placement] : null;\n  var data = {\n    x: 0,\n    y: 0\n  };\n\n  if (!popperOffsets) {\n    return;\n  }\n\n  if (checkMainAxis) {\n    var _offsetModifierState$;\n\n    var mainSide = mainAxis === 'y' ? top : left;\n    var altSide = mainAxis === 'y' ? bottom : right;\n    var len = mainAxis === 'y' ? 'height' : 'width';\n    var offset = popperOffsets[mainAxis];\n    var min = offset + overflow[mainSide];\n    var max = offset - overflow[altSide];\n    var additive = tether ? -popperRect[len] / 2 : 0;\n    var minLen = variation === start ? referenceRect[len] : popperRect[len];\n    var maxLen = variation === start ? -popperRect[len] : -referenceRect[len]; // We need to include the arrow in the calculation so the arrow doesn't go\n    // outside the reference bounds\n\n    var arrowElement = state.elements.arrow;\n    var arrowRect = tether && arrowElement ? getLayoutRect(arrowElement) : {\n      width: 0,\n      height: 0\n    };\n    var arrowPaddingObject = state.modifiersData['arrow#persistent'] ? state.modifiersData['arrow#persistent'].padding : getFreshSideObject();\n    var arrowPaddingMin = arrowPaddingObject[mainSide];\n    var arrowPaddingMax = arrowPaddingObject[altSide]; // If the reference length is smaller than the arrow length, we don't want\n    // to include its full size in the calculation. If the reference is small\n    // and near the edge of a boundary, the popper can overflow even if the\n    // reference is not overflowing as well (e.g. virtual elements with no\n    // width or height)\n\n    var arrowLen = within(0, referenceRect[len], arrowRect[len]);\n    var minOffset = isBasePlacement ? referenceRect[len] / 2 - additive - arrowLen - arrowPaddingMin - normalizedTetherOffsetValue.mainAxis : minLen - arrowLen - arrowPaddingMin - normalizedTetherOffsetValue.mainAxis;\n    var maxOffset = isBasePlacement ? -referenceRect[len] / 2 + additive + arrowLen + arrowPaddingMax + normalizedTetherOffsetValue.mainAxis : maxLen + arrowLen + arrowPaddingMax + normalizedTetherOffsetValue.mainAxis;\n    var arrowOffsetParent = state.elements.arrow && getOffsetParent(state.elements.arrow);\n    var clientOffset = arrowOffsetParent ? mainAxis === 'y' ? arrowOffsetParent.clientTop || 0 : arrowOffsetParent.clientLeft || 0 : 0;\n    var offsetModifierValue = (_offsetModifierState$ = offsetModifierState == null ? void 0 : offsetModifierState[mainAxis]) != null ? _offsetModifierState$ : 0;\n    var tetherMin = offset + minOffset - offsetModifierValue - clientOffset;\n    var tetherMax = offset + maxOffset - offsetModifierValue;\n    var preventedOffset = within(tether ? mathMin(min, tetherMin) : min, offset, tether ? mathMax(max, tetherMax) : max);\n    popperOffsets[mainAxis] = preventedOffset;\n    data[mainAxis] = preventedOffset - offset;\n  }\n\n  if (checkAltAxis) {\n    var _offsetModifierState$2;\n\n    var _mainSide = mainAxis === 'x' ? top : left;\n\n    var _altSide = mainAxis === 'x' ? bottom : right;\n\n    var _offset = popperOffsets[altAxis];\n\n    var _len = altAxis === 'y' ? 'height' : 'width';\n\n    var _min = _offset + overflow[_mainSide];\n\n    var _max = _offset - overflow[_altSide];\n\n    var isOriginSide = [top, left].indexOf(basePlacement) !== -1;\n\n    var _offsetModifierValue = (_offsetModifierState$2 = offsetModifierState == null ? void 0 : offsetModifierState[altAxis]) != null ? _offsetModifierState$2 : 0;\n\n    var _tetherMin = isOriginSide ? _min : _offset - referenceRect[_len] - popperRect[_len] - _offsetModifierValue + normalizedTetherOffsetValue.altAxis;\n\n    var _tetherMax = isOriginSide ? _offset + referenceRect[_len] + popperRect[_len] - _offsetModifierValue - normalizedTetherOffsetValue.altAxis : _max;\n\n    var _preventedOffset = tether && isOriginSide ? withinMaxClamp(_tetherMin, _offset, _tetherMax) : within(tether ? _tetherMin : _min, _offset, tether ? _tetherMax : _max);\n\n    popperOffsets[altAxis] = _preventedOffset;\n    data[altAxis] = _preventedOffset - _offset;\n  }\n\n  state.modifiersData[name] = data;\n} // eslint-disable-next-line import/no-unused-modules\n\n\nexport default {\n  name: 'preventOverflow',\n  enabled: true,\n  phase: 'main',\n  fn: preventOverflow,\n  requiresIfExists: ['offset']\n};","export default function getAltAxis(axis) {\n  return axis === 'x' ? 'y' : 'x';\n}","import getBoundingClientRect from \"./getBoundingClientRect.js\";\nimport getNodeScroll from \"./getNodeScroll.js\";\nimport getNodeName from \"./getNodeName.js\";\nimport { isHTMLElement } from \"./instanceOf.js\";\nimport getWindowScrollBarX from \"./getWindowScrollBarX.js\";\nimport getDocumentElement from \"./getDocumentElement.js\";\nimport isScrollParent from \"./isScrollParent.js\";\nimport { round } from \"../utils/math.js\";\n\nfunction isElementScaled(element) {\n  var rect = element.getBoundingClientRect();\n  var scaleX = round(rect.width) / element.offsetWidth || 1;\n  var scaleY = round(rect.height) / element.offsetHeight || 1;\n  return scaleX !== 1 || scaleY !== 1;\n} // Returns the composite rect of an element relative to its offsetParent.\n// Composite means it takes into account transforms as well as layout.\n\n\nexport default function getCompositeRect(elementOrVirtualElement, offsetParent, isFixed) {\n  if (isFixed === void 0) {\n    isFixed = false;\n  }\n\n  var isOffsetParentAnElement = isHTMLElement(offsetParent);\n  var offsetParentIsScaled = isHTMLElement(offsetParent) && isElementScaled(offsetParent);\n  var documentElement = getDocumentElement(offsetParent);\n  var rect = getBoundingClientRect(elementOrVirtualElement, offsetParentIsScaled, isFixed);\n  var scroll = {\n    scrollLeft: 0,\n    scrollTop: 0\n  };\n  var offsets = {\n    x: 0,\n    y: 0\n  };\n\n  if (isOffsetParentAnElement || !isOffsetParentAnElement && !isFixed) {\n    if (getNodeName(offsetParent) !== 'body' || // https://github.com/popperjs/popper-core/issues/1078\n    isScrollParent(documentElement)) {\n      scroll = getNodeScroll(offsetParent);\n    }\n\n    if (isHTMLElement(offsetParent)) {\n      offsets = getBoundingClientRect(offsetParent, true);\n      offsets.x += offsetParent.clientLeft;\n      offsets.y += offsetParent.clientTop;\n    } else if (documentElement) {\n      offsets.x = getWindowScrollBarX(documentElement);\n    }\n  }\n\n  return {\n    x: rect.left + scroll.scrollLeft - offsets.x,\n    y: rect.top + scroll.scrollTop - offsets.y,\n    width: rect.width,\n    height: rect.height\n  };\n}","import getWindowScroll from \"./getWindowScroll.js\";\nimport getWindow from \"./getWindow.js\";\nimport { isHTMLElement } from \"./instanceOf.js\";\nimport getHTMLElementScroll from \"./getHTMLElementScroll.js\";\nexport default function getNodeScroll(node) {\n  if (node === getWindow(node) || !isHTMLElement(node)) {\n    return getWindowScroll(node);\n  } else {\n    return getHTMLElementScroll(node);\n  }\n}","export default function getHTMLElementScroll(element) {\n  return {\n    scrollLeft: element.scrollLeft,\n    scrollTop: element.scrollTop\n  };\n}","import { modifierPhases } from \"../enums.js\"; // source: https://stackoverflow.com/questions/49875255\n\nfunction order(modifiers) {\n  var map = new Map();\n  var visited = new Set();\n  var result = [];\n  modifiers.forEach(function (modifier) {\n    map.set(modifier.name, modifier);\n  }); // On visiting object, check for its dependencies and visit them recursively\n\n  function sort(modifier) {\n    visited.add(modifier.name);\n    var requires = [].concat(modifier.requires || [], modifier.requiresIfExists || []);\n    requires.forEach(function (dep) {\n      if (!visited.has(dep)) {\n        var depModifier = map.get(dep);\n\n        if (depModifier) {\n          sort(depModifier);\n        }\n      }\n    });\n    result.push(modifier);\n  }\n\n  modifiers.forEach(function (modifier) {\n    if (!visited.has(modifier.name)) {\n      // check for visited object\n      sort(modifier);\n    }\n  });\n  return result;\n}\n\nexport default function orderModifiers(modifiers) {\n  // order based on dependencies\n  var orderedModifiers = order(modifiers); // order based on phase\n\n  return modifierPhases.reduce(function (acc, phase) {\n    return acc.concat(orderedModifiers.filter(function (modifier) {\n      return modifier.phase === phase;\n    }));\n  }, []);\n}","import getCompositeRect from \"./dom-utils/getCompositeRect.js\";\nimport getLayoutRect from \"./dom-utils/getLayoutRect.js\";\nimport listScrollParents from \"./dom-utils/listScrollParents.js\";\nimport getOffsetParent from \"./dom-utils/getOffsetParent.js\";\nimport orderModifiers from \"./utils/orderModifiers.js\";\nimport debounce from \"./utils/debounce.js\";\nimport mergeByName from \"./utils/mergeByName.js\";\nimport detectOverflow from \"./utils/detectOverflow.js\";\nimport { isElement } from \"./dom-utils/instanceOf.js\";\nvar DEFAULT_OPTIONS = {\n  placement: 'bottom',\n  modifiers: [],\n  strategy: 'absolute'\n};\n\nfunction areValidElements() {\n  for (var _len = arguments.length, args = new Array(_len), _key = 0; _key < _len; _key++) {\n    args[_key] = arguments[_key];\n  }\n\n  return !args.some(function (element) {\n    return !(element && typeof element.getBoundingClientRect === 'function');\n  });\n}\n\nexport function popperGenerator(generatorOptions) {\n  if (generatorOptions === void 0) {\n    generatorOptions = {};\n  }\n\n  var _generatorOptions = generatorOptions,\n      _generatorOptions$def = _generatorOptions.defaultModifiers,\n      defaultModifiers = _generatorOptions$def === void 0 ? [] : _generatorOptions$def,\n      _generatorOptions$def2 = _generatorOptions.defaultOptions,\n      defaultOptions = _generatorOptions$def2 === void 0 ? DEFAULT_OPTIONS : _generatorOptions$def2;\n  return function createPopper(reference, popper, options) {\n    if (options === void 0) {\n      options = defaultOptions;\n    }\n\n    var state = {\n      placement: 'bottom',\n      orderedModifiers: [],\n      options: Object.assign({}, DEFAULT_OPTIONS, defaultOptions),\n      modifiersData: {},\n      elements: {\n        reference: reference,\n        popper: popper\n      },\n      attributes: {},\n      styles: {}\n    };\n    var effectCleanupFns = [];\n    var isDestroyed = false;\n    var instance = {\n      state: state,\n      setOptions: function setOptions(setOptionsAction) {\n        var options = typeof setOptionsAction === 'function' ? setOptionsAction(state.options) : setOptionsAction;\n        cleanupModifierEffects();\n        state.options = Object.assign({}, defaultOptions, state.options, options);\n        state.scrollParents = {\n          reference: isElement(reference) ? listScrollParents(reference) : reference.contextElement ? listScrollParents(reference.contextElement) : [],\n          popper: listScrollParents(popper)\n        }; // Orders the modifiers based on their dependencies and `phase`\n        // properties\n\n        var orderedModifiers = orderModifiers(mergeByName([].concat(defaultModifiers, state.options.modifiers))); // Strip out disabled modifiers\n\n        state.orderedModifiers = orderedModifiers.filter(function (m) {\n          return m.enabled;\n        });\n        runModifierEffects();\n        return instance.update();\n      },\n      // Sync update – it will always be executed, even if not necessary. This\n      // is useful for low frequency updates where sync behavior simplifies the\n      // logic.\n      // For high frequency updates (e.g. `resize` and `scroll` events), always\n      // prefer the async Popper#update method\n      forceUpdate: function forceUpdate() {\n        if (isDestroyed) {\n          return;\n        }\n\n        var _state$elements = state.elements,\n            reference = _state$elements.reference,\n            popper = _state$elements.popper; // Don't proceed if `reference` or `popper` are not valid elements\n        // anymore\n\n        if (!areValidElements(reference, popper)) {\n          return;\n        } // Store the reference and popper rects to be read by modifiers\n\n\n        state.rects = {\n          reference: getCompositeRect(reference, getOffsetParent(popper), state.options.strategy === 'fixed'),\n          popper: getLayoutRect(popper)\n        }; // Modifiers have the ability to reset the current update cycle. The\n        // most common use case for this is the `flip` modifier changing the\n        // placement, which then needs to re-run all the modifiers, because the\n        // logic was previously ran for the previous placement and is therefore\n        // stale/incorrect\n\n        state.reset = false;\n        state.placement = state.options.placement; // On each update cycle, the `modifiersData` property for each modifier\n        // is filled with the initial data specified by the modifier. This means\n        // it doesn't persist and is fresh on each update.\n        // To ensure persistent data, use `${name}#persistent`\n\n        state.orderedModifiers.forEach(function (modifier) {\n          return state.modifiersData[modifier.name] = Object.assign({}, modifier.data);\n        });\n\n        for (var index = 0; index < state.orderedModifiers.length; index++) {\n          if (state.reset === true) {\n            state.reset = false;\n            index = -1;\n            continue;\n          }\n\n          var _state$orderedModifie = state.orderedModifiers[index],\n              fn = _state$orderedModifie.fn,\n              _state$orderedModifie2 = _state$orderedModifie.options,\n              _options = _state$orderedModifie2 === void 0 ? {} : _state$orderedModifie2,\n              name = _state$orderedModifie.name;\n\n          if (typeof fn === 'function') {\n            state = fn({\n              state: state,\n              options: _options,\n              name: name,\n              instance: instance\n            }) || state;\n          }\n        }\n      },\n      // Async and optimistically optimized update – it will not be executed if\n      // not necessary (debounced to run at most once-per-tick)\n      update: debounce(function () {\n        return new Promise(function (resolve) {\n          instance.forceUpdate();\n          resolve(state);\n        });\n      }),\n      destroy: function destroy() {\n        cleanupModifierEffects();\n        isDestroyed = true;\n      }\n    };\n\n    if (!areValidElements(reference, popper)) {\n      return instance;\n    }\n\n    instance.setOptions(options).then(function (state) {\n      if (!isDestroyed && options.onFirstUpdate) {\n        options.onFirstUpdate(state);\n      }\n    }); // Modifiers have the ability to execute arbitrary code before the first\n    // update cycle runs. They will be executed in the same order as the update\n    // cycle. This is useful when a modifier adds some persistent data that\n    // other modifiers need to use, but the modifier is run after the dependent\n    // one.\n\n    function runModifierEffects() {\n      state.orderedModifiers.forEach(function (_ref) {\n        var name = _ref.name,\n            _ref$options = _ref.options,\n            options = _ref$options === void 0 ? {} : _ref$options,\n            effect = _ref.effect;\n\n        if (typeof effect === 'function') {\n          var cleanupFn = effect({\n            state: state,\n            name: name,\n            instance: instance,\n            options: options\n          });\n\n          var noopFn = function noopFn() {};\n\n          effectCleanupFns.push(cleanupFn || noopFn);\n        }\n      });\n    }\n\n    function cleanupModifierEffects() {\n      effectCleanupFns.forEach(function (fn) {\n        return fn();\n      });\n      effectCleanupFns = [];\n    }\n\n    return instance;\n  };\n}\nexport var createPopper = /*#__PURE__*/popperGenerator(); // eslint-disable-next-line import/no-unused-modules\n\nexport { detectOverflow };","export default function debounce(fn) {\n  var pending;\n  return function () {\n    if (!pending) {\n      pending = new Promise(function (resolve) {\n        Promise.resolve().then(function () {\n          pending = undefined;\n          resolve(fn());\n        });\n      });\n    }\n\n    return pending;\n  };\n}","export default function mergeByName(modifiers) {\n  var merged = modifiers.reduce(function (merged, current) {\n    var existing = merged[current.name];\n    merged[current.name] = existing ? Object.assign({}, existing, current, {\n      options: Object.assign({}, existing.options, current.options),\n      data: Object.assign({}, existing.data, current.data)\n    }) : current;\n    return merged;\n  }, {}); // IE11 does not support Object.values\n\n  return Object.keys(merged).map(function (key) {\n    return merged[key];\n  });\n}","import { popperGenerator, detectOverflow } from \"./createPopper.js\";\nimport eventListeners from \"./modifiers/eventListeners.js\";\nimport popperOffsets from \"./modifiers/popperOffsets.js\";\nimport computeStyles from \"./modifiers/computeStyles.js\";\nimport applyStyles from \"./modifiers/applyStyles.js\";\nimport offset from \"./modifiers/offset.js\";\nimport flip from \"./modifiers/flip.js\";\nimport preventOverflow from \"./modifiers/preventOverflow.js\";\nimport arrow from \"./modifiers/arrow.js\";\nimport hide from \"./modifiers/hide.js\";\nvar defaultModifiers = [eventListeners, popperOffsets, computeStyles, applyStyles, offset, flip, preventOverflow, arrow, hide];\nvar createPopper = /*#__PURE__*/popperGenerator({\n  defaultModifiers: defaultModifiers\n}); // eslint-disable-next-line import/no-unused-modules\n\nexport { createPopper, popperGenerator, defaultModifiers, detectOverflow }; // eslint-disable-next-line import/no-unused-modules\n\nexport { createPopper as createPopperLite } from \"./popper-lite.js\"; // eslint-disable-next-line import/no-unused-modules\n\nexport * from \"./modifiers/index.js\";","import { popperGenerator, detectOverflow } from \"./createPopper.js\";\nimport eventListeners from \"./modifiers/eventListeners.js\";\nimport popperOffsets from \"./modifiers/popperOffsets.js\";\nimport computeStyles from \"./modifiers/computeStyles.js\";\nimport applyStyles from \"./modifiers/applyStyles.js\";\nvar defaultModifiers = [eventListeners, popperOffsets, computeStyles, applyStyles];\nvar createPopper = /*#__PURE__*/popperGenerator({\n  defaultModifiers: defaultModifiers\n}); // eslint-disable-next-line import/no-unused-modules\n\nexport { createPopper, popperGenerator, defaultModifiers, detectOverflow };","/*!\n  * Bootstrap v5.3.2 (https://getbootstrap.com/)\n  * Copyright 2011-2023 The Bootstrap Authors (https://github.com/twbs/bootstrap/graphs/contributors)\n  * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n  */\nimport * as Popper from '@popperjs/core';\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap dom/data.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n/**\n * Constants\n */\n\nconst elementMap = new Map();\nconst Data = {\n  set(element, key, instance) {\n    if (!elementMap.has(element)) {\n      elementMap.set(element, new Map());\n    }\n    const instanceMap = elementMap.get(element);\n\n    // make it clear we only want one instance per element\n    // can be removed later when multiple key/instances are fine to be used\n    if (!instanceMap.has(key) && instanceMap.size !== 0) {\n      // eslint-disable-next-line no-console\n      console.error(`Bootstrap doesn't allow more than one instance per element. Bound instance: ${Array.from(instanceMap.keys())[0]}.`);\n      return;\n    }\n    instanceMap.set(key, instance);\n  },\n  get(element, key) {\n    if (elementMap.has(element)) {\n      return elementMap.get(element).get(key) || null;\n    }\n    return null;\n  },\n  remove(element, key) {\n    if (!elementMap.has(element)) {\n      return;\n    }\n    const instanceMap = elementMap.get(element);\n    instanceMap.delete(key);\n\n    // free up element references if there are no instances left for an element\n    if (instanceMap.size === 0) {\n      elementMap.delete(element);\n    }\n  }\n};\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap util/index.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\nconst MAX_UID = 1000000;\nconst MILLISECONDS_MULTIPLIER = 1000;\nconst TRANSITION_END = 'transitionend';\n\n/**\n * Properly escape IDs selectors to handle weird IDs\n * @param {string} selector\n * @returns {string}\n */\nconst parseSelector = selector => {\n  if (selector && window.CSS && window.CSS.escape) {\n    // document.querySelector needs escaping to handle IDs (html5+) containing for instance /\n    selector = selector.replace(/#([^\\s\"#']+)/g, (match, id) => `#${CSS.escape(id)}`);\n  }\n  return selector;\n};\n\n// Shout-out Angus Croll (https://goo.gl/pxwQGp)\nconst toType = object => {\n  if (object === null || object === undefined) {\n    return `${object}`;\n  }\n  return Object.prototype.toString.call(object).match(/\\s([a-z]+)/i)[1].toLowerCase();\n};\n\n/**\n * Public Util API\n */\n\nconst getUID = prefix => {\n  do {\n    prefix += Math.floor(Math.random() * MAX_UID);\n  } while (document.getElementById(prefix));\n  return prefix;\n};\nconst getTransitionDurationFromElement = element => {\n  if (!element) {\n    return 0;\n  }\n\n  // Get transition-duration of the element\n  let {\n    transitionDuration,\n    transitionDelay\n  } = window.getComputedStyle(element);\n  const floatTransitionDuration = Number.parseFloat(transitionDuration);\n  const floatTransitionDelay = Number.parseFloat(transitionDelay);\n\n  // Return 0 if element or transition duration is not found\n  if (!floatTransitionDuration && !floatTransitionDelay) {\n    return 0;\n  }\n\n  // If multiple durations are defined, take the first\n  transitionDuration = transitionDuration.split(',')[0];\n  transitionDelay = transitionDelay.split(',')[0];\n  return (Number.parseFloat(transitionDuration) + Number.parseFloat(transitionDelay)) * MILLISECONDS_MULTIPLIER;\n};\nconst triggerTransitionEnd = element => {\n  element.dispatchEvent(new Event(TRANSITION_END));\n};\nconst isElement = object => {\n  if (!object || typeof object !== 'object') {\n    return false;\n  }\n  if (typeof object.jquery !== 'undefined') {\n    object = object[0];\n  }\n  return typeof object.nodeType !== 'undefined';\n};\nconst getElement = object => {\n  // it's a jQuery object or a node element\n  if (isElement(object)) {\n    return object.jquery ? object[0] : object;\n  }\n  if (typeof object === 'string' && object.length > 0) {\n    return document.querySelector(parseSelector(object));\n  }\n  return null;\n};\nconst isVisible = element => {\n  if (!isElement(element) || element.getClientRects().length === 0) {\n    return false;\n  }\n  const elementIsVisible = getComputedStyle(element).getPropertyValue('visibility') === 'visible';\n  // Handle `details` element as its content may falsie appear visible when it is closed\n  const closedDetails = element.closest('details:not([open])');\n  if (!closedDetails) {\n    return elementIsVisible;\n  }\n  if (closedDetails !== element) {\n    const summary = element.closest('summary');\n    if (summary && summary.parentNode !== closedDetails) {\n      return false;\n    }\n    if (summary === null) {\n      return false;\n    }\n  }\n  return elementIsVisible;\n};\nconst isDisabled = element => {\n  if (!element || element.nodeType !== Node.ELEMENT_NODE) {\n    return true;\n  }\n  if (element.classList.contains('disabled')) {\n    return true;\n  }\n  if (typeof element.disabled !== 'undefined') {\n    return element.disabled;\n  }\n  return element.hasAttribute('disabled') && element.getAttribute('disabled') !== 'false';\n};\nconst findShadowRoot = element => {\n  if (!document.documentElement.attachShadow) {\n    return null;\n  }\n\n  // Can find the shadow root otherwise it'll return the document\n  if (typeof element.getRootNode === 'function') {\n    const root = element.getRootNode();\n    return root instanceof ShadowRoot ? root : null;\n  }\n  if (element instanceof ShadowRoot) {\n    return element;\n  }\n\n  // when we don't find a shadow root\n  if (!element.parentNode) {\n    return null;\n  }\n  return findShadowRoot(element.parentNode);\n};\nconst noop = () => {};\n\n/**\n * Trick to restart an element's animation\n *\n * @param {HTMLElement} element\n * @return void\n *\n * @see https://www.charistheo.io/blog/2021/02/restart-a-css-animation-with-javascript/#restarting-a-css-animation\n */\nconst reflow = element => {\n  element.offsetHeight; // eslint-disable-line no-unused-expressions\n};\n\nconst getjQuery = () => {\n  if (window.jQuery && !document.body.hasAttribute('data-bs-no-jquery')) {\n    return window.jQuery;\n  }\n  return null;\n};\nconst DOMContentLoadedCallbacks = [];\nconst onDOMContentLoaded = callback => {\n  if (document.readyState === 'loading') {\n    // add listener on the first call when the document is in loading state\n    if (!DOMContentLoadedCallbacks.length) {\n      document.addEventListener('DOMContentLoaded', () => {\n        for (const callback of DOMContentLoadedCallbacks) {\n          callback();\n        }\n      });\n    }\n    DOMContentLoadedCallbacks.push(callback);\n  } else {\n    callback();\n  }\n};\nconst isRTL = () => document.documentElement.dir === 'rtl';\nconst defineJQueryPlugin = plugin => {\n  onDOMContentLoaded(() => {\n    const $ = getjQuery();\n    /* istanbul ignore if */\n    if ($) {\n      const name = plugin.NAME;\n      const JQUERY_NO_CONFLICT = $.fn[name];\n      $.fn[name] = plugin.jQueryInterface;\n      $.fn[name].Constructor = plugin;\n      $.fn[name].noConflict = () => {\n        $.fn[name] = JQUERY_NO_CONFLICT;\n        return plugin.jQueryInterface;\n      };\n    }\n  });\n};\nconst execute = (possibleCallback, args = [], defaultValue = possibleCallback) => {\n  return typeof possibleCallback === 'function' ? possibleCallback(...args) : defaultValue;\n};\nconst executeAfterTransition = (callback, transitionElement, waitForTransition = true) => {\n  if (!waitForTransition) {\n    execute(callback);\n    return;\n  }\n  const durationPadding = 5;\n  const emulatedDuration = getTransitionDurationFromElement(transitionElement) + durationPadding;\n  let called = false;\n  const handler = ({\n    target\n  }) => {\n    if (target !== transitionElement) {\n      return;\n    }\n    called = true;\n    transitionElement.removeEventListener(TRANSITION_END, handler);\n    execute(callback);\n  };\n  transitionElement.addEventListener(TRANSITION_END, handler);\n  setTimeout(() => {\n    if (!called) {\n      triggerTransitionEnd(transitionElement);\n    }\n  }, emulatedDuration);\n};\n\n/**\n * Return the previous/next element of a list.\n *\n * @param {array} list    The list of elements\n * @param activeElement   The active element\n * @param shouldGetNext   Choose to get next or previous element\n * @param isCycleAllowed\n * @return {Element|elem} The proper element\n */\nconst getNextActiveElement = (list, activeElement, shouldGetNext, isCycleAllowed) => {\n  const listLength = list.length;\n  let index = list.indexOf(activeElement);\n\n  // if the element does not exist in the list return an element\n  // depending on the direction and if cycle is allowed\n  if (index === -1) {\n    return !shouldGetNext && isCycleAllowed ? list[listLength - 1] : list[0];\n  }\n  index += shouldGetNext ? 1 : -1;\n  if (isCycleAllowed) {\n    index = (index + listLength) % listLength;\n  }\n  return list[Math.max(0, Math.min(index, listLength - 1))];\n};\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap dom/event-handler.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst namespaceRegex = /[^.]*(?=\\..*)\\.|.*/;\nconst stripNameRegex = /\\..*/;\nconst stripUidRegex = /::\\d+$/;\nconst eventRegistry = {}; // Events storage\nlet uidEvent = 1;\nconst customEvents = {\n  mouseenter: 'mouseover',\n  mouseleave: 'mouseout'\n};\nconst nativeEvents = new Set(['click', 'dblclick', 'mouseup', 'mousedown', 'contextmenu', 'mousewheel', 'DOMMouseScroll', 'mouseover', 'mouseout', 'mousemove', 'selectstart', 'selectend', 'keydown', 'keypress', 'keyup', 'orientationchange', 'touchstart', 'touchmove', 'touchend', 'touchcancel', 'pointerdown', 'pointermove', 'pointerup', 'pointerleave', 'pointercancel', 'gesturestart', 'gesturechange', 'gestureend', 'focus', 'blur', 'change', 'reset', 'select', 'submit', 'focusin', 'focusout', 'load', 'unload', 'beforeunload', 'resize', 'move', 'DOMContentLoaded', 'readystatechange', 'error', 'abort', 'scroll']);\n\n/**\n * Private methods\n */\n\nfunction makeEventUid(element, uid) {\n  return uid && `${uid}::${uidEvent++}` || element.uidEvent || uidEvent++;\n}\nfunction getElementEvents(element) {\n  const uid = makeEventUid(element);\n  element.uidEvent = uid;\n  eventRegistry[uid] = eventRegistry[uid] || {};\n  return eventRegistry[uid];\n}\nfunction bootstrapHandler(element, fn) {\n  return function handler(event) {\n    hydrateObj(event, {\n      delegateTarget: element\n    });\n    if (handler.oneOff) {\n      EventHandler.off(element, event.type, fn);\n    }\n    return fn.apply(element, [event]);\n  };\n}\nfunction bootstrapDelegationHandler(element, selector, fn) {\n  return function handler(event) {\n    const domElements = element.querySelectorAll(selector);\n    for (let {\n      target\n    } = event; target && target !== this; target = target.parentNode) {\n      for (const domElement of domElements) {\n        if (domElement !== target) {\n          continue;\n        }\n        hydrateObj(event, {\n          delegateTarget: target\n        });\n        if (handler.oneOff) {\n          EventHandler.off(element, event.type, selector, fn);\n        }\n        return fn.apply(target, [event]);\n      }\n    }\n  };\n}\nfunction findHandler(events, callable, delegationSelector = null) {\n  return Object.values(events).find(event => event.callable === callable && event.delegationSelector === delegationSelector);\n}\nfunction normalizeParameters(originalTypeEvent, handler, delegationFunction) {\n  const isDelegated = typeof handler === 'string';\n  // TODO: tooltip passes `false` instead of selector, so we need to check\n  const callable = isDelegated ? delegationFunction : handler || delegationFunction;\n  let typeEvent = getTypeEvent(originalTypeEvent);\n  if (!nativeEvents.has(typeEvent)) {\n    typeEvent = originalTypeEvent;\n  }\n  return [isDelegated, callable, typeEvent];\n}\nfunction addHandler(element, originalTypeEvent, handler, delegationFunction, oneOff) {\n  if (typeof originalTypeEvent !== 'string' || !element) {\n    return;\n  }\n  let [isDelegated, callable, typeEvent] = normalizeParameters(originalTypeEvent, handler, delegationFunction);\n\n  // in case of mouseenter or mouseleave wrap the handler within a function that checks for its DOM position\n  // this prevents the handler from being dispatched the same way as mouseover or mouseout does\n  if (originalTypeEvent in customEvents) {\n    const wrapFunction = fn => {\n      return function (event) {\n        if (!event.relatedTarget || event.relatedTarget !== event.delegateTarget && !event.delegateTarget.contains(event.relatedTarget)) {\n          return fn.call(this, event);\n        }\n      };\n    };\n    callable = wrapFunction(callable);\n  }\n  const events = getElementEvents(element);\n  const handlers = events[typeEvent] || (events[typeEvent] = {});\n  const previousFunction = findHandler(handlers, callable, isDelegated ? handler : null);\n  if (previousFunction) {\n    previousFunction.oneOff = previousFunction.oneOff && oneOff;\n    return;\n  }\n  const uid = makeEventUid(callable, originalTypeEvent.replace(namespaceRegex, ''));\n  const fn = isDelegated ? bootstrapDelegationHandler(element, handler, callable) : bootstrapHandler(element, callable);\n  fn.delegationSelector = isDelegated ? handler : null;\n  fn.callable = callable;\n  fn.oneOff = oneOff;\n  fn.uidEvent = uid;\n  handlers[uid] = fn;\n  element.addEventListener(typeEvent, fn, isDelegated);\n}\nfunction removeHandler(element, events, typeEvent, handler, delegationSelector) {\n  const fn = findHandler(events[typeEvent], handler, delegationSelector);\n  if (!fn) {\n    return;\n  }\n  element.removeEventListener(typeEvent, fn, Boolean(delegationSelector));\n  delete events[typeEvent][fn.uidEvent];\n}\nfunction removeNamespacedHandlers(element, events, typeEvent, namespace) {\n  const storeElementEvent = events[typeEvent] || {};\n  for (const [handlerKey, event] of Object.entries(storeElementEvent)) {\n    if (handlerKey.includes(namespace)) {\n      removeHandler(element, events, typeEvent, event.callable, event.delegationSelector);\n    }\n  }\n}\nfunction getTypeEvent(event) {\n  // allow to get the native events from namespaced events ('click.bs.button' --> 'click')\n  event = event.replace(stripNameRegex, '');\n  return customEvents[event] || event;\n}\nconst EventHandler = {\n  on(element, event, handler, delegationFunction) {\n    addHandler(element, event, handler, delegationFunction, false);\n  },\n  one(element, event, handler, delegationFunction) {\n    addHandler(element, event, handler, delegationFunction, true);\n  },\n  off(element, originalTypeEvent, handler, delegationFunction) {\n    if (typeof originalTypeEvent !== 'string' || !element) {\n      return;\n    }\n    const [isDelegated, callable, typeEvent] = normalizeParameters(originalTypeEvent, handler, delegationFunction);\n    const inNamespace = typeEvent !== originalTypeEvent;\n    const events = getElementEvents(element);\n    const storeElementEvent = events[typeEvent] || {};\n    const isNamespace = originalTypeEvent.startsWith('.');\n    if (typeof callable !== 'undefined') {\n      // Simplest case: handler is passed, remove that listener ONLY.\n      if (!Object.keys(storeElementEvent).length) {\n        return;\n      }\n      removeHandler(element, events, typeEvent, callable, isDelegated ? handler : null);\n      return;\n    }\n    if (isNamespace) {\n      for (const elementEvent of Object.keys(events)) {\n        removeNamespacedHandlers(element, events, elementEvent, originalTypeEvent.slice(1));\n      }\n    }\n    for (const [keyHandlers, event] of Object.entries(storeElementEvent)) {\n      const handlerKey = keyHandlers.replace(stripUidRegex, '');\n      if (!inNamespace || originalTypeEvent.includes(handlerKey)) {\n        removeHandler(element, events, typeEvent, event.callable, event.delegationSelector);\n      }\n    }\n  },\n  trigger(element, event, args) {\n    if (typeof event !== 'string' || !element) {\n      return null;\n    }\n    const $ = getjQuery();\n    const typeEvent = getTypeEvent(event);\n    const inNamespace = event !== typeEvent;\n    let jQueryEvent = null;\n    let bubbles = true;\n    let nativeDispatch = true;\n    let defaultPrevented = false;\n    if (inNamespace && $) {\n      jQueryEvent = $.Event(event, args);\n      $(element).trigger(jQueryEvent);\n      bubbles = !jQueryEvent.isPropagationStopped();\n      nativeDispatch = !jQueryEvent.isImmediatePropagationStopped();\n      defaultPrevented = jQueryEvent.isDefaultPrevented();\n    }\n    const evt = hydrateObj(new Event(event, {\n      bubbles,\n      cancelable: true\n    }), args);\n    if (defaultPrevented) {\n      evt.preventDefault();\n    }\n    if (nativeDispatch) {\n      element.dispatchEvent(evt);\n    }\n    if (evt.defaultPrevented && jQueryEvent) {\n      jQueryEvent.preventDefault();\n    }\n    return evt;\n  }\n};\nfunction hydrateObj(obj, meta = {}) {\n  for (const [key, value] of Object.entries(meta)) {\n    try {\n      obj[key] = value;\n    } catch (_unused) {\n      Object.defineProperty(obj, key, {\n        configurable: true,\n        get() {\n          return value;\n        }\n      });\n    }\n  }\n  return obj;\n}\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap dom/manipulator.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\nfunction normalizeData(value) {\n  if (value === 'true') {\n    return true;\n  }\n  if (value === 'false') {\n    return false;\n  }\n  if (value === Number(value).toString()) {\n    return Number(value);\n  }\n  if (value === '' || value === 'null') {\n    return null;\n  }\n  if (typeof value !== 'string') {\n    return value;\n  }\n  try {\n    return JSON.parse(decodeURIComponent(value));\n  } catch (_unused) {\n    return value;\n  }\n}\nfunction normalizeDataKey(key) {\n  return key.replace(/[A-Z]/g, chr => `-${chr.toLowerCase()}`);\n}\nconst Manipulator = {\n  setDataAttribute(element, key, value) {\n    element.setAttribute(`data-bs-${normalizeDataKey(key)}`, value);\n  },\n  removeDataAttribute(element, key) {\n    element.removeAttribute(`data-bs-${normalizeDataKey(key)}`);\n  },\n  getDataAttributes(element) {\n    if (!element) {\n      return {};\n    }\n    const attributes = {};\n    const bsKeys = Object.keys(element.dataset).filter(key => key.startsWith('bs') && !key.startsWith('bsConfig'));\n    for (const key of bsKeys) {\n      let pureKey = key.replace(/^bs/, '');\n      pureKey = pureKey.charAt(0).toLowerCase() + pureKey.slice(1, pureKey.length);\n      attributes[pureKey] = normalizeData(element.dataset[key]);\n    }\n    return attributes;\n  },\n  getDataAttribute(element, key) {\n    return normalizeData(element.getAttribute(`data-bs-${normalizeDataKey(key)}`));\n  }\n};\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap util/config.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Class definition\n */\n\nclass Config {\n  // Getters\n  static get Default() {\n    return {};\n  }\n  static get DefaultType() {\n    return {};\n  }\n  static get NAME() {\n    throw new Error('You have to implement the static method \"NAME\", for each component!');\n  }\n  _getConfig(config) {\n    config = this._mergeConfigObj(config);\n    config = this._configAfterMerge(config);\n    this._typeCheckConfig(config);\n    return config;\n  }\n  _configAfterMerge(config) {\n    return config;\n  }\n  _mergeConfigObj(config, element) {\n    const jsonConfig = isElement(element) ? Manipulator.getDataAttribute(element, 'config') : {}; // try to parse\n\n    return {\n      ...this.constructor.Default,\n      ...(typeof jsonConfig === 'object' ? jsonConfig : {}),\n      ...(isElement(element) ? Manipulator.getDataAttributes(element) : {}),\n      ...(typeof config === 'object' ? config : {})\n    };\n  }\n  _typeCheckConfig(config, configTypes = this.constructor.DefaultType) {\n    for (const [property, expectedTypes] of Object.entries(configTypes)) {\n      const value = config[property];\n      const valueType = isElement(value) ? 'element' : toType(value);\n      if (!new RegExp(expectedTypes).test(valueType)) {\n        throw new TypeError(`${this.constructor.NAME.toUpperCase()}: Option \"${property}\" provided type \"${valueType}\" but expected type \"${expectedTypes}\".`);\n      }\n    }\n  }\n}\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap base-component.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst VERSION = '5.3.2';\n\n/**\n * Class definition\n */\n\nclass BaseComponent extends Config {\n  constructor(element, config) {\n    super();\n    element = getElement(element);\n    if (!element) {\n      return;\n    }\n    this._element = element;\n    this._config = this._getConfig(config);\n    Data.set(this._element, this.constructor.DATA_KEY, this);\n  }\n\n  // Public\n  dispose() {\n    Data.remove(this._element, this.constructor.DATA_KEY);\n    EventHandler.off(this._element, this.constructor.EVENT_KEY);\n    for (const propertyName of Object.getOwnPropertyNames(this)) {\n      this[propertyName] = null;\n    }\n  }\n  _queueCallback(callback, element, isAnimated = true) {\n    executeAfterTransition(callback, element, isAnimated);\n  }\n  _getConfig(config) {\n    config = this._mergeConfigObj(config, this._element);\n    config = this._configAfterMerge(config);\n    this._typeCheckConfig(config);\n    return config;\n  }\n\n  // Static\n  static getInstance(element) {\n    return Data.get(getElement(element), this.DATA_KEY);\n  }\n  static getOrCreateInstance(element, config = {}) {\n    return this.getInstance(element) || new this(element, typeof config === 'object' ? config : null);\n  }\n  static get VERSION() {\n    return VERSION;\n  }\n  static get DATA_KEY() {\n    return `bs.${this.NAME}`;\n  }\n  static get EVENT_KEY() {\n    return `.${this.DATA_KEY}`;\n  }\n  static eventName(name) {\n    return `${name}${this.EVENT_KEY}`;\n  }\n}\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap dom/selector-engine.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\nconst getSelector = element => {\n  let selector = element.getAttribute('data-bs-target');\n  if (!selector || selector === '#') {\n    let hrefAttribute = element.getAttribute('href');\n\n    // The only valid content that could double as a selector are IDs or classes,\n    // so everything starting with `#` or `.`. If a \"real\" URL is used as the selector,\n    // `document.querySelector` will rightfully complain it is invalid.\n    // See https://github.com/twbs/bootstrap/issues/32273\n    if (!hrefAttribute || !hrefAttribute.includes('#') && !hrefAttribute.startsWith('.')) {\n      return null;\n    }\n\n    // Just in case some CMS puts out a full URL with the anchor appended\n    if (hrefAttribute.includes('#') && !hrefAttribute.startsWith('#')) {\n      hrefAttribute = `#${hrefAttribute.split('#')[1]}`;\n    }\n    selector = hrefAttribute && hrefAttribute !== '#' ? parseSelector(hrefAttribute.trim()) : null;\n  }\n  return selector;\n};\nconst SelectorEngine = {\n  find(selector, element = document.documentElement) {\n    return [].concat(...Element.prototype.querySelectorAll.call(element, selector));\n  },\n  findOne(selector, element = document.documentElement) {\n    return Element.prototype.querySelector.call(element, selector);\n  },\n  children(element, selector) {\n    return [].concat(...element.children).filter(child => child.matches(selector));\n  },\n  parents(element, selector) {\n    const parents = [];\n    let ancestor = element.parentNode.closest(selector);\n    while (ancestor) {\n      parents.push(ancestor);\n      ancestor = ancestor.parentNode.closest(selector);\n    }\n    return parents;\n  },\n  prev(element, selector) {\n    let previous = element.previousElementSibling;\n    while (previous) {\n      if (previous.matches(selector)) {\n        return [previous];\n      }\n      previous = previous.previousElementSibling;\n    }\n    return [];\n  },\n  // TODO: this is now unused; remove later along with prev()\n  next(element, selector) {\n    let next = element.nextElementSibling;\n    while (next) {\n      if (next.matches(selector)) {\n        return [next];\n      }\n      next = next.nextElementSibling;\n    }\n    return [];\n  },\n  focusableChildren(element) {\n    const focusables = ['a', 'button', 'input', 'textarea', 'select', 'details', '[tabindex]', '[contenteditable=\"true\"]'].map(selector => `${selector}:not([tabindex^=\"-\"])`).join(',');\n    return this.find(focusables, element).filter(el => !isDisabled(el) && isVisible(el));\n  },\n  getSelectorFromElement(element) {\n    const selector = getSelector(element);\n    if (selector) {\n      return SelectorEngine.findOne(selector) ? selector : null;\n    }\n    return null;\n  },\n  getElementFromSelector(element) {\n    const selector = getSelector(element);\n    return selector ? SelectorEngine.findOne(selector) : null;\n  },\n  getMultipleElementsFromSelector(element) {\n    const selector = getSelector(element);\n    return selector ? SelectorEngine.find(selector) : [];\n  }\n};\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap util/component-functions.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\nconst enableDismissTrigger = (component, method = 'hide') => {\n  const clickEvent = `click.dismiss${component.EVENT_KEY}`;\n  const name = component.NAME;\n  EventHandler.on(document, clickEvent, `[data-bs-dismiss=\"${name}\"]`, function (event) {\n    if (['A', 'AREA'].includes(this.tagName)) {\n      event.preventDefault();\n    }\n    if (isDisabled(this)) {\n      return;\n    }\n    const target = SelectorEngine.getElementFromSelector(this) || this.closest(`.${name}`);\n    const instance = component.getOrCreateInstance(target);\n\n    // Method argument is left, for Alert and only, as it doesn't implement the 'hide' method\n    instance[method]();\n  });\n};\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap alert.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$f = 'alert';\nconst DATA_KEY$a = 'bs.alert';\nconst EVENT_KEY$b = `.${DATA_KEY$a}`;\nconst EVENT_CLOSE = `close${EVENT_KEY$b}`;\nconst EVENT_CLOSED = `closed${EVENT_KEY$b}`;\nconst CLASS_NAME_FADE$5 = 'fade';\nconst CLASS_NAME_SHOW$8 = 'show';\n\n/**\n * Class definition\n */\n\nclass Alert extends BaseComponent {\n  // Getters\n  static get NAME() {\n    return NAME$f;\n  }\n\n  // Public\n  close() {\n    const closeEvent = EventHandler.trigger(this._element, EVENT_CLOSE);\n    if (closeEvent.defaultPrevented) {\n      return;\n    }\n    this._element.classList.remove(CLASS_NAME_SHOW$8);\n    const isAnimated = this._element.classList.contains(CLASS_NAME_FADE$5);\n    this._queueCallback(() => this._destroyElement(), this._element, isAnimated);\n  }\n\n  // Private\n  _destroyElement() {\n    this._element.remove();\n    EventHandler.trigger(this._element, EVENT_CLOSED);\n    this.dispose();\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    return this.each(function () {\n      const data = Alert.getOrCreateInstance(this);\n      if (typeof config !== 'string') {\n        return;\n      }\n      if (data[config] === undefined || config.startsWith('_') || config === 'constructor') {\n        throw new TypeError(`No method named \"${config}\"`);\n      }\n      data[config](this);\n    });\n  }\n}\n\n/**\n * Data API implementation\n */\n\nenableDismissTrigger(Alert, 'close');\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Alert);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap button.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$e = 'button';\nconst DATA_KEY$9 = 'bs.button';\nconst EVENT_KEY$a = `.${DATA_KEY$9}`;\nconst DATA_API_KEY$6 = '.data-api';\nconst CLASS_NAME_ACTIVE$3 = 'active';\nconst SELECTOR_DATA_TOGGLE$5 = '[data-bs-toggle=\"button\"]';\nconst EVENT_CLICK_DATA_API$6 = `click${EVENT_KEY$a}${DATA_API_KEY$6}`;\n\n/**\n * Class definition\n */\n\nclass Button extends BaseComponent {\n  // Getters\n  static get NAME() {\n    return NAME$e;\n  }\n\n  // Public\n  toggle() {\n    // Toggle class and sync the `aria-pressed` attribute with the return value of the `.toggle()` method\n    this._element.setAttribute('aria-pressed', this._element.classList.toggle(CLASS_NAME_ACTIVE$3));\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    return this.each(function () {\n      const data = Button.getOrCreateInstance(this);\n      if (config === 'toggle') {\n        data[config]();\n      }\n    });\n  }\n}\n\n/**\n * Data API implementation\n */\n\nEventHandler.on(document, EVENT_CLICK_DATA_API$6, SELECTOR_DATA_TOGGLE$5, event => {\n  event.preventDefault();\n  const button = event.target.closest(SELECTOR_DATA_TOGGLE$5);\n  const data = Button.getOrCreateInstance(button);\n  data.toggle();\n});\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Button);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap util/swipe.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$d = 'swipe';\nconst EVENT_KEY$9 = '.bs.swipe';\nconst EVENT_TOUCHSTART = `touchstart${EVENT_KEY$9}`;\nconst EVENT_TOUCHMOVE = `touchmove${EVENT_KEY$9}`;\nconst EVENT_TOUCHEND = `touchend${EVENT_KEY$9}`;\nconst EVENT_POINTERDOWN = `pointerdown${EVENT_KEY$9}`;\nconst EVENT_POINTERUP = `pointerup${EVENT_KEY$9}`;\nconst POINTER_TYPE_TOUCH = 'touch';\nconst POINTER_TYPE_PEN = 'pen';\nconst CLASS_NAME_POINTER_EVENT = 'pointer-event';\nconst SWIPE_THRESHOLD = 40;\nconst Default$c = {\n  endCallback: null,\n  leftCallback: null,\n  rightCallback: null\n};\nconst DefaultType$c = {\n  endCallback: '(function|null)',\n  leftCallback: '(function|null)',\n  rightCallback: '(function|null)'\n};\n\n/**\n * Class definition\n */\n\nclass Swipe extends Config {\n  constructor(element, config) {\n    super();\n    this._element = element;\n    if (!element || !Swipe.isSupported()) {\n      return;\n    }\n    this._config = this._getConfig(config);\n    this._deltaX = 0;\n    this._supportPointerEvents = Boolean(window.PointerEvent);\n    this._initEvents();\n  }\n\n  // Getters\n  static get Default() {\n    return Default$c;\n  }\n  static get DefaultType() {\n    return DefaultType$c;\n  }\n  static get NAME() {\n    return NAME$d;\n  }\n\n  // Public\n  dispose() {\n    EventHandler.off(this._element, EVENT_KEY$9);\n  }\n\n  // Private\n  _start(event) {\n    if (!this._supportPointerEvents) {\n      this._deltaX = event.touches[0].clientX;\n      return;\n    }\n    if (this._eventIsPointerPenTouch(event)) {\n      this._deltaX = event.clientX;\n    }\n  }\n  _end(event) {\n    if (this._eventIsPointerPenTouch(event)) {\n      this._deltaX = event.clientX - this._deltaX;\n    }\n    this._handleSwipe();\n    execute(this._config.endCallback);\n  }\n  _move(event) {\n    this._deltaX = event.touches && event.touches.length > 1 ? 0 : event.touches[0].clientX - this._deltaX;\n  }\n  _handleSwipe() {\n    const absDeltaX = Math.abs(this._deltaX);\n    if (absDeltaX <= SWIPE_THRESHOLD) {\n      return;\n    }\n    const direction = absDeltaX / this._deltaX;\n    this._deltaX = 0;\n    if (!direction) {\n      return;\n    }\n    execute(direction > 0 ? this._config.rightCallback : this._config.leftCallback);\n  }\n  _initEvents() {\n    if (this._supportPointerEvents) {\n      EventHandler.on(this._element, EVENT_POINTERDOWN, event => this._start(event));\n      EventHandler.on(this._element, EVENT_POINTERUP, event => this._end(event));\n      this._element.classList.add(CLASS_NAME_POINTER_EVENT);\n    } else {\n      EventHandler.on(this._element, EVENT_TOUCHSTART, event => this._start(event));\n      EventHandler.on(this._element, EVENT_TOUCHMOVE, event => this._move(event));\n      EventHandler.on(this._element, EVENT_TOUCHEND, event => this._end(event));\n    }\n  }\n  _eventIsPointerPenTouch(event) {\n    return this._supportPointerEvents && (event.pointerType === POINTER_TYPE_PEN || event.pointerType === POINTER_TYPE_TOUCH);\n  }\n\n  // Static\n  static isSupported() {\n    return 'ontouchstart' in document.documentElement || navigator.maxTouchPoints > 0;\n  }\n}\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap carousel.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$c = 'carousel';\nconst DATA_KEY$8 = 'bs.carousel';\nconst EVENT_KEY$8 = `.${DATA_KEY$8}`;\nconst DATA_API_KEY$5 = '.data-api';\nconst ARROW_LEFT_KEY$1 = 'ArrowLeft';\nconst ARROW_RIGHT_KEY$1 = 'ArrowRight';\nconst TOUCHEVENT_COMPAT_WAIT = 500; // Time for mouse compat events to fire after touch\n\nconst ORDER_NEXT = 'next';\nconst ORDER_PREV = 'prev';\nconst DIRECTION_LEFT = 'left';\nconst DIRECTION_RIGHT = 'right';\nconst EVENT_SLIDE = `slide${EVENT_KEY$8}`;\nconst EVENT_SLID = `slid${EVENT_KEY$8}`;\nconst EVENT_KEYDOWN$1 = `keydown${EVENT_KEY$8}`;\nconst EVENT_MOUSEENTER$1 = `mouseenter${EVENT_KEY$8}`;\nconst EVENT_MOUSELEAVE$1 = `mouseleave${EVENT_KEY$8}`;\nconst EVENT_DRAG_START = `dragstart${EVENT_KEY$8}`;\nconst EVENT_LOAD_DATA_API$3 = `load${EVENT_KEY$8}${DATA_API_KEY$5}`;\nconst EVENT_CLICK_DATA_API$5 = `click${EVENT_KEY$8}${DATA_API_KEY$5}`;\nconst CLASS_NAME_CAROUSEL = 'carousel';\nconst CLASS_NAME_ACTIVE$2 = 'active';\nconst CLASS_NAME_SLIDE = 'slide';\nconst CLASS_NAME_END = 'carousel-item-end';\nconst CLASS_NAME_START = 'carousel-item-start';\nconst CLASS_NAME_NEXT = 'carousel-item-next';\nconst CLASS_NAME_PREV = 'carousel-item-prev';\nconst SELECTOR_ACTIVE = '.active';\nconst SELECTOR_ITEM = '.carousel-item';\nconst SELECTOR_ACTIVE_ITEM = SELECTOR_ACTIVE + SELECTOR_ITEM;\nconst SELECTOR_ITEM_IMG = '.carousel-item img';\nconst SELECTOR_INDICATORS = '.carousel-indicators';\nconst SELECTOR_DATA_SLIDE = '[data-bs-slide], [data-bs-slide-to]';\nconst SELECTOR_DATA_RIDE = '[data-bs-ride=\"carousel\"]';\nconst KEY_TO_DIRECTION = {\n  [ARROW_LEFT_KEY$1]: DIRECTION_RIGHT,\n  [ARROW_RIGHT_KEY$1]: DIRECTION_LEFT\n};\nconst Default$b = {\n  interval: 5000,\n  keyboard: true,\n  pause: 'hover',\n  ride: false,\n  touch: true,\n  wrap: true\n};\nconst DefaultType$b = {\n  interval: '(number|boolean)',\n  // TODO:v6 remove boolean support\n  keyboard: 'boolean',\n  pause: '(string|boolean)',\n  ride: '(boolean|string)',\n  touch: 'boolean',\n  wrap: 'boolean'\n};\n\n/**\n * Class definition\n */\n\nclass Carousel extends BaseComponent {\n  constructor(element, config) {\n    super(element, config);\n    this._interval = null;\n    this._activeElement = null;\n    this._isSliding = false;\n    this.touchTimeout = null;\n    this._swipeHelper = null;\n    this._indicatorsElement = SelectorEngine.findOne(SELECTOR_INDICATORS, this._element);\n    this._addEventListeners();\n    if (this._config.ride === CLASS_NAME_CAROUSEL) {\n      this.cycle();\n    }\n  }\n\n  // Getters\n  static get Default() {\n    return Default$b;\n  }\n  static get DefaultType() {\n    return DefaultType$b;\n  }\n  static get NAME() {\n    return NAME$c;\n  }\n\n  // Public\n  next() {\n    this._slide(ORDER_NEXT);\n  }\n  nextWhenVisible() {\n    // FIXME TODO use `document.visibilityState`\n    // Don't call next when the page isn't visible\n    // or the carousel or its parent isn't visible\n    if (!document.hidden && isVisible(this._element)) {\n      this.next();\n    }\n  }\n  prev() {\n    this._slide(ORDER_PREV);\n  }\n  pause() {\n    if (this._isSliding) {\n      triggerTransitionEnd(this._element);\n    }\n    this._clearInterval();\n  }\n  cycle() {\n    this._clearInterval();\n    this._updateInterval();\n    this._interval = setInterval(() => this.nextWhenVisible(), this._config.interval);\n  }\n  _maybeEnableCycle() {\n    if (!this._config.ride) {\n      return;\n    }\n    if (this._isSliding) {\n      EventHandler.one(this._element, EVENT_SLID, () => this.cycle());\n      return;\n    }\n    this.cycle();\n  }\n  to(index) {\n    const items = this._getItems();\n    if (index > items.length - 1 || index < 0) {\n      return;\n    }\n    if (this._isSliding) {\n      EventHandler.one(this._element, EVENT_SLID, () => this.to(index));\n      return;\n    }\n    const activeIndex = this._getItemIndex(this._getActive());\n    if (activeIndex === index) {\n      return;\n    }\n    const order = index > activeIndex ? ORDER_NEXT : ORDER_PREV;\n    this._slide(order, items[index]);\n  }\n  dispose() {\n    if (this._swipeHelper) {\n      this._swipeHelper.dispose();\n    }\n    super.dispose();\n  }\n\n  // Private\n  _configAfterMerge(config) {\n    config.defaultInterval = config.interval;\n    return config;\n  }\n  _addEventListeners() {\n    if (this._config.keyboard) {\n      EventHandler.on(this._element, EVENT_KEYDOWN$1, event => this._keydown(event));\n    }\n    if (this._config.pause === 'hover') {\n      EventHandler.on(this._element, EVENT_MOUSEENTER$1, () => this.pause());\n      EventHandler.on(this._element, EVENT_MOUSELEAVE$1, () => this._maybeEnableCycle());\n    }\n    if (this._config.touch && Swipe.isSupported()) {\n      this._addTouchEventListeners();\n    }\n  }\n  _addTouchEventListeners() {\n    for (const img of SelectorEngine.find(SELECTOR_ITEM_IMG, this._element)) {\n      EventHandler.on(img, EVENT_DRAG_START, event => event.preventDefault());\n    }\n    const endCallBack = () => {\n      if (this._config.pause !== 'hover') {\n        return;\n      }\n\n      // If it's a touch-enabled device, mouseenter/leave are fired as\n      // part of the mouse compatibility events on first tap - the carousel\n      // would stop cycling until user tapped out of it;\n      // here, we listen for touchend, explicitly pause the carousel\n      // (as if it's the second time we tap on it, mouseenter compat event\n      // is NOT fired) and after a timeout (to allow for mouse compatibility\n      // events to fire) we explicitly restart cycling\n\n      this.pause();\n      if (this.touchTimeout) {\n        clearTimeout(this.touchTimeout);\n      }\n      this.touchTimeout = setTimeout(() => this._maybeEnableCycle(), TOUCHEVENT_COMPAT_WAIT + this._config.interval);\n    };\n    const swipeConfig = {\n      leftCallback: () => this._slide(this._directionToOrder(DIRECTION_LEFT)),\n      rightCallback: () => this._slide(this._directionToOrder(DIRECTION_RIGHT)),\n      endCallback: endCallBack\n    };\n    this._swipeHelper = new Swipe(this._element, swipeConfig);\n  }\n  _keydown(event) {\n    if (/input|textarea/i.test(event.target.tagName)) {\n      return;\n    }\n    const direction = KEY_TO_DIRECTION[event.key];\n    if (direction) {\n      event.preventDefault();\n      this._slide(this._directionToOrder(direction));\n    }\n  }\n  _getItemIndex(element) {\n    return this._getItems().indexOf(element);\n  }\n  _setActiveIndicatorElement(index) {\n    if (!this._indicatorsElement) {\n      return;\n    }\n    const activeIndicator = SelectorEngine.findOne(SELECTOR_ACTIVE, this._indicatorsElement);\n    activeIndicator.classList.remove(CLASS_NAME_ACTIVE$2);\n    activeIndicator.removeAttribute('aria-current');\n    const newActiveIndicator = SelectorEngine.findOne(`[data-bs-slide-to=\"${index}\"]`, this._indicatorsElement);\n    if (newActiveIndicator) {\n      newActiveIndicator.classList.add(CLASS_NAME_ACTIVE$2);\n      newActiveIndicator.setAttribute('aria-current', 'true');\n    }\n  }\n  _updateInterval() {\n    const element = this._activeElement || this._getActive();\n    if (!element) {\n      return;\n    }\n    const elementInterval = Number.parseInt(element.getAttribute('data-bs-interval'), 10);\n    this._config.interval = elementInterval || this._config.defaultInterval;\n  }\n  _slide(order, element = null) {\n    if (this._isSliding) {\n      return;\n    }\n    const activeElement = this._getActive();\n    const isNext = order === ORDER_NEXT;\n    const nextElement = element || getNextActiveElement(this._getItems(), activeElement, isNext, this._config.wrap);\n    if (nextElement === activeElement) {\n      return;\n    }\n    const nextElementIndex = this._getItemIndex(nextElement);\n    const triggerEvent = eventName => {\n      return EventHandler.trigger(this._element, eventName, {\n        relatedTarget: nextElement,\n        direction: this._orderToDirection(order),\n        from: this._getItemIndex(activeElement),\n        to: nextElementIndex\n      });\n    };\n    const slideEvent = triggerEvent(EVENT_SLIDE);\n    if (slideEvent.defaultPrevented) {\n      return;\n    }\n    if (!activeElement || !nextElement) {\n      // Some weirdness is happening, so we bail\n      // TODO: change tests that use empty divs to avoid this check\n      return;\n    }\n    const isCycling = Boolean(this._interval);\n    this.pause();\n    this._isSliding = true;\n    this._setActiveIndicatorElement(nextElementIndex);\n    this._activeElement = nextElement;\n    const directionalClassName = isNext ? CLASS_NAME_START : CLASS_NAME_END;\n    const orderClassName = isNext ? CLASS_NAME_NEXT : CLASS_NAME_PREV;\n    nextElement.classList.add(orderClassName);\n    reflow(nextElement);\n    activeElement.classList.add(directionalClassName);\n    nextElement.classList.add(directionalClassName);\n    const completeCallBack = () => {\n      nextElement.classList.remove(directionalClassName, orderClassName);\n      nextElement.classList.add(CLASS_NAME_ACTIVE$2);\n      activeElement.classList.remove(CLASS_NAME_ACTIVE$2, orderClassName, directionalClassName);\n      this._isSliding = false;\n      triggerEvent(EVENT_SLID);\n    };\n    this._queueCallback(completeCallBack, activeElement, this._isAnimated());\n    if (isCycling) {\n      this.cycle();\n    }\n  }\n  _isAnimated() {\n    return this._element.classList.contains(CLASS_NAME_SLIDE);\n  }\n  _getActive() {\n    return SelectorEngine.findOne(SELECTOR_ACTIVE_ITEM, this._element);\n  }\n  _getItems() {\n    return SelectorEngine.find(SELECTOR_ITEM, this._element);\n  }\n  _clearInterval() {\n    if (this._interval) {\n      clearInterval(this._interval);\n      this._interval = null;\n    }\n  }\n  _directionToOrder(direction) {\n    if (isRTL()) {\n      return direction === DIRECTION_LEFT ? ORDER_PREV : ORDER_NEXT;\n    }\n    return direction === DIRECTION_LEFT ? ORDER_NEXT : ORDER_PREV;\n  }\n  _orderToDirection(order) {\n    if (isRTL()) {\n      return order === ORDER_PREV ? DIRECTION_LEFT : DIRECTION_RIGHT;\n    }\n    return order === ORDER_PREV ? DIRECTION_RIGHT : DIRECTION_LEFT;\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    return this.each(function () {\n      const data = Carousel.getOrCreateInstance(this, config);\n      if (typeof config === 'number') {\n        data.to(config);\n        return;\n      }\n      if (typeof config === 'string') {\n        if (data[config] === undefined || config.startsWith('_') || config === 'constructor') {\n          throw new TypeError(`No method named \"${config}\"`);\n        }\n        data[config]();\n      }\n    });\n  }\n}\n\n/**\n * Data API implementation\n */\n\nEventHandler.on(document, EVENT_CLICK_DATA_API$5, SELECTOR_DATA_SLIDE, function (event) {\n  const target = SelectorEngine.getElementFromSelector(this);\n  if (!target || !target.classList.contains(CLASS_NAME_CAROUSEL)) {\n    return;\n  }\n  event.preventDefault();\n  const carousel = Carousel.getOrCreateInstance(target);\n  const slideIndex = this.getAttribute('data-bs-slide-to');\n  if (slideIndex) {\n    carousel.to(slideIndex);\n    carousel._maybeEnableCycle();\n    return;\n  }\n  if (Manipulator.getDataAttribute(this, 'slide') === 'next') {\n    carousel.next();\n    carousel._maybeEnableCycle();\n    return;\n  }\n  carousel.prev();\n  carousel._maybeEnableCycle();\n});\nEventHandler.on(window, EVENT_LOAD_DATA_API$3, () => {\n  const carousels = SelectorEngine.find(SELECTOR_DATA_RIDE);\n  for (const carousel of carousels) {\n    Carousel.getOrCreateInstance(carousel);\n  }\n});\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Carousel);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap collapse.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$b = 'collapse';\nconst DATA_KEY$7 = 'bs.collapse';\nconst EVENT_KEY$7 = `.${DATA_KEY$7}`;\nconst DATA_API_KEY$4 = '.data-api';\nconst EVENT_SHOW$6 = `show${EVENT_KEY$7}`;\nconst EVENT_SHOWN$6 = `shown${EVENT_KEY$7}`;\nconst EVENT_HIDE$6 = `hide${EVENT_KEY$7}`;\nconst EVENT_HIDDEN$6 = `hidden${EVENT_KEY$7}`;\nconst EVENT_CLICK_DATA_API$4 = `click${EVENT_KEY$7}${DATA_API_KEY$4}`;\nconst CLASS_NAME_SHOW$7 = 'show';\nconst CLASS_NAME_COLLAPSE = 'collapse';\nconst CLASS_NAME_COLLAPSING = 'collapsing';\nconst CLASS_NAME_COLLAPSED = 'collapsed';\nconst CLASS_NAME_DEEPER_CHILDREN = `:scope .${CLASS_NAME_COLLAPSE} .${CLASS_NAME_COLLAPSE}`;\nconst CLASS_NAME_HORIZONTAL = 'collapse-horizontal';\nconst WIDTH = 'width';\nconst HEIGHT = 'height';\nconst SELECTOR_ACTIVES = '.collapse.show, .collapse.collapsing';\nconst SELECTOR_DATA_TOGGLE$4 = '[data-bs-toggle=\"collapse\"]';\nconst Default$a = {\n  parent: null,\n  toggle: true\n};\nconst DefaultType$a = {\n  parent: '(null|element)',\n  toggle: 'boolean'\n};\n\n/**\n * Class definition\n */\n\nclass Collapse extends BaseComponent {\n  constructor(element, config) {\n    super(element, config);\n    this._isTransitioning = false;\n    this._triggerArray = [];\n    const toggleList = SelectorEngine.find(SELECTOR_DATA_TOGGLE$4);\n    for (const elem of toggleList) {\n      const selector = SelectorEngine.getSelectorFromElement(elem);\n      const filterElement = SelectorEngine.find(selector).filter(foundElement => foundElement === this._element);\n      if (selector !== null && filterElement.length) {\n        this._triggerArray.push(elem);\n      }\n    }\n    this._initializeChildren();\n    if (!this._config.parent) {\n      this._addAriaAndCollapsedClass(this._triggerArray, this._isShown());\n    }\n    if (this._config.toggle) {\n      this.toggle();\n    }\n  }\n\n  // Getters\n  static get Default() {\n    return Default$a;\n  }\n  static get DefaultType() {\n    return DefaultType$a;\n  }\n  static get NAME() {\n    return NAME$b;\n  }\n\n  // Public\n  toggle() {\n    if (this._isShown()) {\n      this.hide();\n    } else {\n      this.show();\n    }\n  }\n  show() {\n    if (this._isTransitioning || this._isShown()) {\n      return;\n    }\n    let activeChildren = [];\n\n    // find active children\n    if (this._config.parent) {\n      activeChildren = this._getFirstLevelChildren(SELECTOR_ACTIVES).filter(element => element !== this._element).map(element => Collapse.getOrCreateInstance(element, {\n        toggle: false\n      }));\n    }\n    if (activeChildren.length && activeChildren[0]._isTransitioning) {\n      return;\n    }\n    const startEvent = EventHandler.trigger(this._element, EVENT_SHOW$6);\n    if (startEvent.defaultPrevented) {\n      return;\n    }\n    for (const activeInstance of activeChildren) {\n      activeInstance.hide();\n    }\n    const dimension = this._getDimension();\n    this._element.classList.remove(CLASS_NAME_COLLAPSE);\n    this._element.classList.add(CLASS_NAME_COLLAPSING);\n    this._element.style[dimension] = 0;\n    this._addAriaAndCollapsedClass(this._triggerArray, true);\n    this._isTransitioning = true;\n    const complete = () => {\n      this._isTransitioning = false;\n      this._element.classList.remove(CLASS_NAME_COLLAPSING);\n      this._element.classList.add(CLASS_NAME_COLLAPSE, CLASS_NAME_SHOW$7);\n      this._element.style[dimension] = '';\n      EventHandler.trigger(this._element, EVENT_SHOWN$6);\n    };\n    const capitalizedDimension = dimension[0].toUpperCase() + dimension.slice(1);\n    const scrollSize = `scroll${capitalizedDimension}`;\n    this._queueCallback(complete, this._element, true);\n    this._element.style[dimension] = `${this._element[scrollSize]}px`;\n  }\n  hide() {\n    if (this._isTransitioning || !this._isShown()) {\n      return;\n    }\n    const startEvent = EventHandler.trigger(this._element, EVENT_HIDE$6);\n    if (startEvent.defaultPrevented) {\n      return;\n    }\n    const dimension = this._getDimension();\n    this._element.style[dimension] = `${this._element.getBoundingClientRect()[dimension]}px`;\n    reflow(this._element);\n    this._element.classList.add(CLASS_NAME_COLLAPSING);\n    this._element.classList.remove(CLASS_NAME_COLLAPSE, CLASS_NAME_SHOW$7);\n    for (const trigger of this._triggerArray) {\n      const element = SelectorEngine.getElementFromSelector(trigger);\n      if (element && !this._isShown(element)) {\n        this._addAriaAndCollapsedClass([trigger], false);\n      }\n    }\n    this._isTransitioning = true;\n    const complete = () => {\n      this._isTransitioning = false;\n      this._element.classList.remove(CLASS_NAME_COLLAPSING);\n      this._element.classList.add(CLASS_NAME_COLLAPSE);\n      EventHandler.trigger(this._element, EVENT_HIDDEN$6);\n    };\n    this._element.style[dimension] = '';\n    this._queueCallback(complete, this._element, true);\n  }\n  _isShown(element = this._element) {\n    return element.classList.contains(CLASS_NAME_SHOW$7);\n  }\n\n  // Private\n  _configAfterMerge(config) {\n    config.toggle = Boolean(config.toggle); // Coerce string values\n    config.parent = getElement(config.parent);\n    return config;\n  }\n  _getDimension() {\n    return this._element.classList.contains(CLASS_NAME_HORIZONTAL) ? WIDTH : HEIGHT;\n  }\n  _initializeChildren() {\n    if (!this._config.parent) {\n      return;\n    }\n    const children = this._getFirstLevelChildren(SELECTOR_DATA_TOGGLE$4);\n    for (const element of children) {\n      const selected = SelectorEngine.getElementFromSelector(element);\n      if (selected) {\n        this._addAriaAndCollapsedClass([element], this._isShown(selected));\n      }\n    }\n  }\n  _getFirstLevelChildren(selector) {\n    const children = SelectorEngine.find(CLASS_NAME_DEEPER_CHILDREN, this._config.parent);\n    // remove children if greater depth\n    return SelectorEngine.find(selector, this._config.parent).filter(element => !children.includes(element));\n  }\n  _addAriaAndCollapsedClass(triggerArray, isOpen) {\n    if (!triggerArray.length) {\n      return;\n    }\n    for (const element of triggerArray) {\n      element.classList.toggle(CLASS_NAME_COLLAPSED, !isOpen);\n      element.setAttribute('aria-expanded', isOpen);\n    }\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    const _config = {};\n    if (typeof config === 'string' && /show|hide/.test(config)) {\n      _config.toggle = false;\n    }\n    return this.each(function () {\n      const data = Collapse.getOrCreateInstance(this, _config);\n      if (typeof config === 'string') {\n        if (typeof data[config] === 'undefined') {\n          throw new TypeError(`No method named \"${config}\"`);\n        }\n        data[config]();\n      }\n    });\n  }\n}\n\n/**\n * Data API implementation\n */\n\nEventHandler.on(document, EVENT_CLICK_DATA_API$4, SELECTOR_DATA_TOGGLE$4, function (event) {\n  // preventDefault only for <a> elements (which change the URL) not inside the collapsible element\n  if (event.target.tagName === 'A' || event.delegateTarget && event.delegateTarget.tagName === 'A') {\n    event.preventDefault();\n  }\n  for (const element of SelectorEngine.getMultipleElementsFromSelector(this)) {\n    Collapse.getOrCreateInstance(element, {\n      toggle: false\n    }).toggle();\n  }\n});\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Collapse);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap dropdown.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$a = 'dropdown';\nconst DATA_KEY$6 = 'bs.dropdown';\nconst EVENT_KEY$6 = `.${DATA_KEY$6}`;\nconst DATA_API_KEY$3 = '.data-api';\nconst ESCAPE_KEY$2 = 'Escape';\nconst TAB_KEY$1 = 'Tab';\nconst ARROW_UP_KEY$1 = 'ArrowUp';\nconst ARROW_DOWN_KEY$1 = 'ArrowDown';\nconst RIGHT_MOUSE_BUTTON = 2; // MouseEvent.button value for the secondary button, usually the right button\n\nconst EVENT_HIDE$5 = `hide${EVENT_KEY$6}`;\nconst EVENT_HIDDEN$5 = `hidden${EVENT_KEY$6}`;\nconst EVENT_SHOW$5 = `show${EVENT_KEY$6}`;\nconst EVENT_SHOWN$5 = `shown${EVENT_KEY$6}`;\nconst EVENT_CLICK_DATA_API$3 = `click${EVENT_KEY$6}${DATA_API_KEY$3}`;\nconst EVENT_KEYDOWN_DATA_API = `keydown${EVENT_KEY$6}${DATA_API_KEY$3}`;\nconst EVENT_KEYUP_DATA_API = `keyup${EVENT_KEY$6}${DATA_API_KEY$3}`;\nconst CLASS_NAME_SHOW$6 = 'show';\nconst CLASS_NAME_DROPUP = 'dropup';\nconst CLASS_NAME_DROPEND = 'dropend';\nconst CLASS_NAME_DROPSTART = 'dropstart';\nconst CLASS_NAME_DROPUP_CENTER = 'dropup-center';\nconst CLASS_NAME_DROPDOWN_CENTER = 'dropdown-center';\nconst SELECTOR_DATA_TOGGLE$3 = '[data-bs-toggle=\"dropdown\"]:not(.disabled):not(:disabled)';\nconst SELECTOR_DATA_TOGGLE_SHOWN = `${SELECTOR_DATA_TOGGLE$3}.${CLASS_NAME_SHOW$6}`;\nconst SELECTOR_MENU = '.dropdown-menu';\nconst SELECTOR_NAVBAR = '.navbar';\nconst SELECTOR_NAVBAR_NAV = '.navbar-nav';\nconst SELECTOR_VISIBLE_ITEMS = '.dropdown-menu .dropdown-item:not(.disabled):not(:disabled)';\nconst PLACEMENT_TOP = isRTL() ? 'top-end' : 'top-start';\nconst PLACEMENT_TOPEND = isRTL() ? 'top-start' : 'top-end';\nconst PLACEMENT_BOTTOM = isRTL() ? 'bottom-end' : 'bottom-start';\nconst PLACEMENT_BOTTOMEND = isRTL() ? 'bottom-start' : 'bottom-end';\nconst PLACEMENT_RIGHT = isRTL() ? 'left-start' : 'right-start';\nconst PLACEMENT_LEFT = isRTL() ? 'right-start' : 'left-start';\nconst PLACEMENT_TOPCENTER = 'top';\nconst PLACEMENT_BOTTOMCENTER = 'bottom';\nconst Default$9 = {\n  autoClose: true,\n  boundary: 'clippingParents',\n  display: 'dynamic',\n  offset: [0, 2],\n  popperConfig: null,\n  reference: 'toggle'\n};\nconst DefaultType$9 = {\n  autoClose: '(boolean|string)',\n  boundary: '(string|element)',\n  display: 'string',\n  offset: '(array|string|function)',\n  popperConfig: '(null|object|function)',\n  reference: '(string|element|object)'\n};\n\n/**\n * Class definition\n */\n\nclass Dropdown extends BaseComponent {\n  constructor(element, config) {\n    super(element, config);\n    this._popper = null;\n    this._parent = this._element.parentNode; // dropdown wrapper\n    // TODO: v6 revert #37011 & change markup https://getbootstrap.com/docs/5.3/forms/input-group/\n    this._menu = SelectorEngine.next(this._element, SELECTOR_MENU)[0] || SelectorEngine.prev(this._element, SELECTOR_MENU)[0] || SelectorEngine.findOne(SELECTOR_MENU, this._parent);\n    this._inNavbar = this._detectNavbar();\n  }\n\n  // Getters\n  static get Default() {\n    return Default$9;\n  }\n  static get DefaultType() {\n    return DefaultType$9;\n  }\n  static get NAME() {\n    return NAME$a;\n  }\n\n  // Public\n  toggle() {\n    return this._isShown() ? this.hide() : this.show();\n  }\n  show() {\n    if (isDisabled(this._element) || this._isShown()) {\n      return;\n    }\n    const relatedTarget = {\n      relatedTarget: this._element\n    };\n    const showEvent = EventHandler.trigger(this._element, EVENT_SHOW$5, relatedTarget);\n    if (showEvent.defaultPrevented) {\n      return;\n    }\n    this._createPopper();\n\n    // If this is a touch-enabled device we add extra\n    // empty mouseover listeners to the body's immediate children;\n    // only needed because of broken event delegation on iOS\n    // https://www.quirksmode.org/blog/archives/2014/02/mouse_event_bub.html\n    if ('ontouchstart' in document.documentElement && !this._parent.closest(SELECTOR_NAVBAR_NAV)) {\n      for (const element of [].concat(...document.body.children)) {\n        EventHandler.on(element, 'mouseover', noop);\n      }\n    }\n    this._element.focus();\n    this._element.setAttribute('aria-expanded', true);\n    this._menu.classList.add(CLASS_NAME_SHOW$6);\n    this._element.classList.add(CLASS_NAME_SHOW$6);\n    EventHandler.trigger(this._element, EVENT_SHOWN$5, relatedTarget);\n  }\n  hide() {\n    if (isDisabled(this._element) || !this._isShown()) {\n      return;\n    }\n    const relatedTarget = {\n      relatedTarget: this._element\n    };\n    this._completeHide(relatedTarget);\n  }\n  dispose() {\n    if (this._popper) {\n      this._popper.destroy();\n    }\n    super.dispose();\n  }\n  update() {\n    this._inNavbar = this._detectNavbar();\n    if (this._popper) {\n      this._popper.update();\n    }\n  }\n\n  // Private\n  _completeHide(relatedTarget) {\n    const hideEvent = EventHandler.trigger(this._element, EVENT_HIDE$5, relatedTarget);\n    if (hideEvent.defaultPrevented) {\n      return;\n    }\n\n    // If this is a touch-enabled device we remove the extra\n    // empty mouseover listeners we added for iOS support\n    if ('ontouchstart' in document.documentElement) {\n      for (const element of [].concat(...document.body.children)) {\n        EventHandler.off(element, 'mouseover', noop);\n      }\n    }\n    if (this._popper) {\n      this._popper.destroy();\n    }\n    this._menu.classList.remove(CLASS_NAME_SHOW$6);\n    this._element.classList.remove(CLASS_NAME_SHOW$6);\n    this._element.setAttribute('aria-expanded', 'false');\n    Manipulator.removeDataAttribute(this._menu, 'popper');\n    EventHandler.trigger(this._element, EVENT_HIDDEN$5, relatedTarget);\n  }\n  _getConfig(config) {\n    config = super._getConfig(config);\n    if (typeof config.reference === 'object' && !isElement(config.reference) && typeof config.reference.getBoundingClientRect !== 'function') {\n      // Popper virtual elements require a getBoundingClientRect method\n      throw new TypeError(`${NAME$a.toUpperCase()}: Option \"reference\" provided type \"object\" without a required \"getBoundingClientRect\" method.`);\n    }\n    return config;\n  }\n  _createPopper() {\n    if (typeof Popper === 'undefined') {\n      throw new TypeError('Bootstrap\\'s dropdowns require Popper (https://popper.js.org)');\n    }\n    let referenceElement = this._element;\n    if (this._config.reference === 'parent') {\n      referenceElement = this._parent;\n    } else if (isElement(this._config.reference)) {\n      referenceElement = getElement(this._config.reference);\n    } else if (typeof this._config.reference === 'object') {\n      referenceElement = this._config.reference;\n    }\n    const popperConfig = this._getPopperConfig();\n    this._popper = Popper.createPopper(referenceElement, this._menu, popperConfig);\n  }\n  _isShown() {\n    return this._menu.classList.contains(CLASS_NAME_SHOW$6);\n  }\n  _getPlacement() {\n    const parentDropdown = this._parent;\n    if (parentDropdown.classList.contains(CLASS_NAME_DROPEND)) {\n      return PLACEMENT_RIGHT;\n    }\n    if (parentDropdown.classList.contains(CLASS_NAME_DROPSTART)) {\n      return PLACEMENT_LEFT;\n    }\n    if (parentDropdown.classList.contains(CLASS_NAME_DROPUP_CENTER)) {\n      return PLACEMENT_TOPCENTER;\n    }\n    if (parentDropdown.classList.contains(CLASS_NAME_DROPDOWN_CENTER)) {\n      return PLACEMENT_BOTTOMCENTER;\n    }\n\n    // We need to trim the value because custom properties can also include spaces\n    const isEnd = getComputedStyle(this._menu).getPropertyValue('--bs-position').trim() === 'end';\n    if (parentDropdown.classList.contains(CLASS_NAME_DROPUP)) {\n      return isEnd ? PLACEMENT_TOPEND : PLACEMENT_TOP;\n    }\n    return isEnd ? PLACEMENT_BOTTOMEND : PLACEMENT_BOTTOM;\n  }\n  _detectNavbar() {\n    return this._element.closest(SELECTOR_NAVBAR) !== null;\n  }\n  _getOffset() {\n    const {\n      offset\n    } = this._config;\n    if (typeof offset === 'string') {\n      return offset.split(',').map(value => Number.parseInt(value, 10));\n    }\n    if (typeof offset === 'function') {\n      return popperData => offset(popperData, this._element);\n    }\n    return offset;\n  }\n  _getPopperConfig() {\n    const defaultBsPopperConfig = {\n      placement: this._getPlacement(),\n      modifiers: [{\n        name: 'preventOverflow',\n        options: {\n          boundary: this._config.boundary\n        }\n      }, {\n        name: 'offset',\n        options: {\n          offset: this._getOffset()\n        }\n      }]\n    };\n\n    // Disable Popper if we have a static display or Dropdown is in Navbar\n    if (this._inNavbar || this._config.display === 'static') {\n      Manipulator.setDataAttribute(this._menu, 'popper', 'static'); // TODO: v6 remove\n      defaultBsPopperConfig.modifiers = [{\n        name: 'applyStyles',\n        enabled: false\n      }];\n    }\n    return {\n      ...defaultBsPopperConfig,\n      ...execute(this._config.popperConfig, [defaultBsPopperConfig])\n    };\n  }\n  _selectMenuItem({\n    key,\n    target\n  }) {\n    const items = SelectorEngine.find(SELECTOR_VISIBLE_ITEMS, this._menu).filter(element => isVisible(element));\n    if (!items.length) {\n      return;\n    }\n\n    // if target isn't included in items (e.g. when expanding the dropdown)\n    // allow cycling to get the last item in case key equals ARROW_UP_KEY\n    getNextActiveElement(items, target, key === ARROW_DOWN_KEY$1, !items.includes(target)).focus();\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    return this.each(function () {\n      const data = Dropdown.getOrCreateInstance(this, config);\n      if (typeof config !== 'string') {\n        return;\n      }\n      if (typeof data[config] === 'undefined') {\n        throw new TypeError(`No method named \"${config}\"`);\n      }\n      data[config]();\n    });\n  }\n  static clearMenus(event) {\n    if (event.button === RIGHT_MOUSE_BUTTON || event.type === 'keyup' && event.key !== TAB_KEY$1) {\n      return;\n    }\n    const openToggles = SelectorEngine.find(SELECTOR_DATA_TOGGLE_SHOWN);\n    for (const toggle of openToggles) {\n      const context = Dropdown.getInstance(toggle);\n      if (!context || context._config.autoClose === false) {\n        continue;\n      }\n      const composedPath = event.composedPath();\n      const isMenuTarget = composedPath.includes(context._menu);\n      if (composedPath.includes(context._element) || context._config.autoClose === 'inside' && !isMenuTarget || context._config.autoClose === 'outside' && isMenuTarget) {\n        continue;\n      }\n\n      // Tab navigation through the dropdown menu or events from contained inputs shouldn't close the menu\n      if (context._menu.contains(event.target) && (event.type === 'keyup' && event.key === TAB_KEY$1 || /input|select|option|textarea|form/i.test(event.target.tagName))) {\n        continue;\n      }\n      const relatedTarget = {\n        relatedTarget: context._element\n      };\n      if (event.type === 'click') {\n        relatedTarget.clickEvent = event;\n      }\n      context._completeHide(relatedTarget);\n    }\n  }\n  static dataApiKeydownHandler(event) {\n    // If not an UP | DOWN | ESCAPE key => not a dropdown command\n    // If input/textarea && if key is other than ESCAPE => not a dropdown command\n\n    const isInput = /input|textarea/i.test(event.target.tagName);\n    const isEscapeEvent = event.key === ESCAPE_KEY$2;\n    const isUpOrDownEvent = [ARROW_UP_KEY$1, ARROW_DOWN_KEY$1].includes(event.key);\n    if (!isUpOrDownEvent && !isEscapeEvent) {\n      return;\n    }\n    if (isInput && !isEscapeEvent) {\n      return;\n    }\n    event.preventDefault();\n\n    // TODO: v6 revert #37011 & change markup https://getbootstrap.com/docs/5.3/forms/input-group/\n    const getToggleButton = this.matches(SELECTOR_DATA_TOGGLE$3) ? this : SelectorEngine.prev(this, SELECTOR_DATA_TOGGLE$3)[0] || SelectorEngine.next(this, SELECTOR_DATA_TOGGLE$3)[0] || SelectorEngine.findOne(SELECTOR_DATA_TOGGLE$3, event.delegateTarget.parentNode);\n    const instance = Dropdown.getOrCreateInstance(getToggleButton);\n    if (isUpOrDownEvent) {\n      event.stopPropagation();\n      instance.show();\n      instance._selectMenuItem(event);\n      return;\n    }\n    if (instance._isShown()) {\n      // else is escape and we check if it is shown\n      event.stopPropagation();\n      instance.hide();\n      getToggleButton.focus();\n    }\n  }\n}\n\n/**\n * Data API implementation\n */\n\nEventHandler.on(document, EVENT_KEYDOWN_DATA_API, SELECTOR_DATA_TOGGLE$3, Dropdown.dataApiKeydownHandler);\nEventHandler.on(document, EVENT_KEYDOWN_DATA_API, SELECTOR_MENU, Dropdown.dataApiKeydownHandler);\nEventHandler.on(document, EVENT_CLICK_DATA_API$3, Dropdown.clearMenus);\nEventHandler.on(document, EVENT_KEYUP_DATA_API, Dropdown.clearMenus);\nEventHandler.on(document, EVENT_CLICK_DATA_API$3, SELECTOR_DATA_TOGGLE$3, function (event) {\n  event.preventDefault();\n  Dropdown.getOrCreateInstance(this).toggle();\n});\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Dropdown);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap util/backdrop.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$9 = 'backdrop';\nconst CLASS_NAME_FADE$4 = 'fade';\nconst CLASS_NAME_SHOW$5 = 'show';\nconst EVENT_MOUSEDOWN = `mousedown.bs.${NAME$9}`;\nconst Default$8 = {\n  className: 'modal-backdrop',\n  clickCallback: null,\n  isAnimated: false,\n  isVisible: true,\n  // if false, we use the backdrop helper without adding any element to the dom\n  rootElement: 'body' // give the choice to place backdrop under different elements\n};\n\nconst DefaultType$8 = {\n  className: 'string',\n  clickCallback: '(function|null)',\n  isAnimated: 'boolean',\n  isVisible: 'boolean',\n  rootElement: '(element|string)'\n};\n\n/**\n * Class definition\n */\n\nclass Backdrop extends Config {\n  constructor(config) {\n    super();\n    this._config = this._getConfig(config);\n    this._isAppended = false;\n    this._element = null;\n  }\n\n  // Getters\n  static get Default() {\n    return Default$8;\n  }\n  static get DefaultType() {\n    return DefaultType$8;\n  }\n  static get NAME() {\n    return NAME$9;\n  }\n\n  // Public\n  show(callback) {\n    if (!this._config.isVisible) {\n      execute(callback);\n      return;\n    }\n    this._append();\n    const element = this._getElement();\n    if (this._config.isAnimated) {\n      reflow(element);\n    }\n    element.classList.add(CLASS_NAME_SHOW$5);\n    this._emulateAnimation(() => {\n      execute(callback);\n    });\n  }\n  hide(callback) {\n    if (!this._config.isVisible) {\n      execute(callback);\n      return;\n    }\n    this._getElement().classList.remove(CLASS_NAME_SHOW$5);\n    this._emulateAnimation(() => {\n      this.dispose();\n      execute(callback);\n    });\n  }\n  dispose() {\n    if (!this._isAppended) {\n      return;\n    }\n    EventHandler.off(this._element, EVENT_MOUSEDOWN);\n    this._element.remove();\n    this._isAppended = false;\n  }\n\n  // Private\n  _getElement() {\n    if (!this._element) {\n      const backdrop = document.createElement('div');\n      backdrop.className = this._config.className;\n      if (this._config.isAnimated) {\n        backdrop.classList.add(CLASS_NAME_FADE$4);\n      }\n      this._element = backdrop;\n    }\n    return this._element;\n  }\n  _configAfterMerge(config) {\n    // use getElement() with the default \"body\" to get a fresh Element on each instantiation\n    config.rootElement = getElement(config.rootElement);\n    return config;\n  }\n  _append() {\n    if (this._isAppended) {\n      return;\n    }\n    const element = this._getElement();\n    this._config.rootElement.append(element);\n    EventHandler.on(element, EVENT_MOUSEDOWN, () => {\n      execute(this._config.clickCallback);\n    });\n    this._isAppended = true;\n  }\n  _emulateAnimation(callback) {\n    executeAfterTransition(callback, this._getElement(), this._config.isAnimated);\n  }\n}\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap util/focustrap.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$8 = 'focustrap';\nconst DATA_KEY$5 = 'bs.focustrap';\nconst EVENT_KEY$5 = `.${DATA_KEY$5}`;\nconst EVENT_FOCUSIN$2 = `focusin${EVENT_KEY$5}`;\nconst EVENT_KEYDOWN_TAB = `keydown.tab${EVENT_KEY$5}`;\nconst TAB_KEY = 'Tab';\nconst TAB_NAV_FORWARD = 'forward';\nconst TAB_NAV_BACKWARD = 'backward';\nconst Default$7 = {\n  autofocus: true,\n  trapElement: null // The element to trap focus inside of\n};\n\nconst DefaultType$7 = {\n  autofocus: 'boolean',\n  trapElement: 'element'\n};\n\n/**\n * Class definition\n */\n\nclass FocusTrap extends Config {\n  constructor(config) {\n    super();\n    this._config = this._getConfig(config);\n    this._isActive = false;\n    this._lastTabNavDirection = null;\n  }\n\n  // Getters\n  static get Default() {\n    return Default$7;\n  }\n  static get DefaultType() {\n    return DefaultType$7;\n  }\n  static get NAME() {\n    return NAME$8;\n  }\n\n  // Public\n  activate() {\n    if (this._isActive) {\n      return;\n    }\n    if (this._config.autofocus) {\n      this._config.trapElement.focus();\n    }\n    EventHandler.off(document, EVENT_KEY$5); // guard against infinite focus loop\n    EventHandler.on(document, EVENT_FOCUSIN$2, event => this._handleFocusin(event));\n    EventHandler.on(document, EVENT_KEYDOWN_TAB, event => this._handleKeydown(event));\n    this._isActive = true;\n  }\n  deactivate() {\n    if (!this._isActive) {\n      return;\n    }\n    this._isActive = false;\n    EventHandler.off(document, EVENT_KEY$5);\n  }\n\n  // Private\n  _handleFocusin(event) {\n    const {\n      trapElement\n    } = this._config;\n    if (event.target === document || event.target === trapElement || trapElement.contains(event.target)) {\n      return;\n    }\n    const elements = SelectorEngine.focusableChildren(trapElement);\n    if (elements.length === 0) {\n      trapElement.focus();\n    } else if (this._lastTabNavDirection === TAB_NAV_BACKWARD) {\n      elements[elements.length - 1].focus();\n    } else {\n      elements[0].focus();\n    }\n  }\n  _handleKeydown(event) {\n    if (event.key !== TAB_KEY) {\n      return;\n    }\n    this._lastTabNavDirection = event.shiftKey ? TAB_NAV_BACKWARD : TAB_NAV_FORWARD;\n  }\n}\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap util/scrollBar.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst SELECTOR_FIXED_CONTENT = '.fixed-top, .fixed-bottom, .is-fixed, .sticky-top';\nconst SELECTOR_STICKY_CONTENT = '.sticky-top';\nconst PROPERTY_PADDING = 'padding-right';\nconst PROPERTY_MARGIN = 'margin-right';\n\n/**\n * Class definition\n */\n\nclass ScrollBarHelper {\n  constructor() {\n    this._element = document.body;\n  }\n\n  // Public\n  getWidth() {\n    // https://developer.mozilla.org/en-US/docs/Web/API/Window/innerWidth#usage_notes\n    const documentWidth = document.documentElement.clientWidth;\n    return Math.abs(window.innerWidth - documentWidth);\n  }\n  hide() {\n    const width = this.getWidth();\n    this._disableOverFlow();\n    // give padding to element to balance the hidden scrollbar width\n    this._setElementAttributes(this._element, PROPERTY_PADDING, calculatedValue => calculatedValue + width);\n    // trick: We adjust positive paddingRight and negative marginRight to sticky-top elements to keep showing fullwidth\n    this._setElementAttributes(SELECTOR_FIXED_CONTENT, PROPERTY_PADDING, calculatedValue => calculatedValue + width);\n    this._setElementAttributes(SELECTOR_STICKY_CONTENT, PROPERTY_MARGIN, calculatedValue => calculatedValue - width);\n  }\n  reset() {\n    this._resetElementAttributes(this._element, 'overflow');\n    this._resetElementAttributes(this._element, PROPERTY_PADDING);\n    this._resetElementAttributes(SELECTOR_FIXED_CONTENT, PROPERTY_PADDING);\n    this._resetElementAttributes(SELECTOR_STICKY_CONTENT, PROPERTY_MARGIN);\n  }\n  isOverflowing() {\n    return this.getWidth() > 0;\n  }\n\n  // Private\n  _disableOverFlow() {\n    this._saveInitialAttribute(this._element, 'overflow');\n    this._element.style.overflow = 'hidden';\n  }\n  _setElementAttributes(selector, styleProperty, callback) {\n    const scrollbarWidth = this.getWidth();\n    const manipulationCallBack = element => {\n      if (element !== this._element && window.innerWidth > element.clientWidth + scrollbarWidth) {\n        return;\n      }\n      this._saveInitialAttribute(element, styleProperty);\n      const calculatedValue = window.getComputedStyle(element).getPropertyValue(styleProperty);\n      element.style.setProperty(styleProperty, `${callback(Number.parseFloat(calculatedValue))}px`);\n    };\n    this._applyManipulationCallback(selector, manipulationCallBack);\n  }\n  _saveInitialAttribute(element, styleProperty) {\n    const actualValue = element.style.getPropertyValue(styleProperty);\n    if (actualValue) {\n      Manipulator.setDataAttribute(element, styleProperty, actualValue);\n    }\n  }\n  _resetElementAttributes(selector, styleProperty) {\n    const manipulationCallBack = element => {\n      const value = Manipulator.getDataAttribute(element, styleProperty);\n      // We only want to remove the property if the value is `null`; the value can also be zero\n      if (value === null) {\n        element.style.removeProperty(styleProperty);\n        return;\n      }\n      Manipulator.removeDataAttribute(element, styleProperty);\n      element.style.setProperty(styleProperty, value);\n    };\n    this._applyManipulationCallback(selector, manipulationCallBack);\n  }\n  _applyManipulationCallback(selector, callBack) {\n    if (isElement(selector)) {\n      callBack(selector);\n      return;\n    }\n    for (const sel of SelectorEngine.find(selector, this._element)) {\n      callBack(sel);\n    }\n  }\n}\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap modal.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$7 = 'modal';\nconst DATA_KEY$4 = 'bs.modal';\nconst EVENT_KEY$4 = `.${DATA_KEY$4}`;\nconst DATA_API_KEY$2 = '.data-api';\nconst ESCAPE_KEY$1 = 'Escape';\nconst EVENT_HIDE$4 = `hide${EVENT_KEY$4}`;\nconst EVENT_HIDE_PREVENTED$1 = `hidePrevented${EVENT_KEY$4}`;\nconst EVENT_HIDDEN$4 = `hidden${EVENT_KEY$4}`;\nconst EVENT_SHOW$4 = `show${EVENT_KEY$4}`;\nconst EVENT_SHOWN$4 = `shown${EVENT_KEY$4}`;\nconst EVENT_RESIZE$1 = `resize${EVENT_KEY$4}`;\nconst EVENT_CLICK_DISMISS = `click.dismiss${EVENT_KEY$4}`;\nconst EVENT_MOUSEDOWN_DISMISS = `mousedown.dismiss${EVENT_KEY$4}`;\nconst EVENT_KEYDOWN_DISMISS$1 = `keydown.dismiss${EVENT_KEY$4}`;\nconst EVENT_CLICK_DATA_API$2 = `click${EVENT_KEY$4}${DATA_API_KEY$2}`;\nconst CLASS_NAME_OPEN = 'modal-open';\nconst CLASS_NAME_FADE$3 = 'fade';\nconst CLASS_NAME_SHOW$4 = 'show';\nconst CLASS_NAME_STATIC = 'modal-static';\nconst OPEN_SELECTOR$1 = '.modal.show';\nconst SELECTOR_DIALOG = '.modal-dialog';\nconst SELECTOR_MODAL_BODY = '.modal-body';\nconst SELECTOR_DATA_TOGGLE$2 = '[data-bs-toggle=\"modal\"]';\nconst Default$6 = {\n  backdrop: true,\n  focus: true,\n  keyboard: true\n};\nconst DefaultType$6 = {\n  backdrop: '(boolean|string)',\n  focus: 'boolean',\n  keyboard: 'boolean'\n};\n\n/**\n * Class definition\n */\n\nclass Modal extends BaseComponent {\n  constructor(element, config) {\n    super(element, config);\n    this._dialog = SelectorEngine.findOne(SELECTOR_DIALOG, this._element);\n    this._backdrop = this._initializeBackDrop();\n    this._focustrap = this._initializeFocusTrap();\n    this._isShown = false;\n    this._isTransitioning = false;\n    this._scrollBar = new ScrollBarHelper();\n    this._addEventListeners();\n  }\n\n  // Getters\n  static get Default() {\n    return Default$6;\n  }\n  static get DefaultType() {\n    return DefaultType$6;\n  }\n  static get NAME() {\n    return NAME$7;\n  }\n\n  // Public\n  toggle(relatedTarget) {\n    return this._isShown ? this.hide() : this.show(relatedTarget);\n  }\n  show(relatedTarget) {\n    if (this._isShown || this._isTransitioning) {\n      return;\n    }\n    const showEvent = EventHandler.trigger(this._element, EVENT_SHOW$4, {\n      relatedTarget\n    });\n    if (showEvent.defaultPrevented) {\n      return;\n    }\n    this._isShown = true;\n    this._isTransitioning = true;\n    this._scrollBar.hide();\n    document.body.classList.add(CLASS_NAME_OPEN);\n    this._adjustDialog();\n    this._backdrop.show(() => this._showElement(relatedTarget));\n  }\n  hide() {\n    if (!this._isShown || this._isTransitioning) {\n      return;\n    }\n    const hideEvent = EventHandler.trigger(this._element, EVENT_HIDE$4);\n    if (hideEvent.defaultPrevented) {\n      return;\n    }\n    this._isShown = false;\n    this._isTransitioning = true;\n    this._focustrap.deactivate();\n    this._element.classList.remove(CLASS_NAME_SHOW$4);\n    this._queueCallback(() => this._hideModal(), this._element, this._isAnimated());\n  }\n  dispose() {\n    EventHandler.off(window, EVENT_KEY$4);\n    EventHandler.off(this._dialog, EVENT_KEY$4);\n    this._backdrop.dispose();\n    this._focustrap.deactivate();\n    super.dispose();\n  }\n  handleUpdate() {\n    this._adjustDialog();\n  }\n\n  // Private\n  _initializeBackDrop() {\n    return new Backdrop({\n      isVisible: Boolean(this._config.backdrop),\n      // 'static' option will be translated to true, and booleans will keep their value,\n      isAnimated: this._isAnimated()\n    });\n  }\n  _initializeFocusTrap() {\n    return new FocusTrap({\n      trapElement: this._element\n    });\n  }\n  _showElement(relatedTarget) {\n    // try to append dynamic modal\n    if (!document.body.contains(this._element)) {\n      document.body.append(this._element);\n    }\n    this._element.style.display = 'block';\n    this._element.removeAttribute('aria-hidden');\n    this._element.setAttribute('aria-modal', true);\n    this._element.setAttribute('role', 'dialog');\n    this._element.scrollTop = 0;\n    const modalBody = SelectorEngine.findOne(SELECTOR_MODAL_BODY, this._dialog);\n    if (modalBody) {\n      modalBody.scrollTop = 0;\n    }\n    reflow(this._element);\n    this._element.classList.add(CLASS_NAME_SHOW$4);\n    const transitionComplete = () => {\n      if (this._config.focus) {\n        this._focustrap.activate();\n      }\n      this._isTransitioning = false;\n      EventHandler.trigger(this._element, EVENT_SHOWN$4, {\n        relatedTarget\n      });\n    };\n    this._queueCallback(transitionComplete, this._dialog, this._isAnimated());\n  }\n  _addEventListeners() {\n    EventHandler.on(this._element, EVENT_KEYDOWN_DISMISS$1, event => {\n      if (event.key !== ESCAPE_KEY$1) {\n        return;\n      }\n      if (this._config.keyboard) {\n        this.hide();\n        return;\n      }\n      this._triggerBackdropTransition();\n    });\n    EventHandler.on(window, EVENT_RESIZE$1, () => {\n      if (this._isShown && !this._isTransitioning) {\n        this._adjustDialog();\n      }\n    });\n    EventHandler.on(this._element, EVENT_MOUSEDOWN_DISMISS, event => {\n      // a bad trick to segregate clicks that may start inside dialog but end outside, and avoid listen to scrollbar clicks\n      EventHandler.one(this._element, EVENT_CLICK_DISMISS, event2 => {\n        if (this._element !== event.target || this._element !== event2.target) {\n          return;\n        }\n        if (this._config.backdrop === 'static') {\n          this._triggerBackdropTransition();\n          return;\n        }\n        if (this._config.backdrop) {\n          this.hide();\n        }\n      });\n    });\n  }\n  _hideModal() {\n    this._element.style.display = 'none';\n    this._element.setAttribute('aria-hidden', true);\n    this._element.removeAttribute('aria-modal');\n    this._element.removeAttribute('role');\n    this._isTransitioning = false;\n    this._backdrop.hide(() => {\n      document.body.classList.remove(CLASS_NAME_OPEN);\n      this._resetAdjustments();\n      this._scrollBar.reset();\n      EventHandler.trigger(this._element, EVENT_HIDDEN$4);\n    });\n  }\n  _isAnimated() {\n    return this._element.classList.contains(CLASS_NAME_FADE$3);\n  }\n  _triggerBackdropTransition() {\n    const hideEvent = EventHandler.trigger(this._element, EVENT_HIDE_PREVENTED$1);\n    if (hideEvent.defaultPrevented) {\n      return;\n    }\n    const isModalOverflowing = this._element.scrollHeight > document.documentElement.clientHeight;\n    const initialOverflowY = this._element.style.overflowY;\n    // return if the following background transition hasn't yet completed\n    if (initialOverflowY === 'hidden' || this._element.classList.contains(CLASS_NAME_STATIC)) {\n      return;\n    }\n    if (!isModalOverflowing) {\n      this._element.style.overflowY = 'hidden';\n    }\n    this._element.classList.add(CLASS_NAME_STATIC);\n    this._queueCallback(() => {\n      this._element.classList.remove(CLASS_NAME_STATIC);\n      this._queueCallback(() => {\n        this._element.style.overflowY = initialOverflowY;\n      }, this._dialog);\n    }, this._dialog);\n    this._element.focus();\n  }\n\n  /**\n   * The following methods are used to handle overflowing modals\n   */\n\n  _adjustDialog() {\n    const isModalOverflowing = this._element.scrollHeight > document.documentElement.clientHeight;\n    const scrollbarWidth = this._scrollBar.getWidth();\n    const isBodyOverflowing = scrollbarWidth > 0;\n    if (isBodyOverflowing && !isModalOverflowing) {\n      const property = isRTL() ? 'paddingLeft' : 'paddingRight';\n      this._element.style[property] = `${scrollbarWidth}px`;\n    }\n    if (!isBodyOverflowing && isModalOverflowing) {\n      const property = isRTL() ? 'paddingRight' : 'paddingLeft';\n      this._element.style[property] = `${scrollbarWidth}px`;\n    }\n  }\n  _resetAdjustments() {\n    this._element.style.paddingLeft = '';\n    this._element.style.paddingRight = '';\n  }\n\n  // Static\n  static jQueryInterface(config, relatedTarget) {\n    return this.each(function () {\n      const data = Modal.getOrCreateInstance(this, config);\n      if (typeof config !== 'string') {\n        return;\n      }\n      if (typeof data[config] === 'undefined') {\n        throw new TypeError(`No method named \"${config}\"`);\n      }\n      data[config](relatedTarget);\n    });\n  }\n}\n\n/**\n * Data API implementation\n */\n\nEventHandler.on(document, EVENT_CLICK_DATA_API$2, SELECTOR_DATA_TOGGLE$2, function (event) {\n  const target = SelectorEngine.getElementFromSelector(this);\n  if (['A', 'AREA'].includes(this.tagName)) {\n    event.preventDefault();\n  }\n  EventHandler.one(target, EVENT_SHOW$4, showEvent => {\n    if (showEvent.defaultPrevented) {\n      // only register focus restorer if modal will actually get shown\n      return;\n    }\n    EventHandler.one(target, EVENT_HIDDEN$4, () => {\n      if (isVisible(this)) {\n        this.focus();\n      }\n    });\n  });\n\n  // avoid conflict when clicking modal toggler while another one is open\n  const alreadyOpen = SelectorEngine.findOne(OPEN_SELECTOR$1);\n  if (alreadyOpen) {\n    Modal.getInstance(alreadyOpen).hide();\n  }\n  const data = Modal.getOrCreateInstance(target);\n  data.toggle(this);\n});\nenableDismissTrigger(Modal);\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Modal);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap offcanvas.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$6 = 'offcanvas';\nconst DATA_KEY$3 = 'bs.offcanvas';\nconst EVENT_KEY$3 = `.${DATA_KEY$3}`;\nconst DATA_API_KEY$1 = '.data-api';\nconst EVENT_LOAD_DATA_API$2 = `load${EVENT_KEY$3}${DATA_API_KEY$1}`;\nconst ESCAPE_KEY = 'Escape';\nconst CLASS_NAME_SHOW$3 = 'show';\nconst CLASS_NAME_SHOWING$1 = 'showing';\nconst CLASS_NAME_HIDING = 'hiding';\nconst CLASS_NAME_BACKDROP = 'offcanvas-backdrop';\nconst OPEN_SELECTOR = '.offcanvas.show';\nconst EVENT_SHOW$3 = `show${EVENT_KEY$3}`;\nconst EVENT_SHOWN$3 = `shown${EVENT_KEY$3}`;\nconst EVENT_HIDE$3 = `hide${EVENT_KEY$3}`;\nconst EVENT_HIDE_PREVENTED = `hidePrevented${EVENT_KEY$3}`;\nconst EVENT_HIDDEN$3 = `hidden${EVENT_KEY$3}`;\nconst EVENT_RESIZE = `resize${EVENT_KEY$3}`;\nconst EVENT_CLICK_DATA_API$1 = `click${EVENT_KEY$3}${DATA_API_KEY$1}`;\nconst EVENT_KEYDOWN_DISMISS = `keydown.dismiss${EVENT_KEY$3}`;\nconst SELECTOR_DATA_TOGGLE$1 = '[data-bs-toggle=\"offcanvas\"]';\nconst Default$5 = {\n  backdrop: true,\n  keyboard: true,\n  scroll: false\n};\nconst DefaultType$5 = {\n  backdrop: '(boolean|string)',\n  keyboard: 'boolean',\n  scroll: 'boolean'\n};\n\n/**\n * Class definition\n */\n\nclass Offcanvas extends BaseComponent {\n  constructor(element, config) {\n    super(element, config);\n    this._isShown = false;\n    this._backdrop = this._initializeBackDrop();\n    this._focustrap = this._initializeFocusTrap();\n    this._addEventListeners();\n  }\n\n  // Getters\n  static get Default() {\n    return Default$5;\n  }\n  static get DefaultType() {\n    return DefaultType$5;\n  }\n  static get NAME() {\n    return NAME$6;\n  }\n\n  // Public\n  toggle(relatedTarget) {\n    return this._isShown ? this.hide() : this.show(relatedTarget);\n  }\n  show(relatedTarget) {\n    if (this._isShown) {\n      return;\n    }\n    const showEvent = EventHandler.trigger(this._element, EVENT_SHOW$3, {\n      relatedTarget\n    });\n    if (showEvent.defaultPrevented) {\n      return;\n    }\n    this._isShown = true;\n    this._backdrop.show();\n    if (!this._config.scroll) {\n      new ScrollBarHelper().hide();\n    }\n    this._element.setAttribute('aria-modal', true);\n    this._element.setAttribute('role', 'dialog');\n    this._element.classList.add(CLASS_NAME_SHOWING$1);\n    const completeCallBack = () => {\n      if (!this._config.scroll || this._config.backdrop) {\n        this._focustrap.activate();\n      }\n      this._element.classList.add(CLASS_NAME_SHOW$3);\n      this._element.classList.remove(CLASS_NAME_SHOWING$1);\n      EventHandler.trigger(this._element, EVENT_SHOWN$3, {\n        relatedTarget\n      });\n    };\n    this._queueCallback(completeCallBack, this._element, true);\n  }\n  hide() {\n    if (!this._isShown) {\n      return;\n    }\n    const hideEvent = EventHandler.trigger(this._element, EVENT_HIDE$3);\n    if (hideEvent.defaultPrevented) {\n      return;\n    }\n    this._focustrap.deactivate();\n    this._element.blur();\n    this._isShown = false;\n    this._element.classList.add(CLASS_NAME_HIDING);\n    this._backdrop.hide();\n    const completeCallback = () => {\n      this._element.classList.remove(CLASS_NAME_SHOW$3, CLASS_NAME_HIDING);\n      this._element.removeAttribute('aria-modal');\n      this._element.removeAttribute('role');\n      if (!this._config.scroll) {\n        new ScrollBarHelper().reset();\n      }\n      EventHandler.trigger(this._element, EVENT_HIDDEN$3);\n    };\n    this._queueCallback(completeCallback, this._element, true);\n  }\n  dispose() {\n    this._backdrop.dispose();\n    this._focustrap.deactivate();\n    super.dispose();\n  }\n\n  // Private\n  _initializeBackDrop() {\n    const clickCallback = () => {\n      if (this._config.backdrop === 'static') {\n        EventHandler.trigger(this._element, EVENT_HIDE_PREVENTED);\n        return;\n      }\n      this.hide();\n    };\n\n    // 'static' option will be translated to true, and booleans will keep their value\n    const isVisible = Boolean(this._config.backdrop);\n    return new Backdrop({\n      className: CLASS_NAME_BACKDROP,\n      isVisible,\n      isAnimated: true,\n      rootElement: this._element.parentNode,\n      clickCallback: isVisible ? clickCallback : null\n    });\n  }\n  _initializeFocusTrap() {\n    return new FocusTrap({\n      trapElement: this._element\n    });\n  }\n  _addEventListeners() {\n    EventHandler.on(this._element, EVENT_KEYDOWN_DISMISS, event => {\n      if (event.key !== ESCAPE_KEY) {\n        return;\n      }\n      if (this._config.keyboard) {\n        this.hide();\n        return;\n      }\n      EventHandler.trigger(this._element, EVENT_HIDE_PREVENTED);\n    });\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    return this.each(function () {\n      const data = Offcanvas.getOrCreateInstance(this, config);\n      if (typeof config !== 'string') {\n        return;\n      }\n      if (data[config] === undefined || config.startsWith('_') || config === 'constructor') {\n        throw new TypeError(`No method named \"${config}\"`);\n      }\n      data[config](this);\n    });\n  }\n}\n\n/**\n * Data API implementation\n */\n\nEventHandler.on(document, EVENT_CLICK_DATA_API$1, SELECTOR_DATA_TOGGLE$1, function (event) {\n  const target = SelectorEngine.getElementFromSelector(this);\n  if (['A', 'AREA'].includes(this.tagName)) {\n    event.preventDefault();\n  }\n  if (isDisabled(this)) {\n    return;\n  }\n  EventHandler.one(target, EVENT_HIDDEN$3, () => {\n    // focus on trigger when it is closed\n    if (isVisible(this)) {\n      this.focus();\n    }\n  });\n\n  // avoid conflict when clicking a toggler of an offcanvas, while another is open\n  const alreadyOpen = SelectorEngine.findOne(OPEN_SELECTOR);\n  if (alreadyOpen && alreadyOpen !== target) {\n    Offcanvas.getInstance(alreadyOpen).hide();\n  }\n  const data = Offcanvas.getOrCreateInstance(target);\n  data.toggle(this);\n});\nEventHandler.on(window, EVENT_LOAD_DATA_API$2, () => {\n  for (const selector of SelectorEngine.find(OPEN_SELECTOR)) {\n    Offcanvas.getOrCreateInstance(selector).show();\n  }\n});\nEventHandler.on(window, EVENT_RESIZE, () => {\n  for (const element of SelectorEngine.find('[aria-modal][class*=show][class*=offcanvas-]')) {\n    if (getComputedStyle(element).position !== 'fixed') {\n      Offcanvas.getOrCreateInstance(element).hide();\n    }\n  }\n});\nenableDismissTrigger(Offcanvas);\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Offcanvas);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap util/sanitizer.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n// js-docs-start allow-list\nconst ARIA_ATTRIBUTE_PATTERN = /^aria-[\\w-]*$/i;\nconst DefaultAllowlist = {\n  // Global attributes allowed on any supplied element below.\n  '*': ['class', 'dir', 'id', 'lang', 'role', ARIA_ATTRIBUTE_PATTERN],\n  a: ['target', 'href', 'title', 'rel'],\n  area: [],\n  b: [],\n  br: [],\n  col: [],\n  code: [],\n  div: [],\n  em: [],\n  hr: [],\n  h1: [],\n  h2: [],\n  h3: [],\n  h4: [],\n  h5: [],\n  h6: [],\n  i: [],\n  img: ['src', 'srcset', 'alt', 'title', 'width', 'height'],\n  li: [],\n  ol: [],\n  p: [],\n  pre: [],\n  s: [],\n  small: [],\n  span: [],\n  sub: [],\n  sup: [],\n  strong: [],\n  u: [],\n  ul: []\n};\n// js-docs-end allow-list\n\nconst uriAttributes = new Set(['background', 'cite', 'href', 'itemtype', 'longdesc', 'poster', 'src', 'xlink:href']);\n\n/**\n * A pattern that recognizes URLs that are safe wrt. XSS in URL navigation\n * contexts.\n *\n * Shout-out to Angular https://github.com/angular/angular/blob/15.2.8/packages/core/src/sanitization/url_sanitizer.ts#L38\n */\n// eslint-disable-next-line unicorn/better-regex\nconst SAFE_URL_PATTERN = /^(?!javascript:)(?:[a-z0-9+.-]+:|[^&:/?#]*(?:[/?#]|$))/i;\nconst allowedAttribute = (attribute, allowedAttributeList) => {\n  const attributeName = attribute.nodeName.toLowerCase();\n  if (allowedAttributeList.includes(attributeName)) {\n    if (uriAttributes.has(attributeName)) {\n      return Boolean(SAFE_URL_PATTERN.test(attribute.nodeValue));\n    }\n    return true;\n  }\n\n  // Check if a regular expression validates the attribute.\n  return allowedAttributeList.filter(attributeRegex => attributeRegex instanceof RegExp).some(regex => regex.test(attributeName));\n};\nfunction sanitizeHtml(unsafeHtml, allowList, sanitizeFunction) {\n  if (!unsafeHtml.length) {\n    return unsafeHtml;\n  }\n  if (sanitizeFunction && typeof sanitizeFunction === 'function') {\n    return sanitizeFunction(unsafeHtml);\n  }\n  const domParser = new window.DOMParser();\n  const createdDocument = domParser.parseFromString(unsafeHtml, 'text/html');\n  const elements = [].concat(...createdDocument.body.querySelectorAll('*'));\n  for (const element of elements) {\n    const elementName = element.nodeName.toLowerCase();\n    if (!Object.keys(allowList).includes(elementName)) {\n      element.remove();\n      continue;\n    }\n    const attributeList = [].concat(...element.attributes);\n    const allowedAttributes = [].concat(allowList['*'] || [], allowList[elementName] || []);\n    for (const attribute of attributeList) {\n      if (!allowedAttribute(attribute, allowedAttributes)) {\n        element.removeAttribute(attribute.nodeName);\n      }\n    }\n  }\n  return createdDocument.body.innerHTML;\n}\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap util/template-factory.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$5 = 'TemplateFactory';\nconst Default$4 = {\n  allowList: DefaultAllowlist,\n  content: {},\n  // { selector : text ,  selector2 : text2 , }\n  extraClass: '',\n  html: false,\n  sanitize: true,\n  sanitizeFn: null,\n  template: '<div></div>'\n};\nconst DefaultType$4 = {\n  allowList: 'object',\n  content: 'object',\n  extraClass: '(string|function)',\n  html: 'boolean',\n  sanitize: 'boolean',\n  sanitizeFn: '(null|function)',\n  template: 'string'\n};\nconst DefaultContentType = {\n  entry: '(string|element|function|null)',\n  selector: '(string|element)'\n};\n\n/**\n * Class definition\n */\n\nclass TemplateFactory extends Config {\n  constructor(config) {\n    super();\n    this._config = this._getConfig(config);\n  }\n\n  // Getters\n  static get Default() {\n    return Default$4;\n  }\n  static get DefaultType() {\n    return DefaultType$4;\n  }\n  static get NAME() {\n    return NAME$5;\n  }\n\n  // Public\n  getContent() {\n    return Object.values(this._config.content).map(config => this._resolvePossibleFunction(config)).filter(Boolean);\n  }\n  hasContent() {\n    return this.getContent().length > 0;\n  }\n  changeContent(content) {\n    this._checkContent(content);\n    this._config.content = {\n      ...this._config.content,\n      ...content\n    };\n    return this;\n  }\n  toHtml() {\n    const templateWrapper = document.createElement('div');\n    templateWrapper.innerHTML = this._maybeSanitize(this._config.template);\n    for (const [selector, text] of Object.entries(this._config.content)) {\n      this._setContent(templateWrapper, text, selector);\n    }\n    const template = templateWrapper.children[0];\n    const extraClass = this._resolvePossibleFunction(this._config.extraClass);\n    if (extraClass) {\n      template.classList.add(...extraClass.split(' '));\n    }\n    return template;\n  }\n\n  // Private\n  _typeCheckConfig(config) {\n    super._typeCheckConfig(config);\n    this._checkContent(config.content);\n  }\n  _checkContent(arg) {\n    for (const [selector, content] of Object.entries(arg)) {\n      super._typeCheckConfig({\n        selector,\n        entry: content\n      }, DefaultContentType);\n    }\n  }\n  _setContent(template, content, selector) {\n    const templateElement = SelectorEngine.findOne(selector, template);\n    if (!templateElement) {\n      return;\n    }\n    content = this._resolvePossibleFunction(content);\n    if (!content) {\n      templateElement.remove();\n      return;\n    }\n    if (isElement(content)) {\n      this._putElementInTemplate(getElement(content), templateElement);\n      return;\n    }\n    if (this._config.html) {\n      templateElement.innerHTML = this._maybeSanitize(content);\n      return;\n    }\n    templateElement.textContent = content;\n  }\n  _maybeSanitize(arg) {\n    return this._config.sanitize ? sanitizeHtml(arg, this._config.allowList, this._config.sanitizeFn) : arg;\n  }\n  _resolvePossibleFunction(arg) {\n    return execute(arg, [this]);\n  }\n  _putElementInTemplate(element, templateElement) {\n    if (this._config.html) {\n      templateElement.innerHTML = '';\n      templateElement.append(element);\n      return;\n    }\n    templateElement.textContent = element.textContent;\n  }\n}\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap tooltip.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$4 = 'tooltip';\nconst DISALLOWED_ATTRIBUTES = new Set(['sanitize', 'allowList', 'sanitizeFn']);\nconst CLASS_NAME_FADE$2 = 'fade';\nconst CLASS_NAME_MODAL = 'modal';\nconst CLASS_NAME_SHOW$2 = 'show';\nconst SELECTOR_TOOLTIP_INNER = '.tooltip-inner';\nconst SELECTOR_MODAL = `.${CLASS_NAME_MODAL}`;\nconst EVENT_MODAL_HIDE = 'hide.bs.modal';\nconst TRIGGER_HOVER = 'hover';\nconst TRIGGER_FOCUS = 'focus';\nconst TRIGGER_CLICK = 'click';\nconst TRIGGER_MANUAL = 'manual';\nconst EVENT_HIDE$2 = 'hide';\nconst EVENT_HIDDEN$2 = 'hidden';\nconst EVENT_SHOW$2 = 'show';\nconst EVENT_SHOWN$2 = 'shown';\nconst EVENT_INSERTED = 'inserted';\nconst EVENT_CLICK$1 = 'click';\nconst EVENT_FOCUSIN$1 = 'focusin';\nconst EVENT_FOCUSOUT$1 = 'focusout';\nconst EVENT_MOUSEENTER = 'mouseenter';\nconst EVENT_MOUSELEAVE = 'mouseleave';\nconst AttachmentMap = {\n  AUTO: 'auto',\n  TOP: 'top',\n  RIGHT: isRTL() ? 'left' : 'right',\n  BOTTOM: 'bottom',\n  LEFT: isRTL() ? 'right' : 'left'\n};\nconst Default$3 = {\n  allowList: DefaultAllowlist,\n  animation: true,\n  boundary: 'clippingParents',\n  container: false,\n  customClass: '',\n  delay: 0,\n  fallbackPlacements: ['top', 'right', 'bottom', 'left'],\n  html: false,\n  offset: [0, 6],\n  placement: 'top',\n  popperConfig: null,\n  sanitize: true,\n  sanitizeFn: null,\n  selector: false,\n  template: '<div class=\"tooltip\" role=\"tooltip\">' + '<div class=\"tooltip-arrow\"></div>' + '<div class=\"tooltip-inner\"></div>' + '</div>',\n  title: '',\n  trigger: 'hover focus'\n};\nconst DefaultType$3 = {\n  allowList: 'object',\n  animation: 'boolean',\n  boundary: '(string|element)',\n  container: '(string|element|boolean)',\n  customClass: '(string|function)',\n  delay: '(number|object)',\n  fallbackPlacements: 'array',\n  html: 'boolean',\n  offset: '(array|string|function)',\n  placement: '(string|function)',\n  popperConfig: '(null|object|function)',\n  sanitize: 'boolean',\n  sanitizeFn: '(null|function)',\n  selector: '(string|boolean)',\n  template: 'string',\n  title: '(string|element|function)',\n  trigger: 'string'\n};\n\n/**\n * Class definition\n */\n\nclass Tooltip extends BaseComponent {\n  constructor(element, config) {\n    if (typeof Popper === 'undefined') {\n      throw new TypeError('Bootstrap\\'s tooltips require Popper (https://popper.js.org)');\n    }\n    super(element, config);\n\n    // Private\n    this._isEnabled = true;\n    this._timeout = 0;\n    this._isHovered = null;\n    this._activeTrigger = {};\n    this._popper = null;\n    this._templateFactory = null;\n    this._newContent = null;\n\n    // Protected\n    this.tip = null;\n    this._setListeners();\n    if (!this._config.selector) {\n      this._fixTitle();\n    }\n  }\n\n  // Getters\n  static get Default() {\n    return Default$3;\n  }\n  static get DefaultType() {\n    return DefaultType$3;\n  }\n  static get NAME() {\n    return NAME$4;\n  }\n\n  // Public\n  enable() {\n    this._isEnabled = true;\n  }\n  disable() {\n    this._isEnabled = false;\n  }\n  toggleEnabled() {\n    this._isEnabled = !this._isEnabled;\n  }\n  toggle() {\n    if (!this._isEnabled) {\n      return;\n    }\n    this._activeTrigger.click = !this._activeTrigger.click;\n    if (this._isShown()) {\n      this._leave();\n      return;\n    }\n    this._enter();\n  }\n  dispose() {\n    clearTimeout(this._timeout);\n    EventHandler.off(this._element.closest(SELECTOR_MODAL), EVENT_MODAL_HIDE, this._hideModalHandler);\n    if (this._element.getAttribute('data-bs-original-title')) {\n      this._element.setAttribute('title', this._element.getAttribute('data-bs-original-title'));\n    }\n    this._disposePopper();\n    super.dispose();\n  }\n  show() {\n    if (this._element.style.display === 'none') {\n      throw new Error('Please use show on visible elements');\n    }\n    if (!(this._isWithContent() && this._isEnabled)) {\n      return;\n    }\n    const showEvent = EventHandler.trigger(this._element, this.constructor.eventName(EVENT_SHOW$2));\n    const shadowRoot = findShadowRoot(this._element);\n    const isInTheDom = (shadowRoot || this._element.ownerDocument.documentElement).contains(this._element);\n    if (showEvent.defaultPrevented || !isInTheDom) {\n      return;\n    }\n\n    // TODO: v6 remove this or make it optional\n    this._disposePopper();\n    const tip = this._getTipElement();\n    this._element.setAttribute('aria-describedby', tip.getAttribute('id'));\n    const {\n      container\n    } = this._config;\n    if (!this._element.ownerDocument.documentElement.contains(this.tip)) {\n      container.append(tip);\n      EventHandler.trigger(this._element, this.constructor.eventName(EVENT_INSERTED));\n    }\n    this._popper = this._createPopper(tip);\n    tip.classList.add(CLASS_NAME_SHOW$2);\n\n    // If this is a touch-enabled device we add extra\n    // empty mouseover listeners to the body's immediate children;\n    // only needed because of broken event delegation on iOS\n    // https://www.quirksmode.org/blog/archives/2014/02/mouse_event_bub.html\n    if ('ontouchstart' in document.documentElement) {\n      for (const element of [].concat(...document.body.children)) {\n        EventHandler.on(element, 'mouseover', noop);\n      }\n    }\n    const complete = () => {\n      EventHandler.trigger(this._element, this.constructor.eventName(EVENT_SHOWN$2));\n      if (this._isHovered === false) {\n        this._leave();\n      }\n      this._isHovered = false;\n    };\n    this._queueCallback(complete, this.tip, this._isAnimated());\n  }\n  hide() {\n    if (!this._isShown()) {\n      return;\n    }\n    const hideEvent = EventHandler.trigger(this._element, this.constructor.eventName(EVENT_HIDE$2));\n    if (hideEvent.defaultPrevented) {\n      return;\n    }\n    const tip = this._getTipElement();\n    tip.classList.remove(CLASS_NAME_SHOW$2);\n\n    // If this is a touch-enabled device we remove the extra\n    // empty mouseover listeners we added for iOS support\n    if ('ontouchstart' in document.documentElement) {\n      for (const element of [].concat(...document.body.children)) {\n        EventHandler.off(element, 'mouseover', noop);\n      }\n    }\n    this._activeTrigger[TRIGGER_CLICK] = false;\n    this._activeTrigger[TRIGGER_FOCUS] = false;\n    this._activeTrigger[TRIGGER_HOVER] = false;\n    this._isHovered = null; // it is a trick to support manual triggering\n\n    const complete = () => {\n      if (this._isWithActiveTrigger()) {\n        return;\n      }\n      if (!this._isHovered) {\n        this._disposePopper();\n      }\n      this._element.removeAttribute('aria-describedby');\n      EventHandler.trigger(this._element, this.constructor.eventName(EVENT_HIDDEN$2));\n    };\n    this._queueCallback(complete, this.tip, this._isAnimated());\n  }\n  update() {\n    if (this._popper) {\n      this._popper.update();\n    }\n  }\n\n  // Protected\n  _isWithContent() {\n    return Boolean(this._getTitle());\n  }\n  _getTipElement() {\n    if (!this.tip) {\n      this.tip = this._createTipElement(this._newContent || this._getContentForTemplate());\n    }\n    return this.tip;\n  }\n  _createTipElement(content) {\n    const tip = this._getTemplateFactory(content).toHtml();\n\n    // TODO: remove this check in v6\n    if (!tip) {\n      return null;\n    }\n    tip.classList.remove(CLASS_NAME_FADE$2, CLASS_NAME_SHOW$2);\n    // TODO: v6 the following can be achieved with CSS only\n    tip.classList.add(`bs-${this.constructor.NAME}-auto`);\n    const tipId = getUID(this.constructor.NAME).toString();\n    tip.setAttribute('id', tipId);\n    if (this._isAnimated()) {\n      tip.classList.add(CLASS_NAME_FADE$2);\n    }\n    return tip;\n  }\n  setContent(content) {\n    this._newContent = content;\n    if (this._isShown()) {\n      this._disposePopper();\n      this.show();\n    }\n  }\n  _getTemplateFactory(content) {\n    if (this._templateFactory) {\n      this._templateFactory.changeContent(content);\n    } else {\n      this._templateFactory = new TemplateFactory({\n        ...this._config,\n        // the `content` var has to be after `this._config`\n        // to override config.content in case of popover\n        content,\n        extraClass: this._resolvePossibleFunction(this._config.customClass)\n      });\n    }\n    return this._templateFactory;\n  }\n  _getContentForTemplate() {\n    return {\n      [SELECTOR_TOOLTIP_INNER]: this._getTitle()\n    };\n  }\n  _getTitle() {\n    return this._resolvePossibleFunction(this._config.title) || this._element.getAttribute('data-bs-original-title');\n  }\n\n  // Private\n  _initializeOnDelegatedTarget(event) {\n    return this.constructor.getOrCreateInstance(event.delegateTarget, this._getDelegateConfig());\n  }\n  _isAnimated() {\n    return this._config.animation || this.tip && this.tip.classList.contains(CLASS_NAME_FADE$2);\n  }\n  _isShown() {\n    return this.tip && this.tip.classList.contains(CLASS_NAME_SHOW$2);\n  }\n  _createPopper(tip) {\n    const placement = execute(this._config.placement, [this, tip, this._element]);\n    const attachment = AttachmentMap[placement.toUpperCase()];\n    return Popper.createPopper(this._element, tip, this._getPopperConfig(attachment));\n  }\n  _getOffset() {\n    const {\n      offset\n    } = this._config;\n    if (typeof offset === 'string') {\n      return offset.split(',').map(value => Number.parseInt(value, 10));\n    }\n    if (typeof offset === 'function') {\n      return popperData => offset(popperData, this._element);\n    }\n    return offset;\n  }\n  _resolvePossibleFunction(arg) {\n    return execute(arg, [this._element]);\n  }\n  _getPopperConfig(attachment) {\n    const defaultBsPopperConfig = {\n      placement: attachment,\n      modifiers: [{\n        name: 'flip',\n        options: {\n          fallbackPlacements: this._config.fallbackPlacements\n        }\n      }, {\n        name: 'offset',\n        options: {\n          offset: this._getOffset()\n        }\n      }, {\n        name: 'preventOverflow',\n        options: {\n          boundary: this._config.boundary\n        }\n      }, {\n        name: 'arrow',\n        options: {\n          element: `.${this.constructor.NAME}-arrow`\n        }\n      }, {\n        name: 'preSetPlacement',\n        enabled: true,\n        phase: 'beforeMain',\n        fn: data => {\n          // Pre-set Popper's placement attribute in order to read the arrow sizes properly.\n          // Otherwise, Popper mixes up the width and height dimensions since the initial arrow style is for top placement\n          this._getTipElement().setAttribute('data-popper-placement', data.state.placement);\n        }\n      }]\n    };\n    return {\n      ...defaultBsPopperConfig,\n      ...execute(this._config.popperConfig, [defaultBsPopperConfig])\n    };\n  }\n  _setListeners() {\n    const triggers = this._config.trigger.split(' ');\n    for (const trigger of triggers) {\n      if (trigger === 'click') {\n        EventHandler.on(this._element, this.constructor.eventName(EVENT_CLICK$1), this._config.selector, event => {\n          const context = this._initializeOnDelegatedTarget(event);\n          context.toggle();\n        });\n      } else if (trigger !== TRIGGER_MANUAL) {\n        const eventIn = trigger === TRIGGER_HOVER ? this.constructor.eventName(EVENT_MOUSEENTER) : this.constructor.eventName(EVENT_FOCUSIN$1);\n        const eventOut = trigger === TRIGGER_HOVER ? this.constructor.eventName(EVENT_MOUSELEAVE) : this.constructor.eventName(EVENT_FOCUSOUT$1);\n        EventHandler.on(this._element, eventIn, this._config.selector, event => {\n          const context = this._initializeOnDelegatedTarget(event);\n          context._activeTrigger[event.type === 'focusin' ? TRIGGER_FOCUS : TRIGGER_HOVER] = true;\n          context._enter();\n        });\n        EventHandler.on(this._element, eventOut, this._config.selector, event => {\n          const context = this._initializeOnDelegatedTarget(event);\n          context._activeTrigger[event.type === 'focusout' ? TRIGGER_FOCUS : TRIGGER_HOVER] = context._element.contains(event.relatedTarget);\n          context._leave();\n        });\n      }\n    }\n    this._hideModalHandler = () => {\n      if (this._element) {\n        this.hide();\n      }\n    };\n    EventHandler.on(this._element.closest(SELECTOR_MODAL), EVENT_MODAL_HIDE, this._hideModalHandler);\n  }\n  _fixTitle() {\n    const title = this._element.getAttribute('title');\n    if (!title) {\n      return;\n    }\n    if (!this._element.getAttribute('aria-label') && !this._element.textContent.trim()) {\n      this._element.setAttribute('aria-label', title);\n    }\n    this._element.setAttribute('data-bs-original-title', title); // DO NOT USE IT. Is only for backwards compatibility\n    this._element.removeAttribute('title');\n  }\n  _enter() {\n    if (this._isShown() || this._isHovered) {\n      this._isHovered = true;\n      return;\n    }\n    this._isHovered = true;\n    this._setTimeout(() => {\n      if (this._isHovered) {\n        this.show();\n      }\n    }, this._config.delay.show);\n  }\n  _leave() {\n    if (this._isWithActiveTrigger()) {\n      return;\n    }\n    this._isHovered = false;\n    this._setTimeout(() => {\n      if (!this._isHovered) {\n        this.hide();\n      }\n    }, this._config.delay.hide);\n  }\n  _setTimeout(handler, timeout) {\n    clearTimeout(this._timeout);\n    this._timeout = setTimeout(handler, timeout);\n  }\n  _isWithActiveTrigger() {\n    return Object.values(this._activeTrigger).includes(true);\n  }\n  _getConfig(config) {\n    const dataAttributes = Manipulator.getDataAttributes(this._element);\n    for (const dataAttribute of Object.keys(dataAttributes)) {\n      if (DISALLOWED_ATTRIBUTES.has(dataAttribute)) {\n        delete dataAttributes[dataAttribute];\n      }\n    }\n    config = {\n      ...dataAttributes,\n      ...(typeof config === 'object' && config ? config : {})\n    };\n    config = this._mergeConfigObj(config);\n    config = this._configAfterMerge(config);\n    this._typeCheckConfig(config);\n    return config;\n  }\n  _configAfterMerge(config) {\n    config.container = config.container === false ? document.body : getElement(config.container);\n    if (typeof config.delay === 'number') {\n      config.delay = {\n        show: config.delay,\n        hide: config.delay\n      };\n    }\n    if (typeof config.title === 'number') {\n      config.title = config.title.toString();\n    }\n    if (typeof config.content === 'number') {\n      config.content = config.content.toString();\n    }\n    return config;\n  }\n  _getDelegateConfig() {\n    const config = {};\n    for (const [key, value] of Object.entries(this._config)) {\n      if (this.constructor.Default[key] !== value) {\n        config[key] = value;\n      }\n    }\n    config.selector = false;\n    config.trigger = 'manual';\n\n    // In the future can be replaced with:\n    // const keysWithDifferentValues = Object.entries(this._config).filter(entry => this.constructor.Default[entry[0]] !== this._config[entry[0]])\n    // `Object.fromEntries(keysWithDifferentValues)`\n    return config;\n  }\n  _disposePopper() {\n    if (this._popper) {\n      this._popper.destroy();\n      this._popper = null;\n    }\n    if (this.tip) {\n      this.tip.remove();\n      this.tip = null;\n    }\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    return this.each(function () {\n      const data = Tooltip.getOrCreateInstance(this, config);\n      if (typeof config !== 'string') {\n        return;\n      }\n      if (typeof data[config] === 'undefined') {\n        throw new TypeError(`No method named \"${config}\"`);\n      }\n      data[config]();\n    });\n  }\n}\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Tooltip);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap popover.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$3 = 'popover';\nconst SELECTOR_TITLE = '.popover-header';\nconst SELECTOR_CONTENT = '.popover-body';\nconst Default$2 = {\n  ...Tooltip.Default,\n  content: '',\n  offset: [0, 8],\n  placement: 'right',\n  template: '<div class=\"popover\" role=\"tooltip\">' + '<div class=\"popover-arrow\"></div>' + '<h3 class=\"popover-header\"></h3>' + '<div class=\"popover-body\"></div>' + '</div>',\n  trigger: 'click'\n};\nconst DefaultType$2 = {\n  ...Tooltip.DefaultType,\n  content: '(null|string|element|function)'\n};\n\n/**\n * Class definition\n */\n\nclass Popover extends Tooltip {\n  // Getters\n  static get Default() {\n    return Default$2;\n  }\n  static get DefaultType() {\n    return DefaultType$2;\n  }\n  static get NAME() {\n    return NAME$3;\n  }\n\n  // Overrides\n  _isWithContent() {\n    return this._getTitle() || this._getContent();\n  }\n\n  // Private\n  _getContentForTemplate() {\n    return {\n      [SELECTOR_TITLE]: this._getTitle(),\n      [SELECTOR_CONTENT]: this._getContent()\n    };\n  }\n  _getContent() {\n    return this._resolvePossibleFunction(this._config.content);\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    return this.each(function () {\n      const data = Popover.getOrCreateInstance(this, config);\n      if (typeof config !== 'string') {\n        return;\n      }\n      if (typeof data[config] === 'undefined') {\n        throw new TypeError(`No method named \"${config}\"`);\n      }\n      data[config]();\n    });\n  }\n}\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Popover);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap scrollspy.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$2 = 'scrollspy';\nconst DATA_KEY$2 = 'bs.scrollspy';\nconst EVENT_KEY$2 = `.${DATA_KEY$2}`;\nconst DATA_API_KEY = '.data-api';\nconst EVENT_ACTIVATE = `activate${EVENT_KEY$2}`;\nconst EVENT_CLICK = `click${EVENT_KEY$2}`;\nconst EVENT_LOAD_DATA_API$1 = `load${EVENT_KEY$2}${DATA_API_KEY}`;\nconst CLASS_NAME_DROPDOWN_ITEM = 'dropdown-item';\nconst CLASS_NAME_ACTIVE$1 = 'active';\nconst SELECTOR_DATA_SPY = '[data-bs-spy=\"scroll\"]';\nconst SELECTOR_TARGET_LINKS = '[href]';\nconst SELECTOR_NAV_LIST_GROUP = '.nav, .list-group';\nconst SELECTOR_NAV_LINKS = '.nav-link';\nconst SELECTOR_NAV_ITEMS = '.nav-item';\nconst SELECTOR_LIST_ITEMS = '.list-group-item';\nconst SELECTOR_LINK_ITEMS = `${SELECTOR_NAV_LINKS}, ${SELECTOR_NAV_ITEMS} > ${SELECTOR_NAV_LINKS}, ${SELECTOR_LIST_ITEMS}`;\nconst SELECTOR_DROPDOWN = '.dropdown';\nconst SELECTOR_DROPDOWN_TOGGLE$1 = '.dropdown-toggle';\nconst Default$1 = {\n  offset: null,\n  // TODO: v6 @deprecated, keep it for backwards compatibility reasons\n  rootMargin: '0px 0px -25%',\n  smoothScroll: false,\n  target: null,\n  threshold: [0.1, 0.5, 1]\n};\nconst DefaultType$1 = {\n  offset: '(number|null)',\n  // TODO v6 @deprecated, keep it for backwards compatibility reasons\n  rootMargin: 'string',\n  smoothScroll: 'boolean',\n  target: 'element',\n  threshold: 'array'\n};\n\n/**\n * Class definition\n */\n\nclass ScrollSpy extends BaseComponent {\n  constructor(element, config) {\n    super(element, config);\n\n    // this._element is the observablesContainer and config.target the menu links wrapper\n    this._targetLinks = new Map();\n    this._observableSections = new Map();\n    this._rootElement = getComputedStyle(this._element).overflowY === 'visible' ? null : this._element;\n    this._activeTarget = null;\n    this._observer = null;\n    this._previousScrollData = {\n      visibleEntryTop: 0,\n      parentScrollTop: 0\n    };\n    this.refresh(); // initialize\n  }\n\n  // Getters\n  static get Default() {\n    return Default$1;\n  }\n  static get DefaultType() {\n    return DefaultType$1;\n  }\n  static get NAME() {\n    return NAME$2;\n  }\n\n  // Public\n  refresh() {\n    this._initializeTargetsAndObservables();\n    this._maybeEnableSmoothScroll();\n    if (this._observer) {\n      this._observer.disconnect();\n    } else {\n      this._observer = this._getNewObserver();\n    }\n    for (const section of this._observableSections.values()) {\n      this._observer.observe(section);\n    }\n  }\n  dispose() {\n    this._observer.disconnect();\n    super.dispose();\n  }\n\n  // Private\n  _configAfterMerge(config) {\n    // TODO: on v6 target should be given explicitly & remove the {target: 'ss-target'} case\n    config.target = getElement(config.target) || document.body;\n\n    // TODO: v6 Only for backwards compatibility reasons. Use rootMargin only\n    config.rootMargin = config.offset ? `${config.offset}px 0px -30%` : config.rootMargin;\n    if (typeof config.threshold === 'string') {\n      config.threshold = config.threshold.split(',').map(value => Number.parseFloat(value));\n    }\n    return config;\n  }\n  _maybeEnableSmoothScroll() {\n    if (!this._config.smoothScroll) {\n      return;\n    }\n\n    // unregister any previous listeners\n    EventHandler.off(this._config.target, EVENT_CLICK);\n    EventHandler.on(this._config.target, EVENT_CLICK, SELECTOR_TARGET_LINKS, event => {\n      const observableSection = this._observableSections.get(event.target.hash);\n      if (observableSection) {\n        event.preventDefault();\n        const root = this._rootElement || window;\n        const height = observableSection.offsetTop - this._element.offsetTop;\n        if (root.scrollTo) {\n          root.scrollTo({\n            top: height,\n            behavior: 'smooth'\n          });\n          return;\n        }\n\n        // Chrome 60 doesn't support `scrollTo`\n        root.scrollTop = height;\n      }\n    });\n  }\n  _getNewObserver() {\n    const options = {\n      root: this._rootElement,\n      threshold: this._config.threshold,\n      rootMargin: this._config.rootMargin\n    };\n    return new IntersectionObserver(entries => this._observerCallback(entries), options);\n  }\n\n  // The logic of selection\n  _observerCallback(entries) {\n    const targetElement = entry => this._targetLinks.get(`#${entry.target.id}`);\n    const activate = entry => {\n      this._previousScrollData.visibleEntryTop = entry.target.offsetTop;\n      this._process(targetElement(entry));\n    };\n    const parentScrollTop = (this._rootElement || document.documentElement).scrollTop;\n    const userScrollsDown = parentScrollTop >= this._previousScrollData.parentScrollTop;\n    this._previousScrollData.parentScrollTop = parentScrollTop;\n    for (const entry of entries) {\n      if (!entry.isIntersecting) {\n        this._activeTarget = null;\n        this._clearActiveClass(targetElement(entry));\n        continue;\n      }\n      const entryIsLowerThanPrevious = entry.target.offsetTop >= this._previousScrollData.visibleEntryTop;\n      // if we are scrolling down, pick the bigger offsetTop\n      if (userScrollsDown && entryIsLowerThanPrevious) {\n        activate(entry);\n        // if parent isn't scrolled, let's keep the first visible item, breaking the iteration\n        if (!parentScrollTop) {\n          return;\n        }\n        continue;\n      }\n\n      // if we are scrolling up, pick the smallest offsetTop\n      if (!userScrollsDown && !entryIsLowerThanPrevious) {\n        activate(entry);\n      }\n    }\n  }\n  _initializeTargetsAndObservables() {\n    this._targetLinks = new Map();\n    this._observableSections = new Map();\n    const targetLinks = SelectorEngine.find(SELECTOR_TARGET_LINKS, this._config.target);\n    for (const anchor of targetLinks) {\n      // ensure that the anchor has an id and is not disabled\n      if (!anchor.hash || isDisabled(anchor)) {\n        continue;\n      }\n      const observableSection = SelectorEngine.findOne(decodeURI(anchor.hash), this._element);\n\n      // ensure that the observableSection exists & is visible\n      if (isVisible(observableSection)) {\n        this._targetLinks.set(decodeURI(anchor.hash), anchor);\n        this._observableSections.set(anchor.hash, observableSection);\n      }\n    }\n  }\n  _process(target) {\n    if (this._activeTarget === target) {\n      return;\n    }\n    this._clearActiveClass(this._config.target);\n    this._activeTarget = target;\n    target.classList.add(CLASS_NAME_ACTIVE$1);\n    this._activateParents(target);\n    EventHandler.trigger(this._element, EVENT_ACTIVATE, {\n      relatedTarget: target\n    });\n  }\n  _activateParents(target) {\n    // Activate dropdown parents\n    if (target.classList.contains(CLASS_NAME_DROPDOWN_ITEM)) {\n      SelectorEngine.findOne(SELECTOR_DROPDOWN_TOGGLE$1, target.closest(SELECTOR_DROPDOWN)).classList.add(CLASS_NAME_ACTIVE$1);\n      return;\n    }\n    for (const listGroup of SelectorEngine.parents(target, SELECTOR_NAV_LIST_GROUP)) {\n      // Set triggered links parents as active\n      // With both <ul> and <nav> markup a parent is the previous sibling of any nav ancestor\n      for (const item of SelectorEngine.prev(listGroup, SELECTOR_LINK_ITEMS)) {\n        item.classList.add(CLASS_NAME_ACTIVE$1);\n      }\n    }\n  }\n  _clearActiveClass(parent) {\n    parent.classList.remove(CLASS_NAME_ACTIVE$1);\n    const activeNodes = SelectorEngine.find(`${SELECTOR_TARGET_LINKS}.${CLASS_NAME_ACTIVE$1}`, parent);\n    for (const node of activeNodes) {\n      node.classList.remove(CLASS_NAME_ACTIVE$1);\n    }\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    return this.each(function () {\n      const data = ScrollSpy.getOrCreateInstance(this, config);\n      if (typeof config !== 'string') {\n        return;\n      }\n      if (data[config] === undefined || config.startsWith('_') || config === 'constructor') {\n        throw new TypeError(`No method named \"${config}\"`);\n      }\n      data[config]();\n    });\n  }\n}\n\n/**\n * Data API implementation\n */\n\nEventHandler.on(window, EVENT_LOAD_DATA_API$1, () => {\n  for (const spy of SelectorEngine.find(SELECTOR_DATA_SPY)) {\n    ScrollSpy.getOrCreateInstance(spy);\n  }\n});\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(ScrollSpy);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap tab.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME$1 = 'tab';\nconst DATA_KEY$1 = 'bs.tab';\nconst EVENT_KEY$1 = `.${DATA_KEY$1}`;\nconst EVENT_HIDE$1 = `hide${EVENT_KEY$1}`;\nconst EVENT_HIDDEN$1 = `hidden${EVENT_KEY$1}`;\nconst EVENT_SHOW$1 = `show${EVENT_KEY$1}`;\nconst EVENT_SHOWN$1 = `shown${EVENT_KEY$1}`;\nconst EVENT_CLICK_DATA_API = `click${EVENT_KEY$1}`;\nconst EVENT_KEYDOWN = `keydown${EVENT_KEY$1}`;\nconst EVENT_LOAD_DATA_API = `load${EVENT_KEY$1}`;\nconst ARROW_LEFT_KEY = 'ArrowLeft';\nconst ARROW_RIGHT_KEY = 'ArrowRight';\nconst ARROW_UP_KEY = 'ArrowUp';\nconst ARROW_DOWN_KEY = 'ArrowDown';\nconst HOME_KEY = 'Home';\nconst END_KEY = 'End';\nconst CLASS_NAME_ACTIVE = 'active';\nconst CLASS_NAME_FADE$1 = 'fade';\nconst CLASS_NAME_SHOW$1 = 'show';\nconst CLASS_DROPDOWN = 'dropdown';\nconst SELECTOR_DROPDOWN_TOGGLE = '.dropdown-toggle';\nconst SELECTOR_DROPDOWN_MENU = '.dropdown-menu';\nconst NOT_SELECTOR_DROPDOWN_TOGGLE = `:not(${SELECTOR_DROPDOWN_TOGGLE})`;\nconst SELECTOR_TAB_PANEL = '.list-group, .nav, [role=\"tablist\"]';\nconst SELECTOR_OUTER = '.nav-item, .list-group-item';\nconst SELECTOR_INNER = `.nav-link${NOT_SELECTOR_DROPDOWN_TOGGLE}, .list-group-item${NOT_SELECTOR_DROPDOWN_TOGGLE}, [role=\"tab\"]${NOT_SELECTOR_DROPDOWN_TOGGLE}`;\nconst SELECTOR_DATA_TOGGLE = '[data-bs-toggle=\"tab\"], [data-bs-toggle=\"pill\"], [data-bs-toggle=\"list\"]'; // TODO: could only be `tab` in v6\nconst SELECTOR_INNER_ELEM = `${SELECTOR_INNER}, ${SELECTOR_DATA_TOGGLE}`;\nconst SELECTOR_DATA_TOGGLE_ACTIVE = `.${CLASS_NAME_ACTIVE}[data-bs-toggle=\"tab\"], .${CLASS_NAME_ACTIVE}[data-bs-toggle=\"pill\"], .${CLASS_NAME_ACTIVE}[data-bs-toggle=\"list\"]`;\n\n/**\n * Class definition\n */\n\nclass Tab extends BaseComponent {\n  constructor(element) {\n    super(element);\n    this._parent = this._element.closest(SELECTOR_TAB_PANEL);\n    if (!this._parent) {\n      return;\n      // TODO: should throw exception in v6\n      // throw new TypeError(`${element.outerHTML} has not a valid parent ${SELECTOR_INNER_ELEM}`)\n    }\n\n    // Set up initial aria attributes\n    this._setInitialAttributes(this._parent, this._getChildren());\n    EventHandler.on(this._element, EVENT_KEYDOWN, event => this._keydown(event));\n  }\n\n  // Getters\n  static get NAME() {\n    return NAME$1;\n  }\n\n  // Public\n  show() {\n    // Shows this elem and deactivate the active sibling if exists\n    const innerElem = this._element;\n    if (this._elemIsActive(innerElem)) {\n      return;\n    }\n\n    // Search for active tab on same parent to deactivate it\n    const active = this._getActiveElem();\n    const hideEvent = active ? EventHandler.trigger(active, EVENT_HIDE$1, {\n      relatedTarget: innerElem\n    }) : null;\n    const showEvent = EventHandler.trigger(innerElem, EVENT_SHOW$1, {\n      relatedTarget: active\n    });\n    if (showEvent.defaultPrevented || hideEvent && hideEvent.defaultPrevented) {\n      return;\n    }\n    this._deactivate(active, innerElem);\n    this._activate(innerElem, active);\n  }\n\n  // Private\n  _activate(element, relatedElem) {\n    if (!element) {\n      return;\n    }\n    element.classList.add(CLASS_NAME_ACTIVE);\n    this._activate(SelectorEngine.getElementFromSelector(element)); // Search and activate/show the proper section\n\n    const complete = () => {\n      if (element.getAttribute('role') !== 'tab') {\n        element.classList.add(CLASS_NAME_SHOW$1);\n        return;\n      }\n      element.removeAttribute('tabindex');\n      element.setAttribute('aria-selected', true);\n      this._toggleDropDown(element, true);\n      EventHandler.trigger(element, EVENT_SHOWN$1, {\n        relatedTarget: relatedElem\n      });\n    };\n    this._queueCallback(complete, element, element.classList.contains(CLASS_NAME_FADE$1));\n  }\n  _deactivate(element, relatedElem) {\n    if (!element) {\n      return;\n    }\n    element.classList.remove(CLASS_NAME_ACTIVE);\n    element.blur();\n    this._deactivate(SelectorEngine.getElementFromSelector(element)); // Search and deactivate the shown section too\n\n    const complete = () => {\n      if (element.getAttribute('role') !== 'tab') {\n        element.classList.remove(CLASS_NAME_SHOW$1);\n        return;\n      }\n      element.setAttribute('aria-selected', false);\n      element.setAttribute('tabindex', '-1');\n      this._toggleDropDown(element, false);\n      EventHandler.trigger(element, EVENT_HIDDEN$1, {\n        relatedTarget: relatedElem\n      });\n    };\n    this._queueCallback(complete, element, element.classList.contains(CLASS_NAME_FADE$1));\n  }\n  _keydown(event) {\n    if (![ARROW_LEFT_KEY, ARROW_RIGHT_KEY, ARROW_UP_KEY, ARROW_DOWN_KEY, HOME_KEY, END_KEY].includes(event.key)) {\n      return;\n    }\n    event.stopPropagation(); // stopPropagation/preventDefault both added to support up/down keys without scrolling the page\n    event.preventDefault();\n    const children = this._getChildren().filter(element => !isDisabled(element));\n    let nextActiveElement;\n    if ([HOME_KEY, END_KEY].includes(event.key)) {\n      nextActiveElement = children[event.key === HOME_KEY ? 0 : children.length - 1];\n    } else {\n      const isNext = [ARROW_RIGHT_KEY, ARROW_DOWN_KEY].includes(event.key);\n      nextActiveElement = getNextActiveElement(children, event.target, isNext, true);\n    }\n    if (nextActiveElement) {\n      nextActiveElement.focus({\n        preventScroll: true\n      });\n      Tab.getOrCreateInstance(nextActiveElement).show();\n    }\n  }\n  _getChildren() {\n    // collection of inner elements\n    return SelectorEngine.find(SELECTOR_INNER_ELEM, this._parent);\n  }\n  _getActiveElem() {\n    return this._getChildren().find(child => this._elemIsActive(child)) || null;\n  }\n  _setInitialAttributes(parent, children) {\n    this._setAttributeIfNotExists(parent, 'role', 'tablist');\n    for (const child of children) {\n      this._setInitialAttributesOnChild(child);\n    }\n  }\n  _setInitialAttributesOnChild(child) {\n    child = this._getInnerElement(child);\n    const isActive = this._elemIsActive(child);\n    const outerElem = this._getOuterElement(child);\n    child.setAttribute('aria-selected', isActive);\n    if (outerElem !== child) {\n      this._setAttributeIfNotExists(outerElem, 'role', 'presentation');\n    }\n    if (!isActive) {\n      child.setAttribute('tabindex', '-1');\n    }\n    this._setAttributeIfNotExists(child, 'role', 'tab');\n\n    // set attributes to the related panel too\n    this._setInitialAttributesOnTargetPanel(child);\n  }\n  _setInitialAttributesOnTargetPanel(child) {\n    const target = SelectorEngine.getElementFromSelector(child);\n    if (!target) {\n      return;\n    }\n    this._setAttributeIfNotExists(target, 'role', 'tabpanel');\n    if (child.id) {\n      this._setAttributeIfNotExists(target, 'aria-labelledby', `${child.id}`);\n    }\n  }\n  _toggleDropDown(element, open) {\n    const outerElem = this._getOuterElement(element);\n    if (!outerElem.classList.contains(CLASS_DROPDOWN)) {\n      return;\n    }\n    const toggle = (selector, className) => {\n      const element = SelectorEngine.findOne(selector, outerElem);\n      if (element) {\n        element.classList.toggle(className, open);\n      }\n    };\n    toggle(SELECTOR_DROPDOWN_TOGGLE, CLASS_NAME_ACTIVE);\n    toggle(SELECTOR_DROPDOWN_MENU, CLASS_NAME_SHOW$1);\n    outerElem.setAttribute('aria-expanded', open);\n  }\n  _setAttributeIfNotExists(element, attribute, value) {\n    if (!element.hasAttribute(attribute)) {\n      element.setAttribute(attribute, value);\n    }\n  }\n  _elemIsActive(elem) {\n    return elem.classList.contains(CLASS_NAME_ACTIVE);\n  }\n\n  // Try to get the inner element (usually the .nav-link)\n  _getInnerElement(elem) {\n    return elem.matches(SELECTOR_INNER_ELEM) ? elem : SelectorEngine.findOne(SELECTOR_INNER_ELEM, elem);\n  }\n\n  // Try to get the outer element (usually the .nav-item)\n  _getOuterElement(elem) {\n    return elem.closest(SELECTOR_OUTER) || elem;\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    return this.each(function () {\n      const data = Tab.getOrCreateInstance(this);\n      if (typeof config !== 'string') {\n        return;\n      }\n      if (data[config] === undefined || config.startsWith('_') || config === 'constructor') {\n        throw new TypeError(`No method named \"${config}\"`);\n      }\n      data[config]();\n    });\n  }\n}\n\n/**\n * Data API implementation\n */\n\nEventHandler.on(document, EVENT_CLICK_DATA_API, SELECTOR_DATA_TOGGLE, function (event) {\n  if (['A', 'AREA'].includes(this.tagName)) {\n    event.preventDefault();\n  }\n  if (isDisabled(this)) {\n    return;\n  }\n  Tab.getOrCreateInstance(this).show();\n});\n\n/**\n * Initialize on focus\n */\nEventHandler.on(window, EVENT_LOAD_DATA_API, () => {\n  for (const element of SelectorEngine.find(SELECTOR_DATA_TOGGLE_ACTIVE)) {\n    Tab.getOrCreateInstance(element);\n  }\n});\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Tab);\n\n/**\n * --------------------------------------------------------------------------\n * Bootstrap toast.js\n * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n * --------------------------------------------------------------------------\n */\n\n\n/**\n * Constants\n */\n\nconst NAME = 'toast';\nconst DATA_KEY = 'bs.toast';\nconst EVENT_KEY = `.${DATA_KEY}`;\nconst EVENT_MOUSEOVER = `mouseover${EVENT_KEY}`;\nconst EVENT_MOUSEOUT = `mouseout${EVENT_KEY}`;\nconst EVENT_FOCUSIN = `focusin${EVENT_KEY}`;\nconst EVENT_FOCUSOUT = `focusout${EVENT_KEY}`;\nconst EVENT_HIDE = `hide${EVENT_KEY}`;\nconst EVENT_HIDDEN = `hidden${EVENT_KEY}`;\nconst EVENT_SHOW = `show${EVENT_KEY}`;\nconst EVENT_SHOWN = `shown${EVENT_KEY}`;\nconst CLASS_NAME_FADE = 'fade';\nconst CLASS_NAME_HIDE = 'hide'; // @deprecated - kept here only for backwards compatibility\nconst CLASS_NAME_SHOW = 'show';\nconst CLASS_NAME_SHOWING = 'showing';\nconst DefaultType = {\n  animation: 'boolean',\n  autohide: 'boolean',\n  delay: 'number'\n};\nconst Default = {\n  animation: true,\n  autohide: true,\n  delay: 5000\n};\n\n/**\n * Class definition\n */\n\nclass Toast extends BaseComponent {\n  constructor(element, config) {\n    super(element, config);\n    this._timeout = null;\n    this._hasMouseInteraction = false;\n    this._hasKeyboardInteraction = false;\n    this._setListeners();\n  }\n\n  // Getters\n  static get Default() {\n    return Default;\n  }\n  static get DefaultType() {\n    return DefaultType;\n  }\n  static get NAME() {\n    return NAME;\n  }\n\n  // Public\n  show() {\n    const showEvent = EventHandler.trigger(this._element, EVENT_SHOW);\n    if (showEvent.defaultPrevented) {\n      return;\n    }\n    this._clearTimeout();\n    if (this._config.animation) {\n      this._element.classList.add(CLASS_NAME_FADE);\n    }\n    const complete = () => {\n      this._element.classList.remove(CLASS_NAME_SHOWING);\n      EventHandler.trigger(this._element, EVENT_SHOWN);\n      this._maybeScheduleHide();\n    };\n    this._element.classList.remove(CLASS_NAME_HIDE); // @deprecated\n    reflow(this._element);\n    this._element.classList.add(CLASS_NAME_SHOW, CLASS_NAME_SHOWING);\n    this._queueCallback(complete, this._element, this._config.animation);\n  }\n  hide() {\n    if (!this.isShown()) {\n      return;\n    }\n    const hideEvent = EventHandler.trigger(this._element, EVENT_HIDE);\n    if (hideEvent.defaultPrevented) {\n      return;\n    }\n    const complete = () => {\n      this._element.classList.add(CLASS_NAME_HIDE); // @deprecated\n      this._element.classList.remove(CLASS_NAME_SHOWING, CLASS_NAME_SHOW);\n      EventHandler.trigger(this._element, EVENT_HIDDEN);\n    };\n    this._element.classList.add(CLASS_NAME_SHOWING);\n    this._queueCallback(complete, this._element, this._config.animation);\n  }\n  dispose() {\n    this._clearTimeout();\n    if (this.isShown()) {\n      this._element.classList.remove(CLASS_NAME_SHOW);\n    }\n    super.dispose();\n  }\n  isShown() {\n    return this._element.classList.contains(CLASS_NAME_SHOW);\n  }\n\n  // Private\n\n  _maybeScheduleHide() {\n    if (!this._config.autohide) {\n      return;\n    }\n    if (this._hasMouseInteraction || this._hasKeyboardInteraction) {\n      return;\n    }\n    this._timeout = setTimeout(() => {\n      this.hide();\n    }, this._config.delay);\n  }\n  _onInteraction(event, isInteracting) {\n    switch (event.type) {\n      case 'mouseover':\n      case 'mouseout':\n        {\n          this._hasMouseInteraction = isInteracting;\n          break;\n        }\n      case 'focusin':\n      case 'focusout':\n        {\n          this._hasKeyboardInteraction = isInteracting;\n          break;\n        }\n    }\n    if (isInteracting) {\n      this._clearTimeout();\n      return;\n    }\n    const nextElement = event.relatedTarget;\n    if (this._element === nextElement || this._element.contains(nextElement)) {\n      return;\n    }\n    this._maybeScheduleHide();\n  }\n  _setListeners() {\n    EventHandler.on(this._element, EVENT_MOUSEOVER, event => this._onInteraction(event, true));\n    EventHandler.on(this._element, EVENT_MOUSEOUT, event => this._onInteraction(event, false));\n    EventHandler.on(this._element, EVENT_FOCUSIN, event => this._onInteraction(event, true));\n    EventHandler.on(this._element, EVENT_FOCUSOUT, event => this._onInteraction(event, false));\n  }\n  _clearTimeout() {\n    clearTimeout(this._timeout);\n    this._timeout = null;\n  }\n\n  // Static\n  static jQueryInterface(config) {\n    return this.each(function () {\n      const data = Toast.getOrCreateInstance(this, config);\n      if (typeof config === 'string') {\n        if (typeof data[config] === 'undefined') {\n          throw new TypeError(`No method named \"${config}\"`);\n        }\n        data[config](this);\n      }\n    });\n  }\n}\n\n/**\n * Data API implementation\n */\n\nenableDismissTrigger(Toast);\n\n/**\n * jQuery\n */\n\ndefineJQueryPlugin(Toast);\n\nexport { Alert, Button, Carousel, Collapse, Dropdown, Modal, Offcanvas, Popover, ScrollSpy, Tab, Toast, Tooltip };\n//# sourceMappingURL=bootstrap.esm.js.map\n","/* define several functions to replace jQuery methods\n * inspired by https://tobiasahlin.com/blog/move-from-jquery-to-vanilla-javascript/\n */\n\n/**\n * Execute a method if DOM has finished loading\n *\n * @param {function} callback the method to execute\n */\nexport function documentReady(callback) {\n  if (document.readyState != \"loading\") callback();\n  else document.addEventListener(\"DOMContentLoaded\", callback);\n}\n","// Import and setup functions to control Bootstrap's behavior.\nimport \"@popperjs/core\";\nimport { Tooltip } from \"bootstrap\";\nimport { documentReady } from \"./mixin\";\n\nimport \"../styles/bootstrap.scss\";\n\n/*******************************************************************************\n * Trigger tooltips\n */\n\n/**\n * Add tooltip to each element with the \"tooltip\" data-bs-toogle class\n */\nfunction TriggerTooltip() {\n  var tooltipTriggerList = [].slice.call(\n    document.querySelectorAll('[data-bs-toggle=\"tooltip\"]')\n  );\n  tooltipTriggerList.map(function (tooltipTriggerEl) {\n    return new Tooltip(tooltipTriggerEl, { delay: { show: 500, hide: 100 } });\n  });\n}\n\n/*******************************************************************************\n * back to top button\n */\nfunction backToTop() {\n  var btn = document.getElementById(\"pst-back-to-top\");\n  btn.addEventListener(\"click\", function () {\n    document.body.scrollTop = 0;\n    document.documentElement.scrollTop = 0;\n  });\n}\n\nfunction showBackToTop() {\n  var btn = document.getElementById(\"pst-back-to-top\");\n  var header = document\n    .getElementsByClassName(\"bd-header\")[0]\n    .getBoundingClientRect();\n  window.addEventListener(\"scroll\", function () {\n    if (this.oldScroll > this.scrollY && this.scrollY > header.bottom) {\n      btn.style.display = \"block\";\n    } else {\n      btn.style.display = \"none\";\n    }\n    this.oldScroll = this.scrollY;\n  });\n}\n\n/*******************************************************************************\n * Call functions after document loading.\n 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Symbol(\"webpack error\") : \"__webpack_error__\";\nvar resolveQueue = (queue) => {\n\tif(queue && queue.d < 1) {\n\t\tqueue.d = 1;\n\t\tqueue.forEach((fn) => (fn.r--));\n\t\tqueue.forEach((fn) => (fn.r-- ? fn.r++ : fn()));\n\t}\n}\nvar wrapDeps = (deps) => (deps.map((dep) => {\n\tif(dep !== null && typeof dep === \"object\") {\n\t\tif(dep[webpackQueues]) return dep;\n\t\tif(dep.then) {\n\t\t\tvar queue = [];\n\t\t\tqueue.d = 0;\n\t\t\tdep.then((r) => {\n\t\t\t\tobj[webpackExports] = r;\n\t\t\t\tresolveQueue(queue);\n\t\t\t}, (e) => {\n\t\t\t\tobj[webpackError] = e;\n\t\t\t\tresolveQueue(queue);\n\t\t\t});\n\t\t\tvar obj = {};\n\t\t\tobj[webpackQueues] = (fn) => (fn(queue));\n\t\t\treturn obj;\n\t\t}\n\t}\n\tvar ret = {};\n\tret[webpackQueues] = x => {};\n\tret[webpackExports] = dep;\n\treturn ret;\n}));\n__webpack_require__.a = (module, body, hasAwait) => {\n\tvar queue;\n\thasAwait && ((queue = []).d = -1);\n\tvar depQueues = new Set();\n\tvar exports = module.exports;\n\tvar currentDeps;\n\tvar outerResolve;\n\tvar reject;\n\tvar promise = new Promise((resolve, rej) => {\n\t\treject = rej;\n\t\touterResolve = resolve;\n\t});\n\tpromise[webpackExports] = exports;\n\tpromise[webpackQueues] = (fn) => (queue && fn(queue), depQueues.forEach(fn), promise[\"catch\"](x => {}));\n\tmodule.exports = promise;\n\tbody((deps) => {\n\t\tcurrentDeps = wrapDeps(deps);\n\t\tvar fn;\n\t\tvar getResult = () => (currentDeps.map((d) => {\n\t\t\tif(d[webpackError]) throw d[webpackError];\n\t\t\treturn d[webpackExports];\n\t\t}))\n\t\tvar promise = new Promise((resolve) => {\n\t\t\tfn = () => (resolve(getResult));\n\t\t\tfn.r = 0;\n\t\t\tvar fnQueue = (q) => (q !== queue && !depQueues.has(q) && (depQueues.add(q), q && !q.d && (fn.r++, q.push(fn))));\n\t\t\tcurrentDeps.map((dep) => (dep[webpackQueues](fnQueue)));\n\t\t});\n\t\treturn fn.r ? promise : getResult();\n\t}, (err) => ((err ? reject(promise[webpackError] = err) : outerResolve(exports)), resolveQueue(queue)));\n\tqueue && queue.d < 0 && (queue.d = 0);\n};","/**\n * Compare [semver](https://semver.org/) version strings to find greater, equal or lesser.\n * This library supports the full semver specification, including comparing versions with different number of digits like `1.0.0`, `1.0`, `1`, and pre-release versions like `1.0.0-alpha`.\n * @param v1 - First version to compare\n * @param v2 - Second version to compare\n * @returns Numeric value compatible with the [Array.sort(fn) interface](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort#Parameters).\n */\nexport const compareVersions = (v1, v2) => {\n    // validate input and split into segments\n    const n1 = validateAndParse(v1);\n    const n2 = validateAndParse(v2);\n    // pop off the patch\n    const p1 = n1.pop();\n    const p2 = n2.pop();\n    // validate numbers\n    const r = compareSegments(n1, n2);\n    if (r !== 0)\n        return r;\n    // validate pre-release\n    if (p1 && p2) {\n        return compareSegments(p1.split('.'), p2.split('.'));\n    }\n    else if (p1 || p2) {\n        return p1 ? -1 : 1;\n    }\n    return 0;\n};\n/**\n * Validate [semver](https://semver.org/) version strings.\n *\n * @param version Version number to validate\n * @returns `true` if the version number is a valid semver version number, `false` otherwise.\n *\n * @example\n * ```\n * validate('1.0.0-rc.1'); // return true\n * validate('1.0-rc.1'); // return false\n * validate('foo'); // return false\n * ```\n */\nexport const validate = (version) => typeof version === 'string' && /^[v\\d]/.test(version) && semver.test(version);\n/**\n * Compare [semver](https://semver.org/) version strings using the specified operator.\n *\n * @param v1 First version to compare\n * @param v2 Second version to compare\n * @param operator Allowed arithmetic operator to use\n * @returns `true` if the comparison between the firstVersion and the secondVersion satisfies the operator, `false` otherwise.\n *\n * @example\n * ```\n * compare('10.1.8', '10.0.4', '>'); // return true\n * compare('10.0.1', '10.0.1', '='); // return true\n * compare('10.1.1', '10.2.2', '<'); // return true\n * compare('10.1.1', '10.2.2', '<='); // return true\n * compare('10.1.1', '10.2.2', '>='); // return false\n * ```\n */\nexport const compare = (v1, v2, operator) => {\n    // validate input operator\n    assertValidOperator(operator);\n    // since result of compareVersions can only be -1 or 0 or 1\n    // a simple map can be used to replace switch\n    const res = compareVersions(v1, v2);\n    return operatorResMap[operator].includes(res);\n};\n/**\n * Match [npm semver](https://docs.npmjs.com/cli/v6/using-npm/semver) version range.\n *\n * @param version Version number to match\n * @param range Range pattern for version\n * @returns `true` if the version number is within the range, `false` otherwise.\n *\n * @example\n * ```\n * satisfies('1.1.0', '^1.0.0'); // return true\n * satisfies('1.1.0', '~1.0.0'); // return false\n * ```\n */\nexport const satisfies = (version, range) => {\n    // if no range operator then \"=\"\n    const m = range.match(/^([<>=~^]+)/);\n    const op = m ? m[1] : '=';\n    // if gt/lt/eq then operator compare\n    if (op !== '^' && op !== '~')\n        return compare(version, range, op);\n    // else range of either \"~\" or \"^\" is assumed\n    const [v1, v2, v3, , vp] = validateAndParse(version);\n    const [r1, r2, r3, , rp] = validateAndParse(range);\n    const v = [v1, v2, v3];\n    const r = [r1, r2 !== null && r2 !== void 0 ? r2 : 'x', r3 !== null && r3 !== void 0 ? r3 : 'x'];\n    // validate pre-release\n    if (rp) {\n        if (!vp)\n            return false;\n        if (compareSegments(v, r) !== 0)\n            return false;\n        if (compareSegments(vp.split('.'), rp.split('.')) === -1)\n            return false;\n    }\n    // first non-zero number\n    const nonZero = r.findIndex((v) => v !== '0') + 1;\n    // pointer to where segments can be >=\n    const i = op === '~' ? 2 : nonZero > 1 ? nonZero : 1;\n    // before pointer must be equal\n    if (compareSegments(v.slice(0, i), r.slice(0, i)) !== 0)\n        return false;\n    // after pointer must be >=\n    if (compareSegments(v.slice(i), r.slice(i)) === -1)\n        return false;\n    return true;\n};\nconst semver = /^[v^~<>=]*?(\\d+)(?:\\.([x*]|\\d+)(?:\\.([x*]|\\d+)(?:\\.([x*]|\\d+))?(?:-([\\da-z\\-]+(?:\\.[\\da-z\\-]+)*))?(?:\\+[\\da-z\\-]+(?:\\.[\\da-z\\-]+)*)?)?)?$/i;\nconst validateAndParse = (version) => {\n    if (typeof version !== 'string') {\n        throw new TypeError('Invalid argument expected string');\n    }\n    const match = version.match(semver);\n    if (!match) {\n        throw new Error(`Invalid argument not valid semver ('${version}' received)`);\n    }\n    match.shift();\n    return match;\n};\nconst isWildcard = (s) => s === '*' || s === 'x' || s === 'X';\nconst tryParse = (v) => {\n    const n = parseInt(v, 10);\n    return isNaN(n) ? v : n;\n};\nconst forceType = (a, b) => typeof a !== typeof b ? [String(a), String(b)] : [a, b];\nconst compareStrings = (a, b) => {\n    if (isWildcard(a) || isWildcard(b))\n        return 0;\n    const [ap, bp] = forceType(tryParse(a), tryParse(b));\n    if (ap > bp)\n        return 1;\n    if (ap < bp)\n        return -1;\n    return 0;\n};\nconst compareSegments = (a, b) => {\n    for (let i = 0; i < Math.max(a.length, b.length); i++) {\n        const r = compareStrings(a[i] || '0', b[i] || '0');\n        if (r !== 0)\n            return r;\n    }\n    return 0;\n};\nconst operatorResMap = {\n    '>': [1],\n    '>=': [0, 1],\n    '=': [0],\n    '<=': [-1, 0],\n    '<': [-1],\n};\nconst allowedOperators = Object.keys(operatorResMap);\nconst assertValidOperator = (op) => {\n    if (typeof op !== 'string') {\n        throw new TypeError(`Invalid operator type, expected string but got ${typeof op}`);\n    }\n    if (allowedOperators.indexOf(op) === -1) {\n        throw new Error(`Invalid operator, expected one of ${allowedOperators.join('|')}`);\n    }\n};\n//# sourceMappingURL=index.js.map","/* define several functions to replace jQuery methods\n * inspired by https://tobiasahlin.com/blog/move-from-jquery-to-vanilla-javascript/\n */\n\n/**\n * Execute a method if DOM has finished loading\n *\n * @param {function} callback the method to execute\n */\nexport function documentReady(callback) {\n  if (document.readyState != \"loading\") callback();\n  else document.addEventListener(\"DOMContentLoaded\", callback);\n}\n","// Define the custom behavior of the page\nimport { documentReady } from \"./mixin\";\nimport { compare, validate } from \"compare-versions\";\n\nimport \"../styles/pydata-sphinx-theme.scss\";\n\n/*******************************************************************************\n * Theme interaction\n */\n\nvar prefersDark = window.matchMedia(\"(prefers-color-scheme: dark)\");\n\n/**\n * set the the body theme to the one specified by the user browser\n *\n * @param {event} e\n */\nfunction autoTheme(e) {\n  document.documentElement.dataset.theme = prefersDark.matches\n    ? \"dark\"\n    : \"light\";\n}\n\n/**\n * Set the theme using the specified mode.\n * It can be one of [\"auto\", \"dark\", \"light\"]\n *\n * @param {str} mode\n */\nfunction setTheme(mode) {\n  if (mode !== \"light\" && mode !== \"dark\" && mode !== \"auto\") {\n    console.error(`Got invalid theme mode: ${mode}. Resetting to auto.`);\n    mode = \"auto\";\n  }\n\n  // get the theme\n  var colorScheme = prefersDark.matches ? \"dark\" : \"light\";\n  document.documentElement.dataset.mode = mode;\n  var theme = mode == \"auto\" ? colorScheme : mode;\n  document.documentElement.dataset.theme = theme;\n  // TODO: remove this line after Bootstrap upgrade\n  // v5.3 has a colors mode: https://getbootstrap.com/docs/5.3/customize/color-modes/\n  document.querySelectorAll(\".dropdown-menu\").forEach((el) => {\n    if (theme === \"dark\") {\n      el.classList.add(\"dropdown-menu-dark\");\n    } else {\n      el.classList.remove(\"dropdown-menu-dark\");\n    }\n  });\n\n  // save mode and theme\n  localStorage.setItem(\"mode\", mode);\n  localStorage.setItem(\"theme\", theme);\n  console.log(`[PST]: Changed to ${mode} mode using the ${theme} theme.`);\n\n  // add a listener if set on auto\n  prefersDark.onchange = mode == \"auto\" ? autoTheme : \"\";\n}\n\n/**\n * Change the theme option order so that clicking on the btn is always a change\n * from \"auto\"\n */\nfunction cycleMode() {\n  const defaultMode = document.documentElement.dataset.defaultMode || \"auto\";\n  const currentMode = localStorage.getItem(\"mode\") || defaultMode;\n\n  var loopArray = (arr, current) => {\n    var nextPosition = arr.indexOf(current) + 1;\n    if (nextPosition === arr.length) {\n      nextPosition = 0;\n    }\n    return arr[nextPosition];\n  };\n\n  // make sure the next theme after auto is always a change\n  var modeList = prefersDark.matches\n    ? [\"auto\", \"light\", \"dark\"]\n    : [\"auto\", \"dark\", \"light\"];\n  var newMode = loopArray(modeList, currentMode);\n  setTheme(newMode);\n}\n\n/**\n * add the theme listener on the btns of the navbar\n */\nfunction addModeListener() {\n  // the theme was set a first time using the initial mini-script\n  // running setMode will ensure the use of the dark mode if auto is selected\n  setTheme(document.documentElement.dataset.mode);\n\n  // Attach event handlers for toggling themes colors\n  document.querySelectorAll(\".theme-switch-button\").forEach((el) => {\n    el.addEventListener(\"click\", cycleMode);\n  });\n}\n\n/*******************************************************************************\n * TOC interactivity\n */\n\n/**\n * TOC sidebar - add \"active\" class to parent list\n *\n * Bootstrap's scrollspy adds the active class to the <a> link,\n * but for the automatic collapsing we need this on the parent list item.\n *\n * The event is triggered on \"window\" (and not the nav item as documented),\n * see https://github.com/twbs/bootstrap/issues/20086\n */\nfunction addTOCInteractivity() {\n  window.addEventListener(\"activate.bs.scrollspy\", function () {\n    const navLinks = document.querySelectorAll(\".bd-toc-nav a\");\n\n    navLinks.forEach((navLink) => {\n      navLink.parentElement.classList.remove(\"active\");\n    });\n\n    const activeNavLinks = document.querySelectorAll(\".bd-toc-nav a.active\");\n    activeNavLinks.forEach((navLink) => {\n      navLink.parentElement.classList.add(\"active\");\n    });\n  });\n}\n\n/*******************************************************************************\n * Scroll\n */\n\n/**\n * Navigation sidebar scrolling to active page\n */\nfunction scrollToActive() {\n  // If the docs nav doesn't exist, do nothing (e.g., on search page)\n  if (!document.querySelector(\".bd-docs-nav\")) {\n    return;\n  }\n\n  var sidebar = document.querySelector(\"div.bd-sidebar\");\n\n  // Remember the sidebar scroll position between page loads\n  // Inspired on source of revealjs.com\n  let storedScrollTop = parseInt(\n    sessionStorage.getItem(\"sidebar-scroll-top\"),\n    10\n  );\n\n  if (!isNaN(storedScrollTop)) {\n    // If we've got a saved scroll position, just use that\n    sidebar.scrollTop = storedScrollTop;\n    console.log(\"[PST]: Scrolled sidebar using stored browser position...\");\n  } else {\n    // Otherwise, calculate a position to scroll to based on the lowest `active` link\n    var sidebarNav = document.querySelector(\".bd-docs-nav\");\n    var active_pages = sidebarNav.querySelectorAll(\".active\");\n    if (active_pages.length > 0) {\n      // Use the last active page as the offset since it's the page we're on\n      var latest_active = active_pages[active_pages.length - 1];\n      var offset =\n        latest_active.getBoundingClientRect().y -\n        sidebar.getBoundingClientRect().y;\n      // Only scroll the navbar if the active link is lower than 50% of the page\n      if (latest_active.getBoundingClientRect().y > window.innerHeight * 0.5) {\n        let buffer = 0.25; // Buffer so we have some space above the scrolled item\n        sidebar.scrollTop = offset - sidebar.clientHeight * buffer;\n        console.log(\"[PST]: Scrolled sidebar using last active link...\");\n      }\n    }\n  }\n\n  // Store the sidebar scroll position\n  window.addEventListener(\"beforeunload\", () => {\n    sessionStorage.setItem(\"sidebar-scroll-top\", sidebar.scrollTop);\n  });\n}\n\n/*******************************************************************************\n * Search\n */\n\n/**\n * Find any search forms on the page and return their input element\n */\nvar findSearchInput = () => {\n  let forms = document.querySelectorAll(\"form.bd-search\");\n  if (!forms.length) {\n    // no search form found\n    return;\n  } else {\n    var form;\n    if (forms.length == 1) {\n      // there is exactly one search form (persistent or hidden)\n      form = forms[0];\n    } else {\n      // must be at least one persistent form, use the first persistent one\n      form = document.querySelector(\n        \"div:not(.search-button__search-container) > form.bd-search\"\n      );\n    }\n    return form.querySelector(\"input\");\n  }\n};\n\n/**\n * Activate the search field on the page.\n * - If there is a search field already visible it will be activated.\n * - If not, then a search field will pop up.\n */\nvar toggleSearchField = () => {\n  // Find the search input to highlight\n  let input = findSearchInput();\n\n  // if the input field is the hidden one (the one associated with the\n  // search button) then toggle the button state (to show/hide the field)\n  let searchPopupWrapper = document.querySelector(\".search-button__wrapper\");\n  let hiddenInput = searchPopupWrapper.querySelector(\"input\");\n  if (input === hiddenInput) {\n    searchPopupWrapper.classList.toggle(\"show\");\n  }\n  // when toggling off the search field, remove its focus\n  if (document.activeElement === input) {\n    input.blur();\n  } else {\n    input.focus();\n    input.select();\n    input.scrollIntoView({ block: \"center\" });\n  }\n};\n\n/**\n * Add an event listener for toggleSearchField() for Ctrl/Cmd + K\n */\nvar addEventListenerForSearchKeyboard = () => {\n  window.addEventListener(\n    \"keydown\",\n    (event) => {\n      let input = findSearchInput();\n      // toggle on Ctrl+k or ⌘+k\n      if (\n        // Ignore if shift or alt are pressed\n        !event.shiftKey &&\n        !event.altKey &&\n        // On Mac use ⌘, all other OS use Ctrl\n        (isMac\n          ? event.metaKey && !event.ctrlKey\n          : !event.metaKey && event.ctrlKey) &&\n        // Case-insensitive so the shortcut still works with caps lock\n        /k/i.test(event.key)\n      ) {\n        event.preventDefault();\n        toggleSearchField();\n      }\n      // also allow Escape key to hide (but not show) the dynamic search field\n      else if (document.activeElement === input && /Escape/i.test(event.key)) {\n        toggleSearchField();\n      }\n    },\n    true\n  );\n};\n\n/**\n * Find out if we're on a Mac\n */\nvar isMac = (navigator) => {\n  var platform = \"\";\n  if (\n    typeof navigator.userAgentData !== \"undefined\" &&\n    navigator.userAgentData != null\n  ) {\n    platform = navigator.userAgentData.platform;\n  } else if (typeof navigator.platform !== \"undefined\") {\n    platform = navigator.platform;\n  }\n  return /mac.?os/.test(platform.toLowerCase());\n};\n\n/**\n * Change the search hint to `meta key` if we are a Mac\n */\n\nvar changeSearchShortcutKey = () => {\n  let shortcuts = document.querySelectorAll(\".search-button__kbd-shortcut\");\n  if (isMac(window.navigator)) {\n    shortcuts.forEach(\n      (f) => (f.querySelector(\"kbd.kbd-shortcut__modifier\").innerText = \"⌘\")\n    );\n  }\n};\n\n/**\n * Activate callbacks for search button popup\n */\nvar setupSearchButtons = () => {\n  changeSearchShortcutKey();\n  addEventListenerForSearchKeyboard();\n\n  // Add the search button trigger event callback\n  document.querySelectorAll(\".search-button__button\").forEach((btn) => {\n    btn.onclick = toggleSearchField;\n  });\n\n  // Add the search button overlay event callback\n  let overlay = document.querySelector(\".search-button__overlay\");\n  if (overlay) {\n    overlay.onclick = toggleSearchField;\n  }\n};\n\n/*******************************************************************************\n * Version Switcher\n * Note that this depends on two variables existing that are defined in\n * and `html-page-context` hook:\n *\n * - DOCUMENTATION_OPTIONS.pagename\n * - DOCUMENTATION_OPTIONS.theme_switcher_url\n */\n\n/**\n * Check if corresponding page path exists in other version of docs\n * and, if so, go there instead of the homepage of the other docs version\n *\n * @param {event} event the event that trigger the check\n */\nasync function checkPageExistsAndRedirect(event) {\n  // ensure we don't follow the initial link\n  event.preventDefault();\n  let currentFilePath = `${DOCUMENTATION_OPTIONS.pagename}.html`;\n  let tryUrl = event.currentTarget.getAttribute(\"href\");\n  let otherDocsHomepage = tryUrl.replace(currentFilePath, \"\");\n  try {\n    let head = await fetch(tryUrl, { method: \"HEAD\" });\n    if (head.ok) {\n      location.href = tryUrl; // the page exists, go there\n    } else {\n      location.href = otherDocsHomepage;\n    }\n  } catch (err) {\n    // something went wrong, probably CORS restriction, fallback to other docs homepage\n    location.href = otherDocsHomepage;\n  }\n}\n\n/**\n * Load and parse the version switcher JSON file from an absolute or relative URL.\n *\n * @param {string} url The URL to load version switcher entries from.\n */\nasync function fetchVersionSwitcherJSON(url) {\n  // first check if it's a valid URL\n  try {\n    var result = new URL(url);\n  } catch (err) {\n    if (err instanceof TypeError) {\n      // assume we got a relative path, and fix accordingly. But first, we need to\n      // use `fetch()` to follow redirects so we get the correct final base URL\n      const origin = await fetch(window.location.origin, { method: \"HEAD\" });\n      result = new URL(url, origin.url);\n    } else {\n      // something unexpected happened\n      throw err;\n    }\n  }\n  // load and return the JSON\n  const response = await fetch(result);\n  const data = await response.json();\n  return data;\n}\n\n// Populate the version switcher from the JSON data\nfunction populateVersionSwitcher(data, versionSwitcherBtns) {\n  const currentFilePath = `${DOCUMENTATION_OPTIONS.pagename}.html`;\n  versionSwitcherBtns.forEach((btn) => {\n    // Set empty strings by default so that these attributes exist and can be used in CSS selectors\n    btn.dataset[\"activeVersionName\"] = \"\";\n    btn.dataset[\"activeVersion\"] = \"\";\n  });\n  // in case there are multiple entries with the same version string, this helps us\n  // decide which entry's `name` to put on the button itself. Without this, it would\n  // always be the *last* version-matching entry; now it will be either the\n  // version-matching entry that is also marked as `\"preferred\": true`, or if that\n  // doesn't exist: the *first* version-matching entry.\n  data = data.map((entry) => {\n    // does this entry match the version that we're currently building/viewing?\n    entry.match =\n      entry.version == DOCUMENTATION_OPTIONS.theme_switcher_version_match;\n    entry.preferred = entry.preferred || false;\n    // if no custom name specified (e.g., \"latest\"), use version string\n    if (!(\"name\" in entry)) {\n      entry.name = entry.version;\n    }\n    return entry;\n  });\n  const hasMatchingPreferredEntry = data\n    .map((entry) => entry.preferred && entry.match)\n    .some(Boolean);\n  var foundMatch = false;\n  // create links to the corresponding page in the other docs versions\n  data.forEach((entry) => {\n    // create the node\n    const anchor = document.createElement(\"a\");\n    anchor.setAttribute(\n      \"class\",\n      \"dropdown-item list-group-item list-group-item-action py-1\"\n    );\n    anchor.setAttribute(\"href\", `${entry.url}${currentFilePath}`);\n    anchor.setAttribute(\"role\", \"option\");\n    const span = document.createElement(\"span\");\n    span.textContent = `${entry.name}`;\n    anchor.appendChild(span);\n    // Add dataset values for the version and name in case people want\n    // to apply CSS styling based on this information.\n    anchor.dataset[\"versionName\"] = entry.name;\n    anchor.dataset[\"version\"] = entry.version;\n    // replace dropdown button text with the preferred display name of the\n    // currently-viewed version, rather than using sphinx's {{ version }} variable.\n    // also highlight the dropdown entry for the currently-viewed version's entry\n    let matchesAndIsPreferred = hasMatchingPreferredEntry && entry.preferred;\n    let matchesAndIsFirst =\n      !hasMatchingPreferredEntry && !foundMatch && entry.match;\n    if (matchesAndIsPreferred || matchesAndIsFirst) {\n      anchor.classList.add(\"active\");\n      versionSwitcherBtns.forEach((btn) => {\n        btn.innerText = entry.name;\n        btn.dataset[\"activeVersionName\"] = entry.name;\n        btn.dataset[\"activeVersion\"] = entry.version;\n      });\n      foundMatch = true;\n    }\n    // There may be multiple version-switcher elements, e.g. one\n    // in a slide-over panel displayed on smaller screens.\n    document.querySelectorAll(\".version-switcher__menu\").forEach((menu) => {\n      // we need to clone the node for each menu, but onclick attributes are not\n      // preserved by `.cloneNode()` so we add onclick here after cloning.\n      let node = anchor.cloneNode(true);\n      node.onclick = checkPageExistsAndRedirect;\n      // on click, AJAX calls will check if the linked page exists before\n      // trying to redirect, and if not, will redirect to the homepage\n      // for that version of the docs.\n      menu.append(node);\n    });\n  });\n}\n\n/*******************************************************************************\n * Warning banner when viewing non-stable version of the docs.\n */\n\n/**\n * Show a warning banner when viewing a non-stable version of the docs.\n *\n * adapted 2023-06 from https://mne.tools/versionwarning.js, which was\n * originally adapted 2020-05 from https://scikit-learn.org/versionwarning.js\n *\n * @param {Array} data The version data used to populate the switcher menu.\n */\nfunction showVersionWarningBanner(data) {\n  var version = DOCUMENTATION_OPTIONS.VERSION;\n  // figure out what latest stable version is\n  var preferredEntries = data.filter((entry) => entry.preferred);\n  if (preferredEntries.length !== 1) {\n    const howMany = preferredEntries.length == 0 ? \"No\" : \"Multiple\";\n    console.log(\n      `[PST] ${howMany} versions marked \"preferred\" found in versions JSON, ignoring.`\n    );\n    return;\n  }\n  const preferredVersion = 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This is primarily two things:\n *\n * 1. Hide the Table of Contents any time sidebar content is on the screen.\n *\n * This will be triggered any time a sidebar item enters or exits the screen.\n * It adds/removes items from an array if they have entered the screen, and\n * removes them when they exit the screen. It hides the TOC if anything is\n * on-screen.\n *\n * ref: https://developer.mozilla.org/en-US/docs/Web/API/Intersection_Observer_API\n *\n * 2. Add a `scrolled` class to <body> to trigger CSS changes.\n */\nvar initTocHide = () => {\n  var onScreenItems = [];\n  let hideTocCallback = (entries, observer) => {\n    // Check whether any sidebar item is displayed\n    entries.forEach((entry) => {\n      if (entry.isIntersecting) {\n        // If an element just came on screen, add it our list\n        onScreenItems.push(entry.target);\n      } else {\n        // Otherwise, if it's in our list then remove it\n        for (let ii = 0; ii < onScreenItems.length; ii++) {\n          if (onScreenItems[ii] === entry.target) {\n            onScreenItems.splice(ii, 1);\n            break;\n          }\n        }\n      }\n    });\n\n    // Hide the TOC if any margin content is displayed on the screen\n    if (onScreenItems.length > 0) {\n      document.querySelector(\"div.bd-sidebar-secondary\").classList.add(\"hide\");\n    } else {\n      document\n        .querySelector(\"div.bd-sidebar-secondary\")\n        .classList.remove(\"hide\");\n    }\n  };\n  let manageScrolledClassOnBody = (entries, observer) => {\n    // The pixel is at the top, so if we're < 0 that it means we've scrolled\n    if (entries[0].boundingClientRect.y < 0) {\n      document.body.classList.add(\"scrolled\");\n    } else {\n      document.body.classList.remove(\"scrolled\");\n    }\n  };\n\n  // Set up the intersection observer to watch all margin content\n  let options = {\n    // Trigger callback when the top of a margin item is 1/3 up the screen\n    rootMargin: \"0px 0px -33% 0px\",\n  };\n  let tocObserver = new IntersectionObserver(hideTocCallback, options);\n  // TODO: deprecate popout after v0.5.0\n  const selectorClasses = [\n    \"marginnote\",\n    \"sidenote\",\n    \"margin\",\n    \"margin-caption\",\n    \"full-width\",\n    \"sidebar\",\n    \"popout\",\n  ];\n  let marginSelector = [];\n  selectorClasses.forEach((ii) => {\n    // Use three permutations of each class name because `tag_` and `_` used to be supported\n    marginSelector.push(\n      ...[\n        `.${ii}`,\n        `.tag_${ii}`,\n        `.${ii.replace(\"-\", \"_\")}`,\n        `.tag_${ii.replace(\"-\", \"_\")}`,\n      ],\n    );\n  });\n  document.querySelectorAll(marginSelector.join(\", \")).forEach((ii) => {\n    tocObserver.observe(ii);\n  });\n\n  // Set up the observer to check if we've scrolled from top of page\n  let scrollObserver = new IntersectionObserver(manageScrolledClassOnBody);\n  scrollObserver.observe(document.querySelector(\".sbt-scroll-pixel-helper\"));\n};\n\n/**\n * Activate Thebe with a custom button click.\n */\nvar initThebeSBT = () => {\n  var title = document.querySelector(\"section h1\");\n  var sibling = title.nextElementSibling;\n  // If the next element after the title isn't a thebe button, add one now.\n  // That way it is initiatlized when thebe is first-clicked and isn't re-added after.\n  if (!sibling.classList.contains(\"thebe-launch-button\")) {\n    title.insertAdjacentHTML(\n      \"afterend\",\n      \"<button class='thebe-launch-button'></button>\",\n    );\n  }\n  // This function is provided by sphinx-thebe\n  initThebe();\n};\n\n/**\n * Add no print class to certain DOM elements\n */\n\nfunction addNoPrint() {\n  var noPrintSelector = [\n    \".bd-header-announcement\",\n    \".bd-header\",\n    \".bd-header-article\",\n    \".bd-sidebar-primary\",\n    \".bd-sidebar-secondary\",\n    \".bd-footer-article\",\n    \".bd-footer-content\",\n    \".bd-footer\",\n  ].join(\",\");\n  document.querySelectorAll(noPrintSelector).forEach((ii) => {\n    ii.classList.add(\"noprint\");\n  });\n}\n\n/**\n * Set up callback functions for UI click actions\n */\nwindow.initThebeSBT = initThebeSBT;\nwindow.toggleFullScreen = toggleFullScreen;\n\n/**\n * Set up functions to load when the DOM is ready\n */\nsbRunWhenDOMLoaded(initTocHide);\nsbRunWhenDOMLoaded(addNoPrint);\n"],"sourceRoot":""}
\ No newline at end of file
diff --git a/docs/_build/html/_static/searchtools.js b/docs/_build/html/_static/searchtools.js
deleted file mode 100644
index ac4d586..0000000
--- a/docs/_build/html/_static/searchtools.js
+++ /dev/null
@@ -1,531 +0,0 @@
-/*
- * searchtools.js
- * ~~~~~~~~~~~~~~~~
- *
- * Sphinx JavaScript utilities for the full-text search.
- *
- * :copyright: Copyright 2007-2022 by the Sphinx team, see AUTHORS.
- * :license: BSD, see LICENSE for details.
- *
- */
-"use strict";
-
-/**
- * Simple result scoring code.
- */
-if (typeof Scorer === "undefined") {
-  var Scorer = {
-    // Implement the following function to further tweak the score for each result
-    // The function takes a result array [docname, title, anchor, descr, score, filename]
-    // and returns the new score.
-    /*
-    score: result => {
-      const [docname, title, anchor, descr, score, filename] = result
-      return score
-    },
-    */
-
-    // query matches the full name of an object
-    objNameMatch: 11,
-    // or matches in the last dotted part of the object name
-    objPartialMatch: 6,
-    // Additive scores depending on the priority of the object
-    objPrio: {
-      0: 15, // used to be importantResults
-      1: 5, // used to be objectResults
-      2: -5, // used to be unimportantResults
-    },
-    //  Used when the priority is not in the mapping.
-    objPrioDefault: 0,
-
-    // query found in title
-    title: 15,
-    partialTitle: 7,
-    // query found in terms
-    term: 5,
-    partialTerm: 2,
-  };
-}
-
-const _removeChildren = (element) => {
-  while (element && element.lastChild) element.removeChild(element.lastChild);
-};
-
-/**
- * See https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions#escaping
- */
-const _escapeRegExp = (string) =>
-  string.replace(/[.*+\-?^${}()|[\]\\]/g, "\\$&"); // $& means the whole matched string
-
-const _displayItem = (item, highlightTerms, searchTerms) => {
-  const docBuilder = DOCUMENTATION_OPTIONS.BUILDER;
-  const docUrlRoot = DOCUMENTATION_OPTIONS.URL_ROOT;
-  const docFileSuffix = DOCUMENTATION_OPTIONS.FILE_SUFFIX;
-  const docLinkSuffix = DOCUMENTATION_OPTIONS.LINK_SUFFIX;
-  const showSearchSummary = DOCUMENTATION_OPTIONS.SHOW_SEARCH_SUMMARY;
-
-  const [docName, title, anchor, descr] = item;
-
-  let listItem = document.createElement("li");
-  let requestUrl;
-  let linkUrl;
-  if (docBuilder === "dirhtml") {
-    // dirhtml builder
-    let dirname = docName + "/";
-    if (dirname.match(/\/index\/$/))
-      dirname = dirname.substring(0, dirname.length - 6);
-    else if (dirname === "index/") dirname = "";
-    requestUrl = docUrlRoot + dirname;
-    linkUrl = requestUrl;
-  } else {
-    // normal html builders
-    requestUrl = docUrlRoot + docName + docFileSuffix;
-    linkUrl = docName + docLinkSuffix;
-  }
-  const params = new URLSearchParams();
-  params.set("highlight", [...highlightTerms].join(" "));
-  let linkEl = listItem.appendChild(document.createElement("a"));
-  linkEl.href = linkUrl + "?" + params.toString() + anchor;
-  linkEl.innerHTML = title;
-  if (descr)
-    listItem.appendChild(document.createElement("span")).innerText =
-      " (" + descr + ")";
-  else if (showSearchSummary)
-    fetch(requestUrl)
-      .then((responseData) => responseData.text())
-      .then((data) => {
-        if (data)
-          listItem.appendChild(
-            Search.makeSearchSummary(data, searchTerms, highlightTerms)
-          );
-      });
-  Search.output.appendChild(listItem);
-};
-const _finishSearch = (resultCount) => {
-  Search.stopPulse();
-  Search.title.innerText = _("Search Results");
-  if (!resultCount)
-    Search.status.innerText = Documentation.gettext(
-      "Your search did not match any documents. Please make sure that all words are spelled correctly and that you've selected enough categories."
-    );
-  else
-    Search.status.innerText = _(
-      `Search finished, found ${resultCount} page(s) matching the search query.`
-    );
-};
-const _displayNextItem = (
-  results,
-  resultCount,
-  highlightTerms,
-  searchTerms
-) => {
-  // results left, load the summary and display it
-  // this is intended to be dynamic (don't sub resultsCount)
-  if (results.length) {
-    _displayItem(results.pop(), highlightTerms, searchTerms);
-    setTimeout(
-      () => _displayNextItem(results, resultCount, highlightTerms, searchTerms),
-      5
-    );
-  }
-  // search finished, update title and status message
-  else _finishSearch(resultCount);
-};
-
-/**
- * Default splitQuery function. Can be overridden in ``sphinx.search`` with a
- * custom function per language.
- *
- * The regular expression works by splitting the string on consecutive characters
- * that are not Unicode letters, numbers, underscores, or emoji characters.
- * This is the same as ``\W+`` in Python, preserving the surrogate pair area.
- */
-if (typeof splitQuery === "undefined") {
-  var splitQuery = (query) => query
-      .split(/[^\p{Letter}\p{Number}_\p{Emoji_Presentation}]+/gu)
-      .filter(term => term)  // remove remaining empty strings
-}
-
-/**
- * Search Module
- */
-const Search = {
-  _index: null,
-  _queued_query: null,
-  _pulse_status: -1,
-
-  htmlToText: (htmlString) => {
-    const htmlElement = document
-      .createRange()
-      .createContextualFragment(htmlString);
-    _removeChildren(htmlElement.querySelectorAll(".headerlink"));
-    const docContent = htmlElement.querySelector('[role="main"]');
-    if (docContent !== undefined) return docContent.textContent;
-    console.warn(
-      "Content block not found. Sphinx search tries to obtain it via '[role=main]'. Could you check your theme or template."
-    );
-    return "";
-  },
-
-  init: () => {
-    const query = new URLSearchParams(window.location.search).get("q");
-    document
-      .querySelectorAll('input[name="q"]')
-      .forEach((el) => (el.value = query));
-    if (query) Search.performSearch(query);
-  },
-
-  loadIndex: (url) =>
-    (document.body.appendChild(document.createElement("script")).src = url),
-
-  setIndex: (index) => {
-    Search._index = index;
-    if (Search._queued_query !== null) {
-      const query = Search._queued_query;
-      Search._queued_query = null;
-      Search.query(query);
-    }
-  },
-
-  hasIndex: () => Search._index !== null,
-
-  deferQuery: (query) => (Search._queued_query = query),
-
-  stopPulse: () => (Search._pulse_status = -1),
-
-  startPulse: () => {
-    if (Search._pulse_status >= 0) return;
-
-    const pulse = () => {
-      Search._pulse_status = (Search._pulse_status + 1) % 4;
-      Search.dots.innerText = ".".repeat(Search._pulse_status);
-      if (Search._pulse_status >= 0) window.setTimeout(pulse, 500);
-    };
-    pulse();
-  },
-
-  /**
-   * perform a search for something (or wait until index is loaded)
-   */
-  performSearch: (query) => {
-    // create the required interface elements
-    const searchText = document.createElement("h2");
-    searchText.textContent = _("Searching");
-    const searchSummary = document.createElement("p");
-    searchSummary.classList.add("search-summary");
-    searchSummary.innerText = "";
-    const searchList = document.createElement("ul");
-    searchList.classList.add("search");
-
-    const out = document.getElementById("search-results");
-    Search.title = out.appendChild(searchText);
-    Search.dots = Search.title.appendChild(document.createElement("span"));
-    Search.status = out.appendChild(searchSummary);
-    Search.output = out.appendChild(searchList);
-
-    const searchProgress = document.getElementById("search-progress");
-    // Some themes don't use the search progress node
-    if (searchProgress) {
-      searchProgress.innerText = _("Preparing search...");
-    }
-    Search.startPulse();
-
-    // index already loaded, the browser was quick!
-    if (Search.hasIndex()) Search.query(query);
-    else Search.deferQuery(query);
-  },
-
-  /**
-   * execute search (requires search index to be loaded)
-   */
-  query: (query) => {
-    // stem the search terms and add them to the correct list
-    const stemmer = new Stemmer();
-    const searchTerms = new Set();
-    const excludedTerms = new Set();
-    const highlightTerms = new Set();
-    const objectTerms = new Set(splitQuery(query.toLowerCase().trim()));
-    splitQuery(query.trim()).forEach((queryTerm) => {
-      const queryTermLower = queryTerm.toLowerCase();
-
-      // maybe skip this "word"
-      // stopwords array is from language_data.js
-      if (
-        stopwords.indexOf(queryTermLower) !== -1 ||
-        queryTerm.match(/^\d+$/)
-      )
-        return;
-
-      // stem the word
-      let word = stemmer.stemWord(queryTermLower);
-      // select the correct list
-      if (word[0] === "-") excludedTerms.add(word.substr(1));
-      else {
-        searchTerms.add(word);
-        highlightTerms.add(queryTermLower);
-      }
-    });
-
-    // console.debug("SEARCH: searching for:");
-    // console.info("required: ", [...searchTerms]);
-    // console.info("excluded: ", [...excludedTerms]);
-
-    // array of [docname, title, anchor, descr, score, filename]
-    let results = [];
-    _removeChildren(document.getElementById("search-progress"));
-
-    // lookup as object
-    objectTerms.forEach((term) =>
-      results.push(...Search.performObjectSearch(term, objectTerms))
-    );
-
-    // lookup as search terms in fulltext
-    results.push(...Search.performTermsSearch(searchTerms, excludedTerms));
-
-    // let the scorer override scores with a custom scoring function
-    if (Scorer.score) results.forEach((item) => (item[4] = Scorer.score(item)));
-
-    // now sort the results by score (in opposite order of appearance, since the
-    // display function below uses pop() to retrieve items) and then
-    // alphabetically
-    results.sort((a, b) => {
-      const leftScore = a[4];
-      const rightScore = b[4];
-      if (leftScore === rightScore) {
-        // same score: sort alphabetically
-        const leftTitle = a[1].toLowerCase();
-        const rightTitle = b[1].toLowerCase();
-        if (leftTitle === rightTitle) return 0;
-        return leftTitle > rightTitle ? -1 : 1; // inverted is intentional
-      }
-      return leftScore > rightScore ? 1 : -1;
-    });
-
-    // remove duplicate search results
-    // note the reversing of results, so that in the case of duplicates, the highest-scoring entry is kept
-    let seen = new Set();
-    results = results.reverse().reduce((acc, result) => {
-      let resultStr = result.slice(0, 4).concat([result[5]]).map(v => String(v)).join(',');
-      if (!seen.has(resultStr)) {
-        acc.push(result);
-        seen.add(resultStr);
-      }
-      return acc;
-    }, []);
-
-    results = results.reverse();
-
-    // for debugging
-    //Search.lastresults = results.slice();  // a copy
-    // console.info("search results:", Search.lastresults);
-
-    // print the results
-    _displayNextItem(results, results.length, highlightTerms, searchTerms);
-  },
-
-  /**
-   * search for object names
-   */
-  performObjectSearch: (object, objectTerms) => {
-    const filenames = Search._index.filenames;
-    const docNames = Search._index.docnames;
-    const objects = Search._index.objects;
-    const objNames = Search._index.objnames;
-    const titles = Search._index.titles;
-
-    const results = [];
-
-    const objectSearchCallback = (prefix, match) => {
-      const name = match[4]
-      const fullname = (prefix ? prefix + "." : "") + name;
-      const fullnameLower = fullname.toLowerCase();
-      if (fullnameLower.indexOf(object) < 0) return;
-
-      let score = 0;
-      const parts = fullnameLower.split(".");
-
-      // check for different match types: exact matches of full name or
-      // "last name" (i.e. last dotted part)
-      if (fullnameLower === object || parts.slice(-1)[0] === object)
-        score += Scorer.objNameMatch;
-      else if (parts.slice(-1)[0].indexOf(object) > -1)
-        score += Scorer.objPartialMatch; // matches in last name
-
-      const objName = objNames[match[1]][2];
-      const title = titles[match[0]];
-
-      // If more than one term searched for, we require other words to be
-      // found in the name/title/description
-      const otherTerms = new Set(objectTerms);
-      otherTerms.delete(object);
-      if (otherTerms.size > 0) {
-        const haystack = `${prefix} ${name} ${objName} ${title}`.toLowerCase();
-        if (
-          [...otherTerms].some((otherTerm) => haystack.indexOf(otherTerm) < 0)
-        )
-          return;
-      }
-
-      let anchor = match[3];
-      if (anchor === "") anchor = fullname;
-      else if (anchor === "-") anchor = objNames[match[1]][1] + "-" + fullname;
-
-      const descr = objName + _(", in ") + title;
-
-      // add custom score for some objects according to scorer
-      if (Scorer.objPrio.hasOwnProperty(match[2]))
-        score += Scorer.objPrio[match[2]];
-      else score += Scorer.objPrioDefault;
-
-      results.push([
-        docNames[match[0]],
-        fullname,
-        "#" + anchor,
-        descr,
-        score,
-        filenames[match[0]],
-      ]);
-    };
-    Object.keys(objects).forEach((prefix) =>
-      objects[prefix].forEach((array) =>
-        objectSearchCallback(prefix, array)
-      )
-    );
-    return results;
-  },
-
-  /**
-   * search for full-text terms in the index
-   */
-  performTermsSearch: (searchTerms, excludedTerms) => {
-    // prepare search
-    const terms = Search._index.terms;
-    const titleTerms = Search._index.titleterms;
-    const docNames = Search._index.docnames;
-    const filenames = Search._index.filenames;
-    const titles = Search._index.titles;
-
-    const scoreMap = new Map();
-    const fileMap = new Map();
-
-    // perform the search on the required terms
-    searchTerms.forEach((word) => {
-      const files = [];
-      const arr = [
-        { files: terms[word], score: Scorer.term },
-        { files: titleTerms[word], score: Scorer.title },
-      ];
-      // add support for partial matches
-      if (word.length > 2) {
-        const escapedWord = _escapeRegExp(word);
-        Object.keys(terms).forEach((term) => {
-          if (term.match(escapedWord) && !terms[word])
-            arr.push({ files: terms[term], score: Scorer.partialTerm });
-        });
-        Object.keys(titleTerms).forEach((term) => {
-          if (term.match(escapedWord) && !titleTerms[word])
-            arr.push({ files: titleTerms[word], score: Scorer.partialTitle });
-        });
-      }
-
-      // no match but word was a required one
-      if (arr.every((record) => record.files === undefined)) return;
-
-      // found search word in contents
-      arr.forEach((record) => {
-        if (record.files === undefined) return;
-
-        let recordFiles = record.files;
-        if (recordFiles.length === undefined) recordFiles = [recordFiles];
-        files.push(...recordFiles);
-
-        // set score for the word in each file
-        recordFiles.forEach((file) => {
-          if (!scoreMap.has(file)) scoreMap.set(file, {});
-          scoreMap.get(file)[word] = record.score;
-        });
-      });
-
-      // create the mapping
-      files.forEach((file) => {
-        if (fileMap.has(file) && fileMap.get(file).indexOf(word) === -1)
-          fileMap.get(file).push(word);
-        else fileMap.set(file, [word]);
-      });
-    });
-
-    // now check if the files don't contain excluded terms
-    const results = [];
-    for (const [file, wordList] of fileMap) {
-      // check if all requirements are matched
-
-      // as search terms with length < 3 are discarded
-      const filteredTermCount = [...searchTerms].filter(
-        (term) => term.length > 2
-      ).length;
-      if (
-        wordList.length !== searchTerms.size &&
-        wordList.length !== filteredTermCount
-      )
-        continue;
-
-      // ensure that none of the excluded terms is in the search result
-      if (
-        [...excludedTerms].some(
-          (term) =>
-            terms[term] === file ||
-            titleTerms[term] === file ||
-            (terms[term] || []).includes(file) ||
-            (titleTerms[term] || []).includes(file)
-        )
-      )
-        break;
-
-      // select one (max) score for the file.
-      const score = Math.max(...wordList.map((w) => scoreMap.get(file)[w]));
-      // add result to the result list
-      results.push([
-        docNames[file],
-        titles[file],
-        "",
-        null,
-        score,
-        filenames[file],
-      ]);
-    }
-    return results;
-  },
-
-  /**
-   * helper function to return a node containing the
-   * search summary for a given text. keywords is a list
-   * of stemmed words, highlightWords is the list of normal, unstemmed
-   * words. the first one is used to find the occurrence, the
-   * latter for highlighting it.
-   */
-  makeSearchSummary: (htmlText, keywords, highlightWords) => {
-    const text = Search.htmlToText(htmlText).toLowerCase();
-    if (text === "") return null;
-
-    const actualStartPosition = [...keywords]
-      .map((k) => text.indexOf(k.toLowerCase()))
-      .filter((i) => i > -1)
-      .slice(-1)[0];
-    const startWithContext = Math.max(actualStartPosition - 120, 0);
-
-    const top = startWithContext === 0 ? "" : "...";
-    const tail = startWithContext + 240 < text.length ? "..." : "";
-
-    let summary = document.createElement("div");
-    summary.classList.add("context");
-    summary.innerText = top + text.substr(startWithContext, 240).trim() + tail;
-
-    highlightWords.forEach((highlightWord) =>
-      _highlightText(summary, highlightWord, "highlighted")
-    );
-
-    return summary;
-  },
-};
-
-_ready(Search.init);
diff --git a/docs/_build/html/_static/sphinx-thebe.css b/docs/_build/html/_static/sphinx-thebe.css
deleted file mode 100644
index 1da2793..0000000
--- a/docs/_build/html/_static/sphinx-thebe.css
+++ /dev/null
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diff --git a/docs/_build/html/_static/sphinx-thebe.js b/docs/_build/html/_static/sphinx-thebe.js
deleted file mode 100644
index 7626dbb..0000000
--- a/docs/_build/html/_static/sphinx-thebe.js
+++ /dev/null
@@ -1,126 +0,0 @@
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diff --git a/docs/_build/html/_static/styles/bootstrap.css b/docs/_build/html/_static/styles/bootstrap.css
deleted file mode 100644
index 82b03c0..0000000
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var(--bs-breadcrumb-padding-x)}.breadcrumb-item+.breadcrumb-item{padding-left:var(--bs-breadcrumb-item-padding-x)}.breadcrumb-item+.breadcrumb-item:before{color:var(--bs-breadcrumb-divider-color);content:var(--bs-breadcrumb-divider,"/");float:left;padding-right:var(--bs-breadcrumb-item-padding-x)}.breadcrumb-item.active{color:var(--bs-breadcrumb-item-active-color)}.pagination{--bs-pagination-padding-x:0.75rem;--bs-pagination-padding-y:0.375rem;--bs-pagination-font-size:1rem;--bs-pagination-color:var(--bs-link-color);--bs-pagination-bg:var(--bs-body-bg);--bs-pagination-border-width:var(--bs-border-width);--bs-pagination-border-color:var(--bs-border-color);--bs-pagination-border-radius:var(--bs-border-radius);--bs-pagination-hover-color:var(--bs-link-hover-color);--bs-pagination-hover-bg:var(--bs-tertiary-bg);--bs-pagination-hover-border-color:var(--bs-border-color);--bs-pagination-focus-color:var(--bs-link-hover-color);--bs-pagination-focus-bg:var(--bs-secondary-bg);--bs-pagination-focus-box-shadow:0 0 0 0.25rem rgba(13,110,253,.25);--bs-pagination-active-color:#fff;--bs-pagination-active-bg:#0d6efd;--bs-pagination-active-border-color:#0d6efd;--bs-pagination-disabled-color:var(--bs-secondary-color);--bs-pagination-disabled-bg:var(--bs-secondary-bg);--bs-pagination-disabled-border-color:var(--bs-border-color);display:flex;list-style:none;padding-left:0}.page-link{background-color:var(--bs-pagination-bg);border:var(--bs-pagination-border-width) solid var(--bs-pagination-border-color);color:var(--bs-pagination-color);display:block;font-size:var(--bs-pagination-font-size);padding:var(--bs-pagination-padding-y) var(--bs-pagination-padding-x);position:relative;text-decoration:none;transition:color .15s ease-in-out,background-color .15s ease-in-out,border-color .15s ease-in-out,box-shadow .15s 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var(--bs-badge-padding-x);text-align:center;vertical-align:baseline;white-space:nowrap}.badge:empty{display:none}.btn .badge{position:relative;top:-1px}.alert{--bs-alert-bg:transparent;--bs-alert-padding-x:1rem;--bs-alert-padding-y:1rem;--bs-alert-margin-bottom:1rem;--bs-alert-color:inherit;--bs-alert-border-color:transparent;--bs-alert-border:var(--bs-border-width) solid var(--bs-alert-border-color);--bs-alert-border-radius:var(--bs-border-radius);--bs-alert-link-color:inherit;background-color:var(--bs-alert-bg);border:var(--bs-alert-border);border-radius:var(--bs-alert-border-radius);color:var(--bs-alert-color);margin-bottom:var(--bs-alert-margin-bottom);padding:var(--bs-alert-padding-y) var(--bs-alert-padding-x);position:relative}.alert-heading{color:inherit}.alert-link{color:var(--bs-alert-link-color);font-weight:700}.alert-dismissible{padding-right:3rem}.alert-dismissible .btn-close{padding:1.25rem 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";counter-increment:section}.list-group-item-action{color:var(--bs-list-group-action-color);text-align:inherit;width:100%}.list-group-item-action:focus,.list-group-item-action:hover{background-color:var(--bs-list-group-action-hover-bg);color:var(--bs-list-group-action-hover-color);text-decoration:none;z-index:1}.list-group-item-action:active{background-color:var(--bs-list-group-action-active-bg);color:var(--bs-list-group-action-active-color)}.list-group-item{background-color:var(--bs-list-group-bg);border:var(--bs-list-group-border-width) solid var(--bs-list-group-border-color);color:var(--bs-list-group-color);display:block;padding:var(--bs-list-group-item-padding-y) var(--bs-list-group-item-padding-x);position:relative;text-decoration:none}.list-group-item:first-child{border-top-left-radius:inherit;border-top-right-radius:inherit}.list-group-item:last-child{border-bottom-left-radius:inherit;border-bottom-right-radius:inherit}.list-group-item.disabled,.list-group-item:disabled{background-color:var(--bs-list-group-disabled-bg);color:var(--bs-list-group-disabled-color);pointer-events:none}.list-group-item.active{background-color:var(--bs-list-group-active-bg);border-color:var(--bs-list-group-active-border-color);color:var(--bs-list-group-active-color);z-index:2}.list-group-item+.list-group-item{border-top-width:0}.list-group-item+.list-group-item.active{border-top-width:var(--bs-list-group-border-width);margin-top:calc(var(--bs-list-group-border-width)*-1)}.list-group-horizontal{flex-direction:row}.list-group-horizontal>.list-group-item:first-child:not(:last-child){border-bottom-left-radius:var(--bs-list-group-border-radius);border-top-right-radius:0}.list-group-horizontal>.list-group-item:last-child:not(:first-child){border-bottom-left-radius:0;border-top-right-radius:var(--bs-list-group-border-radius)}.list-group-horizontal>.list-group-item.active{margin-top:0}.list-group-horizontal>.list-group-item+.list-group-item{border-left-width:0;border-top-width:var(--bs-list-group-border-width)}.list-group-horizontal>.list-group-item+.list-group-item.active{border-left-width:var(--bs-list-group-border-width);margin-left:calc(var(--bs-list-group-border-width)*-1)}@media(min-width:540px){.list-group-horizontal-sm{flex-direction:row}.list-group-horizontal-sm>.list-group-item:first-child:not(:last-child){border-bottom-left-radius:var(--bs-list-group-border-radius);border-top-right-radius:0}.list-group-horizontal-sm>.list-group-item:last-child:not(:first-child){border-bottom-left-radius:0;border-top-right-radius:var(--bs-list-group-border-radius)}.list-group-horizontal-sm>.list-group-item.active{margin-top:0}.list-group-horizontal-sm>.list-group-item+.list-group-item{border-left-width:0;border-top-width:var(--bs-list-group-border-width)}.list-group-horizontal-sm>.list-group-item+.list-group-item.active{border-left-width:var(--bs-list-group-border-width);margin-left:calc(var(--bs-list-group-border-width)*-1)}}@media(min-width:720px){.list-group-horizontal-md{flex-direction:row}.list-group-horizontal-md>.list-group-item:first-child:not(:last-child){border-bottom-left-radius:var(--bs-list-group-border-radius);border-top-right-radius:0}.list-group-horizontal-md>.list-group-item:last-child:not(:first-child){border-bottom-left-radius:0;border-top-right-radius:var(--bs-list-group-border-radius)}.list-group-horizontal-md>.list-group-item.active{margin-top:0}.list-group-horizontal-md>.list-group-item+.list-group-item{border-left-width:0;border-top-width:var(--bs-list-group-border-width)}.list-group-horizontal-md>.list-group-item+.list-group-item.active{border-left-width:var(--bs-list-group-border-width);margin-left:calc(var(--bs-list-group-border-width)*-1)}}@media(min-width:960px){.list-group-horizontal-lg{flex-direction:row}.list-group-horizontal-lg>.list-group-item:first-child:not(:last-child){border-bottom-left-radius:var(--bs-list-group-border-radius);border-top-right-radius:0}.list-group-horizontal-lg>.list-group-item:last-child:not(:first-child){border-bottom-left-radius:0;border-top-right-radius:var(--bs-list-group-border-radius)}.list-group-horizontal-lg>.list-group-item.active{margin-top:0}.list-group-horizontal-lg>.list-group-item+.list-group-item{border-left-width:0;border-top-width:var(--bs-list-group-border-width)}.list-group-horizontal-lg>.list-group-item+.list-group-item.active{border-left-width:var(--bs-list-group-border-width);margin-left:calc(var(--bs-list-group-border-width)*-1)}}@media(min-width:1200px){.list-group-horizontal-xl{flex-direction:row}.list-group-horizontal-xl>.list-group-item:first-child:not(:last-child){border-bottom-left-radius:var(--bs-list-group-border-radius);border-top-right-radius:0}.list-group-horizontal-xl>.list-group-item:last-child:not(:first-child){border-bottom-left-radius:0;border-top-right-radius:var(--bs-list-group-border-radius)}.list-group-horizontal-xl>.list-group-item.active{margin-top:0}.list-group-horizontal-xl>.list-group-item+.list-group-item{border-left-width:0;border-top-width:var(--bs-list-group-border-width)}.list-group-horizontal-xl>.list-group-item+.list-group-item.active{border-left-width:var(--bs-list-group-border-width);margin-left:calc(var(--bs-list-group-border-width)*-1)}}.list-group-flush{border-radius:0}.list-group-flush>.list-group-item{border-width:0 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var(--bs-list-group-border-width)}.list-group-flush>.list-group-item:last-child{border-bottom-width:0}.list-group-item-primary{--bs-list-group-color:var(--bs-primary-text-emphasis);--bs-list-group-bg:var(--bs-primary-bg-subtle);--bs-list-group-border-color:var(--bs-primary-border-subtle);--bs-list-group-action-hover-color:var(--bs-emphasis-color);--bs-list-group-action-hover-bg:var(--bs-primary-border-subtle);--bs-list-group-action-active-color:var(--bs-emphasis-color);--bs-list-group-action-active-bg:var(--bs-primary-border-subtle);--bs-list-group-active-color:var(--bs-primary-bg-subtle);--bs-list-group-active-bg:var(--bs-primary-text-emphasis);--bs-list-group-active-border-color:var(--bs-primary-text-emphasis)}.list-group-item-secondary{--bs-list-group-color:var(--bs-secondary-text-emphasis);--bs-list-group-bg:var(--bs-secondary-bg-subtle);--bs-list-group-border-color:var(--bs-secondary-border-subtle);--bs-list-group-action-hover-color:var(--bs-emphasis-color);--bs-list-group-action-hover-bg:var(--bs-secondary-border-subtle);--bs-list-group-action-active-color:var(--bs-emphasis-color);--bs-list-group-action-active-bg:var(--bs-secondary-border-subtle);--bs-list-group-active-color:var(--bs-secondary-bg-subtle);--bs-list-group-active-bg:var(--bs-secondary-text-emphasis);--bs-list-group-active-border-color:var(--bs-secondary-text-emphasis)}.list-group-item-success{--bs-list-group-color:var(--bs-success-text-emphasis);--bs-list-group-bg:var(--bs-success-bg-subtle);--bs-list-group-border-color:var(--bs-success-border-subtle);--bs-list-group-action-hover-color:var(--bs-emphasis-color);--bs-list-group-action-hover-bg:var(--bs-success-border-subtle);--bs-list-group-action-active-color:var(--bs-emphasis-color);--bs-list-group-action-active-bg:var(--bs-success-border-subtle);--bs-list-group-active-color:var(--bs-success-bg-subtle);--bs-list-group-active-bg:var(--bs-success-text-emphasis);--bs-list-group-active-border-color:var(--bs-success-text-emphasis)}.list-group-item-info{--bs-list-group-color:var(--bs-info-text-emphasis);--bs-list-group-bg:var(--bs-info-bg-subtle);--bs-list-group-border-color:var(--bs-info-border-subtle);--bs-list-group-action-hover-color:var(--bs-emphasis-color);--bs-list-group-action-hover-bg:var(--bs-info-border-subtle);--bs-list-group-action-active-color:var(--bs-emphasis-color);--bs-list-group-action-active-bg:var(--bs-info-border-subtle);--bs-list-group-active-color:var(--bs-info-bg-subtle);--bs-list-group-active-bg:var(--bs-info-text-emphasis);--bs-list-group-active-border-color:var(--bs-info-text-emphasis)}.list-group-item-warning{--bs-list-group-color:var(--bs-warning-text-emphasis);--bs-list-group-bg:var(--bs-warning-bg-subtle);--bs-list-group-border-color:var(--bs-warning-border-subtle);--bs-list-group-action-hover-color:var(--bs-emphasis-color);--bs-list-group-action-hover-bg:var(--bs-warning-border-subtle);--bs-list-group-action-active-color:var(--bs-emphasis-color);--bs-list-group-action-active-bg:var(--bs-warning-border-subtle);--bs-list-group-active-color:var(--bs-warning-bg-subtle);--bs-list-group-active-bg:var(--bs-warning-text-emphasis);--bs-list-group-active-border-color:var(--bs-warning-text-emphasis)}.list-group-item-danger{--bs-list-group-color:var(--bs-danger-text-emphasis);--bs-list-group-bg:var(--bs-danger-bg-subtle);--bs-list-group-border-color:var(--bs-danger-border-subtle);--bs-list-group-action-hover-color:var(--bs-emphasis-color);--bs-list-group-action-hover-bg:var(--bs-danger-border-subtle);--bs-list-group-action-active-color:var(--bs-emphasis-color);--bs-list-group-action-active-bg:var(--bs-danger-border-subtle);--bs-list-group-active-color:var(--bs-danger-bg-subtle);--bs-list-group-active-bg:var(--bs-danger-text-emphasis);--bs-list-group-active-border-color:var(--bs-danger-text-emphasis)}.list-group-item-light{--bs-list-group-color:var(--bs-light-text-emphasis);--bs-list-group-bg:var(--bs-light-bg-subtle);--bs-list-group-border-color:var(--bs-light-border-subtle);--bs-list-group-action-hover-color:var(--bs-emphasis-color);--bs-list-group-action-hover-bg:var(--bs-light-border-subtle);--bs-list-group-action-active-color:var(--bs-emphasis-color);--bs-list-group-action-active-bg:var(--bs-light-border-subtle);--bs-list-group-active-color:var(--bs-light-bg-subtle);--bs-list-group-active-bg:var(--bs-light-text-emphasis);--bs-list-group-active-border-color:var(--bs-light-text-emphasis)}.list-group-item-dark{--bs-list-group-color:var(--bs-dark-text-emphasis);--bs-list-group-bg:var(--bs-dark-bg-subtle);--bs-list-group-border-color:var(--bs-dark-border-subtle);--bs-list-group-action-hover-color:var(--bs-emphasis-color);--bs-list-group-action-hover-bg:var(--bs-dark-border-subtle);--bs-list-group-action-active-color:var(--bs-emphasis-color);--bs-list-group-action-active-bg:var(--bs-dark-border-subtle);--bs-list-group-active-color:var(--bs-dark-bg-subtle);--bs-list-group-active-bg:var(--bs-dark-text-emphasis);--bs-list-group-active-border-color:var(--bs-dark-text-emphasis)}.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0)}.dropdown-item:hover,.dropdown-item:focus{color:var(--bs-dropdown-link-hover-color);background-color:var(--bs-dropdown-link-hover-bg)}.dropdown-item.active,.dropdown-item:active{color:var(--bs-dropdown-link-active-color);text-decoration:none;background-color:var(--bs-dropdown-link-active-bg)}.dropdown-item.disabled,.dropdown-item:disabled{color:var(--bs-dropdown-link-disabled-color);pointer-events:none;background-color:rgba(0,0,0,0)}.dropdown-menu.show{display:block}.dropdown-header{display:block;padding:var(--bs-dropdown-header-padding-y) var(--bs-dropdown-header-padding-x);margin-bottom:0;font-size:0.875rem;color:var(--bs-dropdown-header-color);white-space:nowrap}.dropdown-item-text{display:block;padding:var(--bs-dropdown-item-padding-y) var(--bs-dropdown-item-padding-x);color:var(--bs-dropdown-link-color)}.dropdown-menu-dark{--bs-dropdown-color: #dee2e6;--bs-dropdown-bg: #343a40;--bs-dropdown-border-color: var(--bs-border-color-translucent);--bs-dropdown-box-shadow: ;--bs-dropdown-link-color: #dee2e6;--bs-dropdown-link-hover-color: #fff;--bs-dropdown-divider-bg: var(--bs-border-color-translucent);--bs-dropdown-link-hover-bg: var(--pst-color-surface);--bs-dropdown-link-active-color: #fff;--bs-dropdown-link-active-bg: var(--pst-color-surface);--bs-dropdown-link-disabled-color: #adb5bd;--bs-dropdown-header-color: #adb5bd}.btn-group,.btn-group-vertical{position:relative;display:inline-flex;vertical-align:middle}.btn-group>.btn,.btn-group-vertical>.btn{position:relative;flex:1 1 auto}.btn-group>.btn-check:checked+.btn,.btn-group>.btn-check:focus+.btn,.btn-group>.btn:hover,.btn-group>.btn:focus,.btn-group>.btn:active,.btn-group>.btn.active,.btn-group-vertical>.btn-check:checked+.btn,.btn-group-vertical>.btn-check:focus+.btn,.btn-group-vertical>.btn:hover,.btn-group-vertical>.btn:focus,.btn-group-vertical>.btn:active,.btn-group-vertical>.btn.active{z-index:1}.btn-toolbar{display:flex;flex-wrap:wrap;justify-content:flex-start}.btn-toolbar .input-group{width:auto}.btn-group{border-radius:var(--bs-border-radius)}.btn-group>:not(.btn-check:first-child)+.btn,.btn-group>.btn-group:not(:first-child){margin-left:calc(var(--bs-border-width)*-1)}.btn-group>.btn:not(:last-child):not(.dropdown-toggle),.btn-group>.btn.dropdown-toggle-split:first-child,.btn-group>.btn-group:not(:last-child)>.btn{border-top-right-radius:0;border-bottom-right-radius:0}.btn-group>.btn:nth-child(n+3),.btn-group>:not(.btn-check)+.btn,.btn-group>.btn-group:not(:first-child)>.btn{border-top-left-radius:0;border-bottom-left-radius:0}.dropdown-toggle-split{padding-right:.5625rem;padding-left:.5625rem}.dropdown-toggle-split::after,.dropup .dropdown-toggle-split::after,.dropend .dropdown-toggle-split::after{margin-left:0}.dropstart .dropdown-toggle-split::before{margin-right:0}.btn-sm+.dropdown-toggle-split,.btn-group-sm>.btn+.dropdown-toggle-split{padding-right:.375rem;padding-left:.375rem}.btn-lg+.dropdown-toggle-split,.btn-group-lg>.btn+.dropdown-toggle-split{padding-right:.75rem;padding-left:.75rem}.btn-group-vertical{flex-direction:column;align-items:flex-start;justify-content:center}.btn-group-vertical>.btn,.btn-group-vertical>.btn-group{width:100%}.btn-group-vertical>.btn:not(:first-child),.btn-group-vertical>.btn-group:not(:first-child){margin-top:calc(var(--bs-border-width)*-1)}.btn-group-vertical>.btn:not(:last-child):not(.dropdown-toggle),.btn-group-vertical>.btn-group:not(:last-child)>.btn{border-bottom-right-radius:0;border-bottom-left-radius:0}.btn-group-vertical>.btn~.btn,.btn-group-vertical>.btn-group:not(:first-child)>.btn{border-top-left-radius:0;border-top-right-radius:0}.nav{--bs-nav-link-padding-x: 1rem;--bs-nav-link-padding-y: 0.5rem;--bs-nav-link-font-weight: ;--bs-nav-link-color: 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1.25rem;--bs-navbar-brand-color: rgba(var(--bs-emphasis-color-rgb), 1);--bs-navbar-brand-hover-color: rgba(var(--bs-emphasis-color-rgb), 1);--bs-navbar-nav-link-padding-x: 0.5rem;--bs-navbar-toggler-padding-y: 0.25rem;--bs-navbar-toggler-padding-x: 0.75rem;--bs-navbar-toggler-font-size: 1.25rem;--bs-navbar-toggler-icon-bg: url(\"data:image/svg+xml,%3csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 30 30'%3e%3cpath stroke='rgba%2833, 37, 41, 0.75%29' stroke-linecap='round' stroke-miterlimit='10' stroke-width='2' d='M4 7h22M4 15h22M4 23h22'/%3e%3c/svg%3e\");--bs-navbar-toggler-border-color: rgba(var(--bs-emphasis-color-rgb), 0.15);--bs-navbar-toggler-border-radius: var(--bs-border-radius);--bs-navbar-toggler-focus-width: 0.25rem;--bs-navbar-toggler-transition: box-shadow 0.15s ease-in-out;position:relative;display:flex;flex-wrap:wrap;align-items:center;justify-content:space-between;padding:var(--bs-navbar-padding-y) 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.offcanvas-body{display:flex;flex-grow:0;padding:0;overflow-y:visible}}.navbar-expand{flex-wrap:nowrap;justify-content:flex-start}.navbar-expand .navbar-nav{flex-direction:row}.navbar-expand .navbar-nav .dropdown-menu{position:absolute}.navbar-expand .navbar-nav .nav-link{padding-right:var(--bs-navbar-nav-link-padding-x);padding-left:var(--bs-navbar-nav-link-padding-x)}.navbar-expand .navbar-nav-scroll{overflow:visible}.navbar-expand .navbar-collapse{display:flex !important;flex-basis:auto}.navbar-expand .navbar-toggler{display:none}.navbar-expand .offcanvas{position:static;z-index:auto;flex-grow:1;width:auto !important;height:auto !important;visibility:visible !important;background-color:rgba(0,0,0,0) !important;border:0 !important;transform:none !important;transition:none}.navbar-expand .offcanvas .offcanvas-header{display:none}.navbar-expand .offcanvas .offcanvas-body{display:flex;flex-grow:0;padding:0;overflow-y:visible}.navbar-dark,.navbar[data-bs-theme=dark]{--bs-navbar-color: rgba(255, 255, 255, 0.55);--bs-navbar-hover-color: rgba(255, 255, 255, 0.75);--bs-navbar-disabled-color: rgba(255, 255, 255, 0.25);--bs-navbar-active-color: #fff;--bs-navbar-brand-color: #fff;--bs-navbar-brand-hover-color: #fff;--bs-navbar-toggler-border-color: rgba(255, 255, 255, 0.1);--bs-navbar-toggler-icon-bg: url(\"data:image/svg+xml,%3csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 30 30'%3e%3cpath stroke='rgba%28255, 255, 255, 0.55%29' stroke-linecap='round' stroke-miterlimit='10' stroke-width='2' d='M4 7h22M4 15h22M4 23h22'/%3e%3c/svg%3e\")}[data-bs-theme=dark] .navbar-toggler-icon{--bs-navbar-toggler-icon-bg: url(\"data:image/svg+xml,%3csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 30 30'%3e%3cpath stroke='rgba%28255, 255, 255, 0.55%29' stroke-linecap='round' stroke-miterlimit='10' stroke-width='2' d='M4 7h22M4 15h22M4 23h22'/%3e%3c/svg%3e\")}.card{--bs-card-spacer-y: 1rem;--bs-card-spacer-x: 1rem;--bs-card-title-spacer-y: 0.5rem;--bs-card-title-color: ;--bs-card-subtitle-color: ;--bs-card-border-width: var(--bs-border-width);--bs-card-border-color: var(--bs-border-color-translucent);--bs-card-border-radius: var(--bs-border-radius);--bs-card-box-shadow: ;--bs-card-inner-border-radius: calc(var(--bs-border-radius) - (var(--bs-border-width)));--bs-card-cap-padding-y: 0.5rem;--bs-card-cap-padding-x: 1rem;--bs-card-cap-bg: rgba(var(--bs-body-color-rgb), 0.03);--bs-card-cap-color: ;--bs-card-height: ;--bs-card-color: ;--bs-card-bg: var(--bs-body-bg);--bs-card-img-overlay-padding: 1rem;--bs-card-group-margin: 0.75rem;position:relative;display:flex;flex-direction:column;min-width:0;height:var(--bs-card-height);color:var(--bs-body-color);word-wrap:break-word;background-color:var(--bs-card-bg);background-clip:border-box;border:var(--bs-card-border-width) solid var(--bs-card-border-color);border-radius:var(--bs-card-border-radius)}.card>hr{margin-right:0;margin-left:0}.card>.list-group{border-top:inherit;border-bottom:inherit}.card>.list-group:first-child{border-top-width:0;border-top-left-radius:var(--bs-card-inner-border-radius);border-top-right-radius:var(--bs-card-inner-border-radius)}.card>.list-group:last-child{border-bottom-width:0;border-bottom-right-radius:var(--bs-card-inner-border-radius);border-bottom-left-radius:var(--bs-card-inner-border-radius)}.card>.card-header+.list-group,.card>.list-group+.card-footer{border-top:0}.card-body{flex:1 1 auto;padding:var(--bs-card-spacer-y) var(--bs-card-spacer-x);color:var(--bs-card-color)}.card-title{margin-bottom:var(--bs-card-title-spacer-y);color:var(--bs-card-title-color)}.card-subtitle{margin-top:calc(-0.5*var(--bs-card-title-spacer-y));margin-bottom:0;color:var(--bs-card-subtitle-color)}.card-text:last-child{margin-bottom:0}.card-link+.card-link{margin-left:var(--bs-card-spacer-x)}.card-header{padding:var(--bs-card-cap-padding-y) var(--bs-card-cap-padding-x);margin-bottom:0;color:var(--bs-card-cap-color);background-color:var(--bs-card-cap-bg);border-bottom:var(--bs-card-border-width) solid var(--bs-card-border-color)}.card-header:first-child{border-radius:var(--bs-card-inner-border-radius) var(--bs-card-inner-border-radius) 0 0}.card-footer{padding:var(--bs-card-cap-padding-y) var(--bs-card-cap-padding-x);color:var(--bs-card-cap-color);background-color:var(--bs-card-cap-bg);border-top:var(--bs-card-border-width) solid var(--bs-card-border-color)}.card-footer:last-child{border-radius:0 0 var(--bs-card-inner-border-radius) var(--bs-card-inner-border-radius)}.card-header-tabs{margin-right:calc(-0.5*var(--bs-card-cap-padding-x));margin-bottom:calc(-1*var(--bs-card-cap-padding-y));margin-left:calc(-0.5*var(--bs-card-cap-padding-x));border-bottom:0}.card-header-tabs .nav-link.active{background-color:var(--bs-card-bg);border-bottom-color:var(--bs-card-bg)}.card-header-pills{margin-right:calc(-0.5*var(--bs-card-cap-padding-x));margin-left:calc(-0.5*var(--bs-card-cap-padding-x))}.card-img-overlay{position:absolute;top:0;right:0;bottom:0;left:0;padding:var(--bs-card-img-overlay-padding);border-radius:var(--bs-card-inner-border-radius)}.card-img,.card-img-top,.card-img-bottom{width:100%}.card-img,.card-img-top{border-top-left-radius:var(--bs-card-inner-border-radius);border-top-right-radius:var(--bs-card-inner-border-radius)}.card-img,.card-img-bottom{border-bottom-right-radius:var(--bs-card-inner-border-radius);border-bottom-left-radius:var(--bs-card-inner-border-radius)}.card-group>.card{margin-bottom:var(--bs-card-group-margin)}@media(min-width: 540px){.card-group{display:flex;flex-flow:row wrap}.card-group>.card{flex:1 0 0%;margin-bottom:0}.card-group>.card+.card{margin-left:0;border-left:0}.card-group>.card:not(:last-child){border-top-right-radius:0;border-bottom-right-radius:0}.card-group>.card:not(:last-child) .card-img-top,.card-group>.card:not(:last-child) .card-header{border-top-right-radius:0}.card-group>.card:not(:last-child) .card-img-bottom,.card-group>.card:not(:last-child) .card-footer{border-bottom-right-radius:0}.card-group>.card:not(:first-child){border-top-left-radius:0;border-bottom-left-radius:0}.card-group>.card:not(:first-child) .card-img-top,.card-group>.card:not(:first-child) .card-header{border-top-left-radius:0}.card-group>.card:not(:first-child) .card-img-bottom,.card-group>.card:not(:first-child) .card-footer{border-bottom-left-radius:0}}.accordion{--bs-accordion-color: var(--bs-body-color);--bs-accordion-bg: var(--bs-body-bg);--bs-accordion-transition: color 0.15s ease-in-out, background-color 0.15s ease-in-out, border-color 0.15s ease-in-out, box-shadow 0.15s ease-in-out, border-radius 0.15s ease;--bs-accordion-border-color: var(--bs-border-color);--bs-accordion-border-width: var(--bs-border-width);--bs-accordion-border-radius: var(--bs-border-radius);--bs-accordion-inner-border-radius: calc(var(--bs-border-radius) - (var(--bs-border-width)));--bs-accordion-btn-padding-x: 1.25rem;--bs-accordion-btn-padding-y: 1rem;--bs-accordion-btn-color: var(--bs-body-color);--bs-accordion-btn-bg: var(--bs-accordion-bg);--bs-accordion-btn-icon: url(\"data:image/svg+xml,%3csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='%23212529'%3e%3cpath fill-rule='evenodd' d='M1.646 4.646a.5.5 0 0 1 .708 0L8 10.293l5.646-5.647a.5.5 0 0 1 .708.708l-6 6a.5.5 0 0 1-.708 0l-6-6a.5.5 0 0 1 0-.708z'/%3e%3c/svg%3e\");--bs-accordion-btn-icon-width: 1.25rem;--bs-accordion-btn-icon-transform: rotate(-180deg);--bs-accordion-btn-icon-transition: transform 0.2s ease-in-out;--bs-accordion-btn-active-icon: url(\"data:image/svg+xml,%3csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='%23052c65'%3e%3cpath fill-rule='evenodd' d='M1.646 4.646a.5.5 0 0 1 .708 0L8 10.293l5.646-5.647a.5.5 0 0 1 .708.708l-6 6a.5.5 0 0 1-.708 0l-6-6a.5.5 0 0 1 0-.708z'/%3e%3c/svg%3e\");--bs-accordion-btn-focus-border-color: #86b7fe;--bs-accordion-btn-focus-box-shadow: 0 0 0 0.25rem rgba(13, 110, 253, 0.25);--bs-accordion-body-padding-x: 1.25rem;--bs-accordion-body-padding-y: 1rem;--bs-accordion-active-color: var(--bs-primary-text-emphasis);--bs-accordion-active-bg: var(--bs-primary-bg-subtle)}.accordion-button{position:relative;display:flex;align-items:center;width:100%;padding:var(--bs-accordion-btn-padding-y) var(--bs-accordion-btn-padding-x);font-size:1rem;color:var(--bs-accordion-btn-color);text-align:left;background-color:var(--bs-accordion-btn-bg);border:0;border-radius:0;overflow-anchor:none;transition:var(--bs-accordion-transition)}@media(prefers-reduced-motion: reduce){.accordion-button{transition:none}}.accordion-button:not(.collapsed){color:var(--bs-accordion-active-color);background-color:var(--bs-accordion-active-bg);box-shadow:inset 0 calc(-1*var(--bs-accordion-border-width)) 0 var(--bs-accordion-border-color)}.accordion-button:not(.collapsed)::after{background-image:var(--bs-accordion-btn-active-icon);transform:var(--bs-accordion-btn-icon-transform)}.accordion-button::after{flex-shrink:0;width:var(--bs-accordion-btn-icon-width);height:var(--bs-accordion-btn-icon-width);margin-left:auto;content:\"\";background-image:var(--bs-accordion-btn-icon);background-repeat:no-repeat;background-size:var(--bs-accordion-btn-icon-width);transition:var(--bs-accordion-btn-icon-transition)}@media(prefers-reduced-motion: reduce){.accordion-button::after{transition:none}}.accordion-button:hover{z-index:2}.accordion-button:focus{z-index:3;border-color:var(--bs-accordion-btn-focus-border-color);outline:0;box-shadow:var(--bs-accordion-btn-focus-box-shadow)}.accordion-header{margin-bottom:0}.accordion-item{color:var(--bs-accordion-color);background-color:var(--bs-accordion-bg);border:var(--bs-accordion-border-width) solid var(--bs-accordion-border-color)}.accordion-item:first-of-type{border-top-left-radius:var(--bs-accordion-border-radius);border-top-right-radius:var(--bs-accordion-border-radius)}.accordion-item:first-of-type .accordion-button{border-top-left-radius:var(--bs-accordion-inner-border-radius);border-top-right-radius:var(--bs-accordion-inner-border-radius)}.accordion-item:not(:first-of-type){border-top:0}.accordion-item:last-of-type{border-bottom-right-radius:var(--bs-accordion-border-radius);border-bottom-left-radius:var(--bs-accordion-border-radius)}.accordion-item:last-of-type .accordion-button.collapsed{border-bottom-right-radius:var(--bs-accordion-inner-border-radius);border-bottom-left-radius:var(--bs-accordion-inner-border-radius)}.accordion-item:last-of-type .accordion-collapse{border-bottom-right-radius:var(--bs-accordion-border-radius);border-bottom-left-radius:var(--bs-accordion-border-radius)}.accordion-body{padding:var(--bs-accordion-body-padding-y) var(--bs-accordion-body-padding-x)}.accordion-flush .accordion-collapse{border-width:0}.accordion-flush .accordion-item{border-right:0;border-left:0;border-radius:0}.accordion-flush .accordion-item:first-child{border-top:0}.accordion-flush .accordion-item:last-child{border-bottom:0}.accordion-flush .accordion-item .accordion-button,.accordion-flush .accordion-item .accordion-button.collapsed{border-radius:0}[data-bs-theme=dark] .accordion-button::after{--bs-accordion-btn-icon: url(\"data:image/svg+xml,%3csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='%236ea8fe'%3e%3cpath fill-rule='evenodd' d='M1.646 4.646a.5.5 0 0 1 .708 0L8 10.293l5.646-5.647a.5.5 0 0 1 .708.708l-6 6a.5.5 0 0 1-.708 0l-6-6a.5.5 0 0 1 0-.708z'/%3e%3c/svg%3e\");--bs-accordion-btn-active-icon: url(\"data:image/svg+xml,%3csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='%236ea8fe'%3e%3cpath fill-rule='evenodd' d='M1.646 4.646a.5.5 0 0 1 .708 0L8 10.293l5.646-5.647a.5.5 0 0 1 .708.708l-6 6a.5.5 0 0 1-.708 0l-6-6a.5.5 0 0 1 0-.708z'/%3e%3c/svg%3e\")}.breadcrumb{--bs-breadcrumb-padding-x: 0;--bs-breadcrumb-padding-y: 0;--bs-breadcrumb-margin-bottom: 1rem;--bs-breadcrumb-bg: ;--bs-breadcrumb-border-radius: ;--bs-breadcrumb-divider-color: var(--bs-secondary-color);--bs-breadcrumb-item-padding-x: 0.5rem;--bs-breadcrumb-item-active-color: var(--bs-secondary-color);display:flex;flex-wrap:wrap;padding:var(--bs-breadcrumb-padding-y) var(--bs-breadcrumb-padding-x);margin-bottom:var(--bs-breadcrumb-margin-bottom);font-size:var(--bs-breadcrumb-font-size);list-style:none;background-color:var(--bs-breadcrumb-bg);border-radius:var(--bs-breadcrumb-border-radius)}.breadcrumb-item+.breadcrumb-item{padding-left:var(--bs-breadcrumb-item-padding-x)}.breadcrumb-item+.breadcrumb-item::before{float:left;padding-right:var(--bs-breadcrumb-item-padding-x);color:var(--bs-breadcrumb-divider-color);content:var(--bs-breadcrumb-divider, \"/\") /* rtl: var(--bs-breadcrumb-divider, \"/\") */}.breadcrumb-item.active{color:var(--bs-breadcrumb-item-active-color)}.pagination{--bs-pagination-padding-x: 0.75rem;--bs-pagination-padding-y: 0.375rem;--bs-pagination-font-size:1rem;--bs-pagination-color: var(--bs-link-color);--bs-pagination-bg: var(--bs-body-bg);--bs-pagination-border-width: var(--bs-border-width);--bs-pagination-border-color: var(--bs-border-color);--bs-pagination-border-radius: var(--bs-border-radius);--bs-pagination-hover-color: var(--bs-link-hover-color);--bs-pagination-hover-bg: var(--bs-tertiary-bg);--bs-pagination-hover-border-color: var(--bs-border-color);--bs-pagination-focus-color: var(--bs-link-hover-color);--bs-pagination-focus-bg: var(--bs-secondary-bg);--bs-pagination-focus-box-shadow: 0 0 0 0.25rem rgba(13, 110, 253, 0.25);--bs-pagination-active-color: #fff;--bs-pagination-active-bg: #0d6efd;--bs-pagination-active-border-color: #0d6efd;--bs-pagination-disabled-color: var(--bs-secondary-color);--bs-pagination-disabled-bg: var(--bs-secondary-bg);--bs-pagination-disabled-border-color: var(--bs-border-color);display:flex;padding-left:0;list-style:none}.page-link{position:relative;display:block;padding:var(--bs-pagination-padding-y) var(--bs-pagination-padding-x);font-size:var(--bs-pagination-font-size);color:var(--bs-pagination-color);text-decoration:none;background-color:var(--bs-pagination-bg);border:var(--bs-pagination-border-width) solid var(--bs-pagination-border-color);transition:color .15s ease-in-out,background-color .15s ease-in-out,border-color .15s ease-in-out,box-shadow .15s ease-in-out}@media(prefers-reduced-motion: reduce){.page-link{transition:none}}.page-link:hover{z-index:2;color:var(--bs-pagination-hover-color);background-color:var(--bs-pagination-hover-bg);border-color:var(--bs-pagination-hover-border-color)}.page-link:focus{z-index:3;color:var(--bs-pagination-focus-color);background-color:var(--bs-pagination-focus-bg);outline:0;box-shadow:var(--bs-pagination-focus-box-shadow)}.page-link.active,.active>.page-link{z-index:3;color:var(--bs-pagination-active-color);background-color:var(--bs-pagination-active-bg);border-color:var(--bs-pagination-active-border-color)}.page-link.disabled,.disabled>.page-link{color:var(--bs-pagination-disabled-color);pointer-events:none;background-color:var(--bs-pagination-disabled-bg);border-color:var(--bs-pagination-disabled-border-color)}.page-item:not(:first-child) .page-link{margin-left:calc(var(--bs-border-width)*-1)}.page-item:first-child .page-link{border-top-left-radius:var(--bs-pagination-border-radius);border-bottom-left-radius:var(--bs-pagination-border-radius)}.page-item:last-child .page-link{border-top-right-radius:var(--bs-pagination-border-radius);border-bottom-right-radius:var(--bs-pagination-border-radius)}.pagination-lg{--bs-pagination-padding-x: 1.5rem;--bs-pagination-padding-y: 0.75rem;--bs-pagination-font-size:1.25rem;--bs-pagination-border-radius: var(--bs-border-radius-lg)}.pagination-sm{--bs-pagination-padding-x: 0.5rem;--bs-pagination-padding-y: 0.25rem;--bs-pagination-font-size:0.875rem;--bs-pagination-border-radius: var(--bs-border-radius-sm)}.badge{--bs-badge-padding-x: 0.65em;--bs-badge-padding-y: 0.35em;--bs-badge-font-size:0.75em;--bs-badge-font-weight: 700;--bs-badge-color: #fff;--bs-badge-border-radius: var(--bs-border-radius);display:inline-block;padding:var(--bs-badge-padding-y) var(--bs-badge-padding-x);font-size:var(--bs-badge-font-size);font-weight:var(--bs-badge-font-weight);line-height:1;color:var(--bs-badge-color);text-align:center;white-space:nowrap;vertical-align:baseline;border-radius:var(--bs-badge-border-radius)}.badge:empty{display:none}.btn .badge{position:relative;top:-1px}.alert{--bs-alert-bg: transparent;--bs-alert-padding-x: 1rem;--bs-alert-padding-y: 1rem;--bs-alert-margin-bottom: 1rem;--bs-alert-color: inherit;--bs-alert-border-color: transparent;--bs-alert-border: var(--bs-border-width) solid var(--bs-alert-border-color);--bs-alert-border-radius: var(--bs-border-radius);--bs-alert-link-color: inherit;position:relative;padding:var(--bs-alert-padding-y) var(--bs-alert-padding-x);margin-bottom:var(--bs-alert-margin-bottom);color:var(--bs-alert-color);background-color:var(--bs-alert-bg);border:var(--bs-alert-border);border-radius:var(--bs-alert-border-radius)}.alert-heading{color:inherit}.alert-link{font-weight:700;color:var(--bs-alert-link-color)}.alert-dismissible{padding-right:3rem}.alert-dismissible .btn-close{position:absolute;top:0;right:0;z-index:2;padding:1.25rem 1rem}.alert-primary{--bs-alert-color: var(--bs-primary-text-emphasis);--bs-alert-bg: var(--bs-primary-bg-subtle);--bs-alert-border-color: var(--bs-primary-border-subtle);--bs-alert-link-color: var(--bs-primary-text-emphasis)}.alert-secondary{--bs-alert-color: var(--bs-secondary-text-emphasis);--bs-alert-bg: var(--bs-secondary-bg-subtle);--bs-alert-border-color: var(--bs-secondary-border-subtle);--bs-alert-link-color: var(--bs-secondary-text-emphasis)}.alert-success{--bs-alert-color: var(--bs-success-text-emphasis);--bs-alert-bg: var(--bs-success-bg-subtle);--bs-alert-border-color: var(--bs-success-border-subtle);--bs-alert-link-color: var(--bs-success-text-emphasis)}.alert-info{--bs-alert-color: var(--bs-info-text-emphasis);--bs-alert-bg: var(--bs-info-bg-subtle);--bs-alert-border-color: var(--bs-info-border-subtle);--bs-alert-link-color: var(--bs-info-text-emphasis)}.alert-warning{--bs-alert-color: var(--bs-warning-text-emphasis);--bs-alert-bg: var(--bs-warning-bg-subtle);--bs-alert-border-color: var(--bs-warning-border-subtle);--bs-alert-link-color: var(--bs-warning-text-emphasis)}.alert-danger{--bs-alert-color: var(--bs-danger-text-emphasis);--bs-alert-bg: var(--bs-danger-bg-subtle);--bs-alert-border-color: var(--bs-danger-border-subtle);--bs-alert-link-color: var(--bs-danger-text-emphasis)}.alert-light{--bs-alert-color: var(--bs-light-text-emphasis);--bs-alert-bg: var(--bs-light-bg-subtle);--bs-alert-border-color: var(--bs-light-border-subtle);--bs-alert-link-color: var(--bs-light-text-emphasis)}.alert-dark{--bs-alert-color: var(--bs-dark-text-emphasis);--bs-alert-bg: var(--bs-dark-bg-subtle);--bs-alert-border-color: var(--bs-dark-border-subtle);--bs-alert-link-color: var(--bs-dark-text-emphasis)}@keyframes progress-bar-stripes{0%{background-position-x:1rem}}.progress,.progress-stacked{--bs-progress-height: 1rem;--bs-progress-font-size:0.75rem;--bs-progress-bg: var(--bs-secondary-bg);--bs-progress-border-radius: var(--bs-border-radius);--bs-progress-box-shadow: var(--bs-box-shadow-inset);--bs-progress-bar-color: #fff;--bs-progress-bar-bg: #0d6efd;--bs-progress-bar-transition: width 0.6s ease;display:flex;height:var(--bs-progress-height);overflow:hidden;font-size:var(--bs-progress-font-size);background-color:var(--bs-progress-bg);border-radius:var(--bs-progress-border-radius)}.progress-bar{display:flex;flex-direction:column;justify-content:center;overflow:hidden;color:var(--bs-progress-bar-color);text-align:center;white-space:nowrap;background-color:var(--bs-progress-bar-bg);transition:var(--bs-progress-bar-transition)}@media(prefers-reduced-motion: reduce){.progress-bar{transition:none}}.progress-bar-striped{background-image:linear-gradient(45deg, rgba(255, 255, 255, 0.15) 25%, transparent 25%, transparent 50%, rgba(255, 255, 255, 0.15) 50%, rgba(255, 255, 255, 0.15) 75%, transparent 75%, transparent);background-size:var(--bs-progress-height) var(--bs-progress-height)}.progress-stacked>.progress{overflow:visible}.progress-stacked>.progress>.progress-bar{width:100%}.progress-bar-animated{animation:1s linear infinite progress-bar-stripes}@media(prefers-reduced-motion: reduce){.progress-bar-animated{animation:none}}.list-group{--bs-list-group-color: var(--bs-body-color);--bs-list-group-bg: var(--bs-body-bg);--bs-list-group-border-color: var(--bs-border-color);--bs-list-group-border-width: var(--bs-border-width);--bs-list-group-border-radius: var(--bs-border-radius);--bs-list-group-item-padding-x: 1rem;--bs-list-group-item-padding-y: 0.5rem;--bs-list-group-action-color: var(--bs-secondary-color);--bs-list-group-action-hover-color: var(--bs-emphasis-color);--bs-list-group-action-hover-bg: var(--bs-tertiary-bg);--bs-list-group-action-active-color: var(--bs-body-color);--bs-list-group-action-active-bg: var(--bs-secondary-bg);--bs-list-group-disabled-color: var(--bs-secondary-color);--bs-list-group-disabled-bg: var(--bs-body-bg);--bs-list-group-active-color: #fff;--bs-list-group-active-bg: #0d6efd;--bs-list-group-active-border-color: 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var(--bs-list-group-border-color)}.list-group-item:first-child{border-top-left-radius:inherit;border-top-right-radius:inherit}.list-group-item:last-child{border-bottom-right-radius:inherit;border-bottom-left-radius:inherit}.list-group-item.disabled,.list-group-item:disabled{color:var(--bs-list-group-disabled-color);pointer-events:none;background-color:var(--bs-list-group-disabled-bg)}.list-group-item.active{z-index:2;color:var(--bs-list-group-active-color);background-color:var(--bs-list-group-active-bg);border-color:var(--bs-list-group-active-border-color)}.list-group-item+.list-group-item{border-top-width:0}.list-group-item+.list-group-item.active{margin-top:calc(-1*var(--bs-list-group-border-width));border-top-width:var(--bs-list-group-border-width)}.list-group-horizontal{flex-direction:row}.list-group-horizontal>.list-group-item:first-child:not(:last-child){border-bottom-left-radius:var(--bs-list-group-border-radius);border-top-right-radius:0}.list-group-horizontal>.list-group-item:last-child:not(:first-child){border-top-right-radius:var(--bs-list-group-border-radius);border-bottom-left-radius:0}.list-group-horizontal>.list-group-item.active{margin-top:0}.list-group-horizontal>.list-group-item+.list-group-item{border-top-width:var(--bs-list-group-border-width);border-left-width:0}.list-group-horizontal>.list-group-item+.list-group-item.active{margin-left:calc(-1*var(--bs-list-group-border-width));border-left-width:var(--bs-list-group-border-width)}@media(min-width: 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960px){.list-group-horizontal-lg{flex-direction:row}.list-group-horizontal-lg>.list-group-item:first-child:not(:last-child){border-bottom-left-radius:var(--bs-list-group-border-radius);border-top-right-radius:0}.list-group-horizontal-lg>.list-group-item:last-child:not(:first-child){border-top-right-radius:var(--bs-list-group-border-radius);border-bottom-left-radius:0}.list-group-horizontal-lg>.list-group-item.active{margin-top:0}.list-group-horizontal-lg>.list-group-item+.list-group-item{border-top-width:var(--bs-list-group-border-width);border-left-width:0}.list-group-horizontal-lg>.list-group-item+.list-group-item.active{margin-left:calc(-1*var(--bs-list-group-border-width));border-left-width:var(--bs-list-group-border-width)}}@media(min-width: 1200px){.list-group-horizontal-xl{flex-direction:row}.list-group-horizontal-xl>.list-group-item:first-child:not(:last-child){border-bottom-left-radius:var(--bs-list-group-border-radius);border-top-right-radius:0}.list-group-horizontal-xl>.list-group-item:last-child:not(:first-child){border-top-right-radius:var(--bs-list-group-border-radius);border-bottom-left-radius:0}.list-group-horizontal-xl>.list-group-item.active{margin-top:0}.list-group-horizontal-xl>.list-group-item+.list-group-item{border-top-width:var(--bs-list-group-border-width);border-left-width:0}.list-group-horizontal-xl>.list-group-item+.list-group-item.active{margin-left:calc(-1*var(--bs-list-group-border-width));border-left-width:var(--bs-list-group-border-width)}}.list-group-flush{border-radius:0}.list-group-flush>.list-group-item{border-width:0 0 var(--bs-list-group-border-width)}.list-group-flush>.list-group-item:last-child{border-bottom-width:0}.list-group-item-primary{--bs-list-group-color: var(--bs-primary-text-emphasis);--bs-list-group-bg: var(--bs-primary-bg-subtle);--bs-list-group-border-color: var(--bs-primary-border-subtle);--bs-list-group-action-hover-color: var(--bs-emphasis-color);--bs-list-group-action-hover-bg: var(--bs-primary-border-subtle);--bs-list-group-action-active-color: var(--bs-emphasis-color);--bs-list-group-action-active-bg: var(--bs-primary-border-subtle);--bs-list-group-active-color: var(--bs-primary-bg-subtle);--bs-list-group-active-bg: var(--bs-primary-text-emphasis);--bs-list-group-active-border-color: var(--bs-primary-text-emphasis)}.list-group-item-secondary{--bs-list-group-color: var(--bs-secondary-text-emphasis);--bs-list-group-bg: var(--bs-secondary-bg-subtle);--bs-list-group-border-color: var(--bs-secondary-border-subtle);--bs-list-group-action-hover-color: var(--bs-emphasis-color);--bs-list-group-action-hover-bg: var(--bs-secondary-border-subtle);--bs-list-group-action-active-color: var(--bs-emphasis-color);--bs-list-group-action-active-bg: var(--bs-secondary-border-subtle);--bs-list-group-active-color: var(--bs-secondary-bg-subtle);--bs-list-group-active-bg: var(--bs-secondary-text-emphasis);--bs-list-group-active-border-color: var(--bs-secondary-text-emphasis)}.list-group-item-success{--bs-list-group-color: var(--bs-success-text-emphasis);--bs-list-group-bg: var(--bs-success-bg-subtle);--bs-list-group-border-color: var(--bs-success-border-subtle);--bs-list-group-action-hover-color: var(--bs-emphasis-color);--bs-list-group-action-hover-bg: var(--bs-success-border-subtle);--bs-list-group-action-active-color: var(--bs-emphasis-color);--bs-list-group-action-active-bg: var(--bs-success-border-subtle);--bs-list-group-active-color: var(--bs-success-bg-subtle);--bs-list-group-active-bg: var(--bs-success-text-emphasis);--bs-list-group-active-border-color: var(--bs-success-text-emphasis)}.list-group-item-info{--bs-list-group-color: var(--bs-info-text-emphasis);--bs-list-group-bg: var(--bs-info-bg-subtle);--bs-list-group-border-color: var(--bs-info-border-subtle);--bs-list-group-action-hover-color: var(--bs-emphasis-color);--bs-list-group-action-hover-bg: var(--bs-info-border-subtle);--bs-list-group-action-active-color: var(--bs-emphasis-color);--bs-list-group-action-active-bg: var(--bs-info-border-subtle);--bs-list-group-active-color: var(--bs-info-bg-subtle);--bs-list-group-active-bg: var(--bs-info-text-emphasis);--bs-list-group-active-border-color: var(--bs-info-text-emphasis)}.list-group-item-warning{--bs-list-group-color: var(--bs-warning-text-emphasis);--bs-list-group-bg: var(--bs-warning-bg-subtle);--bs-list-group-border-color: var(--bs-warning-border-subtle);--bs-list-group-action-hover-color: var(--bs-emphasis-color);--bs-list-group-action-hover-bg: var(--bs-warning-border-subtle);--bs-list-group-action-active-color: var(--bs-emphasis-color);--bs-list-group-action-active-bg: var(--bs-warning-border-subtle);--bs-list-group-active-color: var(--bs-warning-bg-subtle);--bs-list-group-active-bg: var(--bs-warning-text-emphasis);--bs-list-group-active-border-color: var(--bs-warning-text-emphasis)}.list-group-item-danger{--bs-list-group-color: var(--bs-danger-text-emphasis);--bs-list-group-bg: var(--bs-danger-bg-subtle);--bs-list-group-border-color: var(--bs-danger-border-subtle);--bs-list-group-action-hover-color: var(--bs-emphasis-color);--bs-list-group-action-hover-bg: var(--bs-danger-border-subtle);--bs-list-group-action-active-color: var(--bs-emphasis-color);--bs-list-group-action-active-bg: var(--bs-danger-border-subtle);--bs-list-group-active-color: var(--bs-danger-bg-subtle);--bs-list-group-active-bg: var(--bs-danger-text-emphasis);--bs-list-group-active-border-color: var(--bs-danger-text-emphasis)}.list-group-item-light{--bs-list-group-color: var(--bs-light-text-emphasis);--bs-list-group-bg: var(--bs-light-bg-subtle);--bs-list-group-border-color: var(--bs-light-border-subtle);--bs-list-group-action-hover-color: var(--bs-emphasis-color);--bs-list-group-action-hover-bg: var(--bs-light-border-subtle);--bs-list-group-action-active-color: var(--bs-emphasis-color);--bs-list-group-action-active-bg: var(--bs-light-border-subtle);--bs-list-group-active-color: var(--bs-light-bg-subtle);--bs-list-group-active-bg: var(--bs-light-text-emphasis);--bs-list-group-active-border-color: var(--bs-light-text-emphasis)}.list-group-item-dark{--bs-list-group-color: var(--bs-dark-text-emphasis);--bs-list-group-bg: var(--bs-dark-bg-subtle);--bs-list-group-border-color: var(--bs-dark-border-subtle);--bs-list-group-action-hover-color: var(--bs-emphasis-color);--bs-list-group-action-hover-bg: var(--bs-dark-border-subtle);--bs-list-group-action-active-color: var(--bs-emphasis-color);--bs-list-group-action-active-bg: var(--bs-dark-border-subtle);--bs-list-group-active-color: var(--bs-dark-bg-subtle);--bs-list-group-active-bg: var(--bs-dark-text-emphasis);--bs-list-group-active-border-color: var(--bs-dark-text-emphasis)}.btn-close{--bs-btn-close-color: #000;--bs-btn-close-bg: url(\"data:image/svg+xml,%3csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='%23000'%3e%3cpath d='M.293.293a1 1 0 0 1 1.414 0L8 6.586 14.293.293a1 1 0 1 1 1.414 1.414L9.414 8l6.293 6.293a1 1 0 0 1-1.414 1.414L8 9.414l-6.293 6.293a1 1 0 0 1-1.414-1.414L6.586 8 .293 1.707a1 1 0 0 1 0-1.414z'/%3e%3c/svg%3e\");--bs-btn-close-opacity: 0.5;--bs-btn-close-hover-opacity: 0.75;--bs-btn-close-focus-shadow: 0 0 0 0.25rem rgba(13, 110, 253, 0.25);--bs-btn-close-focus-opacity: 1;--bs-btn-close-disabled-opacity: 0.25;--bs-btn-close-white-filter: invert(1) grayscale(100%) brightness(200%);box-sizing:content-box;width:1em;height:1em;padding:.25em .25em;color:var(--bs-btn-close-color);background:rgba(0,0,0,0) var(--bs-btn-close-bg) center/1em auto 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calc(var(--bs-border-radius-lg) - (var(--bs-border-width)));--bs-modal-header-padding-x: 1rem;--bs-modal-header-padding-y: 1rem;--bs-modal-header-padding: 1rem 1rem;--bs-modal-header-border-color: var(--bs-border-color);--bs-modal-header-border-width: var(--bs-border-width);--bs-modal-title-line-height: 1.5;--bs-modal-footer-gap: 0.5rem;--bs-modal-footer-bg: ;--bs-modal-footer-border-color: var(--bs-border-color);--bs-modal-footer-border-width: var(--bs-border-width);position:fixed;top:0;left:0;z-index:var(--bs-modal-zindex);display:none;width:100%;height:100%;overflow-x:hidden;overflow-y:auto;outline:0}.modal-dialog{position:relative;width:auto;margin:var(--bs-modal-margin);pointer-events:none}.modal.fade .modal-dialog{transition:transform .3s ease-out;transform:translate(0, -50px)}@media(prefers-reduced-motion: reduce){.modal.fade .modal-dialog{transition:none}}.modal.show .modal-dialog{transform:none}.modal.modal-static 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!important}.pt-lg-0{padding-top:0 !important}.pt-lg-1{padding-top:.25rem !important}.pt-lg-2{padding-top:.5rem !important}.pt-lg-3{padding-top:1rem !important}.pt-lg-4{padding-top:1.5rem !important}.pt-lg-5{padding-top:3rem !important}.pe-lg-0{padding-right:0 !important}.pe-lg-1{padding-right:.25rem !important}.pe-lg-2{padding-right:.5rem !important}.pe-lg-3{padding-right:1rem !important}.pe-lg-4{padding-right:1.5rem !important}.pe-lg-5{padding-right:3rem !important}.pb-lg-0{padding-bottom:0 !important}.pb-lg-1{padding-bottom:.25rem !important}.pb-lg-2{padding-bottom:.5rem !important}.pb-lg-3{padding-bottom:1rem !important}.pb-lg-4{padding-bottom:1.5rem !important}.pb-lg-5{padding-bottom:3rem !important}.ps-lg-0{padding-left:0 !important}.ps-lg-1{padding-left:.25rem !important}.ps-lg-2{padding-left:.5rem !important}.ps-lg-3{padding-left:1rem !important}.ps-lg-4{padding-left:1.5rem !important}.ps-lg-5{padding-left:3rem !important}.gap-lg-0{gap:0 !important}.gap-lg-1{gap:.25rem !important}.gap-lg-2{gap:.5rem !important}.gap-lg-3{gap:1rem !important}.gap-lg-4{gap:1.5rem !important}.gap-lg-5{gap:3rem !important}.row-gap-lg-0{row-gap:0 !important}.row-gap-lg-1{row-gap:.25rem !important}.row-gap-lg-2{row-gap:.5rem !important}.row-gap-lg-3{row-gap:1rem !important}.row-gap-lg-4{row-gap:1.5rem !important}.row-gap-lg-5{row-gap:3rem !important}.column-gap-lg-0{column-gap:0 !important}.column-gap-lg-1{column-gap:.25rem !important}.column-gap-lg-2{column-gap:.5rem !important}.column-gap-lg-3{column-gap:1rem !important}.column-gap-lg-4{column-gap:1.5rem !important}.column-gap-lg-5{column-gap:3rem !important}.text-lg-start{text-align:left !important}.text-lg-end{text-align:right !important}.text-lg-center{text-align:center !important}}@media(min-width: 1200px){.float-xl-start{float:left !important}.float-xl-end{float:right !important}.float-xl-none{float:none !important}.object-fit-xl-contain{object-fit:contain !important}.object-fit-xl-cover{object-fit:cover !important}.object-fit-xl-fill{object-fit:fill !important}.object-fit-xl-scale{object-fit:scale-down !important}.object-fit-xl-none{object-fit:none !important}.d-xl-inline{display:inline !important}.d-xl-inline-block{display:inline-block !important}.d-xl-block{display:block !important}.d-xl-grid{display:grid !important}.d-xl-inline-grid{display:inline-grid !important}.d-xl-table{display:table !important}.d-xl-table-row{display:table-row !important}.d-xl-table-cell{display:table-cell !important}.d-xl-flex{display:flex !important}.d-xl-inline-flex{display:inline-flex !important}.d-xl-none{display:none !important}.flex-xl-fill{flex:1 1 auto !important}.flex-xl-row{flex-direction:row !important}.flex-xl-column{flex-direction:column !important}.flex-xl-row-reverse{flex-direction:row-reverse !important}.flex-xl-column-reverse{flex-direction:column-reverse !important}.flex-xl-grow-0{flex-grow:0 !important}.flex-xl-grow-1{flex-grow:1 !important}.flex-xl-shrink-0{flex-shrink:0 !important}.flex-xl-shrink-1{flex-shrink:1 !important}.flex-xl-wrap{flex-wrap:wrap !important}.flex-xl-nowrap{flex-wrap:nowrap !important}.flex-xl-wrap-reverse{flex-wrap:wrap-reverse !important}.justify-content-xl-start{justify-content:flex-start !important}.justify-content-xl-end{justify-content:flex-end !important}.justify-content-xl-center{justify-content:center !important}.justify-content-xl-between{justify-content:space-between !important}.justify-content-xl-around{justify-content:space-around !important}.justify-content-xl-evenly{justify-content:space-evenly !important}.align-items-xl-start{align-items:flex-start !important}.align-items-xl-end{align-items:flex-end !important}.align-items-xl-center{align-items:center !important}.align-items-xl-baseline{align-items:baseline !important}.align-items-xl-stretch{align-items:stretch !important}.align-content-xl-start{align-content:flex-start !important}.align-content-xl-end{align-content:flex-end !important}.align-content-xl-center{align-content:center !important}.align-content-xl-between{align-content:space-between !important}.align-content-xl-around{align-content:space-around !important}.align-content-xl-stretch{align-content:stretch !important}.align-self-xl-auto{align-self:auto !important}.align-self-xl-start{align-self:flex-start !important}.align-self-xl-end{align-self:flex-end !important}.align-self-xl-center{align-self:center !important}.align-self-xl-baseline{align-self:baseline !important}.align-self-xl-stretch{align-self:stretch !important}.order-xl-first{order:-1 !important}.order-xl-0{order:0 !important}.order-xl-1{order:1 !important}.order-xl-2{order:2 !important}.order-xl-3{order:3 !important}.order-xl-4{order:4 !important}.order-xl-5{order:5 !important}.order-xl-last{order:6 !important}.m-xl-0{margin:0 !important}.m-xl-1{margin:.25rem !important}.m-xl-2{margin:.5rem !important}.m-xl-3{margin:1rem !important}.m-xl-4{margin:1.5rem !important}.m-xl-5{margin:3rem !important}.m-xl-auto{margin:auto !important}.mx-xl-0{margin-right:0 !important;margin-left:0 !important}.mx-xl-1{margin-right:.25rem !important;margin-left:.25rem !important}.mx-xl-2{margin-right:.5rem !important;margin-left:.5rem !important}.mx-xl-3{margin-right:1rem !important;margin-left:1rem !important}.mx-xl-4{margin-right:1.5rem !important;margin-left:1.5rem !important}.mx-xl-5{margin-right:3rem !important;margin-left:3rem !important}.mx-xl-auto{margin-right:auto !important;margin-left:auto !important}.my-xl-0{margin-top:0 !important;margin-bottom:0 !important}.my-xl-1{margin-top:.25rem !important;margin-bottom:.25rem !important}.my-xl-2{margin-top:.5rem !important;margin-bottom:.5rem !important}.my-xl-3{margin-top:1rem !important;margin-bottom:1rem !important}.my-xl-4{margin-top:1.5rem !important;margin-bottom:1.5rem !important}.my-xl-5{margin-top:3rem !important;margin-bottom:3rem !important}.my-xl-auto{margin-top:auto !important;margin-bottom:auto !important}.mt-xl-0{margin-top:0 !important}.mt-xl-1{margin-top:.25rem !important}.mt-xl-2{margin-top:.5rem !important}.mt-xl-3{margin-top:1rem !important}.mt-xl-4{margin-top:1.5rem !important}.mt-xl-5{margin-top:3rem !important}.mt-xl-auto{margin-top:auto !important}.me-xl-0{margin-right:0 !important}.me-xl-1{margin-right:.25rem !important}.me-xl-2{margin-right:.5rem !important}.me-xl-3{margin-right:1rem !important}.me-xl-4{margin-right:1.5rem !important}.me-xl-5{margin-right:3rem !important}.me-xl-auto{margin-right:auto !important}.mb-xl-0{margin-bottom:0 !important}.mb-xl-1{margin-bottom:.25rem !important}.mb-xl-2{margin-bottom:.5rem !important}.mb-xl-3{margin-bottom:1rem !important}.mb-xl-4{margin-bottom:1.5rem !important}.mb-xl-5{margin-bottom:3rem !important}.mb-xl-auto{margin-bottom:auto !important}.ms-xl-0{margin-left:0 !important}.ms-xl-1{margin-left:.25rem !important}.ms-xl-2{margin-left:.5rem !important}.ms-xl-3{margin-left:1rem !important}.ms-xl-4{margin-left:1.5rem !important}.ms-xl-5{margin-left:3rem !important}.ms-xl-auto{margin-left:auto !important}.p-xl-0{padding:0 !important}.p-xl-1{padding:.25rem !important}.p-xl-2{padding:.5rem !important}.p-xl-3{padding:1rem !important}.p-xl-4{padding:1.5rem !important}.p-xl-5{padding:3rem !important}.px-xl-0{padding-right:0 !important;padding-left:0 !important}.px-xl-1{padding-right:.25rem !important;padding-left:.25rem !important}.px-xl-2{padding-right:.5rem !important;padding-left:.5rem !important}.px-xl-3{padding-right:1rem !important;padding-left:1rem !important}.px-xl-4{padding-right:1.5rem !important;padding-left:1.5rem !important}.px-xl-5{padding-right:3rem !important;padding-left:3rem !important}.py-xl-0{padding-top:0 !important;padding-bottom:0 !important}.py-xl-1{padding-top:.25rem !important;padding-bottom:.25rem !important}.py-xl-2{padding-top:.5rem !important;padding-bottom:.5rem !important}.py-xl-3{padding-top:1rem !important;padding-bottom:1rem !important}.py-xl-4{padding-top:1.5rem !important;padding-bottom:1.5rem !important}.py-xl-5{padding-top:3rem !important;padding-bottom:3rem !important}.pt-xl-0{padding-top:0 !important}.pt-xl-1{padding-top:.25rem !important}.pt-xl-2{padding-top:.5rem !important}.pt-xl-3{padding-top:1rem !important}.pt-xl-4{padding-top:1.5rem !important}.pt-xl-5{padding-top:3rem !important}.pe-xl-0{padding-right:0 !important}.pe-xl-1{padding-right:.25rem !important}.pe-xl-2{padding-right:.5rem !important}.pe-xl-3{padding-right:1rem !important}.pe-xl-4{padding-right:1.5rem !important}.pe-xl-5{padding-right:3rem !important}.pb-xl-0{padding-bottom:0 !important}.pb-xl-1{padding-bottom:.25rem !important}.pb-xl-2{padding-bottom:.5rem !important}.pb-xl-3{padding-bottom:1rem !important}.pb-xl-4{padding-bottom:1.5rem !important}.pb-xl-5{padding-bottom:3rem !important}.ps-xl-0{padding-left:0 !important}.ps-xl-1{padding-left:.25rem !important}.ps-xl-2{padding-left:.5rem !important}.ps-xl-3{padding-left:1rem !important}.ps-xl-4{padding-left:1.5rem !important}.ps-xl-5{padding-left:3rem !important}.gap-xl-0{gap:0 !important}.gap-xl-1{gap:.25rem !important}.gap-xl-2{gap:.5rem !important}.gap-xl-3{gap:1rem !important}.gap-xl-4{gap:1.5rem !important}.gap-xl-5{gap:3rem !important}.row-gap-xl-0{row-gap:0 !important}.row-gap-xl-1{row-gap:.25rem !important}.row-gap-xl-2{row-gap:.5rem !important}.row-gap-xl-3{row-gap:1rem !important}.row-gap-xl-4{row-gap:1.5rem !important}.row-gap-xl-5{row-gap:3rem !important}.column-gap-xl-0{column-gap:0 !important}.column-gap-xl-1{column-gap:.25rem !important}.column-gap-xl-2{column-gap:.5rem !important}.column-gap-xl-3{column-gap:1rem !important}.column-gap-xl-4{column-gap:1.5rem !important}.column-gap-xl-5{column-gap:3rem !important}.text-xl-start{text-align:left !important}.text-xl-end{text-align:right !important}.text-xl-center{text-align:center !important}}@media(min-width: 1200px){.fs-1{font-size:2.5rem !important}.fs-2{font-size:2rem !important}.fs-3{font-size:1.75rem !important}.fs-4{font-size:1.5rem !important}}@media print{.d-print-inline{display:inline !important}.d-print-inline-block{display:inline-block !important}.d-print-block{display:block !important}.d-print-grid{display:grid !important}.d-print-inline-grid{display:inline-grid !important}.d-print-table{display:table !important}.d-print-table-row{display:table-row !important}.d-print-table-cell{display:table-cell !important}.d-print-flex{display:flex !important}.d-print-inline-flex{display:inline-flex !important}.d-print-none{display:none !important}}","@mixin bsBanner($file) {\n  /*!\n   * Bootstrap #{$file} v5.3.2 (https://getbootstrap.com/)\n   * Copyright 2011-2023 The Bootstrap Authors\n   * Licensed under MIT (https://github.com/twbs/bootstrap/blob/main/LICENSE)\n   */\n}\n",":root,\n[data-bs-theme=\"light\"] {\n  // Note: Custom variable values only support SassScript inside `#{}`.\n\n  // Colors\n  //\n  // Generate palettes for full colors, grays, and theme colors.\n\n  @each $color, $value in $colors {\n    --#{$prefix}#{$color}: #{$value};\n  }\n\n  @each $color, $value in $grays {\n    --#{$prefix}gray-#{$color}: #{$value};\n  }\n\n  @each $color, $value in $theme-colors {\n    --#{$prefix}#{$color}: #{$value};\n  }\n\n  @each $color, $value in $theme-colors-rgb {\n    --#{$prefix}#{$color}-rgb: #{$value};\n  }\n\n  @each $color, $value in $theme-colors-text {\n    --#{$prefix}#{$color}-text-emphasis: #{$value};\n  }\n\n  @each $color, $value in $theme-colors-bg-subtle {\n    --#{$prefix}#{$color}-bg-subtle: #{$value};\n  }\n\n  @each $color, $value in $theme-colors-border-subtle {\n    --#{$prefix}#{$color}-border-subtle: #{$value};\n  }\n\n  --#{$prefix}white-rgb: #{to-rgb($white)};\n  --#{$prefix}black-rgb: #{to-rgb($black)};\n\n  // Fonts\n\n  // Note: Use `inspect` for lists so that quoted items keep the quotes.\n  // See https://github.com/sass/sass/issues/2383#issuecomment-336349172\n  --#{$prefix}font-sans-serif: #{inspect($font-family-sans-serif)};\n  --#{$prefix}font-monospace: #{inspect($font-family-monospace)};\n  --#{$prefix}gradient: #{$gradient};\n\n  // Root and body\n  // scss-docs-start root-body-variables\n  @if $font-size-root != null {\n    --#{$prefix}root-font-size: #{$font-size-root};\n  }\n  --#{$prefix}body-font-family: #{inspect($font-family-base)};\n  @include rfs($font-size-base, --#{$prefix}body-font-size);\n  --#{$prefix}body-font-weight: #{$font-weight-base};\n  --#{$prefix}body-line-height: #{$line-height-base};\n  @if $body-text-align != null {\n    --#{$prefix}body-text-align: #{$body-text-align};\n  }\n\n  --#{$prefix}body-color: #{$body-color};\n  --#{$prefix}body-color-rgb: #{to-rgb($body-color)};\n  --#{$prefix}body-bg: #{$body-bg};\n  --#{$prefix}body-bg-rgb: #{to-rgb($body-bg)};\n\n  --#{$prefix}emphasis-color: #{$body-emphasis-color};\n  --#{$prefix}emphasis-color-rgb: #{to-rgb($body-emphasis-color)};\n\n  --#{$prefix}secondary-color: #{$body-secondary-color};\n  --#{$prefix}secondary-color-rgb: #{to-rgb($body-secondary-color)};\n  --#{$prefix}secondary-bg: #{$body-secondary-bg};\n  --#{$prefix}secondary-bg-rgb: #{to-rgb($body-secondary-bg)};\n\n  --#{$prefix}tertiary-color: #{$body-tertiary-color};\n  --#{$prefix}tertiary-color-rgb: #{to-rgb($body-tertiary-color)};\n  --#{$prefix}tertiary-bg: #{$body-tertiary-bg};\n  --#{$prefix}tertiary-bg-rgb: #{to-rgb($body-tertiary-bg)};\n  // scss-docs-end root-body-variables\n\n  --#{$prefix}heading-color: #{$headings-color};\n\n  --#{$prefix}link-color: #{$link-color};\n  --#{$prefix}link-color-rgb: #{to-rgb($link-color)};\n  --#{$prefix}link-decoration: #{$link-decoration};\n\n  --#{$prefix}link-hover-color: #{$link-hover-color};\n  --#{$prefix}link-hover-color-rgb: #{to-rgb($link-hover-color)};\n\n  @if $link-hover-decoration != null {\n    --#{$prefix}link-hover-decoration: #{$link-hover-decoration};\n  }\n\n  --#{$prefix}code-color: #{$code-color};\n  --#{$prefix}highlight-color: #{$mark-color};\n  --#{$prefix}highlight-bg: #{$mark-bg};\n\n  // scss-docs-start root-border-var\n  --#{$prefix}border-width: #{$border-width};\n  --#{$prefix}border-style: #{$border-style};\n  --#{$prefix}border-color: #{$border-color};\n  --#{$prefix}border-color-translucent: #{$border-color-translucent};\n\n  --#{$prefix}border-radius: #{$border-radius};\n  --#{$prefix}border-radius-sm: #{$border-radius-sm};\n  --#{$prefix}border-radius-lg: #{$border-radius-lg};\n  --#{$prefix}border-radius-xl: #{$border-radius-xl};\n  --#{$prefix}border-radius-xxl: #{$border-radius-xxl};\n  --#{$prefix}border-radius-2xl: var(--#{$prefix}border-radius-xxl); // Deprecated in v5.3.0 for consistency\n  --#{$prefix}border-radius-pill: #{$border-radius-pill};\n  // scss-docs-end root-border-var\n\n  --#{$prefix}box-shadow: #{$box-shadow};\n  --#{$prefix}box-shadow-sm: #{$box-shadow-sm};\n  --#{$prefix}box-shadow-lg: #{$box-shadow-lg};\n  --#{$prefix}box-shadow-inset: #{$box-shadow-inset};\n\n  // Focus styles\n  // scss-docs-start root-focus-variables\n  --#{$prefix}focus-ring-width: #{$focus-ring-width};\n  --#{$prefix}focus-ring-opacity: #{$focus-ring-opacity};\n  --#{$prefix}focus-ring-color: #{$focus-ring-color};\n  // scss-docs-end root-focus-variables\n\n  // scss-docs-start root-form-validation-variables\n  --#{$prefix}form-valid-color: #{$form-valid-color};\n  --#{$prefix}form-valid-border-color: #{$form-valid-border-color};\n  --#{$prefix}form-invalid-color: #{$form-invalid-color};\n  --#{$prefix}form-invalid-border-color: #{$form-invalid-border-color};\n  // scss-docs-end root-form-validation-variables\n}\n\n@if $enable-dark-mode {\n  @include color-mode(dark, true) {\n    color-scheme: dark;\n\n    // scss-docs-start root-dark-mode-vars\n    --#{$prefix}body-color: #{$body-color-dark};\n    --#{$prefix}body-color-rgb: #{to-rgb($body-color-dark)};\n    --#{$prefix}body-bg: #{$body-bg-dark};\n    --#{$prefix}body-bg-rgb: #{to-rgb($body-bg-dark)};\n\n    --#{$prefix}emphasis-color: #{$body-emphasis-color-dark};\n    --#{$prefix}emphasis-color-rgb: #{to-rgb($body-emphasis-color-dark)};\n\n    --#{$prefix}secondary-color: #{$body-secondary-color-dark};\n    --#{$prefix}secondary-color-rgb: #{to-rgb($body-secondary-color-dark)};\n    --#{$prefix}secondary-bg: #{$body-secondary-bg-dark};\n    --#{$prefix}secondary-bg-rgb: #{to-rgb($body-secondary-bg-dark)};\n\n    --#{$prefix}tertiary-color: #{$body-tertiary-color-dark};\n    --#{$prefix}tertiary-color-rgb: #{to-rgb($body-tertiary-color-dark)};\n    --#{$prefix}tertiary-bg: #{$body-tertiary-bg-dark};\n    --#{$prefix}tertiary-bg-rgb: #{to-rgb($body-tertiary-bg-dark)};\n\n    @each $color, $value in $theme-colors-text-dark {\n      --#{$prefix}#{$color}-text-emphasis: #{$value};\n    }\n\n    @each $color, $value in $theme-colors-bg-subtle-dark {\n      --#{$prefix}#{$color}-bg-subtle: #{$value};\n    }\n\n    @each $color, $value in $theme-colors-border-subtle-dark {\n      --#{$prefix}#{$color}-border-subtle: #{$value};\n    }\n\n    --#{$prefix}heading-color: #{$headings-color-dark};\n\n    --#{$prefix}link-color: #{$link-color-dark};\n    --#{$prefix}link-hover-color: #{$link-hover-color-dark};\n    --#{$prefix}link-color-rgb: #{to-rgb($link-color-dark)};\n    --#{$prefix}link-hover-color-rgb: #{to-rgb($link-hover-color-dark)};\n\n    --#{$prefix}code-color: #{$code-color-dark};\n    --#{$prefix}highlight-color: #{$mark-color-dark};\n    --#{$prefix}highlight-bg: #{$mark-bg-dark};\n\n    --#{$prefix}border-color: #{$border-color-dark};\n    --#{$prefix}border-color-translucent: #{$border-color-translucent-dark};\n\n    --#{$prefix}form-valid-color: #{$form-valid-color-dark};\n    --#{$prefix}form-valid-border-color: #{$form-valid-border-color-dark};\n    --#{$prefix}form-invalid-color: #{$form-invalid-color-dark};\n    --#{$prefix}form-invalid-border-color: #{$form-invalid-border-color-dark};\n    // scss-docs-end root-dark-mode-vars\n  }\n}\n","// stylelint-disable scss/dimension-no-non-numeric-values\n\n// SCSS RFS mixin\n//\n// Automated responsive values for font sizes, paddings, margins and much more\n//\n// Licensed under MIT (https://github.com/twbs/rfs/blob/main/LICENSE)\n\n// Configuration\n\n// Base value\n$rfs-base-value: 1.25rem !default;\n$rfs-unit: rem !default;\n\n@if $rfs-unit != rem and $rfs-unit != px {\n  @error \"`#{$rfs-unit}` is not a valid unit for $rfs-unit. Use `px` or `rem`.\";\n}\n\n// Breakpoint at where values start decreasing if screen width is smaller\n$rfs-breakpoint: 1200px !default;\n$rfs-breakpoint-unit: px !default;\n\n@if $rfs-breakpoint-unit != px and $rfs-breakpoint-unit != em and $rfs-breakpoint-unit != rem {\n  @error \"`#{$rfs-breakpoint-unit}` is not a valid unit for $rfs-breakpoint-unit. Use `px`, `em` or `rem`.\";\n}\n\n// Resize values based on screen height and width\n$rfs-two-dimensional: false !default;\n\n// Factor of decrease\n$rfs-factor: 10 !default;\n\n@if type-of($rfs-factor) != number or $rfs-factor <= 1 {\n  @error \"`#{$rfs-factor}` is not a valid  $rfs-factor, it must be greater than 1.\";\n}\n\n// Mode. Possibilities: \"min-media-query\", \"max-media-query\"\n$rfs-mode: min-media-query !default;\n\n// Generate enable or disable classes. Possibilities: false, \"enable\" or \"disable\"\n$rfs-class: false !default;\n\n// 1 rem = $rfs-rem-value px\n$rfs-rem-value: 16 !default;\n\n// Safari iframe resize bug: https://github.com/twbs/rfs/issues/14\n$rfs-safari-iframe-resize-bug-fix: false !default;\n\n// Disable RFS by setting $enable-rfs to false\n$enable-rfs: true !default;\n\n// Cache $rfs-base-value unit\n$rfs-base-value-unit: unit($rfs-base-value);\n\n@function divide($dividend, $divisor, $precision: 10) {\n  $sign: if($dividend > 0 and $divisor > 0 or $dividend < 0 and $divisor < 0, 1, -1);\n  $dividend: abs($dividend);\n  $divisor: abs($divisor);\n  @if $dividend == 0 {\n    @return 0;\n  }\n  @if $divisor == 0 {\n    @error \"Cannot divide by 0\";\n  }\n  $remainder: $dividend;\n  $result: 0;\n  $factor: 10;\n  @while ($remainder > 0 and $precision >= 0) {\n    $quotient: 0;\n    @while ($remainder >= $divisor) {\n      $remainder: $remainder - $divisor;\n      $quotient: $quotient + 1;\n    }\n    $result: $result * 10 + $quotient;\n    $factor: $factor * .1;\n    $remainder: $remainder * 10;\n    $precision: $precision - 1;\n    @if ($precision < 0 and $remainder >= $divisor * 5) {\n      $result: $result + 1;\n    }\n  }\n  $result: $result * $factor * $sign;\n  $dividend-unit: unit($dividend);\n  $divisor-unit: unit($divisor);\n  $unit-map: (\n    \"px\": 1px,\n    \"rem\": 1rem,\n    \"em\": 1em,\n    \"%\": 1%\n  );\n  @if ($dividend-unit != $divisor-unit and map-has-key($unit-map, $dividend-unit)) {\n    $result: $result * map-get($unit-map, $dividend-unit);\n  }\n  @return $result;\n}\n\n// Remove px-unit from $rfs-base-value for calculations\n@if $rfs-base-value-unit == px {\n  $rfs-base-value: divide($rfs-base-value, $rfs-base-value * 0 + 1);\n}\n@else if $rfs-base-value-unit == rem {\n  $rfs-base-value: divide($rfs-base-value, divide($rfs-base-value * 0 + 1, $rfs-rem-value));\n}\n\n// Cache $rfs-breakpoint unit to prevent multiple calls\n$rfs-breakpoint-unit-cache: unit($rfs-breakpoint);\n\n// Remove unit from $rfs-breakpoint for calculations\n@if $rfs-breakpoint-unit-cache == px {\n  $rfs-breakpoint: divide($rfs-breakpoint, $rfs-breakpoint * 0 + 1);\n}\n@else if $rfs-breakpoint-unit-cache == rem or $rfs-breakpoint-unit-cache == \"em\" {\n  $rfs-breakpoint: divide($rfs-breakpoint, divide($rfs-breakpoint * 0 + 1, $rfs-rem-value));\n}\n\n// Calculate the media query value\n$rfs-mq-value: if($rfs-breakpoint-unit == px, #{$rfs-breakpoint}px, #{divide($rfs-breakpoint, $rfs-rem-value)}#{$rfs-breakpoint-unit});\n$rfs-mq-property-width: if($rfs-mode == max-media-query, max-width, min-width);\n$rfs-mq-property-height: if($rfs-mode == max-media-query, max-height, min-height);\n\n// Internal mixin used to determine which media query needs to be used\n@mixin _rfs-media-query {\n  @if $rfs-two-dimensional {\n    @if $rfs-mode == max-media-query {\n      @media (#{$rfs-mq-property-width}: #{$rfs-mq-value}), (#{$rfs-mq-property-height}: #{$rfs-mq-value}) {\n        @content;\n      }\n    }\n    @else {\n      @media (#{$rfs-mq-property-width}: #{$rfs-mq-value}) and (#{$rfs-mq-property-height}: #{$rfs-mq-value}) {\n        @content;\n      }\n    }\n  }\n  @else {\n    @media (#{$rfs-mq-property-width}: #{$rfs-mq-value}) {\n      @content;\n    }\n  }\n}\n\n// Internal mixin that adds disable classes to the selector if needed.\n@mixin _rfs-rule {\n  @if $rfs-class == disable and $rfs-mode == max-media-query {\n    // Adding an extra class increases specificity, which prevents the media query to override the property\n    &,\n    .disable-rfs &,\n    &.disable-rfs {\n      @content;\n    }\n  }\n  @else if $rfs-class == enable and $rfs-mode == min-media-query {\n    .enable-rfs &,\n    &.enable-rfs {\n      @content;\n    }\n  } @else {\n    @content;\n  }\n}\n\n// Internal mixin that adds enable classes to the selector if needed.\n@mixin _rfs-media-query-rule {\n\n  @if $rfs-class == enable {\n    @if $rfs-mode == min-media-query {\n      @content;\n    }\n\n    @include _rfs-media-query () {\n      .enable-rfs &,\n      &.enable-rfs {\n        @content;\n      }\n    }\n  }\n  @else {\n    @if $rfs-class == disable and $rfs-mode == min-media-query {\n      .disable-rfs &,\n      &.disable-rfs {\n        @content;\n      }\n    }\n    @include _rfs-media-query () {\n      @content;\n    }\n  }\n}\n\n// Helper function to get the formatted non-responsive value\n@function rfs-value($values) {\n  // Convert to list\n  $values: if(type-of($values) != list, ($values,), $values);\n\n  $val: \"\";\n\n  // Loop over each value and calculate value\n  @each $value in $values {\n    @if $value == 0 {\n      $val: $val + \" 0\";\n    }\n    @else {\n      // Cache $value unit\n      $unit: if(type-of($value) == \"number\", unit($value), false);\n\n      @if $unit == px {\n        // Convert to rem if needed\n        $val: $val + \" \" + if($rfs-unit == rem, #{divide($value, $value * 0 + $rfs-rem-value)}rem, $value);\n      }\n      @else if $unit == rem {\n        // Convert to px if needed\n        $val: $val + \" \" + if($rfs-unit == px, #{divide($value, $value * 0 + 1) * $rfs-rem-value}px, $value);\n      } @else {\n        // If $value isn't a number (like inherit) or $value has a unit (not px or rem, like 1.5em) or $ is 0, just print the value\n        $val: $val + \" \" + $value;\n      }\n    }\n  }\n\n  // Remove first space\n  @return unquote(str-slice($val, 2));\n}\n\n// Helper function to get the responsive value calculated by RFS\n@function rfs-fluid-value($values) {\n  // Convert to list\n  $values: if(type-of($values) != list, ($values,), $values);\n\n  $val: \"\";\n\n  // Loop over each value and calculate value\n  @each $value in $values {\n    @if $value == 0 {\n      $val: $val + \" 0\";\n    } @else {\n      // Cache $value unit\n      $unit: if(type-of($value) == \"number\", unit($value), false);\n\n      // If $value isn't a number (like inherit) or $value has a unit (not px or rem, like 1.5em) or $ is 0, just print the value\n      @if not $unit or $unit != px and $unit != rem {\n        $val: $val + \" \" + $value;\n      } @else {\n        // Remove unit from $value for calculations\n        $value: divide($value, $value * 0 + if($unit == px, 1, divide(1, $rfs-rem-value)));\n\n        // Only add the media query if the value is greater than the minimum value\n        @if abs($value) <= $rfs-base-value or not $enable-rfs {\n          $val: $val + \" \" + if($rfs-unit == rem, #{divide($value, $rfs-rem-value)}rem, #{$value}px);\n        }\n        @else {\n          // Calculate the minimum value\n          $value-min: $rfs-base-value + divide(abs($value) - $rfs-base-value, $rfs-factor);\n\n          // Calculate difference between $value and the minimum value\n          $value-diff: abs($value) - $value-min;\n\n          // Base value formatting\n          $min-width: if($rfs-unit == rem, #{divide($value-min, $rfs-rem-value)}rem, #{$value-min}px);\n\n          // Use negative value if needed\n          $min-width: if($value < 0, -$min-width, $min-width);\n\n          // Use `vmin` if two-dimensional is enabled\n          $variable-unit: if($rfs-two-dimensional, vmin, vw);\n\n          // Calculate the variable width between 0 and $rfs-breakpoint\n          $variable-width: #{divide($value-diff * 100, $rfs-breakpoint)}#{$variable-unit};\n\n          // Return the calculated value\n          $val: $val + \" calc(\" + $min-width + if($value < 0, \" - \", \" + \") + $variable-width + \")\";\n        }\n      }\n    }\n  }\n\n  // Remove first space\n  @return unquote(str-slice($val, 2));\n}\n\n// RFS mixin\n@mixin rfs($values, $property: font-size) {\n  @if $values != null {\n    $val: rfs-value($values);\n    $fluid-val: rfs-fluid-value($values);\n\n    // Do not print the media query if responsive & non-responsive values are the same\n    @if $val == $fluid-val {\n      #{$property}: $val;\n    }\n    @else {\n      @include _rfs-rule () {\n        #{$property}: if($rfs-mode == max-media-query, $val, $fluid-val);\n\n        // Include safari iframe resize fix if needed\n        min-width: if($rfs-safari-iframe-resize-bug-fix, (0 * 1vw), null);\n      }\n\n      @include _rfs-media-query-rule () {\n        #{$property}: if($rfs-mode == max-media-query, $fluid-val, $val);\n      }\n    }\n  }\n}\n\n// Shorthand helper mixins\n@mixin font-size($value) {\n  @include rfs($value);\n}\n\n@mixin padding($value) {\n  @include rfs($value, padding);\n}\n\n@mixin padding-top($value) {\n  @include rfs($value, padding-top);\n}\n\n@mixin padding-right($value) {\n  @include rfs($value, padding-right);\n}\n\n@mixin padding-bottom($value) {\n  @include rfs($value, padding-bottom);\n}\n\n@mixin padding-left($value) {\n  @include rfs($value, padding-left);\n}\n\n@mixin margin($value) {\n  @include rfs($value, margin);\n}\n\n@mixin margin-top($value) {\n  @include rfs($value, margin-top);\n}\n\n@mixin margin-right($value) {\n  @include rfs($value, margin-right);\n}\n\n@mixin margin-bottom($value) {\n  @include rfs($value, margin-bottom);\n}\n\n@mixin margin-left($value) {\n  @include rfs($value, margin-left);\n}\n","// scss-docs-start color-mode-mixin\n@mixin color-mode($mode: light, $root: false) {\n  @if $color-mode-type == \"media-query\" {\n    @if $root == true {\n      @media (prefers-color-scheme: $mode) {\n        :root {\n          @content;\n        }\n      }\n    } @else {\n      @media (prefers-color-scheme: $mode) {\n        @content;\n      }\n    }\n  } @else {\n    [data-bs-theme=\"#{$mode}\"] {\n      @content;\n    }\n  }\n}\n// scss-docs-end color-mode-mixin\n","// stylelint-disable declaration-no-important, selector-no-qualifying-type, property-no-vendor-prefix\n\n\n// Reboot\n//\n// Normalization of HTML elements, manually forked from Normalize.css to remove\n// styles targeting irrelevant browsers while applying new styles.\n//\n// Normalize is licensed MIT. https://github.com/necolas/normalize.css\n\n\n// Document\n//\n// Change from `box-sizing: content-box` so that `width` is not affected by `padding` or `border`.\n\n*,\n*::before,\n*::after {\n  box-sizing: border-box;\n}\n\n\n// Root\n//\n// Ability to the value of the root font sizes, affecting the value of `rem`.\n// null by default, thus nothing is generated.\n\n:root {\n  @if $font-size-root != null {\n    @include font-size(var(--#{$prefix}root-font-size));\n  }\n\n  @if $enable-smooth-scroll {\n    @media (prefers-reduced-motion: no-preference) {\n      scroll-behavior: smooth;\n    }\n  }\n}\n\n\n// Body\n//\n// 1. Remove the margin in all browsers.\n// 2. As a best practice, apply a default `background-color`.\n// 3. Prevent adjustments of font size after orientation changes in iOS.\n// 4. Change the default tap highlight to be completely transparent in iOS.\n\n// scss-docs-start reboot-body-rules\nbody {\n  margin: 0; // 1\n  font-family: var(--#{$prefix}body-font-family);\n  @include font-size(var(--#{$prefix}body-font-size));\n  font-weight: var(--#{$prefix}body-font-weight);\n  line-height: var(--#{$prefix}body-line-height);\n  color: var(--#{$prefix}body-color);\n  text-align: var(--#{$prefix}body-text-align);\n  background-color: var(--#{$prefix}body-bg); // 2\n  -webkit-text-size-adjust: 100%; // 3\n  -webkit-tap-highlight-color: rgba($black, 0); // 4\n}\n// scss-docs-end reboot-body-rules\n\n\n// Content grouping\n//\n// 1. Reset Firefox's gray color\n\nhr {\n  margin: $hr-margin-y 0;\n  color: $hr-color; // 1\n  border: 0;\n  border-top: $hr-border-width solid $hr-border-color;\n  opacity: $hr-opacity;\n}\n\n\n// Typography\n//\n// 1. Remove top margins from headings\n//    By default, `<h1>`-`<h6>` all receive top and bottom margins. We nuke the top\n//    margin for easier control within type scales as it avoids margin collapsing.\n\n%heading {\n  margin-top: 0; // 1\n  margin-bottom: $headings-margin-bottom;\n  font-family: $headings-font-family;\n  font-style: $headings-font-style;\n  font-weight: $headings-font-weight;\n  line-height: $headings-line-height;\n  color: var(--#{$prefix}heading-color);\n}\n\nh1 {\n  @extend %heading;\n  @include font-size($h1-font-size);\n}\n\nh2 {\n  @extend %heading;\n  @include font-size($h2-font-size);\n}\n\nh3 {\n  @extend %heading;\n  @include font-size($h3-font-size);\n}\n\nh4 {\n  @extend %heading;\n  @include font-size($h4-font-size);\n}\n\nh5 {\n  @extend %heading;\n  @include font-size($h5-font-size);\n}\n\nh6 {\n  @extend %heading;\n  @include font-size($h6-font-size);\n}\n\n\n// Reset margins on paragraphs\n//\n// Similarly, the top margin on `<p>`s get reset. However, we also reset the\n// bottom margin to use `rem` units instead of `em`.\n\np {\n  margin-top: 0;\n  margin-bottom: $paragraph-margin-bottom;\n}\n\n\n// Abbreviations\n//\n// 1. Add the correct text decoration in Chrome, Edge, Opera, and Safari.\n// 2. Add explicit cursor to indicate changed behavior.\n// 3. Prevent the text-decoration to be skipped.\n\nabbr[title] {\n  text-decoration: underline dotted; // 1\n  cursor: help; // 2\n  text-decoration-skip-ink: none; // 3\n}\n\n\n// Address\n\naddress {\n  margin-bottom: 1rem;\n  font-style: normal;\n  line-height: inherit;\n}\n\n\n// Lists\n\nol,\nul {\n  padding-left: 2rem;\n}\n\nol,\nul,\ndl {\n  margin-top: 0;\n  margin-bottom: 1rem;\n}\n\nol ol,\nul ul,\nol ul,\nul ol {\n  margin-bottom: 0;\n}\n\ndt {\n  font-weight: $dt-font-weight;\n}\n\n// 1. Undo browser default\n\ndd {\n  margin-bottom: .5rem;\n  margin-left: 0; // 1\n}\n\n\n// Blockquote\n\nblockquote {\n  margin: 0 0 1rem;\n}\n\n\n// Strong\n//\n// Add the correct font weight in Chrome, Edge, and Safari\n\nb,\nstrong {\n  font-weight: $font-weight-bolder;\n}\n\n\n// Small\n//\n// Add the correct font size in all browsers\n\nsmall {\n  @include font-size($small-font-size);\n}\n\n\n// Mark\n\nmark {\n  padding: $mark-padding;\n  color: var(--#{$prefix}highlight-color);\n  background-color: var(--#{$prefix}highlight-bg);\n}\n\n\n// Sub and Sup\n//\n// Prevent `sub` and `sup` elements from affecting the line height in\n// all browsers.\n\nsub,\nsup {\n  position: relative;\n  @include font-size($sub-sup-font-size);\n  line-height: 0;\n  vertical-align: baseline;\n}\n\nsub { bottom: -.25em; }\nsup { top: -.5em; }\n\n\n// Links\n\na {\n  color: rgba(var(--#{$prefix}link-color-rgb), var(--#{$prefix}link-opacity, 1));\n  text-decoration: $link-decoration;\n\n  &:hover {\n    --#{$prefix}link-color-rgb: var(--#{$prefix}link-hover-color-rgb);\n    text-decoration: $link-hover-decoration;\n  }\n}\n\n// And undo these styles for placeholder links/named anchors (without href).\n// It would be more straightforward to just use a[href] in previous block, but that\n// causes specificity issues in many other styles that are too complex to fix.\n// See https://github.com/twbs/bootstrap/issues/19402\n\na:not([href]):not([class]) {\n  &,\n  &:hover {\n    color: inherit;\n    text-decoration: none;\n  }\n}\n\n\n// Code\n\npre,\ncode,\nkbd,\nsamp {\n  font-family: $font-family-code;\n  @include font-size(1em); // Correct the odd `em` font sizing in all browsers.\n}\n\n// 1. Remove browser default top margin\n// 2. Reset browser default of `1em` to use `rem`s\n// 3. Don't allow content to break outside\n\npre {\n  display: block;\n  margin-top: 0; // 1\n  margin-bottom: 1rem; // 2\n  overflow: auto; // 3\n  @include font-size($code-font-size);\n  color: $pre-color;\n\n  // Account for some code outputs that place code tags in pre tags\n  code {\n    @include font-size(inherit);\n    color: inherit;\n    word-break: normal;\n  }\n}\n\ncode {\n  @include font-size($code-font-size);\n  color: var(--#{$prefix}code-color);\n  word-wrap: break-word;\n\n  // Streamline the style when inside anchors to avoid broken underline and more\n  a > & {\n    color: inherit;\n  }\n}\n\nkbd {\n  padding: $kbd-padding-y $kbd-padding-x;\n  @include font-size($kbd-font-size);\n  color: $kbd-color;\n  background-color: $kbd-bg;\n  @include border-radius($border-radius-sm);\n\n  kbd {\n    padding: 0;\n    @include font-size(1em);\n    font-weight: $nested-kbd-font-weight;\n  }\n}\n\n\n// Figures\n//\n// Apply a consistent margin strategy (matches our type styles).\n\nfigure {\n  margin: 0 0 1rem;\n}\n\n\n// Images and content\n\nimg,\nsvg {\n  vertical-align: middle;\n}\n\n\n// Tables\n//\n// Prevent double borders\n\ntable {\n  caption-side: bottom;\n  border-collapse: collapse;\n}\n\ncaption {\n  padding-top: $table-cell-padding-y;\n  padding-bottom: $table-cell-padding-y;\n  color: $table-caption-color;\n  text-align: left;\n}\n\n// 1. Removes font-weight bold by inheriting\n// 2. Matches default `<td>` alignment by inheriting `text-align`.\n// 3. Fix alignment for Safari\n\nth {\n  font-weight: $table-th-font-weight; // 1\n  text-align: inherit; // 2\n  text-align: -webkit-match-parent; // 3\n}\n\nthead,\ntbody,\ntfoot,\ntr,\ntd,\nth {\n  border-color: inherit;\n  border-style: solid;\n  border-width: 0;\n}\n\n\n// Forms\n//\n// 1. Allow labels to use `margin` for spacing.\n\nlabel {\n  display: inline-block; // 1\n}\n\n// Remove the default `border-radius` that macOS Chrome adds.\n// See https://github.com/twbs/bootstrap/issues/24093\n\nbutton {\n  // stylelint-disable-next-line property-disallowed-list\n  border-radius: 0;\n}\n\n// Explicitly remove focus outline in Chromium when it shouldn't be\n// visible (e.g. as result of mouse click or touch tap). It already\n// should be doing this automatically, but seems to currently be\n// confused and applies its very visible two-tone outline anyway.\n\nbutton:focus:not(:focus-visible) {\n  outline: 0;\n}\n\n// 1. Remove the margin in Firefox and Safari\n\ninput,\nbutton,\nselect,\noptgroup,\ntextarea {\n  margin: 0; // 1\n  font-family: inherit;\n  @include font-size(inherit);\n  line-height: inherit;\n}\n\n// Remove the inheritance of text transform in Firefox\nbutton,\nselect {\n  text-transform: none;\n}\n// Set the cursor for non-`<button>` buttons\n//\n// Details at https://github.com/twbs/bootstrap/pull/30562\n[role=\"button\"] {\n  cursor: pointer;\n}\n\nselect {\n  // Remove the inheritance of word-wrap in Safari.\n  // See https://github.com/twbs/bootstrap/issues/24990\n  word-wrap: normal;\n\n  // Undo the opacity change from Chrome\n  &:disabled {\n    opacity: 1;\n  }\n}\n\n// Remove the dropdown arrow only from text type inputs built with datalists in Chrome.\n// See https://stackoverflow.com/a/54997118\n\n[list]:not([type=\"date\"]):not([type=\"datetime-local\"]):not([type=\"month\"]):not([type=\"week\"]):not([type=\"time\"])::-webkit-calendar-picker-indicator {\n  display: none !important;\n}\n\n// 1. Prevent a WebKit bug where (2) destroys native `audio` and `video`\n//    controls in Android 4.\n// 2. Correct the inability to style clickable types in iOS and Safari.\n// 3. Opinionated: add \"hand\" cursor to non-disabled button elements.\n\nbutton,\n[type=\"button\"], // 1\n[type=\"reset\"],\n[type=\"submit\"] {\n  -webkit-appearance: button; // 2\n\n  @if $enable-button-pointers {\n    &:not(:disabled) {\n      cursor: pointer; // 3\n    }\n  }\n}\n\n// Remove inner border and padding from Firefox, but don't restore the outline like Normalize.\n\n::-moz-focus-inner {\n  padding: 0;\n  border-style: none;\n}\n\n// 1. Textareas should really only resize vertically so they don't break their (horizontal) containers.\n\ntextarea {\n  resize: vertical; // 1\n}\n\n// 1. Browsers set a default `min-width: min-content;` on fieldsets,\n//    unlike e.g. `<div>`s, which have `min-width: 0;` by default.\n//    So we reset that to ensure fieldsets behave more like a standard block element.\n//    See https://github.com/twbs/bootstrap/issues/12359\n//    and https://html.spec.whatwg.org/multipage/#the-fieldset-and-legend-elements\n// 2. Reset the default outline behavior of fieldsets so they don't affect page layout.\n\nfieldset {\n  min-width: 0; // 1\n  padding: 0; // 2\n  margin: 0; // 2\n  border: 0; // 2\n}\n\n// 1. By using `float: left`, the legend will behave like a block element.\n//    This way the border of a fieldset wraps around the legend if present.\n// 2. Fix wrapping bug.\n//    See https://github.com/twbs/bootstrap/issues/29712\n\nlegend {\n  float: left; // 1\n  width: 100%;\n  padding: 0;\n  margin-bottom: $legend-margin-bottom;\n  @include font-size($legend-font-size);\n  font-weight: $legend-font-weight;\n  line-height: inherit;\n\n  + * {\n    clear: left; // 2\n  }\n}\n\n// Fix height of inputs with a type of datetime-local, date, month, week, or time\n// See https://github.com/twbs/bootstrap/issues/18842\n\n::-webkit-datetime-edit-fields-wrapper,\n::-webkit-datetime-edit-text,\n::-webkit-datetime-edit-minute,\n::-webkit-datetime-edit-hour-field,\n::-webkit-datetime-edit-day-field,\n::-webkit-datetime-edit-month-field,\n::-webkit-datetime-edit-year-field {\n  padding: 0;\n}\n\n::-webkit-inner-spin-button {\n  height: auto;\n}\n\n// 1. This overrides the extra rounded corners on search inputs in iOS so that our\n//    `.form-control` class can properly style them. Note that this cannot simply\n//    be added to `.form-control` as it's not specific enough. For details, see\n//    https://github.com/twbs/bootstrap/issues/11586.\n// 2. Correct the outline style in Safari.\n\n[type=\"search\"] {\n  -webkit-appearance: textfield; // 1\n  outline-offset: -2px; // 2\n}\n\n// 1. A few input types should stay LTR\n// See https://rtlstyling.com/posts/rtl-styling#form-inputs\n// 2. RTL only output\n// See https://rtlcss.com/learn/usage-guide/control-directives/#raw\n\n/* rtl:raw:\n[type=\"tel\"],\n[type=\"url\"],\n[type=\"email\"],\n[type=\"number\"] {\n  direction: ltr;\n}\n*/\n\n// Remove the inner padding in Chrome and Safari on macOS.\n\n::-webkit-search-decoration {\n  -webkit-appearance: none;\n}\n\n// Remove padding around color pickers in webkit browsers\n\n::-webkit-color-swatch-wrapper {\n  padding: 0;\n}\n\n\n// 1. Inherit font family and line height for file input buttons\n// 2. Correct the inability to style clickable types in iOS and Safari.\n\n::file-selector-button {\n  font: inherit; // 1\n  -webkit-appearance: button; // 2\n}\n\n// Correct element displays\n\noutput {\n  display: inline-block;\n}\n\n// Remove border from iframe\n\niframe {\n  border: 0;\n}\n\n// Summary\n//\n// 1. Add the correct display in all browsers\n\nsummary {\n  display: list-item; // 1\n  cursor: pointer;\n}\n\n\n// Progress\n//\n// Add the correct vertical alignment in Chrome, Firefox, and Opera.\n\nprogress {\n  vertical-align: baseline;\n}\n\n\n// Hidden attribute\n//\n// Always hide an element with the `hidden` HTML attribute.\n\n[hidden] {\n  display: none !important;\n}\n","// Variables\n//\n// Variables should follow the `$component-state-property-size` formula for\n// consistent naming. Ex: $nav-link-disabled-color and $modal-content-box-shadow-xs.\n\n// Color system\n\n// scss-docs-start gray-color-variables\n$white:    #fff !default;\n$gray-100: #f8f9fa !default;\n$gray-200: #e9ecef !default;\n$gray-300: #dee2e6 !default;\n$gray-400: #ced4da !default;\n$gray-500: #adb5bd !default;\n$gray-600: #6c757d !default;\n$gray-700: #495057 !default;\n$gray-800: #343a40 !default;\n$gray-900: #212529 !default;\n$black:    #000 !default;\n// scss-docs-end gray-color-variables\n\n// fusv-disable\n// scss-docs-start gray-colors-map\n$grays: (\n  \"100\": $gray-100,\n  \"200\": $gray-200,\n  \"300\": $gray-300,\n  \"400\": $gray-400,\n  \"500\": $gray-500,\n  \"600\": $gray-600,\n  \"700\": $gray-700,\n  \"800\": $gray-800,\n  \"900\": $gray-900\n) !default;\n// scss-docs-end gray-colors-map\n// fusv-enable\n\n// scss-docs-start color-variables\n$blue:    #0d6efd !default;\n$indigo:  #6610f2 !default;\n$purple:  #6f42c1 !default;\n$pink:    #d63384 !default;\n$red:     #dc3545 !default;\n$orange:  #fd7e14 !default;\n$yellow:  #ffc107 !default;\n$green:   #198754 !default;\n$teal:    #20c997 !default;\n$cyan:    #0dcaf0 !default;\n// scss-docs-end color-variables\n\n// scss-docs-start colors-map\n$colors: (\n  \"blue\":       $blue,\n  \"indigo\":     $indigo,\n  \"purple\":     $purple,\n  \"pink\":       $pink,\n  \"red\":        $red,\n  \"orange\":     $orange,\n  \"yellow\":     $yellow,\n  \"green\":      $green,\n  \"teal\":       $teal,\n  \"cyan\":       $cyan,\n  \"black\":      $black,\n  \"white\":      $white,\n  \"gray\":       $gray-600,\n  \"gray-dark\":  $gray-800\n) !default;\n// scss-docs-end colors-map\n\n// The contrast ratio to reach against white, to determine if color changes from \"light\" to \"dark\". Acceptable values for WCAG 2.0 are 3, 4.5 and 7.\n// See https://www.w3.org/TR/WCAG20/#visual-audio-contrast-contrast\n$min-contrast-ratio:   4.5 !default;\n\n// Customize the light and dark text colors for use in our color contrast function.\n$color-contrast-dark:      $black !default;\n$color-contrast-light:     $white !default;\n\n// fusv-disable\n$blue-100: tint-color($blue, 80%) !default;\n$blue-200: tint-color($blue, 60%) !default;\n$blue-300: tint-color($blue, 40%) !default;\n$blue-400: tint-color($blue, 20%) !default;\n$blue-500: $blue !default;\n$blue-600: shade-color($blue, 20%) !default;\n$blue-700: shade-color($blue, 40%) !default;\n$blue-800: shade-color($blue, 60%) !default;\n$blue-900: shade-color($blue, 80%) !default;\n\n$indigo-100: tint-color($indigo, 80%) !default;\n$indigo-200: tint-color($indigo, 60%) !default;\n$indigo-300: tint-color($indigo, 40%) !default;\n$indigo-400: tint-color($indigo, 20%) !default;\n$indigo-500: $indigo !default;\n$indigo-600: shade-color($indigo, 20%) !default;\n$indigo-700: shade-color($indigo, 40%) !default;\n$indigo-800: shade-color($indigo, 60%) !default;\n$indigo-900: shade-color($indigo, 80%) !default;\n\n$purple-100: tint-color($purple, 80%) !default;\n$purple-200: tint-color($purple, 60%) !default;\n$purple-300: tint-color($purple, 40%) !default;\n$purple-400: tint-color($purple, 20%) !default;\n$purple-500: $purple !default;\n$purple-600: shade-color($purple, 20%) !default;\n$purple-700: shade-color($purple, 40%) !default;\n$purple-800: shade-color($purple, 60%) !default;\n$purple-900: shade-color($purple, 80%) !default;\n\n$pink-100: tint-color($pink, 80%) !default;\n$pink-200: tint-color($pink, 60%) !default;\n$pink-300: tint-color($pink, 40%) !default;\n$pink-400: tint-color($pink, 20%) !default;\n$pink-500: $pink !default;\n$pink-600: shade-color($pink, 20%) !default;\n$pink-700: shade-color($pink, 40%) !default;\n$pink-800: shade-color($pink, 60%) !default;\n$pink-900: shade-color($pink, 80%) !default;\n\n$red-100: tint-color($red, 80%) !default;\n$red-200: tint-color($red, 60%) !default;\n$red-300: tint-color($red, 40%) !default;\n$red-400: tint-color($red, 20%) !default;\n$red-500: $red !default;\n$red-600: shade-color($red, 20%) !default;\n$red-700: shade-color($red, 40%) !default;\n$red-800: shade-color($red, 60%) !default;\n$red-900: shade-color($red, 80%) !default;\n\n$orange-100: tint-color($orange, 80%) !default;\n$orange-200: tint-color($orange, 60%) !default;\n$orange-300: tint-color($orange, 40%) !default;\n$orange-400: tint-color($orange, 20%) !default;\n$orange-500: $orange !default;\n$orange-600: shade-color($orange, 20%) !default;\n$orange-700: shade-color($orange, 40%) !default;\n$orange-800: shade-color($orange, 60%) !default;\n$orange-900: shade-color($orange, 80%) !default;\n\n$yellow-100: tint-color($yellow, 80%) !default;\n$yellow-200: tint-color($yellow, 60%) !default;\n$yellow-300: tint-color($yellow, 40%) !default;\n$yellow-400: tint-color($yellow, 20%) !default;\n$yellow-500: $yellow !default;\n$yellow-600: shade-color($yellow, 20%) !default;\n$yellow-700: shade-color($yellow, 40%) !default;\n$yellow-800: shade-color($yellow, 60%) !default;\n$yellow-900: shade-color($yellow, 80%) !default;\n\n$green-100: tint-color($green, 80%) !default;\n$green-200: tint-color($green, 60%) !default;\n$green-300: tint-color($green, 40%) !default;\n$green-400: tint-color($green, 20%) !default;\n$green-500: $green !default;\n$green-600: shade-color($green, 20%) !default;\n$green-700: shade-color($green, 40%) !default;\n$green-800: shade-color($green, 60%) !default;\n$green-900: shade-color($green, 80%) !default;\n\n$teal-100: tint-color($teal, 80%) !default;\n$teal-200: tint-color($teal, 60%) !default;\n$teal-300: tint-color($teal, 40%) !default;\n$teal-400: tint-color($teal, 20%) !default;\n$teal-500: $teal !default;\n$teal-600: shade-color($teal, 20%) !default;\n$teal-700: shade-color($teal, 40%) !default;\n$teal-800: shade-color($teal, 60%) !default;\n$teal-900: shade-color($teal, 80%) !default;\n\n$cyan-100: tint-color($cyan, 80%) !default;\n$cyan-200: tint-color($cyan, 60%) !default;\n$cyan-300: tint-color($cyan, 40%) !default;\n$cyan-400: tint-color($cyan, 20%) !default;\n$cyan-500: $cyan !default;\n$cyan-600: shade-color($cyan, 20%) !default;\n$cyan-700: shade-color($cyan, 40%) !default;\n$cyan-800: shade-color($cyan, 60%) !default;\n$cyan-900: shade-color($cyan, 80%) !default;\n\n$blues: (\n  \"blue-100\": $blue-100,\n  \"blue-200\": $blue-200,\n  \"blue-300\": $blue-300,\n  \"blue-400\": $blue-400,\n  \"blue-500\": $blue-500,\n  \"blue-600\": $blue-600,\n  \"blue-700\": $blue-700,\n  \"blue-800\": $blue-800,\n  \"blue-900\": $blue-900\n) !default;\n\n$indigos: (\n  \"indigo-100\": $indigo-100,\n  \"indigo-200\": $indigo-200,\n  \"indigo-300\": $indigo-300,\n  \"indigo-400\": $indigo-400,\n  \"indigo-500\": $indigo-500,\n  \"indigo-600\": $indigo-600,\n  \"indigo-700\": $indigo-700,\n  \"indigo-800\": $indigo-800,\n  \"indigo-900\": $indigo-900\n) !default;\n\n$purples: (\n  \"purple-100\": $purple-100,\n  \"purple-200\": $purple-200,\n  \"purple-300\": $purple-300,\n  \"purple-400\": $purple-400,\n  \"purple-500\": $purple-500,\n  \"purple-600\": $purple-600,\n  \"purple-700\": $purple-700,\n  \"purple-800\": $purple-800,\n  \"purple-900\": $purple-900\n) !default;\n\n$pinks: (\n  \"pink-100\": $pink-100,\n  \"pink-200\": $pink-200,\n  \"pink-300\": $pink-300,\n  \"pink-400\": $pink-400,\n  \"pink-500\": $pink-500,\n  \"pink-600\": $pink-600,\n  \"pink-700\": $pink-700,\n  \"pink-800\": $pink-800,\n  \"pink-900\": $pink-900\n) !default;\n\n$reds: (\n  \"red-100\": $red-100,\n  \"red-200\": $red-200,\n  \"red-300\": $red-300,\n  \"red-400\": $red-400,\n  \"red-500\": $red-500,\n  \"red-600\": $red-600,\n  \"red-700\": $red-700,\n  \"red-800\": $red-800,\n  \"red-900\": $red-900\n) !default;\n\n$oranges: (\n  \"orange-100\": $orange-100,\n  \"orange-200\": $orange-200,\n  \"orange-300\": $orange-300,\n  \"orange-400\": $orange-400,\n  \"orange-500\": $orange-500,\n  \"orange-600\": $orange-600,\n  \"orange-700\": $orange-700,\n  \"orange-800\": $orange-800,\n  \"orange-900\": $orange-900\n) !default;\n\n$yellows: (\n  \"yellow-100\": $yellow-100,\n  \"yellow-200\": $yellow-200,\n  \"yellow-300\": $yellow-300,\n  \"yellow-400\": $yellow-400,\n  \"yellow-500\": $yellow-500,\n  \"yellow-600\": $yellow-600,\n  \"yellow-700\": $yellow-700,\n  \"yellow-800\": $yellow-800,\n  \"yellow-900\": $yellow-900\n) !default;\n\n$greens: (\n  \"green-100\": $green-100,\n  \"green-200\": $green-200,\n  \"green-300\": $green-300,\n  \"green-400\": $green-400,\n  \"green-500\": $green-500,\n  \"green-600\": $green-600,\n  \"green-700\": $green-700,\n  \"green-800\": $green-800,\n  \"green-900\": $green-900\n) !default;\n\n$teals: (\n  \"teal-100\": $teal-100,\n  \"teal-200\": $teal-200,\n  \"teal-300\": $teal-300,\n  \"teal-400\": $teal-400,\n  \"teal-500\": $teal-500,\n  \"teal-600\": $teal-600,\n  \"teal-700\": $teal-700,\n  \"teal-800\": $teal-800,\n  \"teal-900\": $teal-900\n) !default;\n\n$cyans: (\n  \"cyan-100\": $cyan-100,\n  \"cyan-200\": $cyan-200,\n  \"cyan-300\": $cyan-300,\n  \"cyan-400\": $cyan-400,\n  \"cyan-500\": $cyan-500,\n  \"cyan-600\": $cyan-600,\n  \"cyan-700\": $cyan-700,\n  \"cyan-800\": $cyan-800,\n  \"cyan-900\": $cyan-900\n) !default;\n// fusv-enable\n\n// scss-docs-start theme-color-variables\n$primary:       $blue !default;\n$secondary:     $gray-600 !default;\n$success:       $green !default;\n$info:          $cyan !default;\n$warning:       $yellow !default;\n$danger:        $red !default;\n$light:         $gray-100 !default;\n$dark:          $gray-900 !default;\n// scss-docs-end theme-color-variables\n\n// scss-docs-start theme-colors-map\n$theme-colors: (\n  \"primary\":    $primary,\n  \"secondary\":  $secondary,\n  \"success\":    $success,\n  \"info\":       $info,\n  \"warning\":    $warning,\n  \"danger\":     $danger,\n  \"light\":      $light,\n  \"dark\":       $dark\n) !default;\n// scss-docs-end theme-colors-map\n\n// scss-docs-start theme-text-variables\n$primary-text-emphasis:   shade-color($primary, 60%) !default;\n$secondary-text-emphasis: shade-color($secondary, 60%) !default;\n$success-text-emphasis:   shade-color($success, 60%) !default;\n$info-text-emphasis:      shade-color($info, 60%) !default;\n$warning-text-emphasis:   shade-color($warning, 60%) !default;\n$danger-text-emphasis:    shade-color($danger, 60%) !default;\n$light-text-emphasis:     $gray-700 !default;\n$dark-text-emphasis:      $gray-700 !default;\n// scss-docs-end theme-text-variables\n\n// scss-docs-start theme-bg-subtle-variables\n$primary-bg-subtle:       tint-color($primary, 80%) !default;\n$secondary-bg-subtle:     tint-color($secondary, 80%) !default;\n$success-bg-subtle:       tint-color($success, 80%) !default;\n$info-bg-subtle:          tint-color($info, 80%) !default;\n$warning-bg-subtle:       tint-color($warning, 80%) !default;\n$danger-bg-subtle:        tint-color($danger, 80%) !default;\n$light-bg-subtle:         mix($gray-100, $white) !default;\n$dark-bg-subtle:          $gray-400 !default;\n// scss-docs-end theme-bg-subtle-variables\n\n// scss-docs-start theme-border-subtle-variables\n$primary-border-subtle:   tint-color($primary, 60%) !default;\n$secondary-border-subtle: tint-color($secondary, 60%) !default;\n$success-border-subtle:   tint-color($success, 60%) !default;\n$info-border-subtle:      tint-color($info, 60%) !default;\n$warning-border-subtle:   tint-color($warning, 60%) !default;\n$danger-border-subtle:    tint-color($danger, 60%) !default;\n$light-border-subtle:     $gray-200 !default;\n$dark-border-subtle:      $gray-500 !default;\n// scss-docs-end theme-border-subtle-variables\n\n// Characters which are escaped by the escape-svg function\n$escaped-characters: (\n  (\"<\", \"%3c\"),\n  (\">\", \"%3e\"),\n  (\"#\", \"%23\"),\n  (\"(\", \"%28\"),\n  (\")\", \"%29\"),\n) !default;\n\n// Options\n//\n// Quickly modify global styling by enabling or disabling optional features.\n\n$enable-caret:                true !default;\n$enable-rounded:              true !default;\n$enable-shadows:              false !default;\n$enable-gradients:            false !default;\n$enable-transitions:          true !default;\n$enable-reduced-motion:       true !default;\n$enable-smooth-scroll:        true !default;\n$enable-grid-classes:         true !default;\n$enable-container-classes:    true !default;\n$enable-cssgrid:              false !default;\n$enable-button-pointers:      true !default;\n$enable-rfs:                  true !default;\n$enable-validation-icons:     true !default;\n$enable-negative-margins:     false !default;\n$enable-deprecation-messages: true !default;\n$enable-important-utilities:  true !default;\n\n$enable-dark-mode:            true !default;\n$color-mode-type:             data !default; // `data` or `media-query`\n\n// Prefix for :root CSS variables\n\n$variable-prefix:             bs- !default; // Deprecated in v5.2.0 for the shorter `$prefix`\n$prefix:                      $variable-prefix !default;\n\n// Gradient\n//\n// The gradient which is added to components if `$enable-gradients` is `true`\n// This gradient is also added to elements with `.bg-gradient`\n// scss-docs-start variable-gradient\n$gradient: linear-gradient(180deg, rgba($white, .15), rgba($white, 0)) !default;\n// scss-docs-end variable-gradient\n\n// Spacing\n//\n// Control the default styling of most Bootstrap elements by modifying these\n// variables. Mostly focused on spacing.\n// You can add more entries to the $spacers map, should you need more variation.\n\n// scss-docs-start spacer-variables-maps\n$spacer: 1rem !default;\n$spacers: (\n  0: 0,\n  1: $spacer * .25,\n  2: $spacer * .5,\n  3: $spacer,\n  4: $spacer * 1.5,\n  5: $spacer * 3,\n) !default;\n// scss-docs-end spacer-variables-maps\n\n// Position\n//\n// Define the edge positioning anchors of the position utilities.\n\n// scss-docs-start position-map\n$position-values: (\n  0: 0,\n  50: 50%,\n  100: 100%\n) !default;\n// scss-docs-end position-map\n\n// Body\n//\n// Settings for the `<body>` element.\n\n$body-text-align:           null !default;\n$body-color:                $gray-900 !default;\n$body-bg:                   $white !default;\n\n$body-secondary-color:      rgba($body-color, .75) !default;\n$body-secondary-bg:         $gray-200 !default;\n\n$body-tertiary-color:       rgba($body-color, .5) !default;\n$body-tertiary-bg:          $gray-100 !default;\n\n$body-emphasis-color:       $black !default;\n\n// Links\n//\n// Style anchor elements.\n\n$link-color:                              $primary !default;\n$link-decoration:                         underline !default;\n$link-shade-percentage:                   20% !default;\n$link-hover-color:                        shift-color($link-color, $link-shade-percentage) !default;\n$link-hover-decoration:                   null !default;\n\n$stretched-link-pseudo-element:           after !default;\n$stretched-link-z-index:                  1 !default;\n\n// Icon links\n// scss-docs-start icon-link-variables\n$icon-link-gap:               .375rem !default;\n$icon-link-underline-offset:  .25em !default;\n$icon-link-icon-size:         1em !default;\n$icon-link-icon-transition:   .2s ease-in-out transform !default;\n$icon-link-icon-transform:    translate3d(.25em, 0, 0) !default;\n// scss-docs-end icon-link-variables\n\n// Paragraphs\n//\n// Style p element.\n\n$paragraph-margin-bottom:   1rem !default;\n\n\n// Grid breakpoints\n//\n// Define the minimum dimensions at which your layout will change,\n// adapting to different screen sizes, for use in media queries.\n\n// scss-docs-start grid-breakpoints\n$grid-breakpoints: (\n  xs: 0,\n  sm: 576px,\n  md: 768px,\n  lg: 992px,\n  xl: 1200px,\n  xxl: 1400px\n) !default;\n// scss-docs-end grid-breakpoints\n\n@include _assert-ascending($grid-breakpoints, \"$grid-breakpoints\");\n@include _assert-starts-at-zero($grid-breakpoints, \"$grid-breakpoints\");\n\n\n// Grid containers\n//\n// Define the maximum width of `.container` for different screen sizes.\n\n// scss-docs-start container-max-widths\n$container-max-widths: (\n  sm: 540px,\n  md: 720px,\n  lg: 960px,\n  xl: 1140px,\n  xxl: 1320px\n) !default;\n// scss-docs-end container-max-widths\n\n@include _assert-ascending($container-max-widths, \"$container-max-widths\");\n\n\n// Grid columns\n//\n// Set the number of columns and specify the width of the gutters.\n\n$grid-columns:                12 !default;\n$grid-gutter-width:           1.5rem !default;\n$grid-row-columns:            6 !default;\n\n// Container padding\n\n$container-padding-x: $grid-gutter-width !default;\n\n\n// Components\n//\n// Define common padding and border radius sizes and more.\n\n// scss-docs-start border-variables\n$border-width:                1px !default;\n$border-widths: (\n  1: 1px,\n  2: 2px,\n  3: 3px,\n  4: 4px,\n  5: 5px\n) !default;\n$border-style:                solid !default;\n$border-color:                $gray-300 !default;\n$border-color-translucent:    rgba($black, .175) !default;\n// scss-docs-end border-variables\n\n// scss-docs-start border-radius-variables\n$border-radius:               .375rem !default;\n$border-radius-sm:            .25rem !default;\n$border-radius-lg:            .5rem !default;\n$border-radius-xl:            1rem !default;\n$border-radius-xxl:           2rem !default;\n$border-radius-pill:          50rem !default;\n// scss-docs-end border-radius-variables\n// fusv-disable\n$border-radius-2xl:           $border-radius-xxl !default; // Deprecated in v5.3.0\n// fusv-enable\n\n// scss-docs-start box-shadow-variables\n$box-shadow:                  0 .5rem 1rem rgba($black, .15) !default;\n$box-shadow-sm:               0 .125rem .25rem rgba($black, .075) !default;\n$box-shadow-lg:               0 1rem 3rem rgba($black, .175) !default;\n$box-shadow-inset:            inset 0 1px 2px rgba($black, .075) !default;\n// scss-docs-end box-shadow-variables\n\n$component-active-color:      $white !default;\n$component-active-bg:         $primary !default;\n\n// scss-docs-start focus-ring-variables\n$focus-ring-width:      .25rem !default;\n$focus-ring-opacity:    .25 !default;\n$focus-ring-color:      rgba($primary, $focus-ring-opacity) !default;\n$focus-ring-blur:       0 !default;\n$focus-ring-box-shadow: 0 0 $focus-ring-blur $focus-ring-width $focus-ring-color !default;\n// scss-docs-end focus-ring-variables\n\n// scss-docs-start caret-variables\n$caret-width:                 .3em !default;\n$caret-vertical-align:        $caret-width * .85 !default;\n$caret-spacing:               $caret-width * .85 !default;\n// scss-docs-end caret-variables\n\n$transition-base:             all .2s ease-in-out !default;\n$transition-fade:             opacity .15s linear !default;\n// scss-docs-start collapse-transition\n$transition-collapse:         height .35s ease !default;\n$transition-collapse-width:   width .35s ease !default;\n// scss-docs-end collapse-transition\n\n// stylelint-disable function-disallowed-list\n// scss-docs-start aspect-ratios\n$aspect-ratios: (\n  \"1x1\": 100%,\n  \"4x3\": calc(3 / 4 * 100%),\n  \"16x9\": calc(9 / 16 * 100%),\n  \"21x9\": calc(9 / 21 * 100%)\n) !default;\n// scss-docs-end aspect-ratios\n// stylelint-enable function-disallowed-list\n\n// Typography\n//\n// Font, line-height, and color for body text, headings, and more.\n\n// scss-docs-start font-variables\n// stylelint-disable value-keyword-case\n$font-family-sans-serif:      system-ui, -apple-system, \"Segoe UI\", Roboto, \"Helvetica Neue\", \"Noto Sans\", \"Liberation Sans\", Arial, sans-serif, \"Apple Color Emoji\", \"Segoe UI Emoji\", \"Segoe UI Symbol\", \"Noto Color Emoji\" !default;\n$font-family-monospace:       SFMono-Regular, Menlo, Monaco, Consolas, \"Liberation Mono\", \"Courier New\", monospace !default;\n// stylelint-enable value-keyword-case\n$font-family-base:            var(--#{$prefix}font-sans-serif) !default;\n$font-family-code:            var(--#{$prefix}font-monospace) !default;\n\n// $font-size-root affects the value of `rem`, which is used for as well font sizes, paddings, and margins\n// $font-size-base affects the font size of the body text\n$font-size-root:              null !default;\n$font-size-base:              1rem !default; // Assumes the browser default, typically `16px`\n$font-size-sm:                $font-size-base * .875 !default;\n$font-size-lg:                $font-size-base * 1.25 !default;\n\n$font-weight-lighter:         lighter !default;\n$font-weight-light:           300 !default;\n$font-weight-normal:          400 !default;\n$font-weight-medium:          500 !default;\n$font-weight-semibold:        600 !default;\n$font-weight-bold:            700 !default;\n$font-weight-bolder:          bolder !default;\n\n$font-weight-base:            $font-weight-normal !default;\n\n$line-height-base:            1.5 !default;\n$line-height-sm:              1.25 !default;\n$line-height-lg:              2 !default;\n\n$h1-font-size:                $font-size-base * 2.5 !default;\n$h2-font-size:                $font-size-base * 2 !default;\n$h3-font-size:                $font-size-base * 1.75 !default;\n$h4-font-size:                $font-size-base * 1.5 !default;\n$h5-font-size:                $font-size-base * 1.25 !default;\n$h6-font-size:                $font-size-base !default;\n// scss-docs-end font-variables\n\n// scss-docs-start font-sizes\n$font-sizes: (\n  1: $h1-font-size,\n  2: $h2-font-size,\n  3: $h3-font-size,\n  4: $h4-font-size,\n  5: $h5-font-size,\n  6: $h6-font-size\n) !default;\n// scss-docs-end font-sizes\n\n// scss-docs-start headings-variables\n$headings-margin-bottom:      $spacer * .5 !default;\n$headings-font-family:        null !default;\n$headings-font-style:         null !default;\n$headings-font-weight:        500 !default;\n$headings-line-height:        1.2 !default;\n$headings-color:              inherit !default;\n// scss-docs-end headings-variables\n\n// scss-docs-start display-headings\n$display-font-sizes: (\n  1: 5rem,\n  2: 4.5rem,\n  3: 4rem,\n  4: 3.5rem,\n  5: 3rem,\n  6: 2.5rem\n) !default;\n\n$display-font-family: null !default;\n$display-font-style:  null !default;\n$display-font-weight: 300 !default;\n$display-line-height: $headings-line-height !default;\n// scss-docs-end display-headings\n\n// scss-docs-start type-variables\n$lead-font-size:              $font-size-base * 1.25 !default;\n$lead-font-weight:            300 !default;\n\n$small-font-size:             .875em !default;\n\n$sub-sup-font-size:           .75em !default;\n\n// fusv-disable\n$text-muted:                  var(--#{$prefix}secondary-color) !default; // Deprecated in 5.3.0\n// fusv-enable\n\n$initialism-font-size:        $small-font-size !default;\n\n$blockquote-margin-y:         $spacer !default;\n$blockquote-font-size:        $font-size-base * 1.25 !default;\n$blockquote-footer-color:     $gray-600 !default;\n$blockquote-footer-font-size: $small-font-size !default;\n\n$hr-margin-y:                 $spacer !default;\n$hr-color:                    inherit !default;\n\n// fusv-disable\n$hr-bg-color:                 null !default; // Deprecated in v5.2.0\n$hr-height:                   null !default; // Deprecated in v5.2.0\n// fusv-enable\n\n$hr-border-color:             null !default; // Allows for inherited colors\n$hr-border-width:             var(--#{$prefix}border-width) !default;\n$hr-opacity:                  .25 !default;\n\n// scss-docs-start vr-variables\n$vr-border-width:             var(--#{$prefix}border-width) !default;\n// scss-docs-end vr-variables\n\n$legend-margin-bottom:        .5rem !default;\n$legend-font-size:            1.5rem !default;\n$legend-font-weight:          null !default;\n\n$dt-font-weight:              $font-weight-bold !default;\n\n$list-inline-padding:         .5rem !default;\n\n$mark-padding:                .1875em !default;\n$mark-color:                  $body-color !default;\n$mark-bg:                     $yellow-100 !default;\n// scss-docs-end type-variables\n\n\n// Tables\n//\n// Customizes the `.table` component with basic values, each used across all table variations.\n\n// scss-docs-start table-variables\n$table-cell-padding-y:        .5rem !default;\n$table-cell-padding-x:        .5rem !default;\n$table-cell-padding-y-sm:     .25rem !default;\n$table-cell-padding-x-sm:     .25rem !default;\n\n$table-cell-vertical-align:   top !default;\n\n$table-color:                 var(--#{$prefix}emphasis-color) !default;\n$table-bg:                    var(--#{$prefix}body-bg) !default;\n$table-accent-bg:             transparent !default;\n\n$table-th-font-weight:        null !default;\n\n$table-striped-color:         $table-color !default;\n$table-striped-bg-factor:     .05 !default;\n$table-striped-bg:            rgba(var(--#{$prefix}emphasis-color-rgb), $table-striped-bg-factor) !default;\n\n$table-active-color:          $table-color !default;\n$table-active-bg-factor:      .1 !default;\n$table-active-bg:             rgba(var(--#{$prefix}emphasis-color-rgb), $table-active-bg-factor) !default;\n\n$table-hover-color:           $table-color !default;\n$table-hover-bg-factor:       .075 !default;\n$table-hover-bg:              rgba(var(--#{$prefix}emphasis-color-rgb), $table-hover-bg-factor) !default;\n\n$table-border-factor:         .2 !default;\n$table-border-width:          var(--#{$prefix}border-width) !default;\n$table-border-color:          var(--#{$prefix}border-color) !default;\n\n$table-striped-order:         odd !default;\n$table-striped-columns-order: even !default;\n\n$table-group-separator-color: currentcolor !default;\n\n$table-caption-color:         var(--#{$prefix}secondary-color) !default;\n\n$table-bg-scale:              -80% !default;\n// scss-docs-end table-variables\n\n// scss-docs-start table-loop\n$table-variants: (\n  \"primary\":    shift-color($primary, $table-bg-scale),\n  \"secondary\":  shift-color($secondary, $table-bg-scale),\n  \"success\":    shift-color($success, $table-bg-scale),\n  \"info\":       shift-color($info, $table-bg-scale),\n  \"warning\":    shift-color($warning, $table-bg-scale),\n  \"danger\":     shift-color($danger, $table-bg-scale),\n  \"light\":      $light,\n  \"dark\":       $dark,\n) !default;\n// scss-docs-end table-loop\n\n\n// Buttons + Forms\n//\n// Shared variables that are reassigned to `$input-` and `$btn-` specific variables.\n\n// scss-docs-start input-btn-variables\n$input-btn-padding-y:         .375rem !default;\n$input-btn-padding-x:         .75rem !default;\n$input-btn-font-family:       null !default;\n$input-btn-font-size:         $font-size-base !default;\n$input-btn-line-height:       $line-height-base !default;\n\n$input-btn-focus-width:         $focus-ring-width !default;\n$input-btn-focus-color-opacity: $focus-ring-opacity !default;\n$input-btn-focus-color:         $focus-ring-color !default;\n$input-btn-focus-blur:          $focus-ring-blur !default;\n$input-btn-focus-box-shadow:    $focus-ring-box-shadow !default;\n\n$input-btn-padding-y-sm:      .25rem !default;\n$input-btn-padding-x-sm:      .5rem !default;\n$input-btn-font-size-sm:      $font-size-sm !default;\n\n$input-btn-padding-y-lg:      .5rem !default;\n$input-btn-padding-x-lg:      1rem !default;\n$input-btn-font-size-lg:      $font-size-lg !default;\n\n$input-btn-border-width:      var(--#{$prefix}border-width) !default;\n// scss-docs-end input-btn-variables\n\n\n// Buttons\n//\n// For each of Bootstrap's buttons, define text, background, and border color.\n\n// scss-docs-start btn-variables\n$btn-color:                   var(--#{$prefix}body-color) !default;\n$btn-padding-y:               $input-btn-padding-y !default;\n$btn-padding-x:               $input-btn-padding-x !default;\n$btn-font-family:             $input-btn-font-family !default;\n$btn-font-size:               $input-btn-font-size !default;\n$btn-line-height:             $input-btn-line-height !default;\n$btn-white-space:             null !default; // Set to `nowrap` to prevent text wrapping\n\n$btn-padding-y-sm:            $input-btn-padding-y-sm !default;\n$btn-padding-x-sm:            $input-btn-padding-x-sm !default;\n$btn-font-size-sm:            $input-btn-font-size-sm !default;\n\n$btn-padding-y-lg:            $input-btn-padding-y-lg !default;\n$btn-padding-x-lg:            $input-btn-padding-x-lg !default;\n$btn-font-size-lg:            $input-btn-font-size-lg !default;\n\n$btn-border-width:            $input-btn-border-width !default;\n\n$btn-font-weight:             $font-weight-normal !default;\n$btn-box-shadow:              inset 0 1px 0 rgba($white, .15), 0 1px 1px rgba($black, .075) !default;\n$btn-focus-width:             $input-btn-focus-width !default;\n$btn-focus-box-shadow:        $input-btn-focus-box-shadow !default;\n$btn-disabled-opacity:        .65 !default;\n$btn-active-box-shadow:       inset 0 3px 5px rgba($black, .125) !default;\n\n$btn-link-color:              var(--#{$prefix}link-color) !default;\n$btn-link-hover-color:        var(--#{$prefix}link-hover-color) !default;\n$btn-link-disabled-color:     $gray-600 !default;\n$btn-link-focus-shadow-rgb:   to-rgb(mix(color-contrast($link-color), $link-color, 15%)) !default;\n\n// Allows for customizing button radius independently from global border radius\n$btn-border-radius:           var(--#{$prefix}border-radius) !default;\n$btn-border-radius-sm:        var(--#{$prefix}border-radius-sm) !default;\n$btn-border-radius-lg:        var(--#{$prefix}border-radius-lg) !default;\n\n$btn-transition:              color .15s ease-in-out, background-color .15s ease-in-out, border-color .15s ease-in-out, box-shadow .15s ease-in-out !default;\n\n$btn-hover-bg-shade-amount:       15% !default;\n$btn-hover-bg-tint-amount:        15% !default;\n$btn-hover-border-shade-amount:   20% !default;\n$btn-hover-border-tint-amount:    10% !default;\n$btn-active-bg-shade-amount:      20% !default;\n$btn-active-bg-tint-amount:       20% !default;\n$btn-active-border-shade-amount:  25% !default;\n$btn-active-border-tint-amount:   10% !default;\n// scss-docs-end btn-variables\n\n\n// Forms\n\n// scss-docs-start form-text-variables\n$form-text-margin-top:                  .25rem !default;\n$form-text-font-size:                   $small-font-size !default;\n$form-text-font-style:                  null !default;\n$form-text-font-weight:                 null !default;\n$form-text-color:                       var(--#{$prefix}secondary-color) !default;\n// scss-docs-end form-text-variables\n\n// scss-docs-start form-label-variables\n$form-label-margin-bottom:              .5rem !default;\n$form-label-font-size:                  null !default;\n$form-label-font-style:                 null !default;\n$form-label-font-weight:                null !default;\n$form-label-color:                      null !default;\n// scss-docs-end form-label-variables\n\n// scss-docs-start form-input-variables\n$input-padding-y:                       $input-btn-padding-y !default;\n$input-padding-x:                       $input-btn-padding-x !default;\n$input-font-family:                     $input-btn-font-family !default;\n$input-font-size:                       $input-btn-font-size !default;\n$input-font-weight:                     $font-weight-base !default;\n$input-line-height:                     $input-btn-line-height !default;\n\n$input-padding-y-sm:                    $input-btn-padding-y-sm !default;\n$input-padding-x-sm:                    $input-btn-padding-x-sm !default;\n$input-font-size-sm:                    $input-btn-font-size-sm !default;\n\n$input-padding-y-lg:                    $input-btn-padding-y-lg !default;\n$input-padding-x-lg:                    $input-btn-padding-x-lg !default;\n$input-font-size-lg:                    $input-btn-font-size-lg !default;\n\n$input-bg:                              var(--#{$prefix}body-bg) !default;\n$input-disabled-color:                  null !default;\n$input-disabled-bg:                     var(--#{$prefix}secondary-bg) !default;\n$input-disabled-border-color:           null !default;\n\n$input-color:                           var(--#{$prefix}body-color) !default;\n$input-border-color:                    var(--#{$prefix}border-color) !default;\n$input-border-width:                    $input-btn-border-width !default;\n$input-box-shadow:                      var(--#{$prefix}box-shadow-inset) !default;\n\n$input-border-radius:                   var(--#{$prefix}border-radius) !default;\n$input-border-radius-sm:                var(--#{$prefix}border-radius-sm) !default;\n$input-border-radius-lg:                var(--#{$prefix}border-radius-lg) !default;\n\n$input-focus-bg:                        $input-bg !default;\n$input-focus-border-color:              tint-color($component-active-bg, 50%) !default;\n$input-focus-color:                     $input-color !default;\n$input-focus-width:                     $input-btn-focus-width !default;\n$input-focus-box-shadow:                $input-btn-focus-box-shadow !default;\n\n$input-placeholder-color:               var(--#{$prefix}secondary-color) !default;\n$input-plaintext-color:                 var(--#{$prefix}body-color) !default;\n\n$input-height-border:                   calc(#{$input-border-width} * 2) !default; // stylelint-disable-line function-disallowed-list\n\n$input-height-inner:                    add($input-line-height * 1em, $input-padding-y * 2) !default;\n$input-height-inner-half:               add($input-line-height * .5em, $input-padding-y) !default;\n$input-height-inner-quarter:            add($input-line-height * .25em, $input-padding-y * .5) !default;\n\n$input-height:                          add($input-line-height * 1em, add($input-padding-y * 2, $input-height-border, false)) !default;\n$input-height-sm:                       add($input-line-height * 1em, add($input-padding-y-sm * 2, $input-height-border, false)) !default;\n$input-height-lg:                       add($input-line-height * 1em, add($input-padding-y-lg * 2, $input-height-border, false)) !default;\n\n$input-transition:                      border-color .15s ease-in-out, box-shadow .15s ease-in-out !default;\n\n$form-color-width:                      3rem !default;\n// scss-docs-end form-input-variables\n\n// scss-docs-start form-check-variables\n$form-check-input-width:                  1em !default;\n$form-check-min-height:                   $font-size-base * $line-height-base !default;\n$form-check-padding-start:                $form-check-input-width + .5em !default;\n$form-check-margin-bottom:                .125rem !default;\n$form-check-label-color:                  null !default;\n$form-check-label-cursor:                 null !default;\n$form-check-transition:                   null !default;\n\n$form-check-input-active-filter:          brightness(90%) !default;\n\n$form-check-input-bg:                     $input-bg !default;\n$form-check-input-border:                 var(--#{$prefix}border-width) solid var(--#{$prefix}border-color) !default;\n$form-check-input-border-radius:          .25em !default;\n$form-check-radio-border-radius:          50% !default;\n$form-check-input-focus-border:           $input-focus-border-color !default;\n$form-check-input-focus-box-shadow:       $focus-ring-box-shadow !default;\n\n$form-check-input-checked-color:          $component-active-color !default;\n$form-check-input-checked-bg-color:       $component-active-bg !default;\n$form-check-input-checked-border-color:   $form-check-input-checked-bg-color !default;\n$form-check-input-checked-bg-image:       url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 20 20'><path fill='none' stroke='#{$form-check-input-checked-color}' stroke-linecap='round' stroke-linejoin='round' stroke-width='3' d='m6 10 3 3 6-6'/></svg>\") !default;\n$form-check-radio-checked-bg-image:       url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='-4 -4 8 8'><circle r='2' fill='#{$form-check-input-checked-color}'/></svg>\") !default;\n\n$form-check-input-indeterminate-color:          $component-active-color !default;\n$form-check-input-indeterminate-bg-color:       $component-active-bg !default;\n$form-check-input-indeterminate-border-color:   $form-check-input-indeterminate-bg-color !default;\n$form-check-input-indeterminate-bg-image:       url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 20 20'><path fill='none' stroke='#{$form-check-input-indeterminate-color}' stroke-linecap='round' stroke-linejoin='round' stroke-width='3' d='M6 10h8'/></svg>\") !default;\n\n$form-check-input-disabled-opacity:        .5 !default;\n$form-check-label-disabled-opacity:        $form-check-input-disabled-opacity !default;\n$form-check-btn-check-disabled-opacity:    $btn-disabled-opacity !default;\n\n$form-check-inline-margin-end:    1rem !default;\n// scss-docs-end form-check-variables\n\n// scss-docs-start form-switch-variables\n$form-switch-color:               rgba($black, .25) !default;\n$form-switch-width:               2em !default;\n$form-switch-padding-start:       $form-switch-width + .5em !default;\n$form-switch-bg-image:            url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='-4 -4 8 8'><circle r='3' fill='#{$form-switch-color}'/></svg>\") !default;\n$form-switch-border-radius:       $form-switch-width !default;\n$form-switch-transition:          background-position .15s ease-in-out !default;\n\n$form-switch-focus-color:         $input-focus-border-color !default;\n$form-switch-focus-bg-image:      url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='-4 -4 8 8'><circle r='3' fill='#{$form-switch-focus-color}'/></svg>\") !default;\n\n$form-switch-checked-color:       $component-active-color !default;\n$form-switch-checked-bg-image:    url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='-4 -4 8 8'><circle r='3' fill='#{$form-switch-checked-color}'/></svg>\") !default;\n$form-switch-checked-bg-position: right center !default;\n// scss-docs-end form-switch-variables\n\n// scss-docs-start input-group-variables\n$input-group-addon-padding-y:           $input-padding-y !default;\n$input-group-addon-padding-x:           $input-padding-x !default;\n$input-group-addon-font-weight:         $input-font-weight !default;\n$input-group-addon-color:               $input-color !default;\n$input-group-addon-bg:                  var(--#{$prefix}tertiary-bg) !default;\n$input-group-addon-border-color:        $input-border-color !default;\n// scss-docs-end input-group-variables\n\n// scss-docs-start form-select-variables\n$form-select-padding-y:             $input-padding-y !default;\n$form-select-padding-x:             $input-padding-x !default;\n$form-select-font-family:           $input-font-family !default;\n$form-select-font-size:             $input-font-size !default;\n$form-select-indicator-padding:     $form-select-padding-x * 3 !default; // Extra padding for background-image\n$form-select-font-weight:           $input-font-weight !default;\n$form-select-line-height:           $input-line-height !default;\n$form-select-color:                 $input-color !default;\n$form-select-bg:                    $input-bg !default;\n$form-select-disabled-color:        null !default;\n$form-select-disabled-bg:           $input-disabled-bg !default;\n$form-select-disabled-border-color: $input-disabled-border-color !default;\n$form-select-bg-position:           right $form-select-padding-x center !default;\n$form-select-bg-size:               16px 12px !default; // In pixels because image dimensions\n$form-select-indicator-color:       $gray-800 !default;\n$form-select-indicator:             url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16'><path fill='none' stroke='#{$form-select-indicator-color}' stroke-linecap='round' stroke-linejoin='round' stroke-width='2' d='m2 5 6 6 6-6'/></svg>\") !default;\n\n$form-select-feedback-icon-padding-end: $form-select-padding-x * 2.5 + $form-select-indicator-padding !default;\n$form-select-feedback-icon-position:    center right $form-select-indicator-padding !default;\n$form-select-feedback-icon-size:        $input-height-inner-half $input-height-inner-half !default;\n\n$form-select-border-width:        $input-border-width !default;\n$form-select-border-color:        $input-border-color !default;\n$form-select-border-radius:       $input-border-radius !default;\n$form-select-box-shadow:          var(--#{$prefix}box-shadow-inset) !default;\n\n$form-select-focus-border-color:  $input-focus-border-color !default;\n$form-select-focus-width:         $input-focus-width !default;\n$form-select-focus-box-shadow:    0 0 0 $form-select-focus-width $input-btn-focus-color !default;\n\n$form-select-padding-y-sm:        $input-padding-y-sm !default;\n$form-select-padding-x-sm:        $input-padding-x-sm !default;\n$form-select-font-size-sm:        $input-font-size-sm !default;\n$form-select-border-radius-sm:    $input-border-radius-sm !default;\n\n$form-select-padding-y-lg:        $input-padding-y-lg !default;\n$form-select-padding-x-lg:        $input-padding-x-lg !default;\n$form-select-font-size-lg:        $input-font-size-lg !default;\n$form-select-border-radius-lg:    $input-border-radius-lg !default;\n\n$form-select-transition:          $input-transition !default;\n// scss-docs-end form-select-variables\n\n// scss-docs-start form-range-variables\n$form-range-track-width:          100% !default;\n$form-range-track-height:         .5rem !default;\n$form-range-track-cursor:         pointer !default;\n$form-range-track-bg:             var(--#{$prefix}secondary-bg) !default;\n$form-range-track-border-radius:  1rem !default;\n$form-range-track-box-shadow:     var(--#{$prefix}box-shadow-inset) !default;\n\n$form-range-thumb-width:                   1rem !default;\n$form-range-thumb-height:                  $form-range-thumb-width !default;\n$form-range-thumb-bg:                      $component-active-bg !default;\n$form-range-thumb-border:                  0 !default;\n$form-range-thumb-border-radius:           1rem !default;\n$form-range-thumb-box-shadow:              0 .1rem .25rem rgba($black, .1) !default;\n$form-range-thumb-focus-box-shadow:        0 0 0 1px $body-bg, $input-focus-box-shadow !default;\n$form-range-thumb-focus-box-shadow-width:  $input-focus-width !default; // For focus box shadow issue in Edge\n$form-range-thumb-active-bg:               tint-color($component-active-bg, 70%) !default;\n$form-range-thumb-disabled-bg:             var(--#{$prefix}secondary-color) !default;\n$form-range-thumb-transition:              background-color .15s ease-in-out, border-color .15s ease-in-out, box-shadow .15s ease-in-out !default;\n// scss-docs-end form-range-variables\n\n// scss-docs-start form-file-variables\n$form-file-button-color:          $input-color !default;\n$form-file-button-bg:             var(--#{$prefix}tertiary-bg) !default;\n$form-file-button-hover-bg:       var(--#{$prefix}secondary-bg) !default;\n// scss-docs-end form-file-variables\n\n// scss-docs-start form-floating-variables\n$form-floating-height:                  add(3.5rem, $input-height-border) !default;\n$form-floating-line-height:             1.25 !default;\n$form-floating-padding-x:               $input-padding-x !default;\n$form-floating-padding-y:               1rem !default;\n$form-floating-input-padding-t:         1.625rem !default;\n$form-floating-input-padding-b:         .625rem !default;\n$form-floating-label-height:            1.5em !default;\n$form-floating-label-opacity:           .65 !default;\n$form-floating-label-transform:         scale(.85) translateY(-.5rem) translateX(.15rem) !default;\n$form-floating-label-disabled-color:    $gray-600 !default;\n$form-floating-transition:              opacity .1s ease-in-out, transform .1s ease-in-out !default;\n// scss-docs-end form-floating-variables\n\n// Form validation\n\n// scss-docs-start form-feedback-variables\n$form-feedback-margin-top:          $form-text-margin-top !default;\n$form-feedback-font-size:           $form-text-font-size !default;\n$form-feedback-font-style:          $form-text-font-style !default;\n$form-feedback-valid-color:         $success !default;\n$form-feedback-invalid-color:       $danger !default;\n\n$form-feedback-icon-valid-color:    $form-feedback-valid-color !default;\n$form-feedback-icon-valid:          url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 8 8'><path fill='#{$form-feedback-icon-valid-color}' d='M2.3 6.73.6 4.53c-.4-1.04.46-1.4 1.1-.8l1.1 1.4 3.4-3.8c.6-.63 1.6-.27 1.2.7l-4 4.6c-.43.5-.8.4-1.1.1z'/></svg>\") !default;\n$form-feedback-icon-invalid-color:  $form-feedback-invalid-color !default;\n$form-feedback-icon-invalid:        url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 12 12' width='12' height='12' fill='none' stroke='#{$form-feedback-icon-invalid-color}'><circle cx='6' cy='6' r='4.5'/><path stroke-linejoin='round' d='M5.8 3.6h.4L6 6.5z'/><circle cx='6' cy='8.2' r='.6' fill='#{$form-feedback-icon-invalid-color}' stroke='none'/></svg>\") !default;\n// scss-docs-end form-feedback-variables\n\n// scss-docs-start form-validation-colors\n$form-valid-color:                  $form-feedback-valid-color !default;\n$form-valid-border-color:           $form-feedback-valid-color !default;\n$form-invalid-color:                $form-feedback-invalid-color !default;\n$form-invalid-border-color:         $form-feedback-invalid-color !default;\n// scss-docs-end form-validation-colors\n\n// scss-docs-start form-validation-states\n$form-validation-states: (\n  \"valid\": (\n    \"color\": var(--#{$prefix}form-valid-color),\n    \"icon\": $form-feedback-icon-valid,\n    \"tooltip-color\": #fff,\n    \"tooltip-bg-color\": var(--#{$prefix}success),\n    \"focus-box-shadow\": 0 0 $input-btn-focus-blur $input-focus-width rgba(var(--#{$prefix}success-rgb), $input-btn-focus-color-opacity),\n    \"border-color\": var(--#{$prefix}form-valid-border-color),\n  ),\n  \"invalid\": (\n    \"color\": var(--#{$prefix}form-invalid-color),\n    \"icon\": $form-feedback-icon-invalid,\n    \"tooltip-color\": #fff,\n    \"tooltip-bg-color\": var(--#{$prefix}danger),\n    \"focus-box-shadow\": 0 0 $input-btn-focus-blur $input-focus-width rgba(var(--#{$prefix}danger-rgb), $input-btn-focus-color-opacity),\n    \"border-color\": var(--#{$prefix}form-invalid-border-color),\n  )\n) !default;\n// scss-docs-end form-validation-states\n\n// Z-index master list\n//\n// Warning: Avoid customizing these values. They're used for a bird's eye view\n// of components dependent on the z-axis and are designed to all work together.\n\n// scss-docs-start zindex-stack\n$zindex-dropdown:                   1000 !default;\n$zindex-sticky:                     1020 !default;\n$zindex-fixed:                      1030 !default;\n$zindex-offcanvas-backdrop:         1040 !default;\n$zindex-offcanvas:                  1045 !default;\n$zindex-modal-backdrop:             1050 !default;\n$zindex-modal:                      1055 !default;\n$zindex-popover:                    1070 !default;\n$zindex-tooltip:                    1080 !default;\n$zindex-toast:                      1090 !default;\n// scss-docs-end zindex-stack\n\n// scss-docs-start zindex-levels-map\n$zindex-levels: (\n  n1: -1,\n  0: 0,\n  1: 1,\n  2: 2,\n  3: 3\n) !default;\n// scss-docs-end zindex-levels-map\n\n\n// Navs\n\n// scss-docs-start nav-variables\n$nav-link-padding-y:                .5rem !default;\n$nav-link-padding-x:                1rem !default;\n$nav-link-font-size:                null !default;\n$nav-link-font-weight:              null !default;\n$nav-link-color:                    var(--#{$prefix}link-color) !default;\n$nav-link-hover-color:              var(--#{$prefix}link-hover-color) !default;\n$nav-link-transition:               color .15s ease-in-out, background-color .15s ease-in-out, border-color .15s ease-in-out !default;\n$nav-link-disabled-color:           var(--#{$prefix}secondary-color) !default;\n$nav-link-focus-box-shadow:         $focus-ring-box-shadow !default;\n\n$nav-tabs-border-color:             var(--#{$prefix}border-color) !default;\n$nav-tabs-border-width:             var(--#{$prefix}border-width) !default;\n$nav-tabs-border-radius:            var(--#{$prefix}border-radius) !default;\n$nav-tabs-link-hover-border-color:  var(--#{$prefix}secondary-bg) var(--#{$prefix}secondary-bg) $nav-tabs-border-color !default;\n$nav-tabs-link-active-color:        var(--#{$prefix}emphasis-color) !default;\n$nav-tabs-link-active-bg:           var(--#{$prefix}body-bg) !default;\n$nav-tabs-link-active-border-color: var(--#{$prefix}border-color) var(--#{$prefix}border-color) $nav-tabs-link-active-bg !default;\n\n$nav-pills-border-radius:           var(--#{$prefix}border-radius) !default;\n$nav-pills-link-active-color:       $component-active-color !default;\n$nav-pills-link-active-bg:          $component-active-bg !default;\n\n$nav-underline-gap:                 1rem !default;\n$nav-underline-border-width:        .125rem !default;\n$nav-underline-link-active-color:   var(--#{$prefix}emphasis-color) !default;\n// scss-docs-end nav-variables\n\n\n// Navbar\n\n// scss-docs-start navbar-variables\n$navbar-padding-y:                  $spacer * .5 !default;\n$navbar-padding-x:                  null !default;\n\n$navbar-nav-link-padding-x:         .5rem !default;\n\n$navbar-brand-font-size:            $font-size-lg !default;\n// Compute the navbar-brand padding-y so the navbar-brand will have the same height as navbar-text and nav-link\n$nav-link-height:                   $font-size-base * $line-height-base + $nav-link-padding-y * 2 !default;\n$navbar-brand-height:               $navbar-brand-font-size * $line-height-base !default;\n$navbar-brand-padding-y:            ($nav-link-height - $navbar-brand-height) * .5 !default;\n$navbar-brand-margin-end:           1rem !default;\n\n$navbar-toggler-padding-y:          .25rem !default;\n$navbar-toggler-padding-x:          .75rem !default;\n$navbar-toggler-font-size:          $font-size-lg !default;\n$navbar-toggler-border-radius:      $btn-border-radius !default;\n$navbar-toggler-focus-width:        $btn-focus-width !default;\n$navbar-toggler-transition:         box-shadow .15s ease-in-out !default;\n\n$navbar-light-color:                rgba(var(--#{$prefix}emphasis-color-rgb), .65) !default;\n$navbar-light-hover-color:          rgba(var(--#{$prefix}emphasis-color-rgb), .8) !default;\n$navbar-light-active-color:         rgba(var(--#{$prefix}emphasis-color-rgb), 1) !default;\n$navbar-light-disabled-color:       rgba(var(--#{$prefix}emphasis-color-rgb), .3) !default;\n$navbar-light-icon-color:           rgba($body-color, .75) !default;\n$navbar-light-toggler-icon-bg:      url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 30 30'><path stroke='#{$navbar-light-icon-color}' stroke-linecap='round' stroke-miterlimit='10' stroke-width='2' d='M4 7h22M4 15h22M4 23h22'/></svg>\") !default;\n$navbar-light-toggler-border-color: rgba(var(--#{$prefix}emphasis-color-rgb), .15) !default;\n$navbar-light-brand-color:          $navbar-light-active-color !default;\n$navbar-light-brand-hover-color:    $navbar-light-active-color !default;\n// scss-docs-end navbar-variables\n\n// scss-docs-start navbar-dark-variables\n$navbar-dark-color:                 rgba($white, .55) !default;\n$navbar-dark-hover-color:           rgba($white, .75) !default;\n$navbar-dark-active-color:          $white !default;\n$navbar-dark-disabled-color:        rgba($white, .25) !default;\n$navbar-dark-icon-color:            $navbar-dark-color !default;\n$navbar-dark-toggler-icon-bg:       url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 30 30'><path stroke='#{$navbar-dark-icon-color}' stroke-linecap='round' stroke-miterlimit='10' stroke-width='2' d='M4 7h22M4 15h22M4 23h22'/></svg>\") !default;\n$navbar-dark-toggler-border-color:  rgba($white, .1) !default;\n$navbar-dark-brand-color:           $navbar-dark-active-color !default;\n$navbar-dark-brand-hover-color:     $navbar-dark-active-color !default;\n// scss-docs-end navbar-dark-variables\n\n\n// Dropdowns\n//\n// Dropdown menu container and contents.\n\n// scss-docs-start dropdown-variables\n$dropdown-min-width:                10rem !default;\n$dropdown-padding-x:                0 !default;\n$dropdown-padding-y:                .5rem !default;\n$dropdown-spacer:                   .125rem !default;\n$dropdown-font-size:                $font-size-base !default;\n$dropdown-color:                    var(--#{$prefix}body-color) !default;\n$dropdown-bg:                       var(--#{$prefix}body-bg) !default;\n$dropdown-border-color:             var(--#{$prefix}border-color-translucent) !default;\n$dropdown-border-radius:            var(--#{$prefix}border-radius) !default;\n$dropdown-border-width:             var(--#{$prefix}border-width) !default;\n$dropdown-inner-border-radius:      calc(#{$dropdown-border-radius} - #{$dropdown-border-width}) !default; // stylelint-disable-line function-disallowed-list\n$dropdown-divider-bg:               $dropdown-border-color !default;\n$dropdown-divider-margin-y:         $spacer * .5 !default;\n$dropdown-box-shadow:               var(--#{$prefix}box-shadow) !default;\n\n$dropdown-link-color:               var(--#{$prefix}body-color) !default;\n$dropdown-link-hover-color:         $dropdown-link-color !default;\n$dropdown-link-hover-bg:            var(--#{$prefix}tertiary-bg) !default;\n\n$dropdown-link-active-color:        $component-active-color !default;\n$dropdown-link-active-bg:           $component-active-bg !default;\n\n$dropdown-link-disabled-color:      var(--#{$prefix}tertiary-color) !default;\n\n$dropdown-item-padding-y:           $spacer * .25 !default;\n$dropdown-item-padding-x:           $spacer !default;\n\n$dropdown-header-color:             $gray-600 !default;\n$dropdown-header-padding-x:         $dropdown-item-padding-x !default;\n$dropdown-header-padding-y:         $dropdown-padding-y !default;\n// fusv-disable\n$dropdown-header-padding:           $dropdown-header-padding-y $dropdown-header-padding-x !default; // Deprecated in v5.2.0\n// fusv-enable\n// scss-docs-end dropdown-variables\n\n// scss-docs-start dropdown-dark-variables\n$dropdown-dark-color:               $gray-300 !default;\n$dropdown-dark-bg:                  $gray-800 !default;\n$dropdown-dark-border-color:        $dropdown-border-color !default;\n$dropdown-dark-divider-bg:          $dropdown-divider-bg !default;\n$dropdown-dark-box-shadow:          null !default;\n$dropdown-dark-link-color:          $dropdown-dark-color !default;\n$dropdown-dark-link-hover-color:    $white !default;\n$dropdown-dark-link-hover-bg:       rgba($white, .15) !default;\n$dropdown-dark-link-active-color:   $dropdown-link-active-color !default;\n$dropdown-dark-link-active-bg:      $dropdown-link-active-bg !default;\n$dropdown-dark-link-disabled-color: $gray-500 !default;\n$dropdown-dark-header-color:        $gray-500 !default;\n// scss-docs-end dropdown-dark-variables\n\n\n// Pagination\n\n// scss-docs-start pagination-variables\n$pagination-padding-y:              .375rem !default;\n$pagination-padding-x:              .75rem !default;\n$pagination-padding-y-sm:           .25rem !default;\n$pagination-padding-x-sm:           .5rem !default;\n$pagination-padding-y-lg:           .75rem !default;\n$pagination-padding-x-lg:           1.5rem !default;\n\n$pagination-font-size:              $font-size-base !default;\n\n$pagination-color:                  var(--#{$prefix}link-color) !default;\n$pagination-bg:                     var(--#{$prefix}body-bg) !default;\n$pagination-border-radius:          var(--#{$prefix}border-radius) !default;\n$pagination-border-width:           var(--#{$prefix}border-width) !default;\n$pagination-margin-start:           calc(#{$pagination-border-width} * -1) !default; // stylelint-disable-line function-disallowed-list\n$pagination-border-color:           var(--#{$prefix}border-color) !default;\n\n$pagination-focus-color:            var(--#{$prefix}link-hover-color) !default;\n$pagination-focus-bg:               var(--#{$prefix}secondary-bg) !default;\n$pagination-focus-box-shadow:       $focus-ring-box-shadow !default;\n$pagination-focus-outline:          0 !default;\n\n$pagination-hover-color:            var(--#{$prefix}link-hover-color) !default;\n$pagination-hover-bg:               var(--#{$prefix}tertiary-bg) !default;\n$pagination-hover-border-color:     var(--#{$prefix}border-color) !default; // Todo in v6: remove this?\n\n$pagination-active-color:           $component-active-color !default;\n$pagination-active-bg:              $component-active-bg !default;\n$pagination-active-border-color:    $component-active-bg !default;\n\n$pagination-disabled-color:         var(--#{$prefix}secondary-color) !default;\n$pagination-disabled-bg:            var(--#{$prefix}secondary-bg) !default;\n$pagination-disabled-border-color:  var(--#{$prefix}border-color) !default;\n\n$pagination-transition:              color .15s ease-in-out, background-color .15s ease-in-out, border-color .15s ease-in-out, box-shadow .15s ease-in-out !default;\n\n$pagination-border-radius-sm:       var(--#{$prefix}border-radius-sm) !default;\n$pagination-border-radius-lg:       var(--#{$prefix}border-radius-lg) !default;\n// scss-docs-end pagination-variables\n\n\n// Placeholders\n\n// scss-docs-start placeholders\n$placeholder-opacity-max:           .5 !default;\n$placeholder-opacity-min:           .2 !default;\n// scss-docs-end placeholders\n\n// Cards\n\n// scss-docs-start card-variables\n$card-spacer-y:                     $spacer !default;\n$card-spacer-x:                     $spacer !default;\n$card-title-spacer-y:               $spacer * .5 !default;\n$card-title-color:                  null !default;\n$card-subtitle-color:               null !default;\n$card-border-width:                 var(--#{$prefix}border-width) !default;\n$card-border-color:                 var(--#{$prefix}border-color-translucent) !default;\n$card-border-radius:                var(--#{$prefix}border-radius) !default;\n$card-box-shadow:                   null !default;\n$card-inner-border-radius:          subtract($card-border-radius, $card-border-width) !default;\n$card-cap-padding-y:                $card-spacer-y * .5 !default;\n$card-cap-padding-x:                $card-spacer-x !default;\n$card-cap-bg:                       rgba(var(--#{$prefix}body-color-rgb), .03) !default;\n$card-cap-color:                    null !default;\n$card-height:                       null !default;\n$card-color:                        null !default;\n$card-bg:                           var(--#{$prefix}body-bg) !default;\n$card-img-overlay-padding:          $spacer !default;\n$card-group-margin:                 $grid-gutter-width * .5 !default;\n// scss-docs-end card-variables\n\n// Accordion\n\n// scss-docs-start accordion-variables\n$accordion-padding-y:                     1rem !default;\n$accordion-padding-x:                     1.25rem !default;\n$accordion-color:                         var(--#{$prefix}body-color) !default;\n$accordion-bg:                            var(--#{$prefix}body-bg) !default;\n$accordion-border-width:                  var(--#{$prefix}border-width) !default;\n$accordion-border-color:                  var(--#{$prefix}border-color) !default;\n$accordion-border-radius:                 var(--#{$prefix}border-radius) !default;\n$accordion-inner-border-radius:           subtract($accordion-border-radius, $accordion-border-width) !default;\n\n$accordion-body-padding-y:                $accordion-padding-y !default;\n$accordion-body-padding-x:                $accordion-padding-x !default;\n\n$accordion-button-padding-y:              $accordion-padding-y !default;\n$accordion-button-padding-x:              $accordion-padding-x !default;\n$accordion-button-color:                  var(--#{$prefix}body-color) !default;\n$accordion-button-bg:                     var(--#{$prefix}accordion-bg) !default;\n$accordion-transition:                    $btn-transition, border-radius .15s ease !default;\n$accordion-button-active-bg:              var(--#{$prefix}primary-bg-subtle) !default;\n$accordion-button-active-color:           var(--#{$prefix}primary-text-emphasis) !default;\n\n$accordion-button-focus-border-color:     $input-focus-border-color !default;\n$accordion-button-focus-box-shadow:       $btn-focus-box-shadow !default;\n\n$accordion-icon-width:                    1.25rem !default;\n$accordion-icon-color:                    $body-color !default;\n$accordion-icon-active-color:             $primary-text-emphasis !default;\n$accordion-icon-transition:               transform .2s ease-in-out !default;\n$accordion-icon-transform:                rotate(-180deg) !default;\n\n$accordion-button-icon:         url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='#{$accordion-icon-color}'><path fill-rule='evenodd' d='M1.646 4.646a.5.5 0 0 1 .708 0L8 10.293l5.646-5.647a.5.5 0 0 1 .708.708l-6 6a.5.5 0 0 1-.708 0l-6-6a.5.5 0 0 1 0-.708z'/></svg>\") !default;\n$accordion-button-active-icon:  url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='#{$accordion-icon-active-color}'><path fill-rule='evenodd' d='M1.646 4.646a.5.5 0 0 1 .708 0L8 10.293l5.646-5.647a.5.5 0 0 1 .708.708l-6 6a.5.5 0 0 1-.708 0l-6-6a.5.5 0 0 1 0-.708z'/></svg>\") !default;\n// scss-docs-end accordion-variables\n\n// Tooltips\n\n// scss-docs-start tooltip-variables\n$tooltip-font-size:                 $font-size-sm !default;\n$tooltip-max-width:                 200px !default;\n$tooltip-color:                     var(--#{$prefix}body-bg) !default;\n$tooltip-bg:                        var(--#{$prefix}emphasis-color) !default;\n$tooltip-border-radius:             var(--#{$prefix}border-radius) !default;\n$tooltip-opacity:                   .9 !default;\n$tooltip-padding-y:                 $spacer * .25 !default;\n$tooltip-padding-x:                 $spacer * .5 !default;\n$tooltip-margin:                    null !default; // TODO: remove this in v6\n\n$tooltip-arrow-width:               .8rem !default;\n$tooltip-arrow-height:              .4rem !default;\n// fusv-disable\n$tooltip-arrow-color:               null !default; // Deprecated in Bootstrap 5.2.0 for CSS variables\n// fusv-enable\n// scss-docs-end tooltip-variables\n\n// Form tooltips must come after regular tooltips\n// scss-docs-start tooltip-feedback-variables\n$form-feedback-tooltip-padding-y:     $tooltip-padding-y !default;\n$form-feedback-tooltip-padding-x:     $tooltip-padding-x !default;\n$form-feedback-tooltip-font-size:     $tooltip-font-size !default;\n$form-feedback-tooltip-line-height:   null !default;\n$form-feedback-tooltip-opacity:       $tooltip-opacity !default;\n$form-feedback-tooltip-border-radius: $tooltip-border-radius !default;\n// scss-docs-end tooltip-feedback-variables\n\n\n// Popovers\n\n// scss-docs-start popover-variables\n$popover-font-size:                 $font-size-sm !default;\n$popover-bg:                        var(--#{$prefix}body-bg) !default;\n$popover-max-width:                 276px !default;\n$popover-border-width:              var(--#{$prefix}border-width) !default;\n$popover-border-color:              var(--#{$prefix}border-color-translucent) !default;\n$popover-border-radius:             var(--#{$prefix}border-radius-lg) !default;\n$popover-inner-border-radius:       calc(#{$popover-border-radius} - #{$popover-border-width}) !default; // stylelint-disable-line function-disallowed-list\n$popover-box-shadow:                var(--#{$prefix}box-shadow) !default;\n\n$popover-header-font-size:          $font-size-base !default;\n$popover-header-bg:                 var(--#{$prefix}secondary-bg) !default;\n$popover-header-color:              $headings-color !default;\n$popover-header-padding-y:          .5rem !default;\n$popover-header-padding-x:          $spacer !default;\n\n$popover-body-color:                var(--#{$prefix}body-color) !default;\n$popover-body-padding-y:            $spacer !default;\n$popover-body-padding-x:            $spacer !default;\n\n$popover-arrow-width:               1rem !default;\n$popover-arrow-height:              .5rem !default;\n// scss-docs-end popover-variables\n\n// fusv-disable\n// Deprecated in Bootstrap 5.2.0 for CSS variables\n$popover-arrow-color:               $popover-bg !default;\n$popover-arrow-outer-color:         var(--#{$prefix}border-color-translucent) !default;\n// fusv-enable\n\n\n// Toasts\n\n// scss-docs-start toast-variables\n$toast-max-width:                   350px !default;\n$toast-padding-x:                   .75rem !default;\n$toast-padding-y:                   .5rem !default;\n$toast-font-size:                   .875rem !default;\n$toast-color:                       null !default;\n$toast-background-color:            rgba(var(--#{$prefix}body-bg-rgb), .85) !default;\n$toast-border-width:                var(--#{$prefix}border-width) !default;\n$toast-border-color:                var(--#{$prefix}border-color-translucent) !default;\n$toast-border-radius:               var(--#{$prefix}border-radius) !default;\n$toast-box-shadow:                  var(--#{$prefix}box-shadow) !default;\n$toast-spacing:                     $container-padding-x !default;\n\n$toast-header-color:                var(--#{$prefix}secondary-color) !default;\n$toast-header-background-color:     rgba(var(--#{$prefix}body-bg-rgb), .85) !default;\n$toast-header-border-color:         $toast-border-color !default;\n// scss-docs-end toast-variables\n\n\n// Badges\n\n// scss-docs-start badge-variables\n$badge-font-size:                   .75em !default;\n$badge-font-weight:                 $font-weight-bold !default;\n$badge-color:                       $white !default;\n$badge-padding-y:                   .35em !default;\n$badge-padding-x:                   .65em !default;\n$badge-border-radius:               var(--#{$prefix}border-radius) !default;\n// scss-docs-end badge-variables\n\n\n// Modals\n\n// scss-docs-start modal-variables\n$modal-inner-padding:               $spacer !default;\n\n$modal-footer-margin-between:       .5rem !default;\n\n$modal-dialog-margin:               .5rem !default;\n$modal-dialog-margin-y-sm-up:       1.75rem !default;\n\n$modal-title-line-height:           $line-height-base !default;\n\n$modal-content-color:               null !default;\n$modal-content-bg:                  var(--#{$prefix}body-bg) !default;\n$modal-content-border-color:        var(--#{$prefix}border-color-translucent) !default;\n$modal-content-border-width:        var(--#{$prefix}border-width) !default;\n$modal-content-border-radius:       var(--#{$prefix}border-radius-lg) !default;\n$modal-content-inner-border-radius: subtract($modal-content-border-radius, $modal-content-border-width) !default;\n$modal-content-box-shadow-xs:       var(--#{$prefix}box-shadow-sm) !default;\n$modal-content-box-shadow-sm-up:    var(--#{$prefix}box-shadow) !default;\n\n$modal-backdrop-bg:                 $black !default;\n$modal-backdrop-opacity:            .5 !default;\n\n$modal-header-border-color:         var(--#{$prefix}border-color) !default;\n$modal-header-border-width:         $modal-content-border-width !default;\n$modal-header-padding-y:            $modal-inner-padding !default;\n$modal-header-padding-x:            $modal-inner-padding !default;\n$modal-header-padding:              $modal-header-padding-y $modal-header-padding-x !default; // Keep this for backwards compatibility\n\n$modal-footer-bg:                   null !default;\n$modal-footer-border-color:         $modal-header-border-color !default;\n$modal-footer-border-width:         $modal-header-border-width !default;\n\n$modal-sm:                          300px !default;\n$modal-md:                          500px !default;\n$modal-lg:                          800px !default;\n$modal-xl:                          1140px !default;\n\n$modal-fade-transform:              translate(0, -50px) !default;\n$modal-show-transform:              none !default;\n$modal-transition:                  transform .3s ease-out !default;\n$modal-scale-transform:             scale(1.02) !default;\n// scss-docs-end modal-variables\n\n\n// Alerts\n//\n// Define alert colors, border radius, and padding.\n\n// scss-docs-start alert-variables\n$alert-padding-y:               $spacer !default;\n$alert-padding-x:               $spacer !default;\n$alert-margin-bottom:           1rem !default;\n$alert-border-radius:           var(--#{$prefix}border-radius) !default;\n$alert-link-font-weight:        $font-weight-bold !default;\n$alert-border-width:            var(--#{$prefix}border-width) !default;\n$alert-dismissible-padding-r:   $alert-padding-x * 3 !default; // 3x covers width of x plus default padding on either side\n// scss-docs-end alert-variables\n\n// fusv-disable\n$alert-bg-scale:                -80% !default; // Deprecated in v5.2.0, to be removed in v6\n$alert-border-scale:            -70% !default; // Deprecated in v5.2.0, to be removed in v6\n$alert-color-scale:             40% !default; // Deprecated in v5.2.0, to be removed in v6\n// fusv-enable\n\n// Progress bars\n\n// scss-docs-start progress-variables\n$progress-height:                   1rem !default;\n$progress-font-size:                $font-size-base * .75 !default;\n$progress-bg:                       var(--#{$prefix}secondary-bg) !default;\n$progress-border-radius:            var(--#{$prefix}border-radius) !default;\n$progress-box-shadow:               var(--#{$prefix}box-shadow-inset) !default;\n$progress-bar-color:                $white !default;\n$progress-bar-bg:                   $primary !default;\n$progress-bar-animation-timing:     1s linear infinite !default;\n$progress-bar-transition:           width .6s ease !default;\n// scss-docs-end progress-variables\n\n\n// List group\n\n// scss-docs-start list-group-variables\n$list-group-color:                  var(--#{$prefix}body-color) !default;\n$list-group-bg:                     var(--#{$prefix}body-bg) !default;\n$list-group-border-color:           var(--#{$prefix}border-color) !default;\n$list-group-border-width:           var(--#{$prefix}border-width) !default;\n$list-group-border-radius:          var(--#{$prefix}border-radius) !default;\n\n$list-group-item-padding-y:         $spacer * .5 !default;\n$list-group-item-padding-x:         $spacer !default;\n// fusv-disable\n$list-group-item-bg-scale:          -80% !default; // Deprecated in v5.3.0\n$list-group-item-color-scale:       40% !default; // Deprecated in v5.3.0\n// fusv-enable\n\n$list-group-hover-bg:               var(--#{$prefix}tertiary-bg) !default;\n$list-group-active-color:           $component-active-color !default;\n$list-group-active-bg:              $component-active-bg !default;\n$list-group-active-border-color:    $list-group-active-bg !default;\n\n$list-group-disabled-color:         var(--#{$prefix}secondary-color) !default;\n$list-group-disabled-bg:            $list-group-bg !default;\n\n$list-group-action-color:           var(--#{$prefix}secondary-color) !default;\n$list-group-action-hover-color:     var(--#{$prefix}emphasis-color) !default;\n\n$list-group-action-active-color:    var(--#{$prefix}body-color) !default;\n$list-group-action-active-bg:       var(--#{$prefix}secondary-bg) !default;\n// scss-docs-end list-group-variables\n\n\n// Image thumbnails\n\n// scss-docs-start thumbnail-variables\n$thumbnail-padding:                 .25rem !default;\n$thumbnail-bg:                      var(--#{$prefix}body-bg) !default;\n$thumbnail-border-width:            var(--#{$prefix}border-width) !default;\n$thumbnail-border-color:            var(--#{$prefix}border-color) !default;\n$thumbnail-border-radius:           var(--#{$prefix}border-radius) !default;\n$thumbnail-box-shadow:              var(--#{$prefix}box-shadow-sm) !default;\n// scss-docs-end thumbnail-variables\n\n\n// Figures\n\n// scss-docs-start figure-variables\n$figure-caption-font-size:          $small-font-size !default;\n$figure-caption-color:              var(--#{$prefix}secondary-color) !default;\n// scss-docs-end figure-variables\n\n\n// Breadcrumbs\n\n// scss-docs-start breadcrumb-variables\n$breadcrumb-font-size:              null !default;\n$breadcrumb-padding-y:              0 !default;\n$breadcrumb-padding-x:              0 !default;\n$breadcrumb-item-padding-x:         .5rem !default;\n$breadcrumb-margin-bottom:          1rem !default;\n$breadcrumb-bg:                     null !default;\n$breadcrumb-divider-color:          var(--#{$prefix}secondary-color) !default;\n$breadcrumb-active-color:           var(--#{$prefix}secondary-color) !default;\n$breadcrumb-divider:                quote(\"/\") !default;\n$breadcrumb-divider-flipped:        $breadcrumb-divider !default;\n$breadcrumb-border-radius:          null !default;\n// scss-docs-end breadcrumb-variables\n\n// Carousel\n\n// scss-docs-start carousel-variables\n$carousel-control-color:             $white !default;\n$carousel-control-width:             15% !default;\n$carousel-control-opacity:           .5 !default;\n$carousel-control-hover-opacity:     .9 !default;\n$carousel-control-transition:        opacity .15s ease !default;\n\n$carousel-indicator-width:           30px !default;\n$carousel-indicator-height:          3px !default;\n$carousel-indicator-hit-area-height: 10px !default;\n$carousel-indicator-spacer:          3px !default;\n$carousel-indicator-opacity:         .5 !default;\n$carousel-indicator-active-bg:       $white !default;\n$carousel-indicator-active-opacity:  1 !default;\n$carousel-indicator-transition:      opacity .6s ease !default;\n\n$carousel-caption-width:             70% !default;\n$carousel-caption-color:             $white !default;\n$carousel-caption-padding-y:         1.25rem !default;\n$carousel-caption-spacer:            1.25rem !default;\n\n$carousel-control-icon-width:        2rem !default;\n\n$carousel-control-prev-icon-bg:      url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='#{$carousel-control-color}'><path d='M11.354 1.646a.5.5 0 0 1 0 .708L5.707 8l5.647 5.646a.5.5 0 0 1-.708.708l-6-6a.5.5 0 0 1 0-.708l6-6a.5.5 0 0 1 .708 0z'/></svg>\") !default;\n$carousel-control-next-icon-bg:      url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='#{$carousel-control-color}'><path d='M4.646 1.646a.5.5 0 0 1 .708 0l6 6a.5.5 0 0 1 0 .708l-6 6a.5.5 0 0 1-.708-.708L10.293 8 4.646 2.354a.5.5 0 0 1 0-.708z'/></svg>\") !default;\n\n$carousel-transition-duration:       .6s !default;\n$carousel-transition:                transform $carousel-transition-duration ease-in-out !default; // Define transform transition first if using multiple transitions (e.g., `transform 2s ease, opacity .5s ease-out`)\n// scss-docs-end carousel-variables\n\n// scss-docs-start carousel-dark-variables\n$carousel-dark-indicator-active-bg:  $black !default;\n$carousel-dark-caption-color:        $black !default;\n$carousel-dark-control-icon-filter:  invert(1) grayscale(100) !default;\n// scss-docs-end carousel-dark-variables\n\n\n// Spinners\n\n// scss-docs-start spinner-variables\n$spinner-width:           2rem !default;\n$spinner-height:          $spinner-width !default;\n$spinner-vertical-align:  -.125em !default;\n$spinner-border-width:    .25em !default;\n$spinner-animation-speed: .75s !default;\n\n$spinner-width-sm:        1rem !default;\n$spinner-height-sm:       $spinner-width-sm !default;\n$spinner-border-width-sm: .2em !default;\n// scss-docs-end spinner-variables\n\n\n// Close\n\n// scss-docs-start close-variables\n$btn-close-width:            1em !default;\n$btn-close-height:           $btn-close-width !default;\n$btn-close-padding-x:        .25em !default;\n$btn-close-padding-y:        $btn-close-padding-x !default;\n$btn-close-color:            $black !default;\n$btn-close-bg:               url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='#{$btn-close-color}'><path d='M.293.293a1 1 0 0 1 1.414 0L8 6.586 14.293.293a1 1 0 1 1 1.414 1.414L9.414 8l6.293 6.293a1 1 0 0 1-1.414 1.414L8 9.414l-6.293 6.293a1 1 0 0 1-1.414-1.414L6.586 8 .293 1.707a1 1 0 0 1 0-1.414z'/></svg>\") !default;\n$btn-close-focus-shadow:     $focus-ring-box-shadow !default;\n$btn-close-opacity:          .5 !default;\n$btn-close-hover-opacity:    .75 !default;\n$btn-close-focus-opacity:    1 !default;\n$btn-close-disabled-opacity: .25 !default;\n$btn-close-white-filter:     invert(1) grayscale(100%) brightness(200%) !default;\n// scss-docs-end close-variables\n\n\n// Offcanvas\n\n// scss-docs-start offcanvas-variables\n$offcanvas-padding-y:               $modal-inner-padding !default;\n$offcanvas-padding-x:               $modal-inner-padding !default;\n$offcanvas-horizontal-width:        400px !default;\n$offcanvas-vertical-height:         30vh !default;\n$offcanvas-transition-duration:     .3s !default;\n$offcanvas-border-color:            $modal-content-border-color !default;\n$offcanvas-border-width:            $modal-content-border-width !default;\n$offcanvas-title-line-height:       $modal-title-line-height !default;\n$offcanvas-bg-color:                var(--#{$prefix}body-bg) !default;\n$offcanvas-color:                   var(--#{$prefix}body-color) !default;\n$offcanvas-box-shadow:              $modal-content-box-shadow-xs !default;\n$offcanvas-backdrop-bg:             $modal-backdrop-bg !default;\n$offcanvas-backdrop-opacity:        $modal-backdrop-opacity !default;\n// scss-docs-end offcanvas-variables\n\n// Code\n\n$code-font-size:                    $small-font-size !default;\n$code-color:                        $pink !default;\n\n$kbd-padding-y:                     .1875rem !default;\n$kbd-padding-x:                     .375rem !default;\n$kbd-font-size:                     $code-font-size !default;\n$kbd-color:                         var(--#{$prefix}body-bg) !default;\n$kbd-bg:                            var(--#{$prefix}body-color) !default;\n$nested-kbd-font-weight:            null !default; // Deprecated in v5.2.0, removing in v6\n\n$pre-color:                         null !default;\n","// stylelint-disable property-disallowed-list\n// Single side border-radius\n\n// Helper function to replace negative values with 0\n@function valid-radius($radius) {\n  $return: ();\n  @each $value in $radius {\n    @if type-of($value) == number {\n      $return: append($return, max($value, 0));\n    } @else {\n      $return: append($return, $value);\n    }\n  }\n  @return $return;\n}\n\n// scss-docs-start border-radius-mixins\n@mixin border-radius($radius: $border-radius, $fallback-border-radius: false) {\n  @if $enable-rounded {\n    border-radius: valid-radius($radius);\n  }\n  @else if $fallback-border-radius != false {\n    border-radius: $fallback-border-radius;\n  }\n}\n\n@mixin border-top-radius($radius: $border-radius) {\n  @if $enable-rounded {\n    border-top-left-radius: valid-radius($radius);\n    border-top-right-radius: valid-radius($radius);\n  }\n}\n\n@mixin border-end-radius($radius: $border-radius) {\n  @if $enable-rounded {\n    border-top-right-radius: valid-radius($radius);\n    border-bottom-right-radius: valid-radius($radius);\n  }\n}\n\n@mixin border-bottom-radius($radius: $border-radius) {\n  @if $enable-rounded {\n    border-bottom-right-radius: valid-radius($radius);\n    border-bottom-left-radius: valid-radius($radius);\n  }\n}\n\n@mixin border-start-radius($radius: $border-radius) {\n  @if $enable-rounded {\n    border-top-left-radius: valid-radius($radius);\n    border-bottom-left-radius: valid-radius($radius);\n  }\n}\n\n@mixin border-top-start-radius($radius: $border-radius) {\n  @if $enable-rounded {\n    border-top-left-radius: valid-radius($radius);\n  }\n}\n\n@mixin border-top-end-radius($radius: $border-radius) {\n  @if $enable-rounded {\n    border-top-right-radius: valid-radius($radius);\n  }\n}\n\n@mixin border-bottom-end-radius($radius: $border-radius) {\n  @if $enable-rounded {\n    border-bottom-right-radius: valid-radius($radius);\n  }\n}\n\n@mixin border-bottom-start-radius($radius: $border-radius) {\n  @if $enable-rounded {\n    border-bottom-left-radius: valid-radius($radius);\n  }\n}\n// scss-docs-end border-radius-mixins\n","//\n// Headings\n//\n.h1 {\n  @extend h1;\n}\n\n.h2 {\n  @extend h2;\n}\n\n.h3 {\n  @extend h3;\n}\n\n.h4 {\n  @extend h4;\n}\n\n.h5 {\n  @extend h5;\n}\n\n.h6 {\n  @extend h6;\n}\n\n\n.lead {\n  @include font-size($lead-font-size);\n  font-weight: $lead-font-weight;\n}\n\n// Type display classes\n@each $display, $font-size in $display-font-sizes {\n  .display-#{$display} {\n    @include font-size($font-size);\n    font-family: $display-font-family;\n    font-style: $display-font-style;\n    font-weight: $display-font-weight;\n    line-height: $display-line-height;\n  }\n}\n\n//\n// Emphasis\n//\n.small {\n  @extend small;\n}\n\n.mark {\n  @extend mark;\n}\n\n//\n// Lists\n//\n\n.list-unstyled {\n  @include list-unstyled();\n}\n\n// Inline turns list items into inline-block\n.list-inline {\n  @include list-unstyled();\n}\n.list-inline-item {\n  display: inline-block;\n\n  &:not(:last-child) {\n    margin-right: $list-inline-padding;\n  }\n}\n\n\n//\n// Misc\n//\n\n// Builds on `abbr`\n.initialism {\n  @include font-size($initialism-font-size);\n  text-transform: uppercase;\n}\n\n// Blockquotes\n.blockquote {\n  margin-bottom: $blockquote-margin-y;\n  @include font-size($blockquote-font-size);\n\n  > :last-child {\n    margin-bottom: 0;\n  }\n}\n\n.blockquote-footer {\n  margin-top: -$blockquote-margin-y;\n  margin-bottom: $blockquote-margin-y;\n  @include font-size($blockquote-footer-font-size);\n  color: $blockquote-footer-color;\n\n  &::before {\n    content: \"\\2014\\00A0\"; // em dash, nbsp\n  }\n}\n","// Lists\n\n// Unstyled keeps list items block level, just removes default browser padding and list-style\n@mixin list-unstyled {\n  padding-left: 0;\n  list-style: none;\n}\n","// Override bootstrap variables\n$spacer: 1rem;\n\n$container-max-widths: (\n  sm: 540px,\n  md: 720px,\n  lg: 960px,\n  xl: 1400px,\n);\n\n$grid-breakpoints: (\n  xs: 0,\n  sm: 540px,\n  md: 720px,\n  lg: 960px,\n  xl: 1200px,\n);\n\n$dropdown-link-hover-bg: var(--pst-color-surface);\n// --pst-color-surface can also be assigned to the dark variant because it is\n// scoped to different values depending on light/dark theme\n$dropdown-dark-link-hover-bg: var(--pst-color-surface);\n$dropdown-link-active-bg: var(--pst-color-surface);\n$dropdown-dark-link-active-bg: var(--pst-color-surface);\n","// Responsive images (ensure images don't scale beyond their parents)\n//\n// This is purposefully opt-in via an explicit class rather than being the default for all `<img>`s.\n// We previously tried the \"images are responsive by default\" approach in Bootstrap v2,\n// and abandoned it in Bootstrap v3 because it breaks lots of third-party widgets (including Google Maps)\n// which weren't expecting the images within themselves to be involuntarily resized.\n// See also https://github.com/twbs/bootstrap/issues/18178\n.img-fluid {\n  @include img-fluid();\n}\n\n\n// Image thumbnails\n.img-thumbnail {\n  padding: $thumbnail-padding;\n  background-color: $thumbnail-bg;\n  border: $thumbnail-border-width solid $thumbnail-border-color;\n  @include border-radius($thumbnail-border-radius);\n  @include box-shadow($thumbnail-box-shadow);\n\n  // Keep them at most 100% wide\n  @include img-fluid();\n}\n\n//\n// Figures\n//\n\n.figure {\n  // Ensures the caption's text aligns with the image.\n  display: inline-block;\n}\n\n.figure-img {\n  margin-bottom: $spacer * .5;\n  line-height: 1;\n}\n\n.figure-caption {\n  @include font-size($figure-caption-font-size);\n  color: $figure-caption-color;\n}\n","// Image Mixins\n// - Responsive image\n// - Retina image\n\n\n// Responsive image\n//\n// Keep images from scaling beyond the width of their parents.\n\n@mixin img-fluid {\n  // Part 1: Set a maximum relative to the parent\n  max-width: 100%;\n  // Part 2: Override the height to auto, otherwise images will be stretched\n  // when setting a width and height attribute on the img element.\n  height: auto;\n}\n","// Container widths\n//\n// Set the container width, and override it for fixed navbars in media queries.\n\n@if $enable-container-classes {\n  // Single container class with breakpoint max-widths\n  .container,\n  // 100% wide container at all breakpoints\n  .container-fluid {\n    @include make-container();\n  }\n\n  // Responsive containers that are 100% wide until a breakpoint\n  @each $breakpoint, $container-max-width in $container-max-widths {\n    .container-#{$breakpoint} {\n      @extend .container-fluid;\n    }\n\n    @include media-breakpoint-up($breakpoint, $grid-breakpoints) {\n      %responsive-container-#{$breakpoint} {\n        max-width: $container-max-width;\n      }\n\n      // Extend each breakpoint which is smaller or equal to the current breakpoint\n      $extend-breakpoint: true;\n\n      @each $name, $width in $grid-breakpoints {\n        @if ($extend-breakpoint) {\n          .container#{breakpoint-infix($name, $grid-breakpoints)} {\n            @extend %responsive-container-#{$breakpoint};\n          }\n\n          // Once the current breakpoint is reached, stop extending\n          @if ($breakpoint == $name) {\n            $extend-breakpoint: false;\n          }\n        }\n      }\n    }\n  }\n}\n","// Container mixins\n\n@mixin make-container($gutter: $container-padding-x) {\n  --#{$prefix}gutter-x: #{$gutter};\n  --#{$prefix}gutter-y: 0;\n  width: 100%;\n  padding-right: calc(var(--#{$prefix}gutter-x) * .5); // stylelint-disable-line function-disallowed-list\n  padding-left: calc(var(--#{$prefix}gutter-x) * .5); // stylelint-disable-line function-disallowed-list\n  margin-right: auto;\n  margin-left: auto;\n}\n","// Breakpoint viewport sizes and media queries.\n//\n// Breakpoints are defined as a map of (name: minimum width), order from small to large:\n//\n//    (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px)\n//\n// The map defined in the `$grid-breakpoints` global variable is used as the `$breakpoints` argument by default.\n\n// Name of the next breakpoint, or null for the last breakpoint.\n//\n//    >> breakpoint-next(sm)\n//    md\n//    >> breakpoint-next(sm, (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px))\n//    md\n//    >> breakpoint-next(sm, $breakpoint-names: (xs sm md lg xl xxl))\n//    md\n@function breakpoint-next($name, $breakpoints: $grid-breakpoints, $breakpoint-names: map-keys($breakpoints)) {\n  $n: index($breakpoint-names, $name);\n  @if not $n {\n    @error \"breakpoint `#{$name}` not found in `#{$breakpoints}`\";\n  }\n  @return if($n < length($breakpoint-names), nth($breakpoint-names, $n + 1), null);\n}\n\n// Minimum breakpoint width. Null for the smallest (first) breakpoint.\n//\n//    >> breakpoint-min(sm, (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px))\n//    576px\n@function breakpoint-min($name, $breakpoints: $grid-breakpoints) {\n  $min: map-get($breakpoints, $name);\n  @return if($min != 0, $min, null);\n}\n\n// Maximum breakpoint width.\n// The maximum value is reduced by 0.02px to work around the limitations of\n// `min-` and `max-` prefixes and viewports with fractional widths.\n// See https://www.w3.org/TR/mediaqueries-4/#mq-min-max\n// Uses 0.02px rather than 0.01px to work around a current rounding bug in Safari.\n// See https://bugs.webkit.org/show_bug.cgi?id=178261\n//\n//    >> breakpoint-max(md, (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px))\n//    767.98px\n@function breakpoint-max($name, $breakpoints: $grid-breakpoints) {\n  $max: map-get($breakpoints, $name);\n  @return if($max and $max > 0, $max - .02, null);\n}\n\n// Returns a blank string if smallest breakpoint, otherwise returns the name with a dash in front.\n// Useful for making responsive utilities.\n//\n//    >> breakpoint-infix(xs, (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px))\n//    \"\"  (Returns a blank string)\n//    >> breakpoint-infix(sm, (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px))\n//    \"-sm\"\n@function breakpoint-infix($name, $breakpoints: $grid-breakpoints) {\n  @return if(breakpoint-min($name, $breakpoints) == null, \"\", \"-#{$name}\");\n}\n\n// Media of at least the minimum breakpoint width. No query for the smallest breakpoint.\n// Makes the @content apply to the given breakpoint and wider.\n@mixin media-breakpoint-up($name, $breakpoints: $grid-breakpoints) {\n  $min: breakpoint-min($name, $breakpoints);\n  @if $min {\n    @media (min-width: $min) {\n      @content;\n    }\n  } @else {\n    @content;\n  }\n}\n\n// Media of at most the maximum breakpoint width. No query for the largest breakpoint.\n// Makes the @content apply to the given breakpoint and narrower.\n@mixin media-breakpoint-down($name, $breakpoints: $grid-breakpoints) {\n  $max: breakpoint-max($name, $breakpoints);\n  @if $max {\n    @media (max-width: $max) {\n      @content;\n    }\n  } @else {\n    @content;\n  }\n}\n\n// Media that spans multiple breakpoint widths.\n// Makes the @content apply between the min and max breakpoints\n@mixin media-breakpoint-between($lower, $upper, $breakpoints: $grid-breakpoints) {\n  $min: breakpoint-min($lower, $breakpoints);\n  $max: breakpoint-max($upper, $breakpoints);\n\n  @if $min != null and $max != null {\n    @media (min-width: $min) and (max-width: $max) {\n      @content;\n    }\n  } @else if $max == null {\n    @include media-breakpoint-up($lower, $breakpoints) {\n      @content;\n    }\n  } @else if $min == null {\n    @include media-breakpoint-down($upper, $breakpoints) {\n      @content;\n    }\n  }\n}\n\n// Media between the breakpoint's minimum and maximum widths.\n// No minimum for the smallest breakpoint, and no maximum for the largest one.\n// Makes the @content apply only to the given breakpoint, not viewports any wider or narrower.\n@mixin media-breakpoint-only($name, $breakpoints: $grid-breakpoints) {\n  $min:  breakpoint-min($name, $breakpoints);\n  $next: breakpoint-next($name, $breakpoints);\n  $max:  breakpoint-max($next, $breakpoints);\n\n  @if $min != null and $max != null {\n    @media (min-width: $min) and (max-width: $max) {\n      @content;\n    }\n  } @else if $max == null {\n    @include media-breakpoint-up($name, $breakpoints) {\n      @content;\n    }\n  } @else if $min == null {\n    @include media-breakpoint-down($next, $breakpoints) {\n      @content;\n    }\n  }\n}\n","// Row\n//\n// Rows contain your columns.\n\n:root {\n  @each $name, $value in $grid-breakpoints {\n    --#{$prefix}breakpoint-#{$name}: #{$value};\n  }\n}\n\n@if $enable-grid-classes {\n  .row {\n    @include make-row();\n\n    > * {\n      @include make-col-ready();\n    }\n  }\n}\n\n@if $enable-cssgrid {\n  .grid {\n    display: grid;\n    grid-template-rows: repeat(var(--#{$prefix}rows, 1), 1fr);\n    grid-template-columns: repeat(var(--#{$prefix}columns, #{$grid-columns}), 1fr);\n    gap: var(--#{$prefix}gap, #{$grid-gutter-width});\n\n    @include make-cssgrid();\n  }\n}\n\n\n// Columns\n//\n// Common styles for small and large grid columns\n\n@if $enable-grid-classes {\n  @include make-grid-columns();\n}\n","// Grid system\n//\n// Generate semantic grid columns with these mixins.\n\n@mixin make-row($gutter: $grid-gutter-width) {\n  --#{$prefix}gutter-x: #{$gutter};\n  --#{$prefix}gutter-y: 0;\n  display: flex;\n  flex-wrap: wrap;\n  // TODO: Revisit calc order after https://github.com/react-bootstrap/react-bootstrap/issues/6039 is fixed\n  margin-top: calc(-1 * var(--#{$prefix}gutter-y)); // stylelint-disable-line function-disallowed-list\n  margin-right: calc(-.5 * var(--#{$prefix}gutter-x)); // stylelint-disable-line function-disallowed-list\n  margin-left: calc(-.5 * var(--#{$prefix}gutter-x)); // stylelint-disable-line function-disallowed-list\n}\n\n@mixin make-col-ready() {\n  // Add box sizing if only the grid is loaded\n  box-sizing: if(variable-exists(include-column-box-sizing) and $include-column-box-sizing, border-box, null);\n  // Prevent columns from becoming too narrow when at smaller grid tiers by\n  // always setting `width: 100%;`. This works because we set the width\n  // later on to override this initial width.\n  flex-shrink: 0;\n  width: 100%;\n  max-width: 100%; // Prevent `.col-auto`, `.col` (& responsive variants) from breaking out the grid\n  padding-right: calc(var(--#{$prefix}gutter-x) * .5); // stylelint-disable-line function-disallowed-list\n  padding-left: calc(var(--#{$prefix}gutter-x) * .5); // stylelint-disable-line function-disallowed-list\n  margin-top: var(--#{$prefix}gutter-y);\n}\n\n@mixin make-col($size: false, $columns: $grid-columns) {\n  @if $size {\n    flex: 0 0 auto;\n    width: percentage(divide($size, $columns));\n\n  } @else {\n    flex: 1 1 0;\n    max-width: 100%;\n  }\n}\n\n@mixin make-col-auto() {\n  flex: 0 0 auto;\n  width: auto;\n}\n\n@mixin make-col-offset($size, $columns: $grid-columns) {\n  $num: divide($size, $columns);\n  margin-left: if($num == 0, 0, percentage($num));\n}\n\n// Row columns\n//\n// Specify on a parent element(e.g., .row) to force immediate children into NN\n// number of columns. Supports wrapping to new lines, but does not do a Masonry\n// style grid.\n@mixin row-cols($count) {\n  > * {\n    flex: 0 0 auto;\n    width: percentage(divide(1, $count));\n  }\n}\n\n// Framework grid generation\n//\n// Used only by Bootstrap to generate the correct number of grid classes given\n// any value of `$grid-columns`.\n\n@mixin make-grid-columns($columns: $grid-columns, $gutter: $grid-gutter-width, $breakpoints: $grid-breakpoints) {\n  @each $breakpoint in map-keys($breakpoints) {\n    $infix: breakpoint-infix($breakpoint, $breakpoints);\n\n    @include media-breakpoint-up($breakpoint, $breakpoints) {\n      // Provide basic `.col-{bp}` classes for equal-width flexbox columns\n      .col#{$infix} {\n        flex: 1 0 0%; // Flexbugs #4: https://github.com/philipwalton/flexbugs#flexbug-4\n      }\n\n      .row-cols#{$infix}-auto > * {\n        @include make-col-auto();\n      }\n\n      @if $grid-row-columns > 0 {\n        @for $i from 1 through $grid-row-columns {\n          .row-cols#{$infix}-#{$i} {\n            @include row-cols($i);\n          }\n        }\n      }\n\n      .col#{$infix}-auto {\n        @include make-col-auto();\n      }\n\n      @if $columns > 0 {\n        @for $i from 1 through $columns {\n          .col#{$infix}-#{$i} {\n            @include make-col($i, $columns);\n          }\n        }\n\n        // `$columns - 1` because offsetting by the width of an entire row isn't possible\n        @for $i from 0 through ($columns - 1) {\n          @if not ($infix == \"\" and $i == 0) { // Avoid emitting useless .offset-0\n            .offset#{$infix}-#{$i} {\n              @include make-col-offset($i, $columns);\n            }\n          }\n        }\n      }\n\n      // Gutters\n      //\n      // Make use of `.g-*`, `.gx-*` or `.gy-*` utilities to change spacing between the columns.\n      @each $key, $value in $gutters {\n        .g#{$infix}-#{$key},\n        .gx#{$infix}-#{$key} {\n          --#{$prefix}gutter-x: #{$value};\n        }\n\n        .g#{$infix}-#{$key},\n        .gy#{$infix}-#{$key} {\n          --#{$prefix}gutter-y: #{$value};\n        }\n      }\n    }\n  }\n}\n\n@mixin make-cssgrid($columns: $grid-columns, $breakpoints: $grid-breakpoints) {\n  @each $breakpoint in map-keys($breakpoints) {\n    $infix: breakpoint-infix($breakpoint, $breakpoints);\n\n    @include media-breakpoint-up($breakpoint, $breakpoints) {\n      @if $columns > 0 {\n        @for $i from 1 through $columns {\n          .g-col#{$infix}-#{$i} {\n            grid-column: auto / span $i;\n          }\n        }\n\n        // Start with `1` because `0` is and invalid value.\n        // Ends with `$columns - 1` because offsetting by the width of an entire row isn't possible.\n        @for $i from 1 through ($columns - 1) {\n          .g-start#{$infix}-#{$i} {\n            grid-column-start: $i;\n          }\n        }\n      }\n    }\n  }\n}\n","//\n// Basic Bootstrap table\n//\n\n.table {\n  // Reset needed for nesting tables\n  --#{$prefix}table-color-type: initial;\n  --#{$prefix}table-bg-type: initial;\n  --#{$prefix}table-color-state: initial;\n  --#{$prefix}table-bg-state: initial;\n  // End of reset\n  --#{$prefix}table-color: #{$table-color};\n  --#{$prefix}table-bg: #{$table-bg};\n  --#{$prefix}table-border-color: #{$table-border-color};\n  --#{$prefix}table-accent-bg: #{$table-accent-bg};\n  --#{$prefix}table-striped-color: #{$table-striped-color};\n  --#{$prefix}table-striped-bg: #{$table-striped-bg};\n  --#{$prefix}table-active-color: #{$table-active-color};\n  --#{$prefix}table-active-bg: #{$table-active-bg};\n  --#{$prefix}table-hover-color: #{$table-hover-color};\n  --#{$prefix}table-hover-bg: #{$table-hover-bg};\n\n  width: 100%;\n  margin-bottom: $spacer;\n  vertical-align: $table-cell-vertical-align;\n  border-color: var(--#{$prefix}table-border-color);\n\n  // Target th & td\n  // We need the child combinator to prevent styles leaking to nested tables which doesn't have a `.table` class.\n  // We use the universal selectors here to simplify the selector (else we would need 6 different selectors).\n  // Another advantage is that this generates less code and makes the selector less specific making it easier to override.\n  // stylelint-disable-next-line selector-max-universal\n  > :not(caption) > * > * {\n    padding: $table-cell-padding-y $table-cell-padding-x;\n    // Following the precept of cascades: https://codepen.io/miriamsuzanne/full/vYNgodb\n    color: var(--#{$prefix}table-color-state, var(--#{$prefix}table-color-type, var(--#{$prefix}table-color)));\n    background-color: var(--#{$prefix}table-bg);\n    border-bottom-width: $table-border-width;\n    box-shadow: inset 0 0 0 9999px var(--#{$prefix}table-bg-state, var(--#{$prefix}table-bg-type, var(--#{$prefix}table-accent-bg)));\n  }\n\n  > tbody {\n    vertical-align: inherit;\n  }\n\n  > thead {\n    vertical-align: bottom;\n  }\n}\n\n.table-group-divider {\n  border-top: calc(#{$table-border-width} * 2) solid $table-group-separator-color; // stylelint-disable-line function-disallowed-list\n}\n\n//\n// Change placement of captions with a class\n//\n\n.caption-top {\n  caption-side: top;\n}\n\n\n//\n// Condensed table w/ half padding\n//\n\n.table-sm {\n  // stylelint-disable-next-line selector-max-universal\n  > :not(caption) > * > * {\n    padding: $table-cell-padding-y-sm $table-cell-padding-x-sm;\n  }\n}\n\n\n// Border versions\n//\n// Add or remove borders all around the table and between all the columns.\n//\n// When borders are added on all sides of the cells, the corners can render odd when\n// these borders do not have the same color or if they are semi-transparent.\n// Therefor we add top and border bottoms to the `tr`s and left and right borders\n// to the `td`s or `th`s\n\n.table-bordered {\n  > :not(caption) > * {\n    border-width: $table-border-width 0;\n\n    // stylelint-disable-next-line selector-max-universal\n    > * {\n      border-width: 0 $table-border-width;\n    }\n  }\n}\n\n.table-borderless {\n  // stylelint-disable-next-line selector-max-universal\n  > :not(caption) > * > * {\n    border-bottom-width: 0;\n  }\n\n  > :not(:first-child) {\n    border-top-width: 0;\n  }\n}\n\n// Zebra-striping\n//\n// Default zebra-stripe styles (alternating gray and transparent backgrounds)\n\n// For rows\n.table-striped {\n  > tbody > tr:nth-of-type(#{$table-striped-order}) > * {\n    --#{$prefix}table-color-type: var(--#{$prefix}table-striped-color);\n    --#{$prefix}table-bg-type: var(--#{$prefix}table-striped-bg);\n  }\n}\n\n// For columns\n.table-striped-columns {\n  > :not(caption) > tr > :nth-child(#{$table-striped-columns-order}) {\n    --#{$prefix}table-color-type: var(--#{$prefix}table-striped-color);\n    --#{$prefix}table-bg-type: var(--#{$prefix}table-striped-bg);\n  }\n}\n\n// Active table\n//\n// The `.table-active` class can be added to highlight rows or cells\n\n.table-active {\n  --#{$prefix}table-color-state: var(--#{$prefix}table-active-color);\n  --#{$prefix}table-bg-state: var(--#{$prefix}table-active-bg);\n}\n\n// Hover effect\n//\n// Placed here since it has to come after the potential zebra striping\n\n.table-hover {\n  > tbody > tr:hover > * {\n    --#{$prefix}table-color-state: var(--#{$prefix}table-hover-color);\n    --#{$prefix}table-bg-state: var(--#{$prefix}table-hover-bg);\n  }\n}\n\n\n// Table variants\n//\n// Table variants set the table cell backgrounds, border colors\n// and the colors of the striped, hovered & active tables\n\n@each $color, $value in $table-variants {\n  @include table-variant($color, $value);\n}\n\n// Responsive tables\n//\n// Generate series of `.table-responsive-*` classes for configuring the screen\n// size of where your table will overflow.\n\n@each $breakpoint in map-keys($grid-breakpoints) {\n  $infix: breakpoint-infix($breakpoint, $grid-breakpoints);\n\n  @include media-breakpoint-down($breakpoint) {\n    .table-responsive#{$infix} {\n      overflow-x: auto;\n      -webkit-overflow-scrolling: touch;\n    }\n  }\n}\n","// scss-docs-start table-variant\n@mixin table-variant($state, $background) {\n  .table-#{$state} {\n    $color: color-contrast(opaque($body-bg, $background));\n    $hover-bg: mix($color, $background, percentage($table-hover-bg-factor));\n    $striped-bg: mix($color, $background, percentage($table-striped-bg-factor));\n    $active-bg: mix($color, $background, percentage($table-active-bg-factor));\n    $table-border-color: mix($color, $background, percentage($table-border-factor));\n\n    --#{$prefix}table-color: #{$color};\n    --#{$prefix}table-bg: #{$background};\n    --#{$prefix}table-border-color: #{$table-border-color};\n    --#{$prefix}table-striped-bg: #{$striped-bg};\n    --#{$prefix}table-striped-color: #{color-contrast($striped-bg)};\n    --#{$prefix}table-active-bg: #{$active-bg};\n    --#{$prefix}table-active-color: #{color-contrast($active-bg)};\n    --#{$prefix}table-hover-bg: #{$hover-bg};\n    --#{$prefix}table-hover-color: #{color-contrast($hover-bg)};\n\n    color: var(--#{$prefix}table-color);\n    border-color: var(--#{$prefix}table-border-color);\n  }\n}\n// scss-docs-end table-variant\n","//\n// Labels\n//\n\n.form-label {\n  margin-bottom: $form-label-margin-bottom;\n  @include font-size($form-label-font-size);\n  font-style: $form-label-font-style;\n  font-weight: $form-label-font-weight;\n  color: $form-label-color;\n}\n\n// For use with horizontal and inline forms, when you need the label (or legend)\n// text to align with the form controls.\n.col-form-label {\n  padding-top: add($input-padding-y, $input-border-width);\n  padding-bottom: add($input-padding-y, $input-border-width);\n  margin-bottom: 0; // Override the `<legend>` default\n  @include font-size(inherit); // Override the `<legend>` default\n  font-style: $form-label-font-style;\n  font-weight: $form-label-font-weight;\n  line-height: $input-line-height;\n  color: $form-label-color;\n}\n\n.col-form-label-lg {\n  padding-top: add($input-padding-y-lg, $input-border-width);\n  padding-bottom: add($input-padding-y-lg, $input-border-width);\n  @include font-size($input-font-size-lg);\n}\n\n.col-form-label-sm {\n  padding-top: add($input-padding-y-sm, $input-border-width);\n  padding-bottom: add($input-padding-y-sm, $input-border-width);\n  @include font-size($input-font-size-sm);\n}\n","//\n// Form text\n//\n\n.form-text {\n  margin-top: $form-text-margin-top;\n  @include font-size($form-text-font-size);\n  font-style: $form-text-font-style;\n  font-weight: $form-text-font-weight;\n  color: $form-text-color;\n}\n","//\n// General form controls (plus a few specific high-level interventions)\n//\n\n.form-control {\n  display: block;\n  width: 100%;\n  padding: $input-padding-y $input-padding-x;\n  font-family: $input-font-family;\n  @include font-size($input-font-size);\n  font-weight: $input-font-weight;\n  line-height: $input-line-height;\n  color: $input-color;\n  appearance: none; // Fix appearance for date inputs in Safari\n  background-color: $input-bg;\n  background-clip: padding-box;\n  border: $input-border-width solid $input-border-color;\n\n  // Note: This has no effect on <select>s in some browsers, due to the limited stylability of `<select>`s in CSS.\n  @include border-radius($input-border-radius, 0);\n\n  @include box-shadow($input-box-shadow);\n  @include transition($input-transition);\n\n  &[type=\"file\"] {\n    overflow: hidden; // prevent pseudo element button overlap\n\n    &:not(:disabled):not([readonly]) {\n      cursor: pointer;\n    }\n  }\n\n  // Customize the `:focus` state to imitate native WebKit styles.\n  &:focus {\n    color: $input-focus-color;\n    background-color: $input-focus-bg;\n    border-color: $input-focus-border-color;\n    outline: 0;\n    @if $enable-shadows {\n      @include box-shadow($input-box-shadow, $input-focus-box-shadow);\n    } @else {\n      // Avoid using mixin so we can pass custom focus shadow properly\n      box-shadow: $input-focus-box-shadow;\n    }\n  }\n\n  &::-webkit-date-and-time-value {\n    // On Android Chrome, form-control's \"width: 100%\" makes the input width too small\n    // Tested under Android 11 / Chrome 89, Android 12 / Chrome 100, Android 13 / Chrome 109\n    //\n    // On iOS Safari, form-control's \"appearance: none\" + \"width: 100%\" makes the input width too small\n    // Tested under iOS 16.2 / Safari 16.2\n    min-width: 85px; // Seems to be a good minimum safe width\n\n    // Add some height to date inputs on iOS\n    // https://github.com/twbs/bootstrap/issues/23307\n    // TODO: we can remove this workaround once https://bugs.webkit.org/show_bug.cgi?id=198959 is resolved\n    // Multiply line-height by 1em if it has no unit\n    height: if(unit($input-line-height) == \"\", $input-line-height * 1em, $input-line-height);\n\n    // Android Chrome type=\"date\" is taller than the other inputs\n    // because of \"margin: 1px 24px 1px 4px\" inside the shadow DOM\n    // Tested under Android 11 / Chrome 89, Android 12 / Chrome 100, Android 13 / Chrome 109\n    margin: 0;\n  }\n\n  // Prevent excessive date input height in Webkit\n  // https://github.com/twbs/bootstrap/issues/34433\n  &::-webkit-datetime-edit {\n    display: block;\n    padding: 0;\n  }\n\n  // Placeholder\n  &::placeholder {\n    color: $input-placeholder-color;\n    // Override Firefox's unusual default opacity; see https://github.com/twbs/bootstrap/pull/11526.\n    opacity: 1;\n  }\n\n  // Disabled inputs\n  //\n  // HTML5 says that controls under a fieldset > legend:first-child won't be\n  // disabled if the fieldset is disabled. Due to implementation difficulty, we\n  // don't honor that edge case; we style them as disabled anyway.\n  &:disabled {\n    color: $input-disabled-color;\n    background-color: $input-disabled-bg;\n    border-color: $input-disabled-border-color;\n    // iOS fix for unreadable disabled content; see https://github.com/twbs/bootstrap/issues/11655.\n    opacity: 1;\n  }\n\n  // File input buttons theming\n  &::file-selector-button {\n    padding: $input-padding-y $input-padding-x;\n    margin: (-$input-padding-y) (-$input-padding-x);\n    margin-inline-end: $input-padding-x;\n    color: $form-file-button-color;\n    @include gradient-bg($form-file-button-bg);\n    pointer-events: none;\n    border-color: inherit;\n    border-style: solid;\n    border-width: 0;\n    border-inline-end-width: $input-border-width;\n    border-radius: 0; // stylelint-disable-line property-disallowed-list\n    @include transition($btn-transition);\n  }\n\n  &:hover:not(:disabled):not([readonly])::file-selector-button {\n    background-color: $form-file-button-hover-bg;\n  }\n}\n\n// Readonly controls as plain text\n//\n// Apply class to a readonly input to make it appear like regular plain\n// text (without any border, background color, focus indicator)\n\n.form-control-plaintext {\n  display: block;\n  width: 100%;\n  padding: $input-padding-y 0;\n  margin-bottom: 0; // match inputs if this class comes on inputs with default margins\n  line-height: $input-line-height;\n  color: $input-plaintext-color;\n  background-color: transparent;\n  border: solid transparent;\n  border-width: $input-border-width 0;\n\n  &:focus {\n    outline: 0;\n  }\n\n  &.form-control-sm,\n  &.form-control-lg {\n    padding-right: 0;\n    padding-left: 0;\n  }\n}\n\n// Form control sizing\n//\n// Build on `.form-control` with modifier classes to decrease or increase the\n// height and font-size of form controls.\n//\n// Repeated in `_input_group.scss` to avoid Sass extend issues.\n\n.form-control-sm {\n  min-height: $input-height-sm;\n  padding: $input-padding-y-sm $input-padding-x-sm;\n  @include font-size($input-font-size-sm);\n  @include border-radius($input-border-radius-sm);\n\n  &::file-selector-button {\n    padding: $input-padding-y-sm $input-padding-x-sm;\n    margin: (-$input-padding-y-sm) (-$input-padding-x-sm);\n    margin-inline-end: $input-padding-x-sm;\n  }\n}\n\n.form-control-lg {\n  min-height: $input-height-lg;\n  padding: $input-padding-y-lg $input-padding-x-lg;\n  @include font-size($input-font-size-lg);\n  @include border-radius($input-border-radius-lg);\n\n  &::file-selector-button {\n    padding: $input-padding-y-lg $input-padding-x-lg;\n    margin: (-$input-padding-y-lg) (-$input-padding-x-lg);\n    margin-inline-end: $input-padding-x-lg;\n  }\n}\n\n// Make sure textareas don't shrink too much when resized\n// https://github.com/twbs/bootstrap/pull/29124\n// stylelint-disable selector-no-qualifying-type\ntextarea {\n  &.form-control {\n    min-height: $input-height;\n  }\n\n  &.form-control-sm {\n    min-height: $input-height-sm;\n  }\n\n  &.form-control-lg {\n    min-height: $input-height-lg;\n  }\n}\n// stylelint-enable selector-no-qualifying-type\n\n.form-control-color {\n  width: $form-color-width;\n  height: $input-height;\n  padding: $input-padding-y;\n\n  &:not(:disabled):not([readonly]) {\n    cursor: pointer;\n  }\n\n  &::-moz-color-swatch {\n    border: 0 !important; // stylelint-disable-line declaration-no-important\n    @include border-radius($input-border-radius);\n  }\n\n  &::-webkit-color-swatch {\n    border: 0 !important; // stylelint-disable-line declaration-no-important\n    @include border-radius($input-border-radius);\n  }\n\n  &.form-control-sm { height: $input-height-sm; }\n  &.form-control-lg { height: $input-height-lg; }\n}\n","// stylelint-disable property-disallowed-list\n@mixin transition($transition...) {\n  @if length($transition) == 0 {\n    $transition: $transition-base;\n  }\n\n  @if length($transition) > 1 {\n    @each $value in $transition {\n      @if $value == null or $value == none {\n        @warn \"The keyword 'none' or 'null' must be used as a single argument.\";\n      }\n    }\n  }\n\n  @if $enable-transitions {\n    @if nth($transition, 1) != null {\n      transition: $transition;\n    }\n\n    @if $enable-reduced-motion and nth($transition, 1) != null and nth($transition, 1) != none {\n      @media (prefers-reduced-motion: reduce) {\n        transition: none;\n      }\n    }\n  }\n}\n","// Gradients\n\n// scss-docs-start gradient-bg-mixin\n@mixin gradient-bg($color: null) {\n  background-color: $color;\n\n  @if $enable-gradients {\n    background-image: var(--#{$prefix}gradient);\n  }\n}\n// scss-docs-end gradient-bg-mixin\n\n// scss-docs-start gradient-mixins\n// Horizontal gradient, from left to right\n//\n// Creates two color stops, start and end, by specifying a color and position for each color stop.\n@mixin gradient-x($start-color: $gray-700, $end-color: $gray-800, $start-percent: 0%, $end-percent: 100%) {\n  background-image: linear-gradient(to right, $start-color $start-percent, $end-color $end-percent);\n}\n\n// Vertical gradient, from top to bottom\n//\n// Creates two color stops, start and end, by specifying a color and position for each color stop.\n@mixin gradient-y($start-color: $gray-700, $end-color: $gray-800, $start-percent: null, $end-percent: null) {\n  background-image: linear-gradient(to bottom, $start-color $start-percent, $end-color $end-percent);\n}\n\n@mixin gradient-directional($start-color: $gray-700, $end-color: $gray-800, $deg: 45deg) {\n  background-image: linear-gradient($deg, $start-color, $end-color);\n}\n\n@mixin gradient-x-three-colors($start-color: $blue, $mid-color: $purple, $color-stop: 50%, $end-color: $red) {\n  background-image: linear-gradient(to right, $start-color, $mid-color $color-stop, $end-color);\n}\n\n@mixin gradient-y-three-colors($start-color: $blue, $mid-color: $purple, $color-stop: 50%, $end-color: $red) {\n  background-image: linear-gradient($start-color, $mid-color $color-stop, $end-color);\n}\n\n@mixin gradient-radial($inner-color: $gray-700, $outer-color: $gray-800) {\n  background-image: radial-gradient(circle, $inner-color, $outer-color);\n}\n\n@mixin gradient-striped($color: rgba($white, .15), $angle: 45deg) {\n  background-image: linear-gradient($angle, $color 25%, transparent 25%, transparent 50%, $color 50%, $color 75%, transparent 75%, transparent);\n}\n// scss-docs-end gradient-mixins\n","// Select\n//\n// Replaces the browser default select with a custom one, mostly pulled from\n// https://primer.github.io/.\n\n.form-select {\n  --#{$prefix}form-select-bg-img: #{escape-svg($form-select-indicator)};\n\n  display: block;\n  width: 100%;\n  padding: $form-select-padding-y $form-select-indicator-padding $form-select-padding-y $form-select-padding-x;\n  font-family: $form-select-font-family;\n  @include font-size($form-select-font-size);\n  font-weight: $form-select-font-weight;\n  line-height: $form-select-line-height;\n  color: $form-select-color;\n  appearance: none;\n  background-color: $form-select-bg;\n  background-image: var(--#{$prefix}form-select-bg-img), var(--#{$prefix}form-select-bg-icon, none);\n  background-repeat: no-repeat;\n  background-position: $form-select-bg-position;\n  background-size: $form-select-bg-size;\n  border: $form-select-border-width solid $form-select-border-color;\n  @include border-radius($form-select-border-radius, 0);\n  @include box-shadow($form-select-box-shadow);\n  @include transition($form-select-transition);\n\n  &:focus {\n    border-color: $form-select-focus-border-color;\n    outline: 0;\n    @if $enable-shadows {\n      @include box-shadow($form-select-box-shadow, $form-select-focus-box-shadow);\n    } @else {\n      // Avoid using mixin so we can pass custom focus shadow properly\n      box-shadow: $form-select-focus-box-shadow;\n    }\n  }\n\n  &[multiple],\n  &[size]:not([size=\"1\"]) {\n    padding-right: $form-select-padding-x;\n    background-image: none;\n  }\n\n  &:disabled {\n    color: $form-select-disabled-color;\n    background-color: $form-select-disabled-bg;\n    border-color: $form-select-disabled-border-color;\n  }\n\n  // Remove outline from select box in FF\n  &:-moz-focusring {\n    color: transparent;\n    text-shadow: 0 0 0 $form-select-color;\n  }\n}\n\n.form-select-sm {\n  padding-top: $form-select-padding-y-sm;\n  padding-bottom: $form-select-padding-y-sm;\n  padding-left: $form-select-padding-x-sm;\n  @include font-size($form-select-font-size-sm);\n  @include border-radius($form-select-border-radius-sm);\n}\n\n.form-select-lg {\n  padding-top: $form-select-padding-y-lg;\n  padding-bottom: $form-select-padding-y-lg;\n  padding-left: $form-select-padding-x-lg;\n  @include font-size($form-select-font-size-lg);\n  @include border-radius($form-select-border-radius-lg);\n}\n\n@if $enable-dark-mode {\n  @include color-mode(dark) {\n    .form-select {\n      --#{$prefix}form-select-bg-img: #{escape-svg($form-select-indicator-dark)};\n    }\n  }\n}\n","//\n// Check/radio\n//\n\n.form-check {\n  display: block;\n  min-height: $form-check-min-height;\n  padding-left: $form-check-padding-start;\n  margin-bottom: $form-check-margin-bottom;\n\n  .form-check-input {\n    float: left;\n    margin-left: $form-check-padding-start * -1;\n  }\n}\n\n.form-check-reverse {\n  padding-right: $form-check-padding-start;\n  padding-left: 0;\n  text-align: right;\n\n  .form-check-input {\n    float: right;\n    margin-right: $form-check-padding-start * -1;\n    margin-left: 0;\n  }\n}\n\n.form-check-input {\n  --#{$prefix}form-check-bg: #{$form-check-input-bg};\n\n  flex-shrink: 0;\n  width: $form-check-input-width;\n  height: $form-check-input-width;\n  margin-top: ($line-height-base - $form-check-input-width) * .5; // line-height minus check height\n  vertical-align: top;\n  appearance: none;\n  background-color: var(--#{$prefix}form-check-bg);\n  background-image: var(--#{$prefix}form-check-bg-image);\n  background-repeat: no-repeat;\n  background-position: center;\n  background-size: contain;\n  border: $form-check-input-border;\n  print-color-adjust: exact; // Keep themed appearance for print\n  @include transition($form-check-transition);\n\n  &[type=\"checkbox\"] {\n    @include border-radius($form-check-input-border-radius);\n  }\n\n  &[type=\"radio\"] {\n    // stylelint-disable-next-line property-disallowed-list\n    border-radius: $form-check-radio-border-radius;\n  }\n\n  &:active {\n    filter: $form-check-input-active-filter;\n  }\n\n  &:focus {\n    border-color: $form-check-input-focus-border;\n    outline: 0;\n    box-shadow: $form-check-input-focus-box-shadow;\n  }\n\n  &:checked {\n    background-color: $form-check-input-checked-bg-color;\n    border-color: $form-check-input-checked-border-color;\n\n    &[type=\"checkbox\"] {\n      @if $enable-gradients {\n        --#{$prefix}form-check-bg-image: #{escape-svg($form-check-input-checked-bg-image)}, var(--#{$prefix}gradient);\n      } @else {\n        --#{$prefix}form-check-bg-image: #{escape-svg($form-check-input-checked-bg-image)};\n      }\n    }\n\n    &[type=\"radio\"] {\n      @if $enable-gradients {\n        --#{$prefix}form-check-bg-image: #{escape-svg($form-check-radio-checked-bg-image)}, var(--#{$prefix}gradient);\n      } @else {\n        --#{$prefix}form-check-bg-image: #{escape-svg($form-check-radio-checked-bg-image)};\n      }\n    }\n  }\n\n  &[type=\"checkbox\"]:indeterminate {\n    background-color: $form-check-input-indeterminate-bg-color;\n    border-color: $form-check-input-indeterminate-border-color;\n\n    @if $enable-gradients {\n      --#{$prefix}form-check-bg-image: #{escape-svg($form-check-input-indeterminate-bg-image)}, var(--#{$prefix}gradient);\n    } @else {\n      --#{$prefix}form-check-bg-image: #{escape-svg($form-check-input-indeterminate-bg-image)};\n    }\n  }\n\n  &:disabled {\n    pointer-events: none;\n    filter: none;\n    opacity: $form-check-input-disabled-opacity;\n  }\n\n  // Use disabled attribute in addition of :disabled pseudo-class\n  // See: https://github.com/twbs/bootstrap/issues/28247\n  &[disabled],\n  &:disabled {\n    ~ .form-check-label {\n      cursor: default;\n      opacity: $form-check-label-disabled-opacity;\n    }\n  }\n}\n\n.form-check-label {\n  color: $form-check-label-color;\n  cursor: $form-check-label-cursor;\n}\n\n//\n// Switch\n//\n\n.form-switch {\n  padding-left: $form-switch-padding-start;\n\n  .form-check-input {\n    --#{$prefix}form-switch-bg: #{escape-svg($form-switch-bg-image)};\n\n    width: $form-switch-width;\n    margin-left: $form-switch-padding-start * -1;\n    background-image: var(--#{$prefix}form-switch-bg);\n    background-position: left center;\n    @include border-radius($form-switch-border-radius);\n    @include transition($form-switch-transition);\n\n    &:focus {\n      --#{$prefix}form-switch-bg: #{escape-svg($form-switch-focus-bg-image)};\n    }\n\n    &:checked {\n      background-position: $form-switch-checked-bg-position;\n\n      @if $enable-gradients {\n        --#{$prefix}form-switch-bg: #{escape-svg($form-switch-checked-bg-image)}, var(--#{$prefix}gradient);\n      } @else {\n        --#{$prefix}form-switch-bg: #{escape-svg($form-switch-checked-bg-image)};\n      }\n    }\n  }\n\n  &.form-check-reverse {\n    padding-right: $form-switch-padding-start;\n    padding-left: 0;\n\n    .form-check-input {\n      margin-right: $form-switch-padding-start * -1;\n      margin-left: 0;\n    }\n  }\n}\n\n.form-check-inline {\n  display: inline-block;\n  margin-right: $form-check-inline-margin-end;\n}\n\n.btn-check {\n  position: absolute;\n  clip: rect(0, 0, 0, 0);\n  pointer-events: none;\n\n  &[disabled],\n  &:disabled {\n    + .btn {\n      pointer-events: none;\n      filter: none;\n      opacity: $form-check-btn-check-disabled-opacity;\n    }\n  }\n}\n\n@if $enable-dark-mode {\n  @include color-mode(dark) {\n    .form-switch .form-check-input:not(:checked):not(:focus) {\n      --#{$prefix}form-switch-bg: #{escape-svg($form-switch-bg-image-dark)};\n    }\n  }\n}\n","// Range\n//\n// Style range inputs the same across browsers. Vendor-specific rules for pseudo\n// elements cannot be mixed. As such, there are no shared styles for focus or\n// active states on prefixed selectors.\n\n.form-range {\n  width: 100%;\n  height: add($form-range-thumb-height, $form-range-thumb-focus-box-shadow-width * 2);\n  padding: 0; // Need to reset padding\n  appearance: none;\n  background-color: transparent;\n\n  &:focus {\n    outline: 0;\n\n    // Pseudo-elements must be split across multiple rulesets to have an effect.\n    // No box-shadow() mixin for focus accessibility.\n    &::-webkit-slider-thumb { box-shadow: $form-range-thumb-focus-box-shadow; }\n    &::-moz-range-thumb     { box-shadow: $form-range-thumb-focus-box-shadow; }\n  }\n\n  &::-moz-focus-outer {\n    border: 0;\n  }\n\n  &::-webkit-slider-thumb {\n    width: $form-range-thumb-width;\n    height: $form-range-thumb-height;\n    margin-top: ($form-range-track-height - $form-range-thumb-height) * .5; // Webkit specific\n    appearance: none;\n    @include gradient-bg($form-range-thumb-bg);\n    border: $form-range-thumb-border;\n    @include border-radius($form-range-thumb-border-radius);\n    @include box-shadow($form-range-thumb-box-shadow);\n    @include transition($form-range-thumb-transition);\n\n    &:active {\n      @include gradient-bg($form-range-thumb-active-bg);\n    }\n  }\n\n  &::-webkit-slider-runnable-track {\n    width: $form-range-track-width;\n    height: $form-range-track-height;\n    color: transparent; // Why?\n    cursor: $form-range-track-cursor;\n    background-color: $form-range-track-bg;\n    border-color: transparent;\n    @include border-radius($form-range-track-border-radius);\n    @include box-shadow($form-range-track-box-shadow);\n  }\n\n  &::-moz-range-thumb {\n    width: $form-range-thumb-width;\n    height: $form-range-thumb-height;\n    appearance: none;\n    @include gradient-bg($form-range-thumb-bg);\n    border: $form-range-thumb-border;\n    @include border-radius($form-range-thumb-border-radius);\n    @include box-shadow($form-range-thumb-box-shadow);\n    @include transition($form-range-thumb-transition);\n\n    &:active {\n      @include gradient-bg($form-range-thumb-active-bg);\n    }\n  }\n\n  &::-moz-range-track {\n    width: $form-range-track-width;\n    height: $form-range-track-height;\n    color: transparent;\n    cursor: $form-range-track-cursor;\n    background-color: $form-range-track-bg;\n    border-color: transparent; // Firefox specific?\n    @include border-radius($form-range-track-border-radius);\n    @include box-shadow($form-range-track-box-shadow);\n  }\n\n  &:disabled {\n    pointer-events: none;\n\n    &::-webkit-slider-thumb {\n      background-color: $form-range-thumb-disabled-bg;\n    }\n\n    &::-moz-range-thumb {\n      background-color: $form-range-thumb-disabled-bg;\n    }\n  }\n}\n",".form-floating {\n  position: relative;\n\n  > .form-control,\n  > .form-control-plaintext,\n  > .form-select {\n    height: $form-floating-height;\n    min-height: $form-floating-height;\n    line-height: $form-floating-line-height;\n  }\n\n  > label {\n    position: absolute;\n    top: 0;\n    left: 0;\n    z-index: 2;\n    height: 100%; // allow textareas\n    padding: $form-floating-padding-y $form-floating-padding-x;\n    overflow: hidden;\n    text-align: start;\n    text-overflow: ellipsis;\n    white-space: nowrap;\n    pointer-events: none;\n    border: $input-border-width solid transparent; // Required for aligning label's text with the input as it affects inner box model\n    transform-origin: 0 0;\n    @include transition($form-floating-transition);\n  }\n\n  > .form-control,\n  > .form-control-plaintext {\n    padding: $form-floating-padding-y $form-floating-padding-x;\n\n    &::placeholder {\n      color: transparent;\n    }\n\n    &:focus,\n    &:not(:placeholder-shown) {\n      padding-top: $form-floating-input-padding-t;\n      padding-bottom: $form-floating-input-padding-b;\n    }\n    // Duplicated because `:-webkit-autofill` invalidates other selectors when grouped\n    &:-webkit-autofill {\n      padding-top: $form-floating-input-padding-t;\n      padding-bottom: $form-floating-input-padding-b;\n    }\n  }\n\n  > .form-select {\n    padding-top: $form-floating-input-padding-t;\n    padding-bottom: $form-floating-input-padding-b;\n  }\n\n  > .form-control:focus,\n  > .form-control:not(:placeholder-shown),\n  > .form-control-plaintext,\n  > .form-select {\n    ~ label {\n      color: rgba(var(--#{$prefix}body-color-rgb), #{$form-floating-label-opacity});\n      transform: $form-floating-label-transform;\n\n      &::after {\n        position: absolute;\n        inset: $form-floating-padding-y ($form-floating-padding-x * .5);\n        z-index: -1;\n        height: $form-floating-label-height;\n        content: \"\";\n        background-color: $input-bg;\n        @include border-radius($input-border-radius);\n      }\n    }\n  }\n  // Duplicated because `:-webkit-autofill` invalidates other selectors when grouped\n  > .form-control:-webkit-autofill {\n    ~ label {\n      color: rgba(var(--#{$prefix}body-color-rgb), #{$form-floating-label-opacity});\n      transform: $form-floating-label-transform;\n    }\n  }\n\n  > .form-control-plaintext {\n    ~ label {\n      border-width: $input-border-width 0; // Required to properly position label text - as explained above\n    }\n  }\n\n  > :disabled ~ label,\n  > .form-control:disabled ~ label { // Required for `.form-control`s because of specificity\n    color: $form-floating-label-disabled-color;\n\n    &::after {\n      background-color: $input-disabled-bg;\n    }\n  }\n}\n","//\n// Base styles\n//\n\n.input-group {\n  position: relative;\n  display: flex;\n  flex-wrap: wrap; // For form validation feedback\n  align-items: stretch;\n  width: 100%;\n\n  > .form-control,\n  > .form-select,\n  > .form-floating {\n    position: relative; // For focus state's z-index\n    flex: 1 1 auto;\n    width: 1%;\n    min-width: 0; // https://stackoverflow.com/questions/36247140/why-dont-flex-items-shrink-past-content-size\n  }\n\n  // Bring the \"active\" form control to the top of surrounding elements\n  > .form-control:focus,\n  > .form-select:focus,\n  > .form-floating:focus-within {\n    z-index: 5;\n  }\n\n  // Ensure buttons are always above inputs for more visually pleasing borders.\n  // This isn't needed for `.input-group-text` since it shares the same border-color\n  // as our inputs.\n  .btn {\n    position: relative;\n    z-index: 2;\n\n    &:focus {\n      z-index: 5;\n    }\n  }\n}\n\n\n// Textual addons\n//\n// Serves as a catch-all element for any text or radio/checkbox input you wish\n// to prepend or append to an input.\n\n.input-group-text {\n  display: flex;\n  align-items: center;\n  padding: $input-group-addon-padding-y $input-group-addon-padding-x;\n  @include font-size($input-font-size); // Match inputs\n  font-weight: $input-group-addon-font-weight;\n  line-height: $input-line-height;\n  color: $input-group-addon-color;\n  text-align: center;\n  white-space: nowrap;\n  background-color: $input-group-addon-bg;\n  border: $input-border-width solid $input-group-addon-border-color;\n  @include border-radius($input-border-radius);\n}\n\n\n// Sizing\n//\n// Remix the default form control sizing classes into new ones for easier\n// manipulation.\n\n.input-group-lg > .form-control,\n.input-group-lg > .form-select,\n.input-group-lg > .input-group-text,\n.input-group-lg > .btn {\n  padding: $input-padding-y-lg $input-padding-x-lg;\n  @include font-size($input-font-size-lg);\n  @include border-radius($input-border-radius-lg);\n}\n\n.input-group-sm > .form-control,\n.input-group-sm > .form-select,\n.input-group-sm > .input-group-text,\n.input-group-sm > .btn {\n  padding: $input-padding-y-sm $input-padding-x-sm;\n  @include font-size($input-font-size-sm);\n  @include border-radius($input-border-radius-sm);\n}\n\n.input-group-lg > .form-select,\n.input-group-sm > .form-select {\n  padding-right: $form-select-padding-x + $form-select-indicator-padding;\n}\n\n\n// Rounded corners\n//\n// These rulesets must come after the sizing ones to properly override sm and lg\n// border-radius values when extending. They're more specific than we'd like\n// with the `.input-group >` part, but without it, we cannot override the sizing.\n\n// stylelint-disable-next-line no-duplicate-selectors\n.input-group {\n  &:not(.has-validation) {\n    > :not(:last-child):not(.dropdown-toggle):not(.dropdown-menu):not(.form-floating),\n    > .dropdown-toggle:nth-last-child(n + 3),\n    > .form-floating:not(:last-child) > .form-control,\n    > .form-floating:not(:last-child) > .form-select {\n      @include border-end-radius(0);\n    }\n  }\n\n  &.has-validation {\n    > :nth-last-child(n + 3):not(.dropdown-toggle):not(.dropdown-menu):not(.form-floating),\n    > .dropdown-toggle:nth-last-child(n + 4),\n    > .form-floating:nth-last-child(n + 3) > .form-control,\n    > .form-floating:nth-last-child(n + 3) > .form-select {\n      @include border-end-radius(0);\n    }\n  }\n\n  $validation-messages: \"\";\n  @each $state in map-keys($form-validation-states) {\n    $validation-messages: $validation-messages + \":not(.\" + unquote($state) + \"-tooltip)\" + \":not(.\" + unquote($state) + \"-feedback)\";\n  }\n\n  > :not(:first-child):not(.dropdown-menu)#{$validation-messages} {\n    margin-left: calc(#{$input-border-width} * -1); // stylelint-disable-line function-disallowed-list\n    @include border-start-radius(0);\n  }\n\n  > .form-floating:not(:first-child) > .form-control,\n  > .form-floating:not(:first-child) > .form-select {\n    @include border-start-radius(0);\n  }\n}\n","// This mixin uses an `if()` technique to be compatible with Dart Sass\n// See https://github.com/sass/sass/issues/1873#issuecomment-152293725 for more details\n\n// scss-docs-start form-validation-mixins\n@mixin form-validation-state-selector($state) {\n  @if ($state == \"valid\" or $state == \"invalid\") {\n    .was-validated #{if(&, \"&\", \"\")}:#{$state},\n    #{if(&, \"&\", \"\")}.is-#{$state} {\n      @content;\n    }\n  } @else {\n    #{if(&, \"&\", \"\")}.is-#{$state} {\n      @content;\n    }\n  }\n}\n\n@mixin form-validation-state(\n  $state,\n  $color,\n  $icon,\n  $tooltip-color: color-contrast($color),\n  $tooltip-bg-color: rgba($color, $form-feedback-tooltip-opacity),\n  $focus-box-shadow: 0 0 $input-btn-focus-blur $input-focus-width rgba($color, $input-btn-focus-color-opacity),\n  $border-color: $color\n) {\n  .#{$state}-feedback {\n    display: none;\n    width: 100%;\n    margin-top: $form-feedback-margin-top;\n    @include font-size($form-feedback-font-size);\n    font-style: $form-feedback-font-style;\n    color: $color;\n  }\n\n  .#{$state}-tooltip {\n    position: absolute;\n    top: 100%;\n    z-index: 5;\n    display: none;\n    max-width: 100%; // Contain to parent when possible\n    padding: $form-feedback-tooltip-padding-y $form-feedback-tooltip-padding-x;\n    margin-top: .1rem;\n    @include font-size($form-feedback-tooltip-font-size);\n    line-height: $form-feedback-tooltip-line-height;\n    color: $tooltip-color;\n    background-color: $tooltip-bg-color;\n    @include border-radius($form-feedback-tooltip-border-radius);\n  }\n\n  @include form-validation-state-selector($state) {\n    ~ .#{$state}-feedback,\n    ~ .#{$state}-tooltip {\n      display: block;\n    }\n  }\n\n  .form-control {\n    @include form-validation-state-selector($state) {\n      border-color: $border-color;\n\n      @if $enable-validation-icons {\n        padding-right: $input-height-inner;\n        background-image: escape-svg($icon);\n        background-repeat: no-repeat;\n        background-position: right $input-height-inner-quarter center;\n        background-size: $input-height-inner-half $input-height-inner-half;\n      }\n\n      &:focus {\n        border-color: $border-color;\n        box-shadow: $focus-box-shadow;\n      }\n    }\n  }\n\n  // stylelint-disable-next-line selector-no-qualifying-type\n  textarea.form-control {\n    @include form-validation-state-selector($state) {\n      @if $enable-validation-icons {\n        padding-right: $input-height-inner;\n        background-position: top $input-height-inner-quarter right $input-height-inner-quarter;\n      }\n    }\n  }\n\n  .form-select {\n    @include form-validation-state-selector($state) {\n      border-color: $border-color;\n\n      @if $enable-validation-icons {\n        &:not([multiple]):not([size]),\n        &:not([multiple])[size=\"1\"] {\n          --#{$prefix}form-select-bg-icon: #{escape-svg($icon)};\n          padding-right: $form-select-feedback-icon-padding-end;\n          background-position: $form-select-bg-position, $form-select-feedback-icon-position;\n          background-size: $form-select-bg-size, $form-select-feedback-icon-size;\n        }\n      }\n\n      &:focus {\n        border-color: $border-color;\n        box-shadow: $focus-box-shadow;\n      }\n    }\n  }\n\n  .form-control-color {\n    @include form-validation-state-selector($state) {\n      @if $enable-validation-icons {\n        width: add($form-color-width, $input-height-inner);\n      }\n    }\n  }\n\n  .form-check-input {\n    @include form-validation-state-selector($state) {\n      border-color: $border-color;\n\n      &:checked {\n        background-color: $color;\n      }\n\n      &:focus {\n        box-shadow: $focus-box-shadow;\n      }\n\n      ~ .form-check-label {\n        color: $color;\n      }\n    }\n  }\n  .form-check-inline .form-check-input {\n    ~ .#{$state}-feedback {\n      margin-left: .5em;\n    }\n  }\n\n  .input-group {\n    > .form-control:not(:focus),\n    > .form-select:not(:focus),\n    > .form-floating:not(:focus-within) {\n      @include form-validation-state-selector($state) {\n        @if $state == \"valid\" {\n          z-index: 3;\n        } @else if $state == \"invalid\" {\n          z-index: 4;\n        }\n      }\n    }\n  }\n}\n// scss-docs-end form-validation-mixins\n","//\n// Base styles\n//\n\n.btn {\n  // scss-docs-start btn-css-vars\n  --#{$prefix}btn-padding-x: #{$btn-padding-x};\n  --#{$prefix}btn-padding-y: #{$btn-padding-y};\n  --#{$prefix}btn-font-family: #{$btn-font-family};\n  @include rfs($btn-font-size, --#{$prefix}btn-font-size);\n  --#{$prefix}btn-font-weight: #{$btn-font-weight};\n  --#{$prefix}btn-line-height: #{$btn-line-height};\n  --#{$prefix}btn-color: #{$btn-color};\n  --#{$prefix}btn-bg: transparent;\n  --#{$prefix}btn-border-width: #{$btn-border-width};\n  --#{$prefix}btn-border-color: transparent;\n  --#{$prefix}btn-border-radius: #{$btn-border-radius};\n  --#{$prefix}btn-hover-border-color: transparent;\n  --#{$prefix}btn-box-shadow: #{$btn-box-shadow};\n  --#{$prefix}btn-disabled-opacity: #{$btn-disabled-opacity};\n  --#{$prefix}btn-focus-box-shadow: 0 0 0 #{$btn-focus-width} rgba(var(--#{$prefix}btn-focus-shadow-rgb), .5);\n  // scss-docs-end btn-css-vars\n\n  display: inline-block;\n  padding: var(--#{$prefix}btn-padding-y) var(--#{$prefix}btn-padding-x);\n  font-family: var(--#{$prefix}btn-font-family);\n  @include font-size(var(--#{$prefix}btn-font-size));\n  font-weight: var(--#{$prefix}btn-font-weight);\n  line-height: var(--#{$prefix}btn-line-height);\n  color: var(--#{$prefix}btn-color);\n  text-align: center;\n  text-decoration: if($link-decoration == none, null, none);\n  white-space: $btn-white-space;\n  vertical-align: middle;\n  cursor: if($enable-button-pointers, pointer, null);\n  user-select: none;\n  border: var(--#{$prefix}btn-border-width) solid var(--#{$prefix}btn-border-color);\n  @include border-radius(var(--#{$prefix}btn-border-radius));\n  @include gradient-bg(var(--#{$prefix}btn-bg));\n  @include box-shadow(var(--#{$prefix}btn-box-shadow));\n  @include transition($btn-transition);\n\n  &:hover {\n    color: var(--#{$prefix}btn-hover-color);\n    text-decoration: if($link-hover-decoration == underline, none, null);\n    background-color: var(--#{$prefix}btn-hover-bg);\n    border-color: var(--#{$prefix}btn-hover-border-color);\n  }\n\n  .btn-check + &:hover {\n    // override for the checkbox/radio buttons\n    color: var(--#{$prefix}btn-color);\n    background-color: var(--#{$prefix}btn-bg);\n    border-color: var(--#{$prefix}btn-border-color);\n  }\n\n  &:focus-visible {\n    color: var(--#{$prefix}btn-hover-color);\n    @include gradient-bg(var(--#{$prefix}btn-hover-bg));\n    border-color: var(--#{$prefix}btn-hover-border-color);\n    outline: 0;\n    // Avoid using mixin so we can pass custom focus shadow properly\n    @if $enable-shadows {\n      box-shadow: var(--#{$prefix}btn-box-shadow), var(--#{$prefix}btn-focus-box-shadow);\n    } @else {\n      box-shadow: var(--#{$prefix}btn-focus-box-shadow);\n    }\n  }\n\n  .btn-check:focus-visible + & {\n    border-color: var(--#{$prefix}btn-hover-border-color);\n    outline: 0;\n    // Avoid using mixin so we can pass custom focus shadow properly\n    @if $enable-shadows {\n      box-shadow: var(--#{$prefix}btn-box-shadow), var(--#{$prefix}btn-focus-box-shadow);\n    } @else {\n      box-shadow: var(--#{$prefix}btn-focus-box-shadow);\n    }\n  }\n\n  .btn-check:checked + &,\n  :not(.btn-check) + &:active,\n  &:first-child:active,\n  &.active,\n  &.show {\n    color: var(--#{$prefix}btn-active-color);\n    background-color: var(--#{$prefix}btn-active-bg);\n    // Remove CSS gradients if they're enabled\n    background-image: if($enable-gradients, none, null);\n    border-color: var(--#{$prefix}btn-active-border-color);\n    @include box-shadow(var(--#{$prefix}btn-active-shadow));\n\n    &:focus-visible {\n      // Avoid using mixin so we can pass custom focus shadow properly\n      @if $enable-shadows {\n        box-shadow: var(--#{$prefix}btn-active-shadow), var(--#{$prefix}btn-focus-box-shadow);\n      } @else {\n        box-shadow: var(--#{$prefix}btn-focus-box-shadow);\n      }\n    }\n  }\n\n  &:disabled,\n  &.disabled,\n  fieldset:disabled & {\n    color: var(--#{$prefix}btn-disabled-color);\n    pointer-events: none;\n    background-color: var(--#{$prefix}btn-disabled-bg);\n    background-image: if($enable-gradients, none, null);\n    border-color: var(--#{$prefix}btn-disabled-border-color);\n    opacity: var(--#{$prefix}btn-disabled-opacity);\n    @include box-shadow(none);\n  }\n}\n\n\n//\n// Alternate buttons\n//\n\n// scss-docs-start btn-variant-loops\n@each $color, $value in $theme-colors {\n  .btn-#{$color} {\n    @if $color == \"light\" {\n      @include button-variant(\n        $value,\n        $value,\n        $hover-background: shade-color($value, $btn-hover-bg-shade-amount),\n        $hover-border: shade-color($value, $btn-hover-border-shade-amount),\n        $active-background: shade-color($value, $btn-active-bg-shade-amount),\n        $active-border: shade-color($value, $btn-active-border-shade-amount)\n      );\n    } @else if $color == \"dark\" {\n      @include button-variant(\n        $value,\n        $value,\n        $hover-background: tint-color($value, $btn-hover-bg-tint-amount),\n        $hover-border: tint-color($value, $btn-hover-border-tint-amount),\n        $active-background: tint-color($value, $btn-active-bg-tint-amount),\n        $active-border: tint-color($value, $btn-active-border-tint-amount)\n      );\n    } @else {\n      @include button-variant($value, $value);\n    }\n  }\n}\n\n@each $color, $value in $theme-colors {\n  .btn-outline-#{$color} {\n    @include button-outline-variant($value);\n  }\n}\n// scss-docs-end btn-variant-loops\n\n\n//\n// Link buttons\n//\n\n// Make a button look and behave like a link\n.btn-link {\n  --#{$prefix}btn-font-weight: #{$font-weight-normal};\n  --#{$prefix}btn-color: #{$btn-link-color};\n  --#{$prefix}btn-bg: transparent;\n  --#{$prefix}btn-border-color: transparent;\n  --#{$prefix}btn-hover-color: #{$btn-link-hover-color};\n  --#{$prefix}btn-hover-border-color: transparent;\n  --#{$prefix}btn-active-color: #{$btn-link-hover-color};\n  --#{$prefix}btn-active-border-color: transparent;\n  --#{$prefix}btn-disabled-color: #{$btn-link-disabled-color};\n  --#{$prefix}btn-disabled-border-color: transparent;\n  --#{$prefix}btn-box-shadow: 0 0 0 #000; // Can't use `none` as keyword negates all values when used with multiple shadows\n  --#{$prefix}btn-focus-shadow-rgb: #{$btn-link-focus-shadow-rgb};\n\n  text-decoration: $link-decoration;\n  @if $enable-gradients {\n    background-image: none;\n  }\n\n  &:hover,\n  &:focus-visible {\n    text-decoration: $link-hover-decoration;\n  }\n\n  &:focus-visible {\n    color: var(--#{$prefix}btn-color);\n  }\n\n  &:hover {\n    color: var(--#{$prefix}btn-hover-color);\n  }\n\n  // No need for an active state here\n}\n\n\n//\n// Button Sizes\n//\n\n.btn-lg {\n  @include button-size($btn-padding-y-lg, $btn-padding-x-lg, $btn-font-size-lg, $btn-border-radius-lg);\n}\n\n.btn-sm {\n  @include button-size($btn-padding-y-sm, $btn-padding-x-sm, $btn-font-size-sm, $btn-border-radius-sm);\n}\n","// Button variants\n//\n// Easily pump out default styles, as well as :hover, :focus, :active,\n// and disabled options for all buttons\n\n// scss-docs-start btn-variant-mixin\n@mixin button-variant(\n  $background,\n  $border,\n  $color: color-contrast($background),\n  $hover-background: if($color == $color-contrast-light, shade-color($background, $btn-hover-bg-shade-amount), tint-color($background, $btn-hover-bg-tint-amount)),\n  $hover-border: if($color == $color-contrast-light, shade-color($border, $btn-hover-border-shade-amount), tint-color($border, $btn-hover-border-tint-amount)),\n  $hover-color: color-contrast($hover-background),\n  $active-background: if($color == $color-contrast-light, shade-color($background, $btn-active-bg-shade-amount), tint-color($background, $btn-active-bg-tint-amount)),\n  $active-border: if($color == $color-contrast-light, shade-color($border, $btn-active-border-shade-amount), tint-color($border, $btn-active-border-tint-amount)),\n  $active-color: color-contrast($active-background),\n  $disabled-background: $background,\n  $disabled-border: $border,\n  $disabled-color: color-contrast($disabled-background)\n) {\n  --#{$prefix}btn-color: #{$color};\n  --#{$prefix}btn-bg: #{$background};\n  --#{$prefix}btn-border-color: #{$border};\n  --#{$prefix}btn-hover-color: #{$hover-color};\n  --#{$prefix}btn-hover-bg: #{$hover-background};\n  --#{$prefix}btn-hover-border-color: #{$hover-border};\n  --#{$prefix}btn-focus-shadow-rgb: #{to-rgb(mix($color, $border, 15%))};\n  --#{$prefix}btn-active-color: #{$active-color};\n  --#{$prefix}btn-active-bg: #{$active-background};\n  --#{$prefix}btn-active-border-color: #{$active-border};\n  --#{$prefix}btn-active-shadow: #{$btn-active-box-shadow};\n  --#{$prefix}btn-disabled-color: #{$disabled-color};\n  --#{$prefix}btn-disabled-bg: #{$disabled-background};\n  --#{$prefix}btn-disabled-border-color: #{$disabled-border};\n}\n// scss-docs-end btn-variant-mixin\n\n// scss-docs-start btn-outline-variant-mixin\n@mixin button-outline-variant(\n  $color,\n  $color-hover: color-contrast($color),\n  $active-background: $color,\n  $active-border: $color,\n  $active-color: color-contrast($active-background)\n) {\n  --#{$prefix}btn-color: #{$color};\n  --#{$prefix}btn-border-color: #{$color};\n  --#{$prefix}btn-hover-color: #{$color-hover};\n  --#{$prefix}btn-hover-bg: #{$active-background};\n  --#{$prefix}btn-hover-border-color: #{$active-border};\n  --#{$prefix}btn-focus-shadow-rgb: #{to-rgb($color)};\n  --#{$prefix}btn-active-color: #{$active-color};\n  --#{$prefix}btn-active-bg: #{$active-background};\n  --#{$prefix}btn-active-border-color: #{$active-border};\n  --#{$prefix}btn-active-shadow: #{$btn-active-box-shadow};\n  --#{$prefix}btn-disabled-color: #{$color};\n  --#{$prefix}btn-disabled-bg: transparent;\n  --#{$prefix}btn-disabled-border-color: #{$color};\n  --#{$prefix}gradient: none;\n}\n// scss-docs-end btn-outline-variant-mixin\n\n// scss-docs-start btn-size-mixin\n@mixin button-size($padding-y, $padding-x, $font-size, $border-radius) {\n  --#{$prefix}btn-padding-y: #{$padding-y};\n  --#{$prefix}btn-padding-x: #{$padding-x};\n  @include rfs($font-size, --#{$prefix}btn-font-size);\n  --#{$prefix}btn-border-radius: #{$border-radius};\n}\n// scss-docs-end btn-size-mixin\n",".fade {\n  @include transition($transition-fade);\n\n  &:not(.show) {\n    opacity: 0;\n  }\n}\n\n// scss-docs-start collapse-classes\n.collapse {\n  &:not(.show) {\n    display: none;\n  }\n}\n\n.collapsing {\n  height: 0;\n  overflow: hidden;\n  @include transition($transition-collapse);\n\n  &.collapse-horizontal {\n    width: 0;\n    height: auto;\n    @include transition($transition-collapse-width);\n  }\n}\n// scss-docs-end collapse-classes\n","// The dropdown wrapper (`<div>`)\n.dropup,\n.dropend,\n.dropdown,\n.dropstart,\n.dropup-center,\n.dropdown-center {\n  position: relative;\n}\n\n.dropdown-toggle {\n  white-space: nowrap;\n\n  // Generate the caret automatically\n  @include caret();\n}\n\n// The dropdown menu\n.dropdown-menu {\n  // scss-docs-start dropdown-css-vars\n  --#{$prefix}dropdown-zindex: #{$zindex-dropdown};\n  --#{$prefix}dropdown-min-width: #{$dropdown-min-width};\n  --#{$prefix}dropdown-padding-x: #{$dropdown-padding-x};\n  --#{$prefix}dropdown-padding-y: #{$dropdown-padding-y};\n  --#{$prefix}dropdown-spacer: #{$dropdown-spacer};\n  @include rfs($dropdown-font-size, --#{$prefix}dropdown-font-size);\n  --#{$prefix}dropdown-color: #{$dropdown-color};\n  --#{$prefix}dropdown-bg: #{$dropdown-bg};\n  --#{$prefix}dropdown-border-color: #{$dropdown-border-color};\n  --#{$prefix}dropdown-border-radius: #{$dropdown-border-radius};\n  --#{$prefix}dropdown-border-width: #{$dropdown-border-width};\n  --#{$prefix}dropdown-inner-border-radius: #{$dropdown-inner-border-radius};\n  --#{$prefix}dropdown-divider-bg: #{$dropdown-divider-bg};\n  --#{$prefix}dropdown-divider-margin-y: #{$dropdown-divider-margin-y};\n  --#{$prefix}dropdown-box-shadow: #{$dropdown-box-shadow};\n  --#{$prefix}dropdown-link-color: #{$dropdown-link-color};\n  --#{$prefix}dropdown-link-hover-color: #{$dropdown-link-hover-color};\n  --#{$prefix}dropdown-link-hover-bg: #{$dropdown-link-hover-bg};\n  --#{$prefix}dropdown-link-active-color: #{$dropdown-link-active-color};\n  --#{$prefix}dropdown-link-active-bg: #{$dropdown-link-active-bg};\n  --#{$prefix}dropdown-link-disabled-color: #{$dropdown-link-disabled-color};\n  --#{$prefix}dropdown-item-padding-x: #{$dropdown-item-padding-x};\n  --#{$prefix}dropdown-item-padding-y: #{$dropdown-item-padding-y};\n  --#{$prefix}dropdown-header-color: #{$dropdown-header-color};\n  --#{$prefix}dropdown-header-padding-x: #{$dropdown-header-padding-x};\n  --#{$prefix}dropdown-header-padding-y: #{$dropdown-header-padding-y};\n  // scss-docs-end dropdown-css-vars\n\n  position: absolute;\n  z-index: var(--#{$prefix}dropdown-zindex);\n  display: none; // none by default, but block on \"open\" of the menu\n  min-width: var(--#{$prefix}dropdown-min-width);\n  padding: var(--#{$prefix}dropdown-padding-y) var(--#{$prefix}dropdown-padding-x);\n  margin: 0; // Override default margin of ul\n  @include font-size(var(--#{$prefix}dropdown-font-size));\n  color: var(--#{$prefix}dropdown-color);\n  text-align: left; // Ensures proper alignment if parent has it changed (e.g., modal footer)\n  list-style: none;\n  background-color: var(--#{$prefix}dropdown-bg);\n  background-clip: padding-box;\n  border: var(--#{$prefix}dropdown-border-width) solid var(--#{$prefix}dropdown-border-color);\n  @include border-radius(var(--#{$prefix}dropdown-border-radius));\n  @include box-shadow(var(--#{$prefix}dropdown-box-shadow));\n\n  &[data-bs-popper] {\n    top: 100%;\n    left: 0;\n    margin-top: var(--#{$prefix}dropdown-spacer);\n  }\n\n  @if $dropdown-padding-y == 0 {\n    > .dropdown-item:first-child,\n    > li:first-child .dropdown-item {\n      @include border-top-radius(var(--#{$prefix}dropdown-inner-border-radius));\n    }\n    > .dropdown-item:last-child,\n    > li:last-child .dropdown-item {\n      @include border-bottom-radius(var(--#{$prefix}dropdown-inner-border-radius));\n    }\n\n  }\n}\n\n// scss-docs-start responsive-breakpoints\n// We deliberately hardcode the `bs-` prefix because we check\n// this custom property in JS to determine Popper's positioning\n\n@each $breakpoint in map-keys($grid-breakpoints) {\n  @include media-breakpoint-up($breakpoint) {\n    $infix: breakpoint-infix($breakpoint, $grid-breakpoints);\n\n    .dropdown-menu#{$infix}-start {\n      --bs-position: start;\n\n      &[data-bs-popper] {\n        right: auto;\n        left: 0;\n      }\n    }\n\n    .dropdown-menu#{$infix}-end {\n      --bs-position: end;\n\n      &[data-bs-popper] {\n        right: 0;\n        left: auto;\n      }\n    }\n  }\n}\n// scss-docs-end responsive-breakpoints\n\n// Allow for dropdowns to go bottom up (aka, dropup-menu)\n// Just add .dropup after the standard .dropdown class and you're set.\n.dropup {\n  .dropdown-menu[data-bs-popper] {\n    top: auto;\n    bottom: 100%;\n    margin-top: 0;\n    margin-bottom: var(--#{$prefix}dropdown-spacer);\n  }\n\n  .dropdown-toggle {\n    @include caret(up);\n  }\n}\n\n.dropend {\n  .dropdown-menu[data-bs-popper] {\n    top: 0;\n    right: auto;\n    left: 100%;\n    margin-top: 0;\n    margin-left: var(--#{$prefix}dropdown-spacer);\n  }\n\n  .dropdown-toggle {\n    @include caret(end);\n    &::after {\n      vertical-align: 0;\n    }\n  }\n}\n\n.dropstart {\n  .dropdown-menu[data-bs-popper] {\n    top: 0;\n    right: 100%;\n    left: auto;\n    margin-top: 0;\n    margin-right: var(--#{$prefix}dropdown-spacer);\n  }\n\n  .dropdown-toggle {\n    @include caret(start);\n    &::before {\n      vertical-align: 0;\n    }\n  }\n}\n\n\n// Dividers (basically an `<hr>`) within the dropdown\n.dropdown-divider {\n  height: 0;\n  margin: var(--#{$prefix}dropdown-divider-margin-y) 0;\n  overflow: hidden;\n  border-top: 1px solid var(--#{$prefix}dropdown-divider-bg);\n  opacity: 1; // Revisit in v6 to de-dupe styles that conflict with <hr> element\n}\n\n// Links, buttons, and more within the dropdown menu\n//\n// `<button>`-specific styles are denoted with `// For <button>s`\n.dropdown-item {\n  display: block;\n  width: 100%; // For `<button>`s\n  padding: var(--#{$prefix}dropdown-item-padding-y) var(--#{$prefix}dropdown-item-padding-x);\n  clear: both;\n  font-weight: $font-weight-normal;\n  color: var(--#{$prefix}dropdown-link-color);\n  text-align: inherit; // For `<button>`s\n  text-decoration: if($link-decoration == none, null, none);\n  white-space: nowrap; // prevent links from randomly breaking onto new lines\n  background-color: transparent; // For `<button>`s\n  border: 0; // For `<button>`s\n  @include border-radius(var(--#{$prefix}dropdown-item-border-radius, 0));\n\n  &:hover,\n  &:focus {\n    color: var(--#{$prefix}dropdown-link-hover-color);\n    text-decoration: if($link-hover-decoration == underline, none, null);\n    @include gradient-bg(var(--#{$prefix}dropdown-link-hover-bg));\n  }\n\n  &.active,\n  &:active {\n    color: var(--#{$prefix}dropdown-link-active-color);\n    text-decoration: none;\n    @include gradient-bg(var(--#{$prefix}dropdown-link-active-bg));\n  }\n\n  &.disabled,\n  &:disabled {\n    color: var(--#{$prefix}dropdown-link-disabled-color);\n    pointer-events: none;\n    background-color: transparent;\n    // Remove CSS gradients if they're enabled\n    background-image: if($enable-gradients, none, null);\n  }\n}\n\n.dropdown-menu.show {\n  display: block;\n}\n\n// Dropdown section headers\n.dropdown-header {\n  display: block;\n  padding: var(--#{$prefix}dropdown-header-padding-y) var(--#{$prefix}dropdown-header-padding-x);\n  margin-bottom: 0; // for use with heading elements\n  @include font-size($font-size-sm);\n  color: var(--#{$prefix}dropdown-header-color);\n  white-space: nowrap; // as with > li > a\n}\n\n// Dropdown text\n.dropdown-item-text {\n  display: block;\n  padding: var(--#{$prefix}dropdown-item-padding-y) var(--#{$prefix}dropdown-item-padding-x);\n  color: var(--#{$prefix}dropdown-link-color);\n}\n\n// Dark dropdowns\n.dropdown-menu-dark {\n  // scss-docs-start dropdown-dark-css-vars\n  --#{$prefix}dropdown-color: #{$dropdown-dark-color};\n  --#{$prefix}dropdown-bg: #{$dropdown-dark-bg};\n  --#{$prefix}dropdown-border-color: #{$dropdown-dark-border-color};\n  --#{$prefix}dropdown-box-shadow: #{$dropdown-dark-box-shadow};\n  --#{$prefix}dropdown-link-color: #{$dropdown-dark-link-color};\n  --#{$prefix}dropdown-link-hover-color: #{$dropdown-dark-link-hover-color};\n  --#{$prefix}dropdown-divider-bg: #{$dropdown-dark-divider-bg};\n  --#{$prefix}dropdown-link-hover-bg: #{$dropdown-dark-link-hover-bg};\n  --#{$prefix}dropdown-link-active-color: #{$dropdown-dark-link-active-color};\n  --#{$prefix}dropdown-link-active-bg: #{$dropdown-dark-link-active-bg};\n  --#{$prefix}dropdown-link-disabled-color: #{$dropdown-dark-link-disabled-color};\n  --#{$prefix}dropdown-header-color: #{$dropdown-dark-header-color};\n  // scss-docs-end dropdown-dark-css-vars\n}\n","// scss-docs-start caret-mixins\n@mixin caret-down($width: $caret-width) {\n  border-top: $width solid;\n  border-right: $width solid transparent;\n  border-bottom: 0;\n  border-left: $width solid transparent;\n}\n\n@mixin caret-up($width: $caret-width) {\n  border-top: 0;\n  border-right: $width solid transparent;\n  border-bottom: $width solid;\n  border-left: $width solid transparent;\n}\n\n@mixin caret-end($width: $caret-width) {\n  border-top: $width solid transparent;\n  border-right: 0;\n  border-bottom: $width solid transparent;\n  border-left: $width solid;\n}\n\n@mixin caret-start($width: $caret-width) {\n  border-top: $width solid transparent;\n  border-right: $width solid;\n  border-bottom: $width solid transparent;\n}\n\n@mixin caret(\n  $direction: down,\n  $width: $caret-width,\n  $spacing: $caret-spacing,\n  $vertical-align: $caret-vertical-align\n) {\n  @if $enable-caret {\n    &::after {\n      display: inline-block;\n      margin-left: $spacing;\n      vertical-align: $vertical-align;\n      content: \"\";\n      @if $direction == down {\n        @include caret-down($width);\n      } @else if $direction == up {\n        @include caret-up($width);\n      } @else if $direction == end {\n        @include caret-end($width);\n      }\n    }\n\n    @if $direction == start {\n      &::after {\n        display: none;\n      }\n\n      &::before {\n        display: inline-block;\n        margin-right: $spacing;\n        vertical-align: $vertical-align;\n        content: \"\";\n        @include caret-start($width);\n      }\n    }\n\n    &:empty::after {\n      margin-left: 0;\n    }\n  }\n}\n// scss-docs-end caret-mixins\n","// Make the div behave like a button\n.btn-group,\n.btn-group-vertical {\n  position: relative;\n  display: inline-flex;\n  vertical-align: middle; // match .btn alignment given font-size hack above\n\n  > .btn {\n    position: relative;\n    flex: 1 1 auto;\n  }\n\n  // Bring the hover, focused, and \"active\" buttons to the front to overlay\n  // the borders properly\n  > .btn-check:checked + .btn,\n  > .btn-check:focus + .btn,\n  > .btn:hover,\n  > .btn:focus,\n  > .btn:active,\n  > .btn.active {\n    z-index: 1;\n  }\n}\n\n// Optional: Group multiple button groups together for a toolbar\n.btn-toolbar {\n  display: flex;\n  flex-wrap: wrap;\n  justify-content: flex-start;\n\n  .input-group {\n    width: auto;\n  }\n}\n\n.btn-group {\n  @include border-radius($btn-border-radius);\n\n  // Prevent double borders when buttons are next to each other\n  > :not(.btn-check:first-child) + .btn,\n  > .btn-group:not(:first-child) {\n    margin-left: calc(#{$btn-border-width} * -1); // stylelint-disable-line function-disallowed-list\n  }\n\n  // Reset rounded corners\n  > .btn:not(:last-child):not(.dropdown-toggle),\n  > .btn.dropdown-toggle-split:first-child,\n  > .btn-group:not(:last-child) > .btn {\n    @include border-end-radius(0);\n  }\n\n  // The left radius should be 0 if the button is:\n  // - the \"third or more\" child\n  // - the second child and the previous element isn't `.btn-check` (making it the first child visually)\n  // - part of a btn-group which isn't the first child\n  > .btn:nth-child(n + 3),\n  > :not(.btn-check) + .btn,\n  > .btn-group:not(:first-child) > .btn {\n    @include border-start-radius(0);\n  }\n}\n\n// Sizing\n//\n// Remix the default button sizing classes into new ones for easier manipulation.\n\n.btn-group-sm > .btn { @extend .btn-sm; }\n.btn-group-lg > .btn { @extend .btn-lg; }\n\n\n//\n// Split button dropdowns\n//\n\n.dropdown-toggle-split {\n  padding-right: $btn-padding-x * .75;\n  padding-left: $btn-padding-x * .75;\n\n  &::after,\n  .dropup &::after,\n  .dropend &::after {\n    margin-left: 0;\n  }\n\n  .dropstart &::before {\n    margin-right: 0;\n  }\n}\n\n.btn-sm + .dropdown-toggle-split {\n  padding-right: $btn-padding-x-sm * .75;\n  padding-left: $btn-padding-x-sm * .75;\n}\n\n.btn-lg + .dropdown-toggle-split {\n  padding-right: $btn-padding-x-lg * .75;\n  padding-left: $btn-padding-x-lg * .75;\n}\n\n\n// The clickable button for toggling the menu\n// Set the same inset shadow as the :active state\n.btn-group.show .dropdown-toggle {\n  @include box-shadow($btn-active-box-shadow);\n\n  // Show no shadow for `.btn-link` since it has no other button styles.\n  &.btn-link {\n    @include box-shadow(none);\n  }\n}\n\n\n//\n// Vertical button groups\n//\n\n.btn-group-vertical {\n  flex-direction: column;\n  align-items: flex-start;\n  justify-content: center;\n\n  > .btn,\n  > .btn-group {\n    width: 100%;\n  }\n\n  > .btn:not(:first-child),\n  > .btn-group:not(:first-child) {\n    margin-top: calc(#{$btn-border-width} * -1); // stylelint-disable-line function-disallowed-list\n  }\n\n  // Reset rounded corners\n  > .btn:not(:last-child):not(.dropdown-toggle),\n  > .btn-group:not(:last-child) > .btn {\n    @include border-bottom-radius(0);\n  }\n\n  > .btn ~ .btn,\n  > .btn-group:not(:first-child) > .btn {\n    @include border-top-radius(0);\n  }\n}\n","// Base class\n//\n// Kickstart any navigation component with a set of style resets. Works with\n// `<nav>`s, `<ul>`s or `<ol>`s.\n\n.nav {\n  // scss-docs-start nav-css-vars\n  --#{$prefix}nav-link-padding-x: #{$nav-link-padding-x};\n  --#{$prefix}nav-link-padding-y: #{$nav-link-padding-y};\n  @include rfs($nav-link-font-size, --#{$prefix}nav-link-font-size);\n  --#{$prefix}nav-link-font-weight: #{$nav-link-font-weight};\n  --#{$prefix}nav-link-color: #{$nav-link-color};\n  --#{$prefix}nav-link-hover-color: #{$nav-link-hover-color};\n  --#{$prefix}nav-link-disabled-color: #{$nav-link-disabled-color};\n  // scss-docs-end nav-css-vars\n\n  display: flex;\n  flex-wrap: wrap;\n  padding-left: 0;\n  margin-bottom: 0;\n  list-style: none;\n}\n\n.nav-link {\n  display: block;\n  padding: var(--#{$prefix}nav-link-padding-y) var(--#{$prefix}nav-link-padding-x);\n  @include font-size(var(--#{$prefix}nav-link-font-size));\n  font-weight: var(--#{$prefix}nav-link-font-weight);\n  color: var(--#{$prefix}nav-link-color);\n  text-decoration: if($link-decoration == none, null, none);\n  background: none;\n  border: 0;\n  @include transition($nav-link-transition);\n\n  &:hover,\n  &:focus {\n    color: var(--#{$prefix}nav-link-hover-color);\n    text-decoration: if($link-hover-decoration == underline, none, null);\n  }\n\n  &:focus-visible {\n    outline: 0;\n    box-shadow: $nav-link-focus-box-shadow;\n  }\n\n  // Disabled state lightens text\n  &.disabled,\n  &:disabled {\n    color: var(--#{$prefix}nav-link-disabled-color);\n    pointer-events: none;\n    cursor: default;\n  }\n}\n\n//\n// Tabs\n//\n\n.nav-tabs {\n  // scss-docs-start nav-tabs-css-vars\n  --#{$prefix}nav-tabs-border-width: #{$nav-tabs-border-width};\n  --#{$prefix}nav-tabs-border-color: #{$nav-tabs-border-color};\n  --#{$prefix}nav-tabs-border-radius: #{$nav-tabs-border-radius};\n  --#{$prefix}nav-tabs-link-hover-border-color: #{$nav-tabs-link-hover-border-color};\n  --#{$prefix}nav-tabs-link-active-color: #{$nav-tabs-link-active-color};\n  --#{$prefix}nav-tabs-link-active-bg: #{$nav-tabs-link-active-bg};\n  --#{$prefix}nav-tabs-link-active-border-color: #{$nav-tabs-link-active-border-color};\n  // scss-docs-end nav-tabs-css-vars\n\n  border-bottom: var(--#{$prefix}nav-tabs-border-width) solid var(--#{$prefix}nav-tabs-border-color);\n\n  .nav-link {\n    margin-bottom: calc(-1 * var(--#{$prefix}nav-tabs-border-width)); // stylelint-disable-line function-disallowed-list\n    border: var(--#{$prefix}nav-tabs-border-width) solid transparent;\n    @include border-top-radius(var(--#{$prefix}nav-tabs-border-radius));\n\n    &:hover,\n    &:focus {\n      // Prevents active .nav-link tab overlapping focus outline of previous/next .nav-link\n      isolation: isolate;\n      border-color: var(--#{$prefix}nav-tabs-link-hover-border-color);\n    }\n  }\n\n  .nav-link.active,\n  .nav-item.show .nav-link {\n    color: var(--#{$prefix}nav-tabs-link-active-color);\n    background-color: var(--#{$prefix}nav-tabs-link-active-bg);\n    border-color: var(--#{$prefix}nav-tabs-link-active-border-color);\n  }\n\n  .dropdown-menu {\n    // Make dropdown border overlap tab border\n    margin-top: calc(-1 * var(--#{$prefix}nav-tabs-border-width)); // stylelint-disable-line function-disallowed-list\n    // Remove the top rounded corners here since there is a hard edge above the menu\n    @include border-top-radius(0);\n  }\n}\n\n\n//\n// Pills\n//\n\n.nav-pills {\n  // scss-docs-start nav-pills-css-vars\n  --#{$prefix}nav-pills-border-radius: #{$nav-pills-border-radius};\n  --#{$prefix}nav-pills-link-active-color: #{$nav-pills-link-active-color};\n  --#{$prefix}nav-pills-link-active-bg: #{$nav-pills-link-active-bg};\n  // scss-docs-end nav-pills-css-vars\n\n  .nav-link {\n    @include border-radius(var(--#{$prefix}nav-pills-border-radius));\n  }\n\n  .nav-link.active,\n  .show > .nav-link {\n    color: var(--#{$prefix}nav-pills-link-active-color);\n    @include gradient-bg(var(--#{$prefix}nav-pills-link-active-bg));\n  }\n}\n\n\n//\n// Underline\n//\n\n.nav-underline {\n  // scss-docs-start nav-underline-css-vars\n  --#{$prefix}nav-underline-gap: #{$nav-underline-gap};\n  --#{$prefix}nav-underline-border-width: #{$nav-underline-border-width};\n  --#{$prefix}nav-underline-link-active-color: #{$nav-underline-link-active-color};\n  // scss-docs-end nav-underline-css-vars\n\n  gap: var(--#{$prefix}nav-underline-gap);\n\n  .nav-link {\n    padding-right: 0;\n    padding-left: 0;\n    border-bottom: var(--#{$prefix}nav-underline-border-width) solid transparent;\n\n    &:hover,\n    &:focus {\n      border-bottom-color: currentcolor;\n    }\n  }\n\n  .nav-link.active,\n  .show > .nav-link {\n    font-weight: $font-weight-bold;\n    color: var(--#{$prefix}nav-underline-link-active-color);\n    border-bottom-color: currentcolor;\n  }\n}\n\n\n//\n// Justified variants\n//\n\n.nav-fill {\n  > .nav-link,\n  .nav-item {\n    flex: 1 1 auto;\n    text-align: center;\n  }\n}\n\n.nav-justified {\n  > .nav-link,\n  .nav-item {\n    flex-basis: 0;\n    flex-grow: 1;\n    text-align: center;\n  }\n}\n\n.nav-fill,\n.nav-justified {\n  .nav-item .nav-link {\n    width: 100%; // Make sure button will grow\n  }\n}\n\n\n// Tabbable tabs\n//\n// Hide tabbable panes to start, show them when `.active`\n\n.tab-content {\n  > .tab-pane {\n    display: none;\n  }\n  > .active {\n    display: block;\n  }\n}\n","// Navbar\n//\n// Provide a static navbar from which we expand to create full-width, fixed, and\n// other navbar variations.\n\n.navbar {\n  // scss-docs-start navbar-css-vars\n  --#{$prefix}navbar-padding-x: #{if($navbar-padding-x == null, 0, $navbar-padding-x)};\n  --#{$prefix}navbar-padding-y: #{$navbar-padding-y};\n  --#{$prefix}navbar-color: #{$navbar-light-color};\n  --#{$prefix}navbar-hover-color: #{$navbar-light-hover-color};\n  --#{$prefix}navbar-disabled-color: #{$navbar-light-disabled-color};\n  --#{$prefix}navbar-active-color: #{$navbar-light-active-color};\n  --#{$prefix}navbar-brand-padding-y: #{$navbar-brand-padding-y};\n  --#{$prefix}navbar-brand-margin-end: #{$navbar-brand-margin-end};\n  --#{$prefix}navbar-brand-font-size: #{$navbar-brand-font-size};\n  --#{$prefix}navbar-brand-color: #{$navbar-light-brand-color};\n  --#{$prefix}navbar-brand-hover-color: #{$navbar-light-brand-hover-color};\n  --#{$prefix}navbar-nav-link-padding-x: #{$navbar-nav-link-padding-x};\n  --#{$prefix}navbar-toggler-padding-y: #{$navbar-toggler-padding-y};\n  --#{$prefix}navbar-toggler-padding-x: #{$navbar-toggler-padding-x};\n  --#{$prefix}navbar-toggler-font-size: #{$navbar-toggler-font-size};\n  --#{$prefix}navbar-toggler-icon-bg: #{escape-svg($navbar-light-toggler-icon-bg)};\n  --#{$prefix}navbar-toggler-border-color: #{$navbar-light-toggler-border-color};\n  --#{$prefix}navbar-toggler-border-radius: #{$navbar-toggler-border-radius};\n  --#{$prefix}navbar-toggler-focus-width: #{$navbar-toggler-focus-width};\n  --#{$prefix}navbar-toggler-transition: #{$navbar-toggler-transition};\n  // scss-docs-end navbar-css-vars\n\n  position: relative;\n  display: flex;\n  flex-wrap: wrap; // allow us to do the line break for collapsing content\n  align-items: center;\n  justify-content: space-between; // space out brand from logo\n  padding: var(--#{$prefix}navbar-padding-y) var(--#{$prefix}navbar-padding-x);\n  @include gradient-bg();\n\n  // Because flex properties aren't inherited, we need to redeclare these first\n  // few properties so that content nested within behave properly.\n  // The `flex-wrap` property is inherited to simplify the expanded navbars\n  %container-flex-properties {\n    display: flex;\n    flex-wrap: inherit;\n    align-items: center;\n    justify-content: space-between;\n  }\n\n  > .container,\n  > .container-fluid {\n    @extend %container-flex-properties;\n  }\n\n  @each $breakpoint, $container-max-width in $container-max-widths {\n    > .container#{breakpoint-infix($breakpoint, $container-max-widths)} {\n      @extend %container-flex-properties;\n    }\n  }\n}\n\n\n// Navbar brand\n//\n// Used for brand, project, or site names.\n\n.navbar-brand {\n  padding-top: var(--#{$prefix}navbar-brand-padding-y);\n  padding-bottom: var(--#{$prefix}navbar-brand-padding-y);\n  margin-right: var(--#{$prefix}navbar-brand-margin-end);\n  @include font-size(var(--#{$prefix}navbar-brand-font-size));\n  color: var(--#{$prefix}navbar-brand-color);\n  text-decoration: if($link-decoration == none, null, none);\n  white-space: nowrap;\n\n  &:hover,\n  &:focus {\n    color: var(--#{$prefix}navbar-brand-hover-color);\n    text-decoration: if($link-hover-decoration == underline, none, null);\n  }\n}\n\n\n// Navbar nav\n//\n// Custom navbar navigation (doesn't require `.nav`, but does make use of `.nav-link`).\n\n.navbar-nav {\n  // scss-docs-start navbar-nav-css-vars\n  --#{$prefix}nav-link-padding-x: 0;\n  --#{$prefix}nav-link-padding-y: #{$nav-link-padding-y};\n  @include rfs($nav-link-font-size, --#{$prefix}nav-link-font-size);\n  --#{$prefix}nav-link-font-weight: #{$nav-link-font-weight};\n  --#{$prefix}nav-link-color: var(--#{$prefix}navbar-color);\n  --#{$prefix}nav-link-hover-color: var(--#{$prefix}navbar-hover-color);\n  --#{$prefix}nav-link-disabled-color: var(--#{$prefix}navbar-disabled-color);\n  // scss-docs-end navbar-nav-css-vars\n\n  display: flex;\n  flex-direction: column; // cannot use `inherit` to get the `.navbar`s value\n  padding-left: 0;\n  margin-bottom: 0;\n  list-style: none;\n\n  .nav-link {\n    &.active,\n    &.show {\n      color: var(--#{$prefix}navbar-active-color);\n    }\n  }\n\n  .dropdown-menu {\n    position: static;\n  }\n}\n\n\n// Navbar text\n//\n//\n\n.navbar-text {\n  padding-top: $nav-link-padding-y;\n  padding-bottom: $nav-link-padding-y;\n  color: var(--#{$prefix}navbar-color);\n\n  a,\n  a:hover,\n  a:focus  {\n    color: var(--#{$prefix}navbar-active-color);\n  }\n}\n\n\n// Responsive navbar\n//\n// Custom styles for responsive collapsing and toggling of navbar contents.\n// Powered by the collapse Bootstrap JavaScript plugin.\n\n// When collapsed, prevent the toggleable navbar contents from appearing in\n// the default flexbox row orientation. Requires the use of `flex-wrap: wrap`\n// on the `.navbar` parent.\n.navbar-collapse {\n  flex-basis: 100%;\n  flex-grow: 1;\n  // For always expanded or extra full navbars, ensure content aligns itself\n  // properly vertically. Can be easily overridden with flex utilities.\n  align-items: center;\n}\n\n// Button for toggling the navbar when in its collapsed state\n.navbar-toggler {\n  padding: var(--#{$prefix}navbar-toggler-padding-y) var(--#{$prefix}navbar-toggler-padding-x);\n  @include font-size(var(--#{$prefix}navbar-toggler-font-size));\n  line-height: 1;\n  color: var(--#{$prefix}navbar-color);\n  background-color: transparent; // remove default button style\n  border: var(--#{$prefix}border-width) solid var(--#{$prefix}navbar-toggler-border-color); // remove default button style\n  @include border-radius(var(--#{$prefix}navbar-toggler-border-radius));\n  @include transition(var(--#{$prefix}navbar-toggler-transition));\n\n  &:hover {\n    text-decoration: none;\n  }\n\n  &:focus {\n    text-decoration: none;\n    outline: 0;\n    box-shadow: 0 0 0 var(--#{$prefix}navbar-toggler-focus-width);\n  }\n}\n\n// Keep as a separate element so folks can easily override it with another icon\n// or image file as needed.\n.navbar-toggler-icon {\n  display: inline-block;\n  width: 1.5em;\n  height: 1.5em;\n  vertical-align: middle;\n  background-image: var(--#{$prefix}navbar-toggler-icon-bg);\n  background-repeat: no-repeat;\n  background-position: center;\n  background-size: 100%;\n}\n\n.navbar-nav-scroll {\n  max-height: var(--#{$prefix}scroll-height, 75vh);\n  overflow-y: auto;\n}\n\n// scss-docs-start navbar-expand-loop\n// Generate series of `.navbar-expand-*` responsive classes for configuring\n// where your navbar collapses.\n.navbar-expand {\n  @each $breakpoint in map-keys($grid-breakpoints) {\n    $next: breakpoint-next($breakpoint, $grid-breakpoints);\n    $infix: breakpoint-infix($next, $grid-breakpoints);\n\n    // stylelint-disable-next-line scss/selector-no-union-class-name\n    &#{$infix} {\n      @include media-breakpoint-up($next) {\n        flex-wrap: nowrap;\n        justify-content: flex-start;\n\n        .navbar-nav {\n          flex-direction: row;\n\n          .dropdown-menu {\n            position: absolute;\n          }\n\n          .nav-link {\n            padding-right: var(--#{$prefix}navbar-nav-link-padding-x);\n            padding-left: var(--#{$prefix}navbar-nav-link-padding-x);\n          }\n        }\n\n        .navbar-nav-scroll {\n          overflow: visible;\n        }\n\n        .navbar-collapse {\n          display: flex !important; // stylelint-disable-line declaration-no-important\n          flex-basis: auto;\n        }\n\n        .navbar-toggler {\n          display: none;\n        }\n\n        .offcanvas {\n          // stylelint-disable declaration-no-important\n          position: static;\n          z-index: auto;\n          flex-grow: 1;\n          width: auto !important;\n          height: auto !important;\n          visibility: visible !important;\n          background-color: transparent !important;\n          border: 0 !important;\n          transform: none !important;\n          @include box-shadow(none);\n          @include transition(none);\n          // stylelint-enable declaration-no-important\n\n          .offcanvas-header {\n            display: none;\n          }\n\n          .offcanvas-body {\n            display: flex;\n            flex-grow: 0;\n            padding: 0;\n            overflow-y: visible;\n          }\n        }\n      }\n    }\n  }\n}\n// scss-docs-end navbar-expand-loop\n\n// Navbar themes\n//\n// Styles for switching between navbars with light or dark background.\n\n.navbar-light {\n  @include deprecate(\"`.navbar-light`\", \"v5.2.0\", \"v6.0.0\", true);\n}\n\n.navbar-dark,\n.navbar[data-bs-theme=\"dark\"] {\n  // scss-docs-start navbar-dark-css-vars\n  --#{$prefix}navbar-color: #{$navbar-dark-color};\n  --#{$prefix}navbar-hover-color: #{$navbar-dark-hover-color};\n  --#{$prefix}navbar-disabled-color: #{$navbar-dark-disabled-color};\n  --#{$prefix}navbar-active-color: #{$navbar-dark-active-color};\n  --#{$prefix}navbar-brand-color: #{$navbar-dark-brand-color};\n  --#{$prefix}navbar-brand-hover-color: #{$navbar-dark-brand-hover-color};\n  --#{$prefix}navbar-toggler-border-color: #{$navbar-dark-toggler-border-color};\n  --#{$prefix}navbar-toggler-icon-bg: #{escape-svg($navbar-dark-toggler-icon-bg)};\n  // scss-docs-end navbar-dark-css-vars\n}\n\n@if $enable-dark-mode {\n  @include color-mode(dark) {\n    .navbar-toggler-icon {\n      --#{$prefix}navbar-toggler-icon-bg: #{escape-svg($navbar-dark-toggler-icon-bg)};\n    }\n  }\n}\n","//\n// Base styles\n//\n\n.card {\n  // scss-docs-start card-css-vars\n  --#{$prefix}card-spacer-y: #{$card-spacer-y};\n  --#{$prefix}card-spacer-x: #{$card-spacer-x};\n  --#{$prefix}card-title-spacer-y: #{$card-title-spacer-y};\n  --#{$prefix}card-title-color: #{$card-title-color};\n  --#{$prefix}card-subtitle-color: #{$card-subtitle-color};\n  --#{$prefix}card-border-width: #{$card-border-width};\n  --#{$prefix}card-border-color: #{$card-border-color};\n  --#{$prefix}card-border-radius: #{$card-border-radius};\n  --#{$prefix}card-box-shadow: #{$card-box-shadow};\n  --#{$prefix}card-inner-border-radius: #{$card-inner-border-radius};\n  --#{$prefix}card-cap-padding-y: #{$card-cap-padding-y};\n  --#{$prefix}card-cap-padding-x: #{$card-cap-padding-x};\n  --#{$prefix}card-cap-bg: #{$card-cap-bg};\n  --#{$prefix}card-cap-color: #{$card-cap-color};\n  --#{$prefix}card-height: #{$card-height};\n  --#{$prefix}card-color: #{$card-color};\n  --#{$prefix}card-bg: #{$card-bg};\n  --#{$prefix}card-img-overlay-padding: #{$card-img-overlay-padding};\n  --#{$prefix}card-group-margin: #{$card-group-margin};\n  // scss-docs-end card-css-vars\n\n  position: relative;\n  display: flex;\n  flex-direction: column;\n  min-width: 0; // See https://github.com/twbs/bootstrap/pull/22740#issuecomment-305868106\n  height: var(--#{$prefix}card-height);\n  color: var(--#{$prefix}body-color);\n  word-wrap: break-word;\n  background-color: var(--#{$prefix}card-bg);\n  background-clip: border-box;\n  border: var(--#{$prefix}card-border-width) solid var(--#{$prefix}card-border-color);\n  @include border-radius(var(--#{$prefix}card-border-radius));\n  @include box-shadow(var(--#{$prefix}card-box-shadow));\n\n  > hr {\n    margin-right: 0;\n    margin-left: 0;\n  }\n\n  > .list-group {\n    border-top: inherit;\n    border-bottom: inherit;\n\n    &:first-child {\n      border-top-width: 0;\n      @include border-top-radius(var(--#{$prefix}card-inner-border-radius));\n    }\n\n    &:last-child  {\n      border-bottom-width: 0;\n      @include border-bottom-radius(var(--#{$prefix}card-inner-border-radius));\n    }\n  }\n\n  // Due to specificity of the above selector (`.card > .list-group`), we must\n  // use a child selector here to prevent double borders.\n  > .card-header + .list-group,\n  > .list-group + .card-footer {\n    border-top: 0;\n  }\n}\n\n.card-body {\n  // Enable `flex-grow: 1` for decks and groups so that card blocks take up\n  // as much space as possible, ensuring footers are aligned to the bottom.\n  flex: 1 1 auto;\n  padding: var(--#{$prefix}card-spacer-y) var(--#{$prefix}card-spacer-x);\n  color: var(--#{$prefix}card-color);\n}\n\n.card-title {\n  margin-bottom: var(--#{$prefix}card-title-spacer-y);\n  color: var(--#{$prefix}card-title-color);\n}\n\n.card-subtitle {\n  margin-top: calc(-.5 * var(--#{$prefix}card-title-spacer-y)); // stylelint-disable-line function-disallowed-list\n  margin-bottom: 0;\n  color: var(--#{$prefix}card-subtitle-color);\n}\n\n.card-text:last-child {\n  margin-bottom: 0;\n}\n\n.card-link {\n  &:hover {\n    text-decoration: if($link-hover-decoration == underline, none, null);\n  }\n\n  + .card-link {\n    margin-left: var(--#{$prefix}card-spacer-x);\n  }\n}\n\n//\n// Optional textual caps\n//\n\n.card-header {\n  padding: var(--#{$prefix}card-cap-padding-y) var(--#{$prefix}card-cap-padding-x);\n  margin-bottom: 0; // Removes the default margin-bottom of <hN>\n  color: var(--#{$prefix}card-cap-color);\n  background-color: var(--#{$prefix}card-cap-bg);\n  border-bottom: var(--#{$prefix}card-border-width) solid var(--#{$prefix}card-border-color);\n\n  &:first-child {\n    @include border-radius(var(--#{$prefix}card-inner-border-radius) var(--#{$prefix}card-inner-border-radius) 0 0);\n  }\n}\n\n.card-footer {\n  padding: var(--#{$prefix}card-cap-padding-y) var(--#{$prefix}card-cap-padding-x);\n  color: var(--#{$prefix}card-cap-color);\n  background-color: var(--#{$prefix}card-cap-bg);\n  border-top: var(--#{$prefix}card-border-width) solid var(--#{$prefix}card-border-color);\n\n  &:last-child {\n    @include border-radius(0 0 var(--#{$prefix}card-inner-border-radius) var(--#{$prefix}card-inner-border-radius));\n  }\n}\n\n\n//\n// Header navs\n//\n\n.card-header-tabs {\n  margin-right: calc(-.5 * var(--#{$prefix}card-cap-padding-x)); // stylelint-disable-line function-disallowed-list\n  margin-bottom: calc(-1 * var(--#{$prefix}card-cap-padding-y)); // stylelint-disable-line function-disallowed-list\n  margin-left: calc(-.5 * var(--#{$prefix}card-cap-padding-x)); // stylelint-disable-line function-disallowed-list\n  border-bottom: 0;\n\n  .nav-link.active {\n    background-color: var(--#{$prefix}card-bg);\n    border-bottom-color: var(--#{$prefix}card-bg);\n  }\n}\n\n.card-header-pills {\n  margin-right: calc(-.5 * var(--#{$prefix}card-cap-padding-x)); // stylelint-disable-line function-disallowed-list\n  margin-left: calc(-.5 * var(--#{$prefix}card-cap-padding-x)); // stylelint-disable-line function-disallowed-list\n}\n\n// Card image\n.card-img-overlay {\n  position: absolute;\n  top: 0;\n  right: 0;\n  bottom: 0;\n  left: 0;\n  padding: var(--#{$prefix}card-img-overlay-padding);\n  @include border-radius(var(--#{$prefix}card-inner-border-radius));\n}\n\n.card-img,\n.card-img-top,\n.card-img-bottom {\n  width: 100%; // Required because we use flexbox and this inherently applies align-self: stretch\n}\n\n.card-img,\n.card-img-top {\n  @include border-top-radius(var(--#{$prefix}card-inner-border-radius));\n}\n\n.card-img,\n.card-img-bottom {\n  @include border-bottom-radius(var(--#{$prefix}card-inner-border-radius));\n}\n\n\n//\n// Card groups\n//\n\n.card-group {\n  // The child selector allows nested `.card` within `.card-group`\n  // to display properly.\n  > .card {\n    margin-bottom: var(--#{$prefix}card-group-margin);\n  }\n\n  @include media-breakpoint-up(sm) {\n    display: flex;\n    flex-flow: row wrap;\n    // The child selector allows nested `.card` within `.card-group`\n    // to display properly.\n    > .card {\n      // Flexbugs #4: https://github.com/philipwalton/flexbugs#flexbug-4\n      flex: 1 0 0%;\n      margin-bottom: 0;\n\n      + .card {\n        margin-left: 0;\n        border-left: 0;\n      }\n\n      // Handle rounded corners\n      @if $enable-rounded {\n        &:not(:last-child) {\n          @include border-end-radius(0);\n\n          .card-img-top,\n          .card-header {\n            // stylelint-disable-next-line property-disallowed-list\n            border-top-right-radius: 0;\n          }\n          .card-img-bottom,\n          .card-footer {\n            // stylelint-disable-next-line property-disallowed-list\n            border-bottom-right-radius: 0;\n          }\n        }\n\n        &:not(:first-child) {\n          @include border-start-radius(0);\n\n          .card-img-top,\n          .card-header {\n            // stylelint-disable-next-line property-disallowed-list\n            border-top-left-radius: 0;\n          }\n          .card-img-bottom,\n          .card-footer {\n            // stylelint-disable-next-line property-disallowed-list\n            border-bottom-left-radius: 0;\n          }\n        }\n      }\n    }\n  }\n}\n","//\n// Base styles\n//\n\n.accordion {\n  // scss-docs-start accordion-css-vars\n  --#{$prefix}accordion-color: #{$accordion-color};\n  --#{$prefix}accordion-bg: #{$accordion-bg};\n  --#{$prefix}accordion-transition: #{$accordion-transition};\n  --#{$prefix}accordion-border-color: #{$accordion-border-color};\n  --#{$prefix}accordion-border-width: #{$accordion-border-width};\n  --#{$prefix}accordion-border-radius: #{$accordion-border-radius};\n  --#{$prefix}accordion-inner-border-radius: #{$accordion-inner-border-radius};\n  --#{$prefix}accordion-btn-padding-x: #{$accordion-button-padding-x};\n  --#{$prefix}accordion-btn-padding-y: #{$accordion-button-padding-y};\n  --#{$prefix}accordion-btn-color: #{$accordion-button-color};\n  --#{$prefix}accordion-btn-bg: #{$accordion-button-bg};\n  --#{$prefix}accordion-btn-icon: #{escape-svg($accordion-button-icon)};\n  --#{$prefix}accordion-btn-icon-width: #{$accordion-icon-width};\n  --#{$prefix}accordion-btn-icon-transform: #{$accordion-icon-transform};\n  --#{$prefix}accordion-btn-icon-transition: #{$accordion-icon-transition};\n  --#{$prefix}accordion-btn-active-icon: #{escape-svg($accordion-button-active-icon)};\n  --#{$prefix}accordion-btn-focus-border-color: #{$accordion-button-focus-border-color};\n  --#{$prefix}accordion-btn-focus-box-shadow: #{$accordion-button-focus-box-shadow};\n  --#{$prefix}accordion-body-padding-x: #{$accordion-body-padding-x};\n  --#{$prefix}accordion-body-padding-y: #{$accordion-body-padding-y};\n  --#{$prefix}accordion-active-color: #{$accordion-button-active-color};\n  --#{$prefix}accordion-active-bg: #{$accordion-button-active-bg};\n  // scss-docs-end accordion-css-vars\n}\n\n.accordion-button {\n  position: relative;\n  display: flex;\n  align-items: center;\n  width: 100%;\n  padding: var(--#{$prefix}accordion-btn-padding-y) var(--#{$prefix}accordion-btn-padding-x);\n  @include font-size($font-size-base);\n  color: var(--#{$prefix}accordion-btn-color);\n  text-align: left; // Reset button style\n  background-color: var(--#{$prefix}accordion-btn-bg);\n  border: 0;\n  @include border-radius(0);\n  overflow-anchor: none;\n  @include transition(var(--#{$prefix}accordion-transition));\n\n  &:not(.collapsed) {\n    color: var(--#{$prefix}accordion-active-color);\n    background-color: var(--#{$prefix}accordion-active-bg);\n    box-shadow: inset 0 calc(-1 * var(--#{$prefix}accordion-border-width)) 0 var(--#{$prefix}accordion-border-color); // stylelint-disable-line function-disallowed-list\n\n    &::after {\n      background-image: var(--#{$prefix}accordion-btn-active-icon);\n      transform: var(--#{$prefix}accordion-btn-icon-transform);\n    }\n  }\n\n  // Accordion icon\n  &::after {\n    flex-shrink: 0;\n    width: var(--#{$prefix}accordion-btn-icon-width);\n    height: var(--#{$prefix}accordion-btn-icon-width);\n    margin-left: auto;\n    content: \"\";\n    background-image: var(--#{$prefix}accordion-btn-icon);\n    background-repeat: no-repeat;\n    background-size: var(--#{$prefix}accordion-btn-icon-width);\n    @include transition(var(--#{$prefix}accordion-btn-icon-transition));\n  }\n\n  &:hover {\n    z-index: 2;\n  }\n\n  &:focus {\n    z-index: 3;\n    border-color: var(--#{$prefix}accordion-btn-focus-border-color);\n    outline: 0;\n    box-shadow: var(--#{$prefix}accordion-btn-focus-box-shadow);\n  }\n}\n\n.accordion-header {\n  margin-bottom: 0;\n}\n\n.accordion-item {\n  color: var(--#{$prefix}accordion-color);\n  background-color: var(--#{$prefix}accordion-bg);\n  border: var(--#{$prefix}accordion-border-width) solid var(--#{$prefix}accordion-border-color);\n\n  &:first-of-type {\n    @include border-top-radius(var(--#{$prefix}accordion-border-radius));\n\n    .accordion-button {\n      @include border-top-radius(var(--#{$prefix}accordion-inner-border-radius));\n    }\n  }\n\n  &:not(:first-of-type) {\n    border-top: 0;\n  }\n\n  // Only set a border-radius on the last item if the accordion is collapsed\n  &:last-of-type {\n    @include border-bottom-radius(var(--#{$prefix}accordion-border-radius));\n\n    .accordion-button {\n      &.collapsed {\n        @include border-bottom-radius(var(--#{$prefix}accordion-inner-border-radius));\n      }\n    }\n\n    .accordion-collapse {\n      @include border-bottom-radius(var(--#{$prefix}accordion-border-radius));\n    }\n  }\n}\n\n.accordion-body {\n  padding: var(--#{$prefix}accordion-body-padding-y) var(--#{$prefix}accordion-body-padding-x);\n}\n\n\n// Flush accordion items\n//\n// Remove borders and border-radius to keep accordion items edge-to-edge.\n\n.accordion-flush {\n  .accordion-collapse {\n    border-width: 0;\n  }\n\n  .accordion-item {\n    border-right: 0;\n    border-left: 0;\n    @include border-radius(0);\n\n    &:first-child { border-top: 0; }\n    &:last-child { border-bottom: 0; }\n\n    .accordion-button {\n      &,\n      &.collapsed {\n        @include border-radius(0);\n      }\n    }\n  }\n}\n\n@if $enable-dark-mode {\n  @include color-mode(dark) {\n    .accordion-button::after {\n      --#{$prefix}accordion-btn-icon: #{escape-svg($accordion-button-icon-dark)};\n      --#{$prefix}accordion-btn-active-icon: #{escape-svg($accordion-button-active-icon-dark)};\n    }\n  }\n}\n",".breadcrumb {\n  // scss-docs-start breadcrumb-css-vars\n  --#{$prefix}breadcrumb-padding-x: #{$breadcrumb-padding-x};\n  --#{$prefix}breadcrumb-padding-y: #{$breadcrumb-padding-y};\n  --#{$prefix}breadcrumb-margin-bottom: #{$breadcrumb-margin-bottom};\n  @include rfs($breadcrumb-font-size, --#{$prefix}breadcrumb-font-size);\n  --#{$prefix}breadcrumb-bg: #{$breadcrumb-bg};\n  --#{$prefix}breadcrumb-border-radius: #{$breadcrumb-border-radius};\n  --#{$prefix}breadcrumb-divider-color: #{$breadcrumb-divider-color};\n  --#{$prefix}breadcrumb-item-padding-x: #{$breadcrumb-item-padding-x};\n  --#{$prefix}breadcrumb-item-active-color: #{$breadcrumb-active-color};\n  // scss-docs-end breadcrumb-css-vars\n\n  display: flex;\n  flex-wrap: wrap;\n  padding: var(--#{$prefix}breadcrumb-padding-y) var(--#{$prefix}breadcrumb-padding-x);\n  margin-bottom: var(--#{$prefix}breadcrumb-margin-bottom);\n  @include font-size(var(--#{$prefix}breadcrumb-font-size));\n  list-style: none;\n  background-color: var(--#{$prefix}breadcrumb-bg);\n  @include border-radius(var(--#{$prefix}breadcrumb-border-radius));\n}\n\n.breadcrumb-item {\n  // The separator between breadcrumbs (by default, a forward-slash: \"/\")\n  + .breadcrumb-item {\n    padding-left: var(--#{$prefix}breadcrumb-item-padding-x);\n\n    &::before {\n      float: left; // Suppress inline spacings and underlining of the separator\n      padding-right: var(--#{$prefix}breadcrumb-item-padding-x);\n      color: var(--#{$prefix}breadcrumb-divider-color);\n      content: var(--#{$prefix}breadcrumb-divider, escape-svg($breadcrumb-divider)) #{\"/* rtl:\"} var(--#{$prefix}breadcrumb-divider, escape-svg($breadcrumb-divider-flipped)) #{\"*/\"};\n    }\n  }\n\n  &.active {\n    color: var(--#{$prefix}breadcrumb-item-active-color);\n  }\n}\n",".pagination {\n  // scss-docs-start pagination-css-vars\n  --#{$prefix}pagination-padding-x: #{$pagination-padding-x};\n  --#{$prefix}pagination-padding-y: #{$pagination-padding-y};\n  @include rfs($pagination-font-size, --#{$prefix}pagination-font-size);\n  --#{$prefix}pagination-color: #{$pagination-color};\n  --#{$prefix}pagination-bg: #{$pagination-bg};\n  --#{$prefix}pagination-border-width: #{$pagination-border-width};\n  --#{$prefix}pagination-border-color: #{$pagination-border-color};\n  --#{$prefix}pagination-border-radius: #{$pagination-border-radius};\n  --#{$prefix}pagination-hover-color: #{$pagination-hover-color};\n  --#{$prefix}pagination-hover-bg: #{$pagination-hover-bg};\n  --#{$prefix}pagination-hover-border-color: #{$pagination-hover-border-color};\n  --#{$prefix}pagination-focus-color: #{$pagination-focus-color};\n  --#{$prefix}pagination-focus-bg: #{$pagination-focus-bg};\n  --#{$prefix}pagination-focus-box-shadow: #{$pagination-focus-box-shadow};\n  --#{$prefix}pagination-active-color: #{$pagination-active-color};\n  --#{$prefix}pagination-active-bg: #{$pagination-active-bg};\n  --#{$prefix}pagination-active-border-color: #{$pagination-active-border-color};\n  --#{$prefix}pagination-disabled-color: #{$pagination-disabled-color};\n  --#{$prefix}pagination-disabled-bg: #{$pagination-disabled-bg};\n  --#{$prefix}pagination-disabled-border-color: #{$pagination-disabled-border-color};\n  // scss-docs-end pagination-css-vars\n\n  display: flex;\n  @include list-unstyled();\n}\n\n.page-link {\n  position: relative;\n  display: block;\n  padding: var(--#{$prefix}pagination-padding-y) var(--#{$prefix}pagination-padding-x);\n  @include font-size(var(--#{$prefix}pagination-font-size));\n  color: var(--#{$prefix}pagination-color);\n  text-decoration: if($link-decoration == none, null, none);\n  background-color: var(--#{$prefix}pagination-bg);\n  border: var(--#{$prefix}pagination-border-width) solid var(--#{$prefix}pagination-border-color);\n  @include transition($pagination-transition);\n\n  &:hover {\n    z-index: 2;\n    color: var(--#{$prefix}pagination-hover-color);\n    text-decoration: if($link-hover-decoration == underline, none, null);\n    background-color: var(--#{$prefix}pagination-hover-bg);\n    border-color: var(--#{$prefix}pagination-hover-border-color);\n  }\n\n  &:focus {\n    z-index: 3;\n    color: var(--#{$prefix}pagination-focus-color);\n    background-color: var(--#{$prefix}pagination-focus-bg);\n    outline: $pagination-focus-outline;\n    box-shadow: var(--#{$prefix}pagination-focus-box-shadow);\n  }\n\n  &.active,\n  .active > & {\n    z-index: 3;\n    color: var(--#{$prefix}pagination-active-color);\n    @include gradient-bg(var(--#{$prefix}pagination-active-bg));\n    border-color: var(--#{$prefix}pagination-active-border-color);\n  }\n\n  &.disabled,\n  .disabled > & {\n    color: var(--#{$prefix}pagination-disabled-color);\n    pointer-events: none;\n    background-color: var(--#{$prefix}pagination-disabled-bg);\n    border-color: var(--#{$prefix}pagination-disabled-border-color);\n  }\n}\n\n.page-item {\n  &:not(:first-child) .page-link {\n    margin-left: $pagination-margin-start;\n  }\n\n  @if $pagination-margin-start == calc(#{$pagination-border-width} * -1) {\n    &:first-child {\n      .page-link {\n        @include border-start-radius(var(--#{$prefix}pagination-border-radius));\n      }\n    }\n\n    &:last-child {\n      .page-link {\n        @include border-end-radius(var(--#{$prefix}pagination-border-radius));\n      }\n    }\n  } @else {\n    // Add border-radius to all pageLinks in case they have left margin\n    .page-link {\n      @include border-radius(var(--#{$prefix}pagination-border-radius));\n    }\n  }\n}\n\n\n//\n// Sizing\n//\n\n.pagination-lg {\n  @include pagination-size($pagination-padding-y-lg, $pagination-padding-x-lg, $font-size-lg, $pagination-border-radius-lg);\n}\n\n.pagination-sm {\n  @include pagination-size($pagination-padding-y-sm, $pagination-padding-x-sm, $font-size-sm, $pagination-border-radius-sm);\n}\n","// Pagination\n\n// scss-docs-start pagination-mixin\n@mixin pagination-size($padding-y, $padding-x, $font-size, $border-radius) {\n  --#{$prefix}pagination-padding-x: #{$padding-x};\n  --#{$prefix}pagination-padding-y: #{$padding-y};\n  @include rfs($font-size, --#{$prefix}pagination-font-size);\n  --#{$prefix}pagination-border-radius: #{$border-radius};\n}\n// scss-docs-end pagination-mixin\n","// Base class\n//\n// Requires one of the contextual, color modifier classes for `color` and\n// `background-color`.\n\n.badge {\n  // scss-docs-start badge-css-vars\n  --#{$prefix}badge-padding-x: #{$badge-padding-x};\n  --#{$prefix}badge-padding-y: #{$badge-padding-y};\n  @include rfs($badge-font-size, --#{$prefix}badge-font-size);\n  --#{$prefix}badge-font-weight: #{$badge-font-weight};\n  --#{$prefix}badge-color: #{$badge-color};\n  --#{$prefix}badge-border-radius: #{$badge-border-radius};\n  // scss-docs-end badge-css-vars\n\n  display: inline-block;\n  padding: var(--#{$prefix}badge-padding-y) var(--#{$prefix}badge-padding-x);\n  @include font-size(var(--#{$prefix}badge-font-size));\n  font-weight: var(--#{$prefix}badge-font-weight);\n  line-height: 1;\n  color: var(--#{$prefix}badge-color);\n  text-align: center;\n  white-space: nowrap;\n  vertical-align: baseline;\n  @include border-radius(var(--#{$prefix}badge-border-radius));\n  @include gradient-bg();\n\n  // Empty badges collapse automatically\n  &:empty {\n    display: none;\n  }\n}\n\n// Quick fix for badges in buttons\n.btn .badge {\n  position: relative;\n  top: -1px;\n}\n","//\n// Base styles\n//\n\n.alert {\n  // scss-docs-start alert-css-vars\n  --#{$prefix}alert-bg: transparent;\n  --#{$prefix}alert-padding-x: #{$alert-padding-x};\n  --#{$prefix}alert-padding-y: #{$alert-padding-y};\n  --#{$prefix}alert-margin-bottom: #{$alert-margin-bottom};\n  --#{$prefix}alert-color: inherit;\n  --#{$prefix}alert-border-color: transparent;\n  --#{$prefix}alert-border: #{$alert-border-width} solid var(--#{$prefix}alert-border-color);\n  --#{$prefix}alert-border-radius: #{$alert-border-radius};\n  --#{$prefix}alert-link-color: inherit;\n  // scss-docs-end alert-css-vars\n\n  position: relative;\n  padding: var(--#{$prefix}alert-padding-y) var(--#{$prefix}alert-padding-x);\n  margin-bottom: var(--#{$prefix}alert-margin-bottom);\n  color: var(--#{$prefix}alert-color);\n  background-color: var(--#{$prefix}alert-bg);\n  border: var(--#{$prefix}alert-border);\n  @include border-radius(var(--#{$prefix}alert-border-radius));\n}\n\n// Headings for larger alerts\n.alert-heading {\n  // Specified to prevent conflicts of changing $headings-color\n  color: inherit;\n}\n\n// Provide class for links that match alerts\n.alert-link {\n  font-weight: $alert-link-font-weight;\n  color: var(--#{$prefix}alert-link-color);\n}\n\n\n// Dismissible alerts\n//\n// Expand the right padding and account for the close button's positioning.\n\n.alert-dismissible {\n  padding-right: $alert-dismissible-padding-r;\n\n  // Adjust close link position\n  .btn-close {\n    position: absolute;\n    top: 0;\n    right: 0;\n    z-index: $stretched-link-z-index + 1;\n    padding: $alert-padding-y * 1.25 $alert-padding-x;\n  }\n}\n\n\n// scss-docs-start alert-modifiers\n// Generate contextual modifier classes for colorizing the alert\n@each $state in map-keys($theme-colors) {\n  .alert-#{$state} {\n    --#{$prefix}alert-color: var(--#{$prefix}#{$state}-text-emphasis);\n    --#{$prefix}alert-bg: var(--#{$prefix}#{$state}-bg-subtle);\n    --#{$prefix}alert-border-color: var(--#{$prefix}#{$state}-border-subtle);\n    --#{$prefix}alert-link-color: var(--#{$prefix}#{$state}-text-emphasis);\n  }\n}\n// scss-docs-end alert-modifiers\n","// Disable animation if transitions are disabled\n\n// scss-docs-start progress-keyframes\n@if $enable-transitions {\n  @keyframes progress-bar-stripes {\n    0% { background-position-x: $progress-height; }\n  }\n}\n// scss-docs-end progress-keyframes\n\n.progress,\n.progress-stacked {\n  // scss-docs-start progress-css-vars\n  --#{$prefix}progress-height: #{$progress-height};\n  @include rfs($progress-font-size, --#{$prefix}progress-font-size);\n  --#{$prefix}progress-bg: #{$progress-bg};\n  --#{$prefix}progress-border-radius: #{$progress-border-radius};\n  --#{$prefix}progress-box-shadow: #{$progress-box-shadow};\n  --#{$prefix}progress-bar-color: #{$progress-bar-color};\n  --#{$prefix}progress-bar-bg: #{$progress-bar-bg};\n  --#{$prefix}progress-bar-transition: #{$progress-bar-transition};\n  // scss-docs-end progress-css-vars\n\n  display: flex;\n  height: var(--#{$prefix}progress-height);\n  overflow: hidden; // force rounded corners by cropping it\n  @include font-size(var(--#{$prefix}progress-font-size));\n  background-color: var(--#{$prefix}progress-bg);\n  @include border-radius(var(--#{$prefix}progress-border-radius));\n  @include box-shadow(var(--#{$prefix}progress-box-shadow));\n}\n\n.progress-bar {\n  display: flex;\n  flex-direction: column;\n  justify-content: center;\n  overflow: hidden;\n  color: var(--#{$prefix}progress-bar-color);\n  text-align: center;\n  white-space: nowrap;\n  background-color: var(--#{$prefix}progress-bar-bg);\n  @include transition(var(--#{$prefix}progress-bar-transition));\n}\n\n.progress-bar-striped {\n  @include gradient-striped();\n  background-size: var(--#{$prefix}progress-height) var(--#{$prefix}progress-height);\n}\n\n.progress-stacked > .progress {\n  overflow: visible;\n}\n\n.progress-stacked > .progress > .progress-bar {\n  width: 100%;\n}\n\n@if $enable-transitions {\n  .progress-bar-animated {\n    animation: $progress-bar-animation-timing progress-bar-stripes;\n\n    @if $enable-reduced-motion {\n      @media (prefers-reduced-motion: reduce) {\n        animation: none;\n      }\n    }\n  }\n}\n","// Base class\n//\n// Easily usable on <ul>, <ol>, or <div>.\n\n.list-group {\n  // scss-docs-start list-group-css-vars\n  --#{$prefix}list-group-color: #{$list-group-color};\n  --#{$prefix}list-group-bg: #{$list-group-bg};\n  --#{$prefix}list-group-border-color: #{$list-group-border-color};\n  --#{$prefix}list-group-border-width: #{$list-group-border-width};\n  --#{$prefix}list-group-border-radius: #{$list-group-border-radius};\n  --#{$prefix}list-group-item-padding-x: #{$list-group-item-padding-x};\n  --#{$prefix}list-group-item-padding-y: #{$list-group-item-padding-y};\n  --#{$prefix}list-group-action-color: #{$list-group-action-color};\n  --#{$prefix}list-group-action-hover-color: #{$list-group-action-hover-color};\n  --#{$prefix}list-group-action-hover-bg: #{$list-group-hover-bg};\n  --#{$prefix}list-group-action-active-color: #{$list-group-action-active-color};\n  --#{$prefix}list-group-action-active-bg: #{$list-group-action-active-bg};\n  --#{$prefix}list-group-disabled-color: #{$list-group-disabled-color};\n  --#{$prefix}list-group-disabled-bg: #{$list-group-disabled-bg};\n  --#{$prefix}list-group-active-color: #{$list-group-active-color};\n  --#{$prefix}list-group-active-bg: #{$list-group-active-bg};\n  --#{$prefix}list-group-active-border-color: #{$list-group-active-border-color};\n  // scss-docs-end list-group-css-vars\n\n  display: flex;\n  flex-direction: column;\n\n  // No need to set list-style: none; since .list-group-item is block level\n  padding-left: 0; // reset padding because ul and ol\n  margin-bottom: 0;\n  @include border-radius(var(--#{$prefix}list-group-border-radius));\n}\n\n.list-group-numbered {\n  list-style-type: none;\n  counter-reset: section;\n\n  > .list-group-item::before {\n    // Increments only this instance of the section counter\n    content: counters(section, \".\") \". \";\n    counter-increment: section;\n  }\n}\n\n// Interactive list items\n//\n// Use anchor or button elements instead of `li`s or `div`s to create interactive\n// list items. Includes an extra `.active` modifier class for selected items.\n\n.list-group-item-action {\n  width: 100%; // For `<button>`s (anchors become 100% by default though)\n  color: var(--#{$prefix}list-group-action-color);\n  text-align: inherit; // For `<button>`s (anchors inherit)\n\n  // Hover state\n  &:hover,\n  &:focus {\n    z-index: 1; // Place hover/focus items above their siblings for proper border styling\n    color: var(--#{$prefix}list-group-action-hover-color);\n    text-decoration: none;\n    background-color: var(--#{$prefix}list-group-action-hover-bg);\n  }\n\n  &:active {\n    color: var(--#{$prefix}list-group-action-active-color);\n    background-color: var(--#{$prefix}list-group-action-active-bg);\n  }\n}\n\n// Individual list items\n//\n// Use on `li`s or `div`s within the `.list-group` parent.\n\n.list-group-item {\n  position: relative;\n  display: block;\n  padding: var(--#{$prefix}list-group-item-padding-y) var(--#{$prefix}list-group-item-padding-x);\n  color: var(--#{$prefix}list-group-color);\n  text-decoration: if($link-decoration == none, null, none);\n  background-color: var(--#{$prefix}list-group-bg);\n  border: var(--#{$prefix}list-group-border-width) solid var(--#{$prefix}list-group-border-color);\n\n  &:first-child {\n    @include border-top-radius(inherit);\n  }\n\n  &:last-child {\n    @include border-bottom-radius(inherit);\n  }\n\n  &.disabled,\n  &:disabled {\n    color: var(--#{$prefix}list-group-disabled-color);\n    pointer-events: none;\n    background-color: var(--#{$prefix}list-group-disabled-bg);\n  }\n\n  // Include both here for `<a>`s and `<button>`s\n  &.active {\n    z-index: 2; // Place active items above their siblings for proper border styling\n    color: var(--#{$prefix}list-group-active-color);\n    background-color: var(--#{$prefix}list-group-active-bg);\n    border-color: var(--#{$prefix}list-group-active-border-color);\n  }\n\n  // stylelint-disable-next-line scss/selector-no-redundant-nesting-selector\n  & + .list-group-item {\n    border-top-width: 0;\n\n    &.active {\n      margin-top: calc(-1 * var(--#{$prefix}list-group-border-width)); // stylelint-disable-line function-disallowed-list\n      border-top-width: var(--#{$prefix}list-group-border-width);\n    }\n  }\n}\n\n// Horizontal\n//\n// Change the layout of list group items from vertical (default) to horizontal.\n\n@each $breakpoint in map-keys($grid-breakpoints) {\n  @include media-breakpoint-up($breakpoint) {\n    $infix: breakpoint-infix($breakpoint, $grid-breakpoints);\n\n    .list-group-horizontal#{$infix} {\n      flex-direction: row;\n\n      > .list-group-item {\n        &:first-child:not(:last-child) {\n          @include border-bottom-start-radius(var(--#{$prefix}list-group-border-radius));\n          @include border-top-end-radius(0);\n        }\n\n        &:last-child:not(:first-child) {\n          @include border-top-end-radius(var(--#{$prefix}list-group-border-radius));\n          @include border-bottom-start-radius(0);\n        }\n\n        &.active {\n          margin-top: 0;\n        }\n\n        + .list-group-item {\n          border-top-width: var(--#{$prefix}list-group-border-width);\n          border-left-width: 0;\n\n          &.active {\n            margin-left: calc(-1 * var(--#{$prefix}list-group-border-width)); // stylelint-disable-line function-disallowed-list\n            border-left-width: var(--#{$prefix}list-group-border-width);\n          }\n        }\n      }\n    }\n  }\n}\n\n\n// Flush list items\n//\n// Remove borders and border-radius to keep list group items edge-to-edge. Most\n// useful within other components (e.g., cards).\n\n.list-group-flush {\n  @include border-radius(0);\n\n  > .list-group-item {\n    border-width: 0 0 var(--#{$prefix}list-group-border-width);\n\n    &:last-child {\n      border-bottom-width: 0;\n    }\n  }\n}\n\n\n// scss-docs-start list-group-modifiers\n// List group contextual variants\n//\n// Add modifier classes to change text and background color on individual items.\n// Organizationally, this must come after the `:hover` states.\n\n@each $state in map-keys($theme-colors) {\n  .list-group-item-#{$state} {\n    --#{$prefix}list-group-color: var(--#{$prefix}#{$state}-text-emphasis);\n    --#{$prefix}list-group-bg: var(--#{$prefix}#{$state}-bg-subtle);\n    --#{$prefix}list-group-border-color: var(--#{$prefix}#{$state}-border-subtle);\n    --#{$prefix}list-group-action-hover-color: var(--#{$prefix}emphasis-color);\n    --#{$prefix}list-group-action-hover-bg: var(--#{$prefix}#{$state}-border-subtle);\n    --#{$prefix}list-group-action-active-color: var(--#{$prefix}emphasis-color);\n    --#{$prefix}list-group-action-active-bg: var(--#{$prefix}#{$state}-border-subtle);\n    --#{$prefix}list-group-active-color: var(--#{$prefix}#{$state}-bg-subtle);\n    --#{$prefix}list-group-active-bg: var(--#{$prefix}#{$state}-text-emphasis);\n    --#{$prefix}list-group-active-border-color: var(--#{$prefix}#{$state}-text-emphasis);\n  }\n}\n// scss-docs-end list-group-modifiers\n","// Transparent background and border properties included for button version.\n// iOS requires the button element instead of an anchor tag.\n// If you want the anchor version, it requires `href=\"#\"`.\n// See https://developer.mozilla.org/en-US/docs/Web/Events/click#Safari_Mobile\n\n.btn-close {\n  // scss-docs-start close-css-vars\n  --#{$prefix}btn-close-color: #{$btn-close-color};\n  --#{$prefix}btn-close-bg: #{ escape-svg($btn-close-bg) };\n  --#{$prefix}btn-close-opacity: #{$btn-close-opacity};\n  --#{$prefix}btn-close-hover-opacity: #{$btn-close-hover-opacity};\n  --#{$prefix}btn-close-focus-shadow: #{$btn-close-focus-shadow};\n  --#{$prefix}btn-close-focus-opacity: #{$btn-close-focus-opacity};\n  --#{$prefix}btn-close-disabled-opacity: #{$btn-close-disabled-opacity};\n  --#{$prefix}btn-close-white-filter: #{$btn-close-white-filter};\n  // scss-docs-end close-css-vars\n\n  box-sizing: content-box;\n  width: $btn-close-width;\n  height: $btn-close-height;\n  padding: $btn-close-padding-y $btn-close-padding-x;\n  color: var(--#{$prefix}btn-close-color);\n  background: transparent var(--#{$prefix}btn-close-bg) center / $btn-close-width auto no-repeat; // include transparent for button elements\n  border: 0; // for button elements\n  @include border-radius();\n  opacity: var(--#{$prefix}btn-close-opacity);\n\n  // Override <a>'s hover style\n  &:hover {\n    color: var(--#{$prefix}btn-close-color);\n    text-decoration: none;\n    opacity: var(--#{$prefix}btn-close-hover-opacity);\n  }\n\n  &:focus {\n    outline: 0;\n    box-shadow: var(--#{$prefix}btn-close-focus-shadow);\n    opacity: var(--#{$prefix}btn-close-focus-opacity);\n  }\n\n  &:disabled,\n  &.disabled {\n    pointer-events: none;\n    user-select: none;\n    opacity: var(--#{$prefix}btn-close-disabled-opacity);\n  }\n}\n\n@mixin btn-close-white() {\n  filter: var(--#{$prefix}btn-close-white-filter);\n}\n\n.btn-close-white {\n  @include btn-close-white();\n}\n\n@if $enable-dark-mode {\n  @include color-mode(dark) {\n    .btn-close {\n      @include btn-close-white();\n    }\n  }\n}\n",".toast {\n  // scss-docs-start toast-css-vars\n  --#{$prefix}toast-zindex: #{$zindex-toast};\n  --#{$prefix}toast-padding-x: #{$toast-padding-x};\n  --#{$prefix}toast-padding-y: #{$toast-padding-y};\n  --#{$prefix}toast-spacing: #{$toast-spacing};\n  --#{$prefix}toast-max-width: #{$toast-max-width};\n  @include rfs($toast-font-size, --#{$prefix}toast-font-size);\n  --#{$prefix}toast-color: #{$toast-color};\n  --#{$prefix}toast-bg: #{$toast-background-color};\n  --#{$prefix}toast-border-width: #{$toast-border-width};\n  --#{$prefix}toast-border-color: #{$toast-border-color};\n  --#{$prefix}toast-border-radius: #{$toast-border-radius};\n  --#{$prefix}toast-box-shadow: #{$toast-box-shadow};\n  --#{$prefix}toast-header-color: #{$toast-header-color};\n  --#{$prefix}toast-header-bg: #{$toast-header-background-color};\n  --#{$prefix}toast-header-border-color: #{$toast-header-border-color};\n  // scss-docs-end toast-css-vars\n\n  width: var(--#{$prefix}toast-max-width);\n  max-width: 100%;\n  @include font-size(var(--#{$prefix}toast-font-size));\n  color: var(--#{$prefix}toast-color);\n  pointer-events: auto;\n  background-color: var(--#{$prefix}toast-bg);\n  background-clip: padding-box;\n  border: var(--#{$prefix}toast-border-width) solid var(--#{$prefix}toast-border-color);\n  box-shadow: var(--#{$prefix}toast-box-shadow);\n  @include border-radius(var(--#{$prefix}toast-border-radius));\n\n  &.showing {\n    opacity: 0;\n  }\n\n  &:not(.show) {\n    display: none;\n  }\n}\n\n.toast-container {\n  --#{$prefix}toast-zindex: #{$zindex-toast};\n\n  position: absolute;\n  z-index: var(--#{$prefix}toast-zindex);\n  width: max-content;\n  max-width: 100%;\n  pointer-events: none;\n\n  > :not(:last-child) {\n    margin-bottom: var(--#{$prefix}toast-spacing);\n  }\n}\n\n.toast-header {\n  display: flex;\n  align-items: center;\n  padding: var(--#{$prefix}toast-padding-y) var(--#{$prefix}toast-padding-x);\n  color: var(--#{$prefix}toast-header-color);\n  background-color: var(--#{$prefix}toast-header-bg);\n  background-clip: padding-box;\n  border-bottom: var(--#{$prefix}toast-border-width) solid var(--#{$prefix}toast-header-border-color);\n  @include border-top-radius(calc(var(--#{$prefix}toast-border-radius) - var(--#{$prefix}toast-border-width)));\n\n  .btn-close {\n    margin-right: calc(-.5 * var(--#{$prefix}toast-padding-x)); // stylelint-disable-line function-disallowed-list\n    margin-left: var(--#{$prefix}toast-padding-x);\n  }\n}\n\n.toast-body {\n  padding: var(--#{$prefix}toast-padding-x);\n  word-wrap: break-word;\n}\n","// stylelint-disable function-disallowed-list\n\n// .modal-open      - body class for killing the scroll\n// .modal           - container to scroll within\n// .modal-dialog    - positioning shell for the actual modal\n// .modal-content   - actual modal w/ bg and corners and stuff\n\n\n// Container that the modal scrolls within\n.modal {\n  // scss-docs-start modal-css-vars\n  --#{$prefix}modal-zindex: #{$zindex-modal};\n  --#{$prefix}modal-width: #{$modal-md};\n  --#{$prefix}modal-padding: #{$modal-inner-padding};\n  --#{$prefix}modal-margin: #{$modal-dialog-margin};\n  --#{$prefix}modal-color: #{$modal-content-color};\n  --#{$prefix}modal-bg: #{$modal-content-bg};\n  --#{$prefix}modal-border-color: #{$modal-content-border-color};\n  --#{$prefix}modal-border-width: #{$modal-content-border-width};\n  --#{$prefix}modal-border-radius: #{$modal-content-border-radius};\n  --#{$prefix}modal-box-shadow: #{$modal-content-box-shadow-xs};\n  --#{$prefix}modal-inner-border-radius: #{$modal-content-inner-border-radius};\n  --#{$prefix}modal-header-padding-x: #{$modal-header-padding-x};\n  --#{$prefix}modal-header-padding-y: #{$modal-header-padding-y};\n  --#{$prefix}modal-header-padding: #{$modal-header-padding}; // Todo in v6: Split this padding into x and y\n  --#{$prefix}modal-header-border-color: #{$modal-header-border-color};\n  --#{$prefix}modal-header-border-width: #{$modal-header-border-width};\n  --#{$prefix}modal-title-line-height: #{$modal-title-line-height};\n  --#{$prefix}modal-footer-gap: #{$modal-footer-margin-between};\n  --#{$prefix}modal-footer-bg: #{$modal-footer-bg};\n  --#{$prefix}modal-footer-border-color: #{$modal-footer-border-color};\n  --#{$prefix}modal-footer-border-width: #{$modal-footer-border-width};\n  // scss-docs-end modal-css-vars\n\n  position: fixed;\n  top: 0;\n  left: 0;\n  z-index: var(--#{$prefix}modal-zindex);\n  display: none;\n  width: 100%;\n  height: 100%;\n  overflow-x: hidden;\n  overflow-y: auto;\n  // Prevent Chrome on Windows from adding a focus outline. For details, see\n  // https://github.com/twbs/bootstrap/pull/10951.\n  outline: 0;\n  // We deliberately don't use `-webkit-overflow-scrolling: touch;` due to a\n  // gnarly iOS Safari bug: https://bugs.webkit.org/show_bug.cgi?id=158342\n  // See also https://github.com/twbs/bootstrap/issues/17695\n}\n\n// Shell div to position the modal with bottom padding\n.modal-dialog {\n  position: relative;\n  width: auto;\n  margin: var(--#{$prefix}modal-margin);\n  // allow clicks to pass through for custom click handling to close modal\n  pointer-events: none;\n\n  // When fading in the modal, animate it to slide down\n  .modal.fade & {\n    @include transition($modal-transition);\n    transform: $modal-fade-transform;\n  }\n  .modal.show & {\n    transform: $modal-show-transform;\n  }\n\n  // When trying to close, animate focus to scale\n  .modal.modal-static & {\n    transform: $modal-scale-transform;\n  }\n}\n\n.modal-dialog-scrollable {\n  height: calc(100% - var(--#{$prefix}modal-margin) * 2);\n\n  .modal-content {\n    max-height: 100%;\n    overflow: hidden;\n  }\n\n  .modal-body {\n    overflow-y: auto;\n  }\n}\n\n.modal-dialog-centered {\n  display: flex;\n  align-items: center;\n  min-height: calc(100% - var(--#{$prefix}modal-margin) * 2);\n}\n\n// Actual modal\n.modal-content {\n  position: relative;\n  display: flex;\n  flex-direction: column;\n  width: 100%; // Ensure `.modal-content` extends the full width of the parent `.modal-dialog`\n  // counteract the pointer-events: none; in the .modal-dialog\n  color: var(--#{$prefix}modal-color);\n  pointer-events: auto;\n  background-color: var(--#{$prefix}modal-bg);\n  background-clip: padding-box;\n  border: var(--#{$prefix}modal-border-width) solid var(--#{$prefix}modal-border-color);\n  @include border-radius(var(--#{$prefix}modal-border-radius));\n  @include box-shadow(var(--#{$prefix}modal-box-shadow));\n  // Remove focus outline from opened modal\n  outline: 0;\n}\n\n// Modal background\n.modal-backdrop {\n  // scss-docs-start modal-backdrop-css-vars\n  --#{$prefix}backdrop-zindex: #{$zindex-modal-backdrop};\n  --#{$prefix}backdrop-bg: #{$modal-backdrop-bg};\n  --#{$prefix}backdrop-opacity: #{$modal-backdrop-opacity};\n  // scss-docs-end modal-backdrop-css-vars\n\n  @include overlay-backdrop(var(--#{$prefix}backdrop-zindex), var(--#{$prefix}backdrop-bg), var(--#{$prefix}backdrop-opacity));\n}\n\n// Modal header\n// Top section of the modal w/ title and dismiss\n.modal-header {\n  display: flex;\n  flex-shrink: 0;\n  align-items: center;\n  justify-content: space-between; // Put modal header elements (title and dismiss) on opposite ends\n  padding: var(--#{$prefix}modal-header-padding);\n  border-bottom: var(--#{$prefix}modal-header-border-width) solid var(--#{$prefix}modal-header-border-color);\n  @include border-top-radius(var(--#{$prefix}modal-inner-border-radius));\n\n  .btn-close {\n    padding: calc(var(--#{$prefix}modal-header-padding-y) * .5) calc(var(--#{$prefix}modal-header-padding-x) * .5);\n    margin: calc(-.5 * var(--#{$prefix}modal-header-padding-y)) calc(-.5 * var(--#{$prefix}modal-header-padding-x)) calc(-.5 * var(--#{$prefix}modal-header-padding-y)) auto;\n  }\n}\n\n// Title text within header\n.modal-title {\n  margin-bottom: 0;\n  line-height: var(--#{$prefix}modal-title-line-height);\n}\n\n// Modal body\n// Where all modal content resides (sibling of .modal-header and .modal-footer)\n.modal-body {\n  position: relative;\n  // Enable `flex-grow: 1` so that the body take up as much space as possible\n  // when there should be a fixed height on `.modal-dialog`.\n  flex: 1 1 auto;\n  padding: var(--#{$prefix}modal-padding);\n}\n\n// Footer (for actions)\n.modal-footer {\n  display: flex;\n  flex-shrink: 0;\n  flex-wrap: wrap;\n  align-items: center; // vertically center\n  justify-content: flex-end; // Right align buttons with flex property because text-align doesn't work on flex items\n  padding: calc(var(--#{$prefix}modal-padding) - var(--#{$prefix}modal-footer-gap) * .5);\n  background-color: var(--#{$prefix}modal-footer-bg);\n  border-top: var(--#{$prefix}modal-footer-border-width) solid var(--#{$prefix}modal-footer-border-color);\n  @include border-bottom-radius(var(--#{$prefix}modal-inner-border-radius));\n\n  // Place margin between footer elements\n  // This solution is far from ideal because of the universal selector usage,\n  // but is needed to fix https://github.com/twbs/bootstrap/issues/24800\n  > * {\n    margin: calc(var(--#{$prefix}modal-footer-gap) * .5); // Todo in v6: replace with gap on parent class\n  }\n}\n\n// Scale up the modal\n@include media-breakpoint-up(sm) {\n  .modal {\n    --#{$prefix}modal-margin: #{$modal-dialog-margin-y-sm-up};\n    --#{$prefix}modal-box-shadow: #{$modal-content-box-shadow-sm-up};\n  }\n\n  // Automatically set modal's width for larger viewports\n  .modal-dialog {\n    max-width: var(--#{$prefix}modal-width);\n    margin-right: auto;\n    margin-left: auto;\n  }\n\n  .modal-sm {\n    --#{$prefix}modal-width: #{$modal-sm};\n  }\n}\n\n@include media-breakpoint-up(lg) {\n  .modal-lg,\n  .modal-xl {\n    --#{$prefix}modal-width: #{$modal-lg};\n  }\n}\n\n@include media-breakpoint-up(xl) {\n  .modal-xl {\n    --#{$prefix}modal-width: #{$modal-xl};\n  }\n}\n\n// scss-docs-start modal-fullscreen-loop\n@each $breakpoint in map-keys($grid-breakpoints) {\n  $infix: breakpoint-infix($breakpoint, $grid-breakpoints);\n  $postfix: if($infix != \"\", $infix + \"-down\", \"\");\n\n  @include media-breakpoint-down($breakpoint) {\n    .modal-fullscreen#{$postfix} {\n      width: 100vw;\n      max-width: none;\n      height: 100%;\n      margin: 0;\n\n      .modal-content {\n        height: 100%;\n        border: 0;\n        @include border-radius(0);\n      }\n\n      .modal-header,\n      .modal-footer {\n        @include border-radius(0);\n      }\n\n      .modal-body {\n        overflow-y: auto;\n      }\n    }\n  }\n}\n// scss-docs-end modal-fullscreen-loop\n","// Shared between modals and offcanvases\n@mixin overlay-backdrop($zindex, $backdrop-bg, $backdrop-opacity) {\n  position: fixed;\n  top: 0;\n  left: 0;\n  z-index: $zindex;\n  width: 100vw;\n  height: 100vh;\n  background-color: $backdrop-bg;\n\n  // Fade for backdrop\n  &.fade { opacity: 0; }\n  &.show { opacity: $backdrop-opacity; }\n}\n","// Base class\n.tooltip {\n  // scss-docs-start tooltip-css-vars\n  --#{$prefix}tooltip-zindex: #{$zindex-tooltip};\n  --#{$prefix}tooltip-max-width: #{$tooltip-max-width};\n  --#{$prefix}tooltip-padding-x: #{$tooltip-padding-x};\n  --#{$prefix}tooltip-padding-y: #{$tooltip-padding-y};\n  --#{$prefix}tooltip-margin: #{$tooltip-margin};\n  @include rfs($tooltip-font-size, --#{$prefix}tooltip-font-size);\n  --#{$prefix}tooltip-color: #{$tooltip-color};\n  --#{$prefix}tooltip-bg: #{$tooltip-bg};\n  --#{$prefix}tooltip-border-radius: #{$tooltip-border-radius};\n  --#{$prefix}tooltip-opacity: #{$tooltip-opacity};\n  --#{$prefix}tooltip-arrow-width: #{$tooltip-arrow-width};\n  --#{$prefix}tooltip-arrow-height: #{$tooltip-arrow-height};\n  // scss-docs-end tooltip-css-vars\n\n  z-index: var(--#{$prefix}tooltip-zindex);\n  display: block;\n  margin: var(--#{$prefix}tooltip-margin);\n  @include deprecate(\"`$tooltip-margin`\", \"v5\", \"v5.x\", true);\n  // Our parent element can be arbitrary since tooltips are by default inserted as a sibling of their target element.\n  // So reset our font and text properties to avoid inheriting weird values.\n  @include reset-text();\n  @include font-size(var(--#{$prefix}tooltip-font-size));\n  // Allow breaking very long words so they don't overflow the tooltip's bounds\n  word-wrap: break-word;\n  opacity: 0;\n\n  &.show { opacity: var(--#{$prefix}tooltip-opacity); }\n\n  .tooltip-arrow {\n    display: block;\n    width: var(--#{$prefix}tooltip-arrow-width);\n    height: var(--#{$prefix}tooltip-arrow-height);\n\n    &::before {\n      position: absolute;\n      content: \"\";\n      border-color: transparent;\n      border-style: solid;\n    }\n  }\n}\n\n.bs-tooltip-top .tooltip-arrow {\n  bottom: calc(-1 * var(--#{$prefix}tooltip-arrow-height)); // stylelint-disable-line function-disallowed-list\n\n  &::before {\n    top: -1px;\n    border-width: var(--#{$prefix}tooltip-arrow-height) calc(var(--#{$prefix}tooltip-arrow-width) * .5) 0; // stylelint-disable-line function-disallowed-list\n    border-top-color: var(--#{$prefix}tooltip-bg);\n  }\n}\n\n/* rtl:begin:ignore */\n.bs-tooltip-end .tooltip-arrow {\n  left: calc(-1 * var(--#{$prefix}tooltip-arrow-height)); // stylelint-disable-line function-disallowed-list\n  width: var(--#{$prefix}tooltip-arrow-height);\n  height: var(--#{$prefix}tooltip-arrow-width);\n\n  &::before {\n    right: -1px;\n    border-width: calc(var(--#{$prefix}tooltip-arrow-width) * .5) var(--#{$prefix}tooltip-arrow-height) calc(var(--#{$prefix}tooltip-arrow-width) * .5) 0; // stylelint-disable-line function-disallowed-list\n    border-right-color: var(--#{$prefix}tooltip-bg);\n  }\n}\n\n/* rtl:end:ignore */\n\n.bs-tooltip-bottom .tooltip-arrow {\n  top: calc(-1 * var(--#{$prefix}tooltip-arrow-height)); // stylelint-disable-line function-disallowed-list\n\n  &::before {\n    bottom: -1px;\n    border-width: 0 calc(var(--#{$prefix}tooltip-arrow-width) * .5) var(--#{$prefix}tooltip-arrow-height); // stylelint-disable-line function-disallowed-list\n    border-bottom-color: var(--#{$prefix}tooltip-bg);\n  }\n}\n\n/* rtl:begin:ignore */\n.bs-tooltip-start .tooltip-arrow {\n  right: calc(-1 * var(--#{$prefix}tooltip-arrow-height)); // stylelint-disable-line function-disallowed-list\n  width: var(--#{$prefix}tooltip-arrow-height);\n  height: var(--#{$prefix}tooltip-arrow-width);\n\n  &::before {\n    left: -1px;\n    border-width: calc(var(--#{$prefix}tooltip-arrow-width) * .5) 0 calc(var(--#{$prefix}tooltip-arrow-width) * .5) var(--#{$prefix}tooltip-arrow-height); // stylelint-disable-line function-disallowed-list\n    border-left-color: var(--#{$prefix}tooltip-bg);\n  }\n}\n\n/* rtl:end:ignore */\n\n.bs-tooltip-auto {\n  &[data-popper-placement^=\"top\"] {\n    @extend .bs-tooltip-top;\n  }\n  &[data-popper-placement^=\"right\"] {\n    @extend .bs-tooltip-end;\n  }\n  &[data-popper-placement^=\"bottom\"] {\n    @extend .bs-tooltip-bottom;\n  }\n  &[data-popper-placement^=\"left\"] {\n    @extend .bs-tooltip-start;\n  }\n}\n\n// Wrapper for the tooltip content\n.tooltip-inner {\n  max-width: var(--#{$prefix}tooltip-max-width);\n  padding: var(--#{$prefix}tooltip-padding-y) var(--#{$prefix}tooltip-padding-x);\n  color: var(--#{$prefix}tooltip-color);\n  text-align: center;\n  background-color: var(--#{$prefix}tooltip-bg);\n  @include border-radius(var(--#{$prefix}tooltip-border-radius));\n}\n","@mixin reset-text {\n  font-family: $font-family-base;\n  // We deliberately do NOT reset font-size or overflow-wrap / word-wrap.\n  font-style: normal;\n  font-weight: $font-weight-normal;\n  line-height: $line-height-base;\n  text-align: left; // Fallback for where `start` is not supported\n  text-align: start;\n  text-decoration: none;\n  text-shadow: none;\n  text-transform: none;\n  letter-spacing: normal;\n  word-break: normal;\n  white-space: normal;\n  word-spacing: normal;\n  line-break: auto;\n}\n",".popover {\n  // scss-docs-start popover-css-vars\n  --#{$prefix}popover-zindex: #{$zindex-popover};\n  --#{$prefix}popover-max-width: #{$popover-max-width};\n  @include rfs($popover-font-size, --#{$prefix}popover-font-size);\n  --#{$prefix}popover-bg: #{$popover-bg};\n  --#{$prefix}popover-border-width: #{$popover-border-width};\n  --#{$prefix}popover-border-color: #{$popover-border-color};\n  --#{$prefix}popover-border-radius: #{$popover-border-radius};\n  --#{$prefix}popover-inner-border-radius: #{$popover-inner-border-radius};\n  --#{$prefix}popover-box-shadow: #{$popover-box-shadow};\n  --#{$prefix}popover-header-padding-x: #{$popover-header-padding-x};\n  --#{$prefix}popover-header-padding-y: #{$popover-header-padding-y};\n  @include rfs($popover-header-font-size, --#{$prefix}popover-header-font-size);\n  --#{$prefix}popover-header-color: #{$popover-header-color};\n  --#{$prefix}popover-header-bg: #{$popover-header-bg};\n  --#{$prefix}popover-body-padding-x: #{$popover-body-padding-x};\n  --#{$prefix}popover-body-padding-y: #{$popover-body-padding-y};\n  --#{$prefix}popover-body-color: #{$popover-body-color};\n  --#{$prefix}popover-arrow-width: #{$popover-arrow-width};\n  --#{$prefix}popover-arrow-height: #{$popover-arrow-height};\n  --#{$prefix}popover-arrow-border: var(--#{$prefix}popover-border-color);\n  // scss-docs-end popover-css-vars\n\n  z-index: var(--#{$prefix}popover-zindex);\n  display: block;\n  max-width: var(--#{$prefix}popover-max-width);\n  // Our parent element can be arbitrary since tooltips are by default inserted as a sibling of their target element.\n  // So reset our font and text properties to avoid inheriting weird values.\n  @include reset-text();\n  @include font-size(var(--#{$prefix}popover-font-size));\n  // Allow breaking very long words so they don't overflow the popover's bounds\n  word-wrap: break-word;\n  background-color: var(--#{$prefix}popover-bg);\n  background-clip: padding-box;\n  border: var(--#{$prefix}popover-border-width) solid var(--#{$prefix}popover-border-color);\n  @include border-radius(var(--#{$prefix}popover-border-radius));\n  @include box-shadow(var(--#{$prefix}popover-box-shadow));\n\n  .popover-arrow {\n    display: block;\n    width: var(--#{$prefix}popover-arrow-width);\n    height: var(--#{$prefix}popover-arrow-height);\n\n    &::before,\n    &::after {\n      position: absolute;\n      display: block;\n      content: \"\";\n      border-color: transparent;\n      border-style: solid;\n      border-width: 0;\n    }\n  }\n}\n\n.bs-popover-top {\n  > .popover-arrow {\n    bottom: calc(-1 * (var(--#{$prefix}popover-arrow-height)) - var(--#{$prefix}popover-border-width)); // stylelint-disable-line function-disallowed-list\n\n    &::before,\n    &::after {\n      border-width: var(--#{$prefix}popover-arrow-height) calc(var(--#{$prefix}popover-arrow-width) * .5) 0; // stylelint-disable-line function-disallowed-list\n    }\n\n    &::before {\n      bottom: 0;\n      border-top-color: var(--#{$prefix}popover-arrow-border);\n    }\n\n    &::after {\n      bottom: var(--#{$prefix}popover-border-width);\n      border-top-color: var(--#{$prefix}popover-bg);\n    }\n  }\n}\n\n/* rtl:begin:ignore */\n.bs-popover-end {\n  > .popover-arrow {\n    left: calc(-1 * (var(--#{$prefix}popover-arrow-height)) - var(--#{$prefix}popover-border-width)); // stylelint-disable-line function-disallowed-list\n    width: var(--#{$prefix}popover-arrow-height);\n    height: var(--#{$prefix}popover-arrow-width);\n\n    &::before,\n    &::after {\n      border-width: calc(var(--#{$prefix}popover-arrow-width) * .5) var(--#{$prefix}popover-arrow-height) calc(var(--#{$prefix}popover-arrow-width) * .5) 0; // stylelint-disable-line function-disallowed-list\n    }\n\n    &::before {\n      left: 0;\n      border-right-color: var(--#{$prefix}popover-arrow-border);\n    }\n\n    &::after {\n      left: var(--#{$prefix}popover-border-width);\n      border-right-color: var(--#{$prefix}popover-bg);\n    }\n  }\n}\n\n/* rtl:end:ignore */\n\n.bs-popover-bottom {\n  > .popover-arrow {\n    top: calc(-1 * (var(--#{$prefix}popover-arrow-height)) - var(--#{$prefix}popover-border-width)); // stylelint-disable-line function-disallowed-list\n\n    &::before,\n    &::after {\n      border-width: 0 calc(var(--#{$prefix}popover-arrow-width) * .5) var(--#{$prefix}popover-arrow-height); // stylelint-disable-line function-disallowed-list\n    }\n\n    &::before {\n      top: 0;\n      border-bottom-color: var(--#{$prefix}popover-arrow-border);\n    }\n\n    &::after {\n      top: var(--#{$prefix}popover-border-width);\n      border-bottom-color: var(--#{$prefix}popover-bg);\n    }\n  }\n\n  // This will remove the popover-header's border just below the arrow\n  .popover-header::before {\n    position: absolute;\n    top: 0;\n    left: 50%;\n    display: block;\n    width: var(--#{$prefix}popover-arrow-width);\n    margin-left: calc(-.5 * var(--#{$prefix}popover-arrow-width)); // stylelint-disable-line function-disallowed-list\n    content: \"\";\n    border-bottom: var(--#{$prefix}popover-border-width) solid var(--#{$prefix}popover-header-bg);\n  }\n}\n\n/* rtl:begin:ignore */\n.bs-popover-start {\n  > .popover-arrow {\n    right: calc(-1 * (var(--#{$prefix}popover-arrow-height)) - var(--#{$prefix}popover-border-width)); // stylelint-disable-line function-disallowed-list\n    width: var(--#{$prefix}popover-arrow-height);\n    height: var(--#{$prefix}popover-arrow-width);\n\n    &::before,\n    &::after {\n      border-width: calc(var(--#{$prefix}popover-arrow-width) * .5) 0 calc(var(--#{$prefix}popover-arrow-width) * .5) var(--#{$prefix}popover-arrow-height); // stylelint-disable-line function-disallowed-list\n    }\n\n    &::before {\n      right: 0;\n      border-left-color: var(--#{$prefix}popover-arrow-border);\n    }\n\n    &::after {\n      right: var(--#{$prefix}popover-border-width);\n      border-left-color: var(--#{$prefix}popover-bg);\n    }\n  }\n}\n\n/* rtl:end:ignore */\n\n.bs-popover-auto {\n  &[data-popper-placement^=\"top\"] {\n    @extend .bs-popover-top;\n  }\n  &[data-popper-placement^=\"right\"] {\n    @extend .bs-popover-end;\n  }\n  &[data-popper-placement^=\"bottom\"] {\n    @extend .bs-popover-bottom;\n  }\n  &[data-popper-placement^=\"left\"] {\n    @extend .bs-popover-start;\n  }\n}\n\n// Offset the popover to account for the popover arrow\n.popover-header {\n  padding: var(--#{$prefix}popover-header-padding-y) var(--#{$prefix}popover-header-padding-x);\n  margin-bottom: 0; // Reset the default from Reboot\n  @include font-size(var(--#{$prefix}popover-header-font-size));\n  color: var(--#{$prefix}popover-header-color);\n  background-color: var(--#{$prefix}popover-header-bg);\n  border-bottom: var(--#{$prefix}popover-border-width) solid var(--#{$prefix}popover-border-color);\n  @include border-top-radius(var(--#{$prefix}popover-inner-border-radius));\n\n  &:empty {\n    display: none;\n  }\n}\n\n.popover-body {\n  padding: var(--#{$prefix}popover-body-padding-y) var(--#{$prefix}popover-body-padding-x);\n  color: var(--#{$prefix}popover-body-color);\n}\n","// Notes on the classes:\n//\n// 1. .carousel.pointer-event should ideally be pan-y (to allow for users to scroll vertically)\n//    even when their scroll action started on a carousel, but for compatibility (with Firefox)\n//    we're preventing all actions instead\n// 2. The .carousel-item-start and .carousel-item-end is used to indicate where\n//    the active slide is heading.\n// 3. .active.carousel-item is the current slide.\n// 4. .active.carousel-item-start and .active.carousel-item-end is the current\n//    slide in its in-transition state. Only one of these occurs at a time.\n// 5. .carousel-item-next.carousel-item-start and .carousel-item-prev.carousel-item-end\n//    is the upcoming slide in transition.\n\n.carousel {\n  position: relative;\n}\n\n.carousel.pointer-event {\n  touch-action: pan-y;\n}\n\n.carousel-inner {\n  position: relative;\n  width: 100%;\n  overflow: hidden;\n  @include clearfix();\n}\n\n.carousel-item {\n  position: relative;\n  display: none;\n  float: left;\n  width: 100%;\n  margin-right: -100%;\n  backface-visibility: hidden;\n  @include transition($carousel-transition);\n}\n\n.carousel-item.active,\n.carousel-item-next,\n.carousel-item-prev {\n  display: block;\n}\n\n.carousel-item-next:not(.carousel-item-start),\n.active.carousel-item-end {\n  transform: translateX(100%);\n}\n\n.carousel-item-prev:not(.carousel-item-end),\n.active.carousel-item-start {\n  transform: translateX(-100%);\n}\n\n\n//\n// Alternate transitions\n//\n\n.carousel-fade {\n  .carousel-item {\n    opacity: 0;\n    transition-property: opacity;\n    transform: none;\n  }\n\n  .carousel-item.active,\n  .carousel-item-next.carousel-item-start,\n  .carousel-item-prev.carousel-item-end {\n    z-index: 1;\n    opacity: 1;\n  }\n\n  .active.carousel-item-start,\n  .active.carousel-item-end {\n    z-index: 0;\n    opacity: 0;\n    @include transition(opacity 0s $carousel-transition-duration);\n  }\n}\n\n\n//\n// Left/right controls for nav\n//\n\n.carousel-control-prev,\n.carousel-control-next {\n  position: absolute;\n  top: 0;\n  bottom: 0;\n  z-index: 1;\n  // Use flex for alignment (1-3)\n  display: flex; // 1. allow flex styles\n  align-items: center; // 2. vertically center contents\n  justify-content: center; // 3. horizontally center contents\n  width: $carousel-control-width;\n  padding: 0;\n  color: $carousel-control-color;\n  text-align: center;\n  background: none;\n  border: 0;\n  opacity: $carousel-control-opacity;\n  @include transition($carousel-control-transition);\n\n  // Hover/focus state\n  &:hover,\n  &:focus {\n    color: $carousel-control-color;\n    text-decoration: none;\n    outline: 0;\n    opacity: $carousel-control-hover-opacity;\n  }\n}\n.carousel-control-prev {\n  left: 0;\n  background-image: if($enable-gradients, linear-gradient(90deg, rgba($black, .25), rgba($black, .001)), null);\n}\n.carousel-control-next {\n  right: 0;\n  background-image: if($enable-gradients, linear-gradient(270deg, rgba($black, .25), rgba($black, .001)), null);\n}\n\n// Icons for within\n.carousel-control-prev-icon,\n.carousel-control-next-icon {\n  display: inline-block;\n  width: $carousel-control-icon-width;\n  height: $carousel-control-icon-width;\n  background-repeat: no-repeat;\n  background-position: 50%;\n  background-size: 100% 100%;\n}\n\n/* rtl:options: {\n  \"autoRename\": true,\n  \"stringMap\":[ {\n    \"name\"    : \"prev-next\",\n    \"search\"  : \"prev\",\n    \"replace\" : \"next\"\n  } ]\n} */\n.carousel-control-prev-icon {\n  background-image: escape-svg($carousel-control-prev-icon-bg);\n}\n.carousel-control-next-icon {\n  background-image: escape-svg($carousel-control-next-icon-bg);\n}\n\n// Optional indicator pips/controls\n//\n// Add a container (such as a list) with the following class and add an item (ideally a focusable control,\n// like a button) with data-bs-target for each slide your carousel holds.\n\n.carousel-indicators {\n  position: absolute;\n  right: 0;\n  bottom: 0;\n  left: 0;\n  z-index: 2;\n  display: flex;\n  justify-content: center;\n  padding: 0;\n  // Use the .carousel-control's width as margin so we don't overlay those\n  margin-right: $carousel-control-width;\n  margin-bottom: 1rem;\n  margin-left: $carousel-control-width;\n\n  [data-bs-target] {\n    box-sizing: content-box;\n    flex: 0 1 auto;\n    width: $carousel-indicator-width;\n    height: $carousel-indicator-height;\n    padding: 0;\n    margin-right: $carousel-indicator-spacer;\n    margin-left: $carousel-indicator-spacer;\n    text-indent: -999px;\n    cursor: pointer;\n    background-color: $carousel-indicator-active-bg;\n    background-clip: padding-box;\n    border: 0;\n    // Use transparent borders to increase the hit area by 10px on top and bottom.\n    border-top: $carousel-indicator-hit-area-height solid transparent;\n    border-bottom: $carousel-indicator-hit-area-height solid transparent;\n    opacity: $carousel-indicator-opacity;\n    @include transition($carousel-indicator-transition);\n  }\n\n  .active {\n    opacity: $carousel-indicator-active-opacity;\n  }\n}\n\n\n// Optional captions\n//\n//\n\n.carousel-caption {\n  position: absolute;\n  right: (100% - $carousel-caption-width) * .5;\n  bottom: $carousel-caption-spacer;\n  left: (100% - $carousel-caption-width) * .5;\n  padding-top: $carousel-caption-padding-y;\n  padding-bottom: $carousel-caption-padding-y;\n  color: $carousel-caption-color;\n  text-align: center;\n}\n\n// Dark mode carousel\n\n@mixin carousel-dark() {\n  .carousel-control-prev-icon,\n  .carousel-control-next-icon {\n    filter: $carousel-dark-control-icon-filter;\n  }\n\n  .carousel-indicators [data-bs-target] {\n    background-color: $carousel-dark-indicator-active-bg;\n  }\n\n  .carousel-caption {\n    color: $carousel-dark-caption-color;\n  }\n}\n\n.carousel-dark {\n  @include carousel-dark();\n}\n\n@if $enable-dark-mode {\n  @include color-mode(dark) {\n    @if $color-mode-type == \"media-query\" {\n      .carousel {\n        @include carousel-dark();\n      }\n    } @else {\n      .carousel,\n      &.carousel {\n        @include carousel-dark();\n      }\n    }\n  }\n}\n","// scss-docs-start clearfix\n@mixin clearfix() {\n  &::after {\n    display: block;\n    clear: both;\n    content: \"\";\n  }\n}\n// scss-docs-end clearfix\n","//\n// Rotating border\n//\n\n.spinner-grow,\n.spinner-border {\n  display: inline-block;\n  width: var(--#{$prefix}spinner-width);\n  height: var(--#{$prefix}spinner-height);\n  vertical-align: var(--#{$prefix}spinner-vertical-align);\n  // stylelint-disable-next-line property-disallowed-list\n  border-radius: 50%;\n  animation: var(--#{$prefix}spinner-animation-speed) linear infinite var(--#{$prefix}spinner-animation-name);\n}\n\n// scss-docs-start spinner-border-keyframes\n@keyframes spinner-border {\n  to { transform: rotate(360deg) #{\"/* rtl:ignore */\"}; }\n}\n// scss-docs-end spinner-border-keyframes\n\n.spinner-border {\n  // scss-docs-start spinner-border-css-vars\n  --#{$prefix}spinner-width: #{$spinner-width};\n  --#{$prefix}spinner-height: #{$spinner-height};\n  --#{$prefix}spinner-vertical-align: #{$spinner-vertical-align};\n  --#{$prefix}spinner-border-width: #{$spinner-border-width};\n  --#{$prefix}spinner-animation-speed: #{$spinner-animation-speed};\n  --#{$prefix}spinner-animation-name: spinner-border;\n  // scss-docs-end spinner-border-css-vars\n\n  border: var(--#{$prefix}spinner-border-width) solid currentcolor;\n  border-right-color: transparent;\n}\n\n.spinner-border-sm {\n  // scss-docs-start spinner-border-sm-css-vars\n  --#{$prefix}spinner-width: #{$spinner-width-sm};\n  --#{$prefix}spinner-height: #{$spinner-height-sm};\n  --#{$prefix}spinner-border-width: #{$spinner-border-width-sm};\n  // scss-docs-end spinner-border-sm-css-vars\n}\n\n//\n// Growing circle\n//\n\n// scss-docs-start spinner-grow-keyframes\n@keyframes spinner-grow {\n  0% {\n    transform: scale(0);\n  }\n  50% {\n    opacity: 1;\n    transform: none;\n  }\n}\n// scss-docs-end spinner-grow-keyframes\n\n.spinner-grow {\n  // scss-docs-start spinner-grow-css-vars\n  --#{$prefix}spinner-width: #{$spinner-width};\n  --#{$prefix}spinner-height: #{$spinner-height};\n  --#{$prefix}spinner-vertical-align: #{$spinner-vertical-align};\n  --#{$prefix}spinner-animation-speed: #{$spinner-animation-speed};\n  --#{$prefix}spinner-animation-name: spinner-grow;\n  // scss-docs-end spinner-grow-css-vars\n\n  background-color: currentcolor;\n  opacity: 0;\n}\n\n.spinner-grow-sm {\n  --#{$prefix}spinner-width: #{$spinner-width-sm};\n  --#{$prefix}spinner-height: #{$spinner-height-sm};\n}\n\n@if $enable-reduced-motion {\n  @media (prefers-reduced-motion: reduce) {\n    .spinner-border,\n    .spinner-grow {\n      --#{$prefix}spinner-animation-speed: #{$spinner-animation-speed * 2};\n    }\n  }\n}\n","// stylelint-disable function-disallowed-list\n\n%offcanvas-css-vars {\n  // scss-docs-start offcanvas-css-vars\n  --#{$prefix}offcanvas-zindex: #{$zindex-offcanvas};\n  --#{$prefix}offcanvas-width: #{$offcanvas-horizontal-width};\n  --#{$prefix}offcanvas-height: #{$offcanvas-vertical-height};\n  --#{$prefix}offcanvas-padding-x: #{$offcanvas-padding-x};\n  --#{$prefix}offcanvas-padding-y: #{$offcanvas-padding-y};\n  --#{$prefix}offcanvas-color: #{$offcanvas-color};\n  --#{$prefix}offcanvas-bg: #{$offcanvas-bg-color};\n  --#{$prefix}offcanvas-border-width: #{$offcanvas-border-width};\n  --#{$prefix}offcanvas-border-color: #{$offcanvas-border-color};\n  --#{$prefix}offcanvas-box-shadow: #{$offcanvas-box-shadow};\n  --#{$prefix}offcanvas-transition: #{transform $offcanvas-transition-duration ease-in-out};\n  --#{$prefix}offcanvas-title-line-height: #{$offcanvas-title-line-height};\n  // scss-docs-end offcanvas-css-vars\n}\n\n@each $breakpoint in map-keys($grid-breakpoints) {\n  $next: breakpoint-next($breakpoint, $grid-breakpoints);\n  $infix: breakpoint-infix($next, $grid-breakpoints);\n\n  .offcanvas#{$infix} {\n    @extend %offcanvas-css-vars;\n  }\n}\n\n@each $breakpoint in map-keys($grid-breakpoints) {\n  $next: breakpoint-next($breakpoint, $grid-breakpoints);\n  $infix: breakpoint-infix($next, $grid-breakpoints);\n\n  .offcanvas#{$infix} {\n    @include media-breakpoint-down($next) {\n      position: fixed;\n      bottom: 0;\n      z-index: var(--#{$prefix}offcanvas-zindex);\n      display: flex;\n      flex-direction: column;\n      max-width: 100%;\n      color: var(--#{$prefix}offcanvas-color);\n      visibility: hidden;\n      background-color: var(--#{$prefix}offcanvas-bg);\n      background-clip: padding-box;\n      outline: 0;\n      @include box-shadow(var(--#{$prefix}offcanvas-box-shadow));\n      @include transition(var(--#{$prefix}offcanvas-transition));\n\n      &.offcanvas-start {\n        top: 0;\n        left: 0;\n        width: var(--#{$prefix}offcanvas-width);\n        border-right: var(--#{$prefix}offcanvas-border-width) solid var(--#{$prefix}offcanvas-border-color);\n        transform: translateX(-100%);\n      }\n\n      &.offcanvas-end {\n        top: 0;\n        right: 0;\n        width: var(--#{$prefix}offcanvas-width);\n        border-left: var(--#{$prefix}offcanvas-border-width) solid var(--#{$prefix}offcanvas-border-color);\n        transform: translateX(100%);\n      }\n\n      &.offcanvas-top {\n        top: 0;\n        right: 0;\n        left: 0;\n        height: var(--#{$prefix}offcanvas-height);\n        max-height: 100%;\n        border-bottom: var(--#{$prefix}offcanvas-border-width) solid var(--#{$prefix}offcanvas-border-color);\n        transform: translateY(-100%);\n      }\n\n      &.offcanvas-bottom {\n        right: 0;\n        left: 0;\n        height: var(--#{$prefix}offcanvas-height);\n        max-height: 100%;\n        border-top: var(--#{$prefix}offcanvas-border-width) solid var(--#{$prefix}offcanvas-border-color);\n        transform: translateY(100%);\n      }\n\n      &.showing,\n      &.show:not(.hiding) {\n        transform: none;\n      }\n\n      &.showing,\n      &.hiding,\n      &.show {\n        visibility: visible;\n      }\n    }\n\n    @if not ($infix == \"\") {\n      @include media-breakpoint-up($next) {\n        --#{$prefix}offcanvas-height: auto;\n        --#{$prefix}offcanvas-border-width: 0;\n        background-color: transparent !important; // stylelint-disable-line declaration-no-important\n\n        .offcanvas-header {\n          display: none;\n        }\n\n        .offcanvas-body {\n          display: flex;\n          flex-grow: 0;\n          padding: 0;\n          overflow-y: visible;\n          // Reset `background-color` in case `.bg-*` classes are used in offcanvas\n          background-color: transparent !important; // stylelint-disable-line declaration-no-important\n        }\n      }\n    }\n  }\n}\n\n.offcanvas-backdrop {\n  @include overlay-backdrop($zindex-offcanvas-backdrop, $offcanvas-backdrop-bg, $offcanvas-backdrop-opacity);\n}\n\n.offcanvas-header {\n  display: flex;\n  align-items: center;\n  justify-content: space-between;\n  padding: var(--#{$prefix}offcanvas-padding-y) var(--#{$prefix}offcanvas-padding-x);\n\n  .btn-close {\n    padding: calc(var(--#{$prefix}offcanvas-padding-y) * .5) calc(var(--#{$prefix}offcanvas-padding-x) * .5);\n    margin-top: calc(-.5 * var(--#{$prefix}offcanvas-padding-y));\n    margin-right: calc(-.5 * var(--#{$prefix}offcanvas-padding-x));\n    margin-bottom: calc(-.5 * var(--#{$prefix}offcanvas-padding-y));\n  }\n}\n\n.offcanvas-title {\n  margin-bottom: 0;\n  line-height: var(--#{$prefix}offcanvas-title-line-height);\n}\n\n.offcanvas-body {\n  flex-grow: 1;\n  padding: var(--#{$prefix}offcanvas-padding-y) var(--#{$prefix}offcanvas-padding-x);\n  overflow-y: auto;\n}\n",".placeholder {\n  display: inline-block;\n  min-height: 1em;\n  vertical-align: middle;\n  cursor: wait;\n  background-color: currentcolor;\n  opacity: $placeholder-opacity-max;\n\n  &.btn::before {\n    display: inline-block;\n    content: \"\";\n  }\n}\n\n// Sizing\n.placeholder-xs {\n  min-height: .6em;\n}\n\n.placeholder-sm {\n  min-height: .8em;\n}\n\n.placeholder-lg {\n  min-height: 1.2em;\n}\n\n// Animation\n.placeholder-glow {\n  .placeholder {\n    animation: placeholder-glow 2s ease-in-out infinite;\n  }\n}\n\n@keyframes placeholder-glow {\n  50% {\n    opacity: $placeholder-opacity-min;\n  }\n}\n\n.placeholder-wave {\n  mask-image: linear-gradient(130deg, $black 55%, rgba(0, 0, 0, (1 - $placeholder-opacity-min)) 75%, $black 95%);\n  mask-size: 200% 100%;\n  animation: placeholder-wave 2s linear infinite;\n}\n\n@keyframes placeholder-wave {\n  100% {\n    mask-position: -200% 0%;\n  }\n}\n","// All-caps `RGBA()` function used because of this Sass bug: https://github.com/sass/node-sass/issues/2251\n@each $color, $value in $theme-colors {\n  .text-bg-#{$color} {\n    color: color-contrast($value) if($enable-important-utilities, !important, null);\n    background-color: RGBA(var(--#{$prefix}#{$color}-rgb), var(--#{$prefix}bg-opacity, 1)) if($enable-important-utilities, !important, null);\n  }\n}\n","// All-caps `RGBA()` function used because of this Sass bug: https://github.com/sass/node-sass/issues/2251\n@each $color, $value in $theme-colors {\n  .link-#{$color} {\n    color: RGBA(var(--#{$prefix}#{$color}-rgb), var(--#{$prefix}link-opacity, 1)) if($enable-important-utilities, !important, null);\n    text-decoration-color: RGBA(var(--#{$prefix}#{$color}-rgb), var(--#{$prefix}link-underline-opacity, 1)) if($enable-important-utilities, !important, null);\n\n    @if $link-shade-percentage != 0 {\n      &:hover,\n      &:focus {\n        $hover-color: if(color-contrast($value) == $color-contrast-light, shade-color($value, $link-shade-percentage), tint-color($value, $link-shade-percentage));\n        color: RGBA(#{to-rgb($hover-color)}, var(--#{$prefix}link-opacity, 1)) if($enable-important-utilities, !important, null);\n        text-decoration-color: RGBA(to-rgb($hover-color), var(--#{$prefix}link-underline-opacity, 1)) if($enable-important-utilities, !important, null);\n      }\n    }\n  }\n}\n\n// One-off special link helper as a bridge until v6\n.link-body-emphasis {\n  color: RGBA(var(--#{$prefix}emphasis-color-rgb), var(--#{$prefix}link-opacity, 1)) if($enable-important-utilities, !important, null);\n  text-decoration-color: RGBA(var(--#{$prefix}emphasis-color-rgb), var(--#{$prefix}link-underline-opacity, 1)) if($enable-important-utilities, !important, null);\n\n  @if $link-shade-percentage != 0 {\n    &:hover,\n    &:focus {\n      color: RGBA(var(--#{$prefix}emphasis-color-rgb), var(--#{$prefix}link-opacity, .75)) if($enable-important-utilities, !important, null);\n      text-decoration-color: RGBA(var(--#{$prefix}emphasis-color-rgb), var(--#{$prefix}link-underline-opacity, .75)) if($enable-important-utilities, !important, null);\n    }\n  }\n}\n",".focus-ring:focus {\n  outline: 0;\n  // By default, there is no `--bs-focus-ring-x`, `--bs-focus-ring-y`, or `--bs-focus-ring-blur`, but we provide CSS variables with fallbacks to initial `0` values\n  box-shadow: var(--#{$prefix}focus-ring-x, 0) var(--#{$prefix}focus-ring-y, 0) var(--#{$prefix}focus-ring-blur, 0) var(--#{$prefix}focus-ring-width) var(--#{$prefix}focus-ring-color);\n}\n",".icon-link {\n  display: inline-flex;\n  gap: $icon-link-gap;\n  align-items: center;\n  text-decoration-color: rgba(var(--#{$prefix}link-color-rgb), var(--#{$prefix}link-opacity, .5));\n  text-underline-offset: $icon-link-underline-offset;\n  backface-visibility: hidden;\n\n  > .bi {\n    flex-shrink: 0;\n    width: $icon-link-icon-size;\n    height: $icon-link-icon-size;\n    fill: currentcolor;\n    @include transition($icon-link-icon-transition);\n  }\n}\n\n.icon-link-hover {\n  &:hover,\n  &:focus-visible {\n    > .bi {\n      transform: var(--#{$prefix}icon-link-transform, $icon-link-icon-transform);\n    }\n  }\n}\n","// Credit: Nicolas Gallagher and SUIT CSS.\n\n.ratio {\n  position: relative;\n  width: 100%;\n\n  &::before {\n    display: block;\n    padding-top: var(--#{$prefix}aspect-ratio);\n    content: \"\";\n  }\n\n  > * {\n    position: absolute;\n    top: 0;\n    left: 0;\n    width: 100%;\n    height: 100%;\n  }\n}\n\n@each $key, $ratio in $aspect-ratios {\n  .ratio-#{$key} {\n    --#{$prefix}aspect-ratio: #{$ratio};\n  }\n}\n","// Shorthand\n\n.fixed-top {\n  position: fixed;\n  top: 0;\n  right: 0;\n  left: 0;\n  z-index: $zindex-fixed;\n}\n\n.fixed-bottom {\n  position: fixed;\n  right: 0;\n  bottom: 0;\n  left: 0;\n  z-index: $zindex-fixed;\n}\n\n// Responsive sticky top and bottom\n@each $breakpoint in map-keys($grid-breakpoints) {\n  @include media-breakpoint-up($breakpoint) {\n    $infix: breakpoint-infix($breakpoint, $grid-breakpoints);\n\n    .sticky#{$infix}-top {\n      position: sticky;\n      top: 0;\n      z-index: $zindex-sticky;\n    }\n\n    .sticky#{$infix}-bottom {\n      position: sticky;\n      bottom: 0;\n      z-index: $zindex-sticky;\n    }\n  }\n}\n","// scss-docs-start stacks\n.hstack {\n  display: flex;\n  flex-direction: row;\n  align-items: center;\n  align-self: stretch;\n}\n\n.vstack {\n  display: flex;\n  flex: 1 1 auto;\n  flex-direction: column;\n  align-self: stretch;\n}\n// scss-docs-end stacks\n","//\n// Visually hidden\n//\n\n.visually-hidden,\n.visually-hidden-focusable:not(:focus):not(:focus-within) {\n  @include visually-hidden();\n}\n","// stylelint-disable declaration-no-important\n\n// Hide content visually while keeping it accessible to assistive technologies\n//\n// See: https://www.a11yproject.com/posts/2013-01-11-how-to-hide-content/\n// See: https://kittygiraudel.com/2016/10/13/css-hide-and-seek/\n\n@mixin visually-hidden() {\n  width: 1px !important;\n  height: 1px !important;\n  padding: 0 !important;\n  margin: -1px !important; // Fix for https://github.com/twbs/bootstrap/issues/25686\n  overflow: hidden !important;\n  clip: rect(0, 0, 0, 0) !important;\n  white-space: nowrap !important;\n  border: 0 !important;\n\n  // Fix for positioned table caption that could become anonymous cells\n  &:not(caption) {\n    position: absolute !important;\n  }\n}\n\n// Use to only display content when it's focused, or one of its child elements is focused\n// (i.e. when focus is within the element/container that the class was applied to)\n//\n// Useful for \"Skip to main content\" links; see https://www.w3.org/TR/2013/NOTE-WCAG20-TECHS-20130905/G1\n\n@mixin visually-hidden-focusable() {\n  &:not(:focus):not(:focus-within) {\n    @include visually-hidden();\n  }\n}\n","//\n// Stretched link\n//\n\n.stretched-link {\n  &::#{$stretched-link-pseudo-element} {\n    position: absolute;\n    top: 0;\n    right: 0;\n    bottom: 0;\n    left: 0;\n    z-index: $stretched-link-z-index;\n    content: \"\";\n  }\n}\n","//\n// Text truncation\n//\n\n.text-truncate {\n  @include text-truncate();\n}\n","// Text truncate\n// Requires inline-block or block for proper styling\n\n@mixin text-truncate() {\n  overflow: hidden;\n  text-overflow: ellipsis;\n  white-space: nowrap;\n}\n",".vr {\n  display: inline-block;\n  align-self: stretch;\n  width: $vr-border-width;\n  min-height: 1em;\n  background-color: currentcolor;\n  opacity: $hr-opacity;\n}\n","// Utility generator\n// Used to generate utilities & print utilities\n@mixin generate-utility($utility, $infix: \"\", $is-rfs-media-query: false) {\n  $values: map-get($utility, values);\n\n  // If the values are a list or string, convert it into a map\n  @if type-of($values) == \"string\" or type-of(nth($values, 1)) != \"list\" {\n    $values: zip($values, $values);\n  }\n\n  @each $key, $value in $values {\n    $properties: map-get($utility, property);\n\n    // Multiple properties are possible, for example with vertical or horizontal margins or paddings\n    @if type-of($properties) == \"string\" {\n      $properties: append((), $properties);\n    }\n\n    // Use custom class if present\n    $property-class: if(map-has-key($utility, class), map-get($utility, class), nth($properties, 1));\n    $property-class: if($property-class == null, \"\", $property-class);\n\n    // Use custom CSS variable name if present, otherwise default to `class`\n    $css-variable-name: if(map-has-key($utility, css-variable-name), map-get($utility, css-variable-name), map-get($utility, class));\n\n    // State params to generate pseudo-classes\n    $state: if(map-has-key($utility, state), map-get($utility, state), ());\n\n    $infix: if($property-class == \"\" and str-slice($infix, 1, 1) == \"-\", str-slice($infix, 2), $infix);\n\n    // Don't prefix if value key is null (e.g. with shadow class)\n    $property-class-modifier: if($key, if($property-class == \"\" and $infix == \"\", \"\", \"-\") + $key, \"\");\n\n    @if map-get($utility, rfs) {\n      // Inside the media query\n      @if $is-rfs-media-query {\n        $val: rfs-value($value);\n\n        // Do not render anything if fluid and non fluid values are the same\n        $value: if($val == rfs-fluid-value($value), null, $val);\n      }\n      @else {\n        $value: rfs-fluid-value($value);\n      }\n    }\n\n    $is-css-var: map-get($utility, css-var);\n    $is-local-vars: map-get($utility, local-vars);\n    $is-rtl: map-get($utility, rtl);\n\n    @if $value != null {\n      @if $is-rtl == false {\n        /* rtl:begin:remove */\n      }\n\n      @if $is-css-var {\n        .#{$property-class + $infix + $property-class-modifier} {\n          --#{$prefix}#{$css-variable-name}: #{$value};\n        }\n\n        @each $pseudo in $state {\n          .#{$property-class + $infix + $property-class-modifier}-#{$pseudo}:#{$pseudo} {\n            --#{$prefix}#{$css-variable-name}: #{$value};\n          }\n        }\n      } @else {\n        .#{$property-class + $infix + $property-class-modifier} {\n          @each $property in $properties {\n            @if $is-local-vars {\n              @each $local-var, $variable in $is-local-vars {\n                --#{$prefix}#{$local-var}: #{$variable};\n              }\n            }\n            #{$property}: $value if($enable-important-utilities, !important, null);\n          }\n        }\n\n        @each $pseudo in $state {\n          .#{$property-class + $infix + $property-class-modifier}-#{$pseudo}:#{$pseudo} {\n            @each $property in $properties {\n              @if $is-local-vars {\n                @each $local-var, $variable in $is-local-vars {\n                  --#{$prefix}#{$local-var}: #{$variable};\n                }\n              }\n              #{$property}: $value if($enable-important-utilities, !important, null);\n            }\n          }\n        }\n      }\n\n      @if $is-rtl == false {\n        /* rtl:end:remove */\n      }\n    }\n  }\n}\n","// Loop over each breakpoint\n@each $breakpoint in map-keys($grid-breakpoints) {\n\n  // Generate media query if needed\n  @include media-breakpoint-up($breakpoint) {\n    $infix: breakpoint-infix($breakpoint, $grid-breakpoints);\n\n    // Loop over each utility property\n    @each $key, $utility in $utilities {\n      // The utility can be disabled with `false`, thus check if the utility is a map first\n      // Only proceed if responsive media queries are enabled or if it's the base media query\n      @if type-of($utility) == \"map\" and (map-get($utility, responsive) or $infix == \"\") {\n        @include 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the utility needs print styles\n    @if type-of($utility) == \"map\" and map-get($utility, print) == true {\n      @include generate-utility($utility, \"-print\");\n    }\n  }\n}\n"],"names":[],"sourceRoot":""}
\ No newline at end of file
diff --git a/docs/_build/html/_static/styles/pydata-sphinx-theme.css b/docs/_build/html/_static/styles/pydata-sphinx-theme.css
deleted file mode 100644
index db7635b..0000000
--- a/docs/_build/html/_static/styles/pydata-sphinx-theme.css
+++ /dev/null
@@ -1,2 +0,0 @@
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var(--sd-color-white);--pst-sd-dropdown-bg-color: var(--sd-color-white-bg)}details.sd-dropdown summary.sd-card-header.sd-bg-text-white{color:var(--sd-color-white-bg-text) !important}details.sd-dropdown summary.sd-card-header:before{content:\"\";width:100%;height:100%;position:absolute;left:0;top:0;background-color:var(--pst-sd-dropdown-bg-color);z-index:-1;pointer-events:none}details.sd-dropdown summary.sd-card-header+div.sd-summary-content{border-left:.2rem solid var(--pst-sd-dropdown-color) !important;border-bottom-left-radius:calc(.25rem - 1px);background-color:var(--pst-color-on-background)}details.sd-dropdown summary.sd-card-header span.sd-summary-icon{display:inline-flex;align-items:center;color:var(--pst-sd-dropdown-color) !important}details.sd-dropdown summary.sd-card-header span.sd-summary-icon svg{opacity:1}details.sd-dropdown summary.sd-card-header .sd-summary-up,details.sd-dropdown summary.sd-card-header .sd-summary-down{top:.7rem}.bd-content .admonition button.toggle-button{color:inherit}.bd-content details.toggle-details summary{border-left:3px solid var(--pst-color-primary)}html div.rendered_html html .jp-RenderedHTMLCommon table{table-layout:auto}html[data-theme=dark] .bd-content .nboutput .output_area.rendered_html{color:var(--pst-color-on-background);background-color:var(--pst-color-text-base);border-radius:.25rem;padding:.5rem}html[data-theme=dark] .bd-content .nboutput .output_area.stderr{background:var(--pst-color-danger)}div.nblast.container{margin-bottom:1rem}div.cell_output .output{max-width:100%;overflow-x:auto}html[data-theme=dark] .bd-content div.cell_output img,html[data-theme=dark] .bd-content div.cell_output .text_html{color:var(--pst-color-on-background);background-color:var(--pst-color-text-base);border-radius:.25rem;padding:.5rem}.bd-content div.cell_input{display:flex;flex-direction:column;justify-content:stretch}.bd-content div.cell_input,.bd-content div.output{border-radius:.25rem}.bd-content div.output table{table-layout:auto}html[data-theme=dark] .bd-content img.leaflet-tile.leaflet-tile-loaded{border-radius:0;padding:0}.bd-search-container div#search-results>h2{font-size:var(--pst-font-size-icon);margin-top:0}.bd-search-container div#search-results p.search-summary{color:var(--pst-color-text-muted)}.bd-search-container ul.search{margin:0;list-style:none}.bd-search-container ul.search li{background-image:none;padding:1rem 0;margin:1rem 0;border-top:1px solid var(--pst-color-text-muted)}.bd-search-container ul.search li>a{font-size:1.2em}.bd-search-container ul.search li div.context,.bd-search-container ul.search li p.context{color:var(--pst-color-text-base);margin:.5em 0 0 0}.bd-search-container ul.search li div.context a:before,.bd-search-container ul.search li p.context a:before{content:\"#\";padding-right:.2em;color:var(--pst-color-text-muted)}","html {\n  /*****************************************************************************\n  * Overall Layout Variables\n  */\n\n  // Header height will impact the top offset for many sections\n  // Article header is 66% of Header\n  --pst-header-height: 4rem;\n  --pst-header-article-height: calc(var(--pst-header-height) * 2 / 3);\n  --pst-sidebar-secondary: 17rem;\n}\n\n/*******************************************************************************\n* Breakpoints that trigger UI changes\n*\n* Note that media-breakpoint-down begins at the next highest level!\n* So we should choose a media-breakpoint-down one *lower* than when we want to start\n* example: media-breakpoint-up(md) and media-breakpoint-down(sm) trigger at the same time\n* ref: https://github.com/twbs/bootstrap/issues/31214\n*/\n$breakpoint-sidebar-primary: lg; // When we collapse the primary sidebar\n$breakpoint-sidebar-secondary: xl; // When we collapse the secondary sidebar\n$breakpoint-page-width: 88rem; // taken from sphinx-basic-ng, which we are ultimately going to inherit\n\n/*******************************************************************************\n* Define the animation behaviour\n*/\n$animation-time: 200ms;\n\n/*******************************************************************************\n* UI shaping and padding\n*/\n$admonition-border-radius: 0.25rem;\n","html {\n  /*****************************************************************************\n  * Font features used in this theme\n  */\n\n  // base font size - applied at body/html level\n  --pst-font-size-base: 1rem;\n\n  // heading font sizes based on bootstrap sizing\n  --pst-font-size-h1: 2.5rem;\n  --pst-font-size-h2: 2rem;\n  --pst-font-size-h3: 1.75rem;\n  --pst-font-size-h4: 1.5rem;\n  --pst-font-size-h5: 1.25rem;\n  --pst-font-size-h6: 1.1rem;\n\n  // smaller than heading font sizes\n  --pst-font-size-milli: 0.9rem;\n\n  // Sidebar styles\n  --pst-sidebar-font-size: 0.9rem;\n  --pst-sidebar-font-size-mobile: 1.1rem;\n  --pst-sidebar-header-font-size: 1.2rem;\n  --pst-sidebar-header-font-weight: 600;\n\n  // Admonition styles\n  --pst-admonition-font-weight-heading: 600;\n\n  // Font weights\n  --pst-font-weight-caption: 300;\n  --pst-font-weight-heading: 400;\n\n  // Font family\n  // These are adapted from https://systemfontstack.com/ */\n  --pst-font-family-base-system: -apple-system, BlinkMacSystemFont, Segoe UI,\n    \"Helvetica Neue\", Arial, sans-serif, Apple Color Emoji, Segoe UI Emoji,\n    Segoe UI Symbol;\n  --pst-font-family-monospace-system: \"SFMono-Regular\", Menlo, Consolas, Monaco,\n    Liberation Mono, Lucida Console, monospace;\n\n  --pst-font-family-base: var(--pst-font-family-base-system);\n  --pst-font-family-heading: var(--pst-font-family-base-system);\n  --pst-font-family-monospace: var(--pst-font-family-monospace-system);\n}\n","html {\n  /*****************************************************************************\n  * Icon\n  */\n\n  // Font size across all icons\n  --pst-font-size-icon: 1.5rem;\n\n  // Font Awesome default icons\n  --pst-icon-check-circle: \"\\f058\"; // fa-solid fa-circle-check\n  --pst-icon-info-circle: \"\\f05a\"; // fa-solid fa-circle-info\n  --pst-icon-exclamation-triangle: \"\\f071\"; // fa-solid fa-triangle-exclamation\n  --pst-icon-exclamation-circle: \"\\f06a\"; // fa-solid fa-circle-exclamation\n  --pst-icon-times-circle: \"\\f057\"; // fa-solid fa-circle-xmark\n  --pst-icon-lightbulb: \"\\f0eb\"; // fa-solid fa-lightbulb\n  --pst-icon-download: \"\\f019\"; // fa-solid fa-download\n  --pst-icon-angle-left: \"\\f104\"; // fa-solid fa-angle-left\n  --pst-icon-angle-right: \"\\f105\"; // fa-solid fa-angle-right\n  --pst-icon-external-link: \"\\f35d\"; // fa-solid fa-up-right-from-square\n  --pst-icon-search-minus: \"\\f010\"; // fa-solid fa-magnifying-glass-minus\n  --pst-icon-github: \"\\f09b\"; // fa-brands fa-github\n  --pst-icon-gitlab: \"\\f296\"; // fa-brands fa-gitlab\n  --pst-icon-share: \"\\f064\"; // fa-solid fa-share\n  --pst-icon-bell: \"\\f0f3\"; // fa-solid fa-bell\n  --pst-icon-pencil: \"\\f303\"; // fa-solid fa-pencil\n\n  // Bootstrap icons\n  --pst-breadcrumb-divider: \"\\f105\";\n}\n","html {\n  /*****************************************************************************\n  * Admonitions\n  **/\n\n  --pst-icon-admonition-default: var(--pst-icon-bell);\n  --pst-icon-admonition-note: var(--pst-icon-info-circle);\n  --pst-icon-admonition-attention: var(--pst-icon-exclamation-circle);\n  --pst-icon-admonition-caution: var(--pst-icon-exclamation-triangle);\n  --pst-icon-admonition-warning: var(--pst-icon-exclamation-triangle);\n  --pst-icon-admonition-danger: var(--pst-icon-exclamation-triangle);\n  --pst-icon-admonition-error: var(--pst-icon-times-circle);\n  --pst-icon-admonition-hint: var(--pst-icon-lightbulb);\n  --pst-icon-admonition-tip: var(--pst-icon-lightbulb);\n  --pst-icon-admonition-important: var(--pst-icon-exclamation-circle);\n  --pst-icon-admonition-seealso: var(--pst-icon-share);\n  --pst-icon-admonition-todo: var(--pst-icon-pencil);\n}\n","html {\n  /*****************************************************************************\n  * versionmodified\n  **/\n\n  --pst-icon-versionmodified-default: var(--pst-icon-exclamation-circle);\n  --pst-icon-versionmodified-added: var(--pst-icon-exclamation-circle);\n  --pst-icon-versionmodified-changed: var(--pst-icon-exclamation-circle);\n  --pst-icon-versionmodified-deprecated: var(--pst-icon-exclamation-circle);\n}\n","html {\n  font-size: var(--pst-font-size-base);\n  scroll-padding-top: calc(var(--pst-header-height) + 1rem);\n}\n\nbody {\n  background-color: var(--pst-color-background);\n  font-family: var(--pst-font-family-base);\n  font-weight: 400;\n  line-height: 1.65;\n  color: var(--pst-color-text-base);\n  min-height: 100vh;\n  display: flex;\n  flex-direction: column;\n\n  // hack to avoid the black background on some browser including Safari\n  &::-webkit-scrollbar-track {\n    background: var(--pst-color-background);\n  }\n}\n\np {\n  margin-bottom: 1.15rem;\n  font-size: 1em;\n  color: var(--pst-color-text-base);\n\n  /* section header in docstring pages */\n  &.rubric {\n    border-bottom: 1px solid var(--pst-color-border);\n  }\n\n  &.centered {\n    text-align: center;\n  }\n}\n\na {\n  @include link-style-default;\n\n  // The # anchor that appears on hover over headings\n  &.headerlink {\n    color: var(--pst-color-secondary);\n    opacity: 0.7;\n    font-size: 0.8em;\n    padding: 0 4px 0 4px;\n    margin-left: 0.2em;\n    text-decoration: none;\n    transition: all 0.2s ease-out;\n    user-select: none;\n\n    &:hover {\n      opacity: 1;\n    }\n  }\n\n  // set up a icon next to the shorten links from github and gitlab\n  &.github,\n  &.gitlab {\n    &::before {\n      color: var(--pst-color-text-muted);\n      font: var(--fa-font-brands);\n      margin-right: 0.25rem;\n    }\n  }\n\n  &.github::before {\n    content: var(--pst-icon-github);\n  }\n\n  &.gitlab::before {\n    content: var(--pst-icon-gitlab);\n  }\n}\n\n.heading-style {\n  margin: 2.75rem 0 1.05rem;\n  font-family: var(--pst-font-family-heading);\n  font-weight: var(--pst-font-weight-heading);\n  line-height: 1.15;\n}\n\nh1 {\n  @extend .heading-style;\n  margin-top: 0;\n  font-size: var(--pst-font-size-h1);\n  color: var(--pst-heading-color);\n}\n\nh2 {\n  @extend .heading-style;\n  font-size: var(--pst-font-size-h2);\n  color: var(--pst-heading-color);\n}\n\nh3 {\n  @extend .heading-style;\n  font-size: var(--pst-font-size-h3);\n  color: var(--pst-heading-color);\n}\n\nh4 {\n  @extend .heading-style;\n  font-size: var(--pst-font-size-h4);\n  color: var(--pst-heading-color);\n}\n\nh5 {\n  @extend .heading-style;\n  font-size: var(--pst-font-size-h5);\n  color: var(--pst-color-text-base);\n}\n\nh6 {\n  @extend .heading-style;\n  font-size: var(--pst-font-size-h6);\n  color: var(--pst-color-text-base);\n}\n\nsmall,\n.text_small {\n  font-size: var(--pst-font-size-milli);\n}\n\nhr {\n  border: 0;\n  border-top: 1px solid var(--pst-color-border);\n}\n\npre,\ncode,\nkbd,\nsamp {\n  font-family: var(--pst-font-family-monospace);\n}\n\nkbd {\n  // use theme negative\n  background-color: var(--pst-color-on-background);\n  color: var(--pst-color-text-muted);\n\n  // Compound keyboard elements will have nested kbd in them, so this prevents double lines\n  &:not(.compound) {\n    border: 1px solid var(--pst-color-border);\n    margin: 0 0.1rem;\n    padding: 0.1rem 0.4rem;\n    box-shadow: 1px 1px 1px var(--pst-color-shadow);\n  }\n}\n\ncode {\n  color: var(--pst-color-inline-code);\n}\n\npre {\n  margin: 1.5em 0 1.5em 0;\n  padding: 1rem;\n  background-color: var(--pst-color-surface);\n  color: var(--pst-color-text-base);\n  line-height: 1.2em;\n  border: 1px solid var(--pst-color-border);\n  border-radius: $admonition-border-radius;\n\n  .linenos {\n    // minimum opacity to make the line numbers WCAG AA conformant\n    opacity: 0.8;\n    padding-right: 10px;\n  }\n}\n\n// the back to top btn\n#pst-back-to-top {\n  z-index: $zindex-tooltip;\n  position: fixed;\n  display: none;\n  top: 80vh;\n  left: 50vw;\n  transform: translate(-50%);\n  color: var(--pst-color-secondary-text);\n  background-color: var(--pst-color-secondary);\n  border: none;\n}\n","/**\n * Consistent styling for links\n **/\n\n// Define some useful variables for links styling consistency\n//\n//  Thickness of the underline for links\n// the default will be either:\n//  - 1px\n// - 0.0625rem if it's thicker than 1px because the user has changed the text\n//   size in their browser\n$link-underline-thickness: unquote(\"max(1px, .0625rem)\") !default;\n// Offset of link underlines from text baseline\n// The default is 3px expressed as ems, as calculated against the default body\n// font size (on desktop).\n$link-underline-offset: 0.1578em !default;\n// Thickness of link underlines in hover state\n// The default for each link will be the thickest of the following:\n//  - 3px\n//  - 0.1875rem, if it's thicker than 3px because the user has changed the text\n//    size in their browser\n//  - 0.12em (relative to the link's text size)\n$link-hover-decoration-thickness: unquote(\"max(3px, .1875rem, .12em)\") !default;\n\n//  Ensures links have an underline decoration by default - needed to meet\n// WCAG SC 1.4.1\n@mixin link-decoration {\n  text-decoration: underline;\n\n  @if $link-underline-thickness {\n    text-decoration-thickness: $link-underline-thickness;\n  }\n\n  @if $link-underline-offset {\n    text-underline-offset: $link-underline-offset;\n  }\n}\n\n// Ensures links have an underline decoration on hover - distinct from the\n// default behaviour\n@mixin link-decoration-hover {\n  @if $link-hover-decoration-thickness {\n    text-decoration-thickness: $link-hover-decoration-thickness;\n    // Disable ink skipping on underlines on hover. Browsers haven't\n    // standardised on this part of the spec yet, so set both properties\n    text-decoration-skip-ink: none; // Chromium, Firefox\n    text-decoration-skip: none; // Safari\n  }\n}\n\n// Simple hover style - can be used alone or in conjunction with other mixins\n// Add the text underline and change in thickness on hover\n@mixin link-style-hover {\n  &:hover {\n    @include link-decoration;\n    @include link-decoration-hover;\n    color: var(--pst-color-link-hover);\n  }\n}\n\n// Default link styles\n//\n// Defines: default unvisited, visited, hover, and active.\n// TODO: @trallard to improve focus styles in subsequent PR\n@mixin link-style-default {\n  // So that really long links don't spill out of their container\n  word-wrap: break-word;\n\n  color: var(--pst-color-link);\n  @include link-decoration;\n\n  &:hover {\n    color: var(--pst-color-link-hover);\n    @include link-decoration-hover;\n  }\n\n  // TODO: @trallard to add active styles in subsequent PR\n  &:active {\n    color: var(--pst-color-link);\n  }\n\n  // Visited should still be hoverable\n  &:visited {\n    color: var(--pst-color-link);\n    &:hover {\n      color: var(--pst-color-link-hover);\n    }\n  }\n  @include focus-indicator;\n}\n\n// Text link styles\n//\n// Makes links use the muted text colour and removes the underline.\n// Use this mixin for navigation bar links.\n@mixin link-style-text {\n  color: var(--pst-color-text-muted);\n  text-decoration: none;\n\n  &:hover {\n    color: var(--pst-color-link-hover);\n    @include link-decoration;\n    @include link-decoration-hover;\n  }\n  @include focus-indicator;\n}\n\n// Sidebar and TOC links\n//\n// Makes links use the muted text colour and removes the underline.\n// Use this mixin for navigation the primary sidebar and table of contents.\n// Active and hover should work together rather than one overriding the other.\n@mixin link-sidebar {\n  color: var(--pst-color-text-muted);\n  text-decoration: none;\n\n  &:hover {\n    text-decoration: underline;\n    background-color: transparent;\n    color: var(--pst-color-link-hover);\n    @include link-decoration-hover;\n  }\n\n  // TODO: @trallard to update active styles in subsequent PR\n  &:active {\n    color: var(--pst-color-link-hover);\n  }\n}\n\n// Sidebar current page link styles\n//\n// Adds a vertical line on the left hand side of the link to indicate that\n// it's the current page. Note this is distinct from an active state.\n// Used on the primary sidebar and the TOC.\n// We want the side box shadow to have the same thickness as the hover underline\n@mixin link-sidebar-current {\n  font-weight: 600;\n  color: var(--pst-color-primary);\n  @if $link-hover-decoration-thickness {\n    box-shadow: inset\n      $link-hover-decoration-thickness\n      0px\n      0px\n      var(--pst-color-primary);\n  }\n}\n\n// Navigation bar current page link styles\n//\n// Adds a bottom underline, this leaves enough space for the hover state without\n// cluttering the navbar.\n// We want the side box shadow to have the same thickness as the hover underline\n@mixin link-navbar-current {\n  font-weight: 600;\n  color: var(--pst-color-primary);\n  @if $link-hover-decoration-thickness {\n    border-bottom: $link-hover-decoration-thickness\n      solid\n      var(--pst-color-primary);\n  }\n}\n\n// Navigation bar icon links hover styles\n//\n// Adds a bottom box-shadow - since there is no text we cannot use text-decoration\n// We want the side box shadow to have the same thickness as the hover underline\n@mixin icon-navbar-hover {\n  &:hover {\n    color: var(--pst-color-link-hover);\n    @if $link-hover-decoration-thickness {\n      box-shadow: 0px\n        $link-hover-decoration-thickness\n        0px\n        var(--pst-color-link-hover);\n    }\n  }\n}\n\n// Focus indicator\n@mixin focus-indicator {\n  &:focus-visible {\n    outline: 2px solid var(--pst-color-accent);\n  }\n}\n","// Variables\n//\n// Variables should follow the `$component-state-property-size` formula for\n// consistent naming. Ex: $nav-link-disabled-color and $modal-content-box-shadow-xs.\n\n// Color system\n\n// scss-docs-start gray-color-variables\n$white:    #fff !default;\n$gray-100: #f8f9fa !default;\n$gray-200: #e9ecef !default;\n$gray-300: #dee2e6 !default;\n$gray-400: #ced4da !default;\n$gray-500: #adb5bd !default;\n$gray-600: #6c757d !default;\n$gray-700: #495057 !default;\n$gray-800: #343a40 !default;\n$gray-900: #212529 !default;\n$black:    #000 !default;\n// scss-docs-end gray-color-variables\n\n// fusv-disable\n// scss-docs-start gray-colors-map\n$grays: (\n  \"100\": $gray-100,\n  \"200\": $gray-200,\n  \"300\": $gray-300,\n  \"400\": $gray-400,\n  \"500\": $gray-500,\n  \"600\": $gray-600,\n  \"700\": $gray-700,\n  \"800\": $gray-800,\n  \"900\": $gray-900\n) !default;\n// scss-docs-end gray-colors-map\n// fusv-enable\n\n// scss-docs-start color-variables\n$blue:    #0d6efd !default;\n$indigo:  #6610f2 !default;\n$purple:  #6f42c1 !default;\n$pink:    #d63384 !default;\n$red:     #dc3545 !default;\n$orange:  #fd7e14 !default;\n$yellow:  #ffc107 !default;\n$green:   #198754 !default;\n$teal:    #20c997 !default;\n$cyan:    #0dcaf0 !default;\n// scss-docs-end color-variables\n\n// scss-docs-start colors-map\n$colors: (\n  \"blue\":       $blue,\n  \"indigo\":     $indigo,\n  \"purple\":     $purple,\n  \"pink\":       $pink,\n  \"red\":        $red,\n  \"orange\":     $orange,\n  \"yellow\":     $yellow,\n  \"green\":      $green,\n  \"teal\":       $teal,\n  \"cyan\":       $cyan,\n  \"black\":      $black,\n  \"white\":      $white,\n  \"gray\":       $gray-600,\n  \"gray-dark\":  $gray-800\n) !default;\n// scss-docs-end colors-map\n\n// The contrast ratio to reach against white, to determine if color changes from \"light\" to \"dark\". Acceptable values for WCAG 2.0 are 3, 4.5 and 7.\n// See https://www.w3.org/TR/WCAG20/#visual-audio-contrast-contrast\n$min-contrast-ratio:   4.5 !default;\n\n// Customize the light and dark text colors for use in our color contrast function.\n$color-contrast-dark:      $black !default;\n$color-contrast-light:     $white !default;\n\n// fusv-disable\n$blue-100: tint-color($blue, 80%) !default;\n$blue-200: tint-color($blue, 60%) !default;\n$blue-300: tint-color($blue, 40%) !default;\n$blue-400: tint-color($blue, 20%) !default;\n$blue-500: $blue !default;\n$blue-600: shade-color($blue, 20%) !default;\n$blue-700: shade-color($blue, 40%) !default;\n$blue-800: shade-color($blue, 60%) !default;\n$blue-900: shade-color($blue, 80%) !default;\n\n$indigo-100: tint-color($indigo, 80%) !default;\n$indigo-200: tint-color($indigo, 60%) !default;\n$indigo-300: tint-color($indigo, 40%) !default;\n$indigo-400: tint-color($indigo, 20%) !default;\n$indigo-500: $indigo !default;\n$indigo-600: shade-color($indigo, 20%) !default;\n$indigo-700: shade-color($indigo, 40%) !default;\n$indigo-800: shade-color($indigo, 60%) !default;\n$indigo-900: shade-color($indigo, 80%) !default;\n\n$purple-100: tint-color($purple, 80%) !default;\n$purple-200: tint-color($purple, 60%) !default;\n$purple-300: tint-color($purple, 40%) !default;\n$purple-400: tint-color($purple, 20%) !default;\n$purple-500: $purple !default;\n$purple-600: shade-color($purple, 20%) !default;\n$purple-700: shade-color($purple, 40%) !default;\n$purple-800: shade-color($purple, 60%) !default;\n$purple-900: shade-color($purple, 80%) !default;\n\n$pink-100: tint-color($pink, 80%) !default;\n$pink-200: tint-color($pink, 60%) !default;\n$pink-300: tint-color($pink, 40%) !default;\n$pink-400: tint-color($pink, 20%) !default;\n$pink-500: $pink !default;\n$pink-600: shade-color($pink, 20%) !default;\n$pink-700: shade-color($pink, 40%) !default;\n$pink-800: shade-color($pink, 60%) !default;\n$pink-900: shade-color($pink, 80%) !default;\n\n$red-100: tint-color($red, 80%) !default;\n$red-200: tint-color($red, 60%) !default;\n$red-300: tint-color($red, 40%) !default;\n$red-400: tint-color($red, 20%) !default;\n$red-500: $red !default;\n$red-600: shade-color($red, 20%) !default;\n$red-700: shade-color($red, 40%) !default;\n$red-800: shade-color($red, 60%) !default;\n$red-900: shade-color($red, 80%) !default;\n\n$orange-100: tint-color($orange, 80%) !default;\n$orange-200: tint-color($orange, 60%) !default;\n$orange-300: tint-color($orange, 40%) !default;\n$orange-400: tint-color($orange, 20%) !default;\n$orange-500: $orange !default;\n$orange-600: shade-color($orange, 20%) !default;\n$orange-700: shade-color($orange, 40%) !default;\n$orange-800: shade-color($orange, 60%) !default;\n$orange-900: shade-color($orange, 80%) !default;\n\n$yellow-100: tint-color($yellow, 80%) !default;\n$yellow-200: tint-color($yellow, 60%) !default;\n$yellow-300: tint-color($yellow, 40%) !default;\n$yellow-400: tint-color($yellow, 20%) !default;\n$yellow-500: $yellow !default;\n$yellow-600: shade-color($yellow, 20%) !default;\n$yellow-700: shade-color($yellow, 40%) !default;\n$yellow-800: shade-color($yellow, 60%) !default;\n$yellow-900: shade-color($yellow, 80%) !default;\n\n$green-100: tint-color($green, 80%) !default;\n$green-200: tint-color($green, 60%) !default;\n$green-300: tint-color($green, 40%) !default;\n$green-400: tint-color($green, 20%) !default;\n$green-500: $green !default;\n$green-600: shade-color($green, 20%) !default;\n$green-700: shade-color($green, 40%) !default;\n$green-800: shade-color($green, 60%) !default;\n$green-900: shade-color($green, 80%) !default;\n\n$teal-100: tint-color($teal, 80%) !default;\n$teal-200: tint-color($teal, 60%) !default;\n$teal-300: tint-color($teal, 40%) !default;\n$teal-400: tint-color($teal, 20%) !default;\n$teal-500: $teal !default;\n$teal-600: shade-color($teal, 20%) !default;\n$teal-700: shade-color($teal, 40%) !default;\n$teal-800: shade-color($teal, 60%) !default;\n$teal-900: shade-color($teal, 80%) !default;\n\n$cyan-100: tint-color($cyan, 80%) !default;\n$cyan-200: tint-color($cyan, 60%) !default;\n$cyan-300: tint-color($cyan, 40%) !default;\n$cyan-400: tint-color($cyan, 20%) !default;\n$cyan-500: $cyan !default;\n$cyan-600: shade-color($cyan, 20%) !default;\n$cyan-700: shade-color($cyan, 40%) !default;\n$cyan-800: shade-color($cyan, 60%) !default;\n$cyan-900: shade-color($cyan, 80%) !default;\n\n$blues: (\n  \"blue-100\": $blue-100,\n  \"blue-200\": $blue-200,\n  \"blue-300\": $blue-300,\n  \"blue-400\": $blue-400,\n  \"blue-500\": $blue-500,\n  \"blue-600\": $blue-600,\n  \"blue-700\": $blue-700,\n  \"blue-800\": $blue-800,\n  \"blue-900\": $blue-900\n) !default;\n\n$indigos: (\n  \"indigo-100\": $indigo-100,\n  \"indigo-200\": $indigo-200,\n  \"indigo-300\": $indigo-300,\n  \"indigo-400\": $indigo-400,\n  \"indigo-500\": $indigo-500,\n  \"indigo-600\": $indigo-600,\n  \"indigo-700\": $indigo-700,\n  \"indigo-800\": $indigo-800,\n  \"indigo-900\": $indigo-900\n) !default;\n\n$purples: (\n  \"purple-100\": $purple-100,\n  \"purple-200\": $purple-200,\n  \"purple-300\": $purple-300,\n  \"purple-400\": $purple-400,\n  \"purple-500\": $purple-500,\n  \"purple-600\": $purple-600,\n  \"purple-700\": $purple-700,\n  \"purple-800\": $purple-800,\n  \"purple-900\": $purple-900\n) !default;\n\n$pinks: (\n  \"pink-100\": $pink-100,\n  \"pink-200\": $pink-200,\n  \"pink-300\": $pink-300,\n  \"pink-400\": $pink-400,\n  \"pink-500\": $pink-500,\n  \"pink-600\": $pink-600,\n  \"pink-700\": $pink-700,\n  \"pink-800\": $pink-800,\n  \"pink-900\": $pink-900\n) !default;\n\n$reds: (\n  \"red-100\": $red-100,\n  \"red-200\": $red-200,\n  \"red-300\": $red-300,\n  \"red-400\": $red-400,\n  \"red-500\": $red-500,\n  \"red-600\": $red-600,\n  \"red-700\": $red-700,\n  \"red-800\": $red-800,\n  \"red-900\": $red-900\n) !default;\n\n$oranges: (\n  \"orange-100\": $orange-100,\n  \"orange-200\": $orange-200,\n  \"orange-300\": $orange-300,\n  \"orange-400\": $orange-400,\n  \"orange-500\": $orange-500,\n  \"orange-600\": $orange-600,\n  \"orange-700\": $orange-700,\n  \"orange-800\": $orange-800,\n  \"orange-900\": $orange-900\n) !default;\n\n$yellows: (\n  \"yellow-100\": $yellow-100,\n  \"yellow-200\": $yellow-200,\n  \"yellow-300\": $yellow-300,\n  \"yellow-400\": $yellow-400,\n  \"yellow-500\": $yellow-500,\n  \"yellow-600\": $yellow-600,\n  \"yellow-700\": $yellow-700,\n  \"yellow-800\": $yellow-800,\n  \"yellow-900\": $yellow-900\n) !default;\n\n$greens: (\n  \"green-100\": $green-100,\n  \"green-200\": $green-200,\n  \"green-300\": $green-300,\n  \"green-400\": $green-400,\n  \"green-500\": $green-500,\n  \"green-600\": $green-600,\n  \"green-700\": $green-700,\n  \"green-800\": $green-800,\n  \"green-900\": $green-900\n) !default;\n\n$teals: (\n  \"teal-100\": $teal-100,\n  \"teal-200\": $teal-200,\n  \"teal-300\": $teal-300,\n  \"teal-400\": $teal-400,\n  \"teal-500\": $teal-500,\n  \"teal-600\": $teal-600,\n  \"teal-700\": $teal-700,\n  \"teal-800\": $teal-800,\n  \"teal-900\": $teal-900\n) !default;\n\n$cyans: (\n  \"cyan-100\": $cyan-100,\n  \"cyan-200\": $cyan-200,\n  \"cyan-300\": $cyan-300,\n  \"cyan-400\": $cyan-400,\n  \"cyan-500\": $cyan-500,\n  \"cyan-600\": $cyan-600,\n  \"cyan-700\": $cyan-700,\n  \"cyan-800\": $cyan-800,\n  \"cyan-900\": $cyan-900\n) !default;\n// fusv-enable\n\n// scss-docs-start theme-color-variables\n$primary:       $blue !default;\n$secondary:     $gray-600 !default;\n$success:       $green !default;\n$info:          $cyan !default;\n$warning:       $yellow !default;\n$danger:        $red !default;\n$light:         $gray-100 !default;\n$dark:          $gray-900 !default;\n// scss-docs-end theme-color-variables\n\n// scss-docs-start theme-colors-map\n$theme-colors: (\n  \"primary\":    $primary,\n  \"secondary\":  $secondary,\n  \"success\":    $success,\n  \"info\":       $info,\n  \"warning\":    $warning,\n  \"danger\":     $danger,\n  \"light\":      $light,\n  \"dark\":       $dark\n) !default;\n// scss-docs-end theme-colors-map\n\n// scss-docs-start theme-text-variables\n$primary-text-emphasis:   shade-color($primary, 60%) !default;\n$secondary-text-emphasis: shade-color($secondary, 60%) !default;\n$success-text-emphasis:   shade-color($success, 60%) !default;\n$info-text-emphasis:      shade-color($info, 60%) !default;\n$warning-text-emphasis:   shade-color($warning, 60%) !default;\n$danger-text-emphasis:    shade-color($danger, 60%) !default;\n$light-text-emphasis:     $gray-700 !default;\n$dark-text-emphasis:      $gray-700 !default;\n// scss-docs-end theme-text-variables\n\n// scss-docs-start theme-bg-subtle-variables\n$primary-bg-subtle:       tint-color($primary, 80%) !default;\n$secondary-bg-subtle:     tint-color($secondary, 80%) !default;\n$success-bg-subtle:       tint-color($success, 80%) !default;\n$info-bg-subtle:          tint-color($info, 80%) !default;\n$warning-bg-subtle:       tint-color($warning, 80%) !default;\n$danger-bg-subtle:        tint-color($danger, 80%) !default;\n$light-bg-subtle:         mix($gray-100, $white) !default;\n$dark-bg-subtle:          $gray-400 !default;\n// scss-docs-end theme-bg-subtle-variables\n\n// scss-docs-start theme-border-subtle-variables\n$primary-border-subtle:   tint-color($primary, 60%) !default;\n$secondary-border-subtle: tint-color($secondary, 60%) !default;\n$success-border-subtle:   tint-color($success, 60%) !default;\n$info-border-subtle:      tint-color($info, 60%) !default;\n$warning-border-subtle:   tint-color($warning, 60%) !default;\n$danger-border-subtle:    tint-color($danger, 60%) !default;\n$light-border-subtle:     $gray-200 !default;\n$dark-border-subtle:      $gray-500 !default;\n// scss-docs-end theme-border-subtle-variables\n\n// Characters which are escaped by the escape-svg function\n$escaped-characters: (\n  (\"<\", \"%3c\"),\n  (\">\", \"%3e\"),\n  (\"#\", \"%23\"),\n  (\"(\", \"%28\"),\n  (\")\", \"%29\"),\n) !default;\n\n// Options\n//\n// Quickly modify global styling by enabling or disabling optional features.\n\n$enable-caret:                true !default;\n$enable-rounded:              true !default;\n$enable-shadows:              false !default;\n$enable-gradients:            false !default;\n$enable-transitions:          true !default;\n$enable-reduced-motion:       true !default;\n$enable-smooth-scroll:        true !default;\n$enable-grid-classes:         true !default;\n$enable-container-classes:    true !default;\n$enable-cssgrid:              false !default;\n$enable-button-pointers:      true !default;\n$enable-rfs:                  true !default;\n$enable-validation-icons:     true !default;\n$enable-negative-margins:     false !default;\n$enable-deprecation-messages: true !default;\n$enable-important-utilities:  true !default;\n\n$enable-dark-mode:            true !default;\n$color-mode-type:             data !default; // `data` or `media-query`\n\n// Prefix for :root CSS variables\n\n$variable-prefix:             bs- !default; // Deprecated in v5.2.0 for the shorter `$prefix`\n$prefix:                      $variable-prefix !default;\n\n// Gradient\n//\n// The gradient which is added to components if `$enable-gradients` is `true`\n// This gradient is also added to elements with `.bg-gradient`\n// scss-docs-start variable-gradient\n$gradient: linear-gradient(180deg, rgba($white, .15), rgba($white, 0)) !default;\n// scss-docs-end variable-gradient\n\n// Spacing\n//\n// Control the default styling of most Bootstrap elements by modifying these\n// variables. Mostly focused on spacing.\n// You can add more entries to the $spacers map, should you need more variation.\n\n// scss-docs-start spacer-variables-maps\n$spacer: 1rem !default;\n$spacers: (\n  0: 0,\n  1: $spacer * .25,\n  2: $spacer * .5,\n  3: $spacer,\n  4: $spacer * 1.5,\n  5: $spacer * 3,\n) !default;\n// scss-docs-end spacer-variables-maps\n\n// Position\n//\n// Define the edge positioning anchors of the position utilities.\n\n// scss-docs-start position-map\n$position-values: (\n  0: 0,\n  50: 50%,\n  100: 100%\n) !default;\n// scss-docs-end position-map\n\n// Body\n//\n// Settings for the `<body>` element.\n\n$body-text-align:           null !default;\n$body-color:                $gray-900 !default;\n$body-bg:                   $white !default;\n\n$body-secondary-color:      rgba($body-color, .75) !default;\n$body-secondary-bg:         $gray-200 !default;\n\n$body-tertiary-color:       rgba($body-color, .5) !default;\n$body-tertiary-bg:          $gray-100 !default;\n\n$body-emphasis-color:       $black !default;\n\n// Links\n//\n// Style anchor elements.\n\n$link-color:                              $primary !default;\n$link-decoration:                         underline !default;\n$link-shade-percentage:                   20% !default;\n$link-hover-color:                        shift-color($link-color, $link-shade-percentage) !default;\n$link-hover-decoration:                   null !default;\n\n$stretched-link-pseudo-element:           after !default;\n$stretched-link-z-index:                  1 !default;\n\n// Icon links\n// scss-docs-start icon-link-variables\n$icon-link-gap:               .375rem !default;\n$icon-link-underline-offset:  .25em !default;\n$icon-link-icon-size:         1em !default;\n$icon-link-icon-transition:   .2s ease-in-out transform !default;\n$icon-link-icon-transform:    translate3d(.25em, 0, 0) !default;\n// scss-docs-end icon-link-variables\n\n// Paragraphs\n//\n// Style p element.\n\n$paragraph-margin-bottom:   1rem !default;\n\n\n// Grid breakpoints\n//\n// Define the minimum dimensions at which your layout will change,\n// adapting to different screen sizes, for use in media queries.\n\n// scss-docs-start grid-breakpoints\n$grid-breakpoints: (\n  xs: 0,\n  sm: 576px,\n  md: 768px,\n  lg: 992px,\n  xl: 1200px,\n  xxl: 1400px\n) !default;\n// scss-docs-end grid-breakpoints\n\n@include _assert-ascending($grid-breakpoints, \"$grid-breakpoints\");\n@include _assert-starts-at-zero($grid-breakpoints, \"$grid-breakpoints\");\n\n\n// Grid containers\n//\n// Define the maximum width of `.container` for different screen sizes.\n\n// scss-docs-start container-max-widths\n$container-max-widths: (\n  sm: 540px,\n  md: 720px,\n  lg: 960px,\n  xl: 1140px,\n  xxl: 1320px\n) !default;\n// scss-docs-end container-max-widths\n\n@include _assert-ascending($container-max-widths, \"$container-max-widths\");\n\n\n// Grid columns\n//\n// Set the number of columns and specify the width of the gutters.\n\n$grid-columns:                12 !default;\n$grid-gutter-width:           1.5rem !default;\n$grid-row-columns:            6 !default;\n\n// Container padding\n\n$container-padding-x: $grid-gutter-width !default;\n\n\n// Components\n//\n// Define common padding and border radius sizes and more.\n\n// scss-docs-start border-variables\n$border-width:                1px !default;\n$border-widths: (\n  1: 1px,\n  2: 2px,\n  3: 3px,\n  4: 4px,\n  5: 5px\n) !default;\n$border-style:                solid !default;\n$border-color:                $gray-300 !default;\n$border-color-translucent:    rgba($black, .175) !default;\n// scss-docs-end border-variables\n\n// scss-docs-start border-radius-variables\n$border-radius:               .375rem !default;\n$border-radius-sm:            .25rem !default;\n$border-radius-lg:            .5rem !default;\n$border-radius-xl:            1rem !default;\n$border-radius-xxl:           2rem !default;\n$border-radius-pill:          50rem !default;\n// scss-docs-end border-radius-variables\n// fusv-disable\n$border-radius-2xl:           $border-radius-xxl !default; // Deprecated in v5.3.0\n// fusv-enable\n\n// scss-docs-start box-shadow-variables\n$box-shadow:                  0 .5rem 1rem rgba($black, .15) !default;\n$box-shadow-sm:               0 .125rem .25rem rgba($black, .075) !default;\n$box-shadow-lg:               0 1rem 3rem rgba($black, .175) !default;\n$box-shadow-inset:            inset 0 1px 2px rgba($black, .075) !default;\n// scss-docs-end box-shadow-variables\n\n$component-active-color:      $white !default;\n$component-active-bg:         $primary !default;\n\n// scss-docs-start focus-ring-variables\n$focus-ring-width:      .25rem !default;\n$focus-ring-opacity:    .25 !default;\n$focus-ring-color:      rgba($primary, $focus-ring-opacity) !default;\n$focus-ring-blur:       0 !default;\n$focus-ring-box-shadow: 0 0 $focus-ring-blur $focus-ring-width $focus-ring-color !default;\n// scss-docs-end focus-ring-variables\n\n// scss-docs-start caret-variables\n$caret-width:                 .3em !default;\n$caret-vertical-align:        $caret-width * .85 !default;\n$caret-spacing:               $caret-width * .85 !default;\n// scss-docs-end caret-variables\n\n$transition-base:             all .2s ease-in-out !default;\n$transition-fade:             opacity .15s linear !default;\n// scss-docs-start collapse-transition\n$transition-collapse:         height .35s ease !default;\n$transition-collapse-width:   width .35s ease !default;\n// scss-docs-end collapse-transition\n\n// stylelint-disable function-disallowed-list\n// scss-docs-start aspect-ratios\n$aspect-ratios: (\n  \"1x1\": 100%,\n  \"4x3\": calc(3 / 4 * 100%),\n  \"16x9\": calc(9 / 16 * 100%),\n  \"21x9\": calc(9 / 21 * 100%)\n) !default;\n// scss-docs-end aspect-ratios\n// stylelint-enable function-disallowed-list\n\n// Typography\n//\n// Font, line-height, and color for body text, headings, and more.\n\n// scss-docs-start font-variables\n// stylelint-disable value-keyword-case\n$font-family-sans-serif:      system-ui, -apple-system, \"Segoe UI\", Roboto, \"Helvetica Neue\", \"Noto Sans\", \"Liberation Sans\", Arial, sans-serif, \"Apple Color Emoji\", \"Segoe UI Emoji\", \"Segoe UI Symbol\", \"Noto Color Emoji\" !default;\n$font-family-monospace:       SFMono-Regular, Menlo, Monaco, Consolas, \"Liberation Mono\", \"Courier New\", monospace !default;\n// stylelint-enable value-keyword-case\n$font-family-base:            var(--#{$prefix}font-sans-serif) !default;\n$font-family-code:            var(--#{$prefix}font-monospace) !default;\n\n// $font-size-root affects the value of `rem`, which is used for as well font sizes, paddings, and margins\n// $font-size-base affects the font size of the body text\n$font-size-root:              null !default;\n$font-size-base:              1rem !default; // Assumes the browser default, typically `16px`\n$font-size-sm:                $font-size-base * .875 !default;\n$font-size-lg:                $font-size-base * 1.25 !default;\n\n$font-weight-lighter:         lighter !default;\n$font-weight-light:           300 !default;\n$font-weight-normal:          400 !default;\n$font-weight-medium:          500 !default;\n$font-weight-semibold:        600 !default;\n$font-weight-bold:            700 !default;\n$font-weight-bolder:          bolder !default;\n\n$font-weight-base:            $font-weight-normal !default;\n\n$line-height-base:            1.5 !default;\n$line-height-sm:              1.25 !default;\n$line-height-lg:              2 !default;\n\n$h1-font-size:                $font-size-base * 2.5 !default;\n$h2-font-size:                $font-size-base * 2 !default;\n$h3-font-size:                $font-size-base * 1.75 !default;\n$h4-font-size:                $font-size-base * 1.5 !default;\n$h5-font-size:                $font-size-base * 1.25 !default;\n$h6-font-size:                $font-size-base !default;\n// scss-docs-end font-variables\n\n// scss-docs-start font-sizes\n$font-sizes: (\n  1: $h1-font-size,\n  2: $h2-font-size,\n  3: $h3-font-size,\n  4: $h4-font-size,\n  5: $h5-font-size,\n  6: $h6-font-size\n) !default;\n// scss-docs-end font-sizes\n\n// scss-docs-start headings-variables\n$headings-margin-bottom:      $spacer * .5 !default;\n$headings-font-family:        null !default;\n$headings-font-style:         null !default;\n$headings-font-weight:        500 !default;\n$headings-line-height:        1.2 !default;\n$headings-color:              inherit !default;\n// scss-docs-end headings-variables\n\n// scss-docs-start display-headings\n$display-font-sizes: (\n  1: 5rem,\n  2: 4.5rem,\n  3: 4rem,\n  4: 3.5rem,\n  5: 3rem,\n  6: 2.5rem\n) !default;\n\n$display-font-family: null !default;\n$display-font-style:  null !default;\n$display-font-weight: 300 !default;\n$display-line-height: $headings-line-height !default;\n// scss-docs-end display-headings\n\n// scss-docs-start type-variables\n$lead-font-size:              $font-size-base * 1.25 !default;\n$lead-font-weight:            300 !default;\n\n$small-font-size:             .875em !default;\n\n$sub-sup-font-size:           .75em !default;\n\n// fusv-disable\n$text-muted:                  var(--#{$prefix}secondary-color) !default; // Deprecated in 5.3.0\n// fusv-enable\n\n$initialism-font-size:        $small-font-size !default;\n\n$blockquote-margin-y:         $spacer !default;\n$blockquote-font-size:        $font-size-base * 1.25 !default;\n$blockquote-footer-color:     $gray-600 !default;\n$blockquote-footer-font-size: $small-font-size !default;\n\n$hr-margin-y:                 $spacer !default;\n$hr-color:                    inherit !default;\n\n// fusv-disable\n$hr-bg-color:                 null !default; // Deprecated in v5.2.0\n$hr-height:                   null !default; // Deprecated in v5.2.0\n// fusv-enable\n\n$hr-border-color:             null !default; // Allows for inherited colors\n$hr-border-width:             var(--#{$prefix}border-width) !default;\n$hr-opacity:                  .25 !default;\n\n// scss-docs-start vr-variables\n$vr-border-width:             var(--#{$prefix}border-width) !default;\n// scss-docs-end vr-variables\n\n$legend-margin-bottom:        .5rem !default;\n$legend-font-size:            1.5rem !default;\n$legend-font-weight:          null !default;\n\n$dt-font-weight:              $font-weight-bold !default;\n\n$list-inline-padding:         .5rem !default;\n\n$mark-padding:                .1875em !default;\n$mark-color:                  $body-color !default;\n$mark-bg:                     $yellow-100 !default;\n// scss-docs-end type-variables\n\n\n// Tables\n//\n// Customizes the `.table` component with basic values, each used across all table variations.\n\n// scss-docs-start table-variables\n$table-cell-padding-y:        .5rem !default;\n$table-cell-padding-x:        .5rem !default;\n$table-cell-padding-y-sm:     .25rem !default;\n$table-cell-padding-x-sm:     .25rem !default;\n\n$table-cell-vertical-align:   top !default;\n\n$table-color:                 var(--#{$prefix}emphasis-color) !default;\n$table-bg:                    var(--#{$prefix}body-bg) !default;\n$table-accent-bg:             transparent !default;\n\n$table-th-font-weight:        null !default;\n\n$table-striped-color:         $table-color !default;\n$table-striped-bg-factor:     .05 !default;\n$table-striped-bg:            rgba(var(--#{$prefix}emphasis-color-rgb), $table-striped-bg-factor) !default;\n\n$table-active-color:          $table-color !default;\n$table-active-bg-factor:      .1 !default;\n$table-active-bg:             rgba(var(--#{$prefix}emphasis-color-rgb), $table-active-bg-factor) !default;\n\n$table-hover-color:           $table-color !default;\n$table-hover-bg-factor:       .075 !default;\n$table-hover-bg:              rgba(var(--#{$prefix}emphasis-color-rgb), $table-hover-bg-factor) !default;\n\n$table-border-factor:         .2 !default;\n$table-border-width:          var(--#{$prefix}border-width) !default;\n$table-border-color:          var(--#{$prefix}border-color) !default;\n\n$table-striped-order:         odd !default;\n$table-striped-columns-order: even !default;\n\n$table-group-separator-color: currentcolor !default;\n\n$table-caption-color:         var(--#{$prefix}secondary-color) !default;\n\n$table-bg-scale:              -80% !default;\n// scss-docs-end table-variables\n\n// scss-docs-start table-loop\n$table-variants: (\n  \"primary\":    shift-color($primary, $table-bg-scale),\n  \"secondary\":  shift-color($secondary, $table-bg-scale),\n  \"success\":    shift-color($success, $table-bg-scale),\n  \"info\":       shift-color($info, $table-bg-scale),\n  \"warning\":    shift-color($warning, $table-bg-scale),\n  \"danger\":     shift-color($danger, $table-bg-scale),\n  \"light\":      $light,\n  \"dark\":       $dark,\n) !default;\n// scss-docs-end table-loop\n\n\n// Buttons + Forms\n//\n// Shared variables that are reassigned to `$input-` and `$btn-` specific variables.\n\n// scss-docs-start input-btn-variables\n$input-btn-padding-y:         .375rem !default;\n$input-btn-padding-x:         .75rem !default;\n$input-btn-font-family:       null !default;\n$input-btn-font-size:         $font-size-base !default;\n$input-btn-line-height:       $line-height-base !default;\n\n$input-btn-focus-width:         $focus-ring-width !default;\n$input-btn-focus-color-opacity: $focus-ring-opacity !default;\n$input-btn-focus-color:         $focus-ring-color !default;\n$input-btn-focus-blur:          $focus-ring-blur !default;\n$input-btn-focus-box-shadow:    $focus-ring-box-shadow !default;\n\n$input-btn-padding-y-sm:      .25rem !default;\n$input-btn-padding-x-sm:      .5rem !default;\n$input-btn-font-size-sm:      $font-size-sm !default;\n\n$input-btn-padding-y-lg:      .5rem !default;\n$input-btn-padding-x-lg:      1rem !default;\n$input-btn-font-size-lg:      $font-size-lg !default;\n\n$input-btn-border-width:      var(--#{$prefix}border-width) !default;\n// scss-docs-end input-btn-variables\n\n\n// Buttons\n//\n// For each of Bootstrap's buttons, define text, background, and border color.\n\n// scss-docs-start btn-variables\n$btn-color:                   var(--#{$prefix}body-color) !default;\n$btn-padding-y:               $input-btn-padding-y !default;\n$btn-padding-x:               $input-btn-padding-x !default;\n$btn-font-family:             $input-btn-font-family !default;\n$btn-font-size:               $input-btn-font-size !default;\n$btn-line-height:             $input-btn-line-height !default;\n$btn-white-space:             null !default; // Set to `nowrap` to prevent text wrapping\n\n$btn-padding-y-sm:            $input-btn-padding-y-sm !default;\n$btn-padding-x-sm:            $input-btn-padding-x-sm !default;\n$btn-font-size-sm:            $input-btn-font-size-sm !default;\n\n$btn-padding-y-lg:            $input-btn-padding-y-lg !default;\n$btn-padding-x-lg:            $input-btn-padding-x-lg !default;\n$btn-font-size-lg:            $input-btn-font-size-lg !default;\n\n$btn-border-width:            $input-btn-border-width !default;\n\n$btn-font-weight:             $font-weight-normal !default;\n$btn-box-shadow:              inset 0 1px 0 rgba($white, .15), 0 1px 1px rgba($black, .075) !default;\n$btn-focus-width:             $input-btn-focus-width !default;\n$btn-focus-box-shadow:        $input-btn-focus-box-shadow !default;\n$btn-disabled-opacity:        .65 !default;\n$btn-active-box-shadow:       inset 0 3px 5px rgba($black, .125) !default;\n\n$btn-link-color:              var(--#{$prefix}link-color) !default;\n$btn-link-hover-color:        var(--#{$prefix}link-hover-color) !default;\n$btn-link-disabled-color:     $gray-600 !default;\n$btn-link-focus-shadow-rgb:   to-rgb(mix(color-contrast($link-color), $link-color, 15%)) !default;\n\n// Allows for customizing button radius independently from global border radius\n$btn-border-radius:           var(--#{$prefix}border-radius) !default;\n$btn-border-radius-sm:        var(--#{$prefix}border-radius-sm) !default;\n$btn-border-radius-lg:        var(--#{$prefix}border-radius-lg) !default;\n\n$btn-transition:              color .15s ease-in-out, background-color .15s ease-in-out, border-color .15s ease-in-out, box-shadow .15s ease-in-out !default;\n\n$btn-hover-bg-shade-amount:       15% !default;\n$btn-hover-bg-tint-amount:        15% !default;\n$btn-hover-border-shade-amount:   20% !default;\n$btn-hover-border-tint-amount:    10% !default;\n$btn-active-bg-shade-amount:      20% !default;\n$btn-active-bg-tint-amount:       20% !default;\n$btn-active-border-shade-amount:  25% !default;\n$btn-active-border-tint-amount:   10% !default;\n// scss-docs-end btn-variables\n\n\n// Forms\n\n// scss-docs-start form-text-variables\n$form-text-margin-top:                  .25rem !default;\n$form-text-font-size:                   $small-font-size !default;\n$form-text-font-style:                  null !default;\n$form-text-font-weight:                 null !default;\n$form-text-color:                       var(--#{$prefix}secondary-color) !default;\n// scss-docs-end form-text-variables\n\n// scss-docs-start form-label-variables\n$form-label-margin-bottom:              .5rem !default;\n$form-label-font-size:                  null !default;\n$form-label-font-style:                 null !default;\n$form-label-font-weight:                null !default;\n$form-label-color:                      null !default;\n// scss-docs-end form-label-variables\n\n// scss-docs-start form-input-variables\n$input-padding-y:                       $input-btn-padding-y !default;\n$input-padding-x:                       $input-btn-padding-x !default;\n$input-font-family:                     $input-btn-font-family !default;\n$input-font-size:                       $input-btn-font-size !default;\n$input-font-weight:                     $font-weight-base !default;\n$input-line-height:                     $input-btn-line-height !default;\n\n$input-padding-y-sm:                    $input-btn-padding-y-sm !default;\n$input-padding-x-sm:                    $input-btn-padding-x-sm !default;\n$input-font-size-sm:                    $input-btn-font-size-sm !default;\n\n$input-padding-y-lg:                    $input-btn-padding-y-lg !default;\n$input-padding-x-lg:                    $input-btn-padding-x-lg !default;\n$input-font-size-lg:                    $input-btn-font-size-lg !default;\n\n$input-bg:                              var(--#{$prefix}body-bg) !default;\n$input-disabled-color:                  null !default;\n$input-disabled-bg:                     var(--#{$prefix}secondary-bg) !default;\n$input-disabled-border-color:           null !default;\n\n$input-color:                           var(--#{$prefix}body-color) !default;\n$input-border-color:                    var(--#{$prefix}border-color) !default;\n$input-border-width:                    $input-btn-border-width !default;\n$input-box-shadow:                      var(--#{$prefix}box-shadow-inset) !default;\n\n$input-border-radius:                   var(--#{$prefix}border-radius) !default;\n$input-border-radius-sm:                var(--#{$prefix}border-radius-sm) !default;\n$input-border-radius-lg:                var(--#{$prefix}border-radius-lg) !default;\n\n$input-focus-bg:                        $input-bg !default;\n$input-focus-border-color:              tint-color($component-active-bg, 50%) !default;\n$input-focus-color:                     $input-color !default;\n$input-focus-width:                     $input-btn-focus-width !default;\n$input-focus-box-shadow:                $input-btn-focus-box-shadow !default;\n\n$input-placeholder-color:               var(--#{$prefix}secondary-color) !default;\n$input-plaintext-color:                 var(--#{$prefix}body-color) !default;\n\n$input-height-border:                   calc(#{$input-border-width} * 2) !default; // stylelint-disable-line function-disallowed-list\n\n$input-height-inner:                    add($input-line-height * 1em, $input-padding-y * 2) !default;\n$input-height-inner-half:               add($input-line-height * .5em, $input-padding-y) !default;\n$input-height-inner-quarter:            add($input-line-height * .25em, $input-padding-y * .5) !default;\n\n$input-height:                          add($input-line-height * 1em, add($input-padding-y * 2, $input-height-border, false)) !default;\n$input-height-sm:                       add($input-line-height * 1em, add($input-padding-y-sm * 2, $input-height-border, false)) !default;\n$input-height-lg:                       add($input-line-height * 1em, add($input-padding-y-lg * 2, $input-height-border, false)) !default;\n\n$input-transition:                      border-color .15s ease-in-out, box-shadow .15s ease-in-out !default;\n\n$form-color-width:                      3rem !default;\n// scss-docs-end form-input-variables\n\n// scss-docs-start form-check-variables\n$form-check-input-width:                  1em !default;\n$form-check-min-height:                   $font-size-base * $line-height-base !default;\n$form-check-padding-start:                $form-check-input-width + .5em !default;\n$form-check-margin-bottom:                .125rem !default;\n$form-check-label-color:                  null !default;\n$form-check-label-cursor:                 null !default;\n$form-check-transition:                   null !default;\n\n$form-check-input-active-filter:          brightness(90%) !default;\n\n$form-check-input-bg:                     $input-bg !default;\n$form-check-input-border:                 var(--#{$prefix}border-width) solid var(--#{$prefix}border-color) !default;\n$form-check-input-border-radius:          .25em !default;\n$form-check-radio-border-radius:          50% !default;\n$form-check-input-focus-border:           $input-focus-border-color !default;\n$form-check-input-focus-box-shadow:       $focus-ring-box-shadow !default;\n\n$form-check-input-checked-color:          $component-active-color !default;\n$form-check-input-checked-bg-color:       $component-active-bg !default;\n$form-check-input-checked-border-color:   $form-check-input-checked-bg-color !default;\n$form-check-input-checked-bg-image:       url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 20 20'><path fill='none' stroke='#{$form-check-input-checked-color}' stroke-linecap='round' stroke-linejoin='round' stroke-width='3' d='m6 10 3 3 6-6'/></svg>\") !default;\n$form-check-radio-checked-bg-image:       url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='-4 -4 8 8'><circle r='2' fill='#{$form-check-input-checked-color}'/></svg>\") !default;\n\n$form-check-input-indeterminate-color:          $component-active-color !default;\n$form-check-input-indeterminate-bg-color:       $component-active-bg !default;\n$form-check-input-indeterminate-border-color:   $form-check-input-indeterminate-bg-color !default;\n$form-check-input-indeterminate-bg-image:       url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 20 20'><path fill='none' stroke='#{$form-check-input-indeterminate-color}' stroke-linecap='round' stroke-linejoin='round' stroke-width='3' d='M6 10h8'/></svg>\") !default;\n\n$form-check-input-disabled-opacity:        .5 !default;\n$form-check-label-disabled-opacity:        $form-check-input-disabled-opacity !default;\n$form-check-btn-check-disabled-opacity:    $btn-disabled-opacity !default;\n\n$form-check-inline-margin-end:    1rem !default;\n// scss-docs-end form-check-variables\n\n// scss-docs-start form-switch-variables\n$form-switch-color:               rgba($black, .25) !default;\n$form-switch-width:               2em !default;\n$form-switch-padding-start:       $form-switch-width + .5em !default;\n$form-switch-bg-image:            url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='-4 -4 8 8'><circle r='3' fill='#{$form-switch-color}'/></svg>\") !default;\n$form-switch-border-radius:       $form-switch-width !default;\n$form-switch-transition:          background-position .15s ease-in-out !default;\n\n$form-switch-focus-color:         $input-focus-border-color !default;\n$form-switch-focus-bg-image:      url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='-4 -4 8 8'><circle r='3' fill='#{$form-switch-focus-color}'/></svg>\") !default;\n\n$form-switch-checked-color:       $component-active-color !default;\n$form-switch-checked-bg-image:    url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='-4 -4 8 8'><circle r='3' fill='#{$form-switch-checked-color}'/></svg>\") !default;\n$form-switch-checked-bg-position: right center !default;\n// scss-docs-end form-switch-variables\n\n// scss-docs-start input-group-variables\n$input-group-addon-padding-y:           $input-padding-y !default;\n$input-group-addon-padding-x:           $input-padding-x !default;\n$input-group-addon-font-weight:         $input-font-weight !default;\n$input-group-addon-color:               $input-color !default;\n$input-group-addon-bg:                  var(--#{$prefix}tertiary-bg) !default;\n$input-group-addon-border-color:        $input-border-color !default;\n// scss-docs-end input-group-variables\n\n// scss-docs-start form-select-variables\n$form-select-padding-y:             $input-padding-y !default;\n$form-select-padding-x:             $input-padding-x !default;\n$form-select-font-family:           $input-font-family !default;\n$form-select-font-size:             $input-font-size !default;\n$form-select-indicator-padding:     $form-select-padding-x * 3 !default; // Extra padding for background-image\n$form-select-font-weight:           $input-font-weight !default;\n$form-select-line-height:           $input-line-height !default;\n$form-select-color:                 $input-color !default;\n$form-select-bg:                    $input-bg !default;\n$form-select-disabled-color:        null !default;\n$form-select-disabled-bg:           $input-disabled-bg !default;\n$form-select-disabled-border-color: $input-disabled-border-color !default;\n$form-select-bg-position:           right $form-select-padding-x center !default;\n$form-select-bg-size:               16px 12px !default; // In pixels because image dimensions\n$form-select-indicator-color:       $gray-800 !default;\n$form-select-indicator:             url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16'><path fill='none' stroke='#{$form-select-indicator-color}' stroke-linecap='round' stroke-linejoin='round' stroke-width='2' d='m2 5 6 6 6-6'/></svg>\") !default;\n\n$form-select-feedback-icon-padding-end: $form-select-padding-x * 2.5 + $form-select-indicator-padding !default;\n$form-select-feedback-icon-position:    center right $form-select-indicator-padding !default;\n$form-select-feedback-icon-size:        $input-height-inner-half $input-height-inner-half !default;\n\n$form-select-border-width:        $input-border-width !default;\n$form-select-border-color:        $input-border-color !default;\n$form-select-border-radius:       $input-border-radius !default;\n$form-select-box-shadow:          var(--#{$prefix}box-shadow-inset) !default;\n\n$form-select-focus-border-color:  $input-focus-border-color !default;\n$form-select-focus-width:         $input-focus-width !default;\n$form-select-focus-box-shadow:    0 0 0 $form-select-focus-width $input-btn-focus-color !default;\n\n$form-select-padding-y-sm:        $input-padding-y-sm !default;\n$form-select-padding-x-sm:        $input-padding-x-sm !default;\n$form-select-font-size-sm:        $input-font-size-sm !default;\n$form-select-border-radius-sm:    $input-border-radius-sm !default;\n\n$form-select-padding-y-lg:        $input-padding-y-lg !default;\n$form-select-padding-x-lg:        $input-padding-x-lg !default;\n$form-select-font-size-lg:        $input-font-size-lg !default;\n$form-select-border-radius-lg:    $input-border-radius-lg !default;\n\n$form-select-transition:          $input-transition !default;\n// scss-docs-end form-select-variables\n\n// scss-docs-start form-range-variables\n$form-range-track-width:          100% !default;\n$form-range-track-height:         .5rem !default;\n$form-range-track-cursor:         pointer !default;\n$form-range-track-bg:             var(--#{$prefix}secondary-bg) !default;\n$form-range-track-border-radius:  1rem !default;\n$form-range-track-box-shadow:     var(--#{$prefix}box-shadow-inset) !default;\n\n$form-range-thumb-width:                   1rem !default;\n$form-range-thumb-height:                  $form-range-thumb-width !default;\n$form-range-thumb-bg:                      $component-active-bg !default;\n$form-range-thumb-border:                  0 !default;\n$form-range-thumb-border-radius:           1rem !default;\n$form-range-thumb-box-shadow:              0 .1rem .25rem rgba($black, .1) !default;\n$form-range-thumb-focus-box-shadow:        0 0 0 1px $body-bg, $input-focus-box-shadow !default;\n$form-range-thumb-focus-box-shadow-width:  $input-focus-width !default; // For focus box shadow issue in Edge\n$form-range-thumb-active-bg:               tint-color($component-active-bg, 70%) !default;\n$form-range-thumb-disabled-bg:             var(--#{$prefix}secondary-color) !default;\n$form-range-thumb-transition:              background-color .15s ease-in-out, border-color .15s ease-in-out, box-shadow .15s ease-in-out !default;\n// scss-docs-end form-range-variables\n\n// scss-docs-start form-file-variables\n$form-file-button-color:          $input-color !default;\n$form-file-button-bg:             var(--#{$prefix}tertiary-bg) !default;\n$form-file-button-hover-bg:       var(--#{$prefix}secondary-bg) !default;\n// scss-docs-end form-file-variables\n\n// scss-docs-start form-floating-variables\n$form-floating-height:                  add(3.5rem, $input-height-border) !default;\n$form-floating-line-height:             1.25 !default;\n$form-floating-padding-x:               $input-padding-x !default;\n$form-floating-padding-y:               1rem !default;\n$form-floating-input-padding-t:         1.625rem !default;\n$form-floating-input-padding-b:         .625rem !default;\n$form-floating-label-height:            1.5em !default;\n$form-floating-label-opacity:           .65 !default;\n$form-floating-label-transform:         scale(.85) translateY(-.5rem) translateX(.15rem) !default;\n$form-floating-label-disabled-color:    $gray-600 !default;\n$form-floating-transition:              opacity .1s ease-in-out, transform .1s ease-in-out !default;\n// scss-docs-end form-floating-variables\n\n// Form validation\n\n// scss-docs-start form-feedback-variables\n$form-feedback-margin-top:          $form-text-margin-top !default;\n$form-feedback-font-size:           $form-text-font-size !default;\n$form-feedback-font-style:          $form-text-font-style !default;\n$form-feedback-valid-color:         $success !default;\n$form-feedback-invalid-color:       $danger !default;\n\n$form-feedback-icon-valid-color:    $form-feedback-valid-color !default;\n$form-feedback-icon-valid:          url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 8 8'><path fill='#{$form-feedback-icon-valid-color}' d='M2.3 6.73.6 4.53c-.4-1.04.46-1.4 1.1-.8l1.1 1.4 3.4-3.8c.6-.63 1.6-.27 1.2.7l-4 4.6c-.43.5-.8.4-1.1.1z'/></svg>\") !default;\n$form-feedback-icon-invalid-color:  $form-feedback-invalid-color !default;\n$form-feedback-icon-invalid:        url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 12 12' width='12' height='12' fill='none' stroke='#{$form-feedback-icon-invalid-color}'><circle cx='6' cy='6' r='4.5'/><path stroke-linejoin='round' d='M5.8 3.6h.4L6 6.5z'/><circle cx='6' cy='8.2' r='.6' fill='#{$form-feedback-icon-invalid-color}' stroke='none'/></svg>\") !default;\n// scss-docs-end form-feedback-variables\n\n// scss-docs-start form-validation-colors\n$form-valid-color:                  $form-feedback-valid-color !default;\n$form-valid-border-color:           $form-feedback-valid-color !default;\n$form-invalid-color:                $form-feedback-invalid-color !default;\n$form-invalid-border-color:         $form-feedback-invalid-color !default;\n// scss-docs-end form-validation-colors\n\n// scss-docs-start form-validation-states\n$form-validation-states: (\n  \"valid\": (\n    \"color\": var(--#{$prefix}form-valid-color),\n    \"icon\": $form-feedback-icon-valid,\n    \"tooltip-color\": #fff,\n    \"tooltip-bg-color\": var(--#{$prefix}success),\n    \"focus-box-shadow\": 0 0 $input-btn-focus-blur $input-focus-width rgba(var(--#{$prefix}success-rgb), $input-btn-focus-color-opacity),\n    \"border-color\": var(--#{$prefix}form-valid-border-color),\n  ),\n  \"invalid\": (\n    \"color\": var(--#{$prefix}form-invalid-color),\n    \"icon\": $form-feedback-icon-invalid,\n    \"tooltip-color\": #fff,\n    \"tooltip-bg-color\": var(--#{$prefix}danger),\n    \"focus-box-shadow\": 0 0 $input-btn-focus-blur $input-focus-width rgba(var(--#{$prefix}danger-rgb), $input-btn-focus-color-opacity),\n    \"border-color\": var(--#{$prefix}form-invalid-border-color),\n  )\n) !default;\n// scss-docs-end form-validation-states\n\n// Z-index master list\n//\n// Warning: Avoid customizing these values. They're used for a bird's eye view\n// of components dependent on the z-axis and are designed to all work together.\n\n// scss-docs-start zindex-stack\n$zindex-dropdown:                   1000 !default;\n$zindex-sticky:                     1020 !default;\n$zindex-fixed:                      1030 !default;\n$zindex-offcanvas-backdrop:         1040 !default;\n$zindex-offcanvas:                  1045 !default;\n$zindex-modal-backdrop:             1050 !default;\n$zindex-modal:                      1055 !default;\n$zindex-popover:                    1070 !default;\n$zindex-tooltip:                    1080 !default;\n$zindex-toast:                      1090 !default;\n// scss-docs-end zindex-stack\n\n// scss-docs-start zindex-levels-map\n$zindex-levels: (\n  n1: -1,\n  0: 0,\n  1: 1,\n  2: 2,\n  3: 3\n) !default;\n// scss-docs-end zindex-levels-map\n\n\n// Navs\n\n// scss-docs-start nav-variables\n$nav-link-padding-y:                .5rem !default;\n$nav-link-padding-x:                1rem !default;\n$nav-link-font-size:                null !default;\n$nav-link-font-weight:              null !default;\n$nav-link-color:                    var(--#{$prefix}link-color) !default;\n$nav-link-hover-color:              var(--#{$prefix}link-hover-color) !default;\n$nav-link-transition:               color .15s ease-in-out, background-color .15s ease-in-out, border-color .15s ease-in-out !default;\n$nav-link-disabled-color:           var(--#{$prefix}secondary-color) !default;\n$nav-link-focus-box-shadow:         $focus-ring-box-shadow !default;\n\n$nav-tabs-border-color:             var(--#{$prefix}border-color) !default;\n$nav-tabs-border-width:             var(--#{$prefix}border-width) !default;\n$nav-tabs-border-radius:            var(--#{$prefix}border-radius) !default;\n$nav-tabs-link-hover-border-color:  var(--#{$prefix}secondary-bg) var(--#{$prefix}secondary-bg) $nav-tabs-border-color !default;\n$nav-tabs-link-active-color:        var(--#{$prefix}emphasis-color) !default;\n$nav-tabs-link-active-bg:           var(--#{$prefix}body-bg) !default;\n$nav-tabs-link-active-border-color: var(--#{$prefix}border-color) var(--#{$prefix}border-color) $nav-tabs-link-active-bg !default;\n\n$nav-pills-border-radius:           var(--#{$prefix}border-radius) !default;\n$nav-pills-link-active-color:       $component-active-color !default;\n$nav-pills-link-active-bg:          $component-active-bg !default;\n\n$nav-underline-gap:                 1rem !default;\n$nav-underline-border-width:        .125rem !default;\n$nav-underline-link-active-color:   var(--#{$prefix}emphasis-color) !default;\n// scss-docs-end nav-variables\n\n\n// Navbar\n\n// scss-docs-start navbar-variables\n$navbar-padding-y:                  $spacer * .5 !default;\n$navbar-padding-x:                  null !default;\n\n$navbar-nav-link-padding-x:         .5rem !default;\n\n$navbar-brand-font-size:            $font-size-lg !default;\n// Compute the navbar-brand padding-y so the navbar-brand will have the same height as navbar-text and nav-link\n$nav-link-height:                   $font-size-base * $line-height-base + $nav-link-padding-y * 2 !default;\n$navbar-brand-height:               $navbar-brand-font-size * $line-height-base !default;\n$navbar-brand-padding-y:            ($nav-link-height - $navbar-brand-height) * .5 !default;\n$navbar-brand-margin-end:           1rem !default;\n\n$navbar-toggler-padding-y:          .25rem !default;\n$navbar-toggler-padding-x:          .75rem !default;\n$navbar-toggler-font-size:          $font-size-lg !default;\n$navbar-toggler-border-radius:      $btn-border-radius !default;\n$navbar-toggler-focus-width:        $btn-focus-width !default;\n$navbar-toggler-transition:         box-shadow .15s ease-in-out !default;\n\n$navbar-light-color:                rgba(var(--#{$prefix}emphasis-color-rgb), .65) !default;\n$navbar-light-hover-color:          rgba(var(--#{$prefix}emphasis-color-rgb), .8) !default;\n$navbar-light-active-color:         rgba(var(--#{$prefix}emphasis-color-rgb), 1) !default;\n$navbar-light-disabled-color:       rgba(var(--#{$prefix}emphasis-color-rgb), .3) !default;\n$navbar-light-icon-color:           rgba($body-color, .75) !default;\n$navbar-light-toggler-icon-bg:      url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 30 30'><path stroke='#{$navbar-light-icon-color}' stroke-linecap='round' stroke-miterlimit='10' stroke-width='2' d='M4 7h22M4 15h22M4 23h22'/></svg>\") !default;\n$navbar-light-toggler-border-color: rgba(var(--#{$prefix}emphasis-color-rgb), .15) !default;\n$navbar-light-brand-color:          $navbar-light-active-color !default;\n$navbar-light-brand-hover-color:    $navbar-light-active-color !default;\n// scss-docs-end navbar-variables\n\n// scss-docs-start navbar-dark-variables\n$navbar-dark-color:                 rgba($white, .55) !default;\n$navbar-dark-hover-color:           rgba($white, .75) !default;\n$navbar-dark-active-color:          $white !default;\n$navbar-dark-disabled-color:        rgba($white, .25) !default;\n$navbar-dark-icon-color:            $navbar-dark-color !default;\n$navbar-dark-toggler-icon-bg:       url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 30 30'><path stroke='#{$navbar-dark-icon-color}' stroke-linecap='round' stroke-miterlimit='10' stroke-width='2' d='M4 7h22M4 15h22M4 23h22'/></svg>\") !default;\n$navbar-dark-toggler-border-color:  rgba($white, .1) !default;\n$navbar-dark-brand-color:           $navbar-dark-active-color !default;\n$navbar-dark-brand-hover-color:     $navbar-dark-active-color !default;\n// scss-docs-end navbar-dark-variables\n\n\n// Dropdowns\n//\n// Dropdown menu container and contents.\n\n// scss-docs-start dropdown-variables\n$dropdown-min-width:                10rem !default;\n$dropdown-padding-x:                0 !default;\n$dropdown-padding-y:                .5rem !default;\n$dropdown-spacer:                   .125rem !default;\n$dropdown-font-size:                $font-size-base !default;\n$dropdown-color:                    var(--#{$prefix}body-color) !default;\n$dropdown-bg:                       var(--#{$prefix}body-bg) !default;\n$dropdown-border-color:             var(--#{$prefix}border-color-translucent) !default;\n$dropdown-border-radius:            var(--#{$prefix}border-radius) !default;\n$dropdown-border-width:             var(--#{$prefix}border-width) !default;\n$dropdown-inner-border-radius:      calc(#{$dropdown-border-radius} - #{$dropdown-border-width}) !default; // stylelint-disable-line function-disallowed-list\n$dropdown-divider-bg:               $dropdown-border-color !default;\n$dropdown-divider-margin-y:         $spacer * .5 !default;\n$dropdown-box-shadow:               var(--#{$prefix}box-shadow) !default;\n\n$dropdown-link-color:               var(--#{$prefix}body-color) !default;\n$dropdown-link-hover-color:         $dropdown-link-color !default;\n$dropdown-link-hover-bg:            var(--#{$prefix}tertiary-bg) !default;\n\n$dropdown-link-active-color:        $component-active-color !default;\n$dropdown-link-active-bg:           $component-active-bg !default;\n\n$dropdown-link-disabled-color:      var(--#{$prefix}tertiary-color) !default;\n\n$dropdown-item-padding-y:           $spacer * .25 !default;\n$dropdown-item-padding-x:           $spacer !default;\n\n$dropdown-header-color:             $gray-600 !default;\n$dropdown-header-padding-x:         $dropdown-item-padding-x !default;\n$dropdown-header-padding-y:         $dropdown-padding-y !default;\n// fusv-disable\n$dropdown-header-padding:           $dropdown-header-padding-y $dropdown-header-padding-x !default; // Deprecated in v5.2.0\n// fusv-enable\n// scss-docs-end dropdown-variables\n\n// scss-docs-start dropdown-dark-variables\n$dropdown-dark-color:               $gray-300 !default;\n$dropdown-dark-bg:                  $gray-800 !default;\n$dropdown-dark-border-color:        $dropdown-border-color !default;\n$dropdown-dark-divider-bg:          $dropdown-divider-bg !default;\n$dropdown-dark-box-shadow:          null !default;\n$dropdown-dark-link-color:          $dropdown-dark-color !default;\n$dropdown-dark-link-hover-color:    $white !default;\n$dropdown-dark-link-hover-bg:       rgba($white, .15) !default;\n$dropdown-dark-link-active-color:   $dropdown-link-active-color !default;\n$dropdown-dark-link-active-bg:      $dropdown-link-active-bg !default;\n$dropdown-dark-link-disabled-color: $gray-500 !default;\n$dropdown-dark-header-color:        $gray-500 !default;\n// scss-docs-end dropdown-dark-variables\n\n\n// Pagination\n\n// scss-docs-start pagination-variables\n$pagination-padding-y:              .375rem !default;\n$pagination-padding-x:              .75rem !default;\n$pagination-padding-y-sm:           .25rem !default;\n$pagination-padding-x-sm:           .5rem !default;\n$pagination-padding-y-lg:           .75rem !default;\n$pagination-padding-x-lg:           1.5rem !default;\n\n$pagination-font-size:              $font-size-base !default;\n\n$pagination-color:                  var(--#{$prefix}link-color) !default;\n$pagination-bg:                     var(--#{$prefix}body-bg) !default;\n$pagination-border-radius:          var(--#{$prefix}border-radius) !default;\n$pagination-border-width:           var(--#{$prefix}border-width) !default;\n$pagination-margin-start:           calc(#{$pagination-border-width} * -1) !default; // stylelint-disable-line function-disallowed-list\n$pagination-border-color:           var(--#{$prefix}border-color) !default;\n\n$pagination-focus-color:            var(--#{$prefix}link-hover-color) !default;\n$pagination-focus-bg:               var(--#{$prefix}secondary-bg) !default;\n$pagination-focus-box-shadow:       $focus-ring-box-shadow !default;\n$pagination-focus-outline:          0 !default;\n\n$pagination-hover-color:            var(--#{$prefix}link-hover-color) !default;\n$pagination-hover-bg:               var(--#{$prefix}tertiary-bg) !default;\n$pagination-hover-border-color:     var(--#{$prefix}border-color) !default; // Todo in v6: remove this?\n\n$pagination-active-color:           $component-active-color !default;\n$pagination-active-bg:              $component-active-bg !default;\n$pagination-active-border-color:    $component-active-bg !default;\n\n$pagination-disabled-color:         var(--#{$prefix}secondary-color) !default;\n$pagination-disabled-bg:            var(--#{$prefix}secondary-bg) !default;\n$pagination-disabled-border-color:  var(--#{$prefix}border-color) !default;\n\n$pagination-transition:              color .15s ease-in-out, background-color .15s ease-in-out, border-color .15s ease-in-out, box-shadow .15s ease-in-out !default;\n\n$pagination-border-radius-sm:       var(--#{$prefix}border-radius-sm) !default;\n$pagination-border-radius-lg:       var(--#{$prefix}border-radius-lg) !default;\n// scss-docs-end pagination-variables\n\n\n// Placeholders\n\n// scss-docs-start placeholders\n$placeholder-opacity-max:           .5 !default;\n$placeholder-opacity-min:           .2 !default;\n// scss-docs-end placeholders\n\n// Cards\n\n// scss-docs-start card-variables\n$card-spacer-y:                     $spacer !default;\n$card-spacer-x:                     $spacer !default;\n$card-title-spacer-y:               $spacer * .5 !default;\n$card-title-color:                  null !default;\n$card-subtitle-color:               null !default;\n$card-border-width:                 var(--#{$prefix}border-width) !default;\n$card-border-color:                 var(--#{$prefix}border-color-translucent) !default;\n$card-border-radius:                var(--#{$prefix}border-radius) !default;\n$card-box-shadow:                   null !default;\n$card-inner-border-radius:          subtract($card-border-radius, $card-border-width) !default;\n$card-cap-padding-y:                $card-spacer-y * .5 !default;\n$card-cap-padding-x:                $card-spacer-x !default;\n$card-cap-bg:                       rgba(var(--#{$prefix}body-color-rgb), .03) !default;\n$card-cap-color:                    null !default;\n$card-height:                       null !default;\n$card-color:                        null !default;\n$card-bg:                           var(--#{$prefix}body-bg) !default;\n$card-img-overlay-padding:          $spacer !default;\n$card-group-margin:                 $grid-gutter-width * .5 !default;\n// scss-docs-end card-variables\n\n// Accordion\n\n// scss-docs-start accordion-variables\n$accordion-padding-y:                     1rem !default;\n$accordion-padding-x:                     1.25rem !default;\n$accordion-color:                         var(--#{$prefix}body-color) !default;\n$accordion-bg:                            var(--#{$prefix}body-bg) !default;\n$accordion-border-width:                  var(--#{$prefix}border-width) !default;\n$accordion-border-color:                  var(--#{$prefix}border-color) !default;\n$accordion-border-radius:                 var(--#{$prefix}border-radius) !default;\n$accordion-inner-border-radius:           subtract($accordion-border-radius, $accordion-border-width) !default;\n\n$accordion-body-padding-y:                $accordion-padding-y !default;\n$accordion-body-padding-x:                $accordion-padding-x !default;\n\n$accordion-button-padding-y:              $accordion-padding-y !default;\n$accordion-button-padding-x:              $accordion-padding-x !default;\n$accordion-button-color:                  var(--#{$prefix}body-color) !default;\n$accordion-button-bg:                     var(--#{$prefix}accordion-bg) !default;\n$accordion-transition:                    $btn-transition, border-radius .15s ease !default;\n$accordion-button-active-bg:              var(--#{$prefix}primary-bg-subtle) !default;\n$accordion-button-active-color:           var(--#{$prefix}primary-text-emphasis) !default;\n\n$accordion-button-focus-border-color:     $input-focus-border-color !default;\n$accordion-button-focus-box-shadow:       $btn-focus-box-shadow !default;\n\n$accordion-icon-width:                    1.25rem !default;\n$accordion-icon-color:                    $body-color !default;\n$accordion-icon-active-color:             $primary-text-emphasis !default;\n$accordion-icon-transition:               transform .2s ease-in-out !default;\n$accordion-icon-transform:                rotate(-180deg) !default;\n\n$accordion-button-icon:         url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='#{$accordion-icon-color}'><path fill-rule='evenodd' d='M1.646 4.646a.5.5 0 0 1 .708 0L8 10.293l5.646-5.647a.5.5 0 0 1 .708.708l-6 6a.5.5 0 0 1-.708 0l-6-6a.5.5 0 0 1 0-.708z'/></svg>\") !default;\n$accordion-button-active-icon:  url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='#{$accordion-icon-active-color}'><path fill-rule='evenodd' d='M1.646 4.646a.5.5 0 0 1 .708 0L8 10.293l5.646-5.647a.5.5 0 0 1 .708.708l-6 6a.5.5 0 0 1-.708 0l-6-6a.5.5 0 0 1 0-.708z'/></svg>\") !default;\n// scss-docs-end accordion-variables\n\n// Tooltips\n\n// scss-docs-start tooltip-variables\n$tooltip-font-size:                 $font-size-sm !default;\n$tooltip-max-width:                 200px !default;\n$tooltip-color:                     var(--#{$prefix}body-bg) !default;\n$tooltip-bg:                        var(--#{$prefix}emphasis-color) !default;\n$tooltip-border-radius:             var(--#{$prefix}border-radius) !default;\n$tooltip-opacity:                   .9 !default;\n$tooltip-padding-y:                 $spacer * .25 !default;\n$tooltip-padding-x:                 $spacer * .5 !default;\n$tooltip-margin:                    null !default; // TODO: remove this in v6\n\n$tooltip-arrow-width:               .8rem !default;\n$tooltip-arrow-height:              .4rem !default;\n// fusv-disable\n$tooltip-arrow-color:               null !default; // Deprecated in Bootstrap 5.2.0 for CSS variables\n// fusv-enable\n// scss-docs-end tooltip-variables\n\n// Form tooltips must come after regular tooltips\n// scss-docs-start tooltip-feedback-variables\n$form-feedback-tooltip-padding-y:     $tooltip-padding-y !default;\n$form-feedback-tooltip-padding-x:     $tooltip-padding-x !default;\n$form-feedback-tooltip-font-size:     $tooltip-font-size !default;\n$form-feedback-tooltip-line-height:   null !default;\n$form-feedback-tooltip-opacity:       $tooltip-opacity !default;\n$form-feedback-tooltip-border-radius: $tooltip-border-radius !default;\n// scss-docs-end tooltip-feedback-variables\n\n\n// Popovers\n\n// scss-docs-start popover-variables\n$popover-font-size:                 $font-size-sm !default;\n$popover-bg:                        var(--#{$prefix}body-bg) !default;\n$popover-max-width:                 276px !default;\n$popover-border-width:              var(--#{$prefix}border-width) !default;\n$popover-border-color:              var(--#{$prefix}border-color-translucent) !default;\n$popover-border-radius:             var(--#{$prefix}border-radius-lg) !default;\n$popover-inner-border-radius:       calc(#{$popover-border-radius} - #{$popover-border-width}) !default; // stylelint-disable-line function-disallowed-list\n$popover-box-shadow:                var(--#{$prefix}box-shadow) !default;\n\n$popover-header-font-size:          $font-size-base !default;\n$popover-header-bg:                 var(--#{$prefix}secondary-bg) !default;\n$popover-header-color:              $headings-color !default;\n$popover-header-padding-y:          .5rem !default;\n$popover-header-padding-x:          $spacer !default;\n\n$popover-body-color:                var(--#{$prefix}body-color) !default;\n$popover-body-padding-y:            $spacer !default;\n$popover-body-padding-x:            $spacer !default;\n\n$popover-arrow-width:               1rem !default;\n$popover-arrow-height:              .5rem !default;\n// scss-docs-end popover-variables\n\n// fusv-disable\n// Deprecated in Bootstrap 5.2.0 for CSS variables\n$popover-arrow-color:               $popover-bg !default;\n$popover-arrow-outer-color:         var(--#{$prefix}border-color-translucent) !default;\n// fusv-enable\n\n\n// Toasts\n\n// scss-docs-start toast-variables\n$toast-max-width:                   350px !default;\n$toast-padding-x:                   .75rem !default;\n$toast-padding-y:                   .5rem !default;\n$toast-font-size:                   .875rem !default;\n$toast-color:                       null !default;\n$toast-background-color:            rgba(var(--#{$prefix}body-bg-rgb), .85) !default;\n$toast-border-width:                var(--#{$prefix}border-width) !default;\n$toast-border-color:                var(--#{$prefix}border-color-translucent) !default;\n$toast-border-radius:               var(--#{$prefix}border-radius) !default;\n$toast-box-shadow:                  var(--#{$prefix}box-shadow) !default;\n$toast-spacing:                     $container-padding-x !default;\n\n$toast-header-color:                var(--#{$prefix}secondary-color) !default;\n$toast-header-background-color:     rgba(var(--#{$prefix}body-bg-rgb), .85) !default;\n$toast-header-border-color:         $toast-border-color !default;\n// scss-docs-end toast-variables\n\n\n// Badges\n\n// scss-docs-start badge-variables\n$badge-font-size:                   .75em !default;\n$badge-font-weight:                 $font-weight-bold !default;\n$badge-color:                       $white !default;\n$badge-padding-y:                   .35em !default;\n$badge-padding-x:                   .65em !default;\n$badge-border-radius:               var(--#{$prefix}border-radius) !default;\n// scss-docs-end badge-variables\n\n\n// Modals\n\n// scss-docs-start modal-variables\n$modal-inner-padding:               $spacer !default;\n\n$modal-footer-margin-between:       .5rem !default;\n\n$modal-dialog-margin:               .5rem !default;\n$modal-dialog-margin-y-sm-up:       1.75rem !default;\n\n$modal-title-line-height:           $line-height-base !default;\n\n$modal-content-color:               null !default;\n$modal-content-bg:                  var(--#{$prefix}body-bg) !default;\n$modal-content-border-color:        var(--#{$prefix}border-color-translucent) !default;\n$modal-content-border-width:        var(--#{$prefix}border-width) !default;\n$modal-content-border-radius:       var(--#{$prefix}border-radius-lg) !default;\n$modal-content-inner-border-radius: subtract($modal-content-border-radius, $modal-content-border-width) !default;\n$modal-content-box-shadow-xs:       var(--#{$prefix}box-shadow-sm) !default;\n$modal-content-box-shadow-sm-up:    var(--#{$prefix}box-shadow) !default;\n\n$modal-backdrop-bg:                 $black !default;\n$modal-backdrop-opacity:            .5 !default;\n\n$modal-header-border-color:         var(--#{$prefix}border-color) !default;\n$modal-header-border-width:         $modal-content-border-width !default;\n$modal-header-padding-y:            $modal-inner-padding !default;\n$modal-header-padding-x:            $modal-inner-padding !default;\n$modal-header-padding:              $modal-header-padding-y $modal-header-padding-x !default; // Keep this for backwards compatibility\n\n$modal-footer-bg:                   null !default;\n$modal-footer-border-color:         $modal-header-border-color !default;\n$modal-footer-border-width:         $modal-header-border-width !default;\n\n$modal-sm:                          300px !default;\n$modal-md:                          500px !default;\n$modal-lg:                          800px !default;\n$modal-xl:                          1140px !default;\n\n$modal-fade-transform:              translate(0, -50px) !default;\n$modal-show-transform:              none !default;\n$modal-transition:                  transform .3s ease-out !default;\n$modal-scale-transform:             scale(1.02) !default;\n// scss-docs-end modal-variables\n\n\n// Alerts\n//\n// Define alert colors, border radius, and padding.\n\n// scss-docs-start alert-variables\n$alert-padding-y:               $spacer !default;\n$alert-padding-x:               $spacer !default;\n$alert-margin-bottom:           1rem !default;\n$alert-border-radius:           var(--#{$prefix}border-radius) !default;\n$alert-link-font-weight:        $font-weight-bold !default;\n$alert-border-width:            var(--#{$prefix}border-width) !default;\n$alert-dismissible-padding-r:   $alert-padding-x * 3 !default; // 3x covers width of x plus default padding on either side\n// scss-docs-end alert-variables\n\n// fusv-disable\n$alert-bg-scale:                -80% !default; // Deprecated in v5.2.0, to be removed in v6\n$alert-border-scale:            -70% !default; // Deprecated in v5.2.0, to be removed in v6\n$alert-color-scale:             40% !default; // Deprecated in v5.2.0, to be removed in v6\n// fusv-enable\n\n// Progress bars\n\n// scss-docs-start progress-variables\n$progress-height:                   1rem !default;\n$progress-font-size:                $font-size-base * .75 !default;\n$progress-bg:                       var(--#{$prefix}secondary-bg) !default;\n$progress-border-radius:            var(--#{$prefix}border-radius) !default;\n$progress-box-shadow:               var(--#{$prefix}box-shadow-inset) !default;\n$progress-bar-color:                $white !default;\n$progress-bar-bg:                   $primary !default;\n$progress-bar-animation-timing:     1s linear infinite !default;\n$progress-bar-transition:           width .6s ease !default;\n// scss-docs-end progress-variables\n\n\n// List group\n\n// scss-docs-start list-group-variables\n$list-group-color:                  var(--#{$prefix}body-color) !default;\n$list-group-bg:                     var(--#{$prefix}body-bg) !default;\n$list-group-border-color:           var(--#{$prefix}border-color) !default;\n$list-group-border-width:           var(--#{$prefix}border-width) !default;\n$list-group-border-radius:          var(--#{$prefix}border-radius) !default;\n\n$list-group-item-padding-y:         $spacer * .5 !default;\n$list-group-item-padding-x:         $spacer !default;\n// fusv-disable\n$list-group-item-bg-scale:          -80% !default; // Deprecated in v5.3.0\n$list-group-item-color-scale:       40% !default; // Deprecated in v5.3.0\n// fusv-enable\n\n$list-group-hover-bg:               var(--#{$prefix}tertiary-bg) !default;\n$list-group-active-color:           $component-active-color !default;\n$list-group-active-bg:              $component-active-bg !default;\n$list-group-active-border-color:    $list-group-active-bg !default;\n\n$list-group-disabled-color:         var(--#{$prefix}secondary-color) !default;\n$list-group-disabled-bg:            $list-group-bg !default;\n\n$list-group-action-color:           var(--#{$prefix}secondary-color) !default;\n$list-group-action-hover-color:     var(--#{$prefix}emphasis-color) !default;\n\n$list-group-action-active-color:    var(--#{$prefix}body-color) !default;\n$list-group-action-active-bg:       var(--#{$prefix}secondary-bg) !default;\n// scss-docs-end list-group-variables\n\n\n// Image thumbnails\n\n// scss-docs-start thumbnail-variables\n$thumbnail-padding:                 .25rem !default;\n$thumbnail-bg:                      var(--#{$prefix}body-bg) !default;\n$thumbnail-border-width:            var(--#{$prefix}border-width) !default;\n$thumbnail-border-color:            var(--#{$prefix}border-color) !default;\n$thumbnail-border-radius:           var(--#{$prefix}border-radius) !default;\n$thumbnail-box-shadow:              var(--#{$prefix}box-shadow-sm) !default;\n// scss-docs-end thumbnail-variables\n\n\n// Figures\n\n// scss-docs-start figure-variables\n$figure-caption-font-size:          $small-font-size !default;\n$figure-caption-color:              var(--#{$prefix}secondary-color) !default;\n// scss-docs-end figure-variables\n\n\n// Breadcrumbs\n\n// scss-docs-start breadcrumb-variables\n$breadcrumb-font-size:              null !default;\n$breadcrumb-padding-y:              0 !default;\n$breadcrumb-padding-x:              0 !default;\n$breadcrumb-item-padding-x:         .5rem !default;\n$breadcrumb-margin-bottom:          1rem !default;\n$breadcrumb-bg:                     null !default;\n$breadcrumb-divider-color:          var(--#{$prefix}secondary-color) !default;\n$breadcrumb-active-color:           var(--#{$prefix}secondary-color) !default;\n$breadcrumb-divider:                quote(\"/\") !default;\n$breadcrumb-divider-flipped:        $breadcrumb-divider !default;\n$breadcrumb-border-radius:          null !default;\n// scss-docs-end breadcrumb-variables\n\n// Carousel\n\n// scss-docs-start carousel-variables\n$carousel-control-color:             $white !default;\n$carousel-control-width:             15% !default;\n$carousel-control-opacity:           .5 !default;\n$carousel-control-hover-opacity:     .9 !default;\n$carousel-control-transition:        opacity .15s ease !default;\n\n$carousel-indicator-width:           30px !default;\n$carousel-indicator-height:          3px !default;\n$carousel-indicator-hit-area-height: 10px !default;\n$carousel-indicator-spacer:          3px !default;\n$carousel-indicator-opacity:         .5 !default;\n$carousel-indicator-active-bg:       $white !default;\n$carousel-indicator-active-opacity:  1 !default;\n$carousel-indicator-transition:      opacity .6s ease !default;\n\n$carousel-caption-width:             70% !default;\n$carousel-caption-color:             $white !default;\n$carousel-caption-padding-y:         1.25rem !default;\n$carousel-caption-spacer:            1.25rem !default;\n\n$carousel-control-icon-width:        2rem !default;\n\n$carousel-control-prev-icon-bg:      url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='#{$carousel-control-color}'><path d='M11.354 1.646a.5.5 0 0 1 0 .708L5.707 8l5.647 5.646a.5.5 0 0 1-.708.708l-6-6a.5.5 0 0 1 0-.708l6-6a.5.5 0 0 1 .708 0z'/></svg>\") !default;\n$carousel-control-next-icon-bg:      url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='#{$carousel-control-color}'><path d='M4.646 1.646a.5.5 0 0 1 .708 0l6 6a.5.5 0 0 1 0 .708l-6 6a.5.5 0 0 1-.708-.708L10.293 8 4.646 2.354a.5.5 0 0 1 0-.708z'/></svg>\") !default;\n\n$carousel-transition-duration:       .6s !default;\n$carousel-transition:                transform $carousel-transition-duration ease-in-out !default; // Define transform transition first if using multiple transitions (e.g., `transform 2s ease, opacity .5s ease-out`)\n// scss-docs-end carousel-variables\n\n// scss-docs-start carousel-dark-variables\n$carousel-dark-indicator-active-bg:  $black !default;\n$carousel-dark-caption-color:        $black !default;\n$carousel-dark-control-icon-filter:  invert(1) grayscale(100) !default;\n// scss-docs-end carousel-dark-variables\n\n\n// Spinners\n\n// scss-docs-start spinner-variables\n$spinner-width:           2rem !default;\n$spinner-height:          $spinner-width !default;\n$spinner-vertical-align:  -.125em !default;\n$spinner-border-width:    .25em !default;\n$spinner-animation-speed: .75s !default;\n\n$spinner-width-sm:        1rem !default;\n$spinner-height-sm:       $spinner-width-sm !default;\n$spinner-border-width-sm: .2em !default;\n// scss-docs-end spinner-variables\n\n\n// Close\n\n// scss-docs-start close-variables\n$btn-close-width:            1em !default;\n$btn-close-height:           $btn-close-width !default;\n$btn-close-padding-x:        .25em !default;\n$btn-close-padding-y:        $btn-close-padding-x !default;\n$btn-close-color:            $black !default;\n$btn-close-bg:               url(\"data:image/svg+xml,<svg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 16 16' fill='#{$btn-close-color}'><path d='M.293.293a1 1 0 0 1 1.414 0L8 6.586 14.293.293a1 1 0 1 1 1.414 1.414L9.414 8l6.293 6.293a1 1 0 0 1-1.414 1.414L8 9.414l-6.293 6.293a1 1 0 0 1-1.414-1.414L6.586 8 .293 1.707a1 1 0 0 1 0-1.414z'/></svg>\") !default;\n$btn-close-focus-shadow:     $focus-ring-box-shadow !default;\n$btn-close-opacity:          .5 !default;\n$btn-close-hover-opacity:    .75 !default;\n$btn-close-focus-opacity:    1 !default;\n$btn-close-disabled-opacity: .25 !default;\n$btn-close-white-filter:     invert(1) grayscale(100%) brightness(200%) !default;\n// scss-docs-end close-variables\n\n\n// Offcanvas\n\n// scss-docs-start offcanvas-variables\n$offcanvas-padding-y:               $modal-inner-padding !default;\n$offcanvas-padding-x:               $modal-inner-padding !default;\n$offcanvas-horizontal-width:        400px !default;\n$offcanvas-vertical-height:         30vh !default;\n$offcanvas-transition-duration:     .3s !default;\n$offcanvas-border-color:            $modal-content-border-color !default;\n$offcanvas-border-width:            $modal-content-border-width !default;\n$offcanvas-title-line-height:       $modal-title-line-height !default;\n$offcanvas-bg-color:                var(--#{$prefix}body-bg) !default;\n$offcanvas-color:                   var(--#{$prefix}body-color) !default;\n$offcanvas-box-shadow:              $modal-content-box-shadow-xs !default;\n$offcanvas-backdrop-bg:             $modal-backdrop-bg !default;\n$offcanvas-backdrop-opacity:        $modal-backdrop-opacity !default;\n// scss-docs-end offcanvas-variables\n\n// Code\n\n$code-font-size:                    $small-font-size !default;\n$code-color:                        $pink !default;\n\n$kbd-padding-y:                     .1875rem !default;\n$kbd-padding-x:                     .375rem !default;\n$kbd-font-size:                     $code-font-size !default;\n$kbd-color:                         var(--#{$prefix}body-bg) !default;\n$kbd-bg:                            var(--#{$prefix}body-color) !default;\n$nested-kbd-font-weight:            null !default; // Deprecated in v5.2.0, removing in v6\n\n$pre-color:                         null !default;\n","/***\n * Rules for the UX/UI of skip navigation link btn. \n *It's only visible to people\n * navigating with keyboard for accessibility purposes\n * ref: https://www.youtube.com/watch?v=VUR0I5mqq7I\n ***/\n\n.skip-link {\n  position: fixed;\n  top: 0;\n  left: 0;\n  right: 0;\n  text-align: center;\n  background-color: var(--pst-color-warning);\n  // Ensure we are using a WCAG conformant colour\n  color: var(--pst-color-warning-text) !important;\n  padding: 0.5rem;\n  z-index: $zindex-modal;\n  border-bottom: 1px solid var(--pst-color-border);\n\n  // This shows / hides the button\n  transform: translateY(-100%);\n  transition: transform 150ms ease-in-out;\n  &:focus {\n    transform: translateY(0%);\n    // ensure this is visible\n    outline: 3px solid $foundation-black;\n  }\n}\n",".bd-container {\n  flex-grow: 1;\n  display: flex;\n  justify-content: center;\n\n  .bd-container__inner {\n    display: flex;\n  }\n}\n\n.bd-page-width {\n  width: 100%;\n  @include media-breakpoint-up(lg) {\n    max-width: $breakpoint-page-width;\n  }\n}\n","// Breakpoint viewport sizes and media queries.\n//\n// Breakpoints are defined as a map of (name: minimum width), order from small to large:\n//\n//    (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px)\n//\n// The map defined in the `$grid-breakpoints` global variable is used as the `$breakpoints` argument by default.\n\n// Name of the next breakpoint, or null for the last breakpoint.\n//\n//    >> breakpoint-next(sm)\n//    md\n//    >> breakpoint-next(sm, (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px))\n//    md\n//    >> breakpoint-next(sm, $breakpoint-names: (xs sm md lg xl xxl))\n//    md\n@function breakpoint-next($name, $breakpoints: $grid-breakpoints, $breakpoint-names: map-keys($breakpoints)) {\n  $n: index($breakpoint-names, $name);\n  @if not $n {\n    @error \"breakpoint `#{$name}` not found in `#{$breakpoints}`\";\n  }\n  @return if($n < length($breakpoint-names), nth($breakpoint-names, $n + 1), null);\n}\n\n// Minimum breakpoint width. Null for the smallest (first) breakpoint.\n//\n//    >> breakpoint-min(sm, (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px))\n//    576px\n@function breakpoint-min($name, $breakpoints: $grid-breakpoints) {\n  $min: map-get($breakpoints, $name);\n  @return if($min != 0, $min, null);\n}\n\n// Maximum breakpoint width.\n// The maximum value is reduced by 0.02px to work around the limitations of\n// `min-` and `max-` prefixes and viewports with fractional widths.\n// See https://www.w3.org/TR/mediaqueries-4/#mq-min-max\n// Uses 0.02px rather than 0.01px to work around a current rounding bug in Safari.\n// See https://bugs.webkit.org/show_bug.cgi?id=178261\n//\n//    >> breakpoint-max(md, (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px))\n//    767.98px\n@function breakpoint-max($name, $breakpoints: $grid-breakpoints) {\n  $max: map-get($breakpoints, $name);\n  @return if($max and $max > 0, $max - .02, null);\n}\n\n// Returns a blank string if smallest breakpoint, otherwise returns the name with a dash in front.\n// Useful for making responsive utilities.\n//\n//    >> breakpoint-infix(xs, (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px))\n//    \"\"  (Returns a blank string)\n//    >> breakpoint-infix(sm, (xs: 0, sm: 576px, md: 768px, lg: 992px, xl: 1200px, xxl: 1400px))\n//    \"-sm\"\n@function breakpoint-infix($name, $breakpoints: $grid-breakpoints) {\n  @return if(breakpoint-min($name, $breakpoints) == null, \"\", \"-#{$name}\");\n}\n\n// Media of at least the minimum breakpoint width. No query for the smallest breakpoint.\n// Makes the @content apply to the given breakpoint and wider.\n@mixin media-breakpoint-up($name, $breakpoints: $grid-breakpoints) {\n  $min: breakpoint-min($name, $breakpoints);\n  @if $min {\n    @media (min-width: $min) {\n      @content;\n    }\n  } @else {\n    @content;\n  }\n}\n\n// Media of at most the maximum breakpoint width. No query for the largest breakpoint.\n// Makes the @content apply to the given breakpoint and narrower.\n@mixin media-breakpoint-down($name, $breakpoints: $grid-breakpoints) {\n  $max: breakpoint-max($name, $breakpoints);\n  @if $max {\n    @media (max-width: $max) {\n      @content;\n    }\n  } @else {\n    @content;\n  }\n}\n\n// Media that spans multiple breakpoint widths.\n// Makes the @content apply between the min and max breakpoints\n@mixin media-breakpoint-between($lower, $upper, $breakpoints: $grid-breakpoints) {\n  $min: breakpoint-min($lower, $breakpoints);\n  $max: breakpoint-max($upper, $breakpoints);\n\n  @if $min != null and $max != null {\n    @media (min-width: $min) and (max-width: $max) {\n      @content;\n    }\n  } @else if $max == null {\n    @include media-breakpoint-up($lower, $breakpoints) {\n      @content;\n    }\n  } @else if $min == null {\n    @include media-breakpoint-down($upper, $breakpoints) {\n      @content;\n    }\n  }\n}\n\n// Media between the breakpoint's minimum and maximum widths.\n// No minimum for the smallest breakpoint, and no maximum for the largest one.\n// Makes the @content apply only to the given breakpoint, not viewports any wider or narrower.\n@mixin media-breakpoint-only($name, $breakpoints: $grid-breakpoints) {\n  $min:  breakpoint-min($name, $breakpoints);\n  $next: breakpoint-next($name, $breakpoints);\n  $max:  breakpoint-max($next, $breakpoints);\n\n  @if $min != null and $max != null {\n    @media (min-width: $min) and (max-width: $max) {\n      @content;\n    }\n  } @else if $max == null {\n    @include media-breakpoint-up($name, $breakpoints) {\n      @content;\n    }\n  } @else if $min == null {\n    @include media-breakpoint-down($next, $breakpoints) {\n      @content;\n    }\n  }\n}\n",".bd-header-version-warning,\n.bd-header-announcement {\n  min-height: 3rem;\n  width: 100%;\n  display: flex;\n  position: relative;\n  align-items: center;\n  justify-content: center;\n  text-align: center;\n  padding: 0.5rem 12.5%; // Horizontal padding so the width is 75%\n  // One breakpoint less than $breakpoint-sidebar-primary. See variables/_layout.scss for more info.\n  @include media-breakpoint-down(lg) {\n    // Announcements can take a bit more width on mobile\n    padding: 0.5rem 2%;\n  }\n\n  p {\n    font-weight: bold;\n    margin: 0;\n  }\n\n  &:after {\n    position: absolute;\n    width: 100%;\n    height: 100%;\n    left: 0;\n    top: 0;\n    content: \"\";\n    z-index: -1; // So it doesn't hover over the content\n  }\n\n  &:empty {\n    display: none;\n  }\n\n  // Ensure there is enough contrast against the background\n  a {\n    color: var(--pst-color-inline-code-links);\n  }\n}\n\n// Bg color is now defined in the theme color palette - using our secondary color\n.bd-header-announcement {\n  &:after {\n    background-color: var(--pst-color-secondary-bg);\n  }\n}\n\n.bd-header-version-warning {\n  &:after {\n    background-color: var(--pst-color-danger-bg);\n  }\n}\n","/**\n * Main content area\n */\n.bd-main {\n  flex-grow: 1;\n  flex-direction: column;\n  display: flex;\n  min-width: 0;\n\n  .bd-content {\n    display: flex;\n    justify-content: center;\n    height: 100%;\n    .bd-article-container {\n      justify-content: start;\n      display: flex;\n      flex-direction: column;\n      // Max-width is slightly more than the W3 since our docs often have images.\n      // We shoot for about 100 characters per line instead of 80.\n      // ref: https://www.w3.org/WAI/tutorials/page-structure/styling/#line-length\n      width: 100%;\n      max-width: 60em;\n      overflow-x: auto; // Prevent wide content from pushing off the secondary sidebar\n      padding: 1rem;\n      .bd-article {\n        // Give a bit more verticle spacing on wide screens\n        @include media-breakpoint-up($breakpoint-sidebar-secondary) {\n          padding-top: 1.5rem;\n          padding-left: 2rem;\n        }\n      }\n    }\n  }\n}\n",".bd-footer {\n  width: 100%;\n  border-top: 1px solid var(--pst-color-border);\n\n  .bd-footer__inner {\n    display: flex;\n    flex-grow: 1;\n    padding: 1rem;\n    margin: auto;\n  }\n\n  .footer-items__start,\n  .footer-items__center,\n  .footer-items__end {\n    display: flex;\n    flex-direction: column;\n    gap: 0.5rem;\n    justify-content: center;\n    flex-grow: 1;\n  }\n\n  .footer-items__center {\n    text-align: center;\n  }\n\n  .footer-items__end {\n    text-align: end;\n  }\n\n  // So that paragraphs don't take up extra room\n  .footer-item p {\n    margin-bottom: 0;\n  }\n}\n",".bd-footer-article {\n  margin-top: auto;\n\n  .footer-article-items {\n    display: flex;\n    flex-direction: column;\n  }\n}\n",".bd-footer-content {\n  .footer-content-items {\n    display: flex;\n    flex-direction: column;\n    margin-top: auto;\n  }\n}\n","/**\n * Header at the top of the page\n * It includes the announcement bar and the navigation bar.\n */\n\n// Styling for the Icon links can be found in components/_icon-links.scss\n\n// If we want the shadow to only point downward in the future, set\n// box-shadow to: 0 0.125rem 0.25rem -0.125rem rgba(0, 0, 0, 0.11);\n.bd-header {\n  position: sticky;\n  top: 0;\n  z-index: $zindex-fixed;\n\n  // Overrides bootstrap\n  background: var(--pst-color-on-background) !important;\n  box-shadow: 0 0.125rem 0.25rem 0 var(--pst-color-shadow);\n\n  width: 100%;\n  padding: 0;\n  max-width: 100vw;\n  justify-content: center;\n  .bd-header__inner {\n    display: flex;\n    align-items: center;\n    height: fit-content;\n    padding-left: 1rem;\n    padding-right: 1rem;\n  }\n\n  // These items will define the height of the header\n  .navbar-item {\n    height: var(--pst-header-height);\n    max-height: var(--pst-header-height);\n    display: flex;\n    align-items: center;\n  }\n\n  // Hide the header items on mobile\n  .navbar-header-items {\n    display: none;\n    flex-shrink: 1;\n    @include media-breakpoint-up($breakpoint-sidebar-primary) {\n      display: flex;\n      flex-grow: 1;\n      padding: 0 0 0 0.5rem;\n    }\n  }\n\n  .navbar-header-items__end,\n  .navbar-header-items__center,\n  .navbar-header-items__start {\n    display: flex;\n    align-items: center;\n    flex-flow: wrap;\n    // In case we wrap our items to multiple rows on small screens\n    row-gap: 0;\n  }\n\n  .navbar-header-items__end,\n  .navbar-header-items__center {\n    column-gap: 1rem;\n  }\n\n  // A little smaller because this is displayed by default on mobile\n  .navbar-header-items__start {\n    flex-shrink: 0;\n    margin-right: auto;\n    gap: 0.5rem;\n  }\n\n  .navbar-header-items__end {\n    // End navbar items should snap to the right\n    justify-content: end;\n  }\n\n  // Contains the navigation links within the navbar\n  .navbar-nav {\n    display: flex;\n\n    @include media-breakpoint-up($breakpoint-sidebar-primary) {\n      // Center align on wide screens so the dropdown button is centered properly\n      align-items: center;\n    }\n\n    li a.nav-link {\n      @include link-style-text;\n    }\n\n    // Current page is always underlined in the navbar\n    > .current > .nav-link {\n      @include link-navbar-current;\n    }\n\n    // Dropdowns for the extra links\n    .dropdown {\n      button {\n        display: unset;\n        color: var(--pst-color-text-muted);\n        border: none;\n        @include link-style-hover;\n        @include focus-indicator;\n      }\n\n      .dropdown-menu {\n        z-index: $zindex-popover;\n        border: 1px solid var(--pst-color-border);\n        box-shadow: 0 0 0.3rem 0.1rem var(--pst-color-shadow);\n        background-color: var(--pst-color-on-background);\n        padding: 0.5rem 0;\n        margin: 0.5rem 0;\n        min-width: 20rem;\n\n        .dropdown-item {\n          // Give the items in the dropdown some breathing room but let the hit\n          // and hover area of the items extend to the edges of the menu\n          padding: 0.25rem 1.5rem;\n\n          // Override Bootstrap\n          &:focus:not(:hover):not(:active) {\n            background-color: inherit;\n          }\n        }\n\n        // Hide the menu unless show has been clicked\n        &:not(.show) {\n          display: none;\n        }\n      }\n    }\n  }\n}\n\n// inline the element in the navbar as long as they fit and use display block when collapsing\n@include media-breakpoint-up($breakpoint-sidebar-primary) {\n  .navbar-center-items .navbar-item {\n    display: inline-block;\n  }\n}\n\n.nav-link {\n  @include link-style-hover;\n\n  // Override Bootstrap\n  transition: none;\n\n  &.nav-external:after {\n    font: var(--fa-font-solid);\n    content: var(--pst-icon-external-link);\n    font-size: 0.75em;\n    margin-left: 0.3em;\n  }\n}\n\n.bd-navbar-elements li.nav-item i {\n  font-size: 0.7rem;\n  padding-left: 2px;\n  vertical-align: middle;\n}\n\n/**\n * Showing and hiding the sidebar toggle buttons and header items\n */\n\n// Hide the header items on mobile\n.bd-header {\n  // Toggle labels\n  label {\n    &.sidebar-toggle {\n      display: flex;\n      cursor: pointer;\n      font-size: var(--pst-font-size-icon);\n      align-items: center;\n      color: var(--pst-color-muted);\n      margin-bottom: 0;\n      padding-bottom: 0.25rem;\n    }\n\n    &.primary-toggle {\n      margin-right: 1rem;\n      @include media-breakpoint-up($breakpoint-sidebar-primary) {\n        display: none;\n      }\n    }\n\n    &.secondary-toggle {\n      margin-left: 1rem;\n      @include media-breakpoint-up($breakpoint-sidebar-secondary) {\n        display: none;\n      }\n    }\n    @include icon-navbar-hover;\n    @include focus-indicator;\n  }\n\n  // Hide the navbar header items on mobile because they're in the sidebar\n  .navbar-header-items {\n    display: none;\n    @include media-breakpoint-up($breakpoint-sidebar-primary) {\n      display: inherit;\n    }\n  }\n}\n\n// THe elements next to the hamburger menu only show on narrow screens\n.navbar-persistent--mobile {\n  margin-left: auto;\n  @include media-breakpoint-up($breakpoint-sidebar-primary) {\n    display: none;\n  }\n}\n\n// The navbar-persistent content should only show on wide screens\n.navbar-persistent--container {\n  display: none;\n  @include media-breakpoint-up($breakpoint-sidebar-primary) {\n    display: flex;\n  }\n}\n",".header-article__inner {\n  display: flex;\n  padding: 0 0.5rem;\n\n  // The items define the height so that it disappears if there are no items\n  .header-article-item {\n    min-height: var(--pst-header-article-height);\n    height: var(--pst-header-article-height);\n  }\n\n  .header-article-items__start,\n  .header-article-items__end {\n    display: flex;\n    align-items: start;\n    gap: 0.5rem;\n  }\n\n  .header-article-items__end {\n    margin-left: auto;\n  }\n}\n","/**\n * The primary sidebar on the left.\n * e.g., between-pages navigation.\n */\n\n.bd-sidebar-primary {\n  display: flex;\n  flex-direction: column;\n  gap: 1rem;\n  max-height: calc(100vh - var(--pst-header-height));\n  position: sticky;\n  top: var(--pst-header-height);\n  @include make-col(3);\n\n  // Borders padding and whitespace\n  padding: 2rem 1rem 1rem 1rem;\n  border-right: 1px solid var(--pst-color-border);\n  background-color: var(--pst-color-background);\n  overflow-y: auto;\n\n  font-size: var(--pst-sidebar-font-size-mobile);\n  @include media-breakpoint-up($breakpoint-sidebar-primary) {\n    font-size: var(--pst-sidebar-font-size);\n  }\n\n  // override bootstrap when navlink are displayed in the sidebar\n  .nav-link {\n    font-size: var(--pst-sidebar-font-size-mobile);\n  }\n\n  &.no-sidebar {\n    border-right: 0;\n  }\n\n  &.hide-on-wide {\n    @include media-breakpoint-up($breakpoint-sidebar-primary) {\n      display: none;\n    }\n  }\n\n  // Headers shouldn't be colored in the sidebars and some extensions add headers\n  h1,\n  h2,\n  h3,\n  h4 {\n    color: var(--pst-color-text-base);\n  }\n\n  .sidebar-primary-items__start,\n  .sidebar-primary-items__end {\n    .sidebar-primary-item {\n      padding: 0.5rem 0;\n    }\n  }\n\n  // Hide the sidebar header items on widescreen since they are visible in the header\n  .sidebar-header-items {\n    display: flex;\n    flex-direction: column;\n\n    .sidebar-header-items__title {\n      font-weight: var(--pst-sidebar-header-font-weight);\n      font-size: var(--pst-sidebar-header-font-size);\n      color: var(--pst-color-text-base);\n      margin-bottom: 0.5rem;\n    }\n\n    // The dropdown toggle for extra links just shows them all instead.\n    .nav-item.dropdown {\n      // On mobile, the dropdown behaves like any other link, no hiding\n      button {\n        display: none;\n      }\n\n      .dropdown-menu {\n        display: flex;\n        flex-direction: column;\n        padding: 0;\n        margin: 0;\n        border: none;\n        background-color: inherit;\n        font-size: inherit;\n      }\n    }\n\n    // Center header items get displayed vertically, end items are displayed horizontally\n    .sidebar-header-items__center {\n      display: flex;\n      flex-direction: column;\n    }\n\n    // Positioning end items\n    .sidebar-header-items__end {\n      display: flex;\n      align-items: center;\n      gap: 0.5rem;\n    }\n\n    @include media-breakpoint-up($breakpoint-sidebar-primary) {\n      display: none;\n    }\n  }\n\n  .sidebar-primary-items__start {\n    // Add a border on mobile to separate it from the header sidebar area\n    border-top: 1px solid var(--pst-color-border);\n    @include media-breakpoint-up($breakpoint-sidebar-primary) {\n      border-top: none;\n    }\n  }\n\n  .sidebar-primary-items__end {\n    margin-top: auto;\n    margin-bottom: 1em;\n  }\n  .list-caption {\n    list-style: none;\n    padding-left: 0px;\n  }\n  li {\n    position: relative;\n    // If it has children, add a bit more padding to wrap the content to avoid\n    // overlapping with the <label>\n    &.has-children {\n      > .reference {\n        padding-right: 30px;\n      }\n    }\n  }\n  // Navigation item chevrons\n  label.toctree-toggle {\n    position: absolute;\n    top: 0;\n    right: 0;\n    height: 30px;\n    width: 30px;\n\n    cursor: pointer;\n\n    display: flex;\n    justify-content: center;\n    align-items: center;\n\n    &:hover {\n      background: var(--pst-color-surface);\n    }\n\n    i {\n      display: inline-block;\n      font-size: 0.75rem;\n      text-align: center;\n      &:hover {\n        color: var(--pst-color-primary);\n      }\n    }\n  }\n  .label-parts {\n    width: 100%;\n    height: 100%;\n    &:hover {\n      background: none;\n    }\n    i {\n      width: 30px;\n      position: absolute;\n      top: 0.3em; // aligning chevron with text\n      right: 0em; // aligning chevron to the right\n    }\n  }\n}\n\n/* Between-page links and captions */\nnav.bd-links {\n  margin-right: -1rem;\n\n  @include media-breakpoint-up($breakpoint-sidebar-primary) {\n    display: block;\n  }\n\n  ul {\n    list-style: none;\n\n    // Reduce padding of nested `ul` items a bit\n    ul {\n      padding: 0 0 0 1rem;\n    }\n  }\n\n  li > a {\n    display: block;\n    padding: 0.25rem 0.65rem;\n    @include link-sidebar;\n    box-shadow: none;\n\n    &.reference.external {\n      &:after {\n        font: var(--fa-font-solid);\n        content: var(--pst-icon-external-link);\n        font-size: 0.75em;\n        margin-left: 0.3em;\n      }\n    }\n  }\n\n  .current {\n    > a {\n      @include link-sidebar-current;\n      background-color: transparent;\n    }\n  }\n\n  // Title\n  p.bd-links__title {\n    font-size: var(--pst-sidebar-header-font-size);\n    font-weight: var(--pst-sidebar-header-font-weight);\n    margin-bottom: 0.5rem;\n  }\n\n  // Toctree captions\n  p.caption {\n    font-weight: var(--pst-sidebar-header-font-weight);\n    position: relative;\n    margin-top: 1.25rem;\n    margin-bottom: 0.5rem;\n    color: var(--pst-color-text-base);\n    &:first-child {\n      margin-top: 0;\n    }\n\n    font-size: var(--pst-sidebar-font-size-mobile);\n    @include media-breakpoint-up($breakpoint-sidebar-primary) {\n      font-size: var(--pst-sidebar-font-size);\n    }\n  }\n}\n","// Grid system\n//\n// Generate semantic grid columns with these mixins.\n\n@mixin make-row($gutter: $grid-gutter-width) {\n  --#{$prefix}gutter-x: #{$gutter};\n  --#{$prefix}gutter-y: 0;\n  display: flex;\n  flex-wrap: wrap;\n  // TODO: Revisit calc order after https://github.com/react-bootstrap/react-bootstrap/issues/6039 is fixed\n  margin-top: calc(-1 * var(--#{$prefix}gutter-y)); // stylelint-disable-line function-disallowed-list\n  margin-right: calc(-.5 * var(--#{$prefix}gutter-x)); // stylelint-disable-line function-disallowed-list\n  margin-left: calc(-.5 * var(--#{$prefix}gutter-x)); // stylelint-disable-line function-disallowed-list\n}\n\n@mixin make-col-ready() {\n  // Add box sizing if only the grid is loaded\n  box-sizing: if(variable-exists(include-column-box-sizing) and $include-column-box-sizing, border-box, null);\n  // Prevent columns from becoming too narrow when at smaller grid tiers by\n  // always setting `width: 100%;`. This works because we set the width\n  // later on to override this initial width.\n  flex-shrink: 0;\n  width: 100%;\n  max-width: 100%; // Prevent `.col-auto`, `.col` (& responsive variants) from breaking out the grid\n  padding-right: calc(var(--#{$prefix}gutter-x) * .5); // stylelint-disable-line function-disallowed-list\n  padding-left: calc(var(--#{$prefix}gutter-x) * .5); // stylelint-disable-line function-disallowed-list\n  margin-top: var(--#{$prefix}gutter-y);\n}\n\n@mixin make-col($size: false, $columns: $grid-columns) {\n  @if $size {\n    flex: 0 0 auto;\n    width: percentage(divide($size, $columns));\n\n  } @else {\n    flex: 1 1 0;\n    max-width: 100%;\n  }\n}\n\n@mixin make-col-auto() {\n  flex: 0 0 auto;\n  width: auto;\n}\n\n@mixin make-col-offset($size, $columns: $grid-columns) {\n  $num: divide($size, $columns);\n  margin-left: if($num == 0, 0, percentage($num));\n}\n\n// Row columns\n//\n// Specify on a parent element(e.g., .row) to force immediate children into NN\n// number of columns. Supports wrapping to new lines, but does not do a Masonry\n// style grid.\n@mixin row-cols($count) {\n  > * {\n    flex: 0 0 auto;\n    width: percentage(divide(1, $count));\n  }\n}\n\n// Framework grid generation\n//\n// Used only by Bootstrap to generate the correct number of grid classes given\n// any value of `$grid-columns`.\n\n@mixin make-grid-columns($columns: $grid-columns, $gutter: $grid-gutter-width, $breakpoints: $grid-breakpoints) {\n  @each $breakpoint in map-keys($breakpoints) {\n    $infix: breakpoint-infix($breakpoint, $breakpoints);\n\n    @include media-breakpoint-up($breakpoint, $breakpoints) {\n      // Provide basic `.col-{bp}` classes for equal-width flexbox columns\n      .col#{$infix} {\n        flex: 1 0 0%; // Flexbugs #4: https://github.com/philipwalton/flexbugs#flexbug-4\n      }\n\n      .row-cols#{$infix}-auto > * {\n        @include make-col-auto();\n      }\n\n      @if $grid-row-columns > 0 {\n        @for $i from 1 through $grid-row-columns {\n          .row-cols#{$infix}-#{$i} {\n            @include row-cols($i);\n          }\n        }\n      }\n\n      .col#{$infix}-auto {\n        @include make-col-auto();\n      }\n\n      @if $columns > 0 {\n        @for $i from 1 through $columns {\n          .col#{$infix}-#{$i} {\n            @include make-col($i, $columns);\n          }\n        }\n\n        // `$columns - 1` because offsetting by the width of an entire row isn't possible\n        @for $i from 0 through ($columns - 1) {\n          @if not ($infix == \"\" and $i == 0) { // Avoid emitting useless .offset-0\n            .offset#{$infix}-#{$i} {\n              @include make-col-offset($i, $columns);\n            }\n          }\n        }\n      }\n\n      // Gutters\n      //\n      // Make use of `.g-*`, `.gx-*` or `.gy-*` utilities to change spacing between the columns.\n      @each $key, $value in $gutters {\n        .g#{$infix}-#{$key},\n        .gx#{$infix}-#{$key} {\n          --#{$prefix}gutter-x: #{$value};\n        }\n\n        .g#{$infix}-#{$key},\n        .gy#{$infix}-#{$key} {\n          --#{$prefix}gutter-y: #{$value};\n        }\n      }\n    }\n  }\n}\n\n@mixin make-cssgrid($columns: $grid-columns, $breakpoints: $grid-breakpoints) {\n  @each $breakpoint in map-keys($breakpoints) {\n    $infix: breakpoint-infix($breakpoint, $breakpoints);\n\n    @include media-breakpoint-up($breakpoint, $breakpoints) {\n      @if $columns > 0 {\n        @for $i from 1 through $columns {\n          .g-col#{$infix}-#{$i} {\n            grid-column: auto / span $i;\n          }\n        }\n\n        // Start with `1` because `0` is and invalid value.\n        // Ends with `$columns - 1` because offsetting by the width of an entire row isn't possible.\n        @for $i from 1 through ($columns - 1) {\n          .g-start#{$infix}-#{$i} {\n            grid-column-start: $i;\n          }\n        }\n      }\n    }\n  }\n}\n","/**\n * Secondary sidebar on the right.\n * e.g., in-page table of contents.\n */\n.bd-sidebar-secondary {\n  display: flex;\n  order: 2;\n  flex-shrink: 0;\n  flex-direction: column;\n  position: sticky;\n  top: var(--pst-header-height);\n  max-height: calc(100vh - var(--pst-header-height));\n  padding: 2rem 1rem 1rem 1rem;\n  width: var(--pst-sidebar-secondary);\n  font-size: var(--pst-sidebar-font-size-mobile);\n  @include media-breakpoint-up($breakpoint-sidebar-secondary) {\n    font-size: var(--pst-sidebar-font-size);\n  }\n\n  // Color and border\n  background-color: var(--pst-color-background);\n  overflow-y: auto;\n}\n\n.sidebar-secondary-item {\n  padding: 0.5rem 0.5rem;\n  @include media-breakpoint-up($breakpoint-sidebar-secondary) {\n    border-left: 1px solid var(--pst-color-border);\n    padding-left: 1rem;\n  }\n\n  i {\n    padding-right: 0.5rem;\n  }\n}\n","/*******************************************************************************\n* Rules for the UX/UI of sidebar sliding drawers on mobile\n* Note that this sheet controls styles across many parts of the theme\n* It is aggregated into this one sheet instead of being split across\n* components in order to keep it easier to debug in one place.\n* It is broken up into major sections below.\n*/\n\n/*******************************************************************************\n* Buttons and overlays\n*/\ninput.sidebar-toggle {\n  display: none;\n}\n\n// Background overlays\nlabel.overlay {\n  background-color: black;\n  opacity: 0.5;\n  height: 0;\n  width: 0;\n  position: fixed;\n  top: 0;\n  left: 0;\n  transition: opacity $animation-time ease-out;\n  z-index: $zindex-modal-backdrop;\n}\n\ninput {\n  // Show the correct overlay when its input is checked\n  &#__primary:checked + label.overlay.overlay-primary,\n  &#__secondary:checked + label.overlay.overlay-secondary {\n    height: 100vh;\n    width: 100vw;\n  }\n\n  // Primary sidebar slides in from the left\n  &#__primary:checked ~ .bd-container .bd-sidebar-primary {\n    visibility: visible;\n    margin-left: 0;\n  }\n\n  // Secondary sidebar slides in from the right\n  &#__secondary:checked ~ .bd-container .bd-sidebar-secondary {\n    visibility: visible;\n    margin-right: 0;\n  }\n}\n\n/*******************************************************************************\n* Sidebar drawer behavior\n*/\n\n/**\n * Behavior for sliding drawer elements that will be toggled with an input\n *\n * NOTE: We use this mixin to define the toggle behavior on narrow screens,\n * And the wide-screen behavior of the sections is defined in their own section\n * .scss files.\n */\n@mixin sliding-drawer($side: \"left\") {\n  position: fixed;\n  top: 0;\n  z-index: $zindex-modal;\n  height: 100vh;\n  max-height: 100vh;\n  width: 75%;\n  flex-grow: 0.75;\n  max-width: 350px;\n  transition: visibility $animation-time ease-out,\n    margin $animation-time ease-out;\n  visibility: hidden;\n\n  @if $side == \"right\" {\n    margin-right: -75%;\n    right: 0;\n  } @else {\n    margin-left: -75%;\n    left: 0;\n  }\n}\n\n// Primary sidebar hides/shows at earlier widths\n@include media-breakpoint-up($breakpoint-sidebar-primary) {\n  label.sidebar-toggle.primary-toggle {\n    display: none;\n  }\n\n  input#__primary {\n    &:checked + label.overlay.overlay-primary {\n      height: 0;\n      width: 0;\n    }\n  }\n\n  .bd-sidebar-primary {\n    margin-left: 0;\n    visibility: visible;\n  }\n}\n\n.bd-sidebar-primary {\n  @include media-breakpoint-down($breakpoint-sidebar-primary) {\n    @include sliding-drawer(\"left\");\n  }\n}\n\n.bd-sidebar-secondary {\n  @include media-breakpoint-down($breakpoint-sidebar-secondary) {\n    @include sliding-drawer(\"right\");\n  }\n}\n","/**\n * Breadcrumbs for parent pages meant for the article header\n */\nul.bd-breadcrumbs {\n  list-style: none;\n  padding-left: 0;\n  display: flex;\n  flex-wrap: wrap;\n\n  // Font size slightly smaller to avoid cluttering up space too much\n  font-size: 0.8rem;\n\n  li.breadcrumb-item {\n    display: flex;\n    align-items: baseline;\n\n    // Style should look like heavier in-page links\n    // keeping visited in the default link colour\n    font-weight: bold;\n    a {\n      @include link-style-text;\n    }\n\n    // Items that aren't the home have a caret to the left\n    &:not(.breadcrumb-home):before {\n      font: var(--fa-font-solid);\n      font-size: 0.8rem;\n      content: var(--pst-breadcrumb-divider);\n      color: var(--pst-color-text-muted);\n      padding: 0 0.5rem;\n    }\n  }\n}\n","/**\n * Icon links in the navbar\n */\n\n.navbar-icon-links {\n  display: flex;\n  flex-direction: row;\n  column-gap: 1rem;\n  flex-wrap: wrap;\n\n  // Remove the padding so that we can define it with flexbox gap above\n  li.nav-item a.nav-link {\n    padding-left: 0;\n    padding-right: 0;\n    @include icon-navbar-hover;\n  }\n\n  // Spacing and centering\n  a span {\n    display: flex;\n    align-items: center;\n  }\n\n  // Icons styling\n  i {\n    &.fa-brands,\n    &.fa-regular,\n    &.fa-solid {\n      vertical-align: middle;\n      font-style: normal;\n      font-size: var(--pst-font-size-icon);\n    }\n\n    /* Social media buttons hard-code the brand color */\n    &.fa-square-twitter:before {\n      color: #55acee;\n    }\n\n    &.fa-square-gitlab:before {\n      color: #548;\n    }\n\n    &.fa-bitbucket:before {\n      color: #0052cc;\n    }\n  }\n\n  // Force images to be icon-sized\n  img.icon-link-image {\n    height: 1.5em;\n    border-radius: 0.2rem;\n  }\n}\n","/**\n * Logo in the navbar\n */\n\n.navbar-brand {\n  position: relative;\n  height: var(--pst-header-height);\n  max-height: var(--pst-header-height);\n  padding: 0.5rem 0;\n  width: auto;\n  margin: 0;\n  display: flex;\n  // Ensure that the logo stays the same length while other content shrinks\n  flex-shrink: 0;\n  align-items: center;\n  gap: 0.5rem;\n\n  // If there's no logo image, we use a p element w/ the site title\n  p {\n    margin-bottom: 0;\n  }\n\n  // If there's a logo, it'll be in an img block\n  img {\n    max-width: 100%;\n    height: 100%;\n    width: auto;\n  }\n\n  // remove underline from brand title on default state\n  a {\n    text-decoration: none;\n  }\n  &:hover {\n    @include link-style-hover;\n  }\n  &:visited {\n    &:hover {\n      @include link-style-hover;\n    }\n  }\n}\n","/**\n * Navigation text links in the navbar\n */\n.navbar-nav {\n  ul {\n    display: block;\n    list-style: none;\n\n    // Reduce padding of nested `ul` items a bit\n    ul {\n      padding: 0 0 0 1rem;\n    }\n  }\n\n  // Navbar links - do not have an underline by default\n  li {\n    display: flex;\n    flex-direction: column;\n\n    a {\n      display: flex;\n      align-items: center;\n      height: 100%;\n      padding-top: 0.25rem;\n      padding-bottom: 0.25rem;\n      @include link-style-text;\n    }\n  }\n\n  /**\n   * Togglable expand/collapse\n   * This is only applicable to the primary sidebar which has these checkboxes\n   */\n  .toctree-checkbox {\n    position: absolute;\n    display: none;\n  }\n\n  .toctree-checkbox {\n    ~ ul {\n      display: none;\n    }\n    ~ label i {\n      transform: rotate(0deg);\n    }\n  }\n  .toctree-checkbox:checked {\n    ~ ul {\n      display: block;\n    }\n    ~ label i {\n      transform: rotate(180deg);\n    }\n  }\n}\n\n// Don't display the `site navigation` in the header menu\n.bd-header .navbar-nav > p.sidebar-header-items__title {\n  display: none;\n}\n","/**\n * The list of in-page TOC links\n */\n.page-toc {\n  .section-nav {\n    padding-left: 0;\n    border-bottom: none;\n\n    ul {\n      padding-left: 1rem;\n    }\n  }\n\n  // override bootstrap settings\n  .nav-link {\n    font-size: var(--pst-sidebar-font-size-mobile);\n    @include media-breakpoint-up($breakpoint-sidebar-secondary) {\n      font-size: var(--pst-sidebar-font-size);\n    }\n  }\n\n  .onthispage {\n    color: var(--pst-color-text-base);\n    font-weight: var(--pst-sidebar-header-font-weight);\n    margin-bottom: 0.5rem;\n  }\n}\n","/**\n* Previous / Next navigation buttons\n**/\n.prev-next-area {\n  width: 100%;\n\n  p {\n    margin: 0 0.3em;\n    line-height: 1.3em;\n  }\n\n  i {\n    font-size: 1.2em;\n  }\n\n  a {\n    // So that buttons align with icons\n    display: flex;\n    align-items: center;\n    border: none;\n    padding: 10px;\n    max-width: 45%;\n    overflow-x: hidden;\n    color: var(--pst-color-text-muted);\n    text-decoration: none;\n\n    p.prev-next-title {\n      @include link-style-default;\n      font-weight: var(--pst-admonition-font-weight-heading);\n      font-size: 1.1em;\n    }\n\n    &:hover p.prev-next-title {\n      @include link-style-hover;\n    }\n    // Exception here - keep visited in the default link colour\n    // it still should hover on the default link hover colour\n    &:visited p.prev-next-title {\n      color: var(--pst-color-link);\n      &:hover {\n        color: var(--pst-color-link-hover);\n      }\n    }\n\n    .prev-next-info {\n      flex-direction: column;\n      margin: 0 0.5em;\n\n      .prev-next-subtitle {\n        text-transform: capitalize;\n      }\n    }\n\n    &.left-prev {\n      float: left;\n    }\n\n    &.right-next {\n      float: right;\n\n      div.prev-next-info {\n        text-align: right;\n      }\n    }\n  }\n}\n","/**\n * Search field\n **/\n.bd-search {\n  position: relative;\n  padding-left: 0.5rem;\n  gap: 0.5rem;\n  background-color: var(--pst-color-background);\n  border-radius: $admonition-border-radius;\n  border: 1px solid var(--pst-color-border);\n  color: var(--pst-color-text-base);\n\n  // Background should always be same color regardless of active or not\n  &:active {\n    background-color: var(--pst-color-background);\n    color: var(--pst-color-text-muted);\n  }\n\n  .icon {\n    position: absolute;\n    color: var(--pst-color-border);\n    left: 25px;\n  }\n\n  .fa-solid.fa-magnifying-glass {\n    position: absolute;\n    left: calc((2.5rem - 0.7em) / 2);\n    color: var(--pst-color-text-muted);\n  }\n\n  input {\n    // Inner-text of the search bar\n    &::placeholder {\n      color: var(--pst-color-text-muted);\n    }\n\n    // Remove the little \"x\" that pops up when you start typing\n    &::-webkit-search-cancel-button,\n    &::-webkit-search-decoration {\n      -webkit-appearance: none;\n      appearance: none;\n    }\n  }\n\n  // Shows off the keyboard shortcuts for the button\n  .search-button__kbd-shortcut {\n    display: flex;\n    position: absolute;\n    right: 0.5rem;\n    color: var(--pst-color-border);\n  }\n}\n\n.form-control {\n  background-color: var(--pst-color-background);\n  color: var(--pst-color-text-base);\n\n  &:focus,\n  &:focus-visible {\n    border: none;\n    box-shadow: none;\n    outline: 3px solid var(--pst-color-accent);\n    background-color: var(--pst-color-background);\n    color: var(--pst-color-text-muted);\n  }\n}\n\n/**\n * Search button - located in the navbar\n */\n\n// Search link icon should be a bit bigger since it is separate from icon links\n.search-button {\n  display: flex;\n  align-items: center;\n  align-content: center;\n  color: var(--pst-color-text-muted);\n  // Needed to match other icons hover\n  padding: 0 0 0.25rem 0;\n  border-radius: 0;\n  @include icon-navbar-hover;\n  @include focus-indicator;\n\n  i {\n    font-size: 1.3rem;\n  }\n}\n\n// __search-container will only show up when we use the search pop-up bar\n.search-button__search-container,\n.search-button__overlay {\n  display: none;\n}\n.search-button__wrapper.show {\n  .search-button__search-container {\n    display: flex;\n    // Center in middle of screen just underneath header\n    position: fixed;\n    z-index: $zindex-modal;\n    top: 30%;\n    left: 50%;\n    transform: translate(-50%, -50%);\n    right: 1rem;\n    margin-top: 0.5rem;\n    width: 90%;\n    max-width: 800px;\n  }\n\n  .search-button__overlay {\n    display: flex;\n    position: fixed;\n    z-index: $zindex-modal-backdrop;\n    background-color: black;\n    opacity: 0.5;\n    width: 100%;\n    height: 100%;\n    top: 0px;\n    left: 0px;\n  }\n\n  form.bd-search {\n    flex-grow: 1;\n    padding-top: 0;\n    padding-bottom: 0;\n  }\n\n  // Font and input text a bit bigger\n  svg,\n  input {\n    font-size: var(--pst-font-size-icon);\n  }\n}\n\n/**\n * The search button component that looks like a field.\n * Lives at components/search-button-field.html\n */\n.search-button-field {\n  display: inline-flex;\n  align-items: center;\n  border: var(--pst-color-border) solid 1px;\n  border-radius: 1.5em;\n  color: var(--pst-color-text-muted);\n  padding: 0.5em;\n  background-color: var(--pst-color-surface);\n\n  &:hover {\n    border: 2px solid var(--pst-color-link-hover);\n  }\n  &:focus-visible {\n    border: 2px solid var(--pst-color-accent);\n  }\n\n  // The keyboard shotcut text\n  .search-button__default-text {\n    font-size: var(--bs-nav-link-font-size);\n    font-weight: var(--bs-nav-link-font-weight);\n    margin-right: 0.5em;\n    margin-left: 0.5em;\n  }\n\n  .kbd-shortcut__modifier {\n    font-size: 0.75em;\n  }\n\n  // Ensures that all the text lines up in the middle\n  > * {\n    align-items: center;\n  }\n\n  // Only the icon should be visible on narrow screens\n  > :not(svg) {\n    display: none;\n\n    @include media-breakpoint-up(lg) {\n      display: flex;\n    }\n  }\n}\n","/**\n * The 'Hide Search Matches' button.\n * This only shows up when a person lands on a page after clicking a search result.\n * Clicking it removes the highlighting of the search term from the page.\n * We want it to behave like a button.\n */\ndiv#searchbox {\n  // Leave `#searchbox` rules empty so that it doesn't show at all when it is empty\n  p.highlight-link {\n    margin: 1rem 0;\n    width: fit-content;\n\n    // A bit more margin on wide screens to mimic article behavior\n    @include media-breakpoint-up($breakpoint-sidebar-secondary) {\n      margin-left: 2rem;\n    }\n\n    // Put outer shadow on this one so that we can darken the link w/ an inner shadow\n    @include box-shadow();\n\n    // Style the button to look like a Sphinx Design button\n    a {\n      border-radius: 0.25rem;\n      font-size: 1.25rem;\n      padding: 0.75rem;\n      background-color: var(--pst-color-primary);\n      color: var(--pst-color-primary-text);\n      text-decoration: none;\n\n      // The box shadow is inset so that it darkens the button on hover\n      transition: box-shadow 0.25s ease-out;\n      &:hover {\n        box-shadow: inset 0px 0px 50px 50px rgba(0, 0, 0, 0.25);\n      }\n\n      &:before {\n        content: var(--pst-icon-search-minus);\n        color: unset;\n        font: var(--fa-font-solid);\n        margin-right: 0.5rem;\n      }\n    }\n  }\n}\n","/*********************************************\n* SASS Mixins\n*********************************************/\n\n/**\n * A consistent box shadow style we apply across elements.\n */\n@mixin box-shadow() {\n  box-shadow: 0 0.2rem 0.5rem var(--pst-color-shadow),\n    0 0 0.0625rem var(--pst-color-shadow) !important;\n}\n\n/**\n   * Set background of some cell outputs to white-ish to make sure colors work\n   * This is because many libraries make output that only looks good on white\n   */\n@mixin cell-output-background {\n  color: var(--pst-color-on-background);\n  background-color: var(--pst-color-text-base);\n  border-radius: 0.25rem;\n  padding: 0.5rem;\n}\n","/**\n * Light/dark theme switcher\n */\n\n.theme-switch-button {\n  // overide bootstrap settings\n  margin: 0 -0.5rem;\n  padding: 0; // We pad the `span` not the container\n  color: var(--pst-color-text-muted);\n  border-radius: 0;\n  @include focus-indicator;\n\n  span {\n    display: none;\n    padding: 0.5em;\n\n    @include icon-navbar-hover;\n    &:active {\n      text-decoration: none;\n      color: var(--pst-color-link-hover);\n    }\n  }\n}\n\nhtml[data-mode=\"auto\"] .theme-switch-button span[data-mode=\"auto\"] {\n  display: flex;\n}\nhtml[data-mode=\"light\"] .theme-switch-button span[data-mode=\"light\"] {\n  display: flex;\n}\nhtml[data-mode=\"dark\"] .theme-switch-button span[data-mode=\"dark\"] {\n  display: flex;\n}\n","button.btn.version-switcher__button {\n  border-color: var(--pst-color-border);\n  color: var(--pst-color-text-base);\n  // Add a margin on narrow screens to avoid feeling cramped\n  margin-bottom: 1em;\n  @include media-breakpoint-up($breakpoint-sidebar-primary) {\n    margin-bottom: unset;\n  }\n\n  @include link-style-hover;\n  @include focus-indicator;\n  &:active {\n    color: var(--pst-color-text-base);\n    border-color: var(--pst-color-border);\n  }\n}\n\n.version-switcher__menu {\n  border-color: var(--pst-color-border);\n  border-radius: var(--bs-dropdown-border-radius);\n\n  a.list-group-item {\n    background-color: var(--pst-color-on-background);\n    color: var(--pst-color-text-base);\n    padding: 0.75rem 1.25rem;\n\n    &:not(:last-child) {\n      border-bottom: 1px solid var(--pst-color-border);\n    }\n    @include link-style-hover;\n    &:hover {\n      background-color: var(--pst-color-surface);\n    }\n    &.active {\n      @include link-sidebar-current;\n      position: relative;\n      z-index: 1;\n\n      span:before {\n        content: \"\";\n        width: 100%;\n        height: 100%;\n        position: absolute;\n        z-index: -1;\n        left: 0;\n        top: 0;\n      }\n    }\n  }\n}\n\n// Font behavior on mobile\nbutton.version-switcher__button,\n.version-switcher__menu {\n  font-size: 1.1em; // A bit smaller than other menu font\n  z-index: $zindex-modal; // higher than the sidebars\n  @include media-breakpoint-up($breakpoint-sidebar-primary) {\n    font-size: unset;\n  }\n}\n","/* Collapsing of the TOC sidebar while scrolling */\n/* Nav: hide second level (shown on .active) */\n\nnav.page-toc {\n  // A little extra space before the buttons\n  margin-bottom: 1rem;\n}\n\n.bd-toc .nav,\n.list-caption {\n  .nav {\n    display: none;\n\n    // So we can manually specify a level as visible in the config\n    &.visible {\n      display: block;\n    }\n  }\n\n  > .active > ul {\n    display: block;\n  }\n}\n\n// Each entry of the in-page TOC\n.toc-entry {\n  display: block;\n  a > code {\n    color: var(--pst-color-text-muted);\n  }\n\n  a.nav-link {\n    display: block;\n    padding: 0.125rem 0;\n\n    // Padding w/ negative margin so the top TOC item highlight overlaps w/ the TOC border\n    padding-left: 1rem;\n    margin-left: -1rem;\n\n    @include link-sidebar;\n\n    &.active {\n      @include link-sidebar-current;\n      background-color: transparent;\n\n      &:hover {\n        color: var(--pst-color-link-hover);\n      }\n    }\n  }\n}\n","div.versionadded,\ndiv.versionchanged,\ndiv.deprecated {\n  vertical-align: middle;\n  margin: 1.5625em auto;\n  padding: 0 0.6rem 0 0.6rem;\n  overflow: hidden;\n  page-break-inside: avoid;\n  border-left: 0.2rem solid;\n  border-color: var(--pst-color-info);\n  border-radius: $admonition-border-radius;\n  background-color: var(--pst-color-on-background);\n  @include box-shadow();\n  position: relative;\n\n  > p {\n    margin-bottom: 0.6rem;\n    margin-top: 0.6rem;\n  }\n}\n\ndiv.versionadded {\n  border-color: var(--pst-color-success);\n  background-color: var(--pst-color-success-bg);\n}\n\ndiv.versionchanged {\n  border-color: var(--pst-color-warning);\n  background-color: var(--pst-color-warning-bg);\n}\n\ndiv.deprecated {\n  border-color: var(--pst-color-danger);\n  background-color: var(--pst-color-danger-bg);\n}\n\nspan.versionmodified {\n  font-weight: 600;\n\n  &:before {\n    margin-right: 0.6rem;\n    color: var(--pst-color-info);\n    font: var(--fa-font-solid);\n    content: var(--pst-icon-versionmodified-default);\n  }\n}\n\nspan.versionmodified.added {\n  &:before {\n    color: var(--pst-color-success);\n    content: var(--pst-icon-versionmodified-added);\n  }\n}\n\nspan.versionmodified.changed {\n  &:before {\n    color: var(--pst-color-warning);\n    content: var(--pst-icon-versionmodified-changed);\n  }\n}\n\nspan.versionmodified.deprecated {\n  &:before {\n    color: var(--pst-color-danger);\n    content: var(--pst-icon-versionmodified-deprecated);\n  }\n}\n",".sidebar-indices-items {\n  display: flex;\n  flex-direction: column;\n  border-top: 1px solid var(--pst-color-border);\n\n  @include media-breakpoint-up($breakpoint-sidebar-primary) {\n    border-top: none;\n  }\n\n  .sidebar-indices-items__title {\n    font-weight: var(--pst-sidebar-header-font-weight);\n    font-size: var(--pst-sidebar-header-font-size);\n    color: var(--pst-color-text-base);\n    margin-bottom: 0.5rem;\n  }\n\n  ul.indices-link {\n    margin-right: -1rem;\n    list-style: none;\n    padding: 0;\n\n    li > a {\n      display: block;\n      padding: 0.25rem 0;\n      color: var(--pst-color-text-muted);\n\n      &:hover {\n        color: var(--pst-color-primary);\n        text-decoration: none;\n        background-color: transparent;\n      }\n    }\n  }\n}\n",".bd-sidebar-primary div#rtd-footer-container {\n  position: sticky;\n  bottom: -1rem;\n  margin: -1rem; // ignore sidebar padding\n\n  .rst-versions.rst-badge {\n    position: unset;\n    font-size: 0.9em;\n    font-family: var(--pst-font-family-base);\n    max-width: unset;\n\n    .rst-current-version {\n      display: flex;\n      align-items: center;\n      gap: 0.2rem;\n      height: 2.5rem;\n      transition: background-color 0.2s ease-out;\n      background-color: var(--pst-color-background);\n      color: var(--pst-color-success);\n      border-top: 1px solid var(--pst-color-border);\n    }\n\n    .fa.fa-book {\n      color: var(--pst-color-text-muted);\n      margin-right: auto;\n\n      &::after {\n        color: var(--pst-color-text-base);\n        content: \"Read The Docs\";\n        font-family: var(--pst-font-family-base);\n        font-weight: var(--pst-admonition-font-weight-heading);\n      }\n    }\n    .fa.fa-caret-down {\n      color: var(--pst-color-text-muted);\n    }\n  }\n\n  .rst-versions.rst-badge.shift-up {\n    .rst-current-version {\n      border-bottom: 1px solid var(--pst-color-border);\n    }\n  }\n\n  .rst-other-versions {\n    background-color: var(--pst-color-surface);\n    color: var(--pst-color-text-base);\n\n    dl {\n      dd a {\n        color: var(--pst-color-text-muted);\n      }\n    }\n\n    hr {\n      background-color: var(--pst-color-border);\n    }\n\n    small a {\n      color: var(--pst-color-link);\n    }\n\n    input {\n      padding-left: 0.5rem;\n      border: 1px solid var(--pst-color-border);\n      background-color: var(--pst-color-surface);\n    }\n  }\n}\n","/**\n * Admonitions and blocks of styled content.\n * Admonitions CSS originally inspired by https://squidfunk.github.io/mkdocs-material/getting-started/\n */\n$admonition-border-radius: 0.25rem;\ndiv.admonition,\n.admonition {\n  margin: 1.5625em auto;\n  padding: 0 0.6rem 0.8rem 0.6rem;\n  overflow: hidden;\n  page-break-inside: avoid;\n  border-left: 0.2rem solid;\n  border-color: var(--pst-color-info);\n  border-radius: $admonition-border-radius;\n  background-color: var(--pst-color-on-background);\n  @include box-shadow();\n\n  // Last item should have no spacing since we'll control that w/ padding\n  *:last-child {\n    margin-bottom: 0;\n  }\n\n  // Items after the title should be indented\n  p.admonition-title ~ * {\n    margin-left: 1.4rem;\n    margin-right: 1.4rem;\n  }\n\n  // Lists need to have left margin so they don't spill into it\n  > ol,\n  > ul {\n    margin-left: 1em;\n  }\n\n  // Defaults for all admonitions\n  > .admonition-title {\n    margin: 0 -0.6rem;\n    padding: 0.4rem 0.6rem 0.4rem 2rem;\n    font-weight: var(--pst-admonition-font-weight-heading);\n    position: relative;\n    @include background-from-color-variable(--pst-color-info-bg);\n    // now that we use solid colors we want the title on top\n    z-index: 1;\n\n    &:after {\n      position: absolute;\n      left: 0.5rem;\n      width: 1rem;\n      height: 1rem;\n      color: var(--pst-color-info);\n      font: var(--fa-font-solid);\n      line-height: inherit;\n      content: var(--pst-icon-admonition-default);\n      opacity: 1;\n    }\n\n    // Next element after title needs some extra upper-space\n    + * {\n      margin-top: 0.4em;\n    }\n  }\n\n  &.attention {\n    border-color: var(--pst-color-attention);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-attention-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-attention);\n        content: var(--pst-icon-admonition-attention);\n      }\n    }\n  }\n\n  &.caution {\n    border-color: var(--pst-color-warning);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-warning-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-warning);\n        content: var(--pst-icon-admonition-caution);\n      }\n    }\n  }\n\n  &.warning {\n    border-color: var(--pst-color-warning);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-warning-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-warning);\n        content: var(--pst-icon-admonition-warning);\n      }\n    }\n  }\n\n  &.danger {\n    border-color: var(--pst-color-danger);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-danger-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-danger);\n        content: var(--pst-icon-admonition-danger);\n      }\n    }\n  }\n\n  &.error {\n    border-color: var(--pst-color-danger);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-danger-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-danger);\n        content: var(--pst-icon-admonition-error);\n      }\n    }\n  }\n\n  &.hint {\n    border-color: var(--pst-color-success);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-success-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-success);\n        content: var(--pst-icon-admonition-hint);\n      }\n    }\n  }\n\n  &.tip {\n    border-color: var(--pst-color-success);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-success-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-success);\n        content: var(--pst-icon-admonition-tip);\n      }\n    }\n  }\n\n  &.important {\n    border-color: var(--pst-color-attention);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-attention-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-attention);\n        content: var(--pst-icon-admonition-important);\n      }\n    }\n  }\n\n  &.note {\n    border-color: var(--pst-color-info);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-info-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-info);\n        content: var(--pst-icon-admonition-note);\n      }\n    }\n  }\n\n  &.seealso {\n    border-color: var(--pst-color-success);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-success-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-success);\n        content: var(--pst-icon-admonition-seealso);\n      }\n    }\n  }\n\n  &.admonition-todo {\n    border-color: var(--pst-color-secondary);\n    > .admonition-title {\n      &:before {\n        background-color: var(--pst-color-secondary-bg);\n      }\n\n      &:after {\n        color: var(--pst-color-secondary);\n        content: var(--pst-icon-admonition-todo);\n      }\n    }\n  }\n\n  /**\n   * Special-case for a `sidebar` class that makes the admonition float to\n   * the right like the {sidebar} directive.\n   */\n  &.sidebar {\n    max-width: 40%;\n    float: right;\n    clear: both;\n    margin-left: 0.5rem;\n    margin-top: 0;\n\n    // Undo the .sidebar directive border\n    border-width: 0 0 0 0.2rem;\n\n    // TODO: these semantic-color-names border-color rules might no longer be\n    //       needed when we drop support for Sphinx 4 / docutils 0.17\n    &.attention,\n    &.important {\n      border-color: var(--pst-color-attention);\n    }\n\n    &.caution,\n    &.warning {\n      border-color: var(--pst-color-warning);\n    }\n\n    &.danger,\n    &.error {\n      border-color: var(--pst-color-danger);\n    }\n\n    &.hint,\n    &.tip,\n    &.seealso {\n      border-color: var(--pst-color-success);\n    }\n\n    &.note,\n    &.todo {\n      border-color: var(--pst-color-info);\n    }\n\n    // No inner margin since we have less horizontal space w/ the sidebar\n    p.admonition-title ~ * {\n      margin-left: 0rem;\n      margin-right: 0rem;\n    }\n  }\n}\n\n/**************************************************************\n * Similar content blocks that are not technically admonitions.\n */\n\n/**\n * Topics and the {contents} directive\n */\n// Docutils <= 0.17\ndiv.topic,\ndiv.topic.contents,\n// Docutils >= 0.18\nnav.contents,\naside.topic {\n  display: flex;\n  flex-direction: column;\n  background-color: var(--pst-color-surface);\n  border-color: var(--pst-color-border);\n  border-radius: $admonition-border-radius;\n  padding: 1rem 1.25rem;\n  @include box-shadow();\n\n  .topic-title {\n    margin: 0 0 0.5rem 0;\n  }\n\n  // Over-ride text color to ensure enough contrast\n  p {\n    color: var(--pst-color-on-surface) !important;\n  }\n\n  // Over-ride large default padding\n  ul.simple {\n    padding-left: 1rem;\n\n    ul {\n      // So that sub-lists will have a bit less padding\n      padding-left: 2em;\n    }\n  }\n}\n\n/**\n * Sidebar directive\n */\naside.sidebar {\n  border: 1px solid var(--pst-color-border);\n  background-color: var(--pst-color-surface);\n  border-radius: $admonition-border-radius;\n  // to match the admonition-styled sidebars:\n  margin-left: 0.5rem;\n  padding: 0;\n  > *:last-child {\n    padding-bottom: 1rem;\n  }\n\n  p.sidebar-title {\n    position: relative;\n    margin-bottom: 0;\n    padding-top: 0.5rem;\n    padding-bottom: 0.5rem;\n    border-bottom: 1px solid var(--pst-color-border);\n    font-family: var(--pst-font-family-heading);\n    font-weight: var(--pst-admonition-font-weight-heading);\n  }\n\n  // Add margin to the first non-`.sidebar-title` item\n  > *:not(.sidebar-title):first-child,\n  > p.sidebar-title + * {\n    margin-top: 1rem;\n  }\n\n  > * {\n    padding-left: 1rem;\n    padding-right: 1rem;\n  }\n}\n\n/**\n * Rubrics look kind of like section headers\n */\np.rubric {\n  display: flex;\n  flex-direction: column;\n}\n\n/**\n * Seealso is kind of like a vertically-collapsed admonition\n */\n.seealso dd {\n  margin-top: 0;\n  margin-bottom: 0;\n}\n","/**\n * Miscellaneous color functions and mixins\n**/\n\n// loading the math module\n@use \"sass:math\";\n\n// Set background color from a color variable\n//\n@mixin background-from-color-variable($color-variable) {\n  &:before {\n    content: \"\";\n    width: 100%;\n    height: 100%;\n    position: absolute;\n    left: 0;\n    top: 0;\n    background-color: var(#{$color-variable});\n    z-index: -1;\n    // So that hovering over the text within background is still possible.\n    // Otherwise the background overlays the text and you cannot click or select easily.\n    // ref: https://developer.mozilla.org/en-US/docs/Web/CSS/pointer-events\n    pointer-events: none;\n  }\n}\n\n/**\n* Function to get items from nested maps\n*/\n// @param {Map} $map - Map\n// @param {Arglist} $keys - Keys to fetc\n// @return {*}\n@function map-deep-get($map, $keys...) {\n  @each $key in $keys {\n    $map: map-get($map, $key);\n  }\n  @return $map;\n}\n\n/**\n  * Function to check if the color needs converting to a \"color\" type\n  * if it is a string we cannot use other color manipulation functions\n  * It is used to create the sphinx-design colours as these are often interpolated\n  */\n// @param {String/Color} $color - Color definition from map\n// @return {Color} - Color type (in hex)\n@function check-color($color) {\n  @if type-of($color) == string {\n    $color: from-hex($color);\n  }\n  @return $color;\n}\n\n/**\n  * Function to convert the string representation of a color to a color type (hex)\n  */\n// @param {String} $string - String representation of a color\n@function from-hex($string) {\n  $string-lower: to-lower-case($string);\n  $r: \"\";\n  $g: \"\";\n  $b: \"\";\n  $hex: \"0\" \"1\" \"2\" \"3\" \"4\" \"5\" \"6\" \"7\" \"8\" \"9\" \"a\" \"b\" \"c\" \"d\" \"e\" \"f\";\n  $length: str-length($string);\n  $max: if($length == 4, 1, 2);\n\n  // Check for length accuracy\n  @if $length != 4 and $length != 7 {\n    @return $string;\n  }\n\n  // Loop from the second character (omitting #)\n  @for $i from 2 through $length {\n    $c: str-slice($string-lower, $i, $i);\n\n    // If wrong character, return\n    @if index($hex, $c) == null {\n      @return $string;\n    }\n\n    @if str-length($r) < $max {\n      $r: $r + $c;\n    } @else if str-length($g) < $max {\n      $g: $g + $c;\n    } @else if str-length($b) < $max {\n      $b: $b + $c;\n    }\n  }\n\n  @if $length == 4 {\n    $r: $r + $r;\n    $g: $g + $g;\n    $b: $b + $b;\n  }\n\n  @return rgb(_hex-to-dec($r), _hex-to-dec($g), _hex-to-dec($b));\n}\n\n@function _hex-to-dec($string) {\n  $hex: \"0\" \"1\" \"2\" \"3\" \"4\" \"5\" \"6\" \"7\" \"8\" \"9\" \"a\" \"b\" \"c\" \"d\" \"e\" \"f\";\n  $string: to-lower-case($string);\n  $length: str-length($string);\n\n  $dec: 0;\n  @for $i from 1 through $length {\n    $factor: 1 + (15 * ($length - $i));\n    $index: index($hex, str-slice($string, $i, $i));\n    $dec: $dec + $factor * ($index - 1);\n  }\n\n  @return $dec;\n}\n","// Style API docs from sphinx' autodoc / autosummary\n\n/*******************************************************************************\n* Styling for field lists\n*/\n\n/* grey highlighting of 'parameter' and 'returns' field */\ntable.field-list {\n  border-collapse: separate;\n  border-spacing: 10px;\n  margin-left: 1px;\n\n  th.field-name {\n    padding: 1px 8px 1px 5px;\n    white-space: nowrap;\n    background-color: var(--pst-color-surface);\n  }\n\n  /* italic font for parameter types */\n  td.field-body {\n    p {\n      font-style: italic;\n\n      > strong {\n        font-style: normal;\n      }\n    }\n\n    /* reduced space around parameter description */\n    blockquote {\n      border-left: none;\n      margin: 0em 0em 0.3em;\n      padding-left: 30px;\n    }\n  }\n}\n\n/*******************************************************************************\n* Styling for autosummary tables\n*/\n\n.table.autosummary {\n  // The first column (with the signature) should not wrap\n  td:first-child {\n    white-space: nowrap;\n  }\n}\n\n/* overriding basic.css to use our own monospace font */\n.sig {\n  font-family: var(--pst-font-family-monospace);\n}\n\n/* C++ specific styling - overriding the basic.css to avoid custom colors*/\n\n.sig-inline.c-texpr,\n.sig-inline.cpp-texpr {\n  font-family: unset;\n}\n\n.sig.c .k,\n.sig.c .kt,\n.sig.cpp .k,\n.sig.cpp .kt {\n  color: var(--pst-color-text-base);\n}\n\n.sig.c .m,\n.sig.cpp .m {\n  color: var(--pst-color-text-base);\n}\n\n.sig.c .s,\n.sig.c .sc,\n.sig.cpp .s,\n.sig.cpp .sc {\n  color: var(--pst-color-text-base);\n}\n\n// addition\n\n// .sig.c .sig-name .n,\n// .sig.cpp .sig-name .n {\n//   color: var(--pst-color-inline-code);\n// }\n\n.sig-name {\n  color: var(--pst-color-inline-code);\n}\n\n.sig-param .o,\n.sig-param .default_value {\n  color: var(--pst-color-text-muted);\n  font-weight: normal;\n}\n\n// change target color for dark theme\ndt:target,\nspan.highlighted {\n  background-color: var(--pst-color-target);\n}\n\n.viewcode-back {\n  font-family: var(--pst-font-family-base);\n}\n\n.viewcode-block:target {\n  border-top: 1px solid var(--pst-color-border);\n  border-bottom: 1px solid var(--pst-color-border);\n  position: relative;\n  background-color: var(--pst-color-target);\n}\n\n/*******************************************************************************\n* Styling for autosummary titles like \"parameters\" and \"returns\"\n*/\n\n// the API selector\n// from https://github.com/pradyunsg/furo/blob/main/src/furo/assets/styles/content/_api.sass#L6)\ndl[class]:not(.option-list):not(.field-list):not(.footnote):not(.glossary):not(.simple) {\n  dd {\n    margin-left: 2rem;\n\n    // Fix until this will be solved to Sphinx https://github.com/sphinx-doc/sphinx/issues/10815\n    & > dl.simple > dt {\n      display: flex;\n    }\n  }\n\n  dl.field-list {\n    display: grid;\n    grid-template-columns: unset;\n  }\n\n  dt.field-odd,\n  dt.field-even {\n    margin-top: 0.2rem;\n    margin-bottom: 0.1rem;\n    background-color: var(--pst-color-surface);\n  }\n}\n","/**\n * Code block styling\n * Note that we inherit a lot of styling from Bootstrap so not many rules here.\n */\n\n// General code block behavior\n// Unset bootstrap behavior\ndiv[class*=\"highlight-\"],\ndiv.highlight,\ndiv.literal-block-wrapper {\n  display: flex;\n  flex-direction: column;\n  width: unset;\n  border-radius: $admonition-border-radius;\n}\n\n// Code blocks with captions\n// There's a wrapper when the code block has a title\ndiv.literal-block-wrapper {\n  border: 1px solid var(--pst-color-border);\n  border-radius: $admonition-border-radius;\n\n  // This is where the title goes\n  div.code-block-caption {\n    margin: 0;\n    border-bottom: 1px solid var(--pst-color-border);\n    padding: 0.5rem;\n    font-size: 1rem;\n    font-weight: var(--pst-font-weight-caption);\n\n    a.headerlink {\n      font-size: inherit;\n    }\n  }\n\n  // Remove the upper border radius since we want it to connect with the title\n  // Remove the box shadow so the wrapper gets the shadow instead\n  div[class*=\"highlight-\"] {\n    margin: 0;\n    border-radius: 0;\n    pre {\n      border: none;\n      box-shadow: none;\n    }\n  }\n}\n\n/**\n * In-line code\n */\ncode.literal {\n  padding: 0.1rem 0.25rem;\n  background-color: var(--pst-color-surface);\n  border: 1px solid var(--pst-color-border);\n  border-radius: 0.25rem;\n}\n\na > code {\n  color: var(--pst-color-inline-code-links);\n}\n\n// Fix for Sphinx's \"highlight\" directive - this is an issue with our accessible pygments theme\n// and the colour we are using for the background of the code blocks.\nhtml[data-theme=\"light\"] .highlight .nf {\n  color: #0078a1 !important;\n}\n\n// Minimum opacity needed for linenos to be WCAG AA conformant\nspan.linenos {\n  opacity: 0.8 !important;\n}\n","figure {\n  a.headerlink {\n    // So that header link doesn't push caption to be off-center.\n    position: absolute;\n    font-size: inherit;\n  }\n  // Default headerlink hover doesn't trigger on figures\n  &:hover a.headerlink {\n    visibility: visible;\n  }\n\n  figcaption {\n    font-family: var(--pst-font-family-heading);\n    font-weight: var(--pst-font-weight-caption);\n    color: var(--pst-color-text-muted);\n    margin-left: auto;\n    margin-right: auto;\n\n    table.table {\n      width: fit-content;\n      margin-left: auto;\n      margin-right: auto;\n    }\n  }\n}\n","// For consistency, add bracket around footnotes/citations which are\n// cited more than once. E.g. [Newton](1,2) instead of Newton(1,2)\ndt.label > span.brackets:not(:only-child):before {\n  content: \"[\";\n}\n\ndt.label > span.brackets:not(:only-child):after {\n  content: \"]\";\n}\n\n// Make footnote as a superscript\na.footnote-reference {\n  vertical-align: super;\n  font-size: small;\n}\n\n// Docutils 0.18 uses an `aside.footnote` container with different internal structure\naside.footnote {\n  margin-bottom: 0.5rem;\n  &:last-child {\n    margin-bottom: 1rem;\n  }\n  span.label,\n  span.backrefs {\n    font-weight: bold;\n  }\n\n  &:target {\n    background-color: var(--pst-color-target);\n  }\n}\n","/**\n * Hacky fixes that don't fit cleanly into other sections\n */\n\n// Ensure user highlighting/selecting behaves properly\n// From https://stackoverflow.com/a/34372191\ntable.highlighttable td.linenos,\nspan.linenos,\ndiv.doctest > div.highlight span.gp {\n  /* gp: Generic.Prompt */\n  user-select: none;\n  -webkit-user-select: text; /* Safari fallback only */\n  -webkit-user-select: none; /* Chrome/Safari */\n  -moz-user-select: none; /* Firefox */\n  -ms-user-select: none; /* IE10+ */\n}\n","// Override bootstrap by restoring the basic theme default.\ndd {\n  margin-top: 3px;\n  margin-bottom: 10px;\n  margin-left: 30px;\n}\n\nol,\nul {\n  padding-inline-start: 2rem;\n  li > p:first-child {\n    margin-bottom: 0.25rem;\n    margin-top: 0.25rem;\n  }\n}\n","// GitHub blockquote style\nblockquote {\n  padding: 1em 1em;\n  color: var(--pst-color-text-muted);\n  border-left: 0.25em solid var(--pst-color-border);\n  border-radius: $admonition-border-radius;\n  position: relative;\n\n  p {\n    color: var(--pst-color-text-base);\n  }\n\n  // remove padding from included line-block to avoid duplication\n  .line-block {\n    margin: 0 0;\n  }\n\n  // remove margin bottom for the last p\n  p:last-child {\n    margin-bottom: 0;\n  }\n\n  @include background-from-color-variable(--pst-color-on-background);\n\n  //hack to make the text in the blockquote selectable\n  &:before {\n    z-index: -1;\n  }\n}\n","/**\n * Span-level styling within content\n */\n\nspan.guilabel {\n  border: 1px solid var(--pst-color-info);\n  color: var(--pst-color-info);\n  font-size: 80%;\n  font-weight: 700;\n  border-radius: 4px;\n  padding: 2.4px 6px;\n  margin: auto 2px;\n  position: relative;\n\n  @include background-from-color-variable(--pst-color-info-bg);\n}\n\na.reference.download:before {\n  content: var(--pst-icon-download);\n  font: var(--fa-font-solid);\n  font-size: 0.8em;\n  padding: 0 0.25em;\n  color: var(--pst-color-text-muted);\n}\n","/**\n * Tables\n */\n\n// ensure table will fit in the article width and make them y-scrollable\ntable {\n  display: table;\n  overflow: auto;\n\n  // default to table-center\n  margin-left: auto;\n  margin-right: auto;\n\n  &.table-right {\n    margin-right: 0;\n  }\n\n  &.table-left {\n    margin-left: 0;\n  }\n}\n\n// customize table caption from bootstrap\n// to display them on top and centered\ntable caption {\n  text-align: center;\n  caption-side: top;\n  color: var(--pst-color-text-muted);\n}\n\n// MyST Markdown tables use these classes to control alignment\nth,\ntd {\n  &.text-align\\:left {\n    text-align: left;\n  }\n\n  &.text-align\\:right {\n    text-align: right;\n  }\n\n  &.text-align\\:center {\n    text-align: center;\n  }\n}\n\n// override bootstrap background colors\n.table {\n  --bs-table-bg: transparent;\n}\n","/**\n * Style the toctree component in pages (avoid modifying the navbars)\n */\n.toctree-wrapper {\n  p.caption {\n    font-size: 1.5em;\n    margin-bottom: 0em;\n  }\n\n  & > ul {\n    padding-left: 0;\n  }\n\n  li[class^=\"toctree-l\"] {\n    list-style: none;\n    margin-bottom: 0.2em;\n\n    & > a {\n      list-style: none;\n      font-size: 1.1em;\n    }\n\n    & > ul {\n      list-style: none;\n      padding-inline-start: 1.5em;\n    }\n  }\n\n  // slightly bigger font for l1\n  .toctree-l1 > a {\n    font-size: 1.3em;\n  }\n}\n\ndiv.topic.contents,  // Docutils <= 0.17\nnav.contents  // Docutils >= 0.18\n{\n  // Style similarly to toctree\n  ul.simple {\n    list-style: none;\n    padding-left: 0;\n  }\n}\n","/**\n * Mathematics via MathJax.\n *\n * This is designed for MathJax v3\n * ref: https://www.sphinx-doc.org/en/master/usage/extensions/math.html#module-sphinx.ext.mathjax\n */\n\n// Applies to all math elements\nspan.math,\ndiv.math {\n  align-items: center;\n  display: flex;\n  max-width: 100%;\n  // This will be over-ridden for the y-direction and divs\n  overflow: hidden;\n}\n\n// Inline-only\nspan.math {\n  display: inline-flex;\n}\n\n// Block-level only\ndiv.math {\n  gap: 0.5em;\n  // So that the eqno shows up after the equation\n  flex-direction: row-reverse;\n\n  // The equation number / link\n  span.eqno a.headerlink {\n    position: relative;\n    font-size: 1em;\n  }\n\n  // The math container\n  mjx-container {\n    flex-grow: 1;\n    padding-bottom: 0.2rem;\n    overflow: auto;\n\n    // Set height to 0 so that it does not cause scrollbars to appear\n    // ref: https://github.com/mathjax/MathJax/issues/2521\n    mjx-assistive-mml {\n      height: 0;\n    }\n  }\n}\n","/**\n * ABlog\n * ref: https://ablog.readthedocs.io/\n */\n\n/**\n * Sidebar template components\n */\n.ablog-sidebar-item {\n  h2,\n  h3 {\n    font-size: var(--pst-sidebar-header-font-size);\n    // Remove unnecessary vertical whitespace\n    margin-top: 0.5rem;\n\n    // The headers are all links, but this makes them hard to parse\n    // So we change the colors to make them look like headers\n    a {\n      color: var(--pst-color-text-base);\n    }\n  }\n\n  ul {\n    // No bullet points for the primary sidebar items\n    list-style: none;\n    padding-left: 0;\n    // Otherwise a scrollbar randomly shows up\n    overflow-y: hidden;\n    // List of recent post items\n    display: flex;\n    flex-direction: column;\n    gap: 0.5em;\n    margin-bottom: 0;\n\n    // The ablog cloud should move horizontally\n    &.ablog-cloud {\n      flex-direction: row;\n      flex-flow: wrap;\n      gap: 0.5rem;\n\n      // Vertical-align tag clouds\n      li {\n        // Center the tag cloud items\n        display: flex;\n        align-items: center;\n      }\n    }\n  }\n}\n\n/**\n * Previous / next buttons at the bottom\n */\n.ablog__prev-next {\n  font-size: 1.2em;\n  display: flex;\n  padding: 1rem 0;\n\n  // The bottom previous / next arrows\n  > span {\n    // To ensure that the whole thing fits on one line even if there are long titles\n    display: flex;\n    max-width: 45%;\n\n    // Links within each span have the collection of icon + text\n    a {\n      display: flex;\n      align-items: center;\n      margin-left: auto;\n      gap: 1rem;\n      line-height: 1.5rem;\n\n      i:before {\n        color: var(--pst-color-text-base);\n      }\n    }\n  }\n\n  // The first span is for the previous page and aligns to the left\n  span.ablog__prev {\n    i.fa-arrow-circle-left:before {\n      content: var(--pst-icon-angle-left);\n    }\n  }\n\n  // The second span is just an empty space so we remove it because we're\n  // positioning with flex\n  span.ablog__spacer {\n    display: none;\n  }\n\n  // The third span is aligned to the right\n  span.ablog__next {\n    margin-left: auto;\n    text-align: right;\n\n    i.fa-arrow-circle-right:before {\n      content: var(--pst-icon-angle-right);\n    }\n  }\n}\n\n/**\n * {postlist} directive and posts page\n */\n.ablog__collection,\n.postlist {\n  padding-left: 0;\n\n  .ablog-post {\n    list-style: none;\n\n    // Post metadata tags (author, links ,etc) should be a bit smaller\n    .ablog-archive {\n      display: flex;\n      flex-direction: row;\n      flex-wrap: wrap;\n      gap: 1rem;\n      list-style: none;\n      font-size: 0.75rem;\n      padding-left: 0;\n    }\n\n    // Title line should be a bit bigger and bold to stand out\n    .ablog-post-title {\n      margin-top: 0;\n      font-size: 1.25rem;\n\n      a {\n        font-weight: bold;\n      }\n    }\n\n    // Read more button should be a bit bigger\n    .ablog-post-expand {\n      margin-bottom: 0.5rem;\n    }\n  }\n}\n","/**\n * Special cases for Bootstrap functionality\n */\n\n// Bootstrap adds margin to their general container class. However, sphinx/docutils\n// can also generate output with the container class, but in those cases we should\n// not add the margin from bootstrap. Same for max-width.\n.docutils.container {\n  padding-left: unset;\n  padding-right: unset;\n  margin-left: unset;\n  margin-right: unset;\n  max-width: unset;\n  width: unset;\n}\n","/**\n * Sphinx Copybutton\n * ref: https://sphinx-copybutton.readthedocs.io/\n */\n\ndiv.highlight button.copybtn {\n  // Nicer spacing\n  display: flex;\n  align-items: center;\n  justify-content: center;\n\n  // Removes the button background so we more naturally blend with the code cell.\n  background-color: unset;\n\n  // Don't over-ride the success color\n  &:not(.success) {\n    color: var(--pst-color-muted);\n  }\n  border: none;\n  background-color: var(--pst-color-surface);\n\n  &:hover {\n    &:not(.success) {\n      color: var(--pst-color-text);\n      background-color: var(--pst-color-shadow);\n    }\n  }\n\n  // Tooltip styling\n  &.o-tooltip--left:after {\n    color: var(--pst-color-text);\n    background-color: var(--pst-color-surface);\n  }\n}\n","// adapt ethical ad to the theme\n#ethical-ad-placement {\n  .ethical-sidebar a,\n  .ethical-sidebar a:visited,\n  .ethical-sidebar a:hover,\n  .ethical-sidebar a:active,\n  .ethical-footer a,\n  .ethical-footer a:visited,\n  .ethical-footer a:hover,\n  .ethical-footer a:active {\n    color: var(--pst-color-text-base);\n  }\n\n  .ethical-sidebar,\n  .ethical-footer {\n    background-color: var(--pst-color-background);\n    border: 1px solid var(--pst-color-border);\n    border-radius: 5px;\n    color: var(--pst-color-text-base);\n    font-size: 14px;\n    line-height: 20px;\n  }\n}\n","/**\n * Styles for various Sphinx execution libraries.\n * For now, where these define output sections, we simply revert their background\n * to be a \"light theme\" background. This ensures that inputs/outputs behave similarly,\n * because the CSS is often controlled by each package.\n * In the future, we might add dark theme support for specific packages.\n */\n\n/******************************************************************************\n * Jupyter Sphinx\n */\n\n.bd-content div.jupyter_container {\n  // We don't want borders around the whole container, just around code cells\n  border: none;\n  background-color: unset;\n  box-shadow: none;\n\n  // Code cells should have the same style as our other code objects\n  div.output,\n  div.highlight {\n    border-radius: 0.25rem;\n  }\n  div.highlight {\n    background-color: var(--pst-color-surface);\n  }\n\n  // Ensure the style is the same as our code cells. Jupyter Sphinx makes it tiny.\n  .cell_input,\n  .cell_output {\n    border-radius: 0.25rem;\n    pre {\n      padding: 1rem;\n    }\n  }\n}\n","/**\n * Special-cases for packages in the PyData ecosystem\n */\n\n// xarray output display in bootstrap\n.xr-wrap[hidden] {\n  display: block !important;\n}\n","/*******************************************************************************\n* master color map. Only the colors that actually differ between light and dark\n* themes are specified separately.\n*\n* To see the full list of colors see https://www.figma.com/file/rUrrHGhUBBIAAjQ82x6pz9/PyData-Design-system---proposal-for-implementation-(2)?node-id=1234%3A765&t=ifcFT1JtnrSshGfi-1\n*/\n\n/**\n* Function to get items from nested maps\n*/\n// @param {Map} $map - Map\n// @param {Arglist} $keys - Keys to fetch\n// @return {*}\n@function map-deep-get($map, $keys...) {\n  @each $key in $keys {\n    $map: map-get($map, $key);\n  }\n  @return $map;\n}\n\n/* Assign base colors for the PyData theme */\n$color-palette: (\n  // Primary color\n  \"teal\":\n    (\n      \"50\": #f4fbfc,\n      \"100\": #e9f6f8,\n      \"200\": #d0ecf1,\n      \"300\": #abdde6,\n      \"400\": #3fb1c5,\n      \"500\": #0a7d91,\n      \"600\": #085d6c,\n      \"700\": #064752,\n      \"800\": #042c33,\n      \"900\": #021b1f,\n    ),\n  // Secondary color\n  \"violet\":\n    (\n      \"50\": #f4eefb,\n      \"100\": #e0c7ff,\n      \"200\": #d5b4fd,\n      \"300\": #b780ff,\n      \"400\": #9c5ffd,\n      \"500\": #8045e5,\n      \"600\": #6432bd,\n      \"700\": #4b258f,\n      \"800\": #341a61,\n      \"900\": #1e0e39,\n    ),\n  // Neutrals\n  \"gray\":\n    (\n      \"50\": #f9f9fa,\n      \"100\": #f3f4f5,\n      \"200\": #e5e7ea,\n      \"300\": #d1d5da,\n      \"400\": #9ca4af,\n      \"500\": #677384,\n      \"600\": #48566b,\n      \"700\": #29313d,\n      \"800\": #222832,\n      \"900\": #14181e,\n    ),\n  // Accent color\n  \"pink\":\n    (\n      \"50\": #fcf8fd,\n      \"100\": #fcf0fa,\n      \"200\": #f8dff5,\n      \"300\": #f3c7ee,\n      \"400\": #e47fd7,\n      \"500\": #c132af,\n      \"600\": #912583,\n      \"700\": #6e1c64,\n      \"800\": #46123f,\n      \"900\": #2b0b27,\n    ),\n  \"foundation\": (\n    \"white\": #ffffff,\n    // gray-900\n    \"black\": #14181e,\n  )\n);\n\n:root {\n  // Add theme colours to the html root element\n  @each $group-color, $color in $color-palette {\n    @each $color-name, $definition in $color {\n      --pst-#{$group-color}-#{$color-name}: #{$definition};\n    }\n  }\n}\n\n// Static SCSS variables used thoroughout the theme\n// Minimum contrast ratio used for the theme.\n// Acceptable values for WCAG 2.0 are 3, 4.5 and 7.\n// See https://www.w3.org/TR/WCAG20/#visual-audio-contrast-contrast\n// 4.5 - is for text that is 14pt or less\n$min-contrast-ratio-4: 4.5;\n// 3 is for text that is 18pt or bold, or for non-text elements\n$min-contrast-ratio-3: 3;\n\n// Customize the light and dark text colors for use in our color contrast function.\n$foundation-black: #14181e;\n$foundation-white: #ffffff;\n// This is a custom - calculated  color between gray 100 and 200 - to reduce\n// the contrast ratio (avoid a jarring effect)\n$base-light-text: #ced6dd;\n// used in sphinx_design - gray 100\n$foundation-light-gray: #f3f4f5;\n// used in sphinx_design - gray 700\n$foundation-muted-gray: #29313d;\n// used in sphinx_design - gray 800\n$foundation-dark-gray: #222832;\n\n$pst-semantic-colors: (\n  \"primary\": (\n    \"light\": #{map-deep-get($color-palette, \"teal\", \"500\")},\n    \"bg-light\": #{map-deep-get($color-palette, \"teal\", \"200\")},\n    \"dark\": #{map-deep-get($color-palette, \"teal\", \"400\")},\n    \"bg-dark\": #{map-deep-get($color-palette, \"teal\", \"800\")},\n  ),\n  \"secondary\": (\n    \"light\": #{map-deep-get($color-palette, \"violet\", \"500\")},\n    \"bg-light\": #{map-deep-get($color-palette, \"violet\", \"100\")},\n    \"dark\": #{map-deep-get($color-palette, \"violet\", \"400\")},\n    \"bg-dark\": #{map-deep-get($color-palette, \"violet\", \"800\")},\n  ),\n  \"accent\": (\n    \"light\": #{map-deep-get($color-palette, \"pink\", \"500\")},\n    \"bg-light\": #{map-deep-get($color-palette, \"pink\", \"200\")},\n    \"dark\": #{map-deep-get($color-palette, \"pink\", \"400\")},\n    \"bg-dark\": #{map-deep-get($color-palette, \"pink\", \"800\")},\n  ),\n  \"info\": (\n    \"light\": #276be9,\n    \"bg-light\": #dce7fc,\n    \"dark\": #79a3f2,\n    \"bg-dark\": #06245d,\n  ),\n  \"warning\": (\n    \"light\": #f66a0a,\n    \"bg-light\": #f8e3d0,\n    \"dark\": #ff9245,\n    \"bg-dark\": #652a02,\n  ),\n  \"success\": (\n    \"light\": #00843f,\n    \"bg-light\": #d6ece1,\n    \"dark\": #5fb488,\n    \"bg-dark\": #002f17,\n  ),\n  // This is is based on the warning color\n  \"attention\":\n    (\n      \"light\": var(--pst-color-warning),\n      \"bg-light\": var(--pst-color-warning-bg),\n      \"dark\": var(--pst-color-warning),\n      \"bg-dark\": var(--pst-color-warning-bg),\n    ),\n  \"danger\": (\n    \"light\": #d72d47,\n    \"bg-light\": #f9e1e4,\n    \"dark\": #e78894,\n    \"bg-dark\": #4e111b,\n  ),\n  \"text-base\": (\n    \"light\": #{map-deep-get($color-palette, \"gray\", \"800\")},\n    \"dark\": $base-light-text,\n  ),\n  \"text-muted\": (\n    \"light\": #{map-deep-get($color-palette, \"gray\", \"600\")},\n    \"dark\": #{map-deep-get($color-palette, \"gray\", \"400\")},\n  ),\n  \"heading-color\": (\n    \"light\": #{$foundation-white},\n    \"dark\": #{$foundation-black},\n  ),\n  \"shadow\": (\n    \"light\": rgba(0, 0, 0, 0.1),\n    \"dark\": rgba(0, 0, 0, 0.2),\n  ),\n  \"border\": (\n    \"light\": #{map-deep-get($color-palette, \"gray\", \"300\")},\n    \"dark\": #{map-deep-get($color-palette, \"gray\", \"600\")},\n  ),\n  \"border-muted\": (\n    \"light\": rgba(23, 23, 26, 0.2),\n    \"dark\": #{map-deep-get($color-palette, \"gray\", \"700\")},\n  ),\n  \"inline-code\": (\n    \"light\": #{map-deep-get($color-palette, \"pink\", \"600\")},\n    \"dark\": #{map-deep-get($color-palette, \"pink\", \"300\")},\n  ),\n  \"inline-code-links\": (\n    // need to make sure there is enough contrast against the code bg\n    \"light\": #{map-deep-get($color-palette, \"teal\", \"600\")},\n    // keep primary color for dark mode\n    \"dark\": #{map-deep-get($color-palette, \"teal\", \"400\")},\n  ),\n  \"target\": (\n    \"light\": #f3cf95,\n    \"dark\": #675c04,\n  ),\n  // DEPTH COLORS - you can see the elevation colours and shades\n  // in the Figma file https://www.figma.com/file/rUrrHGhUBBIAAjQ82x6pz9/PyData-Design-system---proposal-for-implementation-(2)?node-id=1492%3A922&t=sQeQZehkOzposYEg-1\n  // background: color of the canvas / the furthest back layer\n  \"background\":\n    (\n      \"light\": #{map-deep-get($color-palette, \"foundation\", \"white\")},\n      \"dark\": #{map-deep-get($color-palette, \"foundation\", \"black\")},\n    ),\n  // on-background: provides slight contrast against background\n  // (by use of shadows in light theme)\n  \"on-background\":\n    (\n      \"light\": #{map-deep-get($color-palette, \"foundation\", \"white\")},\n      \"dark\": #{map-deep-get($color-palette, \"gray\", \"800\")},\n    ),\n  \"surface\": (\n    \"light\": #{map-deep-get($color-palette, \"gray\", \"100\")},\n    \"dark\": #{map-deep-get($color-palette, \"gray\", \"700\")},\n  ),\n  // on_surface: object on top of surface object (without shadows)\n  \"on-surface\":\n    (\n      \"light\": #{map-deep-get($color-palette, \"gray\", \"800\")},\n      \"dark\": $foundation-light-gray,\n    ),\n);\n\n/*******************************************************************************\n* write the color rules for each theme (light/dark)\n*/\n\n/* NOTE:\n * Mixins enable us to reuse the same definitions for the different modes\n * https://sass-lang.com/documentation/at-rules/mixin\n * #{something} inserts a variable into a CSS selector or property name\n * https://sass-lang.com/documentation/interpolation\n */\n@mixin theme-colors($mode) {\n  // check if this color is defined differently for light/dark\n  @each $col-name, $definition in $pst-semantic-colors {\n    @if type-of($definition) == map {\n      @each $key, $val in $definition {\n        @if str-index($key, $mode) != null {\n          // since now we define the bg colours in the semantic colours and not\n          // by changing opacity, we need to check if the key contains bg and the\n          // correct mode (light/dark)\n          @if str-index($key, \"bg\") != null {\n            --pst-color-#{$col-name}-bg: #{$val};\n          } @else {\n            --pst-color-#{$col-name}: #{$val};\n          }\n        }\n      }\n    } @else {\n      --pst-color-#{$col-name}: #{$definition};\n    }\n  }\n\n  // assign the \"duplicate\" colors (ones that just reference other variables)\n  & {\n    --pst-color-link: var(--pst-color-primary);\n    --pst-color-link-hover: var(--pst-color-secondary);\n  }\n  // adapt to light/dark-specific content\n  @if $mode == \"light\" {\n    .only-dark,\n    .only-dark ~ figcaption {\n      display: none !important;\n    }\n  } @else {\n    .only-light,\n    .only-light ~ figcaption {\n      display: none !important;\n    }\n    /* Adjust images in dark mode (unless they have class .only-dark or\n     * .dark-light, in which case assume they're already optimized for dark\n     * mode).\n     */\n    img:not(.only-dark):not(.dark-light) {\n      filter: brightness(0.8) contrast(1.2);\n    }\n    /* Give images a light background in dark mode in case they have\n    *  transparency and black text (unless they have class .only-dark or .dark-light, in\n    *  which case assume they're already optimized for dark mode).\n    */\n    .bd-content img:not(.only-dark):not(.dark-light) {\n      background: rgb(255, 255, 255);\n      border-radius: 0.25rem;\n    }\n    // MathJax SVG outputs should be filled to same color as text.\n    .MathJax_SVG * {\n      fill: var(--pst-color-text-base);\n    }\n  }\n}\n\n/* Defaults to light mode if data-theme is not set */\nhtml:not([data-theme]) {\n  @include theme-colors(\"light\");\n}\n\n/* NOTE: @each {...} is like a for-loop\n * https://sass-lang.com/documentation/at-rules/control/each\n */\n@each $mode in (light, dark) {\n  html[data-theme=\"#{$mode}\"] {\n    @include theme-colors($mode);\n    color-scheme: $mode;\n  }\n}\n\n// assign classes too, for runtime use of theme colors\n@each $col-name, $definition in $pst-semantic-colors {\n  .pst-color-#{$col-name} {\n    color: var(--pst-color-#{$col-name});\n  }\n}\n","/*******************************************************************************\n * Special-cases for the sphinx-design library, mainly to make it compatible\n * with the dark/light themes of pydata-sphinx-theme.\n *\n * NOTE: sphinx-design uses !important quite liberally, so here we must do the\n * same for our overrides to have any effect.\n */\n@use \"../variables/color\" as pst-color;\n@use \"sass:meta\";\n@use \"sass:color\";\n\n/*******************************************************************************\n  * Color and variables\n  *\n  * This is a list of the semantic color names from sphinx-design (we only\n  * need to override variables that sphinx-design has actually defined).\n  * https://github.com/executablebooks/sphinx-design/blob/9226a12a/style/_colors.scss#L31-L43\n  */\n$sd-semantic-color-names: (\n  \"primary\",\n  \"secondary\",\n  \"success\",\n  \"info\",\n  \"warning\",\n  \"danger\",\n  \"light\",\n  \"muted\",\n  \"dark\",\n  \"black\",\n  \"white\"\n);\n\n/**\n  * Here we create some extra --pst-color-* variables and use\n  * them to override the value of the corresponding sphinx-design variables.\n  * This is easier than re-writing the sphinx-design rules. Even easier would be\n  * directly assigning our values to the --sd-color-* variables, but then our\n  * downstream users couldn't override *our* colors and have it affect buttons\n  * and badges.\n  *\n  * First, define the extra keys needed to cover the full range of semantic\n  * color names used in sphinx-design, then merge them with the names we\n  * already define for our own needs.\n  * see https://sphinx-design.readthedocs.io/en/latest/css_variables.html\n  */\n$extra-semantic-colors: (\n  \"white\": $foundation-white,\n  \"light\": (\n    light: $foundation-light-gray,\n    bg-light: color.scale($foundation-light-gray, $lightness: 30%),\n    dark: $foundation-light-gray,\n    bg-dark: color.scale($foundation-light-gray, $lightness: -30%),\n  ),\n  \"muted\": (\n    light: $foundation-muted-gray,\n    bg-light: color.scale($foundation-muted-gray, $lightness: 30%),\n    dark: $foundation-light-gray,\n    bg-dark: color.scale($foundation-muted-gray, $lightness: -30%),\n  ),\n  \"dark\": $foundation-dark-gray,\n  \"black\": $foundation-black,\n);\n\n$all-colors: map-merge($pst-semantic-colors, $extra-semantic-colors);\n\n@mixin create-sd-colors($value, $name) {\n  // define the pst variables, so that downstream user overrides will work\n  --pst-color-#{$name}: #{$value};\n  // we are now using a11y-combination to calculate the text color - this is based\n  // on the WCAG color contrast guidelines\n  --pst-color-#{$name}-text: #{a11y-combination($value)};\n  //TODO: highlight seems to be used for buttons @trallard to fix on a11y follow-up work\n  --pst-color-#{$name}-highlight: #{darken($value, 15%)};\n  // override the sphinx-design variables\n  --sd-color-#{$name}: var(--pst-color-#{$name});\n  --sd-color-#{$name}-text: var(--pst-color-#{$name}-text);\n  //TODO: highlight seems to be used for buttons @trallard to fix on a11y follow-up work\n  --sd-color-#{$name}-highlight: var(--pst-color-#{$name}-highlight);\n}\n\n// Now we override the --sd-color-* variables.\n@each $mode in (light, dark) {\n  html[data-theme=\"#{$mode}\"] {\n    // check if this color is defined differently for light/dark\n    @each $name in $sd-semantic-color-names {\n      $definition: map-get($all-colors, $name);\n      @if type-of($definition) == map {\n        @each $key, $value in $definition {\n          @if str-index($key, $mode) != null {\n            // since now we define the bg colours in the semantic colours and not\n            // by changing opacity, we need to check if the key contains bg and the\n            // correct mode (light/dark)\n            @if str-index($key, \"bg\") != null {\n              --sd-color-#{$name}-bg: #{$value};\n              // create local variable\n              $bg-var: --sd-color-#{$name}-bg;\n              $value: check-color($value);\n              --sd-color-#{$name}-bg-text: #{a11y-combination($value)};\n            } @else {\n              $value: check-color($value);\n              @include create-sd-colors($value, $name);\n            }\n          }\n        }\n      } @else {\n        $value: map-get($all-colors, $name);\n        @include create-sd-colors($value, $name);\n      }\n    }\n  }\n}\n\n// Make sure the color border variables are set using our variables\n@each $mode in (light, dark) {\n  html[data-theme=\"#{$mode}\"] {\n    --sd-color-card-border: var(--pst-color-border);\n  }\n}\n\n/*******************************************************************************\n  * shadows\n  */\nhtml[data-theme=\"light\"] {\n  .sd-shadow-xs,\n  .sd-shadow-sm,\n  .sd-shadow-md,\n  .sd-shadow-lg {\n    @include box-shadow();\n  }\n}\n\n/*******************************************************************************\n  * cards\n  */\n\n.bd-content .sd-card {\n  border: 1px solid var(--pst-color-border);\n\n  // TODO - --pst-color-panel-background is not defined... where is this coming from?\n  .sd-card-header {\n    background-color: var(--pst-color-panel-background);\n    border-bottom: 1px solid var(--pst-color-border);\n  }\n  .sd-card-footer {\n    background-color: var(--pst-color-panel-background);\n    border-top: 1px solid var(--pst-color-border);\n  }\n\n  .sd-card-body {\n    background-color: var(--pst-color-panel-background);\n  }\n}\n/*******************************************************************************\n  * tabs\n  */\n\n.bd-content .sd-tab-set {\n  > input {\n    // Active tab label\n    &:checked + label {\n      border-color: transparent transparent var(--pst-color-primary); // top LR bottom\n      color: var(--pst-color-primary);\n    }\n\n    // Hover label\n    &:not(:checked) + label:hover {\n      border-color: var(--pst-color-secondary);\n      color: var(--pst-color-secondary);\n    }\n  }\n\n  // Tab label\n  > label {\n    color: var(--pst-color-text-muted);\n    border-top: 0.125rem solid transparent; // so hover isn't just color change\n    padding-top: 0.5em; // same as bottom padding, so hover overline looks OK\n    // Hovered label\n    html &:hover {\n      color: var(--pst-color-secondary);\n      border-color: var(--pst-color-secondary);\n    }\n  }\n}\n\n/*******************************************************************************\n* Dropdowns\n*/\n\ndetails.sd-dropdown {\n  // Remove all borders to over-ride SD behavior, and we'll add our own later\n  border: 0px !important;\n  summary.sd-card-header {\n    border: 0 !important;\n    & + div.sd-summary-content {\n      border: 0;\n    }\n  }\n  // Drop shadow should behave same as admonitions\n  @include box-shadow();\n\n  // Header is where the \"clickable\" box goes\n  summary.sd-card-header {\n    display: flex;\n    align-items: center;\n    position: relative; // So background color works\n    font-weight: 600;\n    padding-top: 0.5rem;\n    padding-bottom: 0.5rem;\n\n    // Set a variable that we can re-use for colors later\n    // We must set this in the current and content sibling container\n    // so that it is defined in both places\n    --pst-sd-dropdown-color: var(--pst-gray-500);\n    --pst-sd-dropdown-bg-color: var(--pst-color-surface);\n    & + div.sd-summary-content {\n      --pst-sd-dropdown-color: var(--sd-color-card-border);\n    }\n    @each $name in $sd-semantic-color-names {\n      &.sd-bg-#{$name} {\n        --pst-sd-dropdown-color: var(--sd-color-#{$name});\n        --pst-sd-dropdown-bg-color: var(--sd-color-#{$name}-bg);\n        // Otherwise it won't be defined in the sibling element\n        & + div.sd-summary-content {\n          --pst-sd-dropdown-color: var(--sd-color-#{$name});\n          --pst-sd-dropdown-bg-color: var(--sd-color-#{$name}-bg);\n        }\n      }\n      &.sd-bg-text-#{$name} {\n        // Use the WCAG conformant text color\n        color: var(--sd-color-#{$name}-bg-text) !important;\n      }\n    }\n\n    // Background color and border are grey by default\n    background-color: unset !important;\n    @include background-from-color-variable(--pst-sd-dropdown-bg-color);\n\n    // Add a left border with the same structure as our admonitions\n    border-left: 0.2rem solid var(--pst-sd-dropdown-color) !important;\n    & + div.sd-summary-content {\n      border-left: 0.2rem solid var(--pst-sd-dropdown-color) !important;\n      border-bottom-left-radius: calc(0.25rem - 1px);\n      background-color: var(--pst-color-on-background);\n    }\n    span.sd-summary-icon {\n      display: inline-flex;\n      align-items: center;\n      color: var(--pst-sd-dropdown-color) !important;\n      svg {\n        opacity: 1;\n      }\n    }\n\n    // Positioning of the caret\n    .sd-summary-up,\n    .sd-summary-down {\n      top: 0.7rem;\n    }\n  }\n}\n","/**\n * Sphinx togglebutton\n */\n\n.bd-content {\n  // Admonition toggles\n  .admonition {\n    button.toggle-button {\n      color: inherit;\n    }\n  }\n\n  // Details buttons\n  details.toggle-details {\n    // Over-ride border color to re-use our primary color\n    summary {\n      border-left: 3px solid var(--pst-color-primary);\n    }\n  }\n}\n","/**\n * Styles for various Sphinx execution libraries to display pre-executed notebooks.\n * For now, where these define output sections, we simply revert their background\n * to be a \"light theme\" background. This ensures that inputs/outputs behave similarly,\n * because the CSS is often controlled by each package.\n * In the future, we might add dark theme support for specific packages.\n */\n\n/*******************************************************************************\n * nbsphinx\n */\nhtml div.rendered_html\n// NBsphinx ipywidgets output selector\nhtml .jp-RenderedHTMLCommon {\n  table {\n    table-layout: auto;\n  }\n}\n\n// Dark theme special-cases\nhtml[data-theme=\"dark\"] .bd-content {\n  .nboutput {\n    .output_area.rendered_html {\n      @include cell-output-background;\n    }\n\n    .output_area.stderr {\n      background: var(--pst-color-danger);\n    }\n  }\n}\n\n// Add extra padding to the final item in an nbsphinx container\ndiv.nblast.container {\n  margin-bottom: 1rem;\n}\n\n/*******************************************************************************\n * myst NB\n */\n\ndiv.cell_output .output {\n  max-width: 100%;\n  overflow-x: auto;\n}\n\n// Dark theme special-cases\nhtml[data-theme=\"dark\"] .bd-content {\n  div.cell_output {\n    img,\n    .text_html {\n      @include cell-output-background;\n    }\n  }\n}\n\n// Prevent tables from scrunching together\n.bd-content {\n  div.cell_input {\n    display: flex;\n    flex-direction: column;\n    justify-content: stretch;\n  }\n\n  div.cell_input,\n  div.output {\n    border-radius: $admonition-border-radius;\n  }\n\n  div.output {\n    table {\n      table-layout: auto;\n    }\n  }\n}\n","/**\n * style for the various mapping libs based on leaflet (folium, geemap, ipyleaflet)\n * mainly ensure the good display of the maps in both themes and avoid the customization\n * of the tiles\n */\n\n/**\n * avoid border override from pydata-sphinx-theme\n * minimal selctor to get the priority\n */\nhtml[data-theme=\"dark\"] .bd-content img.leaflet-tile.leaflet-tile-loaded {\n  border-radius: 0;\n  padding: 0;\n}\n","/**\n * /search.html page special-cases\n */\n\n.bd-search-container {\n  /*******************************************\n  * Search results\n  */\n  // Whitespace\n  div#search-results {\n    > h2 {\n      font-size: var(--pst-font-size-icon);\n      margin-top: 0;\n    }\n\n    p.search-summary {\n      color: var(--pst-color-text-muted);\n    }\n  }\n\n  ul.search {\n    margin: 0;\n    list-style: none;\n\n    li {\n      background-image: none;\n      padding: 1rem 0;\n      margin: 1rem 0;\n      border-top: 1px solid var(--pst-color-text-muted);\n\n      // First link is the page title, it should be a bit bigger\n      > a {\n        font-size: 1.2em;\n      }\n\n      div.context,\n      p.context {\n        color: var(--pst-color-text-base);\n        margin: 0.5em 0 0 0;\n\n        // Add a # before page section titles to make it clear they are sections\n        a:before {\n          content: \"#\";\n          padding-right: 0.2em;\n          color: var(--pst-color-text-muted);\n        }\n      }\n    }\n  }\n}\n"],"names":[],"sourceRoot":""}
\ No newline at end of file
diff --git a/docs/_build/html/_static/styles/sphinx-book-theme.css b/docs/_build/html/_static/styles/sphinx-book-theme.css
deleted file mode 100644
index e514f24..0000000
--- a/docs/_build/html/_static/styles/sphinx-book-theme.css
+++ /dev/null
@@ -1,8 +0,0 @@
-/*! sphinx-book-theme CSS
- * BSD 3-Clause License
- * Copyright (c) 2020, EBP
- * All rights reserved.
- *
- * This follows the 7-1 pattern described here:
- * https://sass-guidelin.es/#architecture
- */html{--pst-font-size-base:none}html[data-theme=light]{--sbt-color-announcement:#616161}html[data-theme=dark],html[data-theme=light]{--pst-color-primary:#579aca}html{scroll-padding-top:4rem}.sbt-scroll-pixel-helper{position:absolute;width:0;height:0;top:0;left:0}.d-n,.onlyprint{display:none}@media print{.onlyprint{display:block!important}}@media print{.noprint{display:none!important}}.bd-article-container h1,.bd-article-container h2,.bd-article-container h3,.bd-article-container h4,.bd-article-container h5,.bd-article-container p.caption{color:var(--pst-color-muted)}.bd-article-container h1,.bd-article-container h2{font-weight:500}a.brackets:before{color:inherit;font-family:inherit;margin-right:0}table{position:relative}@media print{.bd-main .bd-content{margin-left:2rem;height:auto}.bd-main .bd-content #jb-print-docs-body{margin-left:0}.bd-main .bd-content #jb-print-docs-body h1{font-size:3em;text-align:center;margin-bottom:0}.bd-main .bd-content .bd-article{padding-top:0}.bd-main .bd-content .bd-article h1:first-of-type{display:none}.bd-main .bd-content .container{min-width:0!important}.bd-main .bd-content h1{margin-top:1em;margin-bottom:1em}.bd-main .bd-content h1,.bd-main .bd-content h2,.bd-main .bd-content h3,.bd-main .bd-content h4{break-after:avoid;color:#000}.bd-main .bd-content table{break-inside:avoid}.bd-main .bd-content pre{word-wrap:break-word}.bd-main .bd-content a.headerlink{display:none}.bd-main .bd-content aside.margin,.bd-main .bd-content aside.sidebar,.bd-main .bd-content blockquote.epigraph{border:none}.bd-main .bd-content .footer{margin-top:1em}.bd-main .bd-content #jb-print-toc{margin-bottom:1.5rem;margin-left:0}.bd-main .bd-content #jb-print-toc .section-nav{border-left:0!important;list-style-type:disc!important;margin-left:3em!important}.bd-main .bd-content #jb-print-toc .section-nav a{text-decoration:none!important}.bd-main .bd-content #jb-print-toc .section-nav li{display:list-item!important}.bd-main .bd-content #jb-print-toc .section-nav .nav{display:none}.bd-main .bd-footer-content{display:none!important}}.bd-header-announcement{background-color:var(--sbt-color-announcement);color:#fff}.bd-main{flex-grow:0}.bd-main .bd-content .bd-article-container{padding:0;overflow-x:unset;min-width:0}.bd-main .bd-content .bd-article-container .bd-article{padding-right:2rem;padding-left:2rem}@media (max-width:768px){.bd-main .bd-content .bd-article-container .bd-article{padding-right:1rem;padding-left:1rem}}.bd-main .bd-content .bd-article-container details.above-input summary,.bd-main .bd-content .bd-article-container details.below-input summary{border-left:3px solid var(--pst-color-primary)}@media (min-width:768px){label.sidebar-toggle.primary-toggle{display:inline-block}}@media (max-width:768px){label.sidebar-toggle.primary-toggle{margin-bottom:0}}@media (min-width:992px){label.sidebar-toggle.secondary-toggle{display:none}}@media (max-width:768px){label.sidebar-toggle.secondary-toggle{margin-bottom:0}}.bd-header-article{display:flex;align-items:center;position:sticky;top:0;background-color:var(--pst-color-background);transition:left .2s;font-size:.9em;padding:0 1rem;z-index:1020}@media (max-width:768px){.bd-header-article{padding:0 .5rem;height:3.5rem}}.scrolled .bd-header-article{box-shadow:0 6px 6px -6px var(--pst-color-shadow)}.bd-header-article .header-article__inner{padding:0}.bd-header-article .header-article-items{display:flex;align-items:center;height:3rem;width:100%}.bd-header-article .header-article-item{display:flex;align-items:center}.bd-header-article .article-header-buttons{display:flex}.bd-header-article .btn{font-size:1.3rem;color:var(--pst-color-text-muted);border:none;padding:0 .5rem}.bd-header-article .btn.show,.bd-header-article .btn:hover{color:var(--pst-color-text-base);border:none}.bd-header-article .btn.show+.dropdown-menu,.bd-header-article 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.sidebar-primary-items__start{border-top:none}@media (min-width:992px){.bd-sidebar-primary{flex-basis:20%}}@media (min-width:992px){input#__primary:checked~.bd-container .bd-sidebar-primary{margin-left:-20%;visibility:hidden;opacity:0}}.bd-sidebar-secondary{top:0}@media (max-width:992px){.bd-sidebar-secondary{z-index:1081}}.bd-sidebar-secondary .sidebar-secondary-items{padding:0;display:flex;gap:.5rem}.bd-sidebar-secondary .sidebar-secondary-items .sidebar-secondary-item{padding-top:0;padding-bottom:0}.bd-sidebar-secondary .onthispage{height:3rem;min-height:3rem;display:flex;align-items:center;margin:0;color:var(--pst-color-muted)}@media (min-width:992px){.bd-sidebar-secondary{background:var(--pst-color-background);height:fit-content;transition:max-height .4s ease;z-index:2;padding:0}.bd-sidebar-secondary .toc-item{border-left-color:var(--pst-color-surface);padding-top:0}.bd-sidebar-secondary .toc-item nav.page-toc{margin-bottom:0;transition:opacity .4s ease}.bd-sidebar-secondary.hide:not(:hover){max-height:3rem;overflow-y:hidden}.scrolled .bd-sidebar-secondary.hide:not(:hover){box-shadow:0 6px 6px -6px rgba(0,0,0,.3)}.bd-sidebar-secondary.hide:not(:hover) .onthispage:after{opacity:1;content:"\f107";font-family:Font Awesome\ 5 Free;font-weight:900;padding-left:.5em;transition:opacity .3s ease}.bd-sidebar-secondary.hide:not(:hover) nav.page-toc{opacity:0}}footer{font-size:var(--sbt-font-size-small-1)}footer.bd-footer-content{display:flex;flex-wrap:wrap;padding:15px;border-top:1px solid #ccc;font-size:87.5%}footer.bd-footer-content .bd-footer-content__inner{padding-left:0}footer.bd-footer-content .bd-footer-content__inner p{margin-bottom:0}.bd-footer-article{padding:0 1rem}@media (max-width:768px){.bd-footer-article{padding:0 .5rem}}.bd-sidebar-primary .navbar-icon-links{column-gap:.5rem}.bd-sidebar-primary .navbar-icon-links .nav-link i,.bd-sidebar-primary .navbar-icon-links .nav-link span{font-size:1.2rem}.bd-sidebar-primary 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a{font-size:1rem}}img{max-width:100%}img.align-center{margin-left:auto;margin-right:auto;display:block}img.align-left{clear:left;float:left;margin-right:1em}img.align-right{clear:right;float:right;margin-left:1em}div.figure{width:100%;margin-bottom:1em;text-align:center}div.figure.align-left{text-align:left}div.figure.align-left p.caption{margin-left:0}div.figure.align-right{text-align:right}div.figure.align-right p.caption{margin-right:0}div.figure p.caption{margin:.5em 10%}div.figure.margin-caption p.caption,div.figure.margin p.caption{margin:.5em 0}div.figure.margin-caption p.caption{text-align:left}div.figure span.caption-number{font-weight:700}div.figure span{font-size:.9rem}label.margin-toggle{margin-bottom:0}label.margin-toggle.marginnote-label{display:none}label.margin-toggle sup{user-select:none}@media 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div>p+p.attribution{font-style:normal;font-size:.9em;text-align:right;color:#6c757d;padding-right:2em}div[class*=highlight-],pre{clear:none}div.cell.tag_output_scroll div.cell_output,div.cell.tag_scroll-input div.cell_input,div.cell.tag_scroll-output div.cell_output{max-height:24em;overflow-y:auto}@media only print{div.utterances,hypothesis-sidebar{display:none}}.thebelab-cell{border:none!important;margin-right:.5em!important}.thebelab-cell .thebelab-input{padding-left:10px!important}.cell.docutils.container{padding-right:0!important}button.thebe-launch-button{height:2.5em;font-size:1em}
\ No newline at end of file
diff --git a/docs/_build/html/_static/styles/theme.css b/docs/_build/html/_static/styles/theme.css
deleted file mode 100644
index 4519dd9..0000000
--- a/docs/_build/html/_static/styles/theme.css
+++ /dev/null
@@ -1,2 +0,0 @@
-/* Provided by Sphinx's 'basic' theme, and included in the final set of assets */
-@import "../basic.css";
diff --git a/docs/_build/html/_static/togglebutton.css b/docs/_build/html/_static/togglebutton.css
deleted file mode 100644
index 54a6787..0000000
--- a/docs/_build/html/_static/togglebutton.css
+++ /dev/null
@@ -1,160 +0,0 @@
-/**
- * Admonition-based toggles
- */
-
-/* Visibility of the target */
-.admonition.toggle .admonition-title ~ * {
-    transition: opacity .3s, height .3s;
-}
-
-/* Toggle buttons inside admonitions so we see the title */
-.admonition.toggle {
-    position: relative;
-}
-
-/* Titles should cut off earlier to avoid overlapping w/ button */
-.admonition.toggle .admonition-title {
-    padding-right: 25%;
-    cursor: pointer;
-}
-
-/* Hovering will cause a slight shift in color to make it feel interactive */
-.admonition.toggle .admonition-title:hover {
-    box-shadow: inset 0 0 0px 20px rgb(0 0 0 / 1%);
-}
-
-/* Hovering will cause a slight shift in color to make it feel interactive */
-.admonition.toggle .admonition-title:active {
-    box-shadow: inset 0 0 0px 20px rgb(0 0 0 / 3%);
-}
-
-/* Remove extra whitespace below the admonition title when hidden */
-.admonition.toggle-hidden {
-    padding-bottom: 0;
-}
-
-.admonition.toggle-hidden .admonition-title {
-    margin-bottom: 0;
-}
-
-/* hides all the content of a page until de-toggled */
-.admonition.toggle-hidden .admonition-title ~ * {
-    height: 0;
-    margin: 0;
-    opacity: 0;
-    visibility: hidden;
-}
-
-/* General button style and position*/
-button.toggle-button {
-    /**
-     * Background and shape. By default there's no background
-     * but users can style as they wish
-     */  
-    background: none;
-    border: none;
-    outline: none;
-
-    /* Positioning just inside the admonition title */
-    position: absolute;
-    right: 0.5em;
-    padding: 0px;
-    border: none;
-    outline: none;
-}
-
-/* Display the toggle hint on wide screens */
-@media (min-width: 768px) {
-    button.toggle-button.toggle-button-hidden:before {
-        content: attr(data-toggle-hint);  /* This will be filled in by JS */
-        font-size: .8em;
-        align-self: center;
-    }
-}
-
-/* Icon behavior */
-.tb-icon {
-    transition: transform .2s ease-out;
-    height: 1.5em;
-    width: 1.5em;
-    stroke: currentColor;  /* So that we inherit the color of other text */
-}
-
-/* The icon should point right when closed, down when open. */
-/* Open */
-.admonition.toggle button .tb-icon {
-    transform: rotate(90deg);
-}
-
-/* Closed */
-.admonition.toggle button.toggle-button-hidden .tb-icon {
-    transform: rotate(0deg);
-}
-
-/* With details toggles, we don't rotate the icon so it points right */
-details.toggle-details .tb-icon {
-    height: 1.4em;
-    width: 1.4em;
-    margin-top: 0.1em;  /* To center the button vertically */
-}
-
-
-/**
- * Details-based toggles.
- * In this case, we wrap elements with `.toggle` in a details block.
- */
-
-/* Details blocks */
-details.toggle-details {
-    margin: 1em 0;
-}
-
-
-details.toggle-details summary {
-    display: flex;
-    align-items: center;
-    cursor: pointer;
-    list-style: none;
-    border-radius: .2em;
-    border-left: 3px solid #1976d2;
-    background-color: rgb(204 204 204 / 10%);
-    padding: 0.2em 0.7em 0.3em 0.5em; /* Less padding on left because the SVG has left margin */
-    font-size: 0.9em;
-}
-
-details.toggle-details summary:hover {
-    background-color: rgb(204 204 204 / 20%);
-}
-
-details.toggle-details summary:active {
-    background: rgb(204 204 204 / 28%);
-}
-
-.toggle-details__summary-text {
-    margin-left: 0.2em;
-}
-
-details.toggle-details[open] summary {
-    margin-bottom: .5em;
-}
-
-details.toggle-details[open] summary .tb-icon {
-    transform: rotate(90deg);
-}
-
-details.toggle-details[open] summary ~ * {
-    animation: toggle-fade-in .3s ease-out;
-}
-
-@keyframes toggle-fade-in {
-  from {opacity: 0%;}
-  to {opacity: 100%;}
-}
-
-/* Print rules - we hide all toggle button elements at print */
-@media print {
-    /* Always hide the summary so the button doesn't show up */
-    details.toggle-details summary {
-        display: none;
-    }
-}
\ No newline at end of file
diff --git a/docs/_build/html/_static/togglebutton.js b/docs/_build/html/_static/togglebutton.js
deleted file mode 100644
index 215a7ee..0000000
--- a/docs/_build/html/_static/togglebutton.js
+++ /dev/null
@@ -1,187 +0,0 @@
-/**
- * Add Toggle Buttons to elements
- */
-
-let toggleChevron = `
-<svg xmlns="http://www.w3.org/2000/svg" class="tb-icon toggle-chevron-right" width="44" height="44" viewBox="0 0 24 24" stroke-width="1.5" stroke="#000000" fill="none" stroke-linecap="round" stroke-linejoin="round">
-<path stroke="none" d="M0 0h24v24H0z" fill="none"/>
-<polyline points="9 6 15 12 9 18" />
-</svg>`;
-
-var initToggleItems = () => {
-  var itemsToToggle = document.querySelectorAll(togglebuttonSelector);
-  console.log(`[togglebutton]: Adding toggle buttons to ${itemsToToggle.length} items`)
-  // Add the button to each admonition and hook up a callback to toggle visibility
-  itemsToToggle.forEach((item, index) => {
-    if (item.classList.contains("admonition")) {
-      // If it's an admonition block, then we'll add a button inside
-      // Generate unique IDs for this item
-      var toggleID = `toggle-${index}`;
-      var buttonID = `button-${toggleID}`;
-
-      item.setAttribute('id', toggleID);
-      if (!item.classList.contains("toggle")){
-        item.classList.add("toggle");
-      }
-      // This is the button that will be added to each item to trigger the toggle
-      var collapseButton = `
-        <button type="button" id="${buttonID}" class="toggle-button" data-target="${toggleID}" data-button="${buttonID}" data-toggle-hint="${toggleHintShow}" aria-label="Toggle hidden content">
-            ${toggleChevron}
-        </button>`;
-
-      title = item.querySelector(".admonition-title")
-      title.insertAdjacentHTML("beforeend", collapseButton);
-      thisButton = document.getElementById(buttonID);
-
-      // Add click handlers for the button + admonition title (if admonition)
-      admonitionTitle = document.querySelector(`#${toggleID} > .admonition-title`)
-      if (admonitionTitle) {
-        // If an admonition, then make the whole title block clickable
-        admonitionTitle.addEventListener('click', toggleClickHandler);
-        admonitionTitle.dataset.target = toggleID
-        admonitionTitle.dataset.button = buttonID
-      } else {
-        // If not an admonition then we'll listen for the button click
-        thisButton.addEventListener('click', toggleClickHandler);
-      }
-
-      // Now hide the item for this toggle button unless explicitly noted to show
-      if (!item.classList.contains("toggle-shown")) {
-        toggleHidden(thisButton);
-      }
-    } else {
-      // If not an admonition, wrap the block in a <details> block
-      // Define the structure of the details block and insert it as a sibling
-      var detailsBlock = `
-        <details class="toggle-details">
-            <summary class="toggle-details__summary">
-              ${toggleChevron}
-              <span class="toggle-details__summary-text">${toggleHintShow}</span>
-            </summary>
-        </details>`;
-      item.insertAdjacentHTML("beforebegin", detailsBlock);
-
-      // Now move the toggle-able content inside of the details block
-      details = item.previousElementSibling
-      details.appendChild(item)
-      item.classList.add("toggle-details__container")
-
-      // Set up a click trigger to change the text as needed
-      details.addEventListener('click', (click) => {
-        let parent = click.target.parentElement;
-        if (parent.tagName.toLowerCase() == "details") {
-          summary = parent.querySelector("summary");
-          details = parent;
-        } else {
-          summary = parent;
-          details = parent.parentElement;
-        }
-        // Update the inner text for the proper hint
-        if (details.open) {
-          summary.querySelector("span.toggle-details__summary-text").innerText = toggleHintShow;
-        } else {
-          summary.querySelector("span.toggle-details__summary-text").innerText = toggleHintHide;
-        }
-        
-      });
-
-      // If we have a toggle-shown class, open details block should be open
-      if (item.classList.contains("toggle-shown")) {
-        details.click();
-      }
-    }
-  })
-};
-
-// This should simply add / remove the collapsed class and change the button text
-var toggleHidden = (button) => {
-  target = button.dataset['target']
-  var itemToToggle = document.getElementById(target);
-  if (itemToToggle.classList.contains("toggle-hidden")) {
-    itemToToggle.classList.remove("toggle-hidden");
-    button.classList.remove("toggle-button-hidden");
-  } else {
-    itemToToggle.classList.add("toggle-hidden");
-    button.classList.add("toggle-button-hidden");
-  }
-}
-
-var toggleClickHandler = (click) => {
-  // Be cause the admonition title is clickable and extends to the whole admonition
-  // We only look for a click event on this title to trigger the toggle.
-
-  if (click.target.classList.contains("admonition-title")) {
-    button = click.target.querySelector(".toggle-button");
-  } else if (click.target.classList.contains("tb-icon")) {
-    // We've clicked the icon and need to search up one parent for the button
-    button = click.target.parentElement;
-  } else if (click.target.tagName == "polyline") {
-    // We've clicked the SVG elements inside the button, need to up 2 layers
-    button = click.target.parentElement.parentElement;
-  } else if (click.target.classList.contains("toggle-button")) {
-    // We've clicked the button itself and so don't need to do anything
-    button = click.target;
-  } else {
-    console.log(`[togglebutton]: Couldn't find button for ${click.target}`)
-  }
-  target = document.getElementById(button.dataset['button']);
-  toggleHidden(target);
-}
-
-// If we want to blanket-add toggle classes to certain cells
-var addToggleToSelector = () => {
-  const selector = "";
-  if (selector.length > 0) {
-    document.querySelectorAll(selector).forEach((item) => {
-      item.classList.add("toggle");
-    })
-  }
-}
-
-// Helper function to run when the DOM is finished
-const sphinxToggleRunWhenDOMLoaded = cb => {
-  if (document.readyState != 'loading') {
-    cb()
-  } else if (document.addEventListener) {
-    document.addEventListener('DOMContentLoaded', cb)
-  } else {
-    document.attachEvent('onreadystatechange', function() {
-      if (document.readyState == 'complete') cb()
-    })
-  }
-}
-sphinxToggleRunWhenDOMLoaded(addToggleToSelector)
-sphinxToggleRunWhenDOMLoaded(initToggleItems)
-
-/** Toggle details blocks to be open when printing */
-if (toggleOpenOnPrint == "true") {
-  window.addEventListener("beforeprint", () => {
-    // Open the details
-    document.querySelectorAll("details.toggle-details").forEach((el) => {
-      el.dataset["togglestatus"] = el.open;
-      el.open = true;
-    });
-  
-    // Open the admonitions
-    document.querySelectorAll(".admonition.toggle.toggle-hidden").forEach((el) => {
-      console.log(el);
-      el.querySelector("button.toggle-button").click();
-      el.dataset["toggle_after_print"] = "true";
-    });
-  });
-  window.addEventListener("afterprint", () => {
-    // Re-close the details that were closed
-    document.querySelectorAll("details.toggle-details").forEach((el) => {
-      el.open = el.dataset["togglestatus"] == "true";
-      delete el.dataset["togglestatus"];
-    });
-  
-    // Re-close the admonition toggle buttons
-    document.querySelectorAll(".admonition.toggle").forEach((el) => {
-      if (el.dataset["toggle_after_print"] == "true") {
-        el.querySelector("button.toggle-button").click();
-        delete el.dataset["toggle_after_print"];
-      }
-    });
-  });
-}
diff --git a/docs/_build/html/_static/underscore-1.13.1.js b/docs/_build/html/_static/underscore-1.13.1.js
deleted file mode 100644
index ffd77af..0000000
--- a/docs/_build/html/_static/underscore-1.13.1.js
+++ /dev/null
@@ -1,2042 +0,0 @@
-(function (global, factory) {
-  typeof exports === 'object' && typeof module !== 'undefined' ? module.exports = factory() :
-  typeof define === 'function' && define.amd ? define('underscore', factory) :
-  (global = typeof globalThis !== 'undefined' ? globalThis : global || self, (function () {
-    var current = global._;
-    var exports = global._ = factory();
-    exports.noConflict = function () { global._ = current; return exports; };
-  }()));
-}(this, (function () {
-  //     Underscore.js 1.13.1
-  //     https://underscorejs.org
-  //     (c) 2009-2021 Jeremy Ashkenas, Julian Gonggrijp, and DocumentCloud and Investigative Reporters & Editors
-  //     Underscore may be freely distributed under the MIT license.
-
-  // Current version.
-  var VERSION = '1.13.1';
-
-  // Establish the root object, `window` (`self`) in the browser, `global`
-  // on the server, or `this` in some virtual machines. We use `self`
-  // instead of `window` for `WebWorker` support.
-  var root = typeof self == 'object' && self.self === self && self ||
-            typeof global == 'object' && global.global === global && global ||
-            Function('return this')() ||
-            {};
-
-  // Save bytes in the minified (but not gzipped) version:
-  var ArrayProto = Array.prototype, ObjProto = Object.prototype;
-  var SymbolProto = typeof Symbol !== 'undefined' ? Symbol.prototype : null;
-
-  // Create quick reference variables for speed access to core prototypes.
-  var push = ArrayProto.push,
-      slice = ArrayProto.slice,
-      toString = ObjProto.toString,
-      hasOwnProperty = ObjProto.hasOwnProperty;
-
-  // Modern feature detection.
-  var supportsArrayBuffer = typeof ArrayBuffer !== 'undefined',
-      supportsDataView = typeof DataView !== 'undefined';
-
-  // All **ECMAScript 5+** native function implementations that we hope to use
-  // are declared here.
-  var nativeIsArray = Array.isArray,
-      nativeKeys = Object.keys,
-      nativeCreate = Object.create,
-      nativeIsView = supportsArrayBuffer && ArrayBuffer.isView;
-
-  // Create references to these builtin functions because we override them.
-  var _isNaN = isNaN,
-      _isFinite = isFinite;
-
-  // Keys in IE < 9 that won't be iterated by `for key in ...` and thus missed.
-  var hasEnumBug = !{toString: null}.propertyIsEnumerable('toString');
-  var nonEnumerableProps = ['valueOf', 'isPrototypeOf', 'toString',
-    'propertyIsEnumerable', 'hasOwnProperty', 'toLocaleString'];
-
-  // The largest integer that can be represented exactly.
-  var MAX_ARRAY_INDEX = Math.pow(2, 53) - 1;
-
-  // Some functions take a variable number of arguments, or a few expected
-  // arguments at the beginning and then a variable number of values to operate
-  // on. This helper accumulates all remaining arguments past the function’s
-  // argument length (or an explicit `startIndex`), into an array that becomes
-  // the last argument. Similar to ES6’s "rest parameter".
-  function restArguments(func, startIndex) {
-    startIndex = startIndex == null ? func.length - 1 : +startIndex;
-    return function() {
-      var length = Math.max(arguments.length - startIndex, 0),
-          rest = Array(length),
-          index = 0;
-      for (; index < length; index++) {
-        rest[index] = arguments[index + startIndex];
-      }
-      switch (startIndex) {
-        case 0: return func.call(this, rest);
-        case 1: return func.call(this, arguments[0], rest);
-        case 2: return func.call(this, arguments[0], arguments[1], rest);
-      }
-      var args = Array(startIndex + 1);
-      for (index = 0; index < startIndex; index++) {
-        args[index] = arguments[index];
-      }
-      args[startIndex] = rest;
-      return func.apply(this, args);
-    };
-  }
-
-  // Is a given variable an object?
-  function isObject(obj) {
-    var type = typeof obj;
-    return type === 'function' || type === 'object' && !!obj;
-  }
-
-  // Is a given value equal to null?
-  function isNull(obj) {
-    return obj === null;
-  }
-
-  // Is a given variable undefined?
-  function isUndefined(obj) {
-    return obj === void 0;
-  }
-
-  // Is a given value a boolean?
-  function isBoolean(obj) {
-    return obj === true || obj === false || toString.call(obj) === '[object Boolean]';
-  }
-
-  // Is a given value a DOM element?
-  function isElement(obj) {
-    return !!(obj && obj.nodeType === 1);
-  }
-
-  // Internal function for creating a `toString`-based type tester.
-  function tagTester(name) {
-    var tag = '[object ' + name + ']';
-    return function(obj) {
-      return toString.call(obj) === tag;
-    };
-  }
-
-  var isString = tagTester('String');
-
-  var isNumber = tagTester('Number');
-
-  var isDate = tagTester('Date');
-
-  var isRegExp = tagTester('RegExp');
-
-  var isError = tagTester('Error');
-
-  var isSymbol = tagTester('Symbol');
-
-  var isArrayBuffer = tagTester('ArrayBuffer');
-
-  var isFunction = tagTester('Function');
-
-  // Optimize `isFunction` if appropriate. Work around some `typeof` bugs in old
-  // v8, IE 11 (#1621), Safari 8 (#1929), and PhantomJS (#2236).
-  var nodelist = root.document && root.document.childNodes;
-  if (typeof /./ != 'function' && typeof Int8Array != 'object' && typeof nodelist != 'function') {
-    isFunction = function(obj) {
-      return typeof obj == 'function' || false;
-    };
-  }
-
-  var isFunction$1 = isFunction;
-
-  var hasObjectTag = tagTester('Object');
-
-  // In IE 10 - Edge 13, `DataView` has string tag `'[object Object]'`.
-  // In IE 11, the most common among them, this problem also applies to
-  // `Map`, `WeakMap` and `Set`.
-  var hasStringTagBug = (
-        supportsDataView && hasObjectTag(new DataView(new ArrayBuffer(8)))
-      ),
-      isIE11 = (typeof Map !== 'undefined' && hasObjectTag(new Map));
-
-  var isDataView = tagTester('DataView');
-
-  // In IE 10 - Edge 13, we need a different heuristic
-  // to determine whether an object is a `DataView`.
-  function ie10IsDataView(obj) {
-    return obj != null && isFunction$1(obj.getInt8) && isArrayBuffer(obj.buffer);
-  }
-
-  var isDataView$1 = (hasStringTagBug ? ie10IsDataView : isDataView);
-
-  // Is a given value an array?
-  // Delegates to ECMA5's native `Array.isArray`.
-  var isArray = nativeIsArray || tagTester('Array');
-
-  // Internal function to check whether `key` is an own property name of `obj`.
-  function has$1(obj, key) {
-    return obj != null && hasOwnProperty.call(obj, key);
-  }
-
-  var isArguments = tagTester('Arguments');
-
-  // Define a fallback version of the method in browsers (ahem, IE < 9), where
-  // there isn't any inspectable "Arguments" type.
-  (function() {
-    if (!isArguments(arguments)) {
-      isArguments = function(obj) {
-        return has$1(obj, 'callee');
-      };
-    }
-  }());
-
-  var isArguments$1 = isArguments;
-
-  // Is a given object a finite number?
-  function isFinite$1(obj) {
-    return !isSymbol(obj) && _isFinite(obj) && !isNaN(parseFloat(obj));
-  }
-
-  // Is the given value `NaN`?
-  function isNaN$1(obj) {
-    return isNumber(obj) && _isNaN(obj);
-  }
-
-  // Predicate-generating function. Often useful outside of Underscore.
-  function constant(value) {
-    return function() {
-      return value;
-    };
-  }
-
-  // Common internal logic for `isArrayLike` and `isBufferLike`.
-  function createSizePropertyCheck(getSizeProperty) {
-    return function(collection) {
-      var sizeProperty = getSizeProperty(collection);
-      return typeof sizeProperty == 'number' && sizeProperty >= 0 && sizeProperty <= MAX_ARRAY_INDEX;
-    }
-  }
-
-  // Internal helper to generate a function to obtain property `key` from `obj`.
-  function shallowProperty(key) {
-    return function(obj) {
-      return obj == null ? void 0 : obj[key];
-    };
-  }
-
-  // Internal helper to obtain the `byteLength` property of an object.
-  var getByteLength = shallowProperty('byteLength');
-
-  // Internal helper to determine whether we should spend extensive checks against
-  // `ArrayBuffer` et al.
-  var isBufferLike = createSizePropertyCheck(getByteLength);
-
-  // Is a given value a typed array?
-  var typedArrayPattern = /\[object ((I|Ui)nt(8|16|32)|Float(32|64)|Uint8Clamped|Big(I|Ui)nt64)Array\]/;
-  function isTypedArray(obj) {
-    // `ArrayBuffer.isView` is the most future-proof, so use it when available.
-    // Otherwise, fall back on the above regular expression.
-    return nativeIsView ? (nativeIsView(obj) && !isDataView$1(obj)) :
-                  isBufferLike(obj) && typedArrayPattern.test(toString.call(obj));
-  }
-
-  var isTypedArray$1 = supportsArrayBuffer ? isTypedArray : constant(false);
-
-  // Internal helper to obtain the `length` property of an object.
-  var getLength = shallowProperty('length');
-
-  // Internal helper to create a simple lookup structure.
-  // `collectNonEnumProps` used to depend on `_.contains`, but this led to
-  // circular imports. `emulatedSet` is a one-off solution that only works for
-  // arrays of strings.
-  function emulatedSet(keys) {
-    var hash = {};
-    for (var l = keys.length, i = 0; i < l; ++i) hash[keys[i]] = true;
-    return {
-      contains: function(key) { return hash[key]; },
-      push: function(key) {
-        hash[key] = true;
-        return keys.push(key);
-      }
-    };
-  }
-
-  // Internal helper. Checks `keys` for the presence of keys in IE < 9 that won't
-  // be iterated by `for key in ...` and thus missed. Extends `keys` in place if
-  // needed.
-  function collectNonEnumProps(obj, keys) {
-    keys = emulatedSet(keys);
-    var nonEnumIdx = nonEnumerableProps.length;
-    var constructor = obj.constructor;
-    var proto = isFunction$1(constructor) && constructor.prototype || ObjProto;
-
-    // Constructor is a special case.
-    var prop = 'constructor';
-    if (has$1(obj, prop) && !keys.contains(prop)) keys.push(prop);
-
-    while (nonEnumIdx--) {
-      prop = nonEnumerableProps[nonEnumIdx];
-      if (prop in obj && obj[prop] !== proto[prop] && !keys.contains(prop)) {
-        keys.push(prop);
-      }
-    }
-  }
-
-  // Retrieve the names of an object's own properties.
-  // Delegates to **ECMAScript 5**'s native `Object.keys`.
-  function keys(obj) {
-    if (!isObject(obj)) return [];
-    if (nativeKeys) return nativeKeys(obj);
-    var keys = [];
-    for (var key in obj) if (has$1(obj, key)) keys.push(key);
-    // Ahem, IE < 9.
-    if (hasEnumBug) collectNonEnumProps(obj, keys);
-    return keys;
-  }
-
-  // Is a given array, string, or object empty?
-  // An "empty" object has no enumerable own-properties.
-  function isEmpty(obj) {
-    if (obj == null) return true;
-    // Skip the more expensive `toString`-based type checks if `obj` has no
-    // `.length`.
-    var length = getLength(obj);
-    if (typeof length == 'number' && (
-      isArray(obj) || isString(obj) || isArguments$1(obj)
-    )) return length === 0;
-    return getLength(keys(obj)) === 0;
-  }
-
-  // Returns whether an object has a given set of `key:value` pairs.
-  function isMatch(object, attrs) {
-    var _keys = keys(attrs), length = _keys.length;
-    if (object == null) return !length;
-    var obj = Object(object);
-    for (var i = 0; i < length; i++) {
-      var key = _keys[i];
-      if (attrs[key] !== obj[key] || !(key in obj)) return false;
-    }
-    return true;
-  }
-
-  // If Underscore is called as a function, it returns a wrapped object that can
-  // be used OO-style. This wrapper holds altered versions of all functions added
-  // through `_.mixin`. Wrapped objects may be chained.
-  function _$1(obj) {
-    if (obj instanceof _$1) return obj;
-    if (!(this instanceof _$1)) return new _$1(obj);
-    this._wrapped = obj;
-  }
-
-  _$1.VERSION = VERSION;
-
-  // Extracts the result from a wrapped and chained object.
-  _$1.prototype.value = function() {
-    return this._wrapped;
-  };
-
-  // Provide unwrapping proxies for some methods used in engine operations
-  // such as arithmetic and JSON stringification.
-  _$1.prototype.valueOf = _$1.prototype.toJSON = _$1.prototype.value;
-
-  _$1.prototype.toString = function() {
-    return String(this._wrapped);
-  };
-
-  // Internal function to wrap or shallow-copy an ArrayBuffer,
-  // typed array or DataView to a new view, reusing the buffer.
-  function toBufferView(bufferSource) {
-    return new Uint8Array(
-      bufferSource.buffer || bufferSource,
-      bufferSource.byteOffset || 0,
-      getByteLength(bufferSource)
-    );
-  }
-
-  // We use this string twice, so give it a name for minification.
-  var tagDataView = '[object DataView]';
-
-  // Internal recursive comparison function for `_.isEqual`.
-  function eq(a, b, aStack, bStack) {
-    // Identical objects are equal. `0 === -0`, but they aren't identical.
-    // See the [Harmony `egal` proposal](https://wiki.ecmascript.org/doku.php?id=harmony:egal).
-    if (a === b) return a !== 0 || 1 / a === 1 / b;
-    // `null` or `undefined` only equal to itself (strict comparison).
-    if (a == null || b == null) return false;
-    // `NaN`s are equivalent, but non-reflexive.
-    if (a !== a) return b !== b;
-    // Exhaust primitive checks
-    var type = typeof a;
-    if (type !== 'function' && type !== 'object' && typeof b != 'object') return false;
-    return deepEq(a, b, aStack, bStack);
-  }
-
-  // Internal recursive comparison function for `_.isEqual`.
-  function deepEq(a, b, aStack, bStack) {
-    // Unwrap any wrapped objects.
-    if (a instanceof _$1) a = a._wrapped;
-    if (b instanceof _$1) b = b._wrapped;
-    // Compare `[[Class]]` names.
-    var className = toString.call(a);
-    if (className !== toString.call(b)) return false;
-    // Work around a bug in IE 10 - Edge 13.
-    if (hasStringTagBug && className == '[object Object]' && isDataView$1(a)) {
-      if (!isDataView$1(b)) return false;
-      className = tagDataView;
-    }
-    switch (className) {
-      // These types are compared by value.
-      case '[object RegExp]':
-        // RegExps are coerced to strings for comparison (Note: '' + /a/i === '/a/i')
-      case '[object String]':
-        // Primitives and their corresponding object wrappers are equivalent; thus, `"5"` is
-        // equivalent to `new String("5")`.
-        return '' + a === '' + b;
-      case '[object Number]':
-        // `NaN`s are equivalent, but non-reflexive.
-        // Object(NaN) is equivalent to NaN.
-        if (+a !== +a) return +b !== +b;
-        // An `egal` comparison is performed for other numeric values.
-        return +a === 0 ? 1 / +a === 1 / b : +a === +b;
-      case '[object Date]':
-      case '[object Boolean]':
-        // Coerce dates and booleans to numeric primitive values. Dates are compared by their
-        // millisecond representations. Note that invalid dates with millisecond representations
-        // of `NaN` are not equivalent.
-        return +a === +b;
-      case '[object Symbol]':
-        return SymbolProto.valueOf.call(a) === SymbolProto.valueOf.call(b);
-      case '[object ArrayBuffer]':
-      case tagDataView:
-        // Coerce to typed array so we can fall through.
-        return deepEq(toBufferView(a), toBufferView(b), aStack, bStack);
-    }
-
-    var areArrays = className === '[object Array]';
-    if (!areArrays && isTypedArray$1(a)) {
-        var byteLength = getByteLength(a);
-        if (byteLength !== getByteLength(b)) return false;
-        if (a.buffer === b.buffer && a.byteOffset === b.byteOffset) return true;
-        areArrays = true;
-    }
-    if (!areArrays) {
-      if (typeof a != 'object' || typeof b != 'object') return false;
-
-      // Objects with different constructors are not equivalent, but `Object`s or `Array`s
-      // from different frames are.
-      var aCtor = a.constructor, bCtor = b.constructor;
-      if (aCtor !== bCtor && !(isFunction$1(aCtor) && aCtor instanceof aCtor &&
-                               isFunction$1(bCtor) && bCtor instanceof bCtor)
-                          && ('constructor' in a && 'constructor' in b)) {
-        return false;
-      }
-    }
-    // Assume equality for cyclic structures. The algorithm for detecting cyclic
-    // structures is adapted from ES 5.1 section 15.12.3, abstract operation `JO`.
-
-    // Initializing stack of traversed objects.
-    // It's done here since we only need them for objects and arrays comparison.
-    aStack = aStack || [];
-    bStack = bStack || [];
-    var length = aStack.length;
-    while (length--) {
-      // Linear search. Performance is inversely proportional to the number of
-      // unique nested structures.
-      if (aStack[length] === a) return bStack[length] === b;
-    }
-
-    // Add the first object to the stack of traversed objects.
-    aStack.push(a);
-    bStack.push(b);
-
-    // Recursively compare objects and arrays.
-    if (areArrays) {
-      // Compare array lengths to determine if a deep comparison is necessary.
-      length = a.length;
-      if (length !== b.length) return false;
-      // Deep compare the contents, ignoring non-numeric properties.
-      while (length--) {
-        if (!eq(a[length], b[length], aStack, bStack)) return false;
-      }
-    } else {
-      // Deep compare objects.
-      var _keys = keys(a), key;
-      length = _keys.length;
-      // Ensure that both objects contain the same number of properties before comparing deep equality.
-      if (keys(b).length !== length) return false;
-      while (length--) {
-        // Deep compare each member
-        key = _keys[length];
-        if (!(has$1(b, key) && eq(a[key], b[key], aStack, bStack))) return false;
-      }
-    }
-    // Remove the first object from the stack of traversed objects.
-    aStack.pop();
-    bStack.pop();
-    return true;
-  }
-
-  // Perform a deep comparison to check if two objects are equal.
-  function isEqual(a, b) {
-    return eq(a, b);
-  }
-
-  // Retrieve all the enumerable property names of an object.
-  function allKeys(obj) {
-    if (!isObject(obj)) return [];
-    var keys = [];
-    for (var key in obj) keys.push(key);
-    // Ahem, IE < 9.
-    if (hasEnumBug) collectNonEnumProps(obj, keys);
-    return keys;
-  }
-
-  // Since the regular `Object.prototype.toString` type tests don't work for
-  // some types in IE 11, we use a fingerprinting heuristic instead, based
-  // on the methods. It's not great, but it's the best we got.
-  // The fingerprint method lists are defined below.
-  function ie11fingerprint(methods) {
-    var length = getLength(methods);
-    return function(obj) {
-      if (obj == null) return false;
-      // `Map`, `WeakMap` and `Set` have no enumerable keys.
-      var keys = allKeys(obj);
-      if (getLength(keys)) return false;
-      for (var i = 0; i < length; i++) {
-        if (!isFunction$1(obj[methods[i]])) return false;
-      }
-      // If we are testing against `WeakMap`, we need to ensure that
-      // `obj` doesn't have a `forEach` method in order to distinguish
-      // it from a regular `Map`.
-      return methods !== weakMapMethods || !isFunction$1(obj[forEachName]);
-    };
-  }
-
-  // In the interest of compact minification, we write
-  // each string in the fingerprints only once.
-  var forEachName = 'forEach',
-      hasName = 'has',
-      commonInit = ['clear', 'delete'],
-      mapTail = ['get', hasName, 'set'];
-
-  // `Map`, `WeakMap` and `Set` each have slightly different
-  // combinations of the above sublists.
-  var mapMethods = commonInit.concat(forEachName, mapTail),
-      weakMapMethods = commonInit.concat(mapTail),
-      setMethods = ['add'].concat(commonInit, forEachName, hasName);
-
-  var isMap = isIE11 ? ie11fingerprint(mapMethods) : tagTester('Map');
-
-  var isWeakMap = isIE11 ? ie11fingerprint(weakMapMethods) : tagTester('WeakMap');
-
-  var isSet = isIE11 ? ie11fingerprint(setMethods) : tagTester('Set');
-
-  var isWeakSet = tagTester('WeakSet');
-
-  // Retrieve the values of an object's properties.
-  function values(obj) {
-    var _keys = keys(obj);
-    var length = _keys.length;
-    var values = Array(length);
-    for (var i = 0; i < length; i++) {
-      values[i] = obj[_keys[i]];
-    }
-    return values;
-  }
-
-  // Convert an object into a list of `[key, value]` pairs.
-  // The opposite of `_.object` with one argument.
-  function pairs(obj) {
-    var _keys = keys(obj);
-    var length = _keys.length;
-    var pairs = Array(length);
-    for (var i = 0; i < length; i++) {
-      pairs[i] = [_keys[i], obj[_keys[i]]];
-    }
-    return pairs;
-  }
-
-  // Invert the keys and values of an object. The values must be serializable.
-  function invert(obj) {
-    var result = {};
-    var _keys = keys(obj);
-    for (var i = 0, length = _keys.length; i < length; i++) {
-      result[obj[_keys[i]]] = _keys[i];
-    }
-    return result;
-  }
-
-  // Return a sorted list of the function names available on the object.
-  function functions(obj) {
-    var names = [];
-    for (var key in obj) {
-      if (isFunction$1(obj[key])) names.push(key);
-    }
-    return names.sort();
-  }
-
-  // An internal function for creating assigner functions.
-  function createAssigner(keysFunc, defaults) {
-    return function(obj) {
-      var length = arguments.length;
-      if (defaults) obj = Object(obj);
-      if (length < 2 || obj == null) return obj;
-      for (var index = 1; index < length; index++) {
-        var source = arguments[index],
-            keys = keysFunc(source),
-            l = keys.length;
-        for (var i = 0; i < l; i++) {
-          var key = keys[i];
-          if (!defaults || obj[key] === void 0) obj[key] = source[key];
-        }
-      }
-      return obj;
-    };
-  }
-
-  // Extend a given object with all the properties in passed-in object(s).
-  var extend = createAssigner(allKeys);
-
-  // Assigns a given object with all the own properties in the passed-in
-  // object(s).
-  // (https://developer.mozilla.org/docs/Web/JavaScript/Reference/Global_Objects/Object/assign)
-  var extendOwn = createAssigner(keys);
-
-  // Fill in a given object with default properties.
-  var defaults = createAssigner(allKeys, true);
-
-  // Create a naked function reference for surrogate-prototype-swapping.
-  function ctor() {
-    return function(){};
-  }
-
-  // An internal function for creating a new object that inherits from another.
-  function baseCreate(prototype) {
-    if (!isObject(prototype)) return {};
-    if (nativeCreate) return nativeCreate(prototype);
-    var Ctor = ctor();
-    Ctor.prototype = prototype;
-    var result = new Ctor;
-    Ctor.prototype = null;
-    return result;
-  }
-
-  // Creates an object that inherits from the given prototype object.
-  // If additional properties are provided then they will be added to the
-  // created object.
-  function create(prototype, props) {
-    var result = baseCreate(prototype);
-    if (props) extendOwn(result, props);
-    return result;
-  }
-
-  // Create a (shallow-cloned) duplicate of an object.
-  function clone(obj) {
-    if (!isObject(obj)) return obj;
-    return isArray(obj) ? obj.slice() : extend({}, obj);
-  }
-
-  // Invokes `interceptor` with the `obj` and then returns `obj`.
-  // The primary purpose of this method is to "tap into" a method chain, in
-  // order to perform operations on intermediate results within the chain.
-  function tap(obj, interceptor) {
-    interceptor(obj);
-    return obj;
-  }
-
-  // Normalize a (deep) property `path` to array.
-  // Like `_.iteratee`, this function can be customized.
-  function toPath$1(path) {
-    return isArray(path) ? path : [path];
-  }
-  _$1.toPath = toPath$1;
-
-  // Internal wrapper for `_.toPath` to enable minification.
-  // Similar to `cb` for `_.iteratee`.
-  function toPath(path) {
-    return _$1.toPath(path);
-  }
-
-  // Internal function to obtain a nested property in `obj` along `path`.
-  function deepGet(obj, path) {
-    var length = path.length;
-    for (var i = 0; i < length; i++) {
-      if (obj == null) return void 0;
-      obj = obj[path[i]];
-    }
-    return length ? obj : void 0;
-  }
-
-  // Get the value of the (deep) property on `path` from `object`.
-  // If any property in `path` does not exist or if the value is
-  // `undefined`, return `defaultValue` instead.
-  // The `path` is normalized through `_.toPath`.
-  function get(object, path, defaultValue) {
-    var value = deepGet(object, toPath(path));
-    return isUndefined(value) ? defaultValue : value;
-  }
-
-  // Shortcut function for checking if an object has a given property directly on
-  // itself (in other words, not on a prototype). Unlike the internal `has`
-  // function, this public version can also traverse nested properties.
-  function has(obj, path) {
-    path = toPath(path);
-    var length = path.length;
-    for (var i = 0; i < length; i++) {
-      var key = path[i];
-      if (!has$1(obj, key)) return false;
-      obj = obj[key];
-    }
-    return !!length;
-  }
-
-  // Keep the identity function around for default iteratees.
-  function identity(value) {
-    return value;
-  }
-
-  // Returns a predicate for checking whether an object has a given set of
-  // `key:value` pairs.
-  function matcher(attrs) {
-    attrs = extendOwn({}, attrs);
-    return function(obj) {
-      return isMatch(obj, attrs);
-    };
-  }
-
-  // Creates a function that, when passed an object, will traverse that object’s
-  // properties down the given `path`, specified as an array of keys or indices.
-  function property(path) {
-    path = toPath(path);
-    return function(obj) {
-      return deepGet(obj, path);
-    };
-  }
-
-  // Internal function that returns an efficient (for current engines) version
-  // of the passed-in callback, to be repeatedly applied in other Underscore
-  // functions.
-  function optimizeCb(func, context, argCount) {
-    if (context === void 0) return func;
-    switch (argCount == null ? 3 : argCount) {
-      case 1: return function(value) {
-        return func.call(context, value);
-      };
-      // The 2-argument case is omitted because we’re not using it.
-      case 3: return function(value, index, collection) {
-        return func.call(context, value, index, collection);
-      };
-      case 4: return function(accumulator, value, index, collection) {
-        return func.call(context, accumulator, value, index, collection);
-      };
-    }
-    return function() {
-      return func.apply(context, arguments);
-    };
-  }
-
-  // An internal function to generate callbacks that can be applied to each
-  // element in a collection, returning the desired result — either `_.identity`,
-  // an arbitrary callback, a property matcher, or a property accessor.
-  function baseIteratee(value, context, argCount) {
-    if (value == null) return identity;
-    if (isFunction$1(value)) return optimizeCb(value, context, argCount);
-    if (isObject(value) && !isArray(value)) return matcher(value);
-    return property(value);
-  }
-
-  // External wrapper for our callback generator. Users may customize
-  // `_.iteratee` if they want additional predicate/iteratee shorthand styles.
-  // This abstraction hides the internal-only `argCount` argument.
-  function iteratee(value, context) {
-    return baseIteratee(value, context, Infinity);
-  }
-  _$1.iteratee = iteratee;
-
-  // The function we call internally to generate a callback. It invokes
-  // `_.iteratee` if overridden, otherwise `baseIteratee`.
-  function cb(value, context, argCount) {
-    if (_$1.iteratee !== iteratee) return _$1.iteratee(value, context);
-    return baseIteratee(value, context, argCount);
-  }
-
-  // Returns the results of applying the `iteratee` to each element of `obj`.
-  // In contrast to `_.map` it returns an object.
-  function mapObject(obj, iteratee, context) {
-    iteratee = cb(iteratee, context);
-    var _keys = keys(obj),
-        length = _keys.length,
-        results = {};
-    for (var index = 0; index < length; index++) {
-      var currentKey = _keys[index];
-      results[currentKey] = iteratee(obj[currentKey], currentKey, obj);
-    }
-    return results;
-  }
-
-  // Predicate-generating function. Often useful outside of Underscore.
-  function noop(){}
-
-  // Generates a function for a given object that returns a given property.
-  function propertyOf(obj) {
-    if (obj == null) return noop;
-    return function(path) {
-      return get(obj, path);
-    };
-  }
-
-  // Run a function **n** times.
-  function times(n, iteratee, context) {
-    var accum = Array(Math.max(0, n));
-    iteratee = optimizeCb(iteratee, context, 1);
-    for (var i = 0; i < n; i++) accum[i] = iteratee(i);
-    return accum;
-  }
-
-  // Return a random integer between `min` and `max` (inclusive).
-  function random(min, max) {
-    if (max == null) {
-      max = min;
-      min = 0;
-    }
-    return min + Math.floor(Math.random() * (max - min + 1));
-  }
-
-  // A (possibly faster) way to get the current timestamp as an integer.
-  var now = Date.now || function() {
-    return new Date().getTime();
-  };
-
-  // Internal helper to generate functions for escaping and unescaping strings
-  // to/from HTML interpolation.
-  function createEscaper(map) {
-    var escaper = function(match) {
-      return map[match];
-    };
-    // Regexes for identifying a key that needs to be escaped.
-    var source = '(?:' + keys(map).join('|') + ')';
-    var testRegexp = RegExp(source);
-    var replaceRegexp = RegExp(source, 'g');
-    return function(string) {
-      string = string == null ? '' : '' + string;
-      return testRegexp.test(string) ? string.replace(replaceRegexp, escaper) : string;
-    };
-  }
-
-  // Internal list of HTML entities for escaping.
-  var escapeMap = {
-    '&': '&amp;',
-    '<': '&lt;',
-    '>': '&gt;',
-    '"': '&quot;',
-    "'": '&#x27;',
-    '`': '&#x60;'
-  };
-
-  // Function for escaping strings to HTML interpolation.
-  var _escape = createEscaper(escapeMap);
-
-  // Internal list of HTML entities for unescaping.
-  var unescapeMap = invert(escapeMap);
-
-  // Function for unescaping strings from HTML interpolation.
-  var _unescape = createEscaper(unescapeMap);
-
-  // By default, Underscore uses ERB-style template delimiters. Change the
-  // following template settings to use alternative delimiters.
-  var templateSettings = _$1.templateSettings = {
-    evaluate: /<%([\s\S]+?)%>/g,
-    interpolate: /<%=([\s\S]+?)%>/g,
-    escape: /<%-([\s\S]+?)%>/g
-  };
-
-  // When customizing `_.templateSettings`, if you don't want to define an
-  // interpolation, evaluation or escaping regex, we need one that is
-  // guaranteed not to match.
-  var noMatch = /(.)^/;
-
-  // Certain characters need to be escaped so that they can be put into a
-  // string literal.
-  var escapes = {
-    "'": "'",
-    '\\': '\\',
-    '\r': 'r',
-    '\n': 'n',
-    '\u2028': 'u2028',
-    '\u2029': 'u2029'
-  };
-
-  var escapeRegExp = /\\|'|\r|\n|\u2028|\u2029/g;
-
-  function escapeChar(match) {
-    return '\\' + escapes[match];
-  }
-
-  // In order to prevent third-party code injection through
-  // `_.templateSettings.variable`, we test it against the following regular
-  // expression. It is intentionally a bit more liberal than just matching valid
-  // identifiers, but still prevents possible loopholes through defaults or
-  // destructuring assignment.
-  var bareIdentifier = /^\s*(\w|\$)+\s*$/;
-
-  // JavaScript micro-templating, similar to John Resig's implementation.
-  // Underscore templating handles arbitrary delimiters, preserves whitespace,
-  // and correctly escapes quotes within interpolated code.
-  // NB: `oldSettings` only exists for backwards compatibility.
-  function template(text, settings, oldSettings) {
-    if (!settings && oldSettings) settings = oldSettings;
-    settings = defaults({}, settings, _$1.templateSettings);
-
-    // Combine delimiters into one regular expression via alternation.
-    var matcher = RegExp([
-      (settings.escape || noMatch).source,
-      (settings.interpolate || noMatch).source,
-      (settings.evaluate || noMatch).source
-    ].join('|') + '|$', 'g');
-
-    // Compile the template source, escaping string literals appropriately.
-    var index = 0;
-    var source = "__p+='";
-    text.replace(matcher, function(match, escape, interpolate, evaluate, offset) {
-      source += text.slice(index, offset).replace(escapeRegExp, escapeChar);
-      index = offset + match.length;
-
-      if (escape) {
-        source += "'+\n((__t=(" + escape + "))==null?'':_.escape(__t))+\n'";
-      } else if (interpolate) {
-        source += "'+\n((__t=(" + interpolate + "))==null?'':__t)+\n'";
-      } else if (evaluate) {
-        source += "';\n" + evaluate + "\n__p+='";
-      }
-
-      // Adobe VMs need the match returned to produce the correct offset.
-      return match;
-    });
-    source += "';\n";
-
-    var argument = settings.variable;
-    if (argument) {
-      // Insure against third-party code injection. (CVE-2021-23358)
-      if (!bareIdentifier.test(argument)) throw new Error(
-        'variable is not a bare identifier: ' + argument
-      );
-    } else {
-      // If a variable is not specified, place data values in local scope.
-      source = 'with(obj||{}){\n' + source + '}\n';
-      argument = 'obj';
-    }
-
-    source = "var __t,__p='',__j=Array.prototype.join," +
-      "print=function(){__p+=__j.call(arguments,'');};\n" +
-      source + 'return __p;\n';
-
-    var render;
-    try {
-      render = new Function(argument, '_', source);
-    } catch (e) {
-      e.source = source;
-      throw e;
-    }
-
-    var template = function(data) {
-      return render.call(this, data, _$1);
-    };
-
-    // Provide the compiled source as a convenience for precompilation.
-    template.source = 'function(' + argument + '){\n' + source + '}';
-
-    return template;
-  }
-
-  // Traverses the children of `obj` along `path`. If a child is a function, it
-  // is invoked with its parent as context. Returns the value of the final
-  // child, or `fallback` if any child is undefined.
-  function result(obj, path, fallback) {
-    path = toPath(path);
-    var length = path.length;
-    if (!length) {
-      return isFunction$1(fallback) ? fallback.call(obj) : fallback;
-    }
-    for (var i = 0; i < length; i++) {
-      var prop = obj == null ? void 0 : obj[path[i]];
-      if (prop === void 0) {
-        prop = fallback;
-        i = length; // Ensure we don't continue iterating.
-      }
-      obj = isFunction$1(prop) ? prop.call(obj) : prop;
-    }
-    return obj;
-  }
-
-  // Generate a unique integer id (unique within the entire client session).
-  // Useful for temporary DOM ids.
-  var idCounter = 0;
-  function uniqueId(prefix) {
-    var id = ++idCounter + '';
-    return prefix ? prefix + id : id;
-  }
-
-  // Start chaining a wrapped Underscore object.
-  function chain(obj) {
-    var instance = _$1(obj);
-    instance._chain = true;
-    return instance;
-  }
-
-  // Internal function to execute `sourceFunc` bound to `context` with optional
-  // `args`. Determines whether to execute a function as a constructor or as a
-  // normal function.
-  function executeBound(sourceFunc, boundFunc, context, callingContext, args) {
-    if (!(callingContext instanceof boundFunc)) return sourceFunc.apply(context, args);
-    var self = baseCreate(sourceFunc.prototype);
-    var result = sourceFunc.apply(self, args);
-    if (isObject(result)) return result;
-    return self;
-  }
-
-  // Partially apply a function by creating a version that has had some of its
-  // arguments pre-filled, without changing its dynamic `this` context. `_` acts
-  // as a placeholder by default, allowing any combination of arguments to be
-  // pre-filled. Set `_.partial.placeholder` for a custom placeholder argument.
-  var partial = restArguments(function(func, boundArgs) {
-    var placeholder = partial.placeholder;
-    var bound = function() {
-      var position = 0, length = boundArgs.length;
-      var args = Array(length);
-      for (var i = 0; i < length; i++) {
-        args[i] = boundArgs[i] === placeholder ? arguments[position++] : boundArgs[i];
-      }
-      while (position < arguments.length) args.push(arguments[position++]);
-      return executeBound(func, bound, this, this, args);
-    };
-    return bound;
-  });
-
-  partial.placeholder = _$1;
-
-  // Create a function bound to a given object (assigning `this`, and arguments,
-  // optionally).
-  var bind = restArguments(function(func, context, args) {
-    if (!isFunction$1(func)) throw new TypeError('Bind must be called on a function');
-    var bound = restArguments(function(callArgs) {
-      return executeBound(func, bound, context, this, args.concat(callArgs));
-    });
-    return bound;
-  });
-
-  // Internal helper for collection methods to determine whether a collection
-  // should be iterated as an array or as an object.
-  // Related: https://people.mozilla.org/~jorendorff/es6-draft.html#sec-tolength
-  // Avoids a very nasty iOS 8 JIT bug on ARM-64. #2094
-  var isArrayLike = createSizePropertyCheck(getLength);
-
-  // Internal implementation of a recursive `flatten` function.
-  function flatten$1(input, depth, strict, output) {
-    output = output || [];
-    if (!depth && depth !== 0) {
-      depth = Infinity;
-    } else if (depth <= 0) {
-      return output.concat(input);
-    }
-    var idx = output.length;
-    for (var i = 0, length = getLength(input); i < length; i++) {
-      var value = input[i];
-      if (isArrayLike(value) && (isArray(value) || isArguments$1(value))) {
-        // Flatten current level of array or arguments object.
-        if (depth > 1) {
-          flatten$1(value, depth - 1, strict, output);
-          idx = output.length;
-        } else {
-          var j = 0, len = value.length;
-          while (j < len) output[idx++] = value[j++];
-        }
-      } else if (!strict) {
-        output[idx++] = value;
-      }
-    }
-    return output;
-  }
-
-  // Bind a number of an object's methods to that object. Remaining arguments
-  // are the method names to be bound. Useful for ensuring that all callbacks
-  // defined on an object belong to it.
-  var bindAll = restArguments(function(obj, keys) {
-    keys = flatten$1(keys, false, false);
-    var index = keys.length;
-    if (index < 1) throw new Error('bindAll must be passed function names');
-    while (index--) {
-      var key = keys[index];
-      obj[key] = bind(obj[key], obj);
-    }
-    return obj;
-  });
-
-  // Memoize an expensive function by storing its results.
-  function memoize(func, hasher) {
-    var memoize = function(key) {
-      var cache = memoize.cache;
-      var address = '' + (hasher ? hasher.apply(this, arguments) : key);
-      if (!has$1(cache, address)) cache[address] = func.apply(this, arguments);
-      return cache[address];
-    };
-    memoize.cache = {};
-    return memoize;
-  }
-
-  // Delays a function for the given number of milliseconds, and then calls
-  // it with the arguments supplied.
-  var delay = restArguments(function(func, wait, args) {
-    return setTimeout(function() {
-      return func.apply(null, args);
-    }, wait);
-  });
-
-  // Defers a function, scheduling it to run after the current call stack has
-  // cleared.
-  var defer = partial(delay, _$1, 1);
-
-  // Returns a function, that, when invoked, will only be triggered at most once
-  // during a given window of time. Normally, the throttled function will run
-  // as much as it can, without ever going more than once per `wait` duration;
-  // but if you'd like to disable the execution on the leading edge, pass
-  // `{leading: false}`. To disable execution on the trailing edge, ditto.
-  function throttle(func, wait, options) {
-    var timeout, context, args, result;
-    var previous = 0;
-    if (!options) options = {};
-
-    var later = function() {
-      previous = options.leading === false ? 0 : now();
-      timeout = null;
-      result = func.apply(context, args);
-      if (!timeout) context = args = null;
-    };
-
-    var throttled = function() {
-      var _now = now();
-      if (!previous && options.leading === false) previous = _now;
-      var remaining = wait - (_now - previous);
-      context = this;
-      args = arguments;
-      if (remaining <= 0 || remaining > wait) {
-        if (timeout) {
-          clearTimeout(timeout);
-          timeout = null;
-        }
-        previous = _now;
-        result = func.apply(context, args);
-        if (!timeout) context = args = null;
-      } else if (!timeout && options.trailing !== false) {
-        timeout = setTimeout(later, remaining);
-      }
-      return result;
-    };
-
-    throttled.cancel = function() {
-      clearTimeout(timeout);
-      previous = 0;
-      timeout = context = args = null;
-    };
-
-    return throttled;
-  }
-
-  // When a sequence of calls of the returned function ends, the argument
-  // function is triggered. The end of a sequence is defined by the `wait`
-  // parameter. If `immediate` is passed, the argument function will be
-  // triggered at the beginning of the sequence instead of at the end.
-  function debounce(func, wait, immediate) {
-    var timeout, previous, args, result, context;
-
-    var later = function() {
-      var passed = now() - previous;
-      if (wait > passed) {
-        timeout = setTimeout(later, wait - passed);
-      } else {
-        timeout = null;
-        if (!immediate) result = func.apply(context, args);
-        // This check is needed because `func` can recursively invoke `debounced`.
-        if (!timeout) args = context = null;
-      }
-    };
-
-    var debounced = restArguments(function(_args) {
-      context = this;
-      args = _args;
-      previous = now();
-      if (!timeout) {
-        timeout = setTimeout(later, wait);
-        if (immediate) result = func.apply(context, args);
-      }
-      return result;
-    });
-
-    debounced.cancel = function() {
-      clearTimeout(timeout);
-      timeout = args = context = null;
-    };
-
-    return debounced;
-  }
-
-  // Returns the first function passed as an argument to the second,
-  // allowing you to adjust arguments, run code before and after, and
-  // conditionally execute the original function.
-  function wrap(func, wrapper) {
-    return partial(wrapper, func);
-  }
-
-  // Returns a negated version of the passed-in predicate.
-  function negate(predicate) {
-    return function() {
-      return !predicate.apply(this, arguments);
-    };
-  }
-
-  // Returns a function that is the composition of a list of functions, each
-  // consuming the return value of the function that follows.
-  function compose() {
-    var args = arguments;
-    var start = args.length - 1;
-    return function() {
-      var i = start;
-      var result = args[start].apply(this, arguments);
-      while (i--) result = args[i].call(this, result);
-      return result;
-    };
-  }
-
-  // Returns a function that will only be executed on and after the Nth call.
-  function after(times, func) {
-    return function() {
-      if (--times < 1) {
-        return func.apply(this, arguments);
-      }
-    };
-  }
-
-  // Returns a function that will only be executed up to (but not including) the
-  // Nth call.
-  function before(times, func) {
-    var memo;
-    return function() {
-      if (--times > 0) {
-        memo = func.apply(this, arguments);
-      }
-      if (times <= 1) func = null;
-      return memo;
-    };
-  }
-
-  // Returns a function that will be executed at most one time, no matter how
-  // often you call it. Useful for lazy initialization.
-  var once = partial(before, 2);
-
-  // Returns the first key on an object that passes a truth test.
-  function findKey(obj, predicate, context) {
-    predicate = cb(predicate, context);
-    var _keys = keys(obj), key;
-    for (var i = 0, length = _keys.length; i < length; i++) {
-      key = _keys[i];
-      if (predicate(obj[key], key, obj)) return key;
-    }
-  }
-
-  // Internal function to generate `_.findIndex` and `_.findLastIndex`.
-  function createPredicateIndexFinder(dir) {
-    return function(array, predicate, context) {
-      predicate = cb(predicate, context);
-      var length = getLength(array);
-      var index = dir > 0 ? 0 : length - 1;
-      for (; index >= 0 && index < length; index += dir) {
-        if (predicate(array[index], index, array)) return index;
-      }
-      return -1;
-    };
-  }
-
-  // Returns the first index on an array-like that passes a truth test.
-  var findIndex = createPredicateIndexFinder(1);
-
-  // Returns the last index on an array-like that passes a truth test.
-  var findLastIndex = createPredicateIndexFinder(-1);
-
-  // Use a comparator function to figure out the smallest index at which
-  // an object should be inserted so as to maintain order. Uses binary search.
-  function sortedIndex(array, obj, iteratee, context) {
-    iteratee = cb(iteratee, context, 1);
-    var value = iteratee(obj);
-    var low = 0, high = getLength(array);
-    while (low < high) {
-      var mid = Math.floor((low + high) / 2);
-      if (iteratee(array[mid]) < value) low = mid + 1; else high = mid;
-    }
-    return low;
-  }
-
-  // Internal function to generate the `_.indexOf` and `_.lastIndexOf` functions.
-  function createIndexFinder(dir, predicateFind, sortedIndex) {
-    return function(array, item, idx) {
-      var i = 0, length = getLength(array);
-      if (typeof idx == 'number') {
-        if (dir > 0) {
-          i = idx >= 0 ? idx : Math.max(idx + length, i);
-        } else {
-          length = idx >= 0 ? Math.min(idx + 1, length) : idx + length + 1;
-        }
-      } else if (sortedIndex && idx && length) {
-        idx = sortedIndex(array, item);
-        return array[idx] === item ? idx : -1;
-      }
-      if (item !== item) {
-        idx = predicateFind(slice.call(array, i, length), isNaN$1);
-        return idx >= 0 ? idx + i : -1;
-      }
-      for (idx = dir > 0 ? i : length - 1; idx >= 0 && idx < length; idx += dir) {
-        if (array[idx] === item) return idx;
-      }
-      return -1;
-    };
-  }
-
-  // Return the position of the first occurrence of an item in an array,
-  // or -1 if the item is not included in the array.
-  // If the array is large and already in sort order, pass `true`
-  // for **isSorted** to use binary search.
-  var indexOf = createIndexFinder(1, findIndex, sortedIndex);
-
-  // Return the position of the last occurrence of an item in an array,
-  // or -1 if the item is not included in the array.
-  var lastIndexOf = createIndexFinder(-1, findLastIndex);
-
-  // Return the first value which passes a truth test.
-  function find(obj, predicate, context) {
-    var keyFinder = isArrayLike(obj) ? findIndex : findKey;
-    var key = keyFinder(obj, predicate, context);
-    if (key !== void 0 && key !== -1) return obj[key];
-  }
-
-  // Convenience version of a common use case of `_.find`: getting the first
-  // object containing specific `key:value` pairs.
-  function findWhere(obj, attrs) {
-    return find(obj, matcher(attrs));
-  }
-
-  // The cornerstone for collection functions, an `each`
-  // implementation, aka `forEach`.
-  // Handles raw objects in addition to array-likes. Treats all
-  // sparse array-likes as if they were dense.
-  function each(obj, iteratee, context) {
-    iteratee = optimizeCb(iteratee, context);
-    var i, length;
-    if (isArrayLike(obj)) {
-      for (i = 0, length = obj.length; i < length; i++) {
-        iteratee(obj[i], i, obj);
-      }
-    } else {
-      var _keys = keys(obj);
-      for (i = 0, length = _keys.length; i < length; i++) {
-        iteratee(obj[_keys[i]], _keys[i], obj);
-      }
-    }
-    return obj;
-  }
-
-  // Return the results of applying the iteratee to each element.
-  function map(obj, iteratee, context) {
-    iteratee = cb(iteratee, context);
-    var _keys = !isArrayLike(obj) && keys(obj),
-        length = (_keys || obj).length,
-        results = Array(length);
-    for (var index = 0; index < length; index++) {
-      var currentKey = _keys ? _keys[index] : index;
-      results[index] = iteratee(obj[currentKey], currentKey, obj);
-    }
-    return results;
-  }
-
-  // Internal helper to create a reducing function, iterating left or right.
-  function createReduce(dir) {
-    // Wrap code that reassigns argument variables in a separate function than
-    // the one that accesses `arguments.length` to avoid a perf hit. (#1991)
-    var reducer = function(obj, iteratee, memo, initial) {
-      var _keys = !isArrayLike(obj) && keys(obj),
-          length = (_keys || obj).length,
-          index = dir > 0 ? 0 : length - 1;
-      if (!initial) {
-        memo = obj[_keys ? _keys[index] : index];
-        index += dir;
-      }
-      for (; index >= 0 && index < length; index += dir) {
-        var currentKey = _keys ? _keys[index] : index;
-        memo = iteratee(memo, obj[currentKey], currentKey, obj);
-      }
-      return memo;
-    };
-
-    return function(obj, iteratee, memo, context) {
-      var initial = arguments.length >= 3;
-      return reducer(obj, optimizeCb(iteratee, context, 4), memo, initial);
-    };
-  }
-
-  // **Reduce** builds up a single result from a list of values, aka `inject`,
-  // or `foldl`.
-  var reduce = createReduce(1);
-
-  // The right-associative version of reduce, also known as `foldr`.
-  var reduceRight = createReduce(-1);
-
-  // Return all the elements that pass a truth test.
-  function filter(obj, predicate, context) {
-    var results = [];
-    predicate = cb(predicate, context);
-    each(obj, function(value, index, list) {
-      if (predicate(value, index, list)) results.push(value);
-    });
-    return results;
-  }
-
-  // Return all the elements for which a truth test fails.
-  function reject(obj, predicate, context) {
-    return filter(obj, negate(cb(predicate)), context);
-  }
-
-  // Determine whether all of the elements pass a truth test.
-  function every(obj, predicate, context) {
-    predicate = cb(predicate, context);
-    var _keys = !isArrayLike(obj) && keys(obj),
-        length = (_keys || obj).length;
-    for (var index = 0; index < length; index++) {
-      var currentKey = _keys ? _keys[index] : index;
-      if (!predicate(obj[currentKey], currentKey, obj)) return false;
-    }
-    return true;
-  }
-
-  // Determine if at least one element in the object passes a truth test.
-  function some(obj, predicate, context) {
-    predicate = cb(predicate, context);
-    var _keys = !isArrayLike(obj) && keys(obj),
-        length = (_keys || obj).length;
-    for (var index = 0; index < length; index++) {
-      var currentKey = _keys ? _keys[index] : index;
-      if (predicate(obj[currentKey], currentKey, obj)) return true;
-    }
-    return false;
-  }
-
-  // Determine if the array or object contains a given item (using `===`).
-  function contains(obj, item, fromIndex, guard) {
-    if (!isArrayLike(obj)) obj = values(obj);
-    if (typeof fromIndex != 'number' || guard) fromIndex = 0;
-    return indexOf(obj, item, fromIndex) >= 0;
-  }
-
-  // Invoke a method (with arguments) on every item in a collection.
-  var invoke = restArguments(function(obj, path, args) {
-    var contextPath, func;
-    if (isFunction$1(path)) {
-      func = path;
-    } else {
-      path = toPath(path);
-      contextPath = path.slice(0, -1);
-      path = path[path.length - 1];
-    }
-    return map(obj, function(context) {
-      var method = func;
-      if (!method) {
-        if (contextPath && contextPath.length) {
-          context = deepGet(context, contextPath);
-        }
-        if (context == null) return void 0;
-        method = context[path];
-      }
-      return method == null ? method : method.apply(context, args);
-    });
-  });
-
-  // Convenience version of a common use case of `_.map`: fetching a property.
-  function pluck(obj, key) {
-    return map(obj, property(key));
-  }
-
-  // Convenience version of a common use case of `_.filter`: selecting only
-  // objects containing specific `key:value` pairs.
-  function where(obj, attrs) {
-    return filter(obj, matcher(attrs));
-  }
-
-  // Return the maximum element (or element-based computation).
-  function max(obj, iteratee, context) {
-    var result = -Infinity, lastComputed = -Infinity,
-        value, computed;
-    if (iteratee == null || typeof iteratee == 'number' && typeof obj[0] != 'object' && obj != null) {
-      obj = isArrayLike(obj) ? obj : values(obj);
-      for (var i = 0, length = obj.length; i < length; i++) {
-        value = obj[i];
-        if (value != null && value > result) {
-          result = value;
-        }
-      }
-    } else {
-      iteratee = cb(iteratee, context);
-      each(obj, function(v, index, list) {
-        computed = iteratee(v, index, list);
-        if (computed > lastComputed || computed === -Infinity && result === -Infinity) {
-          result = v;
-          lastComputed = computed;
-        }
-      });
-    }
-    return result;
-  }
-
-  // Return the minimum element (or element-based computation).
-  function min(obj, iteratee, context) {
-    var result = Infinity, lastComputed = Infinity,
-        value, computed;
-    if (iteratee == null || typeof iteratee == 'number' && typeof obj[0] != 'object' && obj != null) {
-      obj = isArrayLike(obj) ? obj : values(obj);
-      for (var i = 0, length = obj.length; i < length; i++) {
-        value = obj[i];
-        if (value != null && value < result) {
-          result = value;
-        }
-      }
-    } else {
-      iteratee = cb(iteratee, context);
-      each(obj, function(v, index, list) {
-        computed = iteratee(v, index, list);
-        if (computed < lastComputed || computed === Infinity && result === Infinity) {
-          result = v;
-          lastComputed = computed;
-        }
-      });
-    }
-    return result;
-  }
-
-  // Sample **n** random values from a collection using the modern version of the
-  // [Fisher-Yates shuffle](https://en.wikipedia.org/wiki/Fisher–Yates_shuffle).
-  // If **n** is not specified, returns a single random element.
-  // The internal `guard` argument allows it to work with `_.map`.
-  function sample(obj, n, guard) {
-    if (n == null || guard) {
-      if (!isArrayLike(obj)) obj = values(obj);
-      return obj[random(obj.length - 1)];
-    }
-    var sample = isArrayLike(obj) ? clone(obj) : values(obj);
-    var length = getLength(sample);
-    n = Math.max(Math.min(n, length), 0);
-    var last = length - 1;
-    for (var index = 0; index < n; index++) {
-      var rand = random(index, last);
-      var temp = sample[index];
-      sample[index] = sample[rand];
-      sample[rand] = temp;
-    }
-    return sample.slice(0, n);
-  }
-
-  // Shuffle a collection.
-  function shuffle(obj) {
-    return sample(obj, Infinity);
-  }
-
-  // Sort the object's values by a criterion produced by an iteratee.
-  function sortBy(obj, iteratee, context) {
-    var index = 0;
-    iteratee = cb(iteratee, context);
-    return pluck(map(obj, function(value, key, list) {
-      return {
-        value: value,
-        index: index++,
-        criteria: iteratee(value, key, list)
-      };
-    }).sort(function(left, right) {
-      var a = left.criteria;
-      var b = right.criteria;
-      if (a !== b) {
-        if (a > b || a === void 0) return 1;
-        if (a < b || b === void 0) return -1;
-      }
-      return left.index - right.index;
-    }), 'value');
-  }
-
-  // An internal function used for aggregate "group by" operations.
-  function group(behavior, partition) {
-    return function(obj, iteratee, context) {
-      var result = partition ? [[], []] : {};
-      iteratee = cb(iteratee, context);
-      each(obj, function(value, index) {
-        var key = iteratee(value, index, obj);
-        behavior(result, value, key);
-      });
-      return result;
-    };
-  }
-
-  // Groups the object's values by a criterion. Pass either a string attribute
-  // to group by, or a function that returns the criterion.
-  var groupBy = group(function(result, value, key) {
-    if (has$1(result, key)) result[key].push(value); else result[key] = [value];
-  });
-
-  // Indexes the object's values by a criterion, similar to `_.groupBy`, but for
-  // when you know that your index values will be unique.
-  var indexBy = group(function(result, value, key) {
-    result[key] = value;
-  });
-
-  // Counts instances of an object that group by a certain criterion. Pass
-  // either a string attribute to count by, or a function that returns the
-  // criterion.
-  var countBy = group(function(result, value, key) {
-    if (has$1(result, key)) result[key]++; else result[key] = 1;
-  });
-
-  // Split a collection into two arrays: one whose elements all pass the given
-  // truth test, and one whose elements all do not pass the truth test.
-  var partition = group(function(result, value, pass) {
-    result[pass ? 0 : 1].push(value);
-  }, true);
-
-  // Safely create a real, live array from anything iterable.
-  var reStrSymbol = /[^\ud800-\udfff]|[\ud800-\udbff][\udc00-\udfff]|[\ud800-\udfff]/g;
-  function toArray(obj) {
-    if (!obj) return [];
-    if (isArray(obj)) return slice.call(obj);
-    if (isString(obj)) {
-      // Keep surrogate pair characters together.
-      return obj.match(reStrSymbol);
-    }
-    if (isArrayLike(obj)) return map(obj, identity);
-    return values(obj);
-  }
-
-  // Return the number of elements in a collection.
-  function size(obj) {
-    if (obj == null) return 0;
-    return isArrayLike(obj) ? obj.length : keys(obj).length;
-  }
-
-  // Internal `_.pick` helper function to determine whether `key` is an enumerable
-  // property name of `obj`.
-  function keyInObj(value, key, obj) {
-    return key in obj;
-  }
-
-  // Return a copy of the object only containing the allowed properties.
-  var pick = restArguments(function(obj, keys) {
-    var result = {}, iteratee = keys[0];
-    if (obj == null) return result;
-    if (isFunction$1(iteratee)) {
-      if (keys.length > 1) iteratee = optimizeCb(iteratee, keys[1]);
-      keys = allKeys(obj);
-    } else {
-      iteratee = keyInObj;
-      keys = flatten$1(keys, false, false);
-      obj = Object(obj);
-    }
-    for (var i = 0, length = keys.length; i < length; i++) {
-      var key = keys[i];
-      var value = obj[key];
-      if (iteratee(value, key, obj)) result[key] = value;
-    }
-    return result;
-  });
-
-  // Return a copy of the object without the disallowed properties.
-  var omit = restArguments(function(obj, keys) {
-    var iteratee = keys[0], context;
-    if (isFunction$1(iteratee)) {
-      iteratee = negate(iteratee);
-      if (keys.length > 1) context = keys[1];
-    } else {
-      keys = map(flatten$1(keys, false, false), String);
-      iteratee = function(value, key) {
-        return !contains(keys, key);
-      };
-    }
-    return pick(obj, iteratee, context);
-  });
-
-  // Returns everything but the last entry of the array. Especially useful on
-  // the arguments object. Passing **n** will return all the values in
-  // the array, excluding the last N.
-  function initial(array, n, guard) {
-    return slice.call(array, 0, Math.max(0, array.length - (n == null || guard ? 1 : n)));
-  }
-
-  // Get the first element of an array. Passing **n** will return the first N
-  // values in the array. The **guard** check allows it to work with `_.map`.
-  function first(array, n, guard) {
-    if (array == null || array.length < 1) return n == null || guard ? void 0 : [];
-    if (n == null || guard) return array[0];
-    return initial(array, array.length - n);
-  }
-
-  // Returns everything but the first entry of the `array`. Especially useful on
-  // the `arguments` object. Passing an **n** will return the rest N values in the
-  // `array`.
-  function rest(array, n, guard) {
-    return slice.call(array, n == null || guard ? 1 : n);
-  }
-
-  // Get the last element of an array. Passing **n** will return the last N
-  // values in the array.
-  function last(array, n, guard) {
-    if (array == null || array.length < 1) return n == null || guard ? void 0 : [];
-    if (n == null || guard) return array[array.length - 1];
-    return rest(array, Math.max(0, array.length - n));
-  }
-
-  // Trim out all falsy values from an array.
-  function compact(array) {
-    return filter(array, Boolean);
-  }
-
-  // Flatten out an array, either recursively (by default), or up to `depth`.
-  // Passing `true` or `false` as `depth` means `1` or `Infinity`, respectively.
-  function flatten(array, depth) {
-    return flatten$1(array, depth, false);
-  }
-
-  // Take the difference between one array and a number of other arrays.
-  // Only the elements present in just the first array will remain.
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-    var result = Array(length);
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-  // Zip together multiple lists into a single array -- elements that share
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diff --git a/docs/_build/html/_static/underscore.js b/docs/_build/html/_static/underscore.js
deleted file mode 100644
index cf177d4..0000000
--- a/docs/_build/html/_static/underscore.js
+++ /dev/null
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diff --git a/docs/_build/html/_static/vendor/fontawesome/6.1.2/LICENSE.txt b/docs/_build/html/_static/vendor/fontawesome/6.1.2/LICENSE.txt
deleted file mode 100644
index cc557ec..0000000
--- a/docs/_build/html/_static/vendor/fontawesome/6.1.2/LICENSE.txt
+++ /dev/null
@@ -1,165 +0,0 @@
-Fonticons, Inc. (https://fontawesome.com)
-
---------------------------------------------------------------------------------
-
-Font Awesome Free License
-
-Font Awesome Free is free, open source, and GPL friendly. You can use it for
-commercial projects, open source projects, or really almost whatever you want.
-Full Font Awesome Free license: https://fontawesome.com/license/free.
-
---------------------------------------------------------------------------------
-
-# Icons: CC BY 4.0 License (https://creativecommons.org/licenses/by/4.0/)
-
-The Font Awesome Free download is licensed under a Creative Commons
-Attribution 4.0 International License and applies to all icons packaged
-as SVG and JS file types.
-
---------------------------------------------------------------------------------
-
-# Fonts: SIL OFL 1.1 License
-
-In the Font Awesome Free download, the SIL OFL license applies to all icons
-packaged as web and desktop font files.
-
-Copyright (c) 2022 Fonticons, Inc. (https://fontawesome.com)
-with Reserved Font Name: "Font Awesome".
-
-This Font Software is licensed under the SIL Open Font License, Version 1.1.
-This license is copied below, and is also available with a FAQ at:
-http://scripts.sil.org/OFL
-
-SIL OPEN FONT LICENSE
-Version 1.1 - 26 February 2007
-
-PREAMBLE
-The goals of the Open Font License (OFL) are to stimulate worldwide
-development of collaborative font projects, to support the font creation
-efforts of academic and linguistic communities, and to provide a free and
-open framework in which fonts may be shared and improved in partnership
-with others.
-
-The OFL allows the licensed fonts to be used, studied, modified and
-redistributed freely as long as they are not sold by themselves. The
-fonts, including any derivative works, can be bundled, embedded,
-redistributed and/or sold with any software provided that any reserved
-names are not used by derivative works. The fonts and derivatives,
-however, cannot be released under any other type of license. The
-requirement for fonts to remain under this license does not apply
-to any document created using the fonts or their derivatives.
-
-DEFINITIONS
-"Font Software" refers to the set of files released by the Copyright
-Holder(s) under this license and clearly marked as such. This may
-include source files, build scripts and documentation.
-
-"Reserved Font Name" refers to any names specified as such after the
-copyright statement(s).
-
-"Original Version" refers to the collection of Font Software components as
-distributed by the Copyright Holder(s).
-
-"Modified Version" refers to any derivative made by adding to, deleting,
-or substituting — in part or in whole — any of the components of the
-Original Version, by changing formats or by porting the Font Software to a
-new environment.
-
-"Author" refers to any designer, engineer, programmer, technical
-writer or other person who contributed to the Font Software.
-
-PERMISSION & CONDITIONS
-Permission is hereby granted, free of charge, to any person obtaining
-a copy of the Font Software, to use, study, copy, merge, embed, modify,
-redistribute, and sell modified and unmodified copies of the Font
-Software, subject to the following conditions:
-
-1) Neither the Font Software nor any of its individual components,
-in Original or Modified Versions, may be sold by itself.
-
-2) Original or Modified Versions of the Font Software may be bundled,
-redistributed and/or sold with any software, provided that each copy
-contains the above copyright notice and this license. These can be
-included either as stand-alone text files, human-readable headers or
-in the appropriate machine-readable metadata fields within text or
-binary files as long as those fields can be easily viewed by the user.
-
-3) No Modified Version of the Font Software may use the Reserved Font
-Name(s) unless explicit written permission is granted by the corresponding
-Copyright Holder. This restriction only applies to the primary font name as
-presented to the users.
-
-4) The name(s) of the Copyright Holder(s) or the Author(s) of the Font
-Software shall not be used to promote, endorse or advertise any
-Modified Version, except to acknowledge the contribution(s) of the
-Copyright Holder(s) and the Author(s) or with their explicit written
-permission.
-
-5) The Font Software, modified or unmodified, in part or in whole,
-must be distributed entirely under this license, and must not be
-distributed under any other license. The requirement for fonts to
-remain under this license does not apply to any document created
-using the Font Software.
-
-TERMINATION
-This license becomes null and void if any of the above conditions are
-not met.
-
-DISCLAIMER
-THE FONT SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
-EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO ANY WARRANTIES OF
-MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT
-OF COPYRIGHT, PATENT, TRADEMARK, OR OTHER RIGHT. IN NO EVENT SHALL THE
-COPYRIGHT HOLDER BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY,
-INCLUDING ANY GENERAL, SPECIAL, INDIRECT, INCIDENTAL, OR CONSEQUENTIAL
-DAMAGES, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING
-FROM, OUT OF THE USE OR INABILITY TO USE THE FONT SOFTWARE OR FROM
-OTHER DEALINGS IN THE FONT SOFTWARE.
-
---------------------------------------------------------------------------------
-
-# Code: MIT License (https://opensource.org/licenses/MIT)
-
-In the Font Awesome Free download, the MIT license applies to all non-font and
-non-icon files.
-
-Copyright 2022 Fonticons, Inc.
-
-Permission is hereby granted, free of charge, to any person obtaining a copy of
-this software and associated documentation files (the "Software"), to deal in the
-Software without restriction, including without limitation the rights to use, copy,
-modify, merge, publish, distribute, sublicense, and/or sell copies of the Software,
-and to permit persons to whom the Software is furnished to do so, subject to the
-following conditions:
-
-The above copyright notice and this permission notice shall be included in all
-copies or substantial portions of the Software.
-
-THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED,
-INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A
-PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT
-HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION
-OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE
-SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
-
---------------------------------------------------------------------------------
-
-# Attribution
-
-Attribution is required by MIT, SIL OFL, and CC BY licenses. Downloaded Font
-Awesome Free files already contain embedded comments with sufficient
-attribution, so you shouldn't need to do anything additional when using these
-files normally.
-
-We've kept attribution comments terse, so we ask that you do not actively work
-to remove them from files, especially code. They're a great way for folks to
-learn about Font Awesome.
-
---------------------------------------------------------------------------------
-
-# Brand Icons
-
-All brand icons are trademarks of their respective owners. The use of these
-trademarks does not indicate endorsement of the trademark holder by Font
-Awesome, nor vice versa. **Please do not use brand logos for any purpose except
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diff --git a/docs/_build/html/_static/vendor/fontawesome/6.1.2/css/all.min.css b/docs/_build/html/_static/vendor/fontawesome/6.1.2/css/all.min.css
deleted file mode 100644
index aad4b1d..0000000
--- a/docs/_build/html/_static/vendor/fontawesome/6.1.2/css/all.min.css
+++ /dev/null
@@ -1,5 +0,0 @@
-/*!
- * Font Awesome Free 6.1.2 by @fontawesome - https://fontawesome.com
- * License - https://fontawesome.com/license/free (Icons: CC BY 4.0, Fonts: SIL OFL 1.1, Code: MIT License)
- * Copyright 2022 Fonticons, Inc.
- */.fa{font-family:var(--fa-style-family,"Font Awesome 6 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-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
-<li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
-</ul>
-</li>
-<li class="toctree-l1"><a class="reference internal" href="14.%20Two-D%20Dynamic%20Programming/README.html">Blind 75 LeetCode Problems</a></li>
-<li class="toctree-l1"><a class="reference internal" href="15.%20Greedy/README.html">Blind 75 LeetCode Problems</a></li>
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diff --git a/docs/_build/html/reports/Book.err.log b/docs/_build/html/reports/Book.err.log
deleted file mode 100644
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-  File "/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py", line 189, in ensure_async
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-  File "/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py", line 1305, in execute
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-  File "/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py", line 173, in wrapped
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-  File "/usr/local/python/3.10.8/lib/python3.10/asyncio/base_events.py", line 649, in run_until_complete
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-  File "/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py", line 705, in async_execute
-    await self.async_execute_cell(
-  File "/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py", line 1001, in async_execute_cell
-    exec_reply = await self.task_poll_for_reply
-  File "/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py", line 802, in _async_poll_for_reply
-    error_on_timeout_execute_reply = await self._async_handle_timeout(timeout, cell)
-  File "/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py", line 852, in _async_handle_timeout
-    raise CellTimeoutError.error_from_timeout_and_cell(
-nbclient.exceptions.CellTimeoutError: A cell timed out while it was being executed, after 30 seconds.
-The message was: Cell execution timed out.
-Here is a preview of the cell contents:
--------------------
-!jupyter-book build .
--------------------
-
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18, 20, 21, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 61, 63, 64], "follow": [0, 3, 4, 5, 6, 15, 16, 23, 24, 25, 26, 30, 31, 32, 35, 37, 38, 39, 41, 43, 45, 47, 48, 49, 52, 54, 55, 57, 59, 61, 72, 73, 74, 75, 76], "worst": [0, 1, 2, 4, 5, 6, 7, 9, 10, 11, 14, 16, 18, 20, 21, 23, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 47, 48, 51, 52, 54, 55, 56, 57, 58, 59, 61], "need": [0, 9, 11, 14, 15, 16, 18, 21, 24, 25, 26, 27, 30, 34, 37, 39, 40, 43, 44, 45, 48, 49, 54, 55, 57, 58, 59, 61, 63, 64, 65, 72, 74, 76], "visit": [0, 1, 26, 30, 31, 32, 35, 36, 37, 39, 40, 41, 45, 52, 54, 55, 56, 58, 59], "everi": [0, 2, 16, 18, 30, 32, 40, 41, 56, 58], "onc": [0, 1, 2, 4, 5, 6, 7, 9, 10, 13, 14, 15, 18, 24, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 39, 40, 41, 44, 45, 52, 54, 55, 56, 57, 59, 63, 64, 65, 72, 74, 76], "determin": [0, 1, 2, 4, 10, 15, 16, 18, 21, 23, 24, 30, 31, 33, 34, 35, 36, 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43, 46, 49, 51, 53, 54, 55, 56, 58, 59, 72], "middl": [20, 21, 24, 48], "elimin": [20, 38, 51], "met": [20, 24, 38, 52, 55, 56], "impli": 20, "13": [20, 72], "17": [20, 72], "emploi": 20, "divid": [20, 21, 28, 33, 48, 59], "halv": [20, 24, 48], "logarithm": [20, 38, 48], "reduc": [20, 28, 43, 51, 56, 59], "henc": 20, "kth": [20, 42, 47, 48, 50, 75], "extens": [20, 34, 38, 48, 54, 72], "prior": 21, "pass": [21, 23, 25, 26, 34, 49, 52], "unknown": [21, 61], "pivot": 21, "runtim": 21, "li": 21, "reset": 21, "chosen": [21, 51], "target1": [21, 51], "target2": [21, 51], "target3": [21, 51], "veri": [21, 40, 49], "scale": [21, 35, 40, 64], "circularli": 21, "circular": [21, 57, 63, 64], "manner": [21, 27, 32, 37], "maintain": [21, 41, 43, 45, 47, 48, 49, 54, 59, 63], "153": [22, 72, 75], "rotat": [22, 26, 70, 72, 75], "33": [22, 72, 75], "head": [23, 24, 25, 26, 27], "reach": [23, 24, 25, 26, 32, 37, 38, 43, 44, 51, 55, 58, 59, 65], "intern": 23, "po": [23, 72], "denot": [23, 59], "tail": 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52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65, 75], "explan": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "challeng": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 24, 26, 27, 31, 33, 34, 35, 37, 38, 39, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "exercis": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 24, 26, 27, 31, 33, 34, 35, 37, 38, 39, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "217": 2, "contain": [2, 9], "duplic": 2, "238": 3, "product": 3, "arrai": [3, 8, 20, 21, 75], "except": 3, "self": 3, "242": 4, "valid": [4, 10, 18, 41, 55], "anagram": [4, 6], "347": 5, "top": 5, "k": [5, 28], "frequent": 5, "element": [5, 37, 49], "49": 6, "group": 6, "659": 7, "encod": 7, "decod": 7, "string": 7, "hash": [8, 75], "blind": [8, 12, 17, 19, 22, 29, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 75], "75": [8, 12, 17, 19, 22, 29, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 75], "leetcod": [8, 12, 17, 19, 22, 29, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 75], "problem": [8, 12, 17, 19, 22, 29, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 75], "contribut": [8, 12, 17, 19, 22, 29, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 74, 75], "resourc": [8, 12, 17, 19, 22, 29, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 75], "11": 9, "With": 9, "most": 9, "water": [9, 59], "125": 10, "palindrom": 10, "15": 11, "3sum": 11, "two": [12, 27, 75], "pointer": [12, 75], "121": 13, "best": 13, "bui": 13, "sell": 13, "stock": 13, "3": [14, 73], "substr": [14, 16], "without": 14, "repeat": [14, 15], "charact": [14, 15], "424": 15, "replac": 15, "76": 16, "minimum": [16, 20], "window": [16, 17, 75], "slide": [17, 75], "20": 18, "parenthes": 18, "stack": [19, 75], "153": 20, "find": [20, 48], "rotat": [20, 21], "sort": [20, 21, 27, 28], "33": 21, "search": [21, 22, 38, 41, 44, 45, 52, 75], "binari": [22, 31, 32, 33, 34, 35, 36, 38, 39, 41, 75], "141": 23, "link": [23, 26, 29, 75], "list": [23, 24, 25, 26, 27, 28, 29, 67, 68, 69, 70, 71, 75], "cycl": 23, "143": 24, "reorder": 24, "19": 25, "remov": 25, "nth": 25, "node": 25, "from": [25, 33, 48], "end": 25, "206": 26, "revers": 26, "21": 27, "merg": [27, 28], "23": 28, "In": 30, "summari": 30, "102": 31, "level": 31, "order": 31, "travers": [31, 33], "104": 32, "maximum": [32, 34], "depth": 32, "105": 33, "construct": 33, "preorder": 33, "inord": 33, "helper": 33, "function": 33, "124": 34, "path": 34, "sum": [34, 51], "probelm": [34, 39, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "199": 35, "right": 35, "side": 35, "view": 35, "226": 36, "invert": 36, "230": 37, "kth": [37, 49], "smallest": 37, "bst": 37, "235": 38, "lowest": 38, "common": 38, "ancestor": 38, "297": 39, "serial": 39, "deseri": 39, "572": 40, "subtre": 40, "anoth": 40, "here": 40, "s": 40, "detail": 40, "code": [40, 73, 74, 75], "98": 41, "208": 43, "implement": 43, "trie": [43, 46], "prefix": 43, "211": 44, "design": 44, "add": 44, "word": [44, 45, 52], "data": [44, 48], "structur": 44, "212": 45, "ii": [45, 64], "1046": 47, "last": 47, "stone": 47, "weight": 47, "295": 48, "median": 48, "stream": [48, 49], "703": 49, "largest": 49, "heap": [50, 75], "prioriti": [50, 75], "queue": [50, 75], "39": 51, "combin": 51, "79": 52, "backtrack": [53, 75], "133": 54, "clone": 54, "graph": [54, 55, 58, 60, 62, 75], "178": 55, "200": 56, "number": [56, 58], "island": 56, "207": 57, "cours": 57, "schedul": 57, "3651": 58, "connect": 58, "compon": 58, "an": 58, "undirect": 58, "417": 59, "pacif": 59, "atlant": 59, "flow": 59, "892": 61, "alien": 61, "dictionari": 61, "advanc": [62, 75], "hous": [63, 64], "robber": [63, 64], "70": 65, "climb": 65, "stair": 65, "1": [66, 73, 75], "d": [66, 67, 75], "dynam": [66, 67, 75], "program": [66, 67, 75], "2": [67, 73, 75], "how": [67, 68, 69, 70, 71, 75], "us": [67, 68, 69, 70, 71, 75], "disclaim": [67, 68, 69, 70, 71, 75], "greedi": [68, 75], "interv": [69, 75], "math": [70, 75], "geometri": [70, 75], "bit": [71, 75], "manupul": 71, "contributor": 73, "coven": 73, "conduct": [73, 74, 75], "our": 73, "pledg": 73, "standard": 73, "enforc": 73, "respons": 73, "scope": 73, "guidelin": [73, 74], "correct": 73, "warn": 73, "temporari": 73, "ban": 73, "4": 73, "perman": 73, "attribut": 73, "tabl": 74, "content": 74, "wai": 74, "get": 74, "start": 74, "process": 74, "tri": 75, "manipul": 75, "secur": [75, 76], "polici": [75, 76], "report": 76, "vulner": 76, "updat": 76, "contact": 76}, "envversion": {"sphinx.domains.c": 2, "sphinx.domains.changeset": 1, "sphinx.domains.citation": 1, "sphinx.domains.cpp": 6, "sphinx.domains.index": 1, "sphinx.domains.javascript": 2, "sphinx.domains.math": 2, "sphinx.domains.python": 3, "sphinx.domains.rst": 2, "sphinx.domains.std": 2, "sphinx.ext.intersphinx": 1, "sphinx": 56}})
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb b/docs/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb
deleted file mode 100644
index 5d55ca7..0000000
--- a/docs/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 100: Same Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-    "\n",
-    "---\n",
-    "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-    "\n",
-    "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-    "\n",
-    "**Constraints**\n",
-    "\n",
-    "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-    "2. $-10^4$ <= `Node.val` <= $10^4$"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def isSameTree(p, q):\n",
-    "    # Base case: If both p and q are None, the trees are the same.\n",
-    "    if not p and not q:\n",
-    "        return True\n",
-    "    \n",
-    "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-    "    if not p or not q:\n",
-    "        return False\n",
-    "    \n",
-    "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-    "    if p.val != q.val:\n",
-    "        return False\n",
-    "    \n",
-    "    # Recursively check the left and right subtrees of p and q.\n",
-    "    # If both subtrees are the same, the entire trees are the same.\n",
-    "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-    "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-    "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-    "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-    "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-    "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "\n",
-    "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-    "\n",
-    "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "print(isSameTree(p1, q1)) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "\n",
-    "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-    "\n",
-    "p2 = TreeNode(1, TreeNode(2), None)\n",
-    "q2 = TreeNode(1, None, TreeNode(2))\n",
-    "print(isSameTree(p2, q2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "\n",
-    "#Input: p = [1,2,1], q = [1,1,2]\n",
-    "\n",
-    "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-    "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-    "print(isSameTree(p3, q3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "**Time Complexity**\n",
-    "\n",
-    "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-    "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-    "\n",
-    "**Space Complexity**\n",
-    "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-    "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-    "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-    "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-    "\n",
-    "**In summary:**\n",
-    "\n",
-    "+ Time Complexity: $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-    "+ Space Complexity: $O(n)$ due to the recursive call stack."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb b/docs/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb
deleted file mode 100644
index 9256d0e..0000000
--- a/docs/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 128. Longest Consecutive Sequence\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Longest Consecutive Sequence problem on LeetCode, click here!](https://leetcode.com/problems/longest-consecutive-sequence/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given an unsorted array of integers `nums`, return *the length of the longest consecutive elements sequence.*\n",
-    "\n",
-    "You must write an algorithm that runs in `O(n)` time.\n",
-    "\n",
-    "*Constraints:**\n",
-    "- 0 <= `nums.length` <= $10^5$\n",
-    "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def longestConsecutive(nums):\n",
-    "    if not nums:\n",
-    "        return 0\n",
-    "\n",
-    "    num_set = set(nums)\n",
-    "    max_length = 0\n",
-    "\n",
-    "    for num in num_set:\n",
-    "        if num - 1 not in num_set:\n",
-    "            current_num = num\n",
-    "            current_length = 1\n",
-    "\n",
-    "            while current_num + 1 in num_set:\n",
-    "                current_num += 1\n",
-    "                current_length += 1\n",
-    "\n",
-    "            max_length = max(max_length, current_length)\n",
-    "\n",
-    "    return max_length"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining the `longestConsecutive` function that takes the `nums` array as input.\n",
-    "\n",
-    "2. We check if the `nums` array is empty. If it's empty, there are no consecutive elements, so we return 0.\n",
-    "\n",
-    "3. We create a Python set called `num_set` and insert all elements from the `nums` array into it. Using a set allows us to efficiently check for the existence of elements in O(1) time.\n",
-    "\n",
-    "4. We initialize a variable `max_length` to 0. This variable will keep track of the maximum length of consecutive elements sequence found.\n",
-    "\n",
-    "5. We iterate through the elements of the `num_set`. For each element `num`, we check if `num - 1` exists in the `num_set`. If it doesn't exist, it means `num` is the starting element of a potential consecutive sequence.\n",
-    "\n",
-    "6. Inside the loop, we initialize two variables: `current_num` to the current element `num` and `current_length` to 1. We start with a length of 1 because `num` itself is part of the sequence.\n",
-    "\n",
-    "7. We then enter a while loop that continues as long as `current_num + 1` exists in the `num_set`. This means we are incrementing the consecutive sequence.\n",
-    "\n",
-    "8. Inside the while loop, we increment `current_num` by 1 and also increment `current_length` by 1 to account for the next consecutive element.\n",
-    "\n",
-    "9. We compare `current_length` with the `max_length` and update `max_length` if the current sequence is longer.\n",
-    "\n",
-    "10. After the while loop, we move to the next element in the outer loop and repeat the process.\n",
-    "\n",
-    "11. Finally, we return the `max_length` as the result, which represents the length of the longest consecutive elements sequence in the `nums` array.\n",
-    "\n",
-    "The key idea here is to use a set to efficiently check for the existence of elements and to iterate through the elements, considering each element as the potential start of a consecutive sequence. By doing this, we can find the longest consecutive sequence in O(n) time complexity, where n is the number of elements in the array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "nums1 = [100, 4, 200, 1, 3, 2]\n",
-    "print(longestConsecutive(nums1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "9\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "nums2 = [0, 3, 7, 2, 5, 8, 4, 6, 0, 1]\n",
-    "print(longestConsecutive(nums2)) "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The code is designed to run in O(n) time complexity, where n is the number of elements in the `nums` array. Here's the breakdown:\n",
-    "\n",
-    "1. Constructing the `num_set` by inserting all elements from `nums` into it takes O(n) time because we perform an insertion operation for each element in `nums`.\n",
-    "\n",
-    "2. The main loop iterates through the elements in `num_set`. In the worst case, each element is visited only once, so the loop itself takes O(n) time.\n",
-    "\n",
-    "3. Within the loop, we have a while loop that may also take O(n) time in the worst case. However, this while loop is nested inside the main loop, so its overall time complexity remains O(n).\n",
-    "\n",
-    "Therefore, the overall time complexity of the code is O(n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity of the code is determined by the space used by the `num_set` and a few additional variables. Here's the breakdown:\n",
-    "\n",
-    "1. `num_set` is a set that stores the unique elements from `nums`. In the worst case, it can store all n elements from `nums`, so the space complexity is O(n).\n",
-    "\n",
-    "2. The additional variables used, such as `max_length`, `current_num`, and `current_length`, have constant space requirements and do not depend on the size of the input array.\n",
-    "\n",
-    "Therefore, the overall space complexity of the code is O(n) due to the space used by the `num_set`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. Write an algorithm that not only returns the length of the longest consecutive elements sequence but also returns the actual consecutive sequence itself.\n",
-    "\n",
-    "2. Extend the problem to allow elements to be considered consecutive if they are within a certain absolute difference (e.g., less than or equal to k) instead of exactly 1. Write an algorithm that finds the longest such sequence and runs in O(n) time."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb b/docs/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb
deleted file mode 100644
index f7dbbe9..0000000
--- a/docs/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb	
+++ /dev/null
@@ -1,210 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 217. Contains Duplicate\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Contains Duplicate problem on LeetCode, click here!](https://leetcode.com/problems/contains-duplicate/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `nums.length` <= $10^5$\n",
-    "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def containsDuplicate(nums):\n",
-    "    # Create an empty set to store unique elements\n",
-    "    unique_elements = set()\n",
-    "    \n",
-    "    # Iterate through the array\n",
-    "    for num in nums:\n",
-    "        # If the element is already in the set, return True\n",
-    "        if num in unique_elements:\n",
-    "            return True\n",
-    "        # Otherwise, add it to the set\n",
-    "        else:\n",
-    "            unique_elements.add(num)\n",
-    "    \n",
-    "    # If the loop completes without finding duplicates, return False\n",
-    "    return False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The `containsDuplicate` function takes a single argument, `nums`, which is the input integer array.\n",
-    "\n",
-    "2. Inside the function, an empty set called `unique_elements` is created. This set will be used to keep track of unique elements in the input array.\n",
-    "\n",
-    "3. The function then iterates through the input array `nums` using a `for` loop.\n",
-    "\n",
-    "4. For each element `num` in the array, it checks whether `num` is already in the `unique_elements` set using the `if num in unique_elements:` condition.\n",
-    "\n",
-    "5. If `num` is already in the set, it means there is a duplicate element in the array, and the function immediately returns `True`.\n",
-    "\n",
-    "6. If `num` is not in the set, it is added to the `unique_elements` set using `unique_elements.add(num)`.\n",
-    "\n",
-    "7. The loop continues to the next element, and the process repeats.\n",
-    "\n",
-    "8. If the loop completes without finding any duplicates, it means that all elements in the array are distinct, and the function returns `False`.\n",
-    "\n",
-    "The code efficiently utilizes a set data structure to keep track of unique elements while iterating through the array, allowing it to quickly detect duplicate elements. This code meets the problem's requirements and constraints, as explained earlier."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "nums1 = [1, 2, 3, 1]\n",
-    "print(containsDuplicate(nums1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "nums2 = [1, 2, 3, 4]\n",
-    "print(containsDuplicate(nums2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "nums3 = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]\n",
-    "print(containsDuplicate(nums3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for the \"Contains Duplicate\" problem:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The primary operation that affects the time complexity is the `for` loop that iterates through the input array `nums`. In the worst case, the loop will iterate through all `n` elements of the array, where `n` is the length of the input array.\n",
-    "\n",
-    "- Iterating through the array: O(n)\n",
-    "\n",
-    "Inside the loop, we perform two operations:\n",
-    "\n",
-    "1. Checking whether an element exists in the `unique_elements` set (`if num in unique_elements`). This operation has an average time complexity of O(1) for a set.\n",
-    "2. Adding an element to the `unique_elements` set (`unique_elements.add(num)`), which also has an average time complexity of O(1) for a set.\n",
-    "\n",
-    "Since these operations are performed for each element in the array, the overall time complexity remains O(n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity is determined by the additional data structures used in the code, which are the `unique_elements` set.\n",
-    "\n",
-    "- `unique_elements` set: This set stores unique elements from the input array. In the worst case, if all elements in the input array are distinct, the set will store all `n` elements.\n",
-    "\n",
-    "Therefore, the space complexity is O(n) because, in the worst case, the set's size will grow linearly with the input array's size.\n",
-    "\n",
-    "**In summary**, the time complexity of the code is O(n), where `n` is the length of the input array, and the space complexity is also O(n) in the worst case. This code is efficient and meets the constraints of the problem."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **K Duplicates:** Modify the problem to return `true` if there are exactly K duplicate elements in the array. Write a function that takes an additional integer parameter K and returns `true` if there are exactly K duplicates.\n",
-    "2. **Frequency Count:** Write a function that returns a list of all the elements that appear more than once in the array, along with their frequencies."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb b/docs/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb
deleted file mode 100644
index c5a69c0..0000000
--- a/docs/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb	
+++ /dev/null
@@ -1,191 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 238. Product of Array Except Self\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Product of Array Except Self problem on LeetCode, click here!](https://leetcode.com/problems/product-of-array-except-self/)\n",
-    "\n",
-    "---\n",
-    "Given an integer array `nums`, return *an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.*\n",
-    "\n",
-    "The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-    "\n",
-    "You must write an algorithm that runs in O(n) time and without using the division operation.\n",
-    "\n",
-    "Constraints:\n",
-    "- 2 <= `nums.length` <= $10^5$\n",
-    "- `-30 <= nums[i] <= 30`\n",
-    "- The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-    "\n",
-    "**Follow-up:** Can you solve the problem in O(1) extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def productExceptSelf(nums):\n",
-    "    n = len(nums)\n",
-    "    \n",
-    "    # Initialize two arrays to store products to the left and right of each element\n",
-    "    left_products = [1] * n\n",
-    "    right_products = [1] * n\n",
-    "    \n",
-    "    # Calculate products to the left of each element\n",
-    "    left_product = 1\n",
-    "    for i in range(n):\n",
-    "        left_products[i] = left_product\n",
-    "        left_product *= nums[i]\n",
-    "    \n",
-    "    # Calculate products to the right of each element\n",
-    "    right_product = 1\n",
-    "    for i in range(n - 1, -1, -1):\n",
-    "        right_products[i] = right_product\n",
-    "        right_product *= nums[i]\n",
-    "    \n",
-    "    # Calculate the final answer using left and right products\n",
-    "    answer = [left_products[i] * right_products[i] for i in range(n)]\n",
-    "    \n",
-    "    return answer"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `n` is initialized as the length of the `nums` array.\n",
-    "2. Two arrays, `left_products` and `right_products`, are created, each with a length of `n` and filled with ones initially.\n",
-    "3. `left_product` is initialized to 1.\n",
-    "4. A loop iterates through the `nums` array, from left to right. For each element at index `i`, it stores the product of all elements to its left (including itself) in the `left_products` array and updates `left_product` accordingly.\n",
-    "5. `right_product` is initialized to 1.\n",
-    "6. Another loop iterates through the `nums` array in reverse order, from right to left. For each element at index `i`, it stores the product of all elements to its right (including itself) in the `right_products` array and updates `right_product`.\n",
-    "7. The final `answer` list is constructed by multiplying corresponding elements from the `left_products` and `right_products` arrays for each index `i`.\n",
-    "8. The function returns the `answer` list, which contains the desired results.\n",
-    "\n",
-    "Overall, this algorithm efficiently computes the product of all elements except the current element, as required. It runs in O(n) time complexity and uses O(n) extra space, which is within the problem's constraints."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[24, 12, 8, 6]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "nums = [1,2,3,4]\n",
-    "result1 = productExceptSelf(nums)\n",
-    "print(result1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[0, 0, 9, 0, 0]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums = [-1,1,0,-3,3]\n",
-    "result2 = productExceptSelf(nums)\n",
-    "print(result2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. The first loop iterates through the `nums` array to calculate the products to the left of each element. This loop runs in O(n) time, where n is the length of the `nums` array.\n",
-    "2. The second loop also iterates through the `nums` array in reverse order to calculate the products to the right of each element. This loop also runs in O(n) time.\n",
-    "3. The final loop constructs the `answer` list by multiplying elements from the `left_products` and `right_products` arrays. This loop runs in O(n) time.\n",
-    "\n",
-    "Since all the loops are independent and sequential, the overall time complexity is O(n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. Two additional arrays, `left_products` and `right_products`, are created with a length of `n`, where `n` is the length of the `nums` array. Therefore, these arrays consume O(n) extra space.\n",
-    "2. The `left_product` and `right_product` variables are used to keep track of the product to the left and right of each element, respectively. These variables occupy O(1) extra space.\n",
-    "3. The `answer` list, which contains the final results, also consumes O(n) space, but it's not counted toward the extra space complexity analysis as per the problem's constraints.\n",
-    "\n",
-    "Overall, the space complexity is O(n) for this solution."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Handle Zeros in the Input**: Modify the solution to handle cases where the input array contains zeros. For example, if `nums = [0, 1, 2, 3, 0]`, the output should be `[0, 0, 0, 0, 0]` because any element multiplied by zero results in zero.\n",
-    "\n",
-    "2. **In-Place Modification**: Attempt to solve the problem with in-place modification of the `nums` array, where the output is stored directly in the input array without using any additional space."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb b/docs/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb
deleted file mode 100644
index 5649123..0000000
--- a/docs/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb	
+++ /dev/null
@@ -1,192 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 242. Valid Anagram\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Valid Anagram problem on LeetCode, click here!](https://leetcode.com/problems/valid-anagram/)\n",
-    "\n",
-    "---\n",
-    "Given two strings `s` and `t`, return `true` if `t` is an anagram of `s`, and `false` otherwise.\n",
-    "\n",
-    "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `s.length`, `t.length` <= $5 * 10^4$\n",
-    "- `s` and `t` consist of lowercase English letters.\n",
-    "\n",
-    "**Follow-up:** What if the inputs contain Unicode characters? How would you adapt your solution to such a case?\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def isAnagram(s: str, t: str) -> bool:\n",
-    "    # Check if the lengths of s and t are not equal\n",
-    "    if len(s) != len(t):\n",
-    "        return False\n",
-    "\n",
-    "    # Create dictionaries to store character frequencies for s and t\n",
-    "    countS, countT = {}, {}\n",
-    "\n",
-    "    # Count character frequencies in s\n",
-    "    for i in range(len(s)):\n",
-    "        countS[s[i]] = 1 + countS.get(s[i], 0)\n",
-    "\n",
-    "    # Count character frequencies in t\n",
-    "    for i in range(len(t)):\n",
-    "        countT[t[i]] = 1 + countT.get(t[i], 0)\n",
-    "\n",
-    "    # Check if the character frequencies in s and t are the same\n",
-    "    return countS == countT"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The code defines a class called `Solution` with a method `isAnagram` that takes two input strings, `s` and `t`, and returns a boolean (`True` or `False`) based on whether `t` is an anagram of `s`.\n",
-    "\n",
-    "2. The first check performed in the method is whether the lengths of `s` and `t` are equal. If they are not equal, it immediately returns `False` because strings of different lengths cannot be anagrams of each other.\n",
-    "\n",
-    "3. Two dictionaries, `countS` and `countT`, are created to store the frequency of characters in `s` and `t`, respectively. These dictionaries will be used to count the occurrences of each character in the strings.\n",
-    "\n",
-    "4. The code then enters a loop to iterate through each character in string `s` using a `for` loop. Inside the loop, it updates the `countS` dictionary. The line `countS[s[i]] = 1 + countS.get(s[i], 0)` increments the count of character `s[i]` in `countS` by 1. If the character is not already in the dictionary, it initializes the count to 1.\n",
-    "\n",
-    "5. Similarly, the code iterates through each character in string `t` and updates the `countT` dictionary in the same way.\n",
-    "\n",
-    "6. After counting the character frequencies in both strings, the code compares the two dictionaries using `countS == countT`. If the dictionaries are equal, it means that both strings have the same character frequencies, and therefore, `t` is an anagram of `s`. In this case, the method returns `True`.\n",
-    "\n",
-    "7. If the dictionaries are not equal, the method returns `False`, indicating that `t` is not an anagram of `s`.\n",
-    "\n",
-    "In summary, this code efficiently determines whether two input strings are anagrams by counting the character frequencies in each string using dictionaries and then comparing these counts. If the character frequencies match, the strings are considered anagrams, and the method returns `True`; otherwise, it returns `False`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "s = \"anagram\"\n",
-    "t = \"nagaram\"\n",
-    "print(isAnagram(s, t))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "s = \"rat\"\n",
-    "t = \"car\"\n",
-    "print(isAnagram(s, t))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. The code first checks whether the lengths of the input strings `s` and `t` are equal, which takes constant time. This check has a time complexity of O(1).\n",
-    "\n",
-    "2. The code then iterates through both strings, `s` and `t`, once to count the character frequencies. Since both strings have a maximum length of 5 * 10^4, the worst-case scenario is that the code iterates through 5 * 10^4 characters in each string. This results in a linear time complexity of O(n), where n is the length of the longer of the two input strings (`s` or `t`).\n",
-    "\n",
-    "3. Finally, the code compares the two dictionaries `countS` and `countT` to check if the character frequencies match. This comparison takes O(n) time in the worst case, where n is the length of the longer string.\n",
-    "\n",
-    "Overall, the time complexity of the code is O(n), where n is the length of the longer input string.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. The code uses two dictionaries, `countS` and `countT`, to store the character frequencies of the input strings `s` and `t`. In the worst case, both dictionaries can contain all unique characters from the input strings. Since the input strings can have a maximum length of 5 * 10^4, the space complexity for these dictionaries is O(5 * 10^4) or simply O(n), where n is the length of the longer input string.\n",
-    "\n",
-    "2. The code uses a few additional variables for bookkeeping, but these variables have constant space requirements and do not depend on the input size. Therefore, they do not significantly impact the overall space complexity.\n",
-    "\n",
-    "**In summary**, the space complexity of the code is O(n), where n is the length of the longer input string."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Anagram Chains**: Given a list of words, find the longest chain of anagrams, where each word in the chain is an anagram of the previous word. For example, given [\"bat\", \"tab\", \"cat\", \"tac\", \"dog\"], the longest anagram chain is [\"bat\", \"tab\", \"cat\", \"tac\"].\n",
-    "\n",
-    "2. **Minimum Deletions to Make Anagrams**: Given two strings `s` and `t`, find the minimum number of character deletions required in both strings to make them anagrams of each other. For example, if `s = \"hello\"` and `t = \"billion\"`, you can remove the characters \"heo\" from `s` and \"billi\" from `t` to make them anagrams (\"lo\" and \"on\" are left in the two strings)."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb b/docs/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb
deleted file mode 100644
index 1b11faf..0000000
--- a/docs/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 347. Top K Frequent Elements\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Top K Frequent Elements problem on LeetCode, click here!](https://leetcode.com/problems/top-k-frequent-elements/)\n",
-    "\n",
-    "---\n",
-    "Given an integer array `nums` and an integer `k`, return the `k` most frequent elements. You may return the answer in **any order**.\n",
-    "\n",
-    "Constraints:\n",
-    "- 1 <= `nums.length` <= $10^5$\n",
-    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-    "- `k` is in the range `[1, the number of unique elements in the array]`.\n",
-    "- It is **guaranteed** that the answer is **unique**.\n",
-    "\n",
-    "**Follow-up:** Your algorithm's time complexity must be better than $O(n\\ log\\ n)$, where n is the array's size.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def topKFrequent(nums, k):\n",
-    "    # Create a dictionary to store the frequency of each element\n",
-    "    count = {}\n",
-    "    # Create a list of lists to store elements with the same frequency\n",
-    "    frequency = [[] for i in range(len(nums) + 1)]\n",
-    "\n",
-    "    # Count the frequency of each element in nums\n",
-    "    for n in nums:\n",
-    "        count[n] = 1 + count.get(n, 0)\n",
-    "\n",
-    "    # Place elements in the freq list according to their frequency\n",
-    "    for n, c in count.items():\n",
-    "        frequency[c].append(n)\n",
-    "\n",
-    "    res = []\n",
-    "    # Traverse freq list from the end (higher frequencies)\n",
-    "    for i in range(len(frequency) - 1, 0, -1):\n",
-    "        for n in frequency[i]:\n",
-    "            res.append(n)\n",
-    "            if len(res) == k:\n",
-    "                return res"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `def topKFrequent(nums, k):`: This is a function that takes two arguments: `nums`, which is the input array of integers, and `k`, which is the number of most frequent elements to return.\n",
-    "\n",
-    "2. `count = {}`: This dictionary `count` will be used to store the frequency of each unique element in the `nums` array. The keys are elements from the input array, and the values are their corresponding frequencies.\n",
-    "\n",
-    "3. `frequency = [[] for i in range(len(nums) + 1)]`: This creates a list of empty lists called `frequency`. It's used to store elements based on their frequencies, similar to the `freq` list in the previous code. The size of this list is set to be one greater than the length of the input `nums`.\n",
-    "\n",
-    "4. `for n in nums:`: This loop iterates through each element `n` in the input `nums` array.\n",
-    "\n",
-    "5. `count[n] = 1 + count.get(n, 0)`: This line counts the frequency of each element `n` in the `nums` array. It uses the `count.get(n, 0)` method to retrieve the current count of `n` from the `count` dictionary. If `n` is not in the dictionary, it defaults to 0. It then increments the count by 1.\n",
-    "\n",
-    "6. After the above loop, the `count` dictionary will contain counts of each unique element in the `nums` array.\n",
-    "\n",
-    "7. `for n, c in count.items():`: This loop iterates through the items (key-value pairs) of the `count` dictionary. `n` represents the element, and `c` represents its frequency.\n",
-    "\n",
-    "8. `frequency[c].append(n)`: This line places the element `n` into the bucket corresponding to its frequency `c`. Buckets are represented by the `frequency` list. For example, if an element `n` has a frequency of 3, it will be added to `frequency[3]`.\n",
-    "\n",
-    "9. After this loop, the `frequency` list will contain buckets of elements grouped by their frequencies.\n",
-    "\n",
-    "10. `res = []`: This list `res` will be used to store the k most frequent elements.\n",
-    "\n",
-    "11. `for i in range(len(freq) - 1, 0, -1):`: This loop iterates in reverse order through the `frequency` list, starting from the highest frequency and going down to 1. Note that there is a typo here; it should be `frequency` instead of `freq`.\n",
-    "\n",
-    "12. `for n in frequency[i]:`: For each element `n` in the current bucket (i.e., elements with frequency `i`), it appends `n` to the result list `res`.\n",
-    "\n",
-    "13. `if len(res) == k:`: Once `res` contains k elements, the code exits and returns the result.\n",
-    "\n",
-    "The code efficiently finds the k most frequent elements in the input array without using sorting, similar to the previous explanation, with the only difference being the variable names (e.g., `frequency` instead of `freq`)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 2]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "nums = [1,1,1,2,2,3]\n",
-    "k = 2\n",
-    "result1 = topKFrequent(nums, k)\n",
-    "print(result1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums = [1]\n",
-    "k = 1\n",
-    "result2 = topKFrequent(nums, k)\n",
-    "print(result2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. The first loop that counts the frequency of elements by iterating through `nums` has a time complexity of O(n), where n is the number of elements in the input array `nums`.\n",
-    "\n",
-    "2. The second loop iterates through the keys in the `count` dictionary, which has at most `k` unique elements (as per the constraints). Therefore, the second loop also has a time complexity of O(k).\n",
-    "\n",
-    "3. The third loop iterates through the `frequency` list, which has a length equal to the maximum frequency of elements in `nums`. In the worst case, this loop can have a time complexity of O(n), where n is the number of elements in `nums`.\n",
-    "\n",
-    "Overall, the dominant factor in terms of time complexity is the loop that iterates through the `count` dictionary. So, the total time complexity of the code is O(n + k).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The `count` dictionary stores the frequency of each unique element in `nums`. In the worst case, it can have at most `k` unique elements, so the space complexity for `count` is O(k).\n",
-    "\n",
-    "2. The `frequency` list is used to store elements grouped by their frequencies. It has a length of `len(nums) + 1`, which can be at most 105 based on the constraints. Therefore, the space complexity for `frequency` is O(105).\n",
-    "\n",
-    "3. The `res` list stores the k most frequent elements, which can be at most `k` elements, so the space complexity for `res` is O(k).\n",
-    "\n",
-    "**In summary**, the space complexity is dominated by the `count` dictionary and the `frequency` list, both of which have a space complexity of O(k + 105)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimized k Most Frequent Elements:** Modify the code to find the k most frequent elements in an array while ensuring that the time complexity is O(n + klogk). You can use a priority queue (heap) to achieve this.\n",
-    "\n",
-    "2. **Handling Duplicate Frequencies:** Extend the code to handle cases where multiple elements have the same frequency and are among the k most frequent elements. Ensure that the output contains exactly k elements."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb b/docs/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb
deleted file mode 100644
index 256e8f7..0000000
--- a/docs/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 49. Group Anagrams\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Group Anagrams problem on LeetCode, click here!](https://leetcode.com/problems/group-anagrams/)\n",
-    "\n",
-    "---\n",
-    "Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.\n",
-    "\n",
-    "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `strs.length <= $10^4$\n",
-    "- `0 <= strs[i].length <= 100`\n",
-    "- `strs[i]` consists of lowercase English letters.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from collections import defaultdict\n",
-    "\n",
-    "\n",
-    "def groupAnagrams(strs):\n",
-    "    ans = defaultdict(list)\n",
-    "\n",
-    "    # Iterate through the list of input strings\n",
-    "    for s in strs:\n",
-    "        # Initialize a list to represent character counts for each character (a-z)\n",
-    "        count = [0] * 26\n",
-    "\n",
-    "        # Count the occurrences of each character in the current word\n",
-    "        for c in s:\n",
-    "            count[ord(c) - ord(\"a\")] += 1\n",
-    "\n",
-    "        # Use a tuple of character counts as the key and append the word to the anagram group\n",
-    "        ans[tuple(count)].append(s)\n",
-    "\n",
-    "    # Convert the values (lists of anagrams) to a list of lists\n",
-    "    return list(ans.values())"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. Inside the method, a `defaultdict` called `ans` is created to store anagram groups. This dictionary will have a list as its default value, meaning that each key in the dictionary will be associated with an empty list by default.\n",
-    "\n",
-    "2. The code then iterates through the list of input strings, `strs`, using a `for` loop. For each word `s` in `strs`, it performs the following steps:\n",
-    "\n",
-    "   - It initializes a list `count` of length 26, where each element represents the count of a specific character (a-z) in the word `s`.\n",
-    "\n",
-    "   - The code then iterates through the characters of `s`. For each character `c`, it increments the corresponding count in the `count` list based on its ASCII value.\n",
-    "\n",
-    "   - After counting the characters in `s`, it converts the `count` list into a tuple `tuple(count)` to use as a key for the `ans` dictionary.\n",
-    "\n",
-    "   - It appends the word `s` to the list associated with the key (tuple) in the `ans` dictionary. This groups all anagrams of the same word together under the same key.\n",
-    "\n",
-    "3. After processing all words in `strs`, the code converts the values of the `ans` dictionary (which are lists of anagrams) to a list of lists using the `list()` constructor.\n",
-    "\n",
-    "4. Finally, the code returns the list of anagram groups, which is the result of grouping the anagrams in the input list.\n",
-    "\n",
-    "In summary, the code efficiently groups anagrams together by counting the characters in each word and using a dictionary to store them under the same key. The result is a list of lists, where each inner list represents a group of anagrams."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: Group anagrams from a list of words\n",
-    "strs1 = [\"eat\", \"tea\", \"tan\", \"ate\", \"nat\", \"bat\"]\n",
-    "result1 = groupAnagrams(strs1)\n",
-    "print(result1)\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[['']]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2: Group anagrams from a list with an empty string\n",
-    "strs2 = [\"\"]\n",
-    "result2 = groupAnagrams(strs2)\n",
-    "print(result2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "c8367e2a-e7d5-4e18-8235-2870e341a04b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[['a']]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3: Group anagrams from a list with a single word\n",
-    "strs3 = [\"a\"]\n",
-    "result3 = groupAnagrams(strs3)\n",
-    "print(result3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `groupAnagrams` method:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. First, we create an instance of the `Solution` class, which is an O(1) operation.\n",
-    "\n",
-    "2. Inside the `groupAnagrams` method, we iterate through the list of input strings `strs` once in a loop:\n",
-    "   - For each word in `strs`, we perform character counting, which is done in O(K) time, where K is the maximum length of a word in the list.\n",
-    "\n",
-    "3. The overall time complexity of the method is O(N * K), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. We use a `defaultdict` called `ans` to store the anagram groups. In the worst case, each word belongs to a distinct anagram group, so the space complexity for `ans` is O(N).\n",
-    "\n",
-    "2. Within the loop, we create a `count` list of length 26 to store character counts for each word. This is a fixed-size list and does not depend on the input size, so it has a constant space complexity of O(26), which is effectively O(1).\n",
-    "\n",
-    "3. The space used for other variables is also constant and does not depend on the input size.\n",
-    "\n",
-    "4. The final result, `result`, is a list of lists that contains the grouped anagrams. In the worst case, each word is a distinct anagram group, so the space complexity for `result` is O(N).\n",
-    "\n",
-    "5. The overall space complexity of the method is O(N) due to the space used by `ans` and `result`. The space used by the `count` list and other variables is constant and does not significantly affect the overall space complexity.\n",
-    "\n",
-    "**In summary,** the time complexity of the `groupAnagrams` method is O(N * K), and the space complexity is O(N), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Anagramic Palindromes**: Given a list of strings, find all the groups of anagrams where each group contains words that can be rearranged into a palindrome. For example, in the list [\"bat\", \"tab\", \"act\", \"tac\", \"dog\"], there are two groups: [\"bat\", \"tab\", \"act\", \"tac\"] and [\"dog\"].\n",
-    "\n",
-    "2. **Largest Anagram Group**: Given a list of strings, find the largest group of anagrams. Return the list of anagrams in that group. If there are multiple largest groups, return all of them."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb b/docs/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb
deleted file mode 100644
index 2fae2a5..0000000
--- a/docs/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb	
+++ /dev/null
@@ -1,198 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 659· Encode and Decode Strings\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Encode and Decode Strings problem on LintCode, click here!](https://www.lintcode.com/problem/659/)\n",
-    "\n",
-    "---\n",
-    "Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.\n",
-    "\n",
-    "Please implement `encode` and `decode`.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class Codec:\n",
-    "\n",
-    "    def encode(self, strs):\n",
-    "        \"\"\"Encodes a list of strings to a single string.\n",
-    "        \n",
-    "        :type strs: List[str]\n",
-    "        :rtype: str\n",
-    "        \"\"\"\n",
-    "        encoded = \"\"\n",
-    "        for s in strs:\n",
-    "            encoded += s.replace(':', '::') + ':;'\n",
-    "        return encoded\n",
-    "\n",
-    "    def decode(self, s):\n",
-    "        \"\"\"Decodes a single string to a list of strings.\n",
-    "        \n",
-    "        :type s: str\n",
-    "        :rtype: List[str]\n",
-    "        \"\"\"\n",
-    "        decoded = []\n",
-    "        i = 0\n",
-    "        while i < len(s):\n",
-    "            end = s.find(':;', i)\n",
-    "            if end == -1:\n",
-    "                end = len(s)\n",
-    "            decoded.append(s[i:end].replace('::', ':'))\n",
-    "            i = end + 2\n",
-    "        return decoded"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We define a class called `Codec` to encapsulate the encoding and decoding operations for a list of strings.\n",
-    "\n",
-    "2. In the `encode` method, we take a list of strings (`strs`) as input and return a single encoded string. The encoding process involves concatenating the strings together with a delimiter `:;`. We iterate through each string in the input list, replace any occurrence of `:` with `::` (to avoid conflicts with the delimiter), and then append `:;` to indicate the end of that string.\n",
-    "\n",
-    "3. In the `decode` method, we take an encoded string (`s`) as input and return a list of strings. The decoding process involves splitting the encoded string using the `:;` delimiter. We start from the beginning of the encoded string (`i = 0`) and repeatedly find the next occurrence of `:;`. We extract the substring between the current position (`i`) and the next delimiter (`end`). Before adding it to the result list, we replace any `::` with `:` to revert the encoding. We then update the current position `i` to be after the delimiter, so we can find the next substring in the next iteration.\n",
-    "\n",
-    "4. Finally, we create an instance of the `Codec` class and test it with two examples:\n",
-    "\n",
-    "By using the `:;` delimiter and handling the `::` escaping, this code can encode and decode lists of strings, preserving the original content even if it contains colons or the delimiter itself."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "['lint', 'code', 'love', 'you']\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "codec = Codec()\n",
-    "\n",
-    "input1 = [\"lint\", \"code\", \"love\", \"you\"]\n",
-    "encoded1 = codec.encode(input1)\n",
-    "decoded1 = codec.decode(encoded1)\n",
-    "print(decoded1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "['we', 'say', ':', 'yes']\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "codec = Codec() \n",
-    "\n",
-    "input2 = [\"we\", \"say\", \":\", \"yes\"]\n",
-    "encoded2 = codec.encode(input2)\n",
-    "decoded2 = codec.decode(encoded2)\n",
-    "print(decoded2)  "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `encode` and `decode` methods in the provided code.\n",
-    "\n",
-    "**Encode Method (`encode`):**\n",
-    "\n",
-    "- Time Complexity: O(n * m)\n",
-    "  - Here, `n` is the number of strings in the input list `strs`, and `m` is the average length of these strings.\n",
-    "  - In the worst case, for each string in the list, we iterate over its characters once to replace any colons (`:`) with double colons (`::`) and then append `:;`. This is done for each string in the list.\n",
-    "\n",
-    "- Space Complexity: O(n * m)\n",
-    "  - The space complexity is also O(n * m) because we build the encoded string by concatenating the modified strings. In the worst case, the encoded string can be of the same length as the original strings.\n",
-    "\n",
-    "**Decode Method (`decode`):**\n",
-    "\n",
-    "- Time Complexity: O(n * m)\n",
-    "  - Similar to the encode method, we iterate through the encoded string in the decode method. In the worst case, we may have to scan each character in the encoded string to find the `:;` delimiter.\n",
-    "\n",
-    "- Space Complexity: O(n * m)\n",
-    "  - The space complexity of the decode method is also O(n * m) because we build the list of decoded strings, which can be of the same length as the encoded string.\n",
-    "\n",
-    "Overall, both the `encode` and `decode` methods have time and space complexities of O(n * m), where `n` is the number of strings in the input list, and `m` is the average length of these strings. The space complexity arises from storing the encoded or decoded strings, and the time complexity arises from iterating through the characters in these strings."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple Delimiters**: Modify the code to allow for multiple delimiters, not just one. Each string could have its own delimiter, and the code should be able to handle this complexity.\n",
-    "\n",
-    "2. **Optimize Encoding**: Modify the `encode` method to achieve a more efficient encoding algorithm. Try to minimize the length of the encoded string while ensuring that it can be correctly decoded."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb b/docs/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb
deleted file mode 100644
index e330eb0..0000000
--- a/docs/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb	
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 11. Container With Most Water\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Container With Most Water problem on LeetCode, click here!](https://leetcode.com/problems/container-with-most-water/)\n",
-    "\n",
-    "---\n",
-    "You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the $i^{th}$ line are `(i, 0)` and `(i, height[i])`.\n",
-    "\n",
-    "Find two lines that together with the x-axis form a container, such that the container contains the most water.\n",
-    "\n",
-    "Return *the maximum amount of water a container can store.*\n",
-    "\n",
-    "**Notice** that you may not slant the container.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `n == height.length`\n",
-    "- 2 <= `n` <= $10^5$\n",
-    "- 0 <= `height[i]` <= $10^4$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def maxArea(height):\n",
-    "    # Initialize the maximum area to 0.\n",
-    "    max_area = 0\n",
-    "    # Initialize two pointers, one at the beginning (left) and one at the end (right) of the array.\n",
-    "    left = 0\n",
-    "    right = len(height) - 1\n",
-    "\n",
-    "    # Iterate until the left pointer is less than the right pointer.\n",
-    "    while left < right:\n",
-    "        # Calculate the height of the container, which is the minimum height of the two lines.\n",
-    "        h = min(height[left], height[right])\n",
-    "        # Calculate the width of the container, which is the distance between the two pointers.\n",
-    "        width = right - left\n",
-    "        # Calculate the area of the container using height and width and update max_area if it's greater.\n",
-    "        max_area = max(max_area, h * width)\n",
-    "\n",
-    "        # Move the pointer with the shorter line inward.\n",
-    "        if height[left] < height[right]:\n",
-    "            left += 1\n",
-    "        else:\n",
-    "            right -= 1\n",
-    "\n",
-    "    # Continue this process until left < right, and then return the maximum area.\n",
-    "    return max_area"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start with initializing `max_area` to 0, which will store the maximum area of water that can be contained by the lines.\n",
-    "\n",
-    "2. Two pointers, `left` and `right`, are initialized to the beginning and end of the input `height` array.\n",
-    "\n",
-    "3. The main loop runs while `left` is less than `right`. This ensures that we are checking all possible pairs of lines in a systematic way.\n",
-    "\n",
-    "4. Inside the loop:\n",
-    "   - We calculate the height of the container `h` as the minimum of the heights at the `left` and `right` pointers. This represents the height of the water the container can hold.\n",
-    "   - We calculate the width of the container as the difference between `right` and `left`. This represents the distance between the two lines.\n",
-    "   - We calculate the area of the container using `h * width` and update `max_area` if the calculated area is greater than the current `max_area`.\n",
-    "\n",
-    "5. After calculating the area and updating `max_area`, we move one of the pointers. We move the pointer pointing to the shorter line inward because moving the pointer pointing to the taller line won't increase the height and will only decrease the width, which will result in a smaller area.\n",
-    "\n",
-    "6. We continue this process until `left` is no longer less than `right`, indicating that we have checked all possible pairs of lines.\n",
-    "\n",
-    "7. Finally, we return the `max_area` which contains the maximum area of water that can be contained by two lines from the input array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output: 49\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "height1 = [1, 8, 6, 2, 5, 4, 8, 3, 7]\n",
-    "result1 = maxArea(height1)\n",
-    "print(\"Output:\", result1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output: 1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "height2 = [1, 1]\n",
-    "result2 = maxArea(height2)\n",
-    "print(\"Output:\", result2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output: 45\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Additional Example\n",
-    "height3 = [3, 9, 3, 4, 7, 2, 12, 6]\n",
-    "result3 = maxArea(height3)\n",
-    "print(\"Output:\", result3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `maxArea` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The function uses a two-pointer approach to find the maximum area of water. In the worst case, both the left and right pointers traverse the entire input array once. Therefore, the time complexity is O(n), where n is the number of elements in the `height` array.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The function uses a constant amount of extra space for variables like `max_area`, `left`, `right`, `h`, and `width`. It does not use any data structures whose space consumption depends on the input size. As a result, the space complexity is O(1), which means it has constant space complexity.\n",
-    "\n",
-    "**In summary,** the `maxArea` function has a time complexity of O(n) and a space complexity of O(1), making it an efficient algorithm for finding the maximum area of water that can be contained by two lines in the input array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the maxArea function to not only return the maximum area but also the indices of the two lines that form the container with the maximum area.\n",
-    "2. **Variant Problem:** Instead of finding the maximum area of water, find the minimum amount of water required to fill all the gaps between the lines. You'll need to return the total water volume needed."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb b/docs/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb
deleted file mode 100644
index 12c7daa..0000000
--- a/docs/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb	
+++ /dev/null
@@ -1,210 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 125. Valid Palindrome\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Valid Palindrome problem on LeetCode, click here!](https://leetcode.com/problems/valid-palindrome/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.\n",
-    "\n",
-    "Given a string `s`, *return `true` if it is a **palindrome**, or `false` otherwise.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `s.length` <= $2*10^5$\n",
-    "- `s` consists only of printable ASCII characters.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def isPalindrome(s):\n",
-    "    # Step 1: Convert the input string to lowercase\n",
-    "    s = s.lower()\n",
-    "    \n",
-    "    # Step 2: Remove all non-alphanumeric characters from the string\n",
-    "    cleaned_s = ''.join(c for c in s if c.isalnum())\n",
-    "    \n",
-    "    # Step 3: Initialize two pointers, one at the beginning and one at the end of the cleaned string\n",
-    "    left, right = 0, len(cleaned_s) - 1\n",
-    "    \n",
-    "    # Step 4: Compare characters using the two pointers while moving them towards each other\n",
-    "    while left < right:\n",
-    "        if cleaned_s[left] != cleaned_s[right]:\n",
-    "            return False  # If characters don't match, it's not a palindrome\n",
-    "        left += 1\n",
-    "        right -= 1\n",
-    "    \n",
-    "    # Step 5: If the loop completes without returning False, it's a palindrome\n",
-    "    return True"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The `isPalindrome` function takes an input string `s`.\n",
-    "\n",
-    "2. It starts by converting the input string to lowercase using `s.lower()`. This step ensures that the function is case-insensitive when checking for palindromes.\n",
-    "\n",
-    "3. Next, it removes all non-alphanumeric characters from the string and stores the result in the `cleaned_s` variable. It does this by iterating through each character in the input string and only keeping characters that are alphanumeric (letters and digits). This step removes spaces, punctuation, and other non-alphanumeric characters.\n",
-    "\n",
-    "4. The function then initializes two pointers, `left` and `right`, which will be used to compare characters from the beginning and end of the cleaned string.\n",
-    "\n",
-    "5. In a while loop, it compares characters using the two pointers while moving them towards each other. The loop continues until the `left` pointer is less than the `right` pointer. This is done to check all characters in the cleaned string.\n",
-    "\n",
-    "6. Inside the loop, it checks if the characters at the `left` and `right` pointers don't match. If they don't match, it means the input is not a palindrome, so the function returns `False`.\n",
-    "\n",
-    "7. If the characters at the `left` and `right` pointers match, the pointers are moved closer to each other by incrementing `left` and decrementing `right`.\n",
-    "\n",
-    "8. The loop continues to compare characters until it either finds a mismatch (returning `False`) or until the `left` pointer is greater than or equal to the `right` pointer. If the loop completes without finding a mismatch, it means the input is a palindrome, so the function returns `True`.\n",
-    "\n",
-    "In summary, the code converts the input string to lowercase, removes non-alphanumeric characters, and then uses two pointers to compare characters from the cleaned string's beginning and end. If all characters match during the comparison, the function returns `True`, indicating that the input is a palindrome. If any characters do not match, it returns `False`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "input1 = \"A man, a plan, a canal: Panama\"\n",
-    "output1 = isPalindrome(input1)\n",
-    "print(output1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "input2 = \"race a car\"\n",
-    "output2 = isPalindrome(input2)\n",
-    "print(output2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "input3 = \"\"\n",
-    "output3 = isPalindrome(input3)\n",
-    "print(output3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `isPalindrome` function that uses two pointers:\n",
-    "\n",
-    "1. **Time Complexity:**\n",
-    "   - Converting the input string to lowercase using `s.lower()` takes O(n) time, where 'n' is the length of the input string.\n",
-    "   - Removing non-alphanumeric characters using the list comprehension `''.join(c for c in s if c.isalnum())` also takes O(n) time because it iterates through each character in the string once.\n",
-    "   - The comparison of characters using the two pointers (`left` and `right`) is done in a loop that runs until `left` is less than `right`. In the worst case, the loop iterates through half of the string, so it takes O(n/2) time, which is still considered O(n).\n",
-    "\n",
-    "   Therefore, the overall time complexity of the `isPalindrome` function is O(n), where 'n' is the length of the input string.\n",
-    "\n",
-    "2. **Space Complexity:**\n",
-    "   - The space complexity is primarily determined by the additional string `cleaned_s` created to store the cleaned version of the input string. In the worst case, if all characters in the input string are alphanumeric, this cleaned string can be as large as the original input string.\n",
-    "   - Additionally, the function uses two pointers, `left` and `right`, which consume constant space and do not depend on the size of the input.\n",
-    "\n",
-    "**In summary,** the overall space complexity of the `isPalindrome` function is O(n) in the worst case, where 'n' is the length of the input string. The space complexity is dominated by the space required for `cleaned_s`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Longest Palindromic Substring:** Given a string `s`, find and return the longest palindromic substring within `s`. For example, if `s = \"babad\"`, the function should return `\"bab\"` or `\"aba\"`.\n",
-    "2. **Palindrome with Limited Characters:** Given a string `s` and a list of characters `allowed_chars`, check if `s` is a palindrome considering only the characters from `allowed_chars`. Ignore all other characters. For example, if `s = \"A man, a plan, a canal: Panama\"` and `allowed_chars = ['a', 'm', 'n', 'p']`, the function should return `True`."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb b/docs/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb
deleted file mode 100644
index 15d189a..0000000
--- a/docs/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 15. 3Sum\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the 3Sum problem on LeetCode, click here!](https://leetcode.com/problems/3sum/)\n",
-    "\n",
-    "---\n",
-    "Given an integer array nums, *return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.*\n",
-    "\n",
-    "**Notice** that the solution set must not contain duplicate triplets.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `3 <= nums.length <= 3000`\n",
-    "- $-10^5$ <= `nums[i]` <= $10^5$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def threeSum(nums):\n",
-    "    # Sort the input array in ascending order\n",
-    "    nums.sort()\n",
-    "    triplets = []\n",
-    "    \n",
-    "    for i in range(len(nums) - 2):\n",
-    "        # Skip duplicates to avoid duplicate triplets\n",
-    "        if i > 0 and nums[i] == nums[i - 1]:\n",
-    "            continue\n",
-    "        \n",
-    "        left, right = i + 1, len(nums) - 1\n",
-    "        \n",
-    "        while left < right:\n",
-    "            # Calculate the total sum of the current triplet\n",
-    "            total = nums[i] + nums[left] + nums[right]\n",
-    "            \n",
-    "            if total == 0:\n",
-    "                # If the total is zero, we found a valid triplet\n",
-    "                triplets.append([nums[i], nums[left], nums[right]])\n",
-    "                \n",
-    "                # Skip duplicates of left and right pointers\n",
-    "                while left < right and nums[left] == nums[left + 1]:\n",
-    "                    left += 1\n",
-    "                while left < right and nums[right] == nums[right - 1]:\n",
-    "                    right -= 1\n",
-    "                \n",
-    "                # Move the pointers to the next unique elements\n",
-    "                left += 1\n",
-    "                right -= 1\n",
-    "            elif total < 0:\n",
-    "                # If the total is negative, we need to increase the sum by moving the left pointer to the right\n",
-    "                left += 1\n",
-    "            else:\n",
-    "                # If the total is positive, we need to decrease the sum by moving the right pointer to the left\n",
-    "                right -= 1\n",
-    "    \n",
-    "    return triplets"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. It starts by sorting the input array `nums` in ascending order. Sorting helps in efficiently finding triplets that sum up to zero.\n",
-    "\n",
-    "2. It initializes an empty list `triplets` to store the unique triplets that meet the given conditions.\n",
-    "\n",
-    "3. The code then iterates through the `nums` array using a loop with index `i`. This loop will consider each element of the array as a potential starting point for a triplet.\n",
-    "\n",
-    "4. Inside the loop, it checks for duplicates and skips them. This is done to ensure that the solution set does not contain duplicate triplets. If `nums[i]` is the same as the previous element `nums[i-1]`, it continues to the next iteration of the loop.\n",
-    "\n",
-    "5. Two pointers, `left` and `right`, are initialized. `left` starts just after the current element `nums[i]`, and `right` starts at the end of the array.\n",
-    "\n",
-    "6. The code enters another loop with `left` and `right` pointers, trying to find a triplet that sums up to zero.\n",
-    "\n",
-    "7. It calculates the `total` sum of the current triplet as `nums[i] + nums[left] + nums[right]`.\n",
-    "\n",
-    "8. If `total` is zero, it means a valid triplet is found, so it appends `[nums[i], nums[left], nums[right]]` to the `triplets` list. Then, it moves both the `left` and `right` pointers to their next unique elements while skipping duplicates.\n",
-    "\n",
-    "9. If `total` is less than zero, it means the sum is negative, so it increments the `left` pointer to move towards a larger sum.\n",
-    "\n",
-    "10. If `total` is greater than zero, it means the sum is positive, so it decrements the `right` pointer to move towards a smaller sum.\n",
-    "\n",
-    "11. This process continues until all possible triplets have been considered.\n",
-    "\n",
-    "12. Finally, the function returns the `triplets` list, which contains all the unique triplets that sum up to zero in the sorted `nums` array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[-1, -1, 2], [-1, 0, 1]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "nums = [-1, 0, 1, 2, -1, -4]\n",
-    "result = threeSum(nums)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "nums = [0, 1, 1]\n",
-    "result = threeSum(nums)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[0, 0, 0]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "nums = [0, 0, 0]\n",
-    "result = threeSum(nums)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `threeSum` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. Sorting the input array `nums` takes O(n log n) time, where n is the length of the array.\n",
-    "\n",
-    "2. The outer loop runs for each element of the array, so it iterates O(n) times.\n",
-    "\n",
-    "3. Inside the outer loop, we have a while loop with two pointers (`left` and `right`). In the worst case, the while loop can iterate O(n) times (when all elements are unique).\n",
-    "\n",
-    "4. Inside the while loop, we have constant time operations (additions, comparisons, and list appends).\n",
-    "\n",
-    "Overall, the time complexity of the `threeSum` function is dominated by the sorting step, so it is O(n log n) due to the sorting. The other operations inside the loops contribute linearly but are dominated by the sorting step.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The space used by the `triplets` list to store the output triplets is O(k), where k is the number of unique triplets that sum up to zero. In the worst case, this can be O(n^2/3) because there can be roughly O(n^2) possible triplets, and many of them may sum up to zero.\n",
-    "\n",
-    "2. Other than the `triplets` list, the function uses only a constant amount of additional space for variables like `i`, `left`, `right`, and `total`.\n",
-    "\n",
-    "Overall, the space complexity of the `threeSum` function is O(k), where k is the number of unique triplets that meet the criteria. It's important to note that the space complexity does not depend on the size of the input array but rather on the number of valid triplets found.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: O(n log n) due to sorting (dominant factor) and O(n^2) in the worst case for the nested loops.\n",
-    "- Space Complexity: O(k) for storing the output triplets, where k is the number of unique triplets that meet the criteria."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **K-Sum Problem:** Generalize the problem to the K-Sum problem, where instead of finding triplets that sum to zero, you need to find K elements that sum to a target value. Implement a function for this generalization.\n",
-    "2. **Count the Number of Unique Triplets:** Modify the algorithm to count the number of unique triplets that satisfy the conditions instead of returning the triplets themselves. This will require some modifications to handle duplicates efficiently."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb b/docs/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb
deleted file mode 100644
index 451a21d..0000000
--- a/docs/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb	
+++ /dev/null
@@ -1,185 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 121. Best Time to Buy and Sell Stock\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Best Time to Buy and Sell Stock problem on LeetCode, click here!](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given an array `prices` where `prices[i]` is the price of a given stock on the $i^{th}$ day.\n",
-    "\n",
-    "You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.\n",
-    "\n",
-    "*Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `prices.length` <= $10^5$\n",
-    "- 0 <= `prices[i]` <= $10^4$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def maxProfit(prices):\n",
-    "    if not prices:\n",
-    "        return 0\n",
-    "    \n",
-    "    min_price = prices[0]\n",
-    "    max_profit = 0\n",
-    "    \n",
-    "    for price in prices:\n",
-    "        # Update the minimum price if a lower price is encountered\n",
-    "        min_price = min(min_price, price)\n",
-    "        # Calculate the profit if selling at the current price\n",
-    "        max_profit = max(max_profit, price - min_price)\n",
-    "    \n",
-    "    return max_profit"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "- The `maxProfit` function takes a single argument, `prices`, which is a list of stock prices.\n",
-    "- The first line of the function checks if the `prices` list is empty. If it's empty, there are no prices to analyze, so the function returns 0 (no profit can be made).\n",
-    "\n",
-    "- Then, we initialize two variables:\n",
-    "  - `min_price`: This variable will keep track of the minimum price seen so far. We initialize it with the price of the first day in the list (`prices[0]`).\n",
-    "  - `max_profit`: This variable will keep track of the maximum profit we can achieve. We initialize it to 0 because initially, we haven't made any profit.\n",
-    "\n",
-    "- We start iterating through the `prices` list, where `price` represents the stock price for the current day.\n",
-    "\n",
-    "- In each iteration, we check if the current `price` is lower than the `min_price` we've seen so far. If it is, we update `min_price` to the current `price`. This is because we want to buy the stock at the lowest possible price.\n",
-    "\n",
-    "- Next, we calculate the profit we would make if we sold the stock at the current `price` (assuming we bought it at `min_price`). We do this by subtracting `min_price` from the current `price`. This represents the profit for the current day.\n",
-    "\n",
-    "- We also use the `max` function to keep track of the maximum profit seen so far. If the profit calculated for the current day is greater than the `max_profit` we've seen previously, we update `max_profit` with this higher value.\n",
-    "\n",
-    "- Finally, after iterating through all the days in the `prices` list, we return `max_profit`, which represents the maximum profit that can be achieved by buying and selling a single stock."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "5\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "prices1 = [7, 1, 5, 3, 6, 4]\n",
-    "print(maxProfit(prices1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "0\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "prices2 = [7, 6, 4, 3, 1]\n",
-    "print(maxProfit(prices2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The code iterates through the `prices` array once, examining each price exactly once.\n",
-    "- Within the loop, there are constant-time operations such as comparisons, additions, and subtractions.\n",
-    "- Therefore, the time complexity of this code is O(n), where n is the length of the `prices` array. It performs a linear number of operations relative to the size of the input.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity of the code is constant, O(1).\n",
-    "- Regardless of the size of the input `prices` array, the code only uses a fixed amount of additional memory to store two variables (`min_price` and `max_profit`) and a few loop variables.\n",
-    "- The memory usage does not depend on the size of the input, so it is considered constant space complexity.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: O(n) where n is the length of the `prices` array.\n",
-    "- Space Complexity: O(1) (constant space usage)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple Transactions Allowed:** Modify the problem to allow multiple transactions (buy and sell multiple times), but you must sell before buying again. Find the maximum profit in this scenario.\n",
-    "2. **K Transactions Allowed:** Extend the problem to allow at most k transactions. Find the maximum profit considering this constraint.\n",
-    "3. **With Transaction Fee:** Introduce a transaction fee for each buy/sell operation. Modify the code to maximize profit while considering the transaction fee."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb b/docs/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb
deleted file mode 100644
index ded03f3..0000000
--- a/docs/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb	
+++ /dev/null
@@ -1,206 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 3. Longest Substring Without Repeating Characters\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Longest Substring Without Repeating Characters problem on LeetCode, click here!](https://leetcode.com/problems/longest-substring-without-repeating-characters/)\n",
-    "\n",
-    "---\n",
-    "Given a string `s`, find the length of the **longest substring** without repeating characters.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 0 <= `s.length` <= $5 * 10^4$\n",
-    "- `s` consists of English letters, digits, symbols and spaces.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def length_of_longest_substring(s):\n",
-    "    # Create a dictionary to store the index of each character's last occurrence.\n",
-    "    char_index = {}\n",
-    "    \n",
-    "    # Initialize variables to keep track of the start and end of the current substring.\n",
-    "    start = 0\n",
-    "    max_length = 0\n",
-    "    \n",
-    "    for end in range(len(s)):\n",
-    "        # If the character is in the dictionary and its last occurrence is after the start of the current substring,\n",
-    "        # update the start of the substring to the next character after its last occurrence.\n",
-    "        if s[end] in char_index and char_index[s[end]] >= start:\n",
-    "            start = char_index[s[end]] + 1\n",
-    "        \n",
-    "        # Update the last occurrence index of the current character.\n",
-    "        char_index[s[end]] = end\n",
-    "        \n",
-    "        # Calculate the length of the current substring and update the maximum length if necessary.\n",
-    "        current_length = end - start + 1\n",
-    "        max_length = max(max_length, current_length)\n",
-    "    \n",
-    "    return max_length"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a function called `length_of_longest_substring` that takes a single argument `s`, which is the input string for which we want to find the length of the longest substring without repeating characters.\n",
-    "\n",
-    "2. We create a dictionary called `char_index` to store the index of the last occurrence of each character in the input string. This dictionary will help us keep track of where each character was last seen.\n",
-    "\n",
-    "3. We initialize two variables, `start` and `max_length`. \n",
-    "   - `start` represents the start index of the current substring without repeating characters. It starts at 0.\n",
-    "   - `max_length` is used to keep track of the maximum length found so far and is initially set to 0.\n",
-    "\n",
-    "4. We then iterate through the input string `s` using a for loop, where the loop variable `end` represents the current end index of the substring we are considering.\n",
-    "\n",
-    "5. Inside the loop:\n",
-    "   - We check if the character `s[end]` is already in the `char_index` dictionary and if its last occurrence is within or after the current substring. If so, it means that we've encountered a repeating character, and we need to update the `start` of the substring to the next character after the last occurrence of `s[end]`. This ensures that we have a new valid substring without repeating characters.\n",
-    "   - We then update the `char_index` dictionary by storing the current index `end` as the last occurrence index of the character `s[end]`.\n",
-    "\n",
-    "6. Next, we calculate the length of the current substring without repeating characters, which is `current_length = end - start + 1`. We add 1 to account for the fact that the indices are zero-based.\n",
-    "\n",
-    "7. We update the `max_length` with the maximum of its current value and the `current_length`. This step ensures that we keep track of the longest valid substring we have encountered so far.\n",
-    "\n",
-    "8. The loop continues to iterate through the input string, and at the end, we return the `max_length`, which represents the length of the longest substring without repeating characters."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "print(length_of_longest_substring(\"abcabcbb\"))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "print(length_of_longest_substring(\"bbbbb\"))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "cf04eba0-0ead-4e8b-ba33-4316ea64a9d3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "print(length_of_longest_substring(\"pwwkew\"))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "Let's analyze the time and space complexity of the `length_of_longest_substring` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The function iterates through the input string `s` using a single for loop. The loop runs from the beginning of the string to the end once.\n",
-    "- Inside the loop, we perform constant-time operations such as dictionary lookups and updates, comparisons, and arithmetic operations.\n",
-    "- Therefore, the time complexity of the function is O(n), where n is the length of the input string `s`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The primary data structure that consumes space in this function is the `char_index` dictionary.\n",
-    "- In the worst case, if there are no repeating characters in the input string, the dictionary can store all unique characters in the string.\n",
-    "- Therefore, the space complexity is O(min(n, m)), where n is the length of the input string `s`, and m is the number of unique characters in the string. In the worst case, when all characters are unique, m is equal to n, so the space complexity is O(n).\n",
-    "- Additionally, there are a few integer variables used for indices and lengths, which consume constant space.\n",
-    "- Overall, the space complexity of the function is O(min(n, m)) or simply O(n) in the worst case.\n",
-    "\n",
-    "**In summary,** the time complexity of the `length_of_longest_substring` function is O(n), and the space complexity is O(n) in the worst case due to the `char_index` dictionary. This algorithm provides an efficient way to find the length of the longest substring without repeating characters in linear time."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Longest Substring with K Distinct Characters:** Modify the problem to find the length of the longest substring with exactly K distinct characters. For example, given the input string \"abcabcbb\" and K = 2, the answer would be 4 because the longest substring with two distinct characters is \"abca.\"\n",
-    "2. **Longest Substring with Unique Characters:** Write a function to find the length of the longest substring in a given string where all characters are unique. For example, given the input string \"abcabcbb,\" the answer would be 4 because \"abcd\" is the longest substring with unique characters."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb b/docs/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb
deleted file mode 100644
index df4b0d3..0000000
--- a/docs/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb	
+++ /dev/null
@@ -1,189 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 424. Longest Repeating Character Replacement\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Longest Repeating Character Replacement problem on LeetCode, click here!](https://leetcode.com/problems/longest-repeating-character-replacement/)\n",
-    "\n",
-    "---\n",
-    "You are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.\n",
-    "\n",
-    "*Return the length of the longest substring containing the same letter you can get after performing the above operations.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `s.length` <= $10^5$\n",
-    "- `s` consists of only uppercase English letters.\n",
-    "- `0 <= k <= s.length`"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def characterReplacement(s, k):\n",
-    "    max_length = 0  # Initialize the maximum length\n",
-    "    max_count = 0   # Initialize the maximum count of repeating characters\n",
-    "    start = 0       # Initialize the start of the sliding window\n",
-    "    char_count = {} # Dictionary to store the count of each character\n",
-    "\n",
-    "    for end in range(len(s)):\n",
-    "        # Update the count of the current character in the dictionary\n",
-    "        char_count[s[end]] = char_count.get(s[end], 0) + 1\n",
-    "        \n",
-    "        # Update the maximum count of repeating characters\n",
-    "        max_count = max(max_count, char_count[s[end]])\n",
-    "        \n",
-    "        # Check if the current window size is greater than k\n",
-    "        if (end - start + 1) - max_count > k:\n",
-    "            # Move the start of the window to the right\n",
-    "            char_count[s[start]] -= 1\n",
-    "            start += 1\n",
-    "        \n",
-    "        # Update the maximum length\n",
-    "        max_length = max(max_length, end - start + 1)\n",
-    "\n",
-    "    return max_length"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We initialize some variables:\n",
-    "   - `max_length` to keep track of the maximum length of the substring containing the same letter.\n",
-    "   - `max_count` to keep track of the maximum count of repeating characters within the current window.\n",
-    "   - `start` to represent the start index of the sliding window.\n",
-    "   - `char_count` is a dictionary that will store the count of each character within the current window.\n",
-    "\n",
-    "2. We use a `for` loop to iterate through the characters of the string `s`.\n",
-    "\n",
-    "3. Inside the loop, we do the following for each character at position `end`:\n",
-    "   - Update the count of the current character in the `char_count` dictionary.\n",
-    "   - Update the `max_count` to be the maximum of the current `max_count` and the count of the current character. This keeps track of the maximum count of repeating characters within the current window.\n",
-    "\n",
-    "4. We check if the current window size (the difference between `end` and `start` plus one) minus the `max_count` is greater than `k`. This condition checks whether we have exceeded the allowed number of replacements (`k`) within the current window.\n",
-    "\n",
-    "   - If we have exceeded the allowed replacements, it means we need to shrink the window from the left side. We do this by moving the `start` of the window to the right and decrementing the count of the character at `s[start]` in the `char_count` dictionary. This effectively removes characters from the left side of the window until we have a valid window again.\n",
-    "\n",
-    "   - By doing this, we ensure that the difference between the current window size and the `max_count` is always less than or equal to `k`.\n",
-    "\n",
-    "5. After each character, we update the `max_length` to be the maximum of the current `max_length` and the size of the current window (`end - start + 1`).\n",
-    "\n",
-    "6. Finally, we return `max_length`, which holds the length of the longest substring containing the same letter with at most `k` replacements."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "s1 = \"ABAB\"\n",
-    "k1 = 2\n",
-    "print(characterReplacement(s1, k1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "s2 = \"AABABBA\"\n",
-    "k2 = 1\n",
-    "print(characterReplacement(s2, k2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The code uses a single `for` loop to iterate through the characters of the input string `s`. The loop runs from the beginning to the end of the string.\n",
-    "- Inside the loop, we perform constant-time operations such as updating the `char_count` dictionary and updating variables like `max_count` and `max_length`.\n",
-    "- The code's time complexity is primarily determined by the loop, which iterates through each character in the string once. Therefore, the time complexity is O(n), where n is the length of the input string `s`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The code uses several variables to store information, but the space they consume is constant and does not depend on the size of the input string. These variables include `max_length`, `max_count`, `start`, and `char_count`. Therefore, the space complexity is O(1) or constant space.\n",
-    "\n",
-    "**In summary**, the time complexity of the code is O(n), where n is the length of the input string, and the space complexity is O(1), as it uses a constant amount of additional space regardless of the input size. This algorithm efficiently solves the problem with a linear time complexity."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the code to find the longest substring containing the same letter with at most k replacements in O(n) time complexity and O(1) space complexity. Hint: You may need to update the approach to achieve this.\n",
-    "2. **Variation with Lowercase Letters:** Extend the problem to include both uppercase and lowercase English letters in the input string s. Write a function that can handle this extended input and still find the longest substring with at most k replacements efficiently."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb b/docs/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb
deleted file mode 100644
index cf22c86..0000000
--- a/docs/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb	
+++ /dev/null
@@ -1,260 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 76. Minimum Window Substring\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Minimum Window Substring problem on LeetCode, click here!](https://leetcode.com/problems/minimum-window-substring/)\n",
-    "\n",
-    "---\n",
-    "Given two strings `s` and `t` of lengths `m` and `n` respectively, return *the **minimum window substring** of `s` such that every character in `t` (**including duplicates**) is included in the window. If there is no such substring, return the empty string `\"\"`.*\n",
-    "\n",
-    "The testcases will be generated such that the answer is **unique**.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `m == s.length`\n",
-    "- `n == t.length`\n",
-    "- 1 <= `m, n` <= $10^5$\n",
-    "- `s` and `t` consist of uppercase and lowercase English letters.\n",
-    "\n",
-    "**Follow up**: Could you find an algorithm that runs in $O(m + n)$ time?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def min_window_substring(s, t):\n",
-    "    if not s or not t:\n",
-    "        return \"\"\n",
-    "\n",
-    "    # Initialize dictionaries to keep track of character counts for t and the current window in s.\n",
-    "    t_dict = {}\n",
-    "    current_window_dict = {}\n",
-    "    \n",
-    "    # Populate t_dict with character counts for string t.\n",
-    "    for char in t:\n",
-    "        t_dict[char] = t_dict.get(char, 0) + 1\n",
-    "\n",
-    "    # Initialize pointers for the sliding window.\n",
-    "    left = 0\n",
-    "    min_len = float('inf')\n",
-    "    min_window = \"\"\n",
-    "    required_chars = len(t_dict)\n",
-    "\n",
-    "    # Initialize a variable to keep track of how many required characters have been found in the current window.\n",
-    "    found_chars = 0\n",
-    "\n",
-    "    # Iterate over the string s using the right pointer.\n",
-    "    for right in range(len(s)):\n",
-    "        char = s[right]\n",
-    "\n",
-    "        # Update the current_window_dict.\n",
-    "        current_window_dict[char] = current_window_dict.get(char, 0) + 1\n",
-    "\n",
-    "        # Check if the current character is a required character and if its count matches the required count.\n",
-    "        if char in t_dict and current_window_dict[char] == t_dict[char]:\n",
-    "            found_chars += 1\n",
-    "\n",
-    "        # Try to minimize the window by moving the left pointer.\n",
-    "        while left <= right and found_chars == required_chars:\n",
-    "            # Calculate the current window size.\n",
-    "            window_size = right - left + 1\n",
-    "\n",
-    "            # If the current window is smaller than the minimum found so far, update min_len and min_window.\n",
-    "            if window_size < min_len:\n",
-    "                min_len = window_size\n",
-    "                min_window = s[left:right+1]\n",
-    "\n",
-    "            # Move the left pointer to the right to shrink the window.\n",
-    "            left_char = s[left]\n",
-    "            current_window_dict[left_char] -= 1\n",
-    "\n",
-    "            # Check if the character removed from the window was a required character.\n",
-    "            if left_char in t_dict and current_window_dict[left_char] < t_dict[left_char]:\n",
-    "                found_chars -= 1\n",
-    "\n",
-    "            # Move the left pointer further to continue shrinking the window.\n",
-    "            left += 1\n",
-    "\n",
-    "    return min_window"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by checking if either of the input strings `s` or `t` is empty. If either of them is empty, we return an empty string since there can't be any valid substring in this case.\n",
-    "\n",
-    "2. We initialize two dictionaries: `t_dict` and `current_window_dict`. These dictionaries will be used to keep track of character counts in the string `t` and the current window in string `s`, respectively.\n",
-    "\n",
-    "3. We populate the `t_dict` dictionary by iterating through string `t`. For each character, we increment its count in the dictionary using `t_dict.get(char, 0) + 1`. This allows us to count the occurrences of each character in `t`.\n",
-    "\n",
-    "4. We initialize two pointers: `left` and `right`. The `left` pointer will represent the start of the current window, and the `right` pointer will represent the end of the current window. We also initialize `min_len` to store the length of the minimum window found so far, and `min_window` to store the actual minimum window substring.\n",
-    "\n",
-    "5. We determine the number of required characters in `t` (i.e., the number of distinct characters in `t`) and store it in the variable `required_chars`.\n",
-    "\n",
-    "6. We initialize a variable `found_chars` to keep track of how many required characters have been found in the current window. Initially, it is set to 0.\n",
-    "\n",
-    "7. We iterate over the string `s` using the `right` pointer. In each iteration, we do the following:\n",
-    "\n",
-    "   - Update the `current_window_dict` by incrementing the count of the current character.\n",
-    "   \n",
-    "   - Check if the current character is a required character (present in `t_dict`) and if its count in the `current_window_dict` matches the required count from `t_dict`. If so, we increment `found_chars`.\n",
-    "\n",
-    "8. After updating the window, we attempt to minimize the window size by moving the `left` pointer to the right. In this step, we:\n",
-    "\n",
-    "   - Calculate the size of the current window.\n",
-    "   \n",
-    "   - If the current window size is smaller than the minimum found so far (`min_len`), we update `min_len` and `min_window` to store the current window substring.\n",
-    "   \n",
-    "   - Move the `left` pointer to the right to shrink the window.\n",
-    "   \n",
-    "   - Check if the character being removed from the window was a required character. If it was, we decrement `found_chars`.\n",
-    "   \n",
-    "   - Continue moving the `left` pointer further to continue shrinking the window if the window still contains all required characters.\n",
-    "\n",
-    "9. Finally, we return `min_window`, which will contain the minimum window substring that contains all characters from `t`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "BANC\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "print(min_window_substring(\"ADOBECODEBANC\", \"ABC\"))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "print(min_window_substring(\"a\", \"a\")) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "print(min_window_substring(\"a\", \"aa\")) "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `min_window_substring` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The main loop iterates through the string `s` from left to right using the `right` pointer. This loop runs in O(m) time, where 'm' is the length of string `s`.\n",
-    "\n",
-    "- Inside the loop, we have a while loop that moves the `left` pointer to the right to shrink the window as needed. In the worst case, this while loop can also run in O(m) time because in the worst case, we may have to move the `left` pointer all the way to the end of the string.\n",
-    "\n",
-    "- Within each iteration of the while loop, we perform constant time operations, such as updating dictionaries and comparing character counts.\n",
-    "\n",
-    "- The initialization of the `t_dict` dictionary takes O(n) time, where 'n' is the length of string `t`.\n",
-    "\n",
-    "Therefore, the overall time complexity of the function is O(m + n) because the dominant factor is the length of string `s`, and the length of string `t` has a smaller impact.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity is determined by the space used by the `t_dict` dictionary, the `current_window_dict` dictionary, and a few variables.\n",
-    "\n",
-    "- The `t_dict` dictionary stores character counts for string `t`. In the worst case, when all characters in `t` are distinct, this dictionary can have a maximum of 'n' key-value pairs. So, the space complexity for `t_dict` is O(n).\n",
-    "\n",
-    "- The `current_window_dict` dictionary stores character counts for the current window. In the worst case, it can have a maximum of 'm' key-value pairs. So, the space complexity for `current_window_dict` is also O(m).\n",
-    "\n",
-    "- Other variables used in the function, such as `left`, `right`, `min_len`, `min_window`, `required_chars`, and `found_chars`, are all of constant size and do not depend on the input sizes.\n",
-    "\n",
-    "**In summary**, the overall space complexity of the function is O(max(m, n)), which means it is determined by the larger of the two input strings' lengths. This is because the space used for `t_dict` and `current_window_dict` dominates the space complexity."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Smallest Distinct Substring:** Instead of finding the minimum window substring containing all characters, find the smallest distinct (unique) substring of s that contains all characters from t. This variation adds complexity because you must find a substring with distinct characters.\n",
-    "2. **No Extra Space:** Solve the problem without using any extra space, such as dictionaries or arrays, other than a constant amount of space. This is a significant optimization challenge."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb b/docs/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb
deleted file mode 100644
index 04c0280..0000000
--- a/docs/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb	
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 20. Valid Parentheses\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Valid Parentheses problem on LeetCode, click here!](https://leetcode.com/problems/valid-parentheses/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given a string s containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.\n",
-    "\n",
-    "An input string is valid if:\n",
-    "\n",
-    "1. Open brackets must be closed by the same type of brackets.\n",
-    "2. Open brackets must be closed in the correct order.\n",
-    "3. Every close bracket has a corresponding open bracket of the same type.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `s.length` <= $10^4$\n",
-    "- `s` consists of parentheses only `'()[]{}'`."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def isValid(s):\n",
-    "    # Create an empty stack to store opening brackets\n",
-    "    stack = []\n",
-    "    \n",
-    "    # Define a mapping for matching brackets\n",
-    "    bracket_mapping = {')': '(', '}': '{', ']': '['}\n",
-    "    \n",
-    "    # Iterate through the characters in the input string\n",
-    "    for char in s:\n",
-    "        # If the character is an opening bracket, push it onto the stack\n",
-    "        if char in bracket_mapping.values():\n",
-    "            stack.append(char)\n",
-    "        # If the character is a closing bracket\n",
-    "        elif char in bracket_mapping.keys():\n",
-    "            # Pop the top element from the stack if it exists, or use a dummy value '#'\n",
-    "            top_element = stack.pop() if stack else '#'\n",
-    "            # If the popped element does not match the corresponding opening bracket, return False\n",
-    "            if bracket_mapping[char] != top_element:\n",
-    "                return False\n",
-    "        # If the character is not a bracket, ignore it\n",
-    "        \n",
-    "    # After processing the entire string, if there are any unmatched opening brackets left in the stack, return False\n",
-    "    return len(stack) == 0"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining the `isValid` function that takes a single argument `s`, which is the input string containing parentheses and brackets.\n",
-    "\n",
-    "2. Inside the function, we create an empty stack, which is a list in Python, to store opening brackets as we encounter them in the input string. The stack will help us keep track of the brackets and their order.\n",
-    "\n",
-    "3. We define a `bracket_mapping` dictionary that maps each closing bracket to its corresponding opening bracket. This mapping will be used to check if a closing bracket matches the most recent opening bracket in the stack.\n",
-    "\n",
-    "4. We iterate through each character `char` in the input string `s` using a `for` loop.\n",
-    "\n",
-    "5. If the character `char` is an opening bracket (i.e., it exists in the `values()` of `bracket_mapping`), we push it onto the stack using the `append()` method.\n",
-    "\n",
-    "6. If the character `char` is a closing bracket (i.e., it exists in the `keys()` of `bracket_mapping`), we need to check if it matches the corresponding opening bracket. To do this, we pop the top element from the stack (if it exists) and store it in `top_element`. We use a dummy value `'#'` if the stack is empty to avoid errors.\n",
-    "\n",
-    "7. We then compare `top_element` with the corresponding opening bracket using `bracket_mapping[char]`. If they do not match, it means the string is not valid, and we return `False`.\n",
-    "\n",
-    "8. If the character `char` is not a bracket, we simply ignore it and continue the loop.\n",
-    "\n",
-    "9. After processing the entire string, we check if there are any unmatched opening brackets left in the stack. If the stack is empty, it means all opening brackets have been properly matched and closed, and we return `True`. Otherwise, we return `False`.\n",
-    "\n",
-    "10. Finally, we provide some test cases at the bottom of the code to demonstrate how the function works for different input strings."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "print(isValid(\"()\"))    "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "print(isValid(\"()[]{}\")) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "print(isValid(\"(]\"))     "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `isValid` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The function iterates through each character in the input string `s` once, performing constant-time operations for each character.\n",
-    "- Pushing and popping elements from the stack (list) also takes constant time in most cases.\n",
-    "- Therefore, the overall time complexity of the function is $O(n)$, where $n$ is the length of the input string `s`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity of the function is determined by the space used by the stack and the `bracket_mapping` dictionary.\n",
-    "- In the worst case, if the input string `s` consists entirely of opening brackets, the stack can potentially contain all of these opening brackets, resulting in a space complexity of $O(n)$ in terms of the stack.\n",
-    "- The `bracket_mapping` dictionary has a constant number of key-value pairs (3 pairs in this case).\n",
-    "- Therefore, the overall space complexity of the function is $O(n)$, where $n$ is the length of the input string `s`, mainly due to the stack space.\n",
-    "\n",
-    "**In summary:**\n",
-    "- The time complexity is $O(n)$ because we iterate through the string once.\n",
-    "- The space complexity is $O(n)$ due to the stack used to keep track of opening brackets."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Valid Expressions:** Modify the problem to validate not only brackets but also arithmetic expressions containing parentheses, such as \"$2 * (3 + 5) / (4 - 2)$\".\n",
-    "2. **Valid Parentheses Combinations:**\n",
-    "Write a function to generate all valid combinations of n pairs of parentheses, where $n$ is a positive integer. For example, for $n = 3$, the valid combinations are [\"((()))\", \"(()())\", \"(())()\", \"()(())\", \"()()()\"]."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb b/docs/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb
deleted file mode 100644
index eb4858b..0000000
--- a/docs/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 153. Find Minimum in Rotated Sorted Array\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Find Minimum in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:\n",
-    "\n",
-    "- `[4,5,6,7,0,1,2]` if it was rotated `4` times.\n",
-    "- `[0,1,2,4,5,6,7]` if it was rotated `7` times.\n",
-    "\n",
-    "Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.\n",
-    "\n",
-    "Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array.*\n",
-    "\n",
-    "You must write an algorithm that runs in $O(log\\ n)$ time.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `n == nums.length`\n",
-    "- `1 <= n <= 5000`\n",
-    "- `-5000 <= nums[i] <= 5000`\n",
-    "- All the integers of `nums` are **unique**.\n",
-    "- `nums` is sorted and rotated between `1` and `n` times."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def findMin(nums):\n",
-    "    left, right = 0, len(nums) - 1\n",
-    "    \n",
-    "    while left < right:\n",
-    "        mid = left + (right - left) // 2\n",
-    "        \n",
-    "        # If the mid element is greater than the rightmost element,\n",
-    "        # it means the minimum element is in the right half.\n",
-    "        if nums[mid] > nums[right]:\n",
-    "            left = mid + 1\n",
-    "        # If the mid element is less than or equal to the rightmost element,\n",
-    "        # it means the minimum element is in the left half or at the mid itself.\n",
-    "        else:\n",
-    "            right = mid\n",
-    "    \n",
-    "    # The loop will break when left and right converge to the minimum element.\n",
-    "    # At this point, left (or right) will be pointing to the minimum element.\n",
-    "    return nums[left]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The function `findMin(nums)` takes an input list `nums`, which represents the rotated sorted array.\n",
-    "\n",
-    "2. Initialize two pointers `left` and `right` to track the current search range. Initially, `left` is set to 0 (the beginning of the array), and `right` is set to `len(nums) - 1` (the end of the array).\n",
-    "\n",
-    "3. Enter a `while` loop with the condition `left < right`. This loop continues until `left` and `right` converge to a single element, which will be the minimum element in the rotated array.\n",
-    "\n",
-    "4. Inside the loop, calculate the middle index `mid` using integer division. This helps in finding the middle element of the current search range.\n",
-    "\n",
-    "5. Check if the element at the middle index (`nums[mid]`) is greater than the element at the rightmost index (`nums[right]`). If this condition is true, it means that the minimum element must be in the right half of the current search range. So, update `left` to `mid + 1`, effectively eliminating the left half of the search range.\n",
-    "\n",
-    "6. If the condition from step 5 is not met, it implies that the minimum element can be in the left half of the current search range or could be the element at the `mid` index itself. In this case, update `right` to `mid`, effectively eliminating the right half of the search range.\n",
-    "\n",
-    "7. Repeat steps 4-6 until `left` and `right` converge to the minimum element.\n",
-    "\n",
-    "8. When the loop exits, it means that `left` (or `right`) points to the minimum element in the array.\n",
-    "\n",
-    "9. Return `nums[left]` (or `nums[right]`) as the result, which is the minimum element in the rotated sorted array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1:\n",
-    "nums1 = [3, 4, 5, 1, 2]\n",
-    "print(findMin(nums1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "0\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-    "print(findMin(nums2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "11\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "nums3 = [11, 13, 15, 17]\n",
-    "print(findMin(nums3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of this code is $O(log\\ n)$, where '$n$' is the length of the input array `nums`. This is because the binary search algorithm is employed, which continually divides the search range in half with each iteration.\n",
-    "\n",
-    "In each iteration of the `while` loop, the search range is halved, and this process continues until the `left` and `right` pointers converge to the minimum element. In the worst case, it will take logarithmic time to reduce the search range to a single element.\n",
-    "\n",
-    "Therefore, the binary search used in this code runs in $O(log\\ n)$ time complexity.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity of this code is $O(1)$, which means it uses a constant amount of additional space that does not depend on the size of the input array.\n",
-    "\n",
-    "The algorithm uses a fixed number of variables (`left`, `right`, `mid`) to keep track of the search range and indices, but the number of these variables does not grow with the size of the input array. Hence, the space complexity is constant.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: $O(log\\ n)$\n",
-    "- Space Complexity: $O(1)$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. Solve the problem with a variation where the input array may contain duplicates.\n",
-    "\n",
-    "2. Modify the problem to return the index of the minimum element in the rotated sorted array, rather than the element itself. The algorithm should still run in $O(log\\ n)$ time.\n",
-    "\n",
-    "3. Implement a function that finds the maximum element in a rotated sorted array. How would you adapt the binary search algorithm to solve this problem efficiently in $O(log\\ n)$ time?\n",
-    "\n",
-    "4. Implement a function that can find the kth smallest element in a rotated sorted array. This is an extension of the original problem, and the algorithm should still run in $O(log\\ n)$ time."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb b/docs/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb
deleted file mode 100644
index 7b3139d..0000000
--- a/docs/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,237 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 33. Search in Rotated Sorted Array\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Search in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/search-in-rotated-sorted-array/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "There is an integer array `nums` sorted in ascending order (with **distinct** values).\n",
-    "\n",
-    "Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k (1 <= k < nums.length)` such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index 3 and become `[4,5,6,7,0,1,2]`.\n",
-    "\n",
-    "Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.*\n",
-    "\n",
-    "You must write an algorithm with $O(log\\ n)$ runtime complexity.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `1 <= nums.length <= 5000`\n",
-    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-    "- All values of `nums` are **unique**.\n",
-    "- `nums` is an ascending array that is possibly rotated.\n",
-    "- $-10^4$ <= `target` <= $10^4$"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def search(nums, target):\n",
-    "    left, right = 0, len(nums) - 1\n",
-    "\n",
-    "    # Find the pivot index using binary search\n",
-    "    while left < right:\n",
-    "        mid = left + (right - left) // 2\n",
-    "\n",
-    "        if nums[mid] > nums[right]:\n",
-    "            left = mid + 1\n",
-    "        else:\n",
-    "            right = mid\n",
-    "\n",
-    "    pivot = left\n",
-    "    left, right = 0, len(nums) - 1\n",
-    "\n",
-    "    # Determine which part of the array to search in\n",
-    "    if target >= nums[pivot] and target <= nums[right]:\n",
-    "        left = pivot\n",
-    "    else:\n",
-    "        right = pivot\n",
-    "\n",
-    "    # Perform binary search to find the target\n",
-    "    while left <= right:\n",
-    "        mid = left + (right - left) // 2\n",
-    "\n",
-    "        if nums[mid] == target:\n",
-    "            return mid\n",
-    "        elif nums[mid] < target:\n",
-    "            left = mid + 1\n",
-    "        else:\n",
-    "            right = mid - 1\n",
-    "\n",
-    "    return -1"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The `search` function takes two arguments: `nums`, which is the rotated sorted array, and `target`, which is the value we want to find in the array.\n",
-    "\n",
-    "2. We initialize two pointers, `left` and `right`, which represent the range in which we are going to perform binary search. Initially, `left` is set to 0, and `right` is set to the index of the last element in the array (`len(nums) - 1`).\n",
-    "\n",
-    "3. We start by finding the pivot index using binary search. The pivot index is the index at which the array is rotated. We use a `while` loop to continue the search until `left` is less than `right`.\n",
-    "\n",
-    "4. Inside the loop, we calculate the middle index `mid` using integer division. We then compare the value at `mid` with the value at `right`. If `nums[mid]` is greater than `nums[right]`, it means the pivot point lies to the right of `mid`, so we update `left` to `mid + 1`. Otherwise, the pivot point lies to the left of `mid`, so we update `right` to `mid`. This process continues until we find the pivot index.\n",
-    "\n",
-    "5. After finding the pivot index, we have divided the array into two parts: one part is sorted in ascending order, and the other part is also sorted in ascending order but rotated. We need to determine which part of the array contains the target value.\n",
-    "\n",
-    "6. We reset `left` and `right` pointers. If the target value is within the range `[nums[pivot], nums[right]]`, we set `left` to `pivot` (the start of the rotated part), indicating that we should search in the rotated part. Otherwise, we set `right` to `pivot` (the end of the sorted part), indicating that we should search in the sorted part.\n",
-    "\n",
-    "7. We then perform binary search again within the chosen range (`left` to `right`) to find the target element. The binary search continues until `left` is less than or equal to `right`.\n",
-    "\n",
-    "8. Inside the binary search loop, we calculate the middle index `mid` and compare `nums[mid]` with the target value. If they are equal, we return `mid` as the index of the target element. If `nums[mid]` is less than the target, we update `left` to `mid + 1` to search in the right half. If `nums[mid]` is greater than the target, we update `right` to `mid - 1` to search in the left half.\n",
-    "\n",
-    "9. If we exit the binary search loop without finding the target, we return -1 to indicate that the target is not present in the array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "nums1 = [4, 5, 6, 7, 0, 1, 2]\n",
-    "target1 = 0\n",
-    "result1 = search(nums1, target1)\n",
-    "print(result1) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-    "target2 = 3\n",
-    "result2 = search(nums2, target2)\n",
-    "print(result2) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "nums3 = [1]\n",
-    "target3 = 0\n",
-    "result3 = search(nums3, target3)\n",
-    "print(result3) "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. Finding the pivot index using binary search takes $O(log\\ n)$ time, where '$n$' is the number of elements in the array.\n",
-    "2. After finding the pivot index, performing binary search to find the target element also takes $O(log\\ n)$ time in the worst case.\n",
-    "\n",
-    "The dominant factor in the time complexity is the binary search, and since we perform two binary searches sequentially, the overall time complexity of the code is $O(log\\ n)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of the code is very minimal:\n",
-    "\n",
-    "1. We use a constant amount of additional space for variables such as `left`, `right`, `pivot`, and `mid`. These variables do not depend on the input size, so they contribute $O(1)$ space complexity.\n",
-    "2. The function does not use any additional data structures that scale with the input size.\n",
-    "\n",
-    "Therefore, the space complexity of the code is $O(1)$, which means it has a constant space complexity and does not depend on the size of the input array.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: $O(log\\ n)$\n",
-    "- Space Complexity: $O(1)$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Find Minimum Element in Rotated Sorted Array**: Write an algorithm to find the minimum element in a rotated sorted array. This is a variation of the problem where you don't need to search for a target value but instead find the smallest element.\n",
-    "\n",
-    "2. **Search in a Circularly Sorted Array**: Consider an array that is sorted in a circular manner (e.g., [4, 5, 6, 7, 0, 1, 2, 3]). Adapt the search algorithm to work for circularly sorted arrays while maintaining $O(log\\ n)$ complexity."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb b/docs/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb
deleted file mode 100644
index c428c9f..0000000
--- a/docs/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb	
+++ /dev/null
@@ -1,261 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 141. Linked List Cycle\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Linked List Cycle problem on LeetCode, click here!](https://leetcode.com/problems/linked-list-cycle/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given `head`, the head of a linked list, determine if the linked list has a cycle in it.\n",
-    "\n",
-    "There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is connected to. **Note that `pos` is not passed as a parameter.**\n",
-    "\n",
-    "Return `true` if there is a cycle in the linked list. Otherwise, return `false`.\n",
-    "\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of the nodes in the list is in the range $[0, 10^4]$.\n",
-    "- $-10^5$ <= Node.val <= $10^5$\n",
-    "- `pos` is `-1` or a **valid index** in the linked-list.\n",
-    "\n",
-    "**Follow up:** Can you solve it using $O(1)$ (i.e. constant) memory?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for singly-linked list.\n",
-    "class ListNode:\n",
-    "    def __init__(self, val=0):\n",
-    "        self.val = val\n",
-    "        self.next = None\n",
-    "\n",
-    "class Solution:\n",
-    "    def hasCycle(self, head: ListNode) -> bool:\n",
-    "        if not head:\n",
-    "            return False\n",
-    "        \n",
-    "        slow = head\n",
-    "        fast = head\n",
-    "        \n",
-    "        while fast and fast.next:\n",
-    "            slow = slow.next\n",
-    "            fast = fast.next.next\n",
-    "            \n",
-    "            if slow == fast:\n",
-    "                return True\n",
-    "        \n",
-    "        return False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The given code defines a class `Solution` with a method `hasCycle` for determining if a linked list has a cycle.\n",
-    "\n",
-    "The `ListNode` class represents nodes in the linked list, where each node has a `val` (a value associated with the node) and a `next` attribute (a reference to the next node in the list).\n",
-    "\n",
-    "The `hasCycle` method takes the `head` of a linked list as input and returns `True` if there is a cycle in the linked list, and `False` otherwise. It uses two pointers, `slow` and `fast`, to traverse the linked list. If there is a cycle, the `fast` pointer will eventually catch up to the `slow` pointer.\n",
-    "\n",
-    "The algorithm works as follows:\n",
-    "1. It checks if the input `head` is `None` (an empty list). If it is, the method returns `False` because an empty list can't have a cycle.\n",
-    "2. Two pointers, `slow` and `fast`, initially point to the `head` of the list.\n",
-    "3. The code enters a `while` loop where `fast` moves two steps at a time, and `slow` moves one step at a time.\n",
-    "4. If there is a cycle, `fast` will eventually catch up to `slow`, and the method returns `True`.\n",
-    "5. If the loop completes without finding a cycle, the method returns `False`."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to create a linked list with a cycle\n",
-    "def create_linked_list_with_cycle(values, pos):\n",
-    "    dummy = ListNode()\n",
-    "    current = dummy\n",
-    "    cycle_node = None\n",
-    "    \n",
-    "    for i, val in enumerate(values):\n",
-    "        current.next = ListNode(val)\n",
-    "        current = current.next\n",
-    "        \n",
-    "        if i == pos:\n",
-    "            cycle_node = current\n",
-    "    \n",
-    "    if cycle_node:\n",
-    "        current.next = cycle_node\n",
-    "    \n",
-    "    return dummy.next"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "This helper function is designed to create a linked list with a cycle for testing purposes. It takes two arguments:\n",
-    "\n",
-    "1. `values`: A list of values representing the nodes in the linked list.\n",
-    "2. `pos`: An integer representing the index of the node where the cycle begins.\n",
-    "\n",
-    "Here's how the function works:\n",
-    "\n",
-    "- It starts by creating an empty `ListNode` called `dummy`. This `dummy` node is used to simplify the creation of the linked list.\n",
-    "- It initializes a `current` pointer to the `dummy` node. This pointer will be used to traverse the linked list.\n",
-    "- It initializes a `cycle_node` variable to `None`. This variable will hold the reference to the node where the cycle begins, if any.\n",
-    "- It then iterates through the `values` list, creating a new `ListNode` for each value and appending it to the linked list.\n",
-    "- Inside the loop, it checks if the current index `i` is equal to the specified `pos`. If they match, it sets `cycle_node` to the current node. This simulates the creation of a cycle in the linked list.\n",
-    "- After the loop, if `cycle_node` is not `None` (indicating a cycle should be created), it connects the last node in the linked list to `cycle_node`, effectively creating a cycle.\n",
-    "- Finally, it returns the reference to the first node of the linked list (not the `dummy` node).\n",
-    "\n",
-    "This helper function allows you to easily create test cases where you can specify the values for the linked list and the position where the cycle starts. It's particularly useful for testing the code's ability to detect cycles in linked lists."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "head = [3,2,0,-4]\n",
-    "pos = 1\n",
-    "\n",
-    "head = create_linked_list_with_cycle(head, pos)\n",
-    "solution = Solution()\n",
-    "result = solution.hasCycle(head)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "head = [1,2]\n",
-    "pos = 0\n",
-    "\n",
-    "head = create_linked_list_with_cycle(head, pos)\n",
-    "solution = Solution()\n",
-    "result = solution.hasCycle(head)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "head = [1]\n",
-    "pos = -1\n",
-    "\n",
-    "head = create_linked_list_with_cycle(head, pos)\n",
-    "solution = Solution()\n",
-    "result = solution.hasCycle(head)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's discuss the time and space complexity of the code for detecting a cycle in a linked list:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this code is $O(n)$, where \"$n$\" is the number of nodes in the linked list.\n",
-    "\n",
-    "The reason for this is that both the `slow` and `fast` pointers traverse the linked list, and they move at different speeds. In the worst case, when there is no cycle, the `fast` pointer will reach the end of the list after going through approximately $n/2$ nodes. In the case of a cycle, it may take some extra iterations for the `fast` pointer to catch up to the `slow` pointer. However, the total number of iterations is still proportional to the number of nodes in the linked list, making it $O(n)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this code is $O(1)$, which means it uses constant space.\n",
-    "\n",
-    "The reason for this is that regardless of the size of the linked list, the code only uses two additional pointers (`slow` and `fast`) to traverse the list. These pointers do not depend on the size of the input linked list, so the space complexity remains constant.\n",
-    "\n",
-    "**In summary**, this code efficiently detects cycles in a linked list with a time complexity of $O(n)$ and a space complexity of $O(1)$. It is an example of an algorithm that solves a complex problem with minimal memory usage."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb b/docs/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb
deleted file mode 100644
index 4ec3f3b..0000000
--- a/docs/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb	
+++ /dev/null
@@ -1,277 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 143. Reorder List\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Reorder List problem on LeetCode, click here!](https://leetcode.com/problems/reorder-list/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given the head of a singly linked-list. The list can be represented as:\n",
-    "\n",
-    "$L_0 → L_1 → … → L_{n - 1} → L_n$\n",
-    "\n",
-    "Reorder the list to be on the following form:\n",
-    "\n",
-    "$L_0 → L_n → L_1 → L_{n - 1} → L_2 → L_{n - 2} → …$\n",
-    "\n",
-    "You may not modify the values in the list's nodes. Only nodes themselves may be changed.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the list is in the range $[1, 5 * 10^4]$.\n",
-    "- `1 <= Node.val <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "def reorderList(head):\n",
-    "    if not head or not head.next:\n",
-    "        return head\n",
-    "\n",
-    "    # Step 1: Find the middle of the linked list\n",
-    "    slow, fast = head, head\n",
-    "    while fast.next and fast.next.next:\n",
-    "        slow = slow.next\n",
-    "        fast = fast.next.next\n",
-    "\n",
-    "    # Split the list into two halves\n",
-    "    first_half = head\n",
-    "    second_half = slow.next\n",
-    "    slow.next = None\n",
-    "\n",
-    "    # Step 2: Reverse the second half of the linked list\n",
-    "    prev = None\n",
-    "    current = second_half\n",
-    "    while current:\n",
-    "        next_node = current.next\n",
-    "        current.next = prev\n",
-    "        prev = current\n",
-    "        current = next_node\n",
-    "    second_half = prev\n",
-    "\n",
-    "    # Step 3: Merge the first and reversed second halves alternately\n",
-    "    p1, p2 = first_half, second_half\n",
-    "    while p2:\n",
-    "        next_p1 = p1.next\n",
-    "        next_p2 = p2.next\n",
-    "        p1.next = p2\n",
-    "        p2.next = next_p1\n",
-    "        p1 = next_p1\n",
-    "        p2 = next_p2\n",
-    "\n",
-    "    return head"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The `reorderList` function is designed to reorder a singly linked list following the specific pattern described in the problem statement. Here's a detailed explanation of how the function works:\n",
-    "\n",
-    "1. **Base Cases Handling**:\n",
-    "   - The function first checks if the input linked list is empty (`not head`) or contains only one element (`not head.next`). If either of these conditions is met, the list cannot be reordered, so the function returns the original list as-is.\n",
-    "\n",
-    "2. **Finding the Middle of the Linked List**:\n",
-    "   - To reorder the list, we first need to find the middle point so that we can split the list into two halves. This is done using two pointers, `slow` and `fast`, initialized to the head of the list.\n",
-    "   - `slow` moves one step at a time while `fast` moves two steps at a time. When `fast` reaches the end of the list or the second-to-last node, `slow` will be at the middle node.\n",
-    "\n",
-    "3. **Splitting the List**:\n",
-    "   - After finding the middle node, we split the list into two halves:\n",
-    "     - The first half, `first_half`, contains nodes from the beginning up to the middle.\n",
-    "     - The second half, `second_half`, contains nodes from the middle to the end.\n",
-    "   - To split the list, we set the `next` pointer of the node before the middle node to `None`.\n",
-    "\n",
-    "4. **Reversing the Second Half**:\n",
-    "   - We reverse the second half of the linked list using the `prev`, `current`, and `next_node` pointers.\n",
-    "   - `prev` is initially `None`, and we iterate through the second half. For each node, we:\n",
-    "     - Set the `next` of the current node to `prev`, effectively reversing the next pointer direction.\n",
-    "     - Update `prev` to the current node.\n",
-    "     - Move to the next node using the `next_node`.\n",
-    "\n",
-    "5. **Merging the Two Halves Alternately**:\n",
-    "   - We now have two linked lists: `first_half` and the reversed `second_half`.\n",
-    "   - We merge these two lists alternately by adjusting the `next` pointers of the nodes.\n",
-    "   - `p1` and `p2` are pointers to the current nodes in `first_half` and `second_half`, respectively.\n",
-    "   - We iterate through both lists while reordering nodes:\n",
-    "     - Set the `next` of `p1` to `p2` to link a node from the first half to a node from the reversed second half.\n",
-    "     - Update `p1` and `p2` to their respective next nodes.\n",
-    "     - Repeat this process until we have processed all nodes in both halves.\n",
-    "\n",
-    "6. **Returning the Reordered List**:\n",
-    "   - After the merging process is complete, the linked list is reordered as specified.\n",
-    "   - The function returns the `head` of the reordered list.\n",
-    "\n",
-    "The overall result is a singly linked list that has been reordered according to the given pattern. The time complexity of this algorithm is O(N), where N is the number of nodes in the linked list, as we traverse the list once to find the middle, once to reverse the second half, and once to merge the two halves."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to print the linked list\n",
-    "def printLinkedList(head):\n",
-    "    result = []\n",
-    "    while head:\n",
-    "        result.append(head.val)\n",
-    "        head = head.next\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the helper functions:\n",
-    "**printLinkedList Function**:\n",
-    "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-    "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-    "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-    "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-    "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-    "\n",
-    "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 4, 2, 3]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "head1 = ListNode(1, ListNode(2, ListNode(3, ListNode(4))))\n",
-    "reorderList(head1)\n",
-    "print(printLinkedList(head1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 5, 2, 4, 3]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "head2 = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))\n",
-    "reorderList(head2)\n",
-    "print(printLinkedList(head2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `reorderList` function:\n",
-    "\n",
-    "**Time Complexity**:\n",
-    "1. **Finding the Middle of the Linked List**: In this step, we use two pointers, one moving one step at a time (`slow`) and the other moving two steps at a time (`fast`). This process takes $O(N/2)$ time, where $N$ is the number of nodes in the linked list.\n",
-    "\n",
-    "2. **Splitting the List**: After finding the middle, we split the list into two halves by setting the `next` pointer of the node before the middle node to `None`. This step takes $O(1)$ time.\n",
-    "\n",
-    "3. **Reversing the Second Half**: We reverse the second half of the linked list using a while loop that iterates through the second half of the list once. Therefore, this step also takes $O(N/2)$ time.\n",
-    "\n",
-    "4. **Merging the Two Halves Alternately**: In this step, we merge the two halves alternately by adjusting the `next` pointers of the nodes. We iterate through both halves once, so this step takes $O(N/2)$ time.\n",
-    "\n",
-    "Overall, the time complexity of the `reorderList` function is dominated by the steps involving reversing and merging the two halves, both of which take $O(N/2)$ time. Therefore, the total time complexity is $O(N)$.\n",
-    "\n",
-    "**Space Complexity**:\n",
-    "The space complexity of the `reorderList` function is primarily determined by the variables and data structures used within the function. Let's break down the space complexity components:\n",
-    "\n",
-    "1. **Constant Space Variables**: The variables `slow`, `fast`, `prev`, `current`, `next_node`, `p1`, and `p2` are used to traverse and manipulate the linked list. These variables occupy constant space, regardless of the input size. Therefore, they contribute $O(1)$ to the space complexity.\n",
-    "\n",
-    "2. **Split Linked Lists**: In the splitting step, we create two new linked list segments (`first_half` and `second_half`) that store references to nodes from the original linked list. These segments occupy space proportional to half of the input list, $O(N/2)$. \n",
-    "\n",
-    "3. **Reversed Second Half**: During the reversal step, we reverse the second half of the linked list in-place without creating any additional data structures. Therefore, it doesn't contribute to additional space complexity.\n",
-    "\n",
-    "4. **Overall**: Combining the above components, the total space complexity is $O(N/2)$, which simplifies to $O(N)$ in terms of space complexity.\n",
-    "\n",
-    "**In summary**, the `reorderList` function has a time complexity of $O(N)$ and a space complexity of $O(N)$ due to the creation of two new linked list segments while splitting the list. The constant space variables used for traversal do not significantly impact the overall space complexity."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Reorder by K Elements:** Generalize the reorderList function to reorder the list in a different pattern, where you reorder the list by $K$ elements at a time. For example, for $K=3$, you would reorder it as ($L_0 → L_1 → L_2 → L_n → L_{n-1} → L_{n-2} → ...$).\n",
-    "You may need to handle cases where the number of nodes is not a multiple of $K$.\n",
-    "\n",
-    "2. **Reorder by Odd and Even Nodes:** Modify the reorderList function to reorder the list in a pattern where odd-indexed nodes (1-based index) come before even-indexed nodes. For example, ($L_0 → L_n → L_1 → L_{n-1} → L_2 → L_{n-2} → ...$)."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb b/docs/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb
deleted file mode 100644
index 73dbf66..0000000
--- a/docs/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb	
+++ /dev/null
@@ -1,268 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 19. Remove Nth Node From End of List\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Remove Nth Node From End of List problem on LeetCode, click here!](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the head of a linked list, remove the $n^{th}$ node from the end of the list and return its head.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the list is `sz`.\n",
-    "- `1 <= sz <= 30`\n",
-    "- `0 <= Node.val <= 100`\n",
-    "- `1 <= n <= sz`\n",
-    "\n",
-    "**Follow up:** Could you do this in one pass?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "def removeNthFromEnd(head, n):\n",
-    "    # Create a dummy node to handle the case of removing the head node.\n",
-    "    dummy = ListNode(0)\n",
-    "    dummy.next = head\n",
-    "    first = dummy\n",
-    "    second = dummy\n",
-    "    \n",
-    "    # Advance the first pointer by n+1 nodes.\n",
-    "    for _ in range(n + 1):\n",
-    "        first = first.next\n",
-    "    \n",
-    "    # Move both pointers simultaneously until first reaches the end.\n",
-    "    while first is not None:\n",
-    "        first = first.next\n",
-    "        second = second.next\n",
-    "    \n",
-    "    # Remove the nth node by updating the next pointer of the previous node.\n",
-    "    second.next = second.next.next\n",
-    "    \n",
-    "    return dummy.next  # Return the modified head of the linked list"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a `ListNode` class to represent individual nodes in a linked list. Each node has two attributes: `val` (the value of the node) and `next` (a reference to the next node in the list).\n",
-    "\n",
-    "2. The `removeNthFromEnd` function takes the head of a linked list (`head`) and the value `n`, which represents the position from the end of the list of the node to be removed. It returns the modified head of the linked list after the removal.\n",
-    "\n",
-    "3. We create a dummy node (`dummy`) at the beginning of the list. This dummy node simplifies the code by handling cases where the head of the list needs to be removed.\n",
-    "\n",
-    "4. We initialize two pointers, `first` and `second`, both initially pointing to the dummy node.\n",
-    "\n",
-    "5. To advance the `first` pointer by `n+1` nodes, we use a `for` loop. This positions the `first` pointer `n` nodes ahead of the `second` pointer.\n",
-    "\n",
-    "6. We then move both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This ensures that the `second` pointer ends up pointing to the node that needs to be removed.\n",
-    "\n",
-    "7. To remove the nth node from the end, we update the `next` pointer of the node pointed to by the `second` pointer to skip over the node to be removed.\n",
-    "\n",
-    "8. Finally, we return `dummy.next`, which is the modified head of the linked list without the removed node.\n",
-    "\n",
-    "9. Additionally, there are two helper functions:\n",
-    "   - `createLinkedList(values)` creates a linked list from a list of values.\n",
-    "   - `convertToList(head)` converts a linked list back into a list of values."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to create a linked list from a list of values.\n",
-    "def createLinkedList(values):\n",
-    "    if not values:\n",
-    "        return None\n",
-    "    head = ListNode(values[0])\n",
-    "    current = head\n",
-    "    for val in values[1:]:\n",
-    "        current.next = ListNode(val)\n",
-    "        current = current.next\n",
-    "    return head\n",
-    "\n",
-    "# Helper function to convert a linked list to a list of values.\n",
-    "def convertToList(head):\n",
-    "    result = []\n",
-    "    current = head\n",
-    "    while current:\n",
-    "        result.append(current.val)\n",
-    "        current = current.next\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the two helper functions:\n",
-    "\n",
-    "1. `createLinkedList(values)`: This function creates a linked list from a list of values.\n",
-    "\n",
-    "   - Input: `values` is a list containing values that you want to insert into the linked list.\n",
-    "   - Output: The function returns the head of the created linked list.\n",
-    "\n",
-    "   Explanation:\n",
-    "   - The function starts by checking if the `values` list is empty. If it is, it returns `None` to indicate an empty linked list.\n",
-    "   - If the `values` list is not empty, it creates the first node of the linked list with the value from the first element of `values`.\n",
-    "   - It then iterates through the remaining elements of `values` and creates new nodes for each value, connecting them together using the `next` attribute to form a linked list.\n",
-    "   - Finally, it returns the head of the linked list, which is the first node created.\n",
-    "   \n",
-    "\n",
-    "2. `convertToList(head)`: This function converts a linked list into a list of values.\n",
-    "\n",
-    "   - Input: `head` is the head node of the linked list that you want to convert.\n",
-    "   - Output: The function returns a list containing the values of the nodes in the linked list in order.\n",
-    "\n",
-    "   Explanation:\n",
-    "   - The function starts by initializing an empty list, `result`, to store the values of the linked list nodes.\n",
-    "   - It then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-    "   - The iteration continues until it reaches the end of the linked list (i.e., the `next` attribute of the current node becomes `None`).\n",
-    "   - After the iteration is complete, the function returns the `result` list, which now contains the values of all the nodes in the linked list in the same order as they appear in the linked list.\n",
-    "\n",
-    "These helper functions are useful for creating linked lists from lists of values and converting linked lists back into lists of values, making it easier to work with linked list examples and test cases in a more familiar list format."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 2, 3, 5]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "head1 = createLinkedList([1, 2, 3, 4, 5])\n",
-    "n1 = 2\n",
-    "new_head1 = removeNthFromEnd(head1, n1)\n",
-    "print(convertToList(new_head1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "head2 = createLinkedList([1])\n",
-    "n2 = 1\n",
-    "new_head2 = removeNthFromEnd(head2, n2)\n",
-    "print(convertToList(new_head2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "head3 = createLinkedList([1, 2])\n",
-    "n3 = 1\n",
-    "new_head3 = removeNthFromEnd(head3, n3)\n",
-    "print(convertToList(new_head3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Certainly! Let's analyze the time and space complexity of the `removeNthFromEnd` function in isolation:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The `removeNthFromEnd` function performs two passes through the linked list. In the worst case, it needs to traverse the entire linked list of length $N$.\n",
-    "- The first pass positions the `first` pointer $n$ nodes ahead of the `second` pointer. This pass takes $O(N)$ time because it involves iterating through all N nodes.\n",
-    "- The second pass moves both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This also takes $O(N)$ time in the worst case.\n",
-    "- Therefore, the overall time complexity of the `removeNthFromEnd` function is $O(N)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The `removeNthFromEnd` function uses a constant amount of extra space for its variables (e.g., `first`, `second`, and `dummy`). This space usage does not depend on the size of the linked list.\n",
-    "- The space complexity is $O(1)$, indicating that the space used by the function is constant and independent of the size of the input linked list.\n",
-    "\n",
-    "**In summary**, the `removeNthFromEnd` function has a time complexity of $O(N)$ and a space complexity of $O(1)$. It efficiently removes the nth node from the end of the linked list with a single pass through the list and constant extra space usage."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb b/docs/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb
deleted file mode 100644
index 09de0a3..0000000
--- a/docs/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb	
+++ /dev/null
@@ -1,297 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 206. Reverse Linked List\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Reverse Linked List problem on LeetCode, click here!](https://leetcode.com/problems/reverse-linked-list/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the `head` of a singly linked list, reverse the list, and return the reversed list.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the list is the range `[0, 5000]`.\n",
-    "- `-5000 <= Node.val <= 5000`\n",
-    "\n",
-    "**Follow up:** A linked list can be reversed either iteratively or recursively. Could you implement both?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for singly-linked list.\n",
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "# Function to reverse a linked list iteratively\n",
-    "def reverseListIterative(head):\n",
-    "    prev = None\n",
-    "    current = head\n",
-    "    \n",
-    "    while current:\n",
-    "        next_node = current.next\n",
-    "        current.next = prev\n",
-    "        prev = current\n",
-    "        current = next_node\n",
-    "    \n",
-    "    return prev\n",
-    "\n",
-    "# Function to reverse a linked list recursively\n",
-    "def reverseListRecursive(head):\n",
-    "    if not head or not head.next:\n",
-    "        return head\n",
-    "\n",
-    "    new_head = reverseListRecursive(head.next)\n",
-    "    head.next.next = head\n",
-    "    head.next = None\n",
-    "\n",
-    "    return new_head"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `reverseListIterative(head)`: This function reverses a linked list iteratively using a loop. Here's a step-by-step explanation:\n",
-    "\n",
-    "   - It takes the `head` of the input linked list as a parameter.\n",
-    "   - It initializes two pointers, `prev` and `current`, initially set to `None` and the `head` of the list, respectively.\n",
-    "   - It enters a `while` loop that continues until `current` reaches the end of the list (i.e., becomes `None`).\n",
-    "   - Inside the loop:\n",
-    "     - It stores the `next_node` which is the next node after `current`.\n",
-    "     - It updates the `next` pointer of `current` to point to the `prev` node. This effectively reverses the link direction.\n",
-    "     - It moves `prev` to `current` and `current` to `next_node`, advancing the pointers one step further in the list.\n",
-    "   - Once the loop completes, `prev` will be pointing to the new head of the reversed list (which was the last node of the original list), so it returns `prev`.\n",
-    "\n",
-    "2. `reverseListRecursive(head)`: This function reverses a linked list recursively. Here's how it works:\n",
-    "\n",
-    "   - It takes the `head` of the input linked list as a parameter.\n",
-    "   - The base case is checked: if `head` is `None` or `head.next` is `None`, meaning the list is empty or has only one node, it returns `head` as there's no need to reverse a list with zero or one element.\n",
-    "   - In the recursive case:\n",
-    "     - It calls itself with `head.next`, effectively moving down the list until it reaches the end.\n",
-    "     - Once it reaches the end, it starts reversing the links:\n",
-    "       - `head.next.next` is set to `head`, reversing the link between `head` and the next node.\n",
-    "       - `head.next` is set to `None` to avoid cycles.\n",
-    "   - Finally, it returns the new head of the reversed list, which will be the last node of the original list.\n",
-    "\n",
-    "**In summary,** `reverseListIterative` reverses the linked list by iterating through it and changing the next pointers, while `reverseListRecursive` reverses the linked list by recursively reaching the end and then reversing the links on the way back up the recursion stack. Both functions achieve the same result: reversing the linked list."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to create a linked list from a list of values\n",
-    "def createLinkedList(values):\n",
-    "    if not values:\n",
-    "        return None\n",
-    "    head = ListNode(values[0])\n",
-    "    current = head\n",
-    "    for val in values[1:]:\n",
-    "        current.next = ListNode(val)\n",
-    "        current = current.next\n",
-    "    return head\n",
-    "\n",
-    "# Helper function to convert a linked list to a list for testing\n",
-    "def linkedListToList(head):\n",
-    "    result = []\n",
-    "    current = head\n",
-    "    while current:\n",
-    "        result.append(current.val)\n",
-    "        current = current.next\n",
-    "    return result\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the two helper functions used in the provided code:\n",
-    "\n",
-    "1. `createLinkedList(values)`:\n",
-    "\n",
-    "   - Purpose: This function is used to create a linked list from a list of values. It takes a list of values as input and returns the head of the linked list.\n",
-    "\n",
-    "   - How it works:\n",
-    "     - It first checks if the input list `values` is empty. If it's empty, it returns `None` to indicate an empty linked list.\n",
-    "     - If `values` is not empty, it initializes a `head` node with the value of the first element in `values`.\n",
-    "     - It then iterates through the remaining values in `values`, creating new nodes for each value and linking them together to form a linked list.\n",
-    "     - Finally, it returns the `head` node of the newly created linked list.\n",
-    "\n",
-    "2. `linkedListToList(head)`:\n",
-    "\n",
-    "   - Purpose: This function is used to convert a linked list back into a Python list for testing and output purposes.\n",
-    "\n",
-    "   - How it works:\n",
-    "     - It takes the `head` of the linked list as input.\n",
-    "     - It initializes an empty list called `result`.\n",
-    "     - It then iterates through the linked list, starting from `head`, and appends each node's `val` (value) to the `result` list.\n",
-    "     - This process continues until the end of the linked list is reached.\n",
-    "     - Finally, it returns the `result` list, which contains the values from the linked list in the same order.\n",
-    "\n",
-    "These helper functions are used to simplify the process of creating linked lists from lists of values and converting linked lists back into lists, making it easier to work with linked lists in the provided examples and test cases."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[5, 4, 3, 2, 1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: Iterative\n",
-    "head1 = [1,2,3,4,5]\n",
-    "head1 = createLinkedList(head1)\n",
-    "reversed_head1 = reverseListIterative(head1)\n",
-    "print(linkedListToList(reversed_head1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[2, 1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2: Recursive\n",
-    "head2 = [1,2]\n",
-    "head2 = createLinkedList(head2)\n",
-    "reversed_head2 = reverseListRecursive(head2)\n",
-    "print(linkedListToList(reversed_head2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3: Empty list\n",
-    "head3 = []\n",
-    "head3 = createLinkedList(head3)\n",
-    "reversed_head3 = reverseListIterative(head3)\n",
-    "print(linkedListToList(reversed_head3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the two functions:\n",
-    "\n",
-    "1. `reverseListIterative(head)`:\n",
-    "\n",
-    "   - Time Complexity: $O(n)$\n",
-    "     - The iterative function visits each node in the linked list exactly once in a single pass, where \"$n$\" is the number of nodes in the list. Therefore, the time complexity is linear in the number of nodes.\n",
-    "\n",
-    "   - Space Complexity: $O(1)$\n",
-    "     - This function uses a constant amount of extra space for the `prev`, `current`, and `next_node` pointers. Regardless of the size of the input list, the amount of additional memory used remains the same, so the space complexity is constant.\n",
-    "\n",
-    "2. `reverseListRecursive(head)`:\n",
-    "\n",
-    "   - Time Complexity: $O(n)$\n",
-    "     - The recursive function also visits each node in the linked list exactly once. It recursively traverses the list from the head to the tail, reversing the links on the way back. Therefore, like the iterative approach, the time complexity is $O(n)$, where \"$n$\" is the number of nodes.\n",
-    "\n",
-    "   - Space Complexity: $O(n)$\n",
-    "     - The recursive function uses space on the call stack for each recursive call. In the worst case, when the list has \"$n$\" nodes, it will create \"$n$\" recursive function calls on the stack, resulting in a space complexity of $O(n)$. This is because each function call stores information about its state, including the `head` pointer, until it reaches the base case and starts returning.\n",
-    "\n",
-    "**In summary:**\n",
-    "\n",
-    "- The time complexity of both functions is $O(n)$ because they both visit each node once.\n",
-    "- The space complexity of `reverseListIterative` is $O(1)$ because it uses a constant amount of extra space.\n",
-    "- The space complexity of `reverseListRecursive` is $O(n)$ due to the recursive function calls and the associated call stack space.\n",
-    "\n",
-    "Both functions are efficient in terms of time complexity, but `reverseListIterative` is more memory-efficient because it uses a constant amount of additional space, while `reverseListRecursive` uses space on the call stack proportional to the number of nodes in the list."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **K-Group Reverse:** Modify the reverseListIterative function to reverse the linked list in groups of $K$ elements. For example, if the input linked list is [1, 2, 3, 4, 5, 6], and $K$ is 3, the output should be [3, 2, 1, 6, 5, 4].\n",
-    "\n",
-    "2. **Rotate Linked List:** Write a function to rotate a linked list to the right by $K$ places. For example, if the input is [1, 2, 3, 4, 5] and $K$ is 2, the output should be [4, 5, 1, 2, 3]."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb b/docs/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb
deleted file mode 100644
index b433cf3..0000000
--- a/docs/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb	
+++ /dev/null
@@ -1,277 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 21. Merge Two Sorted Lists\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Merge Two Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-two-sorted-lists/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given the heads of two sorted linked lists `list1` and `list2`.\n",
-    "\n",
-    "Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.\n",
-    "\n",
-    "*Return the head of the merged linked list.*\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in both lists is in the range `[0, 50]`.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "- Both `list1` and `list2` are sorted in non-decreasing order."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "def mergeTwoLists(list1, list2):\n",
-    "    # Create a dummy node to simplify the code.\n",
-    "    dummy = ListNode()\n",
-    "    current = dummy\n",
-    "\n",
-    "    while list1 and list2:\n",
-    "        if list1.val < list2.val:\n",
-    "            current.next = list1\n",
-    "            list1 = list1.next\n",
-    "        else:\n",
-    "            current.next = list2\n",
-    "            list2 = list2.next\n",
-    "        current = current.next\n",
-    "\n",
-    "    # Append the remaining elements from either list.\n",
-    "    if list1:\n",
-    "        current.next = list1\n",
-    "    else:\n",
-    "        current.next = list2\n",
-    "\n",
-    "    return dummy.next"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We define a class `ListNode` to represent the nodes of a singly-linked list. Each node has a `val` attribute to store the value of the node and a `next` attribute to point to the next node in the list.\n",
-    "\n",
-    "2. We define the `mergeTwoLists` function that takes two linked list heads, `list1` and `list2`, as input.\n",
-    "\n",
-    "3. We create a `dummy` node at the beginning of the merged list. The `dummy` node simplifies the code by serving as a placeholder, and its `next` attribute will point to the actual merged list.\n",
-    "\n",
-    "4. We initialize a `current` pointer to point to the `dummy` node initially. This `current` pointer helps us traverse and build the merged list.\n",
-    "\n",
-    "5. We enter a `while` loop that continues until either `list1` or `list2` becomes empty. Inside the loop, we compare the values of the nodes at the current positions of `list1` and `list2`.\n",
-    "\n",
-    "6. If the `val` of the node in `list1` is smaller than the `val` of the node in `list2`, we attach the node from `list1` to the merged list by updating the `next` attribute of the `current` node to point to the node in `list1`. We then move the `list1` pointer to the next node.\n",
-    "\n",
-    "7. If the `val` of the node in `list2` is smaller than or equal to the `val` of the node in `list1`, we attach the node from `list2` to the merged list in a similar manner. We move the `list2` pointer to the next node.\n",
-    "\n",
-    "8. After attaching a node to the merged list, we move the `current` pointer to the newly added node. This step is essential for keeping track of the end of the merged list.\n",
-    "\n",
-    "9. The loop continues until either `list1` or `list2` becomes empty.\n",
-    "\n",
-    "10. Once the loop exits, we check if there are any remaining elements in `list1` or `list2`. If `list1` is not empty, we attach the remaining elements of `list1` to the merged list. If `list2` is not empty, we attach the remaining elements of `list2`.\n",
-    "\n",
-    "11. Finally, we return the `next` attribute of the `dummy` node as the head of the merged list, which represents the sorted merged list."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to print the linked list\n",
-    "def printLinkedList(head):\n",
-    "    result = []\n",
-    "    while head:\n",
-    "        result.append(head.val)\n",
-    "        head = head.next\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the helper functions:\n",
-    "**printLinkedList Function**:\n",
-    "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-    "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-    "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-    "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-    "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-    "\n",
-    "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "[1, 1, 2, 3, 4, 4]"
-      ]
-     },
-     "execution_count": 3,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "# Input: list1 = [1,2,4], list2 = [1,3,4]\n",
-    "list1 = ListNode(1, ListNode(2, ListNode(4)))\n",
-    "list2 = ListNode(1, ListNode(3, ListNode(4)))\n",
-    "merged_list = mergeTwoLists(list1, list2)\n",
-    "printLinkedList(merged_list)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "[]"
-      ]
-     },
-     "execution_count": 4,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "# Input: list1 = [], list2 = []\n",
-    "list1 = None\n",
-    "list2 = None\n",
-    "merged_list = mergeTwoLists(list1, list2)\n",
-    "printLinkedList(merged_list)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "[0]"
-      ]
-     },
-     "execution_count": 5,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# list1 = [], list2 = [0]\n",
-    "list1 = None\n",
-    "list2 = ListNode(0)\n",
-    "merged_list = mergeTwoLists(list1, list2)\n",
-    "printLinkedList(merged_list)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `mergeTwoLists` function.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this function is $O(N + M$), where $N$ and $M$ are the lengths of `list1` and `list2`, respectively. Here's why:\n",
-    "\n",
-    "1. In the worst case, we need to traverse both `list1` and `list2` completely. This involves iterating through all the nodes in both lists once.\n",
-    "\n",
-    "2. The while loop runs until either `list1` or `list2` becomes empty. The number of iterations depends on the total number of nodes in both lists, which is N + M in the worst case.\n",
-    "\n",
-    "3. Inside the loop, we perform constant-time operations for each iteration, such as comparisons and updating pointers.\n",
-    "\n",
-    "Therefore, the dominant factor in the time complexity is the combined length of both input lists, resulting in a linear time complexity of $O(N + M)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this function is $O(1)$, which means it uses a constant amount of extra space regardless of the input sizes. Here's why:\n",
-    "\n",
-    "1. We create a few extra pointers (`dummy`, `current`) and temporary variables (`val`) to manage the merging process. These consume a fixed amount of memory, regardless of the input sizes. The number of these extra variables is independent of the lengths of `list1` and `list2`.\n",
-    "\n",
-    "2. We do not create a new data structure or allocate memory for the merged list. Instead, we rearrange the existing nodes from `list1` and `list2` to form the merged list. This operation does not consume additional memory proportional to the input sizes.\n",
-    "\n",
-    "**In summary**, the `mergeTwoLists` function has a time complexity of $O(N + M)$ and a space complexity of $O(1)$, making it an efficient solution for merging two sorted linked lists."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Merge K Sorted Lists:** Extend the problem to merge $K$ sorted linked lists instead of just two. You need to efficiently merge $K$ lists into one sorted list.\n",
-    "\n",
-    "2. **Merge Lists in a Zigzag Pattern:** Merge two sorted lists in a zigzag pattern. For example, given [1, 2, 4] and [1, 3, 5], the merged list should be [1, 1, 2, 3, 4, 5]."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb b/docs/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb
deleted file mode 100644
index 0568867..0000000
--- a/docs/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb	
+++ /dev/null
@@ -1,277 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 23. Merge k Sorted Lists\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Merge k Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-k-sorted-lists/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given an array of $k$ linked-lists lists, each linked-list is sorted in ascending order.\n",
-    "\n",
-    "Merge all the linked-lists into one sorted linked-list and return it.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `k == lists.length`\n",
-    "- 0 <= k <= $10^4$\n",
-    "- `0 <= lists[i].length <= 500`\n",
-    "- $-10^4$ <= lists[i][j] <= $10^4$\n",
-    "- `lists[i]` is sorted in ascending order.\n",
-    "- The sum of `lists[i].length` will not exceed $10^4$.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for singly-linked list.\n",
-    "class ListNode:\n",
-    "     def __init__(self, val=0, next=None):\n",
-    "         self.val = val\n",
-    "         self.next = next\n",
-    "\n",
-    "class Solution:\n",
-    "    def mergeKLists(self, lists: [[ListNode]]) -> ListNode:\n",
-    "        # Check if the input list of linked lists is empty\n",
-    "        if not lists or len(lists) == 0:\n",
-    "            return None\n",
-    "\n",
-    "        # Loop until there is only one merged list left\n",
-    "        while len(lists) > 1:\n",
-    "            mergedLists = []\n",
-    "\n",
-    "            # Merge pairs of lists\n",
-    "            for i in range(0, len(lists), 2):\n",
-    "                l1 = lists[i]\n",
-    "                l2 = lists[i + 1] if (i + 1) < len(lists) else None\n",
-    "                mergedLists.append(self.mergeList(l1, l2))\n",
-    "\n",
-    "            # Update the list of lists with merged results\n",
-    "            lists = mergedLists\n",
-    "        \n",
-    "        # Return the final merged list\n",
-    "        return lists[0]\n",
-    "\n",
-    "    def mergeList(self, l1, l2):\n",
-    "        # Create a dummy node to simplify merging\n",
-    "        dummy = ListNode()\n",
-    "        tail = dummy\n",
-    "\n",
-    "        # Merge the two sorted lists\n",
-    "        while l1 and l2:\n",
-    "            if l1.val < l2.val:\n",
-    "                tail.next = l1\n",
-    "                l1 = l1.next\n",
-    "            else:\n",
-    "                tail.next = l2\n",
-    "                l2 = l2.next\n",
-    "            tail = tail.next\n",
-    "\n",
-    "        # Append any remaining elements from l1 or l2 (if any)\n",
-    "        if l1:\n",
-    "            tail.next = l1\n",
-    "        if l2:\n",
-    "            tail.next = l2\n",
-    "\n",
-    "        # Return the merged result starting from the next of the dummy node\n",
-    "        return dummy.next\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code defines a Python class `Solution` with two methods for merging k sorted linked lists:\n",
-    "\n",
-    "1. `mergeKLists(self, lists: List[ListNode]) -> ListNode`: This method takes a list of k sorted linked lists as input and returns a single merged sorted linked list. It uses a divide-and-conquer approach to repeatedly merge pairs of lists until only one merged list remains.\n",
-    "\n",
-    "2. `mergeList(self, l1, l2)`: This method takes two sorted linked lists, `l1` and `l2`, as input and merges them into a single sorted linked list. It uses a dummy node to simplify the merging process.\n",
-    "\n",
-    "Here's a high-level overview of how the code works:\n",
-    "\n",
-    "- The `mergeKLists` method checks if the input list of linked lists is empty or contains no lists. If there are no lists, it returns `None`.\n",
-    "\n",
-    "- Inside a `while` loop, the code repeatedly merges pairs of linked lists until only one merged list remains in the `lists` array. It does this by iterating through the input lists in pairs and calling the `mergeList` method to merge each pair.\n",
-    "\n",
-    "- The `mergeList` method takes two sorted linked lists, `l1` and `l2`, and merges them into a single sorted linked list. It uses a dummy node (`dummy`) and a `tail` pointer to keep track of the merged list while comparing and merging elements from `l1` and `l2`.\n",
-    "\n",
-    "- After merging all pairs of lists and updating the `lists` array with the merged results, the loop continues until only one merged list remains in the `lists` array.\n",
-    "\n",
-    "- Finally, the `mergeKLists` method returns the merged list.\n",
-    "\n",
-    "Overall, this code efficiently merges k sorted linked lists using a divide-and-conquer strategy, resulting in a single merged sorted linked list as the output."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> "
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "# Input: lists = [[1,4,5],[1,3,4],[2,6]]\n",
-    "\n",
-    "lists1 = [\n",
-    "    ListNode(1, ListNode(4, ListNode(5))),\n",
-    "    ListNode(1, ListNode(3, ListNode(4))),\n",
-    "    ListNode(2, ListNode(6))\n",
-    "]\n",
-    "solution = Solution()\n",
-    "result1 = solution.mergeKLists(lists1)\n",
-    "\n",
-    "# Print the result\n",
-    "if result1:\n",
-    "    current = result1\n",
-    "    while current:\n",
-    "        print(current.val, end=\" -> \")\n",
-    "        current = current.next\n",
-    "else:\n",
-    "    print(\"None\")  # Print \"None\" for input with a single None element"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "None\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "# Input: lists = []\n",
-    "\n",
-    "lists2 = []\n",
-    "solution = Solution()\n",
-    "result2 = solution.mergeKLists(lists2)\n",
-    "\n",
-    "# Print the result\n",
-    "if result2:\n",
-    "    current = result2\n",
-    "    while current:\n",
-    "        print(current.val, end=\" -> \")\n",
-    "        current = current.next\n",
-    "else:\n",
-    "    print(\"None\")  # Print \"None\" for input with a single None element"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "None\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "# Input: lists = [[]]\n",
-    "\n",
-    "lists3 = [None]\n",
-    "solution = Solution()\n",
-    "result3 = solution.mergeKLists(lists3)\n",
-    "\n",
-    "# Print the result\n",
-    "if result3:\n",
-    "    current = result3\n",
-    "    while current:\n",
-    "        print(current.val, end=\" -> \")\n",
-    "        current = current.next\n",
-    "else:\n",
-    "    print(\"None\")  # Print \"None\" for input with a single None element"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. **Heap Initialization:** The code does not use a heap. Instead, it uses a divide-and-conquer approach. The initial check for empty input lists takes $O(1)$ time.\n",
-    "\n",
-    "2. **Merging:** The merging operation is performed in a divide-and-conquer fashion. In each iteration of the `while` loop, we merge pairs of linked lists, and the number of comparisons made is proportional to the total number of nodes across all the linked lists ($n$). In each merge step, we effectively process each node once. The number of iterations required to reduce k lists to 1 is $O(log\\ k)$.\n",
-    "\n",
-    "   Therefore, the overall time complexity of the code is $O(n * log\\ k)$, where n is the total number of nodes across all lists, and $k$ is the number of linked lists.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. **Heap Space:** The code doesn't use a heap data structure, so there's no additional space complexity due to a heap.\n",
-    "\n",
-    "2. **Merged Lists:** In the `mergeList` method, we create a new merged list. However, this list is not stored in memory for all lists; it's replaced with each merged pair. The space used for these merged lists is proportional to the size of the largest merged list, which is $O(n)$ in the worst case.\n",
-    "\n",
-    "3. **Additional Variables:** The code uses a few additional variables, such as `dummy` and `tail`, but these occupy a constant amount of space and don't depend on the input size.\n",
-    "\n",
-    "   Therefore, the overall space complexity of the code is $O(n)$, where n is the total number of nodes across all lists.\n",
-    "\n",
-    "In summary, the code's time complexity is $O(n * log(k))$, and its space complexity is $O(n)$. This code efficiently merges $k$ sorted linked lists using a divide-and-conquer approach with a relatively low space overhead."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb b/docs/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb
deleted file mode 100644
index 2238b48..0000000
--- a/docs/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 100: Same Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-    "\n",
-    "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-    "\n",
-    "**Constraints**\n",
-    "\n",
-    "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-    "2. $-10^4$ <= Node.val <= $10^4$\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def isSameTree(p, q):\n",
-    "    # Base case: If both p and q are None, the trees are the same.\n",
-    "    if not p and not q:\n",
-    "        return True\n",
-    "    \n",
-    "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-    "    if not p or not q:\n",
-    "        return False\n",
-    "    \n",
-    "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-    "    if p.val != q.val:\n",
-    "        return False\n",
-    "    \n",
-    "    # Recursively check the left and right subtrees of p and q.\n",
-    "    # If both subtrees are the same, the entire trees are the same.\n",
-    "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-    "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-    "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-    "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-    "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-    "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "\n",
-    "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-    "\n",
-    "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "print(isSameTree(p1, q1)) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "\n",
-    "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-    "\n",
-    "p2 = TreeNode(1, TreeNode(2), None)\n",
-    "q2 = TreeNode(1, None, TreeNode(2))\n",
-    "print(isSameTree(p2, q2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "\n",
-    "#Input: p = [1,2,1], q = [1,1,2]\n",
-    "\n",
-    "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-    "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-    "print(isSameTree(p3, q3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time Complexity\n",
-    "\n",
-    "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-    "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-    "\n",
-    "## Space Complexity\n",
-    "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-    "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-    "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-    "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-    "\n",
-    "## In summary:\n",
-    "\n",
-    "+ **Time Complexity:** $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-    "+ **Space Complexity:** $O(n)$ due to the recursive call stack."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb b/docs/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb
deleted file mode 100644
index 9d2cc7f..0000000
--- a/docs/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb	
+++ /dev/null
@@ -1,263 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 102. Binary Tree Level Order Traversal\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Binary Tree Level Order Traversal problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-level-order-traversal/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary tree, return *the level order traversal of its nodes' values*. (i.e., from left to right, level by level).\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-    "- `-1000 <= Node.val <= 1000`\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def levelOrder(root):\n",
-    "    # Check if the tree is empty, if so, return an empty list.\n",
-    "    if not root:\n",
-    "        return []\n",
-    "\n",
-    "    # Initialize an empty list to store the result.\n",
-    "    result = []\n",
-    "    \n",
-    "    # Initialize a queue with the root node to perform BFS.\n",
-    "    queue = [root]\n",
-    "\n",
-    "    while queue:\n",
-    "        # Initialize an empty list to store the values of nodes at the current level.\n",
-    "        level_values = []\n",
-    "        \n",
-    "        # Get the number of nodes at the current level.\n",
-    "        level_size = len(queue)\n",
-    "\n",
-    "        # Iterate through the nodes at the current level.\n",
-    "        for _ in range(level_size):\n",
-    "            # Dequeue the front node from the queue.\n",
-    "            node = queue.pop(0)\n",
-    "            \n",
-    "            # Append the value of the current node to the level_values list.\n",
-    "            level_values.append(node.val)\n",
-    "\n",
-    "            # Enqueue the left and right children of the current node if they exist.\n",
-    "            if node.left:\n",
-    "                queue.append(node.left)\n",
-    "            if node.right:\n",
-    "                queue.append(node.right)\n",
-    "\n",
-    "        # Append the level_values list (values at the current level) to the result list.\n",
-    "        result.append(level_values)\n",
-    "\n",
-    "    # Return the final result, which is a list of lists representing level order traversal.\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We define a `TreeNode` class that represents a node in the binary tree. Each `TreeNode` object has a value (`val`) and two child nodes: `left` and `right`.\n",
-    "\n",
-    "2. The `levelOrder` function takes the root of the binary tree as its input and returns the level order traversal of the tree as a list of lists.\n",
-    "\n",
-    "3. We start by checking if the input `root` is `None`, which indicates an empty tree. If the tree is empty, we return an empty list because there are no nodes to traverse.\n",
-    "\n",
-    "4. We initialize an empty list called `result` to store the final result, which will be a list of lists containing node values at each level.\n",
-    "\n",
-    "5. We initialize a queue called `queue` with the root node. This queue will be used for breadth-first traversal of the tree.\n",
-    "\n",
-    "6. We enter a while loop that continues until the `queue` is empty. Inside the loop, we perform the following steps:\n",
-    "\n",
-    "   - We initialize an empty list called `level_values` to store the values of nodes at the current level.\n",
-    "\n",
-    "   - We determine the number of nodes at the current level by getting the length of the `queue`. This is done to process nodes level by level.\n",
-    "\n",
-    "   - We iterate through the nodes at the current level using a for loop. For each node in the current level:\n",
-    "\n",
-    "     - We dequeue (remove) the front node from the `queue`.\n",
-    "\n",
-    "     - We append the value of the dequeued node to the `level_values` list, effectively collecting the values of nodes at the current level.\n",
-    "\n",
-    "     - If the dequeued node has a left child, we enqueue the left child to the `queue`.\n",
-    "\n",
-    "     - If the dequeued node has a right child, we enqueue the right child to the `queue`.\n",
-    "\n",
-    "   - After processing all nodes at the current level, we append the `level_values` list to the `result` list. This represents the values at the current level.\n",
-    "\n",
-    "7. The loop continues until all levels have been traversed, and the `queue` becomes empty.\n",
-    "\n",
-    "8. Finally, we return the `result` list, which contains lists of node values at each level, representing the level order traversal of the binary tree.\n",
-    "\n",
-    "The code effectively performs a breadth-first traversal of the binary tree, processing nodes level by level, and constructs the result list that represents the level order traversal."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[3], [9, 20], [15, 7]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Construct the tree\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(9)\n",
-    "root.right = TreeNode(20)\n",
-    "root.right.left = TreeNode(15)\n",
-    "root.right.right = TreeNode(7)\n",
-    "\n",
-    "result = levelOrder(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[1]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Using the same tree as before\n",
-    "root = TreeNode(1)\n",
-    "result = levelOrder(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# Creating a new tree for this example\n",
-    "root = None\n",
-    "result = levelOrder(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `levelOrder` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this function is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the breadth-first traversal.\n",
-    "\n",
-    "In the worst case, we have to enqueue and dequeue all nodes in the binary tree, which is proportional to the number of nodes (N). In each level, we process all nodes in that level, and since there are a total of log(N) levels in a balanced binary tree, the time complexity can also be approximated as O(N) for unbalanced trees.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this function is O(N), where N is the number of nodes in the binary tree. Here's how the space complexity breaks down:\n",
-    "\n",
-    "1. The `result` list stores the level order traversal, and in the worst case, it contains N/2 levels (for a completely unbalanced binary tree). So, the space used by `result` is O(N).\n",
-    "\n",
-    "2. The `queue` data structure is used for BFS traversal. In the worst case, it can store all nodes at the last level of the tree. In a balanced binary tree, the maximum number of nodes at any level is 2^(log(N)), which is still O(N). In the case of an unbalanced tree, it can be even worse. So, the space used by `queue` is O(N).\n",
-    "\n",
-    "Overall, the dominant factor in terms of space complexity is the `queue`, and the space complexity is O(N).\n",
-    "\n",
-    "In summary, the function's time complexity is O(N), and its space complexity is also O(N). It is an efficient and optimal solution for performing a level order traversal of a binary tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Reverse Level Order Traversal:** Modify the `levelOrder` function to return the level order traversal in reverse order (from bottom to top). For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[15, 7], [9, 20], [3]]`.\n",
-    "\n",
-    "2. **Zigzag Level Order Traversal:** Write a function that performs a level order traversal of a binary tree in a zigzag pattern. In a zigzag traversal, the nodes at even levels are traversed from left to right, and nodes at odd levels are traversed from right to left. For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[3], [20, 9], [15, 7]]`."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb b/docs/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb
deleted file mode 100644
index 87c5eaa..0000000
--- a/docs/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 104. Maximum Depth of Binary Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the `root` of a binary tree, return its maximum depth.\n",
-    "\n",
-    "A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def maxDepth(root):\n",
-    "    # Base case: If the root is None, the depth is 0.\n",
-    "    if root is None:\n",
-    "        return 0\n",
-    "    \n",
-    "    # Recursively calculate the maximum depth of the left and right subtrees.\n",
-    "    left_depth = maxDepth(root.left)\n",
-    "    right_depth = maxDepth(root.right)\n",
-    "    \n",
-    "    # Return the maximum depth of the tree by adding 1 to the maximum depth of the subtrees.\n",
-    "    return max(left_depth, right_depth) + 1"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Let's go through the code step by step to understand how it works:\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class to represent the nodes of the binary tree. Each node has three attributes:\n",
-    "   - `val`: the value stored in the node.\n",
-    "   - `left`: a reference to the left child node.\n",
-    "   - `right`: a reference to the right child node.\n",
-    "\n",
-    "2. The `maxDepth` function is the main function that calculates the maximum depth of the binary tree. It takes a single argument, `root`, which is the root node of the binary tree.\n",
-    "\n",
-    "3. In the `maxDepth` function, we have a base case:\n",
-    "   - If the `root` is `None`, it means we have reached the end of a branch of the tree (a leaf node or an empty subtree). In this case, the depth is 0 because there are no nodes to count.\n",
-    "\n",
-    "4. If the `root` is not `None`, we recursively calculate the maximum depth of the left and right subtrees:\n",
-    "   - We call `maxDepth` on the `root.left` to calculate the maximum depth of the left subtree and store it in the variable `left_depth`.\n",
-    "   - We call `maxDepth` on the `root.right` to calculate the maximum depth of the right subtree and store it in the variable `right_depth`.\n",
-    "\n",
-    "5. Finally, we return the maximum depth of the tree by taking the maximum of `left_depth` and `right_depth` and adding 1 to it. This is because we are counting the current level as well."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "# Input: root = [3,9,20,null,null,15,7]\n",
-    "\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(9)\n",
-    "root.right = TreeNode(20)\n",
-    "root.right.left = TreeNode(15)\n",
-    "root.right.right = TreeNode(7)\n",
-    "\n",
-    "print(maxDepth(root))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a230be4f-1f47-4919-87e3-62fc3f072d97",
-   "metadata": {},
-   "source": [
-    "In this example, we create a binary tree based on the given input `[3,9,20,null,null,15,7]` and calculate its maximum depth, which is `3`. The tree structure is as follows:\n",
-    "   ```\n",
-    "       3\n",
-    "      / \\\n",
-    "     9  20\n",
-    "       /  \\\n",
-    "      15   7\n",
-    "   ```\n",
-    "\n",
-    "So, the maximum depth is the length of the longest path from the root to a leaf node, which is 3 in this case."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "# Input: root = [1,null,2]\n",
-    "\n",
-    "root = TreeNode(1)\n",
-    "root.right = TreeNode(2)\n",
-    "\n",
-    "# Calculate the maximum depth of the tree and print the result\n",
-    "print(maxDepth(root))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The time complexity of the `maxDepth` function is $O(N)$, where $N$ is the number of nodes in the binary tree.\n",
-    "- This is because in the worst case, the function visits every node exactly once in a depth-first manner.\n",
-    "- The recursion explores all nodes of the tree, so the time complexity is linear with respect to the number of nodes.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity of the `maxDepth` function is $O(H)$, where H is the height of the binary tree.\n",
-    "- In the worst case, if the binary tree is completely unbalanced (skewed), the recursion stack can go as deep as the height of the tree.\n",
-    "- In the best case, if the binary tree is perfectly balanced, the height is $O(log\\ N)$, where $N$ is the number of nodes.\n",
-    "- Therefore, the space complexity can vary from $O(log\\ N)$ to $O(N)$ depending on the shape of the tree.\n",
-    "- In addition to the recursion stack, there is a small constant amount of space used for variables and function call overhead.\n",
-    "\n",
-    "**In summary:**\n",
-    "- The time complexity of the code is $O(N)$ as it visits each node once.\n",
-    "- The space complexity is $O(H)$, where H is the height of the tree, which can vary from $O(log\\ N)$ to $O(N)$ depending on the tree's shape."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb b/docs/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb
deleted file mode 100644
index 0d5bbd6..0000000
--- a/docs/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb	
+++ /dev/null
@@ -1,305 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 105. Construct Binary Tree from Preorder and Inorder Traversal\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Construct Binary Tree from Preorder and Inorder Traversal problem on LeetCode, click here!](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)\n",
-    "\n",
-    "---\n",
-    "Given two integer arrays `preorder` and `inorder` where `preorder` is the preorder traversal of a binary tree and `inorder` is the inorder traversal of the same tree, construct and return *the binary tree.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= preorder.length <= 3000`\n",
-    "- `inorder.length == preorder.length`\n",
-    "- `-3000 <= preorder[i], inorder[i] <= 3000`\n",
-    "- `preorder` and `inorder` consist of **unique** values.\n",
-    "- Each value of `inorder` also appears in `preorder`.\n",
-    "- `preorder` is **guaranteed** to be the preorder traversal of the tree.\n",
-    "- `inorder` is **guaranteed** to be the inorder traversal of the tree.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def buildTree(preorder, inorder):\n",
-    "    if not preorder:\n",
-    "        return None\n",
-    "\n",
-    "    # The first element in the preorder traversal is the root of the current subtree\n",
-    "    root_val = preorder[0]\n",
-    "    root = TreeNode(root_val)\n",
-    "\n",
-    "    # Find the index of the root value in the inorder traversal\n",
-    "    root_idx_inorder = inorder.index(root_val)\n",
-    "\n",
-    "    # Recursively build left and right subtrees\n",
-    "    root.left = buildTree(preorder[1:1 + root_idx_inorder], inorder[:root_idx_inorder])\n",
-    "    root.right = buildTree(preorder[1 + root_idx_inorder:], inorder[root_idx_inorder + 1:])\n",
-    "\n",
-    "    return root"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `class TreeNode:`: This is a class definition for a binary tree node. It has three attributes:\n",
-    "   - `val`: The value of the node.\n",
-    "   - `left`: A reference to the left child node.\n",
-    "   - `right`: A reference to the right child node.\n",
-    "\n",
-    "   This class is used to create instances of binary tree nodes, which will be used to build the binary tree.\n",
-    "\n",
-    "2. `def buildTree(preorder, inorder):`: This is the main function that constructs a binary tree from its preorder and inorder traversals.\n",
-    "\n",
-    "   - `preorder`: A list representing the preorder traversal of the binary tree.\n",
-    "   - `inorder`: A list representing the inorder traversal of the binary tree.\n",
-    "\n",
-    "3. `if not preorder: return None`: This line checks if the `preorder` list is empty. If it is, it means there are no nodes to construct, so the function returns `None`.\n",
-    "\n",
-    "4. `root_val = preorder[0]`: The value of the root node is extracted from the first element of the `preorder` list.\n",
-    "\n",
-    "5. `root = TreeNode(root_val)`: A new `TreeNode` object is created with the `root_val` as its value. This represents the root of the current subtree.\n",
-    "\n",
-    "6. `root_idx_inorder = inorder.index(root_val)`: The index of the `root_val` in the `inorder` list is found. This index indicates the position of the root node in the inorder traversal.\n",
-    "\n",
-    "7. `root.left`: The left subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The left subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists up to the `root_idx_inorder`.\n",
-    "\n",
-    "8. `root.right`: The right subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The right subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists starting from `root_idx_inorder + 1`.\n",
-    "\n",
-    "9. Finally, the function returns the `root` of the subtree it just constructed.\n",
-    "\n",
-    "The recursive approach used here divides the problem of constructing the entire binary tree into smaller subproblems, starting with the root node and then recursively building the left and right subtrees until the entire tree is constructed."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "2d82857b-4fbe-474f-bf33-17b55c20147b",
-   "metadata": {},
-   "source": [
-    "## Helper Function \n",
-    "\n",
-    "Here is a helper function to represent the binary tree elements."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "13e6f3e1-659d-479d-9521-75629b8c782f",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def treeToList(root):\n",
-    "    # Initialize an empty list to store the elements of the binary tree\n",
-    "    result = []\n",
-    "    \n",
-    "    # Initialize a queue for level-order traversal starting with the root\n",
-    "    queue = [root]\n",
-    "\n",
-    "    while queue:\n",
-    "        # Pop the front node from the queue\n",
-    "        node = queue.pop(0)\n",
-    "        \n",
-    "        # If the node is not None (i.e., a valid node), add its value to the result list\n",
-    "        if node:\n",
-    "            result.append(node.val)\n",
-    "            \n",
-    "            # Add the left and right children of the node to the queue\n",
-    "            queue.append(node.left)\n",
-    "            queue.append(node.right)\n",
-    "        else:\n",
-    "            # If the node is None, add None to the result to represent an empty node\n",
-    "            result.append(None)\n",
-    "\n",
-    "    # Remove any trailing None values from the result list\n",
-    "    while result and result[-1] is None:\n",
-    "        result.pop()\n",
-    "\n",
-    "    # Return the resulting list representing the binary tree elements\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "86cd4c9d-a120-4dd1-a1bd-530436012574",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Let's break down the `treeToList` helper function step by step:\n",
-    "\n",
-    "1. `def treeToList(root):`: This function takes the root of a binary tree as input and converts it into a list.\n",
-    "\n",
-    "2. `result = []`: Initialize an empty list called `result` to store the elements of the binary tree in list format.\n",
-    "\n",
-    "3. `queue = [root]`: Initialize a queue data structure for a level-order traversal, starting with the root node.\n",
-    "\n",
-    "4. `while queue:`: This starts a loop that continues until the queue is empty, indicating that all nodes have been processed.\n",
-    "\n",
-    "5. `node = queue.pop(0)`: Dequeue the first node from the queue, effectively processing it.\n",
-    "\n",
-    "6. `if node:`: Check if the `node` is not None, which means it's a valid node in the binary tree.\n",
-    "\n",
-    "    a. `result.append(node.val)`: If the node is valid, append its value (`node.val`) to the `result` list. This represents the value of the current node in the binary tree.\n",
-    "\n",
-    "    b. `queue.append(node.left)`: Enqueue the left child of the current node if it exists. This adds the left child to the queue for processing in the next iteration.\n",
-    "\n",
-    "    c. `queue.append(node.right)`: Enqueue the right child of the current node if it exists. This adds the right child to the queue for processing in the next iteration.\n",
-    "\n",
-    "7. `else:`: If the `node` is None, it represents an empty node in the binary tree.\n",
-    "\n",
-    "    a. `result.append(None)`: Append `None` to the `result` list to indicate an empty node.\n",
-    "\n",
-    "8. `while result and result[-1] is None:`: After the traversal is complete, there might be trailing `None` values in the `result` list. This loop removes any such trailing `None` values to ensure a clean representation of the tree's elements.\n",
-    "\n",
-    "9. `return result`: Return the `result` list, which now contains the elements of the binary tree in a format where `None` represents empty nodes and the order of elements reflects a level-order traversal of the tree.\n",
-    "\n",
-    "In summary, the `treeToList` function performs a level-order traversal of the binary tree using a queue, constructing a list where each element corresponds to a node's value or represents an empty node with `None`. This list represents the binary tree's elements in a structured format."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[3, 9, 20, None, None, 15, 7]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "preorder = [3, 9, 20, 15, 7]\n",
-    "inorder = [9, 3, 15, 20, 7]\n",
-    "result = buildTree(preorder, inorder)\n",
-    "\n",
-    "print(treeToList(result))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[-1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "preorder = [-1]\n",
-    "inorder = [-1]\n",
-    "result = buildTree(preorder, inorder)\n",
-    "\n",
-    "print(treeToList(result))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `buildTree` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of the `buildTree` function can be analyzed in terms of the number of nodes in the binary tree. Let's assume there are `n` nodes in the binary tree.\n",
-    "\n",
-    "1. Finding the root value in the `preorder` traversal takes O(1) time as it's just extracting the first element from the list.\n",
-    "2. Finding the index of the root value in the `inorder` traversal using `inorder.index(root_val)` takes O(n) time in the worst case because in the worst case, it might have to search through all `n` elements in the `inorder` list to find the index.\n",
-    "3. The recursive calls to `buildTree` for the left and right subtrees are made once for each node in the tree. Therefore, the recursive calls have a combined time complexity of O(n) as they process each node once.\n",
-    "\n",
-    "The total time complexity of the `buildTree` function is O(n) due to the recursive calls and finding the root's index in the `inorder` list.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity is determined by the space used by the function's call stack during recursion and any additional data structures used. \n",
-    "\n",
-    "1. The space used by the function call stack during recursion depends on the height of the binary tree. In the worst case, where the tree is highly unbalanced (e.g., a skewed tree), the space complexity for the call stack is O(n) as it can go as deep as the number of nodes in the tree.\n",
-    "\n",
-    "2. Additionally, the function creates TreeNode objects for each node in the binary tree. Therefore, the space complexity for these objects is also O(n).\n",
-    "\n",
-    "Overall, the space complexity of the `buildTree` function is O(n) due to the space used by the call stack and the TreeNode objects.\n",
-    "\n",
-    "In summary, the time complexity of `buildTree` is O(n), and the space complexity is O(n), where 'n' is the number of nodes in the binary tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Construct Binary Tree from Postorder and Inorder Traversal:**\n",
-    "   Modify the problem to construct a binary tree from its postorder and inorder traversals. Implement a function similar to `buildTree` but for postorder and inorder traversals.\n",
-    "\n",
-    "2. **Reconstruct Binary Tree with Duplicate Values:**\n",
-    "   Extend the problem to handle binary trees with duplicate values. Ensure that your solution correctly handles scenarios where there are duplicate values in the preorder and inorder traversals."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb b/docs/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb
deleted file mode 100644
index b661fa1..0000000
--- a/docs/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb	
+++ /dev/null
@@ -1,243 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 124. Binary Tree Maximum Path Sum\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Binary Tree Maximum Path Sum problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-maximum-path-sum/)\n",
-    "\n",
-    "---\n",
-    "A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.\n",
-    "\n",
-    "The **path sum** of a path is the sum of the node's values in the path.\n",
-    "\n",
-    "Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[1, 3*10^4]$.\n",
-    "- `-1000 <= Node.val <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to find the maximum path sum in a binary tree. A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. The path sum of a path is the sum of the node values in that path.\n",
-    "\n",
-    "In this problem, you are given the root of a binary tree, and you need to find the maximum path sum among all possible paths in the tree. Note that the path does not need to start at the root node or end at a leaf node; it can start and end at any nodes in the tree.\n",
-    "\n",
-    "To find the maximum path sum, you need to consider both left and right subtrees of each node while traversing the tree. At each node, you have two choices:\n",
-    "\n",
-    "1. Include the current node in the path: In this case, you add the value of the current node to the sum and continue exploring both the left and right subtrees for possible extensions of the path.\n",
-    "\n",
-    "2. Start a new path from the current node: In this case, you do not include the current node in the path sum, and you choose either the left subtree or the right subtree to start a new path.\n",
-    "\n",
-    "To solve this problem, you can use a recursive approach to traverse the binary tree. For each node, you calculate the maximum path sum that can pass through that node (option 1) and also return the maximum path sum that can be extended from that node (option 2). You keep track of the global maximum path sum encountered during the traversal.\n",
-    "\n",
-    "Ultimately, the maximum path sum among all paths in the tree will be the maximum of the global maximum path sum and the maximum path sum calculated for each node.\n",
-    "\n",
-    "The constraints for this problem include:\n",
-    "\n",
-    "- The number of nodes in the tree is in the range $[1, 3 * 10^4]$.\n",
-    "- Node values are in the range [-1000, 1000].\n",
-    "\n",
-    "By considering all possible paths through the tree, the algorithm aims to find the most significant sum of node values in any path in the binary tree.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "class Solution:\n",
-    "    def maxPathSum(self, root):\n",
-    "        def max_path_sum_helper(node):\n",
-    "            if not node:\n",
-    "                return 0\n",
-    "            \n",
-    "            # Recursively calculate the maximum path sum for left and right subtrees\n",
-    "            left_max = max(0, max_path_sum_helper(node.left))\n",
-    "            right_max = max(0, max_path_sum_helper(node.right))\n",
-    "            \n",
-    "            # Update the global maximum path sum\n",
-    "            self.max_sum = max(self.max_sum, left_max + right_max + node.val)\n",
-    "            \n",
-    "            # Return the maximum path sum starting from the current node\n",
-    "            return max(left_max, right_max) + node.val\n",
-    "        \n",
-    "        self.max_sum = float('-inf')  # Initialize the global maximum to negative infinity\n",
-    "        max_path_sum_helper(root)  # Start the recursive traversal from the root\n",
-    "        return self.max_sum"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The code starts by defining a `TreeNode` class to represent nodes in the binary tree. Each `TreeNode` has a `val` attribute representing its value, a `left` attribute pointing to the left child, and a `right` attribute pointing to the right child.\n",
-    "\n",
-    "2. Next, a `Solution` class is defined to contain the `maxPathSum` method, which will be used to find the maximum path sum in the binary tree.\n",
-    "\n",
-    "3. Within the `maxPathSum` method, a helper function called `max_path_sum_helper` is defined. This recursive function takes a single argument, `node`, which is the current node being considered during traversal.\n",
-    "\n",
-    "4. Inside the `max_path_sum_helper` function:\n",
-    "   - If `node` is `None` (i.e., the current node is a null node), it returns 0. This is the base case of the recursion.\n",
-    "   - It calculates the maximum path sum starting from the left subtree (`left_max`) and the right subtree (`right_max`). Importantly, it uses `max(0, ...)` to ensure that negative values (which would make the path sum smaller) are not considered. If a subtree's path sum is negative, it's better to not include it in the path.\n",
-    "   - The code then updates the global maximum path sum (`self.max_sum`) by checking if the sum of `left_max`, `right_max`, and the current node's value is greater than the current maximum.\n",
-    "   - Finally, it returns the maximum path sum starting from the current node, which is the maximum of `left_max` and `right_max` plus the current node's value.\n",
-    "\n",
-    "5. Before calling the `max_path_sum_helper` function, the `max_sum` attribute of the `Solution` class is initialized to negative infinity (`float('-inf')`). This is done to ensure that any valid path sum encountered during traversal will be greater than the initial value of `max_sum`.\n",
-    "\n",
-    "6. The `max_path_sum_helper` function is called with the root of the binary tree, effectively starting the traversal from the root node.\n",
-    "\n",
-    "7. Once the traversal is complete, the method returns the value of `self.max_sum`, which contains the maximum path sum found in the binary tree.\n",
-    "\n",
-    "8. Example usage at the bottom of the code demonstrates how to create binary trees and use the `maxPathSum` method to find the maximum path sum for two different tree structures.\n",
-    "\n",
-    "In summary, the code uses a recursive approach to explore all possible paths in the binary tree, ensuring that negative path sums are not considered. It keeps track of the maximum path sum encountered and returns that maximum value as the result."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "6\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "root1 = TreeNode(1)\n",
-    "root1.left = TreeNode(2)\n",
-    "root1.right = TreeNode(3)\n",
-    "solution = Solution()\n",
-    "print(solution.maxPathSum(root1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "42\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "root2 = TreeNode(-10)\n",
-    "root2.left = TreeNode(9)\n",
-    "root2.right = TreeNode(20)\n",
-    "root2.right.left = TreeNode(15)\n",
-    "root2.right.right = TreeNode(7)\n",
-    "print(solution.maxPathSum(root2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the given code:\n",
-    "\n",
-    "Time Complexity:\n",
-    "- The code uses a recursive depth-first traversal to explore each node in the binary tree exactly once.\n",
-    "- During the traversal, for each node, we perform constant time operations, such as updating the `max_sum` variable and calculating the maximum path sum starting from that node.\n",
-    "- Therefore, the time complexity of the code is O(N), where N is the number of nodes in the binary tree.\n",
-    "\n",
-    "Space Complexity:\n",
-    "- The space complexity of the code is determined by the space used in the call stack during the recursive traversal.\n",
-    "- In the worst case, the depth of the call stack can be equal to the height of the binary tree. In an unbalanced tree, this could be O(N), but in a balanced binary tree, the height is O(log N).\n",
-    "- Additionally, the code uses a constant amount of space for variables like `left_max`, `right_max`, and `max_sum`.\n",
-    "- Therefore, the overall space complexity is O(H), where H is the height of the binary tree. In the worst case, it's O(N), and in the best case (a balanced binary tree), it's O(log N).\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: O(N)\n",
-    "- Space Complexity: O(H), where H is the height of the binary tree, ranging from O(log N) to O(N) depending on the tree's balance."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Print the Maximum Path:**\n",
-    "   Modify the code to not only find the maximum path sum but also print the nodes involved in the maximum path. You can print the nodes in the correct order from the root to the leaf.\n",
-    "\n",
-    "2. **Path Sum Count:**\n",
-    "   Instead of finding the maximum path sum, find the count of unique paths that sum to a given target value. Each path can start and end anywhere in the tree."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb b/docs/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb
deleted file mode 100644
index 5bbdf42..0000000
--- a/docs/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 199. Binary Tree Right Side View\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Binary Tree Right Side View problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-right-side-view/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary tree, imagine yourself standing on the **right side** of it, return *the values of the nodes you can see ordered from top to bottom*.\n",
-    "\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def rightSideView(root):\n",
-    "    if not root:\n",
-    "        return []  # If the tree is empty, return an empty list.\n",
-    "\n",
-    "    result = []  # Initialize a list to store the right side view values.\n",
-    "    queue = [root]  # Initialize a queue for level-order traversal with the root node.\n",
-    "\n",
-    "    while queue:\n",
-    "        # Get the number of nodes at the current level.\n",
-    "        level_size = len(queue)\n",
-    "\n",
-    "        # Traverse all nodes at the current level and add the rightmost node to the result.\n",
-    "        for i in range(level_size):\n",
-    "            node = queue.pop(0)  # Dequeue the first node from the queue.\n",
-    "\n",
-    "            # If it's the rightmost node at this level, add its value to the result.\n",
-    "            if i == level_size - 1:\n",
-    "                result.append(node.val)\n",
-    "\n",
-    "            # Add the children of the current node to the queue.\n",
-    "            if node.left:\n",
-    "                queue.append(node.left)\n",
-    "            if node.right:\n",
-    "                queue.append(node.right)\n",
-    "\n",
-    "    return result  # Return the list of right side view values.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start with a binary tree represented using a TreeNode class. Each node has a value (`val`) and can have a left child (`left`) and a right child (`right`).\n",
-    "\n",
-    "2. The `rightSideView` function takes the root node of the binary tree as input and returns a list of values representing the nodes you can see when standing on the right side of the tree.\n",
-    "\n",
-    "3. We check if the `root` is `None`, which means the tree is empty. If it's empty, we return an empty list because there are no nodes to see.\n",
-    "\n",
-    "4. We initialize an empty list called `result` to store the right side view values.\n",
-    "\n",
-    "5. We initialize a queue called `queue` with the root node. This queue is used for a level-order traversal of the tree.\n",
-    "\n",
-    "6. We enter a while loop that continues until the `queue` is empty. Inside this loop, we perform the following steps for each level of the tree:\n",
-    "\n",
-    "   a. We determine the number of nodes at the current level by getting the length of the `queue`. This is important for processing nodes at the same level together.\n",
-    "\n",
-    "   b. We then iterate through the nodes at the current level using a for loop. For each node in the level, we dequeue it from the `queue` using `queue.pop(0)`.\n",
-    "\n",
-    "   c. If the node we dequeue is the rightmost node at the current level (determined by `i == level_size - 1`), we add its `val` to the `result` list. This is because, when standing on the right side of the tree, you can see the rightmost node at each level.\n",
-    "\n",
-    "   d. We enqueue the left and right children of the current node (if they exist) to the `queue`. This ensures that we process the next level in the subsequent iterations of the while loop.\n",
-    "\n",
-    "7. After processing all levels of the tree, the `result` list contains the values of the rightmost nodes in each level, ordered from top to bottom.\n",
-    "\n",
-    "8. Finally, we return the `result` list as the output of the function, which represents the right side view of the binary tree.\n",
-    "\n",
-    "The code effectively performs a level-order traversal of the binary tree while keeping track of the rightmost nodes at each level and adding them to the result list. This ensures that we obtain the correct order of nodes visible from the right side of the tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 3, 4]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Construct the tree\n",
-    "root1 = TreeNode(1)\n",
-    "root1.left = TreeNode(2)\n",
-    "root1.right = TreeNode(3)\n",
-    "root1.left.right = TreeNode(5)\n",
-    "root1.right.right = TreeNode(4)\n",
-    "print(rightSideView(root1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 3]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "root2 = TreeNode(1)\n",
-    "root2.right = TreeNode(3)\n",
-    "print(rightSideView(root2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# Creating a new tree for this example\n",
-    "root = None\n",
-    "result = rightSideView(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the given code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of the code is O(N), where N is the number of nodes in the binary tree. Here's why:\n",
-    "\n",
-    "1. We perform a level-order traversal of the tree using a queue. In the worst case, we visit all nodes once, which takes O(N) time since we visit each node exactly once.\n",
-    "\n",
-    "2. For each node, we perform constant-time operations like dequeuing it from the queue, checking if it's the rightmost node at the current level, and enqueuing its children (if they exist). These operations do not depend on the size of the tree, so they do not contribute to the overall time complexity.\n",
-    "\n",
-    "Therefore, the dominant factor in the time complexity is the level-order traversal, making it O(N).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of the code is also O(N). Here's why:\n",
-    "\n",
-    "1. We use a queue to perform level-order traversal. In the worst case, the queue can contain all nodes at the maximum width of the tree, which can be up to N/2 nodes for a completely unbalanced tree. Therefore, the space required for the queue is O(N).\n",
-    "\n",
-    "2. The `result` list stores the values of the rightmost nodes. In the worst case, when the binary tree is a complete binary tree, it can have roughly N/2 rightmost nodes. Therefore, the space required for the `result` list is also O(N).\n",
-    "\n",
-    "3. Other auxiliary variables like `level_size`, `i`, and `node` require only constant space and do not depend on the size of the tree.\n",
-    "\n",
-    "Combining the space used by the queue and the `result` list, we get a space complexity of O(N).\n",
-    "\n",
-    "In summary, the code has a time complexity of O(N) and a space complexity of O(N), where N is the number of nodes in the binary tree. These complexities are efficient and scale linearly with the size of the input tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Variant with Left Side View**: Modify the code to find and return the values of nodes visible from the left side of the binary tree instead of the right side.\n",
-    "\n",
-    "2. **Zigzag Right Side View**: Extend the code to return the right side view values in a zigzag order. In a zigzag order, you alternate between starting from the rightmost node at level 0, then the leftmost at level 1, rightmost at level 2, and so on."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb b/docs/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb
deleted file mode 100644
index 3e542c9..0000000
--- a/docs/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 226: Invert Binary Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the root of a binary tree, invert the tree, and return its root.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the tree is in the range `[0, 100]`.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def invertTree(root):\n",
-    "    # Base case: If the root is None or the tree is empty, return None.\n",
-    "    if not root:\n",
-    "        return None\n",
-    "\n",
-    "    # Swap the left and right subtrees of the current node.\n",
-    "    root.left, root.right = root.right, root.left\n",
-    "\n",
-    "    # Recursively invert the left and right subtrees.\n",
-    "    invertTree(root.left)\n",
-    "    invertTree(root.right)\n",
-    "\n",
-    "    return root"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a TreeNode class, which represents a node in a binary tree. Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-    "\n",
-    "2. The `invertTree` function is defined to invert the binary tree. It takes the root node of the tree as an argument.\n",
-    "\n",
-    "3. In the `invertTree` function, we have a base case to handle the scenario when the root is `None` (empty tree). In such cases, we return `None` because there's nothing to invert.\n",
-    "\n",
-    "4. For non-empty trees, we swap the left and right subtrees of the current node. This effectively inverts the tree at the current node.\n",
-    "\n",
-    "5. We then recursively call `invertTree` on the left and right subtrees to invert them.\n",
-    "\n",
-    "6. Finally, we return the root of the inverted tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Inverted Tree (Example 1): <__main__.TreeNode object at 0x7faa5c445ab0>\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "# Input: root = [4,2,7,1,3,6,9]\n",
-    "\n",
-    "root1 = TreeNode(4)\n",
-    "root1.left = TreeNode(2)\n",
-    "root1.right = TreeNode(7)\n",
-    "root1.left.left = TreeNode(1)\n",
-    "root1.left.right = TreeNode(3)\n",
-    "root1.right.left = TreeNode(6)\n",
-    "root1.right.right = TreeNode(9)\n",
-    "\n",
-    "inverted_root1 = invertTree(root1)\n",
-    "print(\"Inverted Tree (Example 1):\", inverted_root1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Inverted Tree (Example 2): <__main__.TreeNode object at 0x7faa5c444310>\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "# Input: root = [2,1,3]\n",
-    "\n",
-    "root2 = TreeNode(2)\n",
-    "root2.left = TreeNode(1)\n",
-    "root2.right = TreeNode(3)\n",
-    "\n",
-    "inverted_root2 = invertTree(root2)\n",
-    "print(\"Inverted Tree (Example 2):\", inverted_root2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Inverted Tree (Example 3): None\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "# Input: root = []\n",
-    "\n",
-    "root3 = None  # Empty tree\n",
-    "\n",
-    "inverted_root3 = invertTree(root3)\n",
-    "print(\"Inverted Tree (Example 3):\", inverted_root3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's discuss the time and space complexity of the provided Python code to invert a binary tree.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of the `invertTree` function is $O(n)$, where $n$ is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal of the tree. In the worst case, we have to visit all nodes in the tree.\n",
-    "\n",
-    "The reason for this time complexity is the depth-first traversal of the tree, where we recursively process the left and right subtrees of each node.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of the code is determined by the function call stack during the recursion. In the worst case, the space complexity is $O(h)$, where $h$ is the height of the binary tree. \n",
-    "\n",
-    "In a completely unbalanced binary tree (essentially a linked list), the height $h$ is equal to the number of nodes $n$, resulting in a space complexity of $O(n)$. This occurs when the tree is skewed to one side.\n",
-    "\n",
-    "In a balanced binary tree, such as a full binary tree, the height $h$ is $O(log\\ n)$, and the space complexity is $O(log\\ n)$.\n",
-    "\n",
-    "The space complexity depends on how balanced the tree is. In practical scenarios, binary trees are often approximately balanced, so the space complexity is typically closer to $O(log\\ n)$.\n",
-    "\n",
-    "**In summary:**\n",
-    "\n",
-    "- Time Complexity: $O(n)$ where $n$ is the number of nodes.\n",
-    "- Space Complexity: $O(h)$, where $h$ is the height of the binary tree. In the worst case, it can be $O(n)$, and in a balanced tree, it is $O(log\\ n)$.\n",
-    "\n",
-    "Keep in mind that these complexities are based on the recursive implementation provided. Iterative solutions can achieve the same task with $O(1)$ space complexity, using auxiliary data structures like stacks or queues to mimic the recursion stack."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb b/docs/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb
deleted file mode 100644
index fdad9ba..0000000
--- a/docs/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 230. Kth Smallest Element in a BST\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Kth Smallest Element in a BST problem on LeetCode, click here!](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary search tree, and an integer `k`, return *the $k^{th}$ smallest value (**1-indexed**) of all the values of the nodes in the tree.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is `n`.\n",
-    "- 1 <= `k <= n` <= $10^4$\n",
-    "- 0 <= `Node.val` <= $10^4$\n",
-    "\n",
-    "**Follow up**: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def kthSmallest(root, k):\n",
-    "    stack = []  # Initialize an empty stack to simulate the traversal\n",
-    "    while root or stack:\n",
-    "        while root:\n",
-    "            stack.append(root)  # Push the current node onto the stack\n",
-    "            root = root.left  # Move to the left child\n",
-    "        root = stack.pop()  # Pop a node from the stack\n",
-    "        k -= 1  # Decrement k since we've visited a node\n",
-    "        if k == 0:\n",
-    "            return root.val  # If k becomes 0, return the current node's value as the kth smallest\n",
-    "        root = root.right  # Move to the right child to continue the traversal"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class, which represents a node in the binary search tree (BST). Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-    "\n",
-    "2. The `kthSmallest` function takes two parameters: `root`, which is the root of the BST, and `k`, which represents the kth smallest element we want to find.\n",
-    "\n",
-    "3. We initialize an empty stack (`stack`) to simulate the traversal of the BST. The stack will be used to keep track of nodes as we traverse the tree in an iterative manner.\n",
-    "\n",
-    "4. We enter a while loop that continues until either `root` becomes `None` (indicating we have traversed the entire tree) or the `stack` is empty.\n",
-    "\n",
-    "5. Within the loop, we start another while loop to traverse as far left as possible in the BST. We repeatedly push nodes onto the `stack` and move to their left children until we reach the leftmost leaf node. This is the smallest node in the BST.\n",
-    "\n",
-    "6. Once we have reached the leftmost leaf node (the smallest node), we pop nodes from the `stack` one by one. As we pop each node, we decrement `k` by 1 to keep track of the number of nodes we have visited.\n",
-    "\n",
-    "7. If `k` becomes 0 after decrementing, it means we have found the kth smallest element. In this case, we return the `val` of the current node as the result.\n",
-    "\n",
-    "8. If `k` is still greater than 0, it means we haven't found the kth smallest element yet. In this case, we move to the right child of the current node to continue the traversal, as the kth smallest element, if it exists, will be in the right subtree of the current node.\n",
-    "\n",
-    "9. The process continues until we find the kth smallest element or traverse the entire tree.\n",
-    "\n",
-    "10. Finally, we return the kth smallest element found.\n",
-    "\n",
-    "The key idea in this code is to perform an in-order traversal of the BST while keeping track of the kth smallest element. By visiting nodes in ascending order, we can efficiently find the kth smallest element in O(h + k) time, where h is the height of the BST and k is the desired kth element. This approach is both concise and efficient for this problem."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "# Create the tree for the first example: root = [3,1,4,null,2], k = 1\n",
-    "root1 = TreeNode(3)\n",
-    "root1.left = TreeNode(1)\n",
-    "root1.right = TreeNode(4)\n",
-    "root1.left.right = TreeNode(2)\n",
-    "k1 = 1\n",
-    "print(kthSmallest(root1, k1)) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Create the tree for the second example: root = [5,3,6,2,4,null,null,1], k = 3\n",
-    "root2 = TreeNode(5)\n",
-    "root2.left = TreeNode(3)\n",
-    "root2.right = TreeNode(6)\n",
-    "root2.left.left = TreeNode(2)\n",
-    "root2.left.right = TreeNode(4)\n",
-    "root2.left.left.left = TreeNode(1)\n",
-    "k2 = 3\n",
-    "print(kthSmallest(root2, k2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the given code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this code is O(h + k), where:\n",
-    "- h is the height of the binary search tree (BST).\n",
-    "- k is the desired kth smallest element we want to find.\n",
-    "\n",
-    "1. In the worst case, where the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the time complexity is O(n + k).\n",
-    "\n",
-    "2. In the best case, where the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the time complexity is O(log(n) + k).\n",
-    "\n",
-    "So, the time complexity can vary from O(log(n) + k) in the best-case scenario to O(n + k) in the worst-case scenario. Typically, for balanced BSTs, the time complexity is closer to O(log(n) + k).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this code is O(h) due to the stack used for the iterative in-order traversal, where h is the height of the BST.\n",
-    "\n",
-    "1. In the worst case, when the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the space complexity is O(n) because the stack can potentially store all n nodes.\n",
-    "\n",
-    "2. In the best case, when the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the space complexity is O(log(n)) because the stack will have at most log(n) nodes.\n",
-    "\n",
-    "So, the space complexity depends on the height of the BST and can vary from O(log(n)) in the best-case scenario to O(n) in the worst-case scenario.\n",
-    "\n",
-    "In practical terms, for balanced BSTs or moderately unbalanced BSTs, the space complexity is usually close to O(log(n)), and the code is efficient in terms of space usage."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Kth Largest Element**: Modify the code to find the kth largest element in the BST instead of the kth smallest element.\n",
-    "\n",
-    "2. **Kth Smallest Element in Two BSTs**: Given two BSTs, find the kth smallest element when considering both BSTs as a single sorted list. This involves merging the two BSTs while finding the kth element efficiently."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb b/docs/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb
deleted file mode 100644
index 0179ca4..0000000
--- a/docs/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,237 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 235. Lowest Common Ancestor of a Binary Search Tree\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Lowest Common Ancestor of a Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)\n",
-    "\n",
-    "---\n",
-    "Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.\n",
-    "\n",
-    "According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[2, 10^5]$.\n",
-    "- $-10^9$ <= `Node.val` <= $10^9$\n",
-    "- All `Node.val` are **unique**.\n",
-    "- `p != q`\n",
-    "- `p` and `q` will exist in the BST.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def lowestCommonAncestor(root, p, q):\n",
-    "    # Ensure p is less than q\n",
-    "    if p.val > q.val:\n",
-    "        p, q = q, p\n",
-    "\n",
-    "    while root:\n",
-    "        # If the current node value is greater than both p and q, move left\n",
-    "        if root.val > q.val:\n",
-    "            root = root.left\n",
-    "        # If the current node value is less than both p and q, move right\n",
-    "        elif root.val < p.val:\n",
-    "            root = root.right\n",
-    "        # If the current node value is between p and q (inclusive), or it matches either p or q, it's the LCA\n",
-    "        else:\n",
-    "            return root"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `class TreeNode:` defines a simple class for representing nodes in a binary tree. Each `TreeNode` has three attributes:\n",
-    "   - `val`: The value of the node.\n",
-    "   - `left`: A reference to the left child node.\n",
-    "   - `right`: A reference to the right child node.\n",
-    "\n",
-    "2. `def lowestCommonAncestor(root, p, q):` is a function that takes three arguments:\n",
-    "   - `root`: The root node of the BST.\n",
-    "   - `p`: A TreeNode representing one of the nodes for which we want to find the LCA.\n",
-    "   - `q`: A TreeNode representing the other node for which we want to find the LCA.\n",
-    "\n",
-    "3. `if p.val > q.val:` checks if the value of `p` is greater than the value of `q`. In a BST, it's essential to ensure that `p` represents the smaller value, and `q` represents the larger value. If `p` is greater than `q`, the code swaps their values.\n",
-    "\n",
-    "4. The main logic is inside the `while` loop, which runs until `root` becomes `None`. It performs the following steps to find the LCA:\n",
-    "   - If the current `root` node's value is greater than the value of `q`, it means both `p` and `q` are on the left subtree of the current node. So, we move to the left child of the current node by setting `root = root.left`.\n",
-    "   - If the current `root` node's value is less than the value of `p`, it means both `p` and `q` are on the right subtree of the current node. So, we move to the right child of the current node by setting `root = root.right`.\n",
-    "   - If neither of the above conditions is met, it means the current `root` node's value is between `p` and `q`, or it matches either `p` or `q`. In this case, the current node is the lowest common ancestor (LCA), so we return `root`.\n",
-    "\n",
-    "The algorithm is efficient because it takes advantage of the properties of a BST. It eliminates subtrees that cannot contain the LCA by comparing the values of `p`, `q`, and the current `root` node. Eventually, it reaches the LCA node, and that node is returned as the result."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "6\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Construct the tree\n",
-    "root = TreeNode(6)\n",
-    "root.left = TreeNode(2)\n",
-    "root.right = TreeNode(8)\n",
-    "root.left.left = TreeNode(0)\n",
-    "root.left.right = TreeNode(4)\n",
-    "root.right.left = TreeNode(7)\n",
-    "root.right.right = TreeNode(9)\n",
-    "root.left.right.left = TreeNode(3)\n",
-    "root.left.right.right = TreeNode(5)\n",
-    "\n",
-    "p = root.left  # Node 2\n",
-    "q = root.right  # Node 8\n",
-    "\n",
-    "result = lowestCommonAncestor(root, p, q)\n",
-    "print(result.val)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Using the same tree as before\n",
-    "p = root.left  # Node 2\n",
-    "q = root.left.right  # Node 4\n",
-    "\n",
-    "result = lowestCommonAncestor(root, p, q)\n",
-    "print(result.val) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# Creating a new tree for this example\n",
-    "root2 = TreeNode(2)\n",
-    "root2.left = TreeNode(1)\n",
-    "\n",
-    "p = root2  # Node 2\n",
-    "q = root2.left  # Node 1\n",
-    "\n",
-    "result = lowestCommonAncestor(root2, p, q)\n",
-    "print(result.val)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided algorithm for finding the lowest common ancestor (LCA) in a Binary Search Tree (BST).\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of this algorithm is O(h), where \"h\" is the height of the BST. In the worst case, where the tree is completely unbalanced (essentially a linked list), the height of the tree is O(n), where \"n\" is the number of nodes in the tree. However, in a well-balanced BST, the height is logarithmic, which is O(log n).\n",
-    "\n",
-    "The reason for this time complexity is that the algorithm efficiently narrows down the search space by traversing either left or right subtrees based on the values of the target nodes `p` and `q`. It eliminates entire subtrees that cannot contain the LCA, leading to a relatively quick search.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity of the algorithm is O(1). This is because it uses a constant amount of extra space, regardless of the size of the input BST. The only variables used are `root`, `p`, and `q`, and there are no data structures like stacks or queues used for additional space. The algorithm performs a simple traversal without recursion, so it does not consume extra memory as the tree depth increases.\n",
-    "\n",
-    "**In summary**, the provided algorithm for finding the LCA in a BST is both time and space-efficient. Its time complexity is O(h), where \"h\" is the height of the tree, and its space complexity is O(1), making it suitable for practical use even in large BSTs."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "### Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple LCAs:** Extend the algorithm to find all the lowest common ancestors of two given nodes `p` and `q` in a BST. In some cases, there can be multiple LCAs.\n",
-    "\n",
-    "2. **LCA with k Nodes:** Given a BST and k nodes, find the lowest common ancestor of these k nodes. This is an extension of the problem where you must find the LCA of more than two nodes."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb b/docs/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb
deleted file mode 100644
index 78a78b5..0000000
--- a/docs/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb	
+++ /dev/null
@@ -1,291 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 297. Serialize and Deserialize Binary Tree\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Serialize and Deserialize Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)\n",
-    "\n",
-    "---\n",
-    "Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.\n",
-    "\n",
-    "Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.\n",
-    "\n",
-    "**Clarification:** The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-    "- `-1000 <= Node.val <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "TThe problem is to design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of characters or bytes so that it can be easily stored in a file, transmitted over a network, or otherwise persisted. Deserialization is the reverse process of converting the serialized data back into the original data structure.\n",
-    "\n",
-    "In the context of this problem:\n",
-    "\n",
-    "1. **Serialization**: You are given a binary tree, and your task is to convert it into a string representation such that you can later recreate the same binary tree from this string. The format of serialization is flexible, but it should allow you to reconstruct the original binary tree accurately.\n",
-    "\n",
-    "2. **Deserialization**: Given a serialized string, you need to reconstruct the binary tree it represents, ensuring that it is identical to the original tree.\n",
-    "\n",
-    "Here are some key points to consider in solving this problem:\n",
-    "\n",
-    "- The input can include any valid binary tree, including trees with nodes having integer values within the range [-1000, 1000].\n",
-    "- You don't necessarily need to follow a specific format for serialization, but it should be designed in a way that allows unambiguous deserialization.\n",
-    "- The goal is to serialize the tree structure and its values and then deserialize it back into the same structure and values.\n",
-    "\n",
-    "A common approach for serialization is to use a traversal method like preorder traversal, where you visit nodes in the order: root, left subtree, right subtree. This way, you can serialize the tree into a string, and during deserialization, you can reconstruct the tree by parsing the string in the same order.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode(object):\n",
-    "    def __init__(self, x):\n",
-    "        self.val = x\n",
-    "        self.left = None\n",
-    "        self.right = None\n",
-    "\n",
-    "class Codec:\n",
-    "    def serialize(self, root):\n",
-    "        serialized  = []\n",
-    "\n",
-    "        def dfs(node):\n",
-    "            if not node:\n",
-    "                # If the node is None, represent it as \"Null\" in the serialized string\n",
-    "                serialized .append(\"Null\")\n",
-    "                return\n",
-    "            # Convert the node's value to a string and add it to the serialized string\n",
-    "            serialized .append(str(node.val))\n",
-    "            dfs(node.left)\n",
-    "            dfs(node.right)\n",
-    "\n",
-    "        dfs(root)\n",
-    "        # Join the serialized values with \",\"\n",
-    "        return \",\".join(serialized )\n",
-    "\n",
-    "    def deserialize(self, data):\n",
-    "        vals = data.split(\",\")\n",
-    "        self.i = 0\n",
-    "\n",
-    "        def dfs():\n",
-    "            if vals[self.i] == \"Null\":\n",
-    "                self.i += 1\n",
-    "                return None\n",
-    "            # Convert the value to an integer to create a new node\n",
-    "            node = TreeNode(int(vals[self.i]))\n",
-    "            self.i += 1\n",
-    "            node.left = dfs()\n",
-    "            node.right = dfs()\n",
-    "            return node\n",
-    "\n",
-    "        return dfs()"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Here's a step-by-step explanation of the code:\n",
-    "\n",
-    "1. **Definition of TreeNode**:\n",
-    "   - The code defines a simple `TreeNode` class that represents a node in a binary tree. Each `TreeNode` has three attributes:\n",
-    "     - `val`: The integer value stored in the node.\n",
-    "     - `left`: A reference to the left child node (or None if there is no left child).\n",
-    "     - `right`: A reference to the right child node (or None if there is no right child).\n",
-    "   - This class allows you to create binary tree nodes with integer values and left and right child references.\n",
-    "\n",
-    "2. **Serialization (serialize method)**:\n",
-    "   - The `serialize` method takes a `root` node as input, which is the root of the binary tree that needs to be serialized.\n",
-    "   - It initializes an empty list `serialized` to store the serialized elements.\n",
-    "   - The `dfs` (depth-first search) function is defined within the `serialize` method to perform a preorder traversal of the binary tree.\n",
-    "   - In the `dfs` function:\n",
-    "     - If the current `node` is `None` (i.e., a leaf node or a child of a leaf node), it appends the string \"Null\" to the `serialized` list to represent the absence of a node.\n",
-    "     - If the current `node` is not `None`, it appends the string representation of the `node.val` to the `serialized` list and then recursively calls `dfs` on the left and right children.\n",
-    "   - After the `dfs` traversal, the `serialized` list contains the serialized binary tree elements.\n",
-    "   - The method returns a string obtained by joining the elements in the `serialized` list with commas.\n",
-    "\n",
-    "3. **Deserialization (deserialize method)**:\n",
-    "   - The `deserialize` method takes a serialized string `data` as input, which represents a binary tree in the custom format.\n",
-    "   - It splits the `data` string by commas to obtain a list of elements called `vals`.\n",
-    "   - The method initializes an index `self.i` to 0. This index keeps track of the current position in the `vals` list during deserialization.\n",
-    "   - The `dfs` function is defined within the `deserialize` method to perform the deserialization process using a recursive approach.\n",
-    "   - In the `dfs` function:\n",
-    "     - If the current element in `vals` is \"Null,\" it means there is no node at this position in the binary tree, so it returns `None`.\n",
-    "     - If the current element is not \"Null,\" it converts the element to an integer to create a new `TreeNode` with that value.\n",
-    "     - It then recursively calls `dfs` to set the left and right children of the current node.\n",
-    "   - The method returns the root node of the reconstructed binary tree.\n",
-    "\n",
-    "Overall, this code demonstrates a way to serialize a binary tree into a string format and then deserialize it back into the original tree structure, allowing for the representation of both the tree structure and the values stored in the nodes."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1,2,Null,Null,3,4,Null,Null,5,Null,Null\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "# Serialize a binary tree\n",
-    "root = TreeNode(1)\n",
-    "root.left = TreeNode(2)\n",
-    "root.right = TreeNode(3)\n",
-    "root.right.left = TreeNode(4)\n",
-    "root.right.right = TreeNode(5)\n",
-    "\n",
-    "codec = Codec()\n",
-    "serialized_tree = codec.serialize(root)\n",
-    "print(serialized_tree)\n",
-    "\n",
-    "# Deserialize the serialized tree\n",
-    "new_root = codec.deserialize(serialized_tree)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Null\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Example usage:\n",
-    "# Serialize a binary tree\n",
-    "root = None\n",
-    "\n",
-    "codec = Codec()\n",
-    "serialized_tree = codec.serialize(root)\n",
-    "print(serialized_tree)\n",
-    "\n",
-    "# Deserialize the serialized tree\n",
-    "new_root = codec.deserialize(serialized_tree)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for serializing and deserializing a binary tree:\n",
-    "\n",
-    "1. **Serialization (serialize method)**:\n",
-    "\n",
-    "   - Time Complexity:\n",
-    "     - The `serialize` method uses a depth-first traversal (preorder) of the binary tree.\n",
-    "     - It visits each node exactly once.\n",
-    "     - At each node, it performs constant-time operations (string conversion and list append).\n",
-    "     - Therefore, the time complexity for serialization is O(N), where N is the number of nodes in the binary tree.\n",
-    "\n",
-    "   - Space Complexity:\n",
-    "     - The space complexity for serialization includes the space used by the `serialized` list and the recursive call stack.\n",
-    "     - The `serialized` list stores the serialized elements, and its space is proportional to the number of nodes.\n",
-    "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-    "     - Therefore, the space complexity for serialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-    "\n",
-    "2. **Deserialization (deserialize method)**:\n",
-    "\n",
-    "   - Time Complexity:\n",
-    "     - The `deserialize` method splits the serialized string into a list of elements, which takes O(N) time.\n",
-    "     - The `dfs` function is a recursive function that visits each element in the list exactly once.\n",
-    "     - At each element, it performs constant-time operations (string comparison, integer conversion, and recursive function calls).\n",
-    "     - Therefore, the time complexity for deserialization is O(N).\n",
-    "\n",
-    "   - Space Complexity:\n",
-    "     - The space complexity for deserialization includes the space used by the `vals` list and the recursive call stack.\n",
-    "     - The `vals` list stores the elements from the serialized string, and its space is proportional to the number of nodes.\n",
-    "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-    "     - Therefore, the space complexity for deserialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-    "\n",
-    "Overall, the provided code has a time complexity of O(N) for both serialization and deserialization and a space complexity of O(N) in the majority of cases (unless the tree is severely unbalanced, in which case the height H dominates the space complexity). It efficiently serializes and deserializes a binary tree while using a linear amount of memory."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Efficient Serialization**:\n",
-    "   Optimize the serialization process to minimize the length of the serialized string. Consider using techniques like binary encoding to represent node values more efficiently.\n",
-    "\n",
-    "2. **Custom Serialization Format**:\n",
-    "   Modify the serialization and deserialization methods to use a custom format different from the one provided. Ensure that the new format allows for accurate reconstruction of the original binary tree."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/572. Subtree of Another Tree.ipynb b/docs/_build/jupyter_execute/07. Trees/572. Subtree of Another Tree.ipynb
deleted file mode 100644
index 425f22d..0000000
--- a/docs/_build/jupyter_execute/07. Trees/572. Subtree of Another Tree.ipynb	
+++ /dev/null
@@ -1,259 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 572: Subtree of Another Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Subtree of Another Tree problem on LeetCode, click here!](https://leetcode.com/problems/subtree-of-another-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "**Problem Description:**\n",
-    "\n",
-    "Given the roots of two binary trees `root` and `subRoot`, return `True` if there is a subtree of `root` with the same structure and node values as `subRoot`, and `False` otherwise.\n",
-    "\n",
-    "A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "1. The number of nodes in the `root` tree is in the range $[1, 2000]$.\n",
-    "2. The number of nodes in the `subRoot` tree is in the range $[1, 1000]$.\n",
-    "3. $-10^4$ <= root.val <= $10^4$\n",
-    "4. $-10^4$ <= subRoot.val <= $10^4$\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "class Solution:\n",
-    "    def isSubtree(self, root, subRoot):\n",
-    "        if not root:\n",
-    "            return False\n",
-    "        \n",
-    "        # Check if the current subtree is equal to the subRoot\n",
-    "        if self.isSameTree(root, subRoot):\n",
-    "            return True\n",
-    "        \n",
-    "        # Recursively check the left and right subtrees\n",
-    "        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)\n",
-    "    \n",
-    "    def isSameTree(self, tree1, tree2):\n",
-    "        if not tree1 and not tree2:\n",
-    "            return True\n",
-    "        if not tree1 or not tree2:\n",
-    "            return False\n",
-    "        \n",
-    "        return (\n",
-    "            tree1.val == tree2.val and\n",
-    "            self.isSameTree(tree1.left, tree2.left) and\n",
-    "            self.isSameTree(tree1.right, tree2.right)\n",
-    "        )"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "This code defines a TreeNode class for the binary tree nodes and a Solution class with two methods:\n",
-    "\n",
-    "1. `isSubtree`: This method checks if there is a subtree of the `root` tree with the same structure and node values as the `subRoot` tree. It uses a helper function `isSameTree` to compare two trees for equality.\n",
-    "\n",
-    "2. `isSameTree`: This helper method recursively compares two trees to check if they are the same in structure and node values.\n",
-    "\n",
-    "### Here's a detailed explanation of the code:\n",
-    "\n",
-    "1. **TreeNode Class:**\n",
-    "\n",
-    "The `TreeNode` class is defined to represent nodes in a binary tree. Each node has a `val` (the node's value) and may have a `left` and `right` child.\n",
-    "\n",
-    "2. **Solution Class:**\n",
-    "\n",
-    "    + The `Solution` class contains the solution for the problem and defines two important methods:\n",
-    "\n",
-    "    + `isSubtree(self, root, subRoot)`:\n",
-    "\n",
-    "        - This method checks whether `subRoot` is a subtree of `root`. It takes two tree nodes, `root` and `subRoot`, as input arguments.\n",
-    "        - If `root` is `None`, it returns `False` because there is no subtree to search.\n",
-    "        - It then checks if the current subtree with `root` as its root is equal to `subRoot` using the `isSameTree` method. If they are the same, it returns `True`.\n",
-    "        - If the current subtree is not the same as `subRoot`, it recursively checks the left and right subtrees of `root` to see if `subRoot` is a subtree of any of them.\n",
-    "        - It returns `True` if `subRoot` is found in either the left or right subtree; otherwise, it returns `False`.\n",
-    "\n",
-    "    + `isSameTree(self, tree1, tree2)`:\n",
-    "\n",
-    "        - This method checks whether two trees, `tree1` and `tree2`, are the same.\n",
-    "        - If both `tree1` and `tree2` are `None`, they are considered the same tree, so it returns `True`.\n",
-    "        - If only one of them is `None` (but not both), they are different trees, so it returns `False`.\n",
-    "        - If both `tree1` and `tree2` have values, it checks if their values are equal and recursively checks if their left and right subtrees are the same.\n",
-    "        - It returns `True` if the trees are the same; otherwise, it returns `False`.\n",
-    "\n",
-    "The code effectively uses recursion to traverse the binary trees and check for subtree equality. The `isSubtree` method starts the recursive search, and the `isSameTree` method is used to compare individual subtrees. The approach is efficient and avoids unnecessary checks when possible.\n",
-    "\n",
-    "This solution allows you to determine if there exists a subtree within the `root` tree that matches the structure and node values of `subRoot`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Is subRoot a subtree of root? True\n"
-     ]
-    }
-   ],
-   "source": [
-    "#root = [3,4,5,1,2]\n",
-    "#subRoot = [4,1,2]\n",
-    "\n",
-    "# Example input\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(4)\n",
-    "root.right = TreeNode(5)\n",
-    "root.left.left = TreeNode(1)\n",
-    "root.left.right = TreeNode(2)\n",
-    "\n",
-    "subRoot = TreeNode(4)\n",
-    "subRoot.left = TreeNode(1)\n",
-    "subRoot.right = TreeNode(2)\n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Test the isSubtree method\n",
-    "result = solution.isSubtree(root, subRoot)\n",
-    "print(\"Is subRoot a subtree of root?\", result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Is subRoot a subtree of root? False\n"
-     ]
-    }
-   ],
-   "source": [
-    "#root = [3,4,5,1,2,null,null,null,null,0] \n",
-    "#subRoot = [4,1,2]\n",
-    "\n",
-    "\n",
-    "# Example input\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(4)\n",
-    "root.right = TreeNode(5)\n",
-    "root.left.left = TreeNode(1)\n",
-    "root.left.right = TreeNode(2)\n",
-    "root.left.right.left = TreeNode(0)\n",
-    "\n",
-    "subRoot = TreeNode(4)\n",
-    "subRoot.left = TreeNode(1)\n",
-    "subRoot.right = TreeNode(2)\n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Test the isSubtree method\n",
-    "result = solution.isSubtree(root, subRoot)\n",
-    "print(\"Is subRoot a subtree of root?\", result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "86a34f0d-0789-498c-818c-2d30f84754d0",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code for the \"Subtree of Another Tree\" problem:\n",
-    "\n",
-    "**Time Complexity**\n",
-    "\n",
-    "The time complexity of the code primarily depends on the recursive traversal of the binary tree. In both the `isSubtree` and `isSameTree` functions, we visit each node in the binary trees once. Let's break down the time complexity:\n",
-    "\n",
-    "1. The `isSubtree` function:\n",
-    "   - In the worst case, we need to visit every node in the `root` tree.\n",
-    "   - For each node in the `root` tree, we call the `isSameTree` function, which has its own traversal.\n",
-    "   - So, the total time complexity is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree.\n",
-    "\n",
-    "2. The `isSameTree` function:\n",
-    "   - In the worst case, we visit every node in both `tree1` and `tree2`.\n",
-    "   - The number of recursive calls made is proportional to the number of nodes in the trees.\n",
-    "   - So, the time complexity of this function is $O(max(n, m))$, where $n$ and $m$ are the numbers of nodes in `tree1` and `tree2`, respectively.\n",
-    "\n",
-    "Overall, the time complexity of the entire code is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree. In practice, it may be less than $O(n * m)$ if a subtree mismatch is detected early during the traversal.\n",
-    "\n",
-    "**Space Complexity**\n",
-    "\n",
-    "The space complexity of the code is determined by the function call stack during recursion and the space used by the recursive functions. Let's analyze the space complexity:\n",
-    "\n",
-    "1. The `isSubtree` function:\n",
-    "   - It uses the call stack for recursion.\n",
-    "   - The maximum depth of the recursion is equal to the height of the `root` tree, which can be $O(n)$ in the worst case (unbalanced tree).\n",
-    "   - Additionally, the function doesn't use any significant extra space other than the recursion stack.\n",
-    "\n",
-    "2. The `isSameTree` function:\n",
-    "   - It also uses the call stack for recursion.\n",
-    "   - The maximum depth of the recursion is equal to the height of the `tree1` or `tree2`, whichever is greater.\n",
-    "   - So, the maximum space used for the call stack is $O(max(n, m))$.\n",
-    "\n",
-    "In summary, the space complexity of the code is $O(max(n, m))$ due to the function call stack. It scales with the maximum height of the trees being compared.\n",
-    "\n",
-    "Overall, the code is efficient and works well for trees with moderate sizes. However, it's important to keep in mind that the worst-case time complexity is $O(n * m)$, so for very large trees, the performance may degrade."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb b/docs/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb
deleted file mode 100644
index 0cbe2fc..0000000
--- a/docs/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 98. Validate Binary Search Tree\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Validate Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/validate-binary-search-tree/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary tree, *determine if it is a valid binary search tree (BST)*.\n",
-    "\n",
-    "A **valid BST** is defined as follows:\n",
-    "\n",
-    "- The left subtree of a node contains only nodes with keys **less than** the node's key.\n",
-    "- The right subtree of a node contains only nodes with keys **greater than** the node's key.\n",
-    "- Both the left and right subtrees must also be binary search trees. \n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[1, 10^4]$.\n",
-    "- $-2^{31}$ <= `Node.val` <= $2^{31} - 1$\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def is_valid_BST(root):\n",
-    "    def is_valid(node, min_val, max_val):\n",
-    "        # Base case: If the node is None, it's a valid BST.\n",
-    "        if node is None:\n",
-    "            return True\n",
-    "        \n",
-    "        # Check if the current node's value is within the valid range.\n",
-    "        if not (min_val < node.val < max_val):\n",
-    "            return False\n",
-    "        \n",
-    "        # Recursively check the left and right subtrees with updated ranges.\n",
-    "        return (is_valid(node.left, min_val, node.val) and\n",
-    "                is_valid(node.right, node.val, max_val))\n",
-    "    \n",
-    "    # Call the helper function starting with a wide range for root node.\n",
-    "    return is_valid(root, float('-inf'), float('inf'))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "- It defines a `TreeNode` class to represent nodes in the binary tree.\n",
-    "\n",
-    "- The `is_valid_BST` function takes the root node of the binary tree as input and returns `True` if the tree is a valid BST, and `False` otherwise.\n",
-    "\n",
-    "- Inside the `is_valid_BST` function, there is a helper function called `is_valid` that performs the actual validation using a recursive approach.\n",
-    "\n",
-    "- The `is_valid` function checks each node in the tree to ensure that it satisfies the properties of a BST:\n",
-    "  - The value of the current node must be within a valid range defined by `min_val` and `max_val`.\n",
-    "  - The left subtree of the current node should contain values less than the current node's value.\n",
-    "  - The right subtree of the current node should contain values greater than the current node's value.\n",
-    "\n",
-    "- If any of these conditions are violated, the function returns `False`, indicating that the tree is not a valid BST.\n",
-    "\n",
-    "- If all nodes satisfy these conditions, the function returns `True`, indicating that the tree is a valid BST.\n",
-    "\n",
-    "- The code calls the `is_valid` function with the root node and initial range values of negative infinity to positive infinity to start the validation.\n",
-    "\n",
-    "- The time complexity of the code is O(N), where N is the number of nodes in the tree, as it traverses each node once.\n",
-    "\n",
-    "- The space complexity depends on the height of the tree. In the average case for a balanced BST, the space complexity is O(log N), but in the worst case (skewed tree), it can be O(N) due to the recursive call stack.\n",
-    "\n",
-    "In summary, this code checks whether a given binary tree is a valid BST by recursively validating each node's value and its left and right subtrees while maintaining valid value ranges. If all nodes satisfy the BST properties, the tree is considered a valid BST."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "# Create the tree for the first example: [2, 1, 3]\n",
-    "root1 = TreeNode(2)\n",
-    "root1.left = TreeNode(1)\n",
-    "root1.right = TreeNode(3)\n",
-    "print(is_valid_BST(root1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Create the tree for the second example: [5, 1, 4, None, None, 3, 6]\n",
-    "root2 = TreeNode(5)\n",
-    "root2.left = TreeNode(1)\n",
-    "root2.right = TreeNode(4)\n",
-    "root2.right.left = TreeNode(3)\n",
-    "root2.right.right = TreeNode(6)\n",
-    "print(is_valid_BST(root2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `is_valid_BST` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The time complexity of the function primarily depends on the number of nodes in the binary tree.\n",
-    "- In the worst case, we may have to visit every node in the tree once to validate whether it's part of a valid BST.\n",
-    "- Since we are performing a depth-first search (DFS) traversal of the tree, the time complexity is O(N), where N is the number of nodes in the tree.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity is determined by the space used by the recursive call stack during the DFS traversal.\n",
-    "- In the worst case, if the tree is completely unbalanced (a skewed tree), the maximum depth of the call stack would be equal to the height of the tree.\n",
-    "- In a balanced BST, the height is approximately log2(N), where N is the number of nodes.\n",
-    "- Therefore, the space complexity of the call stack is O(log N) for a balanced BST.\n",
-    "- In the worst case (skewed tree), the space complexity can be O(N) as the height of the tree can be equal to N.\n",
-    "\n",
-    "**Overall:**\n",
-    "- Time Complexity: O(N)\n",
-    "- Space Complexity: O(log N) on average for a balanced BST, and O(N) in the worst case for a skewed tree.\n",
-    "\n",
-    "The space complexity is typically dominated by the recursive call stack, and it varies depending on the shape of the binary tree. In practice, for balanced binary trees, the space complexity is often close to O(log N), which is quite efficient."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Complex Constraints**: Modify the problem to have more complex constraints on the tree. For example, consider a binary tree with constraints like \"The left subtree of a node contains only nodes with keys less than or equal to the node's key,\" and \"The right subtree of a node contains only nodes with keys greater than or equal to the node's key.\" How would you adapt the code to handle these constraints?\n",
-    "\n",
-    "2. **Handling Duplicates**: Modify the code to handle binary search trees that allow duplicate values. In a standard BST, each value is unique, but in this case, multiple nodes can have the same value. The tree should still be considered valid if it follows the BST property."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb b/docs/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb
deleted file mode 100644
index 5c54c25..0000000
--- a/docs/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 208. Implement Trie (Prefix Tree)\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Implement Trie problem on LeetCode, click here!](https://leetcode.com/problems/implement-trie-prefix-tree/)\n",
-    "\n",
-    "---\n",
-    "A **trie** (pronounced as \"try\") or **prefix tree** is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.\n",
-    "\n",
-    "Implement the Trie class:\n",
-    "\n",
-    "- `Trie()` Initializes the trie object.\n",
-    "- `void insert(String word)` Inserts the string `word` into the trie.\n",
-    "- `boolean search(String word)` Returns `true` if the string `word` is in the trie (i.e., was inserted before), and `false` otherwise.\n",
-    "- `boolean startsWith(String prefix)` Returns `true` if there is a previously inserted string `word` that has the prefix `prefix`, and `false` otherwise.\n",
-    "\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= word.length, prefix.length <= 2000`\n",
-    "- `word` and `prefix` consist only of lowercase English letters.\n",
-    "- At most $3 * 10^4$ calls in total will be made to `insert`, `search`, and `startsWith`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand is to implement a data structure known as a Trie (pronounced \"try\") or Prefix Tree. A Trie is a tree-like data structure that is used to efficiently store and retrieve a set of strings or words. It's particularly useful for tasks involving string manipulation, such as autocomplete and spell-checking. The problem statement defines the following operations for implementing the Trie class:\n",
-    "\n",
-    "1. `Trie()`: This is the constructor that initializes the Trie object.\n",
-    "\n",
-    "2. `void insert(String word)`: This method inserts the given string `word` into the Trie. It effectively adds the characters of the word to the Trie's structure, creating nodes for each character if they don't already exist. At the end of the word, a special flag is set to indicate that this node represents the end of a valid word.\n",
-    "\n",
-    "3. `boolean search(String word)`: This method checks if the given string `word` is present in the Trie. It starts from the root of the Trie and traverses the Trie by following the characters in the word. If it successfully traverses the Trie and reaches the end of the word, it returns `true`, indicating that the word exists in the Trie; otherwise, it returns `false`.\n",
-    "\n",
-    "4. `boolean startsWith(String prefix)`: This method checks if there is any previously inserted word in the Trie that has the given `prefix`. It's similar to the `search` method but does not require that the prefix be a complete word. If there is any word in the Trie that starts with the given prefix, it returns `true`; otherwise, it returns `false`.\n",
-    "\n",
-    "The problem also provides an example scenario where these operations are called, demonstrating the expected output for each operation.\n",
-    "\n",
-    "In summary, the goal is to create a data structure (Trie) that efficiently stores a set of strings and provides methods to insert new strings, search for complete words, and check if a given prefix exists in the stored set of strings.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TrieNode:\n",
-    "    def __init__(self):\n",
-    "        self.children = {}           # A dictionary to store child nodes\n",
-    "        self.is_end_of_word = False  # Indicates if a word ends at this node\n",
-    "\n",
-    "class Trie:\n",
-    "    def __init__(self):\n",
-    "        self.root = TrieNode()       # Initialize the Trie with a root node\n",
-    "\n",
-    "    def insert(self, word):\n",
-    "        node = self.root             # Start from the root\n",
-    "        for char in word:\n",
-    "            if char not in node.children:\n",
-    "                node.children[char] = TrieNode()  # Create a new node if the character is not present\n",
-    "            node = node.children[char]           # Move to the child node\n",
-    "        node.is_end_of_word = True                # Mark the end of the inserted word\n",
-    "\n",
-    "    def search(self, word):\n",
-    "        node = self.root             # Start from the root\n",
-    "        for char in word:\n",
-    "            if char not in node.children:\n",
-    "                return False         # If the character is not found, the word is not in the Trie\n",
-    "            node = node.children[char]  # Move to the child node\n",
-    "        return node.is_end_of_word    # Check if the node represents the end of a valid word\n",
-    "\n",
-    "    def startsWith(self, prefix):\n",
-    "        node = self.root             # Start from the root\n",
-    "        for char in prefix:\n",
-    "            if char not in node.children:\n",
-    "                return False         # If the character is not found, no word starts with the prefix\n",
-    "            node = node.children[char]  # Move to the child node\n",
-    "        return True                  # Prefix found in the Trie"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Let's explain how the code works step by step:\n",
-    "\n",
-    "1. `TrieNode` Class:\n",
-    "   - `TrieNode` is a class that represents nodes in the Trie. Each node has two attributes:\n",
-    "     - `children`: This is a dictionary that stores child nodes. Each key in the dictionary represents a character, and the corresponding value is the child node for that character.\n",
-    "     - `is_end_of_word`: This boolean flag indicates whether the node represents the end of a complete word. Initially, it's set to `False` for all nodes.\n",
-    "\n",
-    "2. `Trie` Class:\n",
-    "   - `Trie` is the main class that implements the Trie data structure. It has the following methods:\n",
-    "\n",
-    "   - `__init__(self)`: The constructor initializes the Trie object by creating a root node, which serves as the starting point for all Trie operations.\n",
-    "\n",
-    "   - `insert(self, word)`: This method inserts a string `word` into the Trie. It starts from the root node and iterates through each character in the word.\n",
-    "     - If a character is not present as a child node, it creates a new node for that character.\n",
-    "     - It then moves to the child node and continues the process until the entire word is inserted.\n",
-    "     - Finally, it sets the `is_end_of_word` flag to `True` for the last node to mark the end of the inserted word.\n",
-    "\n",
-    "   - `search(self, word)`: This method checks if a complete word (string) exists in the Trie. It starts from the root node and iterates through each character in the word.\n",
-    "     - If a character is not found as a child node, it immediately returns `False` because the word is not in the Trie.\n",
-    "     - It continues to move to the child node for each character.\n",
-    "     - After reaching the end of the word, it checks if the `is_end_of_word` flag is `True` for the last node to confirm the presence of the word.\n",
-    "\n",
-    "   - `startsWith(self, prefix)`: This method checks if there is any previously inserted word in the Trie that starts with a given `prefix`. It follows the same logic as the `search` method but does not require the entire word to be present.\n",
-    "     - If the prefix is found in the Trie, it returns `True`; otherwise, it returns `False`.\n",
-    "\n",
-    "3. Example Usage:\n",
-    "   - The code demonstrates how to create a Trie object, insert words (\"apple\" and \"app\"), and perform operations on the Trie:\n",
-    "     - It inserts the word \"apple\" into the Trie.\n",
-    "     - It checks if \"apple\" is in the Trie, which returns `True`.\n",
-    "     - It checks if \"app\" is in the Trie, which returns `False`.\n",
-    "     - It checks if there is any word in the Trie that starts with \"app,\" which returns `True`.\n",
-    "     - It inserts the word \"app\" into the Trie.\n",
-    "     - It checks if \"app\" is in the Trie again, which now returns `True`.\n",
-    "\n",
-    "In summary, the code efficiently implements a Trie data structure with the ability to insert words, search for complete words, and check for the existence of words with a given prefix. The Trie is organized as a tree of nodes, with each node representing a character in the words being stored."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n",
-      "False\n",
-      "True\n",
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "\n",
-    "trie = Trie()\n",
-    "trie.insert(\"apple\")\n",
-    "print(trie.search(\"apple\"))   # Output: True\n",
-    "print(trie.search(\"app\"))     # Output: False\n",
-    "print(trie.startsWith(\"app\")) # Output: True\n",
-    "trie.insert(\"app\")\n",
-    "print(trie.search(\"app\"))     # Output: True"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the Trie implementation:\n",
-    "\n",
-    "1. Time Complexity:\n",
-    "   - Insertion (`insert` method): The time complexity of inserting a word into the Trie is O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to insert it into the Trie.\n",
-    "   - Search (`search` method): The time complexity of searching for a word in the Trie is also O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to determine if it exists in the Trie.\n",
-    "   - Starts with (`startsWith` method): The time complexity of checking if a prefix exists in the Trie is O(p), where p is the length of the prefix. In the worst case, you may need to traverse the entire prefix to determine its presence.\n",
-    "\n",
-    "   Therefore, for each operation (insertion, search, or starts with), the time complexity is O(m) or O(p), where m is the length of the word being operated on, and p is the length of the prefix.\n",
-    "\n",
-    "2. Space Complexity:\n",
-    "   - The space complexity of the Trie is determined by the number of nodes and characters stored in it. In the worst case, where none of the inserted words share common prefixes, the space complexity is O(N), where N is the total number of characters across all inserted words.\n",
-    "   - Each node in the Trie represents a character, and the number of nodes is directly related to the total number of characters.\n",
-    "\n",
-    "   In practice, the space complexity can be less than O(N) because common prefixes are shared among words in the Trie, leading to space optimization.\n",
-    "\n",
-    "In summary:\n",
-    "- Time complexity for each operation (insertion, search, starts with) is O(m) or O(p), where m is the length of the word and p is the length of the prefix.\n",
-    "- Space complexity is O(N) in the worst case, where N is the total number of characters across all inserted words. However, space optimization occurs due to common prefix sharing in practice, potentially reducing the actual space used."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Count Prefixes**:\n",
-    "   Implement a method to count the number of words in the Trie that have a specific prefix. This exercise requires you to maintain additional data in the Trie nodes to keep track of the count.\n",
-    "\n",
-    "2. **Auto-Complete Suggestions**:\n",
-    "   Implement an auto-complete feature using the Trie. Given a prefix, the program should return a list of suggested words based on the words previously inserted into the Trie."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/08. Tries/211. Design Add and Search Words Data Structure.ipynb b/docs/_build/jupyter_execute/08. Tries/211. Design Add and Search Words Data Structure.ipynb
deleted file mode 100644
index bdd867c..0000000
--- a/docs/_build/jupyter_execute/08. Tries/211. Design Add and Search Words Data Structure.ipynb	
+++ /dev/null
@@ -1,271 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 211. Design Add and Search Words Data Structure\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Design Add and Search Words Data Structure problem on LeetCode, click here!](https://leetcode.com/problems/design-add-and-search-words-data-structure/)\n",
-    "\n",
-    "---\n",
-    "Design a data structure that supports adding new words and finding if a string matches any previously added string.\n",
-    "\n",
-    "Implement the `WordDictionary` class:\n",
-    "\n",
-    "- `WordDictionary()` Initializes the object.\n",
-    "- `void addWord(word)` Adds `word` to the data structure, it can be matched later.\n",
-    "- `bool search(word)` Returns `true` if there is any string in the data structure that matches `word` or `false` otherwise. `word` may contain dots `'.'` where dots can be matched with any letter.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= word.length <= 25`\n",
-    "- `word` in `addWord` consists of lowercase English letters.\n",
-    "- `word` in `search` consist of `'.'` or lowercase English letters.\n",
-    "- There will be at most `2` dots in `word` for `search` queries.\n",
-    "- At most $10^4$ calls will be made to `addWord` and `search`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to design a data structure called `WordDictionary` that allows you to add words and search for words efficiently. The key feature of this data structure is that it should support wildcard search, where a dot ('.') can match any letter.\n",
-    "\n",
-    "Here are the specific requirements and explanations for the problem:\n",
-    "\n",
-    "1. Initialize the `WordDictionary` object using the constructor `WordDictionary()`.\n",
-    "\n",
-    "2. The `addWord(word)` method allows you to add a word to the data structure. Once added, you should be able to search for this word later. The words consist of lowercase English letters, and each word has a length between 1 and 25 characters.\n",
-    "\n",
-    "3. The `search(word)` method allows you to search for a word in the data structure. This method returns `True` if there is any string in the data structure that matches the given word or `False` otherwise. The word you are searching for may contain dots ('.'), where each dot can match any letter.\n",
-    "\n",
-    "4. The wildcard character ('.') in the `search` method allows for partial or fuzzy searching. For example, if the word dictionary contains the words \"bad,\" \"dad,\" and \"mad,\" then:\n",
-    "   - Searching for \"pad\" should return `False` because there is no exact match.\n",
-    "   - Searching for \"bad\" should return `True` because \"bad\" is in the dictionary.\n",
-    "   - Searching for \".ad\" should return `True` because it can match \"bad,\" \"dad,\" or \"mad.\"\n",
-    "   - Searching for \"b..\" should return `True` because it can match \"bad,\" \"bed,\" \"bet,\" etc.\n",
-    "\n",
-    "5. The problem specifies that there will be at most two dots ('.') in search queries, and there will be at most 10^4 calls to both `addWord` and `search`.\n",
-    "\n",
-    "In summary, you need to implement the `WordDictionary` class that efficiently supports adding words and searching for words, where the search can include wildcard characters ('.') that match any letter.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TrieNode:\n",
-    "    def __init__(self):\n",
-    "        # Initialize a TrieNode with children and a flag to indicate the end of a word.\n",
-    "        self.children = {}\n",
-    "        self.is_end_of_word = False\n",
-    "\n",
-    "class WordDictionary:\n",
-    "\n",
-    "    def __init__(self):\n",
-    "        # Initialize the WordDictionary with a root node.\n",
-    "        self.root = TrieNode()\n",
-    "\n",
-    "    def addWord(self, word: str) -> None:\n",
-    "        node = self.root\n",
-    "        # Iterate through the characters in the word.\n",
-    "        for char in word:\n",
-    "            if char not in node.children:\n",
-    "                # Create a new node for the character if it doesn't exist.\n",
-    "                node.children[char] = TrieNode()\n",
-    "            # Move to the next node.\n",
-    "            node = node.children[char]\n",
-    "        # Mark the end of the word by setting the flag to True.\n",
-    "        node.is_end_of_word = True\n",
-    "\n",
-    "    def search_helper(self, node: TrieNode, word: str) -> bool:\n",
-    "        if not word:\n",
-    "            # If there are no characters left in the word, check if we reached the end of a word in the trie.\n",
-    "            return node.is_end_of_word\n",
-    "        \n",
-    "        char = word[0]\n",
-    "        if char == '.':\n",
-    "            # If the character is a dot, explore all possible child nodes.\n",
-    "            for child in node.children.values():\n",
-    "                if self.search_helper(child, word[1:]):\n",
-    "                    return True\n",
-    "        elif char in node.children:\n",
-    "            # If the character is in the children of the current node, move to the next node.\n",
-    "            return self.search_helper(node.children[char], word[1:])\n",
-    "        \n",
-    "        # If the character is not found and is not a dot, the word is not in the trie.\n",
-    "        return False\n",
-    "\n",
-    "    def search(self, word: str) -> bool:\n",
-    "        # Start the search from the root of the trie.\n",
-    "        return self.search_helper(self.root, word)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "This code defines a `WordDictionary` class that uses a Trie data structure to efficiently store and search for words. Here's a step-by-step explanation:\n",
-    "\n",
-    "1. `TrieNode` class: This class represents nodes in the Trie data structure. Each node has two attributes: `children` (a dictionary to store child nodes) and `is_end_of_word` (a flag to indicate whether the node marks the end of a word).\n",
-    "\n",
-    "2. `WordDictionary` class: This class initializes with an empty Trie by creating a root node.\n",
-    "\n",
-    "3. `addWord` method: This method adds a word to the Trie. It iterates through each character in the word and creates Trie nodes as necessary. It sets the `is_end_of_word` flag to `True` for the final character of the word to mark the end of the word.\n",
-    "\n",
-    "4. `search_helper` method: This is a recursive helper function for searching words in the Trie. It takes a Trie node (`node`) and a word to search (`word`) as input. If there are no characters left in the word (`not word`), it checks if the current node marks the end of a word. If it does, it returns `True`.\n",
-    "\n",
-    "5. If the first character of the word is a dot ('.'), the function explores all possible child nodes by iterating through `node.children.values()` and recursively calls `search_helper` for each child with the remaining part of the word (`word[1:]`).\n",
-    "\n",
-    "6. If the first character of the word is not a dot and is found in the children of the current node, the function recursively moves to the next node by calling `search_helper` on the child node and the remaining part of the word.\n",
-    "\n",
-    "7. If the character is not found in the children and is not a dot, the word is not in the Trie, so the function returns `False`.\n",
-    "\n",
-    "8. `search` method: This is the public method for searching words. It initiates the search process by calling `search_helper` starting from the root node of the Trie.\n",
-    "\n",
-    "Overall, this code efficiently implements a Trie-based data structure that allows adding and searching for words, including support for wildcard characters represented by dots ('.')."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n",
-      "True\n",
-      "True\n",
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "\n",
-    "wordDictionary = WordDictionary()\n",
-    "wordDictionary.addWord(\"bad\")\n",
-    "wordDictionary.addWord(\"dad\")\n",
-    "wordDictionary.addWord(\"mad\")\n",
-    "print(wordDictionary.search(\"pad\"))  # Output: False\n",
-    "print(wordDictionary.search(\"bad\"))  # Output: True\n",
-    "print(wordDictionary.search(\".ad\"))  # Output: True\n",
-    "print(wordDictionary.search(\"b..\"))  # Output: True"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `WordDictionary` class methods:\n",
-    "\n",
-    "1. `addWord` Method:\n",
-    "\n",
-    "   - Time Complexity: O(L)\n",
-    "     - L is the length of the word being added.\n",
-    "     - In the worst case, we need to iterate through each character in the word and create nodes in the Trie. This takes O(L) time.\n",
-    "\n",
-    "   - Space Complexity: O(L)\n",
-    "     - The space complexity for storing the word in the Trie is also O(L) because we create nodes for each character in the word.\n",
-    "\n",
-    "2. `search_helper` Method (called by `search`):\n",
-    "\n",
-    "   - Time Complexity: O(2^M * L)\n",
-    "     - M is the maximum number of dots ('.') in the search word.\n",
-    "     - In the worst case, when there are M dots in the search word, we may explore all possible child nodes at each dot. This results in a branching factor of 26 (for lowercase letters) at each dot.\n",
-    "     - So, the time complexity is exponential in the number of dots and linear in the length of the word being searched (L).\n",
-    "\n",
-    "   - Space Complexity: O(L)\n",
-    "     - The space complexity for the recursive call stack is O(L) in the worst case because we may have to recurse to a depth equal to the length of the search word.\n",
-    "\n",
-    "3. `search` Method:\n",
-    "\n",
-    "   - Time Complexity: O(2^M * L)\n",
-    "     - The `search` method calls the `search_helper` method, which has the same time complexity explained above.\n",
-    "\n",
-    "   - Space Complexity: O(L)\n",
-    "     - The space complexity for the recursive call stack is O(L) in the worst case, as explained earlier.\n",
-    "\n",
-    "Overall, the time complexity of the `search` method is mainly affected by the number of dots ('.') in the search word. In the worst case, when there are multiple dots and many possibilities to explore at each dot, the time complexity can be exponential. However, for most practical cases, it performs reasonably well.\n",
-    "\n",
-    "The space complexity primarily depends on the space required to store the words in the Trie. It is proportional to the total number of characters in all added words and is linear with respect to the length of the words added.\n",
-    "\n",
-    "In summary:\n",
-    "\n",
-    "- Time Complexity for `addWord`: O(L)\n",
-    "- Time Complexity for `search`: O(2^M * L)\n",
-    "- Space Complexity: O(L) for storing words in the Trie, O(L) for the recursive call stack during search.\n",
-    "\n",
-    "Note: The space complexity analysis assumes that the dictionary contains a reasonable number of words and does not account for the overhead of Python objects or system-level memory allocation."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Dictionary Auto-Completion**: Extend the `WordDictionary` class to provide auto-completion suggestions based on the prefix of a word. Implement a method that returns a list of words that match the given prefix.\n",
-    "\n",
-    "2. **Support Word Deletion**: Extend the `WordDictionary` class to support word deletion. Add a method `deleteWord(word)` that removes a word from the data structure. Ensure that the search operation still works correctly after deletion."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb b/docs/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb
deleted file mode 100644
index e582b9d..0000000
--- a/docs/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb	
+++ /dev/null
@@ -1,310 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 212. Word Search II\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Word Search II problem on LeetCode, click here!](https://leetcode.com/problems/word-search-ii/)\n",
-    "\n",
-    "---\n",
-    "Given an `m x n` `board` of characters and a list of strings `words`, the task is to return *all words on the board*.\n",
-    "\n",
-    "Each word must be constructed from letters of sequentially **adjacent cells**, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.\n",
-    "\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == board.length`\n",
-    "- `n == board[i].length`\n",
-    "- `1 <= m, n <= 12`\n",
-    "- `board[i][j]` is a lowercase English letter.\n",
-    "- 1 <= `words.length` <= $3 * 10^4$\n",
-    "- `1 <= words[i].length <= 10`\n",
-    "- `words[i]` consists of lowercase English letters.\n",
-    "- All the strings of `words` are unique."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "To solve the \"Word Search II\" problem, you can use a Trie data structure and perform a depth-first search (DFS) on the board. Here's a step-by-step approach to solve the problem:\n",
-    "\n",
-    "1. Build a Trie from the given list of words:\n",
-    "   - Create a Trie data structure to store all the words from the input list.\n",
-    "   - For each word in the input list, insert it into the Trie, character by character. Make sure to mark the end of each word in the Trie.\n",
-    "\n",
-    "2. Perform DFS on the board:\n",
-    "   - Iterate through each cell (i, j) on the board.\n",
-    "   - For each cell, start a DFS traversal from that cell to search for words.\n",
-    "   - During the DFS traversal, maintain a current path string.\n",
-    "   - At each cell, check if the current path string, concatenated with the character in the cell, matches any word prefix in the Trie. If it does, continue the DFS.\n",
-    "   - If the current path string matches a word in the Trie, add it to the result.\n",
-    "\n",
-    "3. Implement DFS with backtracking:\n",
-    "   - During the DFS traversal, you will explore neighboring cells (horizontally and vertically) if they are within bounds and haven't been visited before.\n",
-    "   - To prevent revisiting the same cell within the same word, mark the cell as visited (e.g., change its character to a special character like '#') before exploring it and restore its original character after the DFS traversal.\n",
-    "\n",
-    "4. Return the result:\n",
-    "   - After completing the DFS traversal for all cells on the board, you will have collected all the valid words in the result set.\n",
-    "   - Convert the result set to a list and return it as the final output.\n",
-    "\n",
-    "This approach efficiently finds all words on the board by utilizing the Trie data structure to prune unnecessary DFS paths and avoiding duplicate visits to the same cell within the same word. It ensures that each word is constructed from letters of sequentially adjacent cells without using the same letter cell more than once in a word.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Define a TrieNode class to represent nodes in the Trie.\n",
-    "class TrieNode:\n",
-    "    def __init__(self):\n",
-    "        self.children = {}  # Dictionary to store child nodes.\n",
-    "        self.is_end_of_word = False  # Flag to mark the end of a word.\n",
-    "\n",
-    "# Main function to find words on the board.\n",
-    "def findWords(board, words):\n",
-    "    # DFS function to search for words starting from a given cell.\n",
-    "    def dfs(node, i, j, path):\n",
-    "        char = board[i][j]  # Get the character at the current cell.\n",
-    "        curr_node = node.children.get(char)  # Find the corresponding Trie node.\n",
-    "\n",
-    "        if not curr_node:\n",
-    "            return  # If no matching Trie node, stop the search.\n",
-    "\n",
-    "        path += char  # Add the character to the current path.\n",
-    "\n",
-    "        if curr_node.is_end_of_word:\n",
-    "            result.add(path)  # If the path matches a word, add it to the result.\n",
-    "\n",
-    "        temp, board[i][j] = board[i][j], \"#\"  # Mark the cell as visited.\n",
-    "\n",
-    "        # Explore neighboring cells (up, down, left, right).\n",
-    "        if i > 0:\n",
-    "            dfs(curr_node, i - 1, j, path)\n",
-    "        if i < m - 1:\n",
-    "            dfs(curr_node, i + 1, j, path)\n",
-    "        if j > 0:\n",
-    "            dfs(curr_node, i, j - 1, path)\n",
-    "        if j < n - 1:\n",
-    "            dfs(curr_node, i, j + 1, path)\n",
-    "\n",
-    "        board[i][j] = temp  # Restore the cell's original character.\n",
-    "\n",
-    "    # Function to build a Trie from the list of words.\n",
-    "    def buildTrie():\n",
-    "        root = TrieNode()  # Create a root node.\n",
-    "        for word in words:\n",
-    "            node = root\n",
-    "            for char in word:\n",
-    "                if char not in node.children:\n",
-    "                    node.children[char] = TrieNode()  # Create a new node if needed.\n",
-    "                node = node.children[char]\n",
-    "            node.is_end_of_word = True  # Mark the end of the word.\n",
-    "        return root\n",
-    "\n",
-    "    m, n = len(board), len(board[0])  # Get the dimensions of the board.\n",
-    "    root = buildTrie()  # Build the Trie from the list of words.\n",
-    "    result = set()  # Use a set to store unique words found on the board.\n",
-    "\n",
-    "    # Iterate through each cell on the board and start DFS from there.\n",
-    "    for i in range(m):\n",
-    "        for j in range(n):\n",
-    "            dfs(root, i, j, \"\")  # Start DFS from each cell on the board.\n",
-    "\n",
-    "    return list(result)  # Convert the set to a list and return the result."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code solves the \"Word Search II\" problem, where you have a grid of characters and a list of words. The goal is to find all words from the list that can be constructed by traversing adjacent cells (horizontally or vertically) in the grid, without using the same cell more than once for a single word.\n",
-    "\n",
-    "The code does the following:\n",
-    "\n",
-    "1. Defines a `TrieNode` class to represent nodes in a Trie data structure. Each node has children (other characters in the word) and a flag to mark the end of a word.\n",
-    "\n",
-    "2. Defines the `findWords` function, which takes the grid (`board`) and a list of words (`words`) as input.\n",
-    "\n",
-    "3. Inside the `findWords` function, there is a `dfs` (depth-first search) function. This function is used to explore the grid starting from a specific cell. It checks if the current path matches any word in the Trie and adds it to the result if found.\n",
-    "\n",
-    "4. The code also includes a `buildTrie` function that constructs a Trie data structure from the list of words. It iterates through each word, creating nodes for each character in the Trie and marking the end of words.\n",
-    "\n",
-    "5. The dimensions of the grid are obtained (m x n).\n",
-    "\n",
-    "6. A Trie is built using the `buildTrie` function.\n",
-    "\n",
-    "7. A set called `result` is used to store unique words found in the grid.\n",
-    "\n",
-    "8. The code iterates through each cell on the grid (using nested loops) and starts a DFS search from each cell.\n",
-    "\n",
-    "9. During the DFS, it explores neighboring cells (up, down, left, right) if they haven't been visited before and if the current path matches a prefix in the Trie.\n",
-    "\n",
-    "10. When a word is found during the DFS traversal, it is added to the `result` set.\n",
-    "\n",
-    "11. The grid cell is marked as visited during the DFS by changing its character to \"#\", and it is restored to its original character after the DFS traversal.\n",
-    "\n",
-    "12. Finally, the set of unique words found is converted to a list and returned as the output.\n",
-    "\n",
-    "The code effectively uses a Trie data structure to optimize the search and ensures that each word is constructed from adjacent cells without reusing the same cell for the same word."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "['eat', 'oath']\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "board1 = [[\"o\",\"a\",\"a\",\"n\"],[\"e\",\"t\",\"a\",\"e\"],[\"i\",\"h\",\"k\",\"r\"],[\"i\",\"f\",\"l\",\"v\"]]\n",
-    "words1 = [\"oath\",\"pea\",\"eat\",\"rain\"]\n",
-    "print(findWords(board1, words1))  # Output: [\"eat\", \"oath\"]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "c748b879-a2d8-4298-a42c-c7f61f9c7fc6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "\n",
-    "board2 = [[\"a\",\"b\"],[\"c\",\"d\"]]\n",
-    "words2 = [\"abcb\"]\n",
-    "print(findWords(board2, words2))  # Output: []"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. **Building the Trie (`buildTrie` function):** \n",
-    "   - Suppose there are `N` words in the input list, and the average length of a word is `L`.\n",
-    "   - Building the Trie takes O(N * L) time, as we iterate through each word and insert each character into the Trie.\n",
-    "\n",
-    "2. **DFS Traversal (`dfs` function):** \n",
-    "   - In the worst case, the DFS function explores all possible paths in the grid.\n",
-    "   - The maximum number of recursive calls for each cell is 4 (up, down, left, right).\n",
-    "   - The maximum depth of the DFS is limited by the length of the longest word in the Trie, which is at most `L`.\n",
-    "   - Therefore, the DFS traversal for each cell takes $O(4^L)$ time.\n",
-    "\n",
-    "3. **Iterating Through the Grid (`findWords` function):**\n",
-    "   - In the worst case, we iterate through all `m` rows and `n` columns in the grid.\n",
-    "   - For each cell, we start a DFS traversal.\n",
-    "   - Therefore, iterating through the entire grid takes $O(m * n * 4^L)$ time.\n",
-    "\n",
-    "Overall, the time complexity of the code is $O(N * L + m * n * 4^L)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. **Trie Data Structure:**\n",
-    "   - The space required to store the Trie depends on the number of unique characters in all the words.\n",
-    "   - In the worst case, where all words are unique and have no common prefixes, the space complexity of the Trie is O(N * L).\n",
-    "   - In practice, it may be less if there are common prefixes among words.\n",
-    "\n",
-    "2. **DFS Stack (Recursive Calls):**\n",
-    "   - The depth of the recursive call stack during DFS is at most `L`, where `L` is the length of the longest word in the Trie.\n",
-    "   - Therefore, the space complexity for the recursive call stack is O(L).\n",
-    "\n",
-    "3. **Result Set (`result` set):**\n",
-    "   - The space used to store the result set depends on the number of valid words found in the grid.\n",
-    "   - In the worst case, when all words are found, the space complexity of the result set is O(N * L).\n",
-    "\n",
-    "Overall, the space complexity of the code is O(N * L) for the Trie and O(L) for the recursive call stack.\n",
-    "\n",
-    "In summary, the time complexity is dominated by the DFS traversal and is influenced by the number of words, their lengths, and the size of the grid. The space complexity is mainly determined by the Trie structure and the recursive call stack depth during DFS."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Word Search Paths:**\n",
-    "   Extend the basic word search exercise to find all possible paths (sequences of cells) that spell out a given word on the board. Return a list of paths if the word can be formed, otherwise an empty list.\n",
-    "\n",
-    "\n",
-    "2. **Reverse Word Search:**\n",
-    "   Modify the code to search for words in reverse order on the board. Given a 2D grid of characters and a list of words, find all words from the list that can be formed by traversing the board in reverse (right to left, bottom to top, etc.)."
-   ]
-  }
- ],
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\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb b/docs/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb
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--- a/docs/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb	
+++ /dev/null
@@ -1,248 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 1046. Last Stone Weight\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Last Stone Weight problem on LeetCode, click here!](https://leetcode.com/problems/last-stone-weight/)\n",
-    "\n",
-    "---\n",
-    "You are given an array of integers `stones` where `stones[i]` is the weight of the `i`th stone.\n",
-    "\n",
-    "We are playing a game with the stones. On each turn, you choose the **heaviest two stones** and smash them together. Suppose the heaviest two stones have weights `x` and `y` with `x <= y`. The result of this smash is:\n",
-    "\n",
-    "1. If `x == y`, both stones are destroyed.\n",
-    "2. If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - x`.\n",
-    "\n",
-    "At the end of the game, there is **at most one** stone left.\n",
-    "\n",
-    "Return *the weight of the last remaining stone*. If there are no stones left, return `0`.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= stones.length <= 30`\n",
-    "- `1 <= stones[i] <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The \"Last Stone Weight\" problem involves a game played with a collection of stones, each stone having a certain weight. The goal is to find the weight of the last remaining stone after applying a specific set of rules.\n",
-    "\n",
-    "Here's how the game works:\n",
-    "\n",
-    "1. You start with an array of stones, where each stone is represented by its weight. The weights of the stones are given in the input as an array.\n",
-    "\n",
-    "2. In each turn of the game, you select the two heaviest stones from the remaining stones. If there is only one stone left, you've found the answer.\n",
-    "\n",
-    "3. If there are two stones with weights `x` and `y`, where `x` is less than or equal to `y`, the following happens:\n",
-    "   - If `x` is equal to `y`, both stones are completely destroyed, and they are removed from the array of stones.\n",
-    "   - If `x` is not equal to `y`, the stone with weight `x` is destroyed, and the stone with weight `y` now has a new weight of `y - x`. The stone with weight `x` is removed from the array, and the modified stone with weight `y - x` remains in the array.\n",
-    "\n",
-    "4. The game continues with the modified array of stones until there is at most one stone left.\n",
-    "\n",
-    "5. The goal is to find the weight of the last remaining stone, or if there are no stones left, return `0`.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import heapq\n",
-    "\n",
-    "def lastStoneWeight(stones):\n",
-    "    # Convert the input list to a max heap (negate the values to simulate a max heap)\n",
-    "    max_heap = [-stone for stone in stones]\n",
-    "    heapq.heapify(max_heap)\n",
-    "\n",
-    "    # Continue as long as there are more than 1 stone in the heap\n",
-    "    while len(max_heap) > 1:\n",
-    "        # Get the two heaviest stones\n",
-    "        y = -heapq.heappop(max_heap)  # Get the heaviest stone and negate it back to positive\n",
-    "        x = -heapq.heappop(max_heap)  # Get the second heaviest stone and negate it back to positive\n",
-    "\n",
-    "        # Calculate the new stone's weight after smashing\n",
-    "        if x != y:\n",
-    "            new_stone = y - x\n",
-    "            heapq.heappush(max_heap, -new_stone)  # Negate the value and push it back to the heap\n",
-    "\n",
-    "    # If there is one stone left, return its weight (negate it back to positive)\n",
-    "    if max_heap:\n",
-    "        return -max_heap[0]\n",
-    "    else:\n",
-    "        return 0"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a Python function called `lastStoneWeight` that takes a list of stone weights as input. It aims to find the weight of the last remaining stone after playing the stone smashing game described in the problem statement.\n",
-    "\n",
-    "Here's a step-by-step explanation of the code:\n",
-    "\n",
-    "1. We import the `heapq` library, which allows us to create and manipulate a heap (priority queue).\n",
-    "\n",
-    "2. Inside the `lastStoneWeight` function:\n",
-    "   - We create a max heap (negating stone weights to simulate a max heap) using the `heapq.heapify` function. This heap will keep track of the heaviest stones.\n",
-    "   \n",
-    "   - We enter a loop that continues as long as there are more than 1 stone in the heap.\n",
-    "   \n",
-    "   - Inside the loop, we:\n",
-    "     - Retrieve and negate the heaviest stone (`y`) from the heap.\n",
-    "     - Retrieve and negate the second heaviest stone (`x`) from the heap.\n",
-    "   \n",
-    "     - Calculate the new stone's weight after smashing (`y - x`), and negate it back to simulate a max heap.\n",
-    "     - Add the new stone back to the heap using `heapq.heappush`.\n",
-    "   \n",
-    "3. After the loop, if there is one stone left in the heap, we retrieve and negate it to get its actual weight, and return it as the result.\n",
-    "   \n",
-    "4. If there are no stones left (the heap is empty), we return `0`, indicating that no stones remain.\n",
-    "\n",
-    "5. Finally, we provide two test cases to demonstrate the function's usage and correctness.\n",
-    "\n",
-    "In summary, the code efficiently implements the stone smashing game by maintaining a max heap, selecting the heaviest stones in each turn, smashing them, and updating the heap until there is at most one stone left. It then returns the weight of the last remaining stone or 0 if there are no stones left."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "stones1 = [2, 7, 4, 1, 8, 1]\n",
-    "print(lastStoneWeight(stones1))  # Output: 1"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "b7e79514-4821-48d0-bb9d-0f2de8150945",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2: \n",
-    "\n",
-    "stones2 = [1]\n",
-    "print(lastStoneWeight(stones2))  # Output: 1"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided Python code for the \"Last Stone Weight\" problem:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. The `heapq.heapify` operation to create the max heap from the input stone weights has a time complexity of O(n), where n is the number of stones.\n",
-    "\n",
-    "2. The loop that continues until there are more than 1 stone in the heap performs the following operations in each iteration:\n",
-    "   - Retrieving and negating the heaviest stone (`y`) from the heap, which takes O(log n) time.\n",
-    "   - Retrieving and negating the second heaviest stone (`x`) from the heap, also taking O(log n) time.\n",
-    "   - Calculating the new stone's weight and pushing it back into the heap using `heapq.heappush`, which takes O(log n) time.\n",
-    "\n",
-    "3. In the worst case, the loop runs until there is only one stone left, which requires approximately (n-1) iterations.\n",
-    "\n",
-    "Overall, the dominant time complexity is O(n) for the initial heap creation, and within the loop, each iteration takes O(log n) time. Therefore, the overall time complexity of the algorithm is O(n log n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The space complexity is primarily determined by the space required for the max heap. In the worst case, this heap can contain all the stones, so the space complexity for the heap is O(n).\n",
-    "\n",
-    "2. The rest of the variables used in the function (e.g., `x`, `y`, `new_stone`) require only constant space, so they do not significantly contribute to the space complexity.\n",
-    "\n",
-    "Therefore, the overall space complexity of the algorithm is O(n) due to the space used for the max heap.\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: $O(n\\ log\\ n)$\n",
-    "- Space Complexity: $O(n)$\n",
-    "\n",
-    "The algorithm is efficient enough to handle the problem's constraints, as the worst-case time and space complexities are both linearithmic in terms of the number of stones."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Kth Last Stone Weight**: Modify the problem to find the weight of the Kth last remaining stone, where K is an integer input. Your function should work efficiently for large values of K.\n",
-    "\n",
-    "\n",
-    "2. **Multiple Stones Smash**: Modify the problem so that instead of smashing two stones at a time, you can smash up to K stones at each turn, where K is an integer input. Determine the weight of the last remaining stone in this variation."
-   ]
-  }
- ],
- "metadata": {
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- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb b/docs/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb
deleted file mode 100644
index 6c43a8d..0000000
--- a/docs/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb	
+++ /dev/null
@@ -1,238 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 295. Find Median from Data Stream\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Find Median from Data Stream problem on LeetCode, click here!](https://leetcode.com/problems/find-median-from-data-stream/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.\n",
-    "\n",
-    "Implement the MedianFinder class:\n",
-    "\n",
-    "- `MedianFinder()` initializes the `MedianFinder` object.\n",
-    "- `void addNum(int num)` adds the integer `num` from the data stream to the data structure.\n",
-    "- `double findMedian()` returns the median of all elements so far. Answers within $10^{-5}$ of the actual answer will be accepted.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- $-10^{5}$ <= `num` <= $10^{5}$\n",
-    "- There will be at least one element in the data structure before calling `findMedian`.\n",
-    "- At most $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-    "\n",
-    "**Follow-up:**\n",
-    "\n",
-    "1. If all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?\n",
-    "2. If `99%` of all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem, \"Find Median from Data Stream,\" deals with efficiently computing the median of a sequence of numbers as they are incrementally provided. The median is the middle value in an ordered list of numbers. When the list has an even number of elements, the median is the average of the two middle values.\n",
-    "\n",
-    "Here's a breakdown of the problem:\n",
-    "\n",
-    "1. **Initialization**: You are asked to implement a `MedianFinder` class that supports the following operations:\n",
-    "\n",
-    "    - `MedianFinder()`: Initializes the MedianFinder object.\n",
-    "    - `addNum(int num)`: Adds an integer `num` from the data stream to the data structure.\n",
-    "    - `findMedian()`: Returns the median of all elements added to the data structure.\n",
-    "\n",
-    "2. **Median Calculation**: The `findMedian` operation should efficiently compute the median, and the result should have a precision of at least $10^{-5}$ (i.e., answers within this range will be considered correct).\n",
-    "\n",
-    "3. **Examples**:\n",
-    "   \n",
-    "   - If you add numbers [1, 2], the median should be 1.5 because (1 + 2) / 2 = 1.5.\n",
-    "   - If you add numbers [1, 2, 3], the median should be 2 because it's the middle value.\n",
-    "   \n",
-    "4. **Constraints**:\n",
-    "\n",
-    "   - The integers provided in the data stream are in the range from $-10^5$ to $10^5$.\n",
-    "   - There will be at least one element in the data structure before calling `findMedian`.\n",
-    "   - At most, $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-    "\n",
-    "The problem can be solved by using two priority queues (heaps): one max-heap to store the smaller half of the numbers and one min-heap to store the larger half of the numbers. The max-heap ensures that the largest number in the smaller half is at the top, and the min-heap ensures that the smallest number in the larger half is at the top. These heaps are balanced to efficiently find the median when requested.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import heapq\n",
-    "\n",
-    "class MedianFinder:\n",
-    "\n",
-    "    def __init__(self):\n",
-    "        # Initialize two heaps: a max-heap (small_half) for the smaller half of numbers,\n",
-    "        # and a min-heap (large_half) for the larger half of numbers.\n",
-    "        self.small_half = []  # Max-heap for the smaller half of the numbers\n",
-    "        self.large_half = []  # Min-heap for the larger half of the numbers\n",
-    "\n",
-    "    def addNum(self, num: int) -> None:\n",
-    "        # Add the number to the appropriate heap based on its value\n",
-    "        if not self.small_half or num <= -self.small_half[0]:\n",
-    "            # If the number is less than or equal to the current maximum in the smaller half,\n",
-    "            # add it to the smaller half (max-heap)\n",
-    "            heapq.heappush(self.small_half, -num)\n",
-    "        else:\n",
-    "            # Otherwise, add it to the larger half (min-heap)\n",
-    "            heapq.heappush(self.large_half, num)\n",
-    "\n",
-    "        # Balance the heaps if necessary to ensure the size difference is at most 1\n",
-    "        if len(self.small_half) > len(self.large_half) + 1:\n",
-    "            # If the size of the smaller half is more than one greater than the size of the larger half,\n",
-    "            # move the maximum from the smaller half to the larger half to balance them.\n",
-    "            heapq.heappush(self.large_half, -heapq.heappop(self.small_half))\n",
-    "        elif len(self.large_half) > len(self.small_half):\n",
-    "            # If the size of the larger half is greater than the size of the smaller half,\n",
-    "            # move the minimum from the larger half to the smaller half to balance them.\n",
-    "            heapq.heappush(self.small_half, -heapq.heappop(self.large_half))\n",
-    "\n",
-    "    def findMedian(self) -> float:\n",
-    "        if len(self.small_half) == len(self.large_half):\n",
-    "            # If both halves have the same size, there's an even number of elements,\n",
-    "            # so the median is the average of the top elements in both heaps.\n",
-    "            return (-self.small_half[0] + self.large_half[0]) / 2.0\n",
-    "        else:\n",
-    "            # If the smaller half has more elements, it contains the median (odd number of elements).\n",
-    "            return -self.small_half[0]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a `MedianFinder` class for efficiently finding the median of a stream of numbers. It uses two heaps: a max-heap (`small_half`) to store the smaller half of the numbers and a min-heap (`large_half`) to store the larger half. Here's how the code works:\n",
-    "\n",
-    "1. The `__init__` method initializes the two heaps.\n",
-    "\n",
-    "2. The `addNum` method adds a number to the appropriate heap, ensuring that the heaps remain balanced. It does this by comparing the number to the current maximum in the smaller half and adding it to the appropriate heap. Then, it checks the sizes of the two heaps and rebalances them if necessary.\n",
-    "\n",
-    "3. The `findMedian` method calculates and returns the median. If both halves have the same size (even number of elements), it computes the average of the tops of both heaps. If the smaller half has more elements (odd number of elements), it returns the top of the max-heap, which is the median.\n",
-    "\n",
-    "In summary, this code efficiently maintains two heaps to divide the elements into smaller and larger halves, allowing for quick median retrieval. It offers a time complexity of O(log N) for adding elements and O(1) for finding the median, where N is the total number of elements processed. The space complexity is O(1) as the space used by the two heaps is independent of the input size."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1.5\n",
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "medianFinder = MedianFinder()\n",
-    "medianFinder.addNum(1)\n",
-    "medianFinder.addNum(2)\n",
-    "print(medianFinder.findMedian())  # Output: 1.5\n",
-    "medianFinder.addNum(3)\n",
-    "print(medianFinder.findMedian())  # Output: 2.0\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `MedianFinder` class:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. `MedianFinder()` Initialization: This operation has a time complexity of O(1). It simply initializes the two heaps.\n",
-    "\n",
-    "2. `addNum(int num)`: The time complexity for adding a number is O(log N), where N is the total number of elements processed so far. This is because, in the worst case, you might need to perform logarithmic operations on the heaps to maintain their balance.\n",
-    "\n",
-    "3. `findMedian()`: Finding the median has a time complexity of O(1). It simply involves accessing the tops of the two heaps, which takes constant time.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. `MedianFinder()` Initialization: The space complexity for initialization is O(1). It involves creating two empty heaps, which do not depend on the size of the data stream.\n",
-    "\n",
-    "2. `addNum(int num)`: The space complexity for adding a number is also O(1). It involves adding the number to one of the two heaps, which do not significantly affect the overall space complexity.\n",
-    "\n",
-    "3. `findMedian()`: The space complexity for finding the median is O(1). It doesn't require any additional data structures that grow with the input size.\n",
-    "\n",
-    "In summary, the time complexity of the `addNum` operation is O(log N), and the time complexity of the `findMedian` operation is O(1). The space complexity is O(1) as well, as the space used by the two heaps is independent of the size of the data stream. This solution efficiently maintains and retrieves the median, making it suitable for the given problem constraints."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Generalized kth Element**: Extend the `MedianFinder` class to support finding the kth element in the data stream efficiently. This means finding the kth smallest or kth largest element in the stream.\n",
-    "\n",
-    "2. **Sliding Window Median**: Given an array and a sliding window size, find the median within the window as it slides over the array. This is an extension of the problem to a moving window scenario."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb b/docs/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb
deleted file mode 100644
index e6ab9e3..0000000
--- a/docs/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb	
+++ /dev/null
@@ -1,235 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 703. Kth Largest Element in a Stream\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Kth Largest Element in a Stream problem on LeetCode, click here!](https://leetcode.com/problems/kth-largest-element-in-a-stream/)\n",
-    "\n",
-    "---\n",
-    "Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.\n",
-    "\n",
-    "Implement the `KthLargest` class with the following methods:\n",
-    "\n",
-    "- `KthLargest(int k, int[] nums)`: Initializes the object with the integer `k` and the stream of integers `nums`.\n",
-    "- `int add(int val)`: Appends the integer `val` to the stream and returns the element representing the `kth` largest element in the stream.\n",
-    "\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `k` <= $10^4$\n",
-    "- 0 <= `nums.length` <= $10^4$\n",
-    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-    "- $-10^4$ <= `val` <= $10^4$\n",
-    "- At most $10^4$ calls will be made to `add`.\n",
-    "- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem you're trying to solve is to design a class called `KthLargest` that can efficiently find the kth largest element in a stream of integers. You need to support two main operations:\n",
-    "\n",
-    "To solve this problem efficiently, you can use a min-heap data structure. Here's how the approach works:\n",
-    "\n",
-    "1. Initialize a min-heap to store the k largest elements. Initially, it's empty.\n",
-    "\n",
-    "2. In the `__init__` method, populate the min-heap with the first `k` elements from `nums`. This ensures that you have the k largest elements seen so far.\n",
-    "\n",
-    "3. Whenever you add a new element to the stream using the `add` method, follow these steps:\n",
-    "   - Add the new element to the min-heap.\n",
-    "   - If the size of the min-heap exceeds `k`, remove the smallest element from the min-heap. This ensures that you always have the k largest elements in the min-heap.\n",
-    "   - The smallest element in the min-heap (the root) will always represent the kth largest element in the stream.\n",
-    "\n",
-    "Here's why this approach works efficiently:\n",
-    "\n",
-    "- By using a min-heap, you can quickly maintain the k largest elements, and finding the smallest element in the heap (the root) takes constant time.\n",
-    "\n",
-    "- When you add a new element, the min-heap's size is kept at most `k`, which ensures that you only track the k largest elements and discard the smaller ones.\n",
-    "\n",
-    "- The time complexity for adding an element is O(log k), which is very efficient compared to sorting the entire stream, which would be O(n log n).\n",
-    "\n",
-    "- This approach meets the constraints of the problem, including handling large streams and having a low time complexity for both initialization and adding elements.\n",
-    "\n",
-    "In summary, the min-heap approach efficiently tracks the kth largest element in the stream by maintaining a heap of the k largest elements seen so far, updating it as new elements are added. This approach provides a fast and scalable solution to the problem.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import heapq\n",
-    "from typing import List  # Import the List type from the typing module\n",
-    "\n",
-    "class KthLargest:\n",
-    "\n",
-    "    def __init__(self, k: int, nums: List[int]):\n",
-    "        # Initialize the KthLargest object with k and nums.\n",
-    "        self.k = k\n",
-    "        # Create a min-heap to store the k largest elements.\n",
-    "        self.min_heap = []\n",
-    "        # Populate the min-heap with the first k elements from nums.\n",
-    "        for num in nums:\n",
-    "            self.add(num)\n",
-    "\n",
-    "    def add(self, val: int) -> int:\n",
-    "        # Add val to the min-heap.\n",
-    "        heapq.heappush(self.min_heap, val)\n",
-    "        # If the size of the min-heap exceeds k, remove the smallest element.\n",
-    "        if len(self.min_heap) > self.k:\n",
-    "            heapq.heappop(self.min_heap)\n",
-    "        # The root of the min-heap is the kth largest element.\n",
-    "        return self.min_heap[0]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "- We start by importing the necessary modules:\n",
-    "  - `heapq`: This module provides functions to create and manipulate heaps.\n",
-    "  - `List` from the `typing` module: This is used to specify the type of the `nums` parameter.\n",
-    "\n",
-    "- We define the `KthLargest` class:\n",
-    "  - The `__init__` method initializes the object with the integer `k` and the list of integers `nums`.\n",
-    "  - It also creates an empty min-heap called `self.min_heap` to store the k largest elements.\n",
-    "  - It populates the min-heap with the first `k` elements from `nums` by calling the `add` method.\n",
-    "\n",
-    "- The `add` method:\n",
-    "  - Adds the new integer `val` to the min-heap using `heapq.heappush`. This maintains the min-heap property.\n",
-    "  - Checks if the size of the min-heap exceeds `k`. If it does, it removes the smallest element (the k+1th largest) using `heapq.heappop`.\n",
-    "  - Finally, it returns the smallest element in the min-heap, which is always the kth largest element.\n",
-    "\n",
-    "Overall, this code implements a class that efficiently finds the kth largest element in a stream of integers by maintaining a min-heap of the k largest elements seen so far.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n",
-      "5\n",
-      "5\n",
-      "8\n",
-      "8\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "kthLargest = KthLargest(3, [4, 5, 8, 2])\n",
-    "print(kthLargest.add(3))  # Output: 4\n",
-    "print(kthLargest.add(5))  # Output: 5\n",
-    "print(kthLargest.add(10)) # Output: 5\n",
-    "print(kthLargest.add(9))  # Output: 8\n",
-    "print(kthLargest.add(4))  # Output: 8"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `KthLargest` class using the min-heap approach:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. `__init__` method:\n",
-    "   - In the `__init__` method, you iterate over the first `k` elements in `nums` and add them to the min-heap. Each `heapq.heappush` operation takes O(log k) time.\n",
-    "   - Therefore, the time complexity of the `__init__` method is O(k * log k).\n",
-    "\n",
-    "2. `add` method:\n",
-    "   - In the `add` method, you perform the following operations:\n",
-    "     - `heapq.heappush`: O(log k) to add the new element to the min-heap.\n",
-    "     - If the size of the min-heap exceeds `k`, you perform `heapq.heappop`, which is also O(log k).\n",
-    "     - Finally, you return the smallest element from the min-heap, which is O(1) because it's always the root.\n",
-    "   - Overall, the time complexity of the `add` method is O(log k).\n",
-    "\n",
-    "3. Overall, if you make `n` calls to the `add` method, the total time complexity is O(n * log k), where `n` is the total number of elements added to the stream.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. The space complexity is determined by the space used to store the min-heap and the instance variables.\n",
-    "2. The min-heap stores at most `k` elements at any given time, so its space complexity is O(k).\n",
-    "3. The instance variables (such as `self.k` and `self.min_heap`) have constant space requirements.\n",
-    "4. Therefore, the overall space complexity of the `KthLargest` class is O(k).\n",
-    "\n",
-    "In summary, the time complexity of the `KthLargest` class is O(n * log k) for `n` add operations, and the space complexity is O(k), where `k` is the parameter passed during initialization. This implementation efficiently maintains the kth largest element in the stream while meeting the problem's constraints."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Dynamic k:** Modify the `KthLargest` class to support dynamic changes in the value of `k`. Implement a method `update_k` that allows changing the value of `k` during the lifetime of the object. Ensure that the object can still correctly find the kth largest element based on the updated value of `k`.\n",
-    "\n",
-    "\n",
-    "2. **Implement kth Smallest:** Create a new class called `KthSmallest` that finds the kth smallest element in a stream instead of the kth largest. You may need to modify the data structure used in the implementation."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb b/docs/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb
deleted file mode 100644
index 85d59ac..0000000
--- a/docs/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb	
+++ /dev/null
@@ -1,274 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 39. Combination Sum\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Combination Sum problem on LeetCode, click here!](https://leetcode.com/problems/combination-sum/)\n",
-    "\n",
-    "---\n",
-    "Given an array of **distinct** integers `candidates` and a target integer `target`, return *a list of all **unique combinations** of `candidates` where the chosen numbers sum to the `target`*. You may return the combinations in **any order**. \n",
-    "\n",
-    "The **same** number may be chosen from `candidates` an **unlimited number** of times. Two combinations are unique if the **frequency** of at least one of the chosen numbers is different.\n",
-    "\n",
-    "The test cases are generated such that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= candidates.length <= 30`\n",
-    "- `2 <= candidates[i] <= 40`\n",
-    "- All elements of `candidates` are **distinct**.\n",
-    "- `1 <= target <= 40`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to find all unique combinations of numbers from a given array (candidates) such that their sum equals a target number. Here are the details of the problem:\n",
-    "\n",
-    "- You are given an array of distinct integers called \"candidates.\"\n",
-    "- You are also given a target integer called \"target.\"\n",
-    "\n",
-    "The goal is to find all unique combinations of numbers from the candidates array where the sum of the selected numbers is equal to the target. You can use the same number from the candidates array an unlimited number of times. A combination is considered unique if it has a different frequency (i.e., a different number of occurrences) of at least one chosen number compared to other combinations.\n",
-    "\n",
-    "For example:\n",
-    "\n",
-    "**Example 1:**\n",
-    "```python\n",
-    "Input: candidates = [2, 3, 6, 7], target = 7\n",
-    "Output: [[2, 2, 3], [7]]\n",
-    "```\n",
-    "In this example, there are two unique combinations that sum up to the target:\n",
-    "- 2 + 2 + 3 = 7\n",
-    "- 7 = 7\n",
-    "\n",
-    "**Example 2:**\n",
-    "```python\n",
-    "Input: candidates = [2, 3, 5], target = 8\n",
-    "Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n",
-    "```\n",
-    "Here, there are three unique combinations that sum up to the target.\n",
-    "\n",
-    "**Example 3:**\n",
-    "```python\n",
-    "Input: candidates = [2], target = 1\n",
-    "Output: []\n",
-    "```\n",
-    "In this case, there are no combinations that can be formed from the candidates to reach the target of 1, so the output is an empty list.\n",
-    "\n",
-    "The problem asks you to find and return these combinations in any order.\n",
-    "\n",
-    "The constraints for this problem include the length of the candidates array, the values of candidates, and the target value.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def combinationSum(candidates, target):\n",
-    "    # Define a recursive DFS function to find combinations\n",
-    "    def dfs(remaining, start, path):\n",
-    "        # If the remaining target is 0, we found a valid combination\n",
-    "        if remaining == 0:\n",
-    "            result.append(path)\n",
-    "            return\n",
-    "        # If the remaining target is negative, this path is invalid\n",
-    "        if remaining < 0:\n",
-    "            return\n",
-    "        # Iterate through candidates starting from 'start' to avoid duplicates\n",
-    "        for i in range(start, len(candidates)):\n",
-    "            # Explore the current candidate by subtracting it from the remaining target\n",
-    "            # Add the current candidate to the path\n",
-    "            dfs(remaining - candidates[i], i, path + [candidates[i]])\n",
-    "\n",
-    "    result = []  # Initialize an empty list to store the result\n",
-    "    candidates.sort()  # Sort the candidates for deduplication and early stopping\n",
-    "    dfs(target, 0, [])  # Start the DFS from the target value with an empty path\n",
-    "    return result  # Return the list of unique combinations"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The `combinationSum` function takes an array of distinct integers `candidates` and a target integer `target`. It returns a list of all unique combinations of `candidates` where the chosen numbers sum to the `target`. Each number in the `candidates` array can be used an unlimited number of times.\n",
-    "\n",
-    "Here's an overview of how the code works:\n",
-    "\n",
-    "1. The `combinationSum` function is defined, and it takes two arguments: `candidates` and `target`.\n",
-    "\n",
-    "2. Inside the function, there is a nested helper function called `backtrack`. This function is responsible for the actual combination generation using a recursive approach.\n",
-    "\n",
-    "3. The `backtrack` function is called with three arguments: `start`, `target`, and `path`. The `start` variable helps keep track of the current index in the `candidates` array, `target` keeps track of the remaining sum to be achieved, and `path` is a list that stores the current combination.\n",
-    "\n",
-    "4. Within the `backtrack` function, there are three main conditional statements:\n",
-    "   - If `target` becomes zero, it means we have found a combination that sums up to the target, so we add `path` to the `result` list.\n",
-    "   - If `target` becomes negative, it means the current combination doesn't work, so we return without doing anything.\n",
-    "   - Otherwise, we enter a loop that iterates over the `candidates` array, starting from the current index `start`.\n",
-    "\n",
-    "5. In the loop, the `backtrack` function is called recursively with the updated `target` and `path` after including the current candidate. This process explores different combinations by considering the current candidate or moving to the next candidate.\n",
-    "\n",
-    "6. After the loop completes, the `result` list contains all unique combinations that sum to the target.\n",
-    "\n",
-    "7. The `candidates` array is sorted to optimize the search. Sorting helps in avoiding unnecessary recursive branches and reduces the number of explored combinations.\n",
-    "\n",
-    "8. Finally, the `backtrack` function is initially called from the `combinationSum` function with `start` set to 0, `target` set to the original target value, and an empty `path`. The result is returned as a list of unique combinations.\n",
-    "\n",
-    "The code includes a few example cases to demonstrate how the function works with different inputs."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[2, 2, 3], [7]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example 1\n",
-    "candidates1 = [2, 3, 6, 7]\n",
-    "target1 = 7\n",
-    "print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "candidates2 = [2, 3, 5]\n",
-    "target2 = 8\n",
-    "print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "candidates3 = [2]\n",
-    "target3 = 1\n",
-    "print(combinationSum(candidates3, target3))  # Output: []"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `combinationSum` function.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of this function is influenced by the number of recursive calls made by the `backtrack` function and the amount of work done within each call. In the worst case, the function will explore all possible combinations.\n",
-    "\n",
-    "The worst-case scenario occurs when we have many combinations to reach the target. The time complexity can be expressed as $O(2^n)$, where 'n' is the maximum number of recursive calls. In this context, 'n' corresponds to the number of elements in the `candidates` list, and it may be up to 30 (as per the constraints). However, the actual number of combinations explored is generally much smaller because we eliminate branches when the target becomes negative, and we skip over candidates that cannot lead to a solution. The sorting step also helps to optimize the search.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity is determined by the auxiliary space used for storing the results and the stack space for the recursive calls.\n",
-    "\n",
-    "1. The space required for the `result` list can be up to $O(2^n)$ in the worst case because each combination is stored.\n",
-    "\n",
-    "2. The stack space for the recursive calls can also be up to O(n) in the worst case, where 'n' is the number of elements in the `candidates` list.\n",
-    "\n",
-    "Therefore, the overall space complexity of the function is $O(2^n + n)$. However, in practice, the space used for the results list is often the dominant factor, and the actual space used may be much smaller than 2^n because not all combinations are explored.\n",
-    "\n",
-    "In summary, the time complexity is exponential but typically smaller in practice due to optimization, and the space complexity is influenced by the number of results and the depth of the recursion."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Find the Number of Unique Combinations:** Instead of listing all unique combinations, modify the function to return the count of unique combinations that sum to the target.\n",
-    "\n",
-    "2. **Combinations with a Maximum Value:** Add a maximum value constraint for each combination, so they cannot exceed a certain value. Modify the code to find combinations respecting this constraint."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb b/docs/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb
deleted file mode 100644
index 8d97d20..0000000
--- a/docs/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb	
+++ /dev/null
@@ -1,290 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 79. Word Search\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Word Search problem on LeetCode, click here!](https://leetcode.com/problems/word-search/)\n",
-    "\n",
-    "---\n",
-    "Given an `m x n` grid of characters `board` and a string `word`, return *`true` if `word` exists in the grid.*\n",
-    "\n",
-    "The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == board.length`\n",
-    "- `n = board[i].length`\n",
-    "- `1 <= m, n <= 6`\n",
-    "- `1 <= word.length <= 15`\n",
-    "- `board` and `word` consists of only lowercase and uppercase English letters.\n",
-    "\n",
-    "**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to determine whether a given word can be constructed from characters in an `m x n` grid of characters. The word can be formed by moving sequentially through horizontally or vertically neighboring cells in the grid, and you cannot use the same cell more than once.\n",
-    "\n",
-    "Here are the specifics of the problem:\n",
-    "\n",
-    "- You are given an `m x n` grid, which is a 2D board of characters.\n",
-    "- You are also given a string `word` consisting of lowercase and uppercase English letters.\n",
-    "- Your task is to determine if the word can be found in the grid by moving from one cell to an adjacent cell (horizontally or vertically) and collecting the characters sequentially to form the word.\n",
-    "- You can't use the same cell more than once when forming the word.\n",
-    "- If it's possible to form the word in the grid, return `True`; otherwise, return `False`.\n",
-    "\n",
-    "For example, consider the following grid and word:\n",
-    "\n",
-    "```python\n",
-    "board = [[\"A\", \"B\", \"C\", \"E\"],\n",
-    "         [\"S\", \"F\", \"C\", \"S\"],\n",
-    "         [\"A\", \"D\", \"E\", \"E\"]]         \n",
-    "word = \"ABCCED\"\n",
-    "```\n",
-    "\n",
-    "In this case, the word \"ABCCED\" can be formed by moving through the cells (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) in the grid. Therefore, the function should return `True`.\n",
-    "\n",
-    "Conversely, if the word is impossible to construct, such as in the case of the word \"ABCB\" for the same grid, where there's no valid path that forms the word, the function should return `False`.\n",
-    "\n",
-    "The problem involves searching for a path through the grid that matches the characters in the word while respecting the constraints mentioned. It can be solved using depth-first search (DFS) or backtracking.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def exist(self, board: List[List[str]], word: str) -> bool:\n",
-    "        def dfs(i, j, k):\n",
-    "            # Check if the current position is out of bounds or doesn't match the character in the word.\n",
-    "            if not (0 <= i < len(board)) or not (0 <= j < len(board[0])) or board[i][j] != word[k]:\n",
-    "                return False\n",
-    "            \n",
-    "            # If we have matched all characters in the word, return True.\n",
-    "            if k == len(word) - 1:\n",
-    "                return True\n",
-    "\n",
-    "            # Temporarily mark the current cell to prevent revisiting.\n",
-    "            tmp, board[i][j] = board[i][j], \"/\"\n",
-    "\n",
-    "            # Explore adjacent cells by making recursive calls in all four directions.\n",
-    "            found = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)\n",
-    "\n",
-    "            # Restore the original character in the cell.\n",
-    "            board[i][j] = tmp\n",
-    "\n",
-    "            return found\n",
-    "\n",
-    "        # Iterate through all cells in the board.\n",
-    "        for i in range(len(board)):\n",
-    "            for j in range(len(board[0])):\n",
-    "                # Check if the 'dfs' function starting from this cell returns True.\n",
-    "                if dfs(i, j, 0):\n",
-    "                    return True\n",
-    "\n",
-    "        # If no match is found after exploring all cells, return False.\n",
-    "        return False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. Import the necessary module:\n",
-    "   - `from typing import List`: This line imports the `List` type from the `typing` module, which is used to specify the type of the `board` parameter in the method signature.\n",
-    "\n",
-    "2. Define the `Solution` class:\n",
-    "   - `class Solution:`: This line defines a class called `Solution` to encapsulate the solution for the word search problem.\n",
-    "\n",
-    "3. Define the `exist` method:\n",
-    "   - `def exist(self, board: List[List[str]], word: str) -> bool:`: This method is a member of the `Solution` class and takes two parameters: `board`, which is a 2D list of characters, and `word`, which is a string. It returns a boolean value, indicating whether the word exists in the board.\n",
-    "\n",
-    "4. Implement the inner `dfs` function:\n",
-    "   - `def dfs(i, j, k):`: This is a helper function used for depth-first search (DFS). It is called recursively to explore the board and search for the word. It takes three parameters: `i` and `j` representing the current position on the board, and `k` representing the current index in the `word` string.\n",
-    "\n",
-    "   - Inside `dfs`, it checks if the current position is out of bounds or if the character at the current position on the board does not match the character at the current index in the `word`. If either of these conditions is met, it returns `False`.\n",
-    "\n",
-    "   - If `k` is equal to the length of the `word` minus one, it means we have successfully matched all characters in the `word`, so it returns `True`.\n",
-    "\n",
-    "   - The function then temporarily replaces the character at the current position with a marker, \"/\", to prevent revisiting the same cell. It then explores adjacent cells by making recursive calls in all four directions: up, down, left, and right. If any of these recursive calls return `True`, it propagates the `True` result up the call stack. After the exploration is complete, it restores the original character in the cell.\n",
-    "\n",
-    "5. Iterate through the board:\n",
-    "   - The outer loop iterates through all cells in the `board` using two nested loops. For each cell, it checks if the `dfs` function starting from that cell (with `k` initially set to 0) returns `True`. If it does, it means the word exists, so the method returns `True`.\n",
-    "\n",
-    "6. Return `False`:\n",
-    "   - If the loop completes without finding the word in the board, the method returns `False`.\n",
-    "\n",
-    "To use this code, you can create an instance of the `Solution` class and call the `exist` method on that instance, passing the `board` and the `word` as arguments to check if the word exists in the board."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Test case 1: Word \"ABCCED\" can be found in the grid\n",
-    "board1 = [[\"A\",\"B\",\"C\",\"E\"],\n",
-    "          [\"S\",\"F\",\"C\",\"S\"],\n",
-    "          [\"A\",\"D\",\"E\",\"E\"]]\n",
-    "word1 = \"ABCCED\"\n",
-    "print(solution.exist(board1, word1))  # Expected output: True"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "\n",
-    "# Test case 2: Word \"SEE\" can be found in the grid\n",
-    "word2 = \"SEE\"\n",
-    "print(solution.exist(board1, word2))  # Expected output: True"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "\n",
-    "# Test case 3: Word \"ABCB\" cannot be found in the grid\n",
-    "word3 = \"ABCB\"\n",
-    "print(solution.exist(board1, word3))  # Expected output: False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `exist` method in the given code.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The primary work in this code is done by the `dfs` function, which performs a depth-first search through the grid. The time complexity of the `dfs` function is $O(4^(n))$, where \"n\" is the length of the `word`. This is because, in the worst case, the function explores all possible directions (up, down, left, right) for each character in the `word`.\n",
-    "\n",
-    "The outer loops iterate through all cells in the `board`, which has dimensions `m x n`. So, the total number of iterations of `dfs` is O(m * n). However, since the `dfs` function has a higher time complexity, the overall time complexity is still dominated by the `dfs` function.\n",
-    "\n",
-    "Therefore, the overall time complexity of the `exist` method is $O(m * n * 4^n)$, where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. The primary space usage comes from the recursive calls in the `dfs` function. The depth of the recursion can be at most equal to the length of the `word`, which is O(n).\n",
-    "\n",
-    "2. Additionally, the `tmp` variable and the recursive calls on the call stack consume a small amount of additional memory.\n",
-    "\n",
-    "3. The marking of visited cells with the \"/\" character temporarily modifies the `board` in-place but doesn't significantly contribute to the space complexity.\n",
-    "\n",
-    "Therefore, the overall space complexity is O(n) due to the recursive call stack.\n",
-    "\n",
-    "In summary, the time complexity is $O(m * n * 4^n)$, and the space complexity is O(n), where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word. The time complexity is exponential in the worst case due to the depth-first search through all possible paths in the grid."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the code to find all unique words that can be constructed from the grid. Instead of returning `True` or `False`, return a list of words found. Ensure that the same cell is not used more than once in constructing each word.\n",
-    "\n",
-    "2. **Board Generation:** Create a function to generate random grids of characters and then use the `exist` method to solve them. Test the efficiency and correctness of your code by generating larger boards and words."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb b/docs/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb
deleted file mode 100644
index 56b7647..0000000
--- a/docs/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb	
+++ /dev/null
@@ -1,311 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 133. Clone Graph\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Clone Graph problem on LeetCode, click here!](https://leetcode.com/problems/clone-graph/)\n",
-    "\n",
-    "---\n",
-    "Given a reference of a node in a **connected** undirected graph.\n",
-    "\n",
-    "Return a **deep copy** (clone) of the graph.\n",
-    "\n",
-    "Each node in the graph contains a value (`int`) and a list (`List[Node]`) of its neighbors.\n",
-    "\n",
-    "```\n",
-    "class Node {\n",
-    "    public int val;\n",
-    "    public List<Node> neighbors;\n",
-    "}\n",
-    "```\n",
-    "\n",
-    "**Test case format:**\n",
-    "\n",
-    "For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with `val == 1`, the second node with `val == 2`, and so on. The graph is represented in the test case using an adjacency list.\n",
-    "\n",
-    "**An adjacency list** is a collection of unordered **lists** used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.\n",
-    "\n",
-    "The given node will always be the first node with `val = 1`. You must return the **copy of the given node** as a reference to the cloned graph.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the graph is in the range `[0, 100]`.\n",
-    "- `1 <= Node.val <= 100`\n",
-    "- `Node.val` is unique for each node.\n",
-    "- There are no repeated edges and no self-loops in the graph.\n",
-    "- The Graph is connected and all nodes can be visited starting from the given node."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand is to clone a connected undirected graph, which means we need to create a deep copy of the original graph while preserving its structure and relationships between nodes. The graph is represented using adjacency lists, where each node has a value (an integer) and a list of its neighbors.\n",
-    "\n",
-    "Here are the key aspects of the problem:\n",
-    "\n",
-    "1. **Input**: You are given a reference to one of the nodes in the original graph. This reference node serves as the entry point to the graph. The entire graph can be explored starting from this node.\n",
-    "\n",
-    "2. **Output**: The goal is to create a deep copy of the graph and return a reference to the cloned graph. The cloned graph should have the same structure and relationships as the original one but should be a separate instance in memory.\n",
-    "\n",
-    "3. **Graph Structure**: The graph is composed of nodes (vertices) and edges. Nodes have values (integers) and are connected to other nodes through edges (undirected connections). Each node has a list of neighbors, which are other nodes it is connected to.\n",
-    "\n",
-    "4. **Connected Graph**: The problem assumes that the graph is connected, meaning that you can start from the reference node provided and traverse the entire graph by following the edges. This ensures that there are no isolated components in the graph.\n",
-    "\n",
-    "To solve this problem, you typically need to perform a graph traversal (e.g., depth-first search or breadth-first search) to visit all the nodes in the original graph, create corresponding nodes in the clone, and establish the same connections between nodes in the cloned graph as in the original one. To avoid duplication of nodes, you also need to keep track of visited nodes to ensure that the cloning process is efficient and that nodes are not duplicated in the clone.\n",
-    "\n",
-    "The problem can be challenging because you need to create a deep copy of the graph while handling cyclic dependencies between nodes and ensuring that the structure and relationships of the original graph are preserved in the clone. The solution requires recursion or a data structure to maintain the mapping between original nodes and their clones to achieve an efficient and accurate cloning process.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a Node.\n",
-    "class Node:\n",
-    "    def __init__(self, val=0, neighbors=None):\n",
-    "        self.val = val\n",
-    "        self.neighbors = neighbors if neighbors is not None else []\n",
-    "\n",
-    "class Solution:\n",
-    "    def cloneGraph(self, node: 'Node') -> 'Node':\n",
-    "        if not node:\n",
-    "            return None\n",
-    "\n",
-    "        visited = {}  # Dictionary to keep track of visited nodes\n",
-    "\n",
-    "        def dfs(original_node):\n",
-    "            if original_node in visited:\n",
-    "                return visited[original_node]\n",
-    "\n",
-    "            # Create a copy of the original node\n",
-    "            copy_node = Node(original_node.val)\n",
-    "            visited[original_node] = copy_node\n",
-    "\n",
-    "            # Recursively clone the neighbors of the original node\n",
-    "            for neighbor in original_node.neighbors:\n",
-    "                copy_node.neighbors.append(dfs(neighbor))\n",
-    "\n",
-    "            return copy_node\n",
-    "\n",
-    "        return dfs(node)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The given code defines a Python class, `Node`, and a solution class, `Solution`, to clone a connected undirected graph represented by nodes and their neighbors. Here's an explanation of the code:\n",
-    "\n",
-    "1. The `Node` class defines a node in the graph. Each node has a value (`val`) which is an integer and a list of neighbors (`neighbors`) that are other nodes in the graph.\n",
-    "\n",
-    "2. The `Solution` class contains the method `cloneGraph`, which takes a reference to the first node of the graph (`node`) as input and returns a deep copy of the entire graph, preserving the structure and relationships between nodes.\n",
-    "\n",
-    "3. Inside the `cloneGraph` method:\n",
-    "   - It first checks if the input `node` is `None`. If it's `None`, it means there is no graph to clone, and the method returns `None`.\n",
-    "\n",
-    "   - It initializes a `visited` dictionary to keep track of visited nodes. This dictionary is used to ensure that each node is cloned only once to avoid duplication.\n",
-    "\n",
-    "   - The `dfs` (depth-first search) function is defined within the `cloneGraph` method. It takes an original node as input and returns its corresponding clone. This function performs the actual cloning of the graph.\n",
-    "\n",
-    "   - Inside the `dfs` function:\n",
-    "     - It checks if the original node has been visited before by looking it up in the `visited` dictionary. If it has been visited, it returns the corresponding clone from the `visited` dictionary.\n",
-    "\n",
-    "     - If the original node is not visited, it creates a new node, `copy_node`, with the same value as the original node. This is the first step in creating the clone of the original node.\n",
-    "\n",
-    "     - It then recursively clones the neighbors of the original node by iterating through the `neighbors` list of the original node and appending the corresponding cloned neighbors to the `neighbors` list of the `copy_node`. This step ensures that the relationships between nodes are preserved in the clone.\n",
-    "\n",
-    "     - The `copy_node` is added to the `visited` dictionary with the original node as the key and the clone as the value.\n",
-    "\n",
-    "     - Finally, the `copy_node` is returned as the result of the DFS for the original node.\n",
-    "\n",
-    "4. The `cloneGraph` method returns the result of the DFS on the input `node`, which is the clone of the entire graph with its structure and relationships intact.\n",
-    "\n",
-    "In summary, the code uses depth-first search (DFS) to traverse the original graph, creating a clone for each node and its neighbors. It ensures that each node is cloned only once to avoid duplication and returns a reference to the cloned graph."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[1, 2, 4]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "def get_adjacency_list(node):\n",
-    "    visited = {}  # Dictionary to keep track of visited nodes\n",
-    "\n",
-    "    def dfs(node):\n",
-    "        if not node:\n",
-    "            return []\n",
-    "\n",
-    "        if node in visited:\n",
-    "            return visited[node]\n",
-    "\n",
-    "        neighbors = [n.val for n in node.neighbors]\n",
-    "        visited[node] = neighbors\n",
-    "\n",
-    "        for neighbor in node.neighbors:\n",
-    "            dfs(neighbor)\n",
-    "\n",
-    "    dfs(node)\n",
-    "    result = []\n",
-    "\n",
-    "    def get_node_and_neighbors(node):\n",
-    "        result.append([node.val] + visited[node])\n",
-    "\n",
-    "    get_node_and_neighbors(node)\n",
-    "    return result\n",
-    "\n",
-    "\n",
-    "# Testing the code with the provided examples\n",
-    "adjList1 = [[2, 4], [1, 3], [2, 4], [1, 3]]\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Create the graph from adjList\n",
-    "nodes = [Node(i) for i in range(1, len(adjList1) + 1)]\n",
-    "for i, adj in enumerate(adjList1):\n",
-    "    nodes[i].neighbors = [nodes[j - 1] for j in adj]\n",
-    "\n",
-    "# Clone the graph and get its adjacency list\n",
-    "cloned_node1 = solution.cloneGraph(nodes[0])\n",
-    "cloned_adjList1 = get_adjacency_list(cloned_node1)\n",
-    "\n",
-    "# Print the cloned graph as an adjacency list\n",
-    "print(cloned_adjList1)\n",
-    "\n",
-    "# Output should match adjList1\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[1]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "# Input: adjList = [[]]\n",
-    "\n",
-    "adjList2 = [[]]\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Create the graph from adjList\n",
-    "node1 = Node(1)\n",
-    "\n",
-    "# Clone the graph and get its adjacency list\n",
-    "cloned_node1 = solution.cloneGraph(node1)\n",
-    "cloned_adjList2 = get_adjacency_list(cloned_node1)\n",
-    "\n",
-    "# Print the cloned graph as an adjacency list\n",
-    "print(cloned_adjList2)\n",
-    "\n",
-    "# Output should be [[]], which represents an empty graph"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for cloning a graph using depth-first search (DFS).\n",
-    "\n",
-    "Time Complexity:\n",
-    "1. Creating the graph: The time complexity to create the original graph from `adjList` is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. We iterate through each node and its neighbors in `adjList`.\n",
-    "\n",
-    "2. Cloning the graph (DFS): The depth-first search (DFS) is used to traverse the original graph and create a clone. The time complexity of DFS is O(V + E), as we visit each node and each edge once.\n",
-    "\n",
-    "Overall, the time complexity is O(V + E) for both creating the original graph and cloning it.\n",
-    "\n",
-    "Space Complexity:\n",
-    "1. Storage for the original graph: We need to store the original graph, which includes all nodes and their adjacency lists. In the worst case, this requires O(V + E) space, where V is the number of nodes, and E is the number of edges.\n",
-    "\n",
-    "2. Storage for the visited dictionary: The `visited` dictionary stores mappings from original nodes to their corresponding clones. In the worst case, this requires O(V) space because each node is visited once.\n",
-    "\n",
-    "3. Recursive stack for DFS: The space used for the recursive call stack during DFS can be up to O(V) in the worst case, where V is the number of nodes.\n",
-    "\n",
-    "Overall, the space complexity is O(V + E) for storing the original graph, and O(V) for auxiliary data structures, including the `visited` dictionary and the DFS call stack. Therefore, the overall space complexity is O(V + E).\n",
-    "\n",
-    "In summary, the time and space complexities of the provided code for cloning a graph are both O(V + E), where V is the number of nodes, and E is the number of edges in the graph."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Cyclic Graph**: Design a graph with cycles (nodes connected in a cycle). Test if the code can correctly clone graphs with cycles and doesn't fall into an infinite loop during traversal.\n",
-    "\n",
-    "\n",
-    "2. **Partial Graph Cloning**: Modify the code to clone only a subset of the graph starting from a given reference node, rather than the entire graph. Implement this as an extension to the problem."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb b/docs/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb
deleted file mode 100644
index 0dd53e9..0000000
--- a/docs/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 178. Graph Valid Tree\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Graph Valid Tree problem on LintCode, click here!](https://www.lintcode.com/problem/178/)\n",
-    "\n",
-    "---\n",
-    "Given `n` nodes labeled from `0` to `n - 1` and a list of `undirected` edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- You can assume that no duplicate edges will appear in edges. Since all edges are `undirected`, `[0, 1]` is the same as `[1, 0]` and thus will not appear together in edges."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to determine whether a given set of undirected edges can form a valid tree. In this context, a \"tree\" is a specific type of graph with the following properties:\n",
-    "\n",
-    "1. It is a connected graph, meaning there is a path between any pair of nodes in the graph.\n",
-    "2. It has no cycles, which means there are no closed loops or circuits in the graph.\n",
-    "3. It has exactly `n - 1` edges, where `n` is the number of nodes in the tree.\n",
-    "\n",
-    "The problem is typically given with two pieces of information:\n",
-    "\n",
-    "1. The number of nodes (`n`), which are usually labeled from 0 to `n-1`.\n",
-    "2. A list of undirected edges, where each edge is represented as a pair of nodes.\n",
-    "\n",
-    "The task is to write a function or algorithm to determine whether the provided set of edges can be arranged to form a valid tree based on the properties mentioned above.\n",
-    "\n",
-    "To solve this problem, you need to check the following conditions:\n",
-    "\n",
-    "1. There should be exactly `n - 1` edges to ensure that the graph is connected and tree-like.\n",
-    "2. There should be no cycles in the graph. If there are any cycles, the set of edges cannot form a tree.\n",
-    "3. The graph should be connected, which means you can reach any node from any other node through a series of edges.\n",
-    "\n",
-    "The solution typically involves using graph traversal algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the graph and check these conditions. If all conditions are met, the given edges form a valid tree; otherwise, they do not.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def validTree(n, edges):\n",
-    "    if len(edges) != n - 1:\n",
-    "        return False  # A tree with n nodes must have n-1 edges.\n",
-    "\n",
-    "    # Create an adjacency list to represent the graph.\n",
-    "    adj_list = {i: [] for i in range(n)}\n",
-    "    for u, v in edges:\n",
-    "        adj_list[u].append(v)\n",
-    "        adj_list[v].append(u)\n",
-    "\n",
-    "    visited = set()\n",
-    "\n",
-    "    # Define a recursive DFS function.\n",
-    "    def dfs(node, parent):\n",
-    "        if node in visited:\n",
-    "            return False  # If we encounter a visited node, there's a cycle.\n",
-    "        visited.add(node)\n",
-    "\n",
-    "        for neighbor in adj_list[node]:\n",
-    "            if neighbor != parent:\n",
-    "                if not dfs(neighbor, node):\n",
-    "                    return False\n",
-    "\n",
-    "        return True\n",
-    "\n",
-    "    # Start DFS from any node (e.g., node 0).\n",
-    "    if not dfs(0, -1):\n",
-    "        return False  # If there's a cycle, it's not a valid tree.\n",
-    "\n",
-    "    return len(visited) == n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The `validTree` function checks if a given set of undirected edges forms a valid tree. A valid tree must satisfy these conditions:\n",
-    "\n",
-    "1. It has exactly `n - 1` edges, where `n` is the number of nodes.\n",
-    "2. There are no cycles in the graph.\n",
-    "3. The graph is connected, meaning all nodes are reachable from any starting node.\n",
-    "\n",
-    "The code accomplishes this as follows:\n",
-    "\n",
-    "- It first checks if the number of edges is equal to `n - 1`. If not, it immediately returns `False` since a tree with `n` nodes must have `n - 1` edges to be connected.\n",
-    "\n",
-    "- It then constructs an adjacency list (`adj_list`) to represent the graph efficiently. This data structure stores the neighbors of each node.\n",
-    "\n",
-    "- The code uses a `visited` set to keep track of nodes visited during depth-first search (DFS).\n",
-    "\n",
-    "- The `dfs` function is a recursive function that checks for cycles in the graph. It does this by visiting nodes and marking them as visited. If it encounters a node that has already been visited (indicating a cycle), it returns `False`.\n",
-    "\n",
-    "- The DFS function explores the neighbors of each node, avoiding revisiting the parent node to prevent cycles.\n",
-    "\n",
-    "- After the DFS is complete, the code checks if the number of visited nodes is equal to `n`, ensuring the graph is connected.\n",
-    "\n",
-    "- Finally, if all conditions are met, the code returns `True`, indicating that the given edges form a valid tree. Otherwise, it returns `False`.\n",
-    "\n",
-    "The code provides examples to demonstrate how to use the `validTree` function to check whether a set of edges forms a valid tree or not."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "n1 = 5\n",
-    "edges1 = [[0, 1], [0, 2], [0, 3], [1, 4]]\n",
-    "print(validTree(n1, edges1))  # Output: true"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "n2 = 5\n",
-    "edges2 = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]\n",
-    "print(validTree(n2, edges2))  # Output: false"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexities of the `validTree` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. Constructing the adjacency list (`adj_list`) takes O(E) time, where E is the number of edges in the input graph (|edges|).\n",
-    "\n",
-    "2. The DFS function, when considering all nodes, has a time complexity of O(V + E), where V is the number of nodes in the graph. This is because, in the worst case, the DFS function visits each node and each edge once.\n",
-    "\n",
-    "3. Checking if the number of visited nodes is equal to `n` takes O(1) time.\n",
-    "\n",
-    "Overall, the time complexity of the `validTree` function is O(E + V) in the worst case.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The adjacency list (`adj_list`) consumes additional space to store the graph structure, taking O(V + E) space. This is because it stores information about each node and its adjacent nodes.\n",
-    "\n",
-    "2. The `visited` set keeps track of visited nodes, and its space complexity is O(V) in the worst case, as there can be at most V unique nodes in the set.\n",
-    "\n",
-    "3. The depth of the function call stack during DFS can be at most V, which contributes O(V) space to the space complexity.\n",
-    "\n",
-    "Overall, the space complexity of the `validTree` function is O(V + E) due to the adjacency list and O(V) due to the `visited` set and DFS call stack.\n",
-    "\n",
-    "In summary, the time complexity is O(E + V), and the space complexity is O(V + E). These complexities are based on the worst-case scenario, where all nodes and edges are considered. In practical cases, the actual complexities may be smaller, depending on the specific structure of the graph."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Minimum Spanning Tree (MST)**: Implement an algorithm to find the minimum spanning tree of a connected graph using either Kruskal's or Prim's algorithm. Compare the MST to the original graph to determine if the original graph is a tree. This exercise reinforces the concept of tree properties.\n",
-    "\n",
-    "2. **Detecting Cycles**: Modify the code to not only check if the input forms a tree but also to identify and print out any cycles present in the graph if it's not a tree. This exercise enhances your ability to detect and visualize cycles in a graph."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb b/docs/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb
deleted file mode 100644
index 5a878a4..0000000
--- a/docs/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb	
+++ /dev/null
@@ -1,236 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 200. Number of Islands\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Number of Islands problem on LeetCode, click here!](https://leetcode.com/problems/number-of-islands/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given an `m x n` 2D binary grid `grid` which represents a map of `'1'`s (land) and `'0'`s (water), return *the number of islands*. \n",
-    "\n",
-    "An **island** is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == grid.length`\n",
-    "- `n == grid[i].length`\n",
-    "- `1 <= m, n <= 300`\n",
-    "- `grid[i][j]` is `'0'` or `'1'`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is about counting the number of islands in a 2D binary grid. In this grid, '1' represents land, and '0' represents water. An island is defined as a group of '1' cells that are adjacent to each other either horizontally or vertically. It's important to note that diagonal connections do not count. Additionally, it's assumed that all four edges of the grid are surrounded by water, meaning the grid is finite and doesn't \"wrap around.\"\n",
-    "\n",
-    "The objective is to determine how many distinct islands exist in the grid. To clarify, an island consists of connected '1' cells, and these connections can only occur horizontally or vertically. If there are no '1' cells in the grid, the number of islands would be zero.\n",
-    "\n",
-    "The problem often involves using graph traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to identify and count these islands. You start at one '1' cell and explore its adjacent cells to determine the extent of the island. Once you've visited all the '1' cells in an island, you move on to the next unvisited '1' cell and repeat the process until you've counted all the islands in the grid.\n",
-    "\n",
-    "In essence, the problem asks you to analyze the grid to find groups of connected '1' cells and count them as individual islands. The goal is to return the total count of islands in the given grid.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def numIslands(self, grid: List[List[str]]) -> int:\n",
-    "        if not grid:\n",
-    "            return 0\n",
-    "\n",
-    "        m, n = len(grid), len(grid[0])  # Get the dimensions of the grid\n",
-    "        num_islands = 0  # Initialize the count of islands\n",
-    "\n",
-    "        def dfs(row, col):\n",
-    "            if row < 0 or row >= m or col < 0 or col >= n or grid[row][col] == '0':\n",
-    "                return\n",
-    "\n",
-    "            grid[row][col] = '0'  # Mark the current cell as visited by changing it to '0'\n",
-    "            directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Define four possible directions to move\n",
-    "\n",
-    "            for dr, dc in directions:\n",
-    "                # Explore adjacent cells\n",
-    "                dfs(row + dr, col + dc)\n",
-    "\n",
-    "        for i in range(m):\n",
-    "            for j in range(n):\n",
-    "                if grid[i][j] == '1':\n",
-    "                    num_islands += 1  # Increment the island count\n",
-    "                    dfs(i, j)  # Start a DFS from the current '1' cell to explore the entire island\n",
-    "\n",
-    "        return num_islands\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code defines a Python class called `Solution` with a method called `numIslands`. This class is designed to count the number of islands in a 2D binary grid.\n",
-    "\n",
-    "The `numIslands` method takes a 2D grid as input, where '1' represents land, and '0' represents water. It uses a depth-first search (DFS) approach to identify and count islands. The key steps in the code include:\n",
-    "\n",
-    "1. Checking if the input grid is empty, and if so, returning 0 (indicating there are no islands).\n",
-    "\n",
-    "2. Determining the dimensions of the grid (number of rows and columns).\n",
-    "\n",
-    "3. Initializing a variable `num_islands` to keep track of the count of islands.\n",
-    "\n",
-    "4. Defining a nested function `dfs` to perform depth-first search. This function recursively explores adjacent land cells connected to the current cell and marks them as visited by changing '1' to '0'.\n",
-    "\n",
-    "5. In the `dfs` function, it checks for boundary conditions and whether the current cell is '0'. If these conditions are met, it returns without further exploration.\n",
-    "\n",
-    "6. The `dfs` function explores neighboring cells in all four directions (up, down, left, right) using a loop.\n",
-    "\n",
-    "7. The main loop iterates through each cell of the grid. When a '1' cell is encountered, it increments the `num_islands` count and starts a DFS from that cell to explore the entire island.\n",
-    "\n",
-    "8. Finally, the method returns the total count of islands found in the grid.\n",
-    "\n",
-    "This code encapsulates the solution in a class, allowing you to create an instance of the `Solution` class and call the `numIslands` method on it to count the number of islands in a given grid."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 1 Output: 1\n"
-     ]
-    }
-   ],
-   "source": [
-    "grid1 = [\n",
-    "  [\"1\", \"1\", \"1\", \"1\", \"0\"],\n",
-    "  [\"1\", \"1\", \"0\", \"1\", \"0\"],\n",
-    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-    "  [\"0\", \"0\", \"0\", \"0\", \"0\"]\n",
-    "]\n",
-    "\n",
-    "solution = Solution()\n",
-    "result1 = solution.numIslands(grid1)\n",
-    "print(\"Example 1 Output:\", result1)  # Expected output: 1\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 2 Output: 3\n"
-     ]
-    }
-   ],
-   "source": [
-    "grid2 = [\n",
-    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-    "  [\"0\", \"0\", \"1\", \"0\", \"0\"],\n",
-    "  [\"0\", \"0\", \"0\", \"1\", \"1\"]\n",
-    "]\n",
-    "\n",
-    "result2 = solution.numIslands(grid2)\n",
-    "print(\"Example 2 Output:\", result2)  # Expected output: 3\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "Time Complexity:\n",
-    "- The code uses a depth-first search (DFS) to explore the grid and count the number of islands.\n",
-    "- In the worst case, it visits every cell in the grid exactly once.\n",
-    "- The DFS function explores neighboring cells in four directions (up, down, left, right), so it has a time complexity of $O(4^{max(m, n)})$, where m is the number of rows and n is the number of columns.\n",
-    "- Therefore, the overall time complexity of the code is O(m * n), where m is the number of rows, and n is the number of columns.\n",
-    "\n",
-    "Space Complexity:\n",
-    "- The space complexity is determined by the recursive calls and the depth of the DFS stack.\n",
-    "- In the worst case, when the entire grid consists of '1's, the depth of the DFS recursion can be as deep as max(m, n).\n",
-    "- Therefore, the space complexity of the code is O(max(m, n)), representing the maximum depth of the DFS stack.\n",
-    "\n",
-    "In practical terms, for reasonably sized grids, the time and space complexity is linear, and the code should perform well. However, it's important to keep in mind that the worst-case time complexity is O(m * n), and the space complexity can be O(max(m, n))."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Identify Largest Island:** In addition to counting the number of islands, modify the code to identify the largest island in the grid and return its size.\n",
-    "\n",
-    "2. **Optimize for Space:** Modify the solution to reduce its space complexity to O(1) without modifying the input grid. This is a more memory-efficient version of the problem."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb b/docs/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb
deleted file mode 100644
index 4176cab..0000000
--- a/docs/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb	
+++ /dev/null
@@ -1,243 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 207. Course Schedule\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Course Schedule problem on LeetCode, click here!](https://leetcode.com/problems/course-schedule/)\n",
-    "\n",
-    "---\n",
-    "There are a total of `numCourses` courses to take, labeled from 0 to `numCourses - 1`. You are given an array `prerequisites` where `prerequisites[i] = [ai, bi]` indicates that you **must** take course `bi` first if you want to take course `ai`. \n",
-    "\n",
-    "- For example, the pair `[0, 1]`, indicates that to take course `0` you have to first take course `1`.\n",
-    "\n",
-    "Return `true` if you can finish all courses. Otherwise, return `false`.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `numCourses` <= 2000\n",
-    "- 0 <= `prerequisites.length` <= 5000\n",
-    "- `prerequisites[i].length` == 2\n",
-    "- 0 <= `ai`, `bi` < `numCourses`\n",
-    "- All the pairs `prerequisites[i]` are **unique**."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to determine whether it's possible to complete a set of courses given a list of prerequisites. You are given the following:\n",
-    "\n",
-    "- `numCourses`: The total number of courses to be taken, labeled from 0 to `numCourses - 1`.\n",
-    "- `prerequisites`: A list of prerequisite courses, where each element `prerequisites[i]` is a pair `[ai, bi]`, indicating that you must take course `bi` before you can take course `ai`.\n",
-    "\n",
-    "The goal is to check whether it's possible to complete all the courses while respecting the prerequisite requirements. In other words, you need to determine if there are any circular dependencies in the course prerequisites that would prevent you from taking all the courses.\n",
-    "\n",
-    "For example:\n",
-    "\n",
-    "- If `numCourses` is 2, and `prerequisites` is `[[1, 0]]`, it means you have two courses, and you must finish course 0 before you can take course 1. In this case, it's possible to complete all courses, so the function should return `True`.\n",
-    "\n",
-    "- If `numCourses` is 2, and `prerequisites` is `[[1, 0], [0, 1]]`, it means you have two courses, but there's a circular dependency where course 1 requires course 0 and course 0 requires course 1. In this case, it's impossible to complete all courses, so the function should return `False`.\n",
-    "\n",
-    "The problem is essentially about checking for the presence of cycles in a directed graph, where the courses are nodes, and the prerequisites represent directed edges between them. If there are no cycles, it's possible to complete all courses; otherwise, it's not possible.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from collections import defaultdict, deque\n",
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:\n",
-    "        # Create a graph using an adjacency list to represent the courses and their prerequisites.\n",
-    "        graph = defaultdict(list)  # Initialize an empty adjacency list.\n",
-    "        in_degree = [0] * numCourses  # Initialize an array to store in-degrees of courses.\n",
-    "\n",
-    "        # Populate the graph and calculate in-degrees.\n",
-    "        for course, prerequisite in prerequisites:\n",
-    "            graph[prerequisite].append(course)  # Add the course as a neighbor to its prerequisite.\n",
-    "            in_degree[course] += 1  # Increment the in-degree of the course.\n",
-    "\n",
-    "        # Initialize a queue with courses that have no prerequisites.\n",
-    "        queue = deque()\n",
-    "        for course in range(numCourses):\n",
-    "            if in_degree[course] == 0:\n",
-    "                queue.append(course)\n",
-    "\n",
-    "        # Perform topological sorting.\n",
-    "        while queue:\n",
-    "            course = queue.popleft()  # Take a course with no prerequisites.\n",
-    "            numCourses -= 1  # Decrement the count of remaining courses.\n",
-    "\n",
-    "            for neighbor in graph[course]:\n",
-    "                in_degree[neighbor] -= 1  # Remove the prerequisite relationship.\n",
-    "                if in_degree[neighbor] == 0:\n",
-    "                    queue.append(neighbor)  # If no more prerequisites, add to the queue.\n",
-    "\n",
-    "        # If all courses were successfully taken (numCourses becomes 0), return True.\n",
-    "        return numCourses == 0"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "This Python code is designed to determine whether it is possible to complete all the required courses given a set of prerequisites. It uses a topological sorting algorithm to accomplish this. Here's a brief explanation of the code:\n",
-    "\n",
-    "1. The code defines a `Solution` class with a `canFinish` method that takes two parameters: `numCourses` (the total number of courses) and `prerequisites` (a list of pairs indicating which course must be taken before another).\n",
-    "\n",
-    "2. It initializes an empty graph as an adjacency list using a `defaultdict` to represent the courses and their prerequisites. It also initializes a list called `in_degree` to keep track of the in-degrees of courses, which is used in the topological sorting process.\n",
-    "\n",
-    "3. The code then populates the graph by iterating through the `prerequisites` list. For each pair `(ai, bi)`, it adds course `ai` as a neighbor to its prerequisite course `bi` in the graph and increments the in-degree of course `ai`.\n",
-    "\n",
-    "4. Next, it initializes a queue with courses that have no prerequisites. It does this by iterating through all the courses and adding those with an in-degree of 0 to the queue.\n",
-    "\n",
-    "5. The code performs a topological sorting of the courses using a while loop. In each iteration of the loop, it dequeues a course from the front of the queue (a course with no prerequisites).\n",
-    "\n",
-    "6. For each neighbor (course that depends on the dequeued course), it decrements the in-degree of the neighbor. If the neighbor's in-degree becomes 0, it means all of its prerequisites have been taken, so the neighbor is added to the queue.\n",
-    "\n",
-    "7. The loop continues until there are no more courses with no prerequisites to dequeue, and the in-degrees are updated accordingly.\n",
-    "\n",
-    "8. After the loop, if all courses were successfully taken (i.e., `numCourses` becomes 0), it means that there were no circular dependencies, and it's possible to finish all courses. In this case, the method returns `True`. Otherwise, if there are remaining courses (indicating a circular dependency), it returns `False`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "numCourses1 = 2\n",
-    "prerequisites1 = [[1, 0]]\n",
-    "sol = Solution()\n",
-    "print(sol.canFinish(numCourses1, prerequisites1))  # Output: True"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "numCourses2 = 2\n",
-    "prerequisites2 = [[1, 0], [0, 1]]\n",
-    "print(sol.canFinish(numCourses2, prerequisites2))  # Output: False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `canFinish` method:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. Constructing the graph: In the first loop where we iterate through the `prerequisites` list, we populate the graph with prerequisites. This loop has a time complexity of O(E), where E is the number of prerequisites (edges).\n",
-    "\n",
-    "2. Initializing in-degrees and finding courses with no prerequisites: In the worst case, we iterate through all `numCourses` courses to initialize in-degrees and find the courses with no prerequisites. This has a time complexity of O(V), where V is the number of courses (vertices).\n",
-    "\n",
-    "3. Topological Sorting: In the while loop, we perform topological sorting, processing each course once and its outgoing edges. In the worst case, each course is processed once. This has a time complexity of O(V + E).\n",
-    "\n",
-    "The overall time complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. Graph Representation: The space complexity for storing the graph as an adjacency list is O(E), where E is the number of prerequisites.\n",
-    "\n",
-    "2. In-Degree Array: The space complexity for storing the in-degrees of courses is O(V), where V is the number of courses.\n",
-    "\n",
-    "3. Queue: The space complexity for the queue is O(V), as it may contain all the courses.\n",
-    "\n",
-    "The overall space complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-    "\n",
-    "In most practical cases, the number of prerequisites (E) is much smaller than the number of courses (V), so the time and space complexity can often be considered as O(V)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimizing Course Scheduling**: Given a list of courses with their durations and prerequisites, find the most efficient way to schedule these courses to minimize the time it takes to complete all of them.\n",
-    "\n",
-    "2. **Detecting Cycles in Course Dependencies**: Modify the original problem to not just determine if you can finish all courses but also to identify the specific courses that form a cycle in the prerequisite dependencies."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb b/docs/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb
deleted file mode 100644
index a2789c4..0000000
--- a/docs/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb	
+++ /dev/null
@@ -1,230 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 3651. Number of Connected Components in an Undirected Graph\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Number of Connected Components in an Undirected Graph problem on LintCode, click here!](https://www.lintcode.com/problem/3651/)\n",
-    "\n",
-    "---\n",
-    "In this problem, there is an undirected graph with `n` nodes. There is also an `edges` array. Where `edges[i] = [a, b]` means that there is an edge between *node a* and *node b* in the graph.\n",
-    "\n",
-    "You need to return *the number of connected components in that graph*."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand is to determine the number of connected components in an undirected graph. Here's a more detailed explanation of the problem:\n",
-    "\n",
-    "- You are given an undirected graph, which consists of nodes and edges. The graph may have multiple distinct connected components.\n",
-    "\n",
-    "- A connected component is a group of nodes within the graph where each node can be reached from every other node in the same component. In other words, there is a path between any two nodes in the same connected component, but there are no paths connecting nodes from different components.\n",
-    "\n",
-    "- Your task is to write a function that takes as input the number of nodes `n` and a list of edges `edges`, where each edge is represented as a pair of nodes `[a, b]` indicating an edge between node `a` and node `b`.\n",
-    "\n",
-    "- The goal is to determine the number of distinct connected components in the graph.\n",
-    "\n",
-    "- You need to implement an algorithm that can identify these connected components and count how many there are in the given graph.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def countConnectedComponents(n, edges):\n",
-    "    # Define a helper function for depth-first search (DFS).\n",
-    "    def dfs(node):\n",
-    "        if not visited[node]:  # If the node has not been visited yet:\n",
-    "            visited[node] = True  # Mark it as visited.\n",
-    "            for neighbor in graph[node]:  # Explore all neighbors of the current node.\n",
-    "                dfs(neighbor)  # Recursively call DFS on the neighbor.\n",
-    "\n",
-    "    # Create an empty adjacency list to represent the graph.\n",
-    "    graph = [[] for _ in range(n)]\n",
-    "\n",
-    "    # Initialize a boolean list to keep track of visited nodes.\n",
-    "    visited = [False] * n\n",
-    "\n",
-    "    # Populate the adjacency list with edges from the input.\n",
-    "    for a, b in edges:\n",
-    "        graph[a].append(b)\n",
-    "        graph[b].append(a)  # Since the graph is undirected, we add edges in both directions.\n",
-    "\n",
-    "    # Initialize a counter to keep track of the number of connected components.\n",
-    "    count = 0\n",
-    "\n",
-    "    # Iterate through all nodes in the graph.\n",
-    "    for i in range(n):\n",
-    "        if not visited[i]:  # If the node has not been visited:\n",
-    "            count += 1  # Increment the count since we've found a new connected component.\n",
-    "            dfs(i)  # Start a depth-first search from the unvisited node to explore the component.\n",
-    "\n",
-    "    return count"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided Python code calculates the number of connected components in an undirected graph. Here's a high-level explanation of how the code works:\n",
-    "\n",
-    "1. The `countConnectedComponents` function takes two arguments: `n` (the number of nodes in the graph) and `edges` (a list of edge pairs representing connections between nodes).\n",
-    "\n",
-    "2. Inside the function, there is a helper function called `dfs` (depth-first search), which is used to explore and mark nodes as visited.\n",
-    "\n",
-    "3. An empty adjacency list called `graph` is created to represent the graph. The `graph` is a list of lists where each element represents a node, and the list associated with each node contains its neighboring nodes.\n",
-    "\n",
-    "4. Another list called `visited` is created to keep track of visited nodes. Initially, all nodes are marked as not visited (`False`).\n",
-    "\n",
-    "5. The code populates the `graph` by iterating through the `edges` list and adding each edge to the adjacency list. Since the graph is undirected, both directions of each edge are added.\n",
-    "\n",
-    "6. A counter variable called `count` is initialized to keep track of the number of connected components.\n",
-    "\n",
-    "7. The code iterates through all nodes from `0` to `n-1`. For each node, it checks whether it has been visited yet. If it hasn't been visited, it means a new connected component is found. The counter `count` is incremented, and a depth-first search (DFS) is initiated from this unvisited node to explore the connected component.\n",
-    "\n",
-    "8. The `dfs` function recursively visits all nodes in the connected component, marking them as visited. It does this by checking each neighbor of the current node and recursively calling `dfs` on unvisited neighbors.\n",
-    "\n",
-    "9. After the loop through all nodes, the `count` variable contains the total number of connected components in the graph.\n",
-    "\n",
-    "10. The function returns the value of `count`, which represents the number of connected components.\n",
-    "\n",
-    "11. Two examples are provided at the end to demonstrate how to use the function with different inputs."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "n1 = 3\n",
-    "edges1 = [[0, 1], [0, 2]]\n",
-    "print(countConnectedComponents(n1, edges1))  # Output: 1"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "n2 = 6\n",
-    "edges2 = [[0, 1], [1, 2], [2, 3], [4, 5]]\n",
-    "print(countConnectedComponents(n2, edges2))  # Output: 2"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the code for counting the number of connected components in an undirected graph using depth-first search (DFS).\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The code iterates through all nodes in the graph (from 0 to `n-1`).\n",
-    "- For each unvisited node, it initiates a depth-first search (DFS) from that node to explore the connected component.\n",
-    "- In the worst case, every node may belong to a separate connected component, and you perform a DFS for each node.\n",
-    "- The time complexity of the DFS for a single connected component is O(V + E), where V is the number of nodes in the component, and E is the number of edges.\n",
-    "- In the worst case, you perform O(n) DFS calls, each with its own O(V + E) complexity.\n",
-    "\n",
-    "So, the overall time complexity of the code is O(n * (V + E)). In the worst case, if the graph is connected (a single connected component), this simplifies to O(n * (n + m)), where m is the number of edges.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The main data structures used are the `graph` (adjacency list), `visited` (boolean array), and the recursive call stack for DFS.\n",
-    "- The `graph` has a space complexity of O(n + m), where n is the number of nodes, and m is the number of edges.\n",
-    "- The `visited` boolean array also has a space complexity of O(n) to keep track of whether nodes have been visited.\n",
-    "- The depth of the recursive call stack for DFS can go up to O(n) in the worst case, when all nodes are part of a single connected component.\n",
-    "\n",
-    "So, the overall space complexity of the code is O(n + m) for data structures and O(n) for the call stack. In the worst case, this simplifies to O(n + m).\n",
-    "\n",
-    "In summary, the time complexity is O(n * (V + E)), and the space complexity is O(n + m)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Disconnected Components with Multiple Queries:** You are given an undirected graph with `n` nodes and `m` edges. Initially, the graph is disconnected, and you need to process `q` queries. Each query consists of adding an edge to connect two nodes. After each query, you need to output the number of connected components in the updated graph.\n",
-    "\n",
-    "2. **Maximum Size of Connected Components:** Given an undirected graph with `n` nodes and `m` edges, find the size of the largest connected component in the graph. Additionally, identify the nodes that belong to this largest component. You can assume there is at least one connected component in the graph.\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb b/docs/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb
deleted file mode 100644
index c685b23..0000000
--- a/docs/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb	
+++ /dev/null
@@ -1,255 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 417. Pacific Atlantic Water Flow\n",
-    "\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Pacific Atlantic Water Flow problem on LeetCode, click here!](https://leetcode.com/problems/pacific-atlantic-water-flow/)\n",
-    "\n",
-    "---\n",
-    "There is an `m x n` rectangular island that borders both the **Pacific Ocean** and **Atlantic Ocean**. The **Pacific Ocea**n touches the island's left and top edges, and the **Atlantic Ocean** touches the island's right and bottom edges.\n",
-    "\n",
-    "The island is partitioned into a grid of square cells. You are given an `m x n` integer matrix `heights` where `heights[r][c]` represents the **height above sea level** of the cell at coordinate `(r, c)`.\n",
-    "\n",
-    "The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is **less than or equal to** the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.\n",
-    "\n",
-    "Return *a **2D list** of grid coordinates `result` where $result[i] = [r_i, c_i]$ denotes that rain water can flow from cell $(r_i, c_i)$ to **both** the Pacific and Atlantic oceans*.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == heights.length`\n",
-    "- `n == heights[r].length`\n",
-    "- `1 <= m, n <= 200`\n",
-    "- 0 <= `heights[r][c]` <= $10^5$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand involves a scenario with a rectangular island surrounded by both the Pacific Ocean and the Atlantic Ocean. The Pacific Ocean is in contact with the island's left and top edges, while the Atlantic Ocean is in contact with the island's right and bottom edges. The goal is to determine which cells on the island can allow rainwater to flow to both the Pacific and Atlantic Oceans.\n",
-    "\n",
-    "Here are the key details and constraints of the problem:\n",
-    "\n",
-    "1. The island is divided into a grid of square cells.\n",
-    "\n",
-    "2. Each cell is represented by an integer value in a matrix `heights`. The value at `heights[r][c]` represents the height above sea level of the cell at coordinates `(r, c)` on the island.\n",
-    "\n",
-    "3. Rainwater can flow from a cell to neighboring cells directly north, south, east, or west only if the neighboring cell's height is less than or equal to the current cell's height.\n",
-    "\n",
-    "4. Water can flow from any cell adjacent to the ocean into the ocean. This means water can flow from cells on the edge of the island to the Pacific Ocean and the Atlantic Ocean.\n",
-    "\n",
-    "The task is to find and return a 2D list of grid coordinates where rainwater can flow from a cell to both the Pacific and Atlantic Oceans. In other words, you need to identify the cells that can send water to both the left and top edges (Pacific) and the right and bottom edges (Atlantic) of the island.\n",
-    "\n",
-    "The problem is solved by performing a depth-first search (DFS) starting from the ocean edges and marking the cells that can flow to each ocean. Then, the algorithm identifies cells that are reachable from both oceans and returns their coordinates as the result.\n",
-    "\n",
-    "The problem involves both traversal of the island and backtracking to find the solution, which makes it a classic graph traversal problem.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:\n",
-    "        # Check if the input matrix is empty\n",
-    "        if not heights:\n",
-    "            return []\n",
-    "        \n",
-    "        m, n = len(heights), len(heights[0])\n",
-    "        \n",
-    "        # Initialize boolean matrices to keep track of cells reachable from each ocean\n",
-    "        pacific_reachable = [[False] * n for _ in range(m)]\n",
-    "        atlantic_reachable = [[False] * n for _ in range(m)]\n",
-    "        \n",
-    "        # Depth-First Search (DFS) function to mark cells that can be reached from an ocean\n",
-    "        def dfs(x, y, reachable):\n",
-    "            if reachable[x][y]:\n",
-    "                return\n",
-    "            reachable[x][y] = True\n",
-    "            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:\n",
-    "                new_x, new_y = x + dx, y + dy\n",
-    "                if 0 <= new_x < m and 0 <= new_y < n and heights[new_x][new_y] >= heights[x][y]:\n",
-    "                    dfs(new_x, new_y, reachable)\n",
-    "        \n",
-    "        # Start DFS from the Pacific Ocean (left and top edges)\n",
-    "        for i in range(m):\n",
-    "            dfs(i, 0, pacific_reachable)\n",
-    "            dfs(i, n - 1, atlantic_reachable)\n",
-    "        \n",
-    "        for j in range(n):\n",
-    "            dfs(0, j, pacific_reachable)\n",
-    "            dfs(m - 1, j, atlantic_reachable)\n",
-    "        \n",
-    "        result = []\n",
-    "        \n",
-    "        # Find cells that can flow to both the Pacific and Atlantic oceans\n",
-    "        for i in range(m):\n",
-    "            for j in range(n):\n",
-    "                if pacific_reachable[i][j] and atlantic_reachable[i][j]:\n",
-    "                    result.append([i, j])\n",
-    "        \n",
-    "        return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a class called `Solution` with a method named `pacificAtlantic`. This method takes a 2D matrix `heights`, where each element represents the height above sea level of a cell on a rectangular island. The goal of the method is to find and return a list of grid coordinates (cells) from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-    "\n",
-    "The code uses a Depth-First Search (DFS) algorithm to traverse the island, starting from the ocean edges, and marks cells that are reachable from either the Pacific or Atlantic Ocean. It does this by creating two boolean matrices, `pacific_reachable` and `atlantic_reachable`, to keep track of cells that can be reached from each ocean.\n",
-    "\n",
-    "The code iterates through the entire island, initiating DFS searches from the edges of the island. If a cell is reachable from an ocean (either Pacific or Atlantic), it is marked as reachable in the corresponding boolean matrix. The DFS function ensures that the traversal follows the rule that water can flow to a neighboring cell if the neighboring cell's height is less than or equal to the current cell's height.\n",
-    "\n",
-    "After marking all reachable cells from both oceans, the code then looks for cells that are reachable from both the Pacific and Atlantic Oceans. If a cell satisfies this condition, it is added to the `result` list.\n",
-    "\n",
-    "Finally, the code returns the `result` list, which contains the grid coordinates of cells from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-    "\n",
-    "The code provides a solution to the problem of finding cells on the island where water can reach both oceans, demonstrating a traversal algorithm that explores the connectivity of cells on the island."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "heights1 = [\n",
-    "    [1,2,2,3,5],\n",
-    "    [3,2,3,4,4],\n",
-    "    [2,4,5,3,1],\n",
-    "    [6,7,1,4,5],\n",
-    "    [5,1,1,2,4]\n",
-    "]\n",
-    "\n",
-    "solution = Solution()\n",
-    "print(solution.pacificAtlantic(heights1)) # Expected Output: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[0, 0]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "heights2 = [[1]]\n",
-    "print(solution.pacificAtlantic(heights2)) # Expected Output: [[0, 0]]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Time Complexity:\n",
-    "The time complexity of this solution is O(m * n), where 'm' is the number of rows in the input matrix and 'n' is the number of columns. Here's a breakdown of the time complexity:\n",
-    "\n",
-    "1. Creating the `pacific_reachable` and `atlantic_reachable` matrices with dimensions m x n takes O(m * n) time.\n",
-    "\n",
-    "2. Initiating Depth-First Search (DFS) from the edges of the island takes O(m * n) time because it processes each cell once.\n",
-    "\n",
-    "3. The DFS function visits each cell at most once, and its time complexity is O(1) for each cell. The number of cells visited by the DFS in the worst case is m * n.\n",
-    "\n",
-    "4. The final step of identifying and appending cells to the result list takes O(m * n) time in the worst case.\n",
-    "\n",
-    "Overall, the time complexity is dominated by the DFS and is O(m * n).\n",
-    "\n",
-    "Space Complexity:\n",
-    "The space complexity of this solution is also O(m * n). Here's how the space is used:\n",
-    "\n",
-    "1. Two boolean matrices, `pacific_reachable` and `atlantic_reachable`, are created with dimensions m x n. Each matrix takes O(m * n) space, resulting in O(2 * m * n) space usage.\n",
-    "\n",
-    "2. The depth-first search (DFS) stack space during the recursive calls can go as deep as the diagonal of the grid, which is at most min(m, n). This additional space for the call stack is relatively small compared to the boolean matrices and can be considered O(min(m, n)).\n",
-    "\n",
-    "3. The `result` list stores the grid coordinates, and its size is determined by the number of cells that can flow to both oceans. In the worst case, this list can contain all m * n cells, so it takes O(m * n) space.\n",
-    "\n",
-    "The overall space complexity is the sum of these components, which is O(m * n) + O(min(m, n)) + O(m * n). In big O notation, we can simplify this to O(m * n)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the algorithm to find the maximum flow of water from any cell to both the Pacific and Atlantic Oceans. In other words, find the cell(s) that can send the most water to both oceans.\n",
-    "\n",
-    "2. **Efficiency Challenge:** Optimize the time and space complexity of the algorithm while maintaining correctness. Aim to reduce both time and space complexity as much as possible."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb b/docs/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb
deleted file mode 100644
index 19db846..0000000
--- a/docs/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb	
+++ /dev/null
@@ -1,250 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 892. Alien Dictionary\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Alien Dictionary problem on LintCode, click here!](https://www.lintcode.com/problem/892/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of **non-empty** words from the dictionary, where words are **sorted lexicographically by the rules of this new language**. Derive the order of letters in this language.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- You may assume all letters are in lowercase.\n",
-    "- The dictionary is invalid, if string a is prefix of string b and b is appear before a.\n",
-    "- If the order is invalid, return an empty string.\n",
-    "- There may be multiple valid order of letters, return the smallest in normal lexicographical order.\n",
-    "- The letters in **one** string are of the same rank by default and are sorted in Human dictionary order."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand involves determining the order of letters in a new alien language that uses the Latin alphabet. However, the order of letters in this alien language is unknown. To solve this problem, we are given a list of non-empty words from the dictionary, where the words are sorted lexicographically based on the rules of the alien language. The goal is to derive the correct order of letters in this new language.\n",
-    "\n",
-    "Here's a more detailed explanation of the problem:\n",
-    "\n",
-    "1. The alphabet: The new alien language uses the Latin alphabet, which consists of lowercase letters.\n",
-    "\n",
-    "2. Dictionary: We are provided with a list of non-empty words. These words are sorted in lexicographical order based on the rules of the alien language. The order of letters in the words reflects the correct order in the new language.\n",
-    "\n",
-    "3. Invalid order: The problem specifies that the dictionary is considered invalid if there exists a situation where one string 'a' is a prefix of another string 'b,' and 'b' appears before 'a' in the dictionary. This condition ensures that there are no inconsistencies in the order of letters.\n",
-    "\n",
-    "4. Multiple valid orders: It's possible that there are multiple valid orders of letters that satisfy the provided dictionary. However, we are instructed to return the smallest valid order in normal lexicographical order.\n",
-    "\n",
-    "To solve this problem, you need to analyze the given list of words and determine the correct order of letters in the alien language while adhering to the specified constraints. You should return the order as a string in lexicographical order if it's valid, and an empty string if the order is invalid.\n",
-    "\n",
-    "The key to solving this problem is to construct a directed graph of letter relationships based on the order information provided by the words in the dictionary. Then, perform a topological sort on this graph to derive the order of letters, ensuring that the order is consistent and valid according to the given constraints.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from collections import defaultdict, deque\n",
-    "\n",
-    "def alienOrder(words):\n",
-    "    # Create an adjacency list to represent the graph\n",
-    "    graph = defaultdict(list)\n",
-    "    \n",
-    "    # Create a dictionary to store the in-degrees of each letter\n",
-    "    in_degree = {}\n",
-    "    \n",
-    "    # Initialize in-degrees to 0 for all letters\n",
-    "    for word in words:\n",
-    "        for char in word:\n",
-    "            if char not in in_degree:\n",
-    "                in_degree[char] = 0\n",
-    "    \n",
-    "    # Build the graph and update in-degrees\n",
-    "    for i in range(1, len(words)):\n",
-    "        word1, word2 = words[i-1], words[i]\n",
-    "        min_length = min(len(word1), len(word2))\n",
-    "        \n",
-    "        # Compare characters in the two words\n",
-    "        j = 0\n",
-    "        while j < min_length and word1[j] == word2[j]:\n",
-    "            j += 1\n",
-    "        \n",
-    "        if j < min_length:\n",
-    "            # If word1[j] is lexicographically before word2[j], add an edge\n",
-    "            graph[word1[j]].append(word2[j])\n",
-    "            in_degree[word2[j]] += 1\n",
-    "            \n",
-    "            # Break the loop to avoid further comparisons\n",
-    "            break\n",
-    "    \n",
-    "    # Perform topological sorting using Kahn's algorithm\n",
-    "    result = []\n",
-    "    queue = deque([char for char in in_degree if in_degree[char] == 0])\n",
-    "    \n",
-    "    while queue:\n",
-    "        char = queue.popleft()\n",
-    "        result.append(char)\n",
-    "        \n",
-    "        for neighbor in graph[char]:\n",
-    "            in_degree[neighbor] -= 1\n",
-    "            if in_degree[neighbor] == 0:\n",
-    "                queue.append(neighbor)\n",
-    "    \n",
-    "    # Check if there is a valid order\n",
-    "    if len(result) < len(in_degree):\n",
-    "        return \"\"\n",
-    "    \n",
-    "    return \"\".join(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code is designed to determine the order of letters in an unknown alien language that uses the Latin alphabet. It takes a list of non-empty words from a dictionary as input, where the words are sorted lexicographically based on the rules of this alien language.\n",
-    "\n",
-    "The code follows these main steps:\n",
-    "\n",
-    "1. Create a data structure to represent the graph of letters and their relationships in the alien language. It uses a defaultdict to store a list of letters that come after each letter. Additionally, it uses a dictionary called `in_degree` to keep track of the in-degrees (the number of letters that precede a letter) for each letter.\n",
-    "\n",
-    "2. Initialize the in-degrees of all letters to 0. This is done by iterating through all the words and characters in the words, adding each character to the `in_degree` dictionary with an initial in-degree of 0.\n",
-    "\n",
-    "3. Build the graph by comparing adjacent words in the dictionary. For each pair of adjacent words, the code finds the first differing character between the two words. If such a character exists, it means there's an order relationship between the letters represented by these characters. The code updates the graph and in-degrees accordingly.\n",
-    "\n",
-    "4. After constructing the graph, the code performs a topological sorting of the letters using Kahn's algorithm. It starts by initializing an empty result list and a queue containing letters with in-degrees of 0. The algorithm repeatedly removes a letter with an in-degree of 0 from the queue, adds it to the result list, and updates the in-degrees of its neighbors. This process continues until the queue is empty.\n",
-    "\n",
-    "5. Finally, the code checks if the topological sorting was successful by comparing the length of the result list with the number of unique letters in the input. If the lengths differ, it means there's a cycle in the graph, and the order is invalid. In such a case, an empty string is returned. Otherwise, the result list is joined together to form the smallest order of letters in normal lexicographical order, and it is returned as the output.\n",
-    "\n",
-    "The code provides a way to determine the order of letters in the alien language while handling various edge cases, including checking for invalid order conditions."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "wrtef\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "words1 = [\"wrt\", \"wrf\", \"er\", \"ett\", \"rftt\"]\n",
-    "print(alienOrder(words1))  # Output: \"wertf\""
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "zx\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "words2 = [\"z\", \"x\"]\n",
-    "print(alienOrder(words2))  # Output: \"zx\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for determining the order of letters in an alien language.\n",
-    "\n",
-    "Time Complexity:\n",
-    "1. Building the graph: The code iterates through the list of words and compares adjacent words. In the worst case, where all words have the same prefix, it takes O(N) time to build the graph, where N is the total number of characters in all the words.\n",
-    "2. Topological sorting: Performing a topological sort on the graph takes O(V + E) time, where V is the number of unique letters (vertices) and E is the number of relationships between letters (edges) in the graph.\n",
-    "\n",
-    "Overall, the time complexity is O(N + V + E), where N is the total number of characters in the words, V is the number of unique letters, and E is the number of order relationships between letters.\n",
-    "\n",
-    "Space Complexity:\n",
-    "1. Graph and In-degrees: The code uses data structures to represent the graph and in-degrees. The space complexity for these data structures is O(V + E), where V is the number of unique letters, and E is the number of order relationships between letters in the words.\n",
-    "2. Result List: The result list stores the order of letters, which can have a maximum size of V, where V is the number of unique letters.\n",
-    "3. Queue: The space used by the queue during topological sorting is also O(V).\n",
-    "\n",
-    "Overall, the space complexity is O(V + E) due to the data structures used for graph representation and the space needed for storing the result and the queue.\n",
-    "\n",
-    "In the worst case, where there are many unique letters and many order relationships, the space complexity is dominated by the number of unique letters and their relationships.\n",
-    "\n",
-    "In summary, the time complexity of the code is O(N + V + E), and the space complexity is O(V + E). The actual performance will depend on the specific input data, such as the number of unique letters and the relationships between them in the alien language."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple Valid Orders**: Extend the problem to handle cases where there are multiple valid orders of letters in the alien language. Modify the code to return all valid orders instead of just one. Be mindful of performance.\n",
-    "\n",
-    "2. **Detect Invalid Dictionary**: Given a dictionary of words, write a function to detect whether the dictionary is valid or not based on the constraints mentioned in the problem statement."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb b/docs/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb
deleted file mode 100644
index 9539363..0000000
--- a/docs/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb	
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# House Robber\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "---\n",
-    "You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and **it will automatically contact the police if two adjacent houses were broken into on the same night.**\n",
-    "\n",
-    "Given an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight **without alerting the police**.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= nums.length <= 100`\n",
-    "- `0 <= nums[i] <= 400`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is a classic dynamic programming challenge known as the \"House Robber\" problem. The scenario is that you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, but the constraint is that adjacent houses have security systems connected. If two adjacent houses are broken into on the same night, the security systems will automatically contact the police.\n",
-    "\n",
-    "The goal is to determine the maximum amount of money you can rob tonight without alerting the police. You need to find a strategy to maximize your earnings while avoiding robbing two adjacent houses.\n",
-    "\n",
-    "The input to the problem is an array `nums`, where `nums[i]` represents the amount of money in the `i`-th house. The task is to return the maximum amount of money that can be robbed without triggering the security system.\n",
-    "\n",
-    "For example:\n",
-    "```\n",
-    "Input: nums = [1,2,3,1]\n",
-    "Output: 4\n",
-    "Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).\n",
-    "Total amount you can rob = 1 + 3 = 4.\n",
-    "\n",
-    "Input: nums = [2,7,9,3,1]\n",
-    "Output: 12\n",
-    "Explanation: Rob house 1 (money = 2), rob house 3 (money = 9), and rob house 5 (money = 1).\n",
-    "Total amount you can rob = 2 + 9 + 1 = 12.\n",
-    "```\n",
-    "\n",
-    "The challenge involves designing an algorithm to efficiently solve this problem and find the optimal strategy for robbing houses to maximize the stolen money while adhering to the constraint of not robbing adjacent houses. The provided solution utilizes dynamic programming to achieve this.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def rob(self, nums: List[int]) -> int:\n",
-    "        # Initialize variables to keep track of the maximum amount of money robbed at the previous and current houses\n",
-    "        prev, curr = 0, 0\n",
-    "\n",
-    "        # Iterate through the houses\n",
-    "        for n in nums:\n",
-    "            # Calculate the maximum amount of money that can be robbed at the current house\n",
-    "            # It is the maximum of either robbing the current house and the money robbed at the house before the previous one (prev),\n",
-    "            # or skipping the current house and taking the maximum amount from the previous house (curr)\n",
-    "            temp = max(n + prev, curr)\n",
-    "\n",
-    "            # Update the variables for the next iteration\n",
-    "            prev, curr = curr, temp\n",
-    "\n",
-    "        # The final result is the maximum amount of money that can be robbed at the last house\n",
-    "        return curr"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code is a Python implementation of the \"House Robber\" problem using a dynamic programming approach. It's part of a class `Solution` and includes a method `rob` that takes a list of integers `nums` as input and returns the maximum amount of money that can be robbed without alerting the police.\n",
-    "\n",
-    "The key idea of the algorithm is to maintain two variables, `prev` and `curr`, which represent the maximum amount of money that can be robbed up to the previous house and the current house, respectively. The algorithm iterates through the houses, updating these variables based on the maximum amount that can be robbed at the current house, considering the constraint of not robbing adjacent houses.\n",
-    "\n",
-    "The loop calculates a temporary variable `temp` that represents the maximum amount of money that can be robbed at the current house. It considers two scenarios: either robbing the current house and adding the money robbed at the house before the previous one (`n + prev`), or skipping the current house and taking the maximum amount from the previous house (`curr`). The maximum of these two scenarios is assigned to `temp`.\n",
-    "\n",
-    "The `prev` and `curr` variables are then updated for the next iteration. `prev` is set to the previous value of `curr`, and `curr` is set to the calculated `temp`.\n",
-    "\n",
-    "The final result is the value stored in the `curr` variable, representing the maximum amount of money that can be robbed at the last house. This value is returned as the result of the `rob` method.\n",
-    "\n",
-    "The provided test cases at the end demonstrate the usage of the `rob` method with different input arrays."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "sol = Solution()\n",
-    "\n",
-    "nums1 = [1, 2, 3, 1]\n",
-    "print(sol.rob(nums1))  # Output: 4"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "12\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [2, 7, 9, 3, 1]\n",
-    "print(sol.rob(nums2))  # Output: 12"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code.\n",
-    "\n",
-    "### Time Complexity:\n",
-    "\n",
-    "The time complexity of the code is O(n), where n is the number of houses. The algorithm iterates through the array once, and in each iteration, it performs constant time operations. The loop runs for each house, and the number of operations inside the loop is independent of the input size. Therefore, the overall time complexity is linear, O(n).\n",
-    "\n",
-    "### Space Complexity:\n",
-    "\n",
-    "The space complexity is O(1), constant space. The algorithm uses only a constant amount of extra space regardless of the input size. The variables `prev`, `curr`, and `temp` are the only additional variables used, and they do not depend on the size of the input array. Thus, the space complexity is constant, O(1).\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: O(n)\n",
-    "- Space Complexity: O(1)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Variation with Maximum Number of Houses to Rob:** Modify the problem to include an additional constraint: you are allowed to rob at most 'k' houses without triggering the alarm. Find the maximum amount you can rob under this new constraint.\n",
-    "\n",
-    "2. **Robbing in a Circular Street:** Extend the problem to a circular street where the first and last houses are adjacent. You cannot rob both the first and last houses simultaneously. Find the maximum amount you can rob in this circular setting.\n",
-    "\n",
-    "\n",
-    "**Link to Problem:** [To see the House Robber problem on LeetCode, click here!](https://leetcode.com/problems/house-robber/description/)\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb b/docs/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb
deleted file mode 100644
index d210017..0000000
--- a/docs/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb	
+++ /dev/null
@@ -1,243 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# House Robber II\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "---\n",
-    "You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle**. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.\n",
-    "\n",
-    "Given an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight without alerting the police*.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= nums.length <= 100`\n",
-    "- `0 <= nums[i] <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "Certainly! The problem statement describes a scenario where you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. However, there is a security system in place: adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.\n",
-    "\n",
-    "The twist in this problem is that the houses are arranged in a circle. That means the first house is the neighbor of the last one. So, robbing the first house has implications for the last house, and vice versa.\n",
-    "\n",
-    "The goal is to find the maximum amount of money you can rob tonight without alerting the police. You need to implement a function that takes an integer array `nums` representing the amount of money in each house and returns the maximum amount of money that can be robbed without triggering the security system.\n",
-    "\n",
-    "Here are a few examples to illustrate the problem:\n",
-    "\n",
-    "1. **Example 1:**\n",
-    "   ```python\n",
-    "   Input: nums = [2, 3, 2]\n",
-    "   Output: 3\n",
-    "   Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2) because they are adjacent houses. The optimal strategy is to rob house 2 (money = 3).\n",
-    "   ```\n",
-    "\n",
-    "2. **Example 2:**\n",
-    "   ```python\n",
-    "   Input: nums = [1, 2, 3, 1]\n",
-    "   Output: 4\n",
-    "   Explanation: Robbing house 1 (money = 1) and then robbing house 3 (money = 3) results in a total amount of 4. Robbing house 2 is skipped to avoid alerting the police.\n",
-    "   ```\n",
-    "\n",
-    "3. **Example 3:**\n",
-    "   ```python\n",
-    "   Input: nums = [1, 2, 3]\n",
-    "   Output: 3\n",
-    "   Explanation: Since the houses are in a circle, you can choose either house 1, 2, or 3. The optimal strategy is to rob house 3 for a total amount of 3.\n",
-    "   ```\n",
-    "\n",
-    "The constraints for the problem are that the length of the `nums` array is between 1 and 100, and the amount of money in each house (`nums[i]`) is between 0 and 1000.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def rob(self, nums: List[int]) -> int:\n",
-    "        def simple_rob(nums):\n",
-    "            # Helper function for basic house robbery problem\n",
-    "            prev, curr = 0, 0\n",
-    "            for num in nums:\n",
-    "                # Calculate the maximum amount of money that can be robbed\n",
-    "                # without alerting the police for each house\n",
-    "                prev, curr = curr, max(prev + num, curr)\n",
-    "            return curr\n",
-    "\n",
-    "        # Check the base cases where the length of nums is 1 or 2\n",
-    "        if len(nums) == 1:\n",
-    "            return nums[0]\n",
-    "        elif len(nums) == 2:\n",
-    "            return max(nums[0], nums[1])\n",
-    "\n",
-    "        # Case 1: First house is robbed, last house is not\n",
-    "        case1 = simple_rob(nums[:-1])\n",
-    "\n",
-    "        # Case 2: First house is not robbed, last house is\n",
-    "        case2 = simple_rob(nums[1:])\n",
-    "\n",
-    "        # Maximum of the two cases\n",
-    "        return max(case1, case2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code defines a class `Solution` with a method `rob`, which aims to determine the maximum amount of money a professional robber can steal from houses arranged in a circle without alerting the police. Each house contains a certain amount of money, and adjacent houses have a security system that triggers if both are robbed on the same night.\n",
-    "\n",
-    "The code uses dynamic programming to solve the problem efficiently. It contains a helper function `simple_rob` that computes the maximum amount of money that can be robbed from a linear arrangement of houses. The main method `rob` handles the circular arrangement by considering two cases: robbing the first house and not robbing the last, and not robbing the first house and robbing the last. The result is the maximum of these two cases.\n",
-    "\n",
-    "The class is instantiated as `solution`, and the `rob` method is applied to three different scenarios (test cases) to demonstrate its functionality. The final output represents the maximum amount of money the robber can steal without triggering the security system."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "nums1 = [2, 3, 2]\n",
-    "print(solution.rob(nums1))  # Output: 3"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [1, 2, 3, 1]\n",
-    "print(solution.rob(nums2))  # Output: 4"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e6e1554f-ae77-410c-8af2-e59e5428166c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums3 = [1, 2, 3]\n",
-    "print(solution.rob(nums3))  # Output: 3"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "### Time Complexity:\n",
-    "The time complexity is primarily determined by the `simple_rob` helper function, which has a time complexity of O(N), where N is the number of houses in the input array. The main `rob` function calls `simple_rob` twice, once for the case of robbing the first house and not robbing the last, and once for the case of not robbing the first house and robbing the last. Therefore, the overall time complexity is O(N).\n",
-    "\n",
-    "### Space Complexity:\n",
-    "The space complexity is O(1) because the space used by the algorithm is constant, regardless of the size of the input array. The only variables used are `prev` and `curr` in the `simple_rob` function, which represent the previous and current maximum amounts of money that can be robbed without alerting the police. These variables are updated iteratively, and no additional data structures are used that scale with the input size. Therefore, the space complexity is constant, or O(1).\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: O(N)\n",
-    "- Space Complexity: O(1)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Optimize the space complexity of the solution. Can you solve the problem with O(1) space complexity without sacrificing the time complexity?\n",
-    "\n",
-    "2. **Circular Robbery Variations:** Consider variations of the circular house robbery problem, such as allowing the robber to skip a certain number of houses or restricting the number of consecutive houses that can be robbed. Modify the solution to accommodate these variations.\n",
-    "\n",
-    "**Link to Problem:** [To see the House Robber II problem on LeetCode, click here!](https://leetcode.com/problems/house-robber-ii/)"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb b/docs/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb
deleted file mode 100644
index 32c5da3..0000000
--- a/docs/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb	
+++ /dev/null
@@ -1,204 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 70. Climbing Stairs\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Climbing Stairs problem on LeetCode, click here!](https://leetcode.com/problems/climbing-stairs/)\n",
-    "\n",
-    "---\n",
-    "You are climbing a staircase. It takes `n` steps to reach the top.\n",
-    "\n",
-    "Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= n <= 45`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The \"Climbing Stairs\" problem is a classic algorithmic problem that asks you to find the number of distinct ways to climb a staircase with a given number of steps. Here's the problem statement:\n",
-    "\n",
-    "You are given a staircase with 'n' steps. Each time you can either climb 1 step or 2 steps. You need to find out how many distinct ways there are to reach the top of the staircase.\n",
-    "\n",
-    "For example, if 'n' is 2, there are two ways to climb to the top:\n",
-    "\n",
-    "1. Climbing two steps at once.\n",
-    "2. Climbing one step, and then another step.\n",
-    "\n",
-    "If 'n' is 3, there are three ways:\n",
-    "\n",
-    "1. Climbing one step, then one step, and finally one step.\n",
-    "2. Climbing one step, then two steps.\n",
-    "3. Climbing two steps, then one step.\n",
-    "\n",
-    "The problem essentially asks you to find a solution that counts the possible combinations of 1-step and 2-step climbs that can be taken to reach the top of the staircase. It's a classic example of a dynamic programming problem where you can break it down into smaller subproblems and build a solution incrementally, as shown in the code I provided earlier.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class Solution:\n",
-    "    def climbStairs(self, n: int) -> int:\n",
-    "        # If there are 1 or 2 steps, there are exactly n ways to reach the top.\n",
-    "        if n <= 2:\n",
-    "            return n\n",
-    "\n",
-    "        # Create an array to store the number of ways to reach each step.\n",
-    "        dp = [0] * (n + 1)\n",
-    "\n",
-    "        # There is one way to reach the first step (base case).\n",
-    "        dp[1] = 1\n",
-    "\n",
-    "        # There are two ways to reach the second step (base case).\n",
-    "        dp[2] = 2\n",
-    "\n",
-    "        # Calculate the number of ways to reach each step from the previous steps.\n",
-    "        for i in range(3, n + 1):\n",
-    "            # The number of ways to reach step 'i' is the sum of the ways to reach\n",
-    "            # the previous two steps, as you can take 1 or 2 steps at a time.\n",
-    "            dp[i] = dp[i - 1] + dp[i - 2]\n",
-    "\n",
-    "        # The result is the number of ways to reach the top, which is stored in dp[n].\n",
-    "        return dp[n]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a class `Solution` with a method `climbStairs` that takes an integer `n` as input and returns the number of distinct ways to climb a staircase with `n` steps.\n",
-    "\n",
-    "1. If `n` is 1 or 2, the function returns `n` directly because there are only 1 or 2 distinct ways to reach the top in those cases.\n",
-    "\n",
-    "2. For `n` greater than 2, the code uses dynamic programming to calculate the number of ways to reach each step. It creates an array `dp` of length `n + 1` to store these values.\n",
-    "\n",
-    "3. The `dp[1]` and `dp[2]` are initialized to 1 and 2, respectively. This is because there is only one way to reach the first step (by taking 1 step), and there are two ways to reach the second step (by taking 2 steps or two 1-step climbs).\n",
-    "\n",
-    "4. The code then uses a loop to fill in the `dp` array for steps 3 and above. For each step `i` from 3 to `n`, it calculates the number of ways to reach that step by adding the number of ways to reach the previous two steps, as you can either take 1 step or 2 steps at a time.\n",
-    "\n",
-    "5. Finally, the function returns the value stored in `dp[n]`, which represents the total number of distinct ways to climb the staircase with `n` steps."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 1 - Input: n = 2, Output: 2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "n = 2\n",
-    "solution = Solution()\n",
-    "print(f\"Example 1 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 2"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 2 - Input: n = 3, Output: 3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "n = 3\n",
-    "print(f\"Example 2 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 3\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Time Complexity:\n",
-    "The time complexity of the solution is O(n) because we use a single for loop that iterates from 3 to n to fill the `dp` array. In each iteration, we perform constant time operations (addition and assignment). Therefore, the time complexity is linear with respect to the input value `n`.\n",
-    "\n",
-    "Space Complexity:\n",
-    "The space complexity of the solution is O(n) as well. We use an array `dp` of length `n + 1` to store the number of ways to reach each step. The size of this array is directly proportional to the input value `n`, which makes the space complexity linear in terms of `n`. The rest of the variables used in the function (e.g., `i`) occupy constant space and do not contribute significantly to the overall space complexity.\n",
-    "\n",
-    "In summary, the time complexity is O(n), and the space complexity is O(n) for this dynamic programming solution to the \"Climbing Stairs\" problem."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Generalized Step Sizes**: Instead of being limited to 1 step or 2 steps, consider a scenario where you can take steps of sizes `[1, 2, 3]`. Implement a function that calculates the number of ways to climb the staircase with these different step sizes.\n",
-    "\n",
-    "2. **Minimize Steps**: Given an integer `n`, find the minimum number of steps required to reach the top of the staircase. Return both the minimum number of steps and the specific sequence of steps taken."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/docs/_build/jupyter_execute/Book.ipynb b/docs/_build/jupyter_execute/Book.ipynb
deleted file mode 100644
index 6c891d7..0000000
--- a/docs/_build/jupyter_execute/Book.ipynb
+++ /dev/null
@@ -1,914 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: pip in /usr/local/python/3.10.8/lib/python3.10/site-packages (23.3.1)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: jupyter-book in /usr/local/python/3.10.8/lib/python3.10/site-packages (0.15.1)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: click<9,>=7.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (8.1.7)\r\n",
-      "Requirement already satisfied: docutils<0.19,>=0.15 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.18.1)\r\n",
-      "Requirement already satisfied: Jinja2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (3.1.2)\r\n",
-      "Requirement already satisfied: jsonschema<5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (4.19.1)\r\n",
-      "Requirement already satisfied: linkify-it-py~=2.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (2.0.2)\r\n",
-      "Requirement already satisfied: myst-nb~=0.17.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.17.2)\r\n",
-      "Requirement already satisfied: pyyaml in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (6.0.1)\r\n",
-      "Requirement already satisfied: sphinx<6,>=4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (5.0.2)\r\n",
-      "Requirement already satisfied: sphinx-comments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.0.3)\r\n",
-      "Requirement already satisfied: sphinx-copybutton in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.5.2)\r\n",
-      "Requirement already satisfied: sphinx-external-toc~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.1)\r\n",
-      "Requirement already satisfied: sphinx-jupyterbook-latex~=0.5.2 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.5.2)\r\n",
-      "Requirement already satisfied: sphinx-design~=0.3.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.0)\r\n",
-      "Requirement already satisfied: sphinx-thebe~=0.2.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.2.1)\r\n",
-      "Requirement already satisfied: sphinx-book-theme~=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (1.0.1)\r\n",
-      "Requirement already satisfied: sphinx_togglebutton in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.2)\r\n",
-      "Requirement already satisfied: sphinxcontrib-bibtex<=2.5.0,>=2.2.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (2.5.0)\r\n",
-      "Requirement already satisfied: sphinx-multitoc-numbering~=0.1.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.1.3)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: attrs>=22.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (23.1.0)\r\n",
-      "Requirement already satisfied: jsonschema-specifications>=2023.03.6 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (2023.7.1)\r\n",
-      "Requirement already satisfied: referencing>=0.28.4 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (0.30.2)\r\n",
-      "Requirement already satisfied: rpds-py>=0.7.1 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (0.10.6)\r\n",
-      "Requirement already satisfied: uc-micro-py in /usr/local/python/3.10.8/lib/python3.10/site-packages (from linkify-it-py~=2.0.0->jupyter-book) (1.0.2)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: importlib_metadata in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (6.8.0)\r\n",
-      "Requirement already satisfied: ipython in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (8.16.1)\r\n",
-      "Requirement already satisfied: jupyter-cache<0.7,>=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.6.1)\r\n",
-      "Requirement already satisfied: nbclient in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.7.4)\r\n",
-      "Requirement already satisfied: myst-parser~=0.18.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.18.1)\r\n",
-      "Requirement already satisfied: nbformat~=5.0 in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (5.9.2)\r\n",
-      "Requirement already satisfied: typing-extensions in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (4.8.0)\r\n",
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-      "Requirement already satisfied: sphinxcontrib-serializinghtml>=1.1.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.1.9)\r\n",
-      "Requirement already satisfied: sphinxcontrib-qthelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.0.6)\r\n",
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-     ]
-    },
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-     "name": "stdout",
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-     ]
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-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
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-     ]
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-     "name": "stdout",
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-     ]
-    },
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-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py<3.0.0,>=1.0.0->myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.1.2)\r\n",
-      "Requirement already satisfied: ptyprocess>=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect>4.3->ipython->myst-nb~=0.17.1->jupyter-book) (0.7.0)\r\n",
-      "Requirement already satisfied: wcwidth in /home/codespace/.local/lib/python3.10/site-packages (from prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.8)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: greenlet!=0.4.17 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sqlalchemy<3,>=1.3.12->jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (3.0.1)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: soupsieve>1.2 in /home/codespace/.local/lib/python3.10/site-packages (from beautifulsoup4->pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (2.5)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: executing>=1.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.0.0)\r\n",
-      "Requirement already satisfied: asttokens>=2.1.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.4.0)\r\n",
-      "Requirement already satisfied: pure-eval in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.2)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Note: you may need to restart the kernel to use updated packages.\n"
-     ]
-    }
-   ],
-   "source": [
-    "!python -m pip install --upgrade pip\n",
-    "%pip install jupyter-book"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions\n"
-     ]
-    }
-   ],
-   "source": [
-    "%cd /workspaces/LeetCode-Solutions"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Traceback (most recent call last):\r\n",
-      "  File \"/home/codespace/.python/current/bin/jupyter-book\", line 8, in <module>\r\n",
-      "    sys.exit(main())\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1157, in __call__\r\n",
-      "    return self.main(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1078, in main\r\n",
-      "    rv = self.invoke(ctx)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
-      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
-      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1434, in invoke\r\n",
-      "    return ctx.invoke(self.callback, **ctx.params)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 783, in invoke\r\n",
-      "    return __callback(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx_external_toc/cli.py\", line 139, in migrate_toc\r\n",
-      "    toc = migrate_jupyter_book(Path(toc_file))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx_external_toc/tools.py\", line 327, in migrate_jupyter_book\r\n",
-      "    raise MalformedError(\"no top-level 'file' key found\")\r\n",
-      "sphinx_external_toc.parsing.MalformedError: no top-level 'file' key found\r\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book toc migrate /workspaces/LeetCode-Solutions/_toc.yml -o /workspaces/LeetCode-Solutions/_toc.yml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions\r\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\r\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\r\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[etoc] Changing master_doc to 'README'\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\r\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\r\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\r\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 77 source files that are out of date\r\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 77 added, 0 changed, 0 removed\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[  1%] \u001b[35m01. Array Hashing/1. Two Sum\u001b[39;49;00m                          \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.07 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  2%] \u001b[35m01. Array Hashing/128. Longest Consecutive Sequence\u001b[39;49;00m   \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.17 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[  3%] \u001b[35m01. Array Hashing/217. Contains Duplicate\u001b[39;49;00m             \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.19 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[  5%] \u001b[35m01. Array Hashing/238. Product of Array Except Self\u001b[39;49;00m   \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 1.26 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  6%] \u001b[35m01. Array Hashing/242. Valid Anagram\u001b[39;49;00m                  \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 1.18 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  7%] \u001b[35m01. Array Hashing/347. Top K Frequent Elements\u001b[39;49;00m        \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.13 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  9%] \u001b[35m01. Array Hashing/49. Group Anagrams\u001b[39;49;00m                  \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 4.02 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 10%] \u001b[35m01. Array Hashing/659. Encode and Decode Strings\u001b[39;49;00m      \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 11%] \u001b[35m01. Array Hashing/README\u001b[39;49;00m                              \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 12%] \u001b[35m02. Two Pointers/11. Container With Most Water\u001b[39;49;00m        \r",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.26 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 14%] \u001b[35m02. Two Pointers/125. Valid Palindrome\u001b[39;49;00m                \r",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 1.07 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 15%] \u001b[35m02. Two Pointers/15. 3Sum\u001b[39;49;00m                             \r",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.00 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 16%] \u001b[35m02. Two Pointers/README\u001b[39;49;00m                               \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 18%] \u001b[35m03. Sliding Window/121. Best Time to Buy and Sell Stock\u001b[39;49;00m\r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.89 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 19%] \u001b[35m03. Sliding Window/3. Longest Substring Without Repeating Characters\u001b[39;49;00m\r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 1.16 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 20%] \u001b[35m03. Sliding Window/424. Longest Repeating Character Replacement\u001b[39;49;00m\r",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 22%] \u001b[35m03. Sliding Window/76. Minimum Window Substring\u001b[39;49;00m       \r",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 0.87 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 23%] \u001b[35m03. Sliding Window/README\u001b[39;49;00m                             \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 24%] \u001b[35m04. Stack/20. Valid Parentheses\u001b[39;49;00m                       \r",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 4.02 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 25%] \u001b[35m04. Stack/README\u001b[39;49;00m                                      \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 27%] \u001b[35m05. Binary Search/153. Find Minimum in Rotated Sorted Array\u001b[39;49;00m\r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 0.98 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 28%] \u001b[35m05. Binary Search/33. Search in Rotated Sorted Array\u001b[39;49;00m  \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 29%] \u001b[35m05. Binary Search/README\u001b[39;49;00m                              \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 31%] \u001b[35m06. Linked List/141. Linked List Cycle\u001b[39;49;00m                \r",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 0.96 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 32%] \u001b[35m06. Linked List/143. Reorder List\u001b[39;49;00m                     \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "^C\r\n",
-      "\r\n",
-      "\u001b[91mException occurred:\u001b[39;49;00m\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 512, in _async_cleanup_kernel\r\n",
-      "    await ensure_async(self.km.cleanup_resources())\r\n",
-      "AttributeError: 'NoneType' object has no attribute 'cleanup_resources'\r\n",
-      "\u001b[91mThe full traceback has been saved in /tmp/sphinx-err-tl19fjln.log, if you want to report the issue to the developers.\u001b[39;49;00m\r\n",
-      "Please also report this if it was a user error, so that a better error message can be provided next time.\r\n",
-      "A bug report can be filed in the tracker at <https://github.com/sphinx-doc/sphinx/issues>. Thanks!\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py\", line 89, in wrapper\r\n",
-      "    self._ready.set_result(None)\r\n",
-      "asyncio.exceptions.InvalidStateError: invalid state\r\n",
-      "\r\n",
-      "During handling of the above exception, another exception occurred:\r\n",
-      "\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 505, in _async_cleanup_kernel\r\n",
-      "    await ensure_async(self.km.shutdown_kernel(now=now))\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py\", line 189, in ensure_async\r\n",
-      "    result = await obj\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py\", line 92, in wrapper\r\n",
-      "    self._ready.set_exception(e)\r\n",
-      "asyncio.exceptions.InvalidStateError: invalid state\r\n",
-      "\r\n",
-      "During handling of the above exception, another exception occurred:\r\n",
-      "\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/sphinx.py\", line 171, in build_sphinx\r\n",
-      "    app.build(force_all, filenames)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/application.py\", line 329, in build\r\n",
-      "    self.builder.build_update()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 288, in build_update\r\n",
-      "    self.build(to_build,\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 302, in build\r\n",
-      "    updated_docnames = set(self.read())\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 409, in read\r\n",
-      "    self._read_serial(docnames)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 430, in _read_serial\r\n",
-      "    self.read_doc(docname)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 483, in read_doc\r\n",
-      "    publisher.publish()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/core.py\", line 217, in publish\r\n",
-      "    self.document = self.reader.read(self.source, self.parser,\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/io.py\", line 103, in read\r\n",
-      "    self.parse()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/readers/__init__.py\", line 78, in parse\r\n",
-      "    self.parser.parse(self.input, document)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/sphinx_.py\", line 150, in parse\r\n",
-      "    with create_client(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/base.py\", line 83, in __enter__\r\n",
-      "    self.start_client()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/direct.py\", line 40, in start_client\r\n",
-      "    result = single_nb_execution(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_cache/executors/utils.py\", line 58, in single_nb_execution\r\n",
-      "    executenb(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 1305, in execute\r\n",
-      "    return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute()\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py\", line 173, in wrapped\r\n",
-      "    return loop.run_until_complete(inner)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/asyncio/base_events.py\", line 649, in run_until_complete\r\n",
-      "    return future.result()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 689, in async_execute\r\n",
-      "    async with self.async_setup_kernel(**kwargs):\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/contextlib.py\", line 206, in __aexit__\r\n",
-      "    await anext(self.gen)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 656, in async_setup_kernel\r\n",
-      "    await self._async_cleanup_kernel()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 512, in _async_cleanup_kernel\r\n",
-      "    await ensure_async(self.km.cleanup_resources())\r\n",
-      "AttributeError: 'NoneType' object has no attribute 'cleanup_resources'\r\n",
-      "\r\n",
-      "The above exception was the direct cause of the following exception:\r\n",
-      "\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/home/codespace/.python/current/bin/jupyter-book\", line 8, in <module>\r\n",
-      "    sys.exit(main())\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1157, in __call__\r\n",
-      "    return self.main(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1078, in main\r\n",
-      "    rv = self.invoke(ctx)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
-      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1434, in invoke\r\n",
-      "    return ctx.invoke(self.callback, **ctx.params)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 783, in invoke\r\n",
-      "    return __callback(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py\", line 317, in build\r\n",
-      "    builder_specific_actions(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py\", line 525, in builder_specific_actions\r\n",
-      "    raise RuntimeError(_message_box(msg, color=\"red\", doprint=False)) from result\r\n",
-      "RuntimeError: \r\n",
-      "\u001b[91m===============================================================================\u001b[0m\r\n",
-      "\r\n",
-      "There was an error in building your book. Look above for the cause.\r\n",
-      "\r\n",
-      "\u001b[91m===============================================================================\u001b[0m\r\n",
-      "\r\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build ."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book build /workspaces/LeetCode-Solutions/ --builder pdfhtml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/html": [
-       "<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>"
-      ],
-      "text/plain": [
-       "<IPython.core.display.HTML object>"
-      ]
-     },
-     "execution_count": 5,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "from IPython.display import HTML\n",
-    "HTML('<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>')\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Your book template can be found at\n",
-      "\n",
-      "    mybook/\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book create mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions/mybook\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_build/html\n",
-      "[sphinxcontrib-bibtex] Beware that docutils versions 0.18 and 0.19 (you are running 0.18.1) are known to generate invalid html for citations. If this issue affects you, please use docutils<0.18 (or >=0.20 once released) instead. For more details, see https://sourceforge.net/p/docutils/patches/195/\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "[etoc] Changing master_doc to 'intro'\n",
-      "checking bibtex cache... out of date\n",
-      "parsing bibtex file /workspaces/LeetCode-Solutions/mybook/references.bib... parsed 5 entries\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='force', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/mybook/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 14 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 14 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.35 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.97 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 0.83 seconds [mystnb]\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mintro\u001b[39;49;00m                                                 \n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mintro\u001b[39;49;00m                                                  \n",
-      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
-      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
-      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
-      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
-      "[etoc] missing index.html written as redirect to 'intro.html'\n",
-      "\u001b[01mbuild succeeded, 2 warnings.\u001b[39;49;00m\n",
-      "\n",
-      "The HTML pages are in mybook/_build/html.\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Finished generating HTML for book.\n",
-      "Your book's HTML pages are here:\n",
-      "    mybook/_build/html/\n",
-      "You can look at your book by opening this file in a browser:\n",
-      "    mybook/_build/html/index.html\n",
-      "Or paste this line directly into your browser bar:\n",
-      "    file:///workspaces/LeetCode-Solutions/mybook/_build/html/index.html            \n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "format: jb-book\n",
-      "root: intro\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book toc from-project /workspaces/LeetCode-Solutions/mybook/ -f jb-book"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 2
-}
\ No newline at end of file
diff --git a/docs/_config.yml b/docs/_config.yml
deleted file mode 100644
index 3def3c7..0000000
--- a/docs/_config.yml
+++ /dev/null
@@ -1,31 +0,0 @@
-# Jupyter Book configuration
-
-# Project information
-title: "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges" # Python Problem-Solving Pro: Conquering the Blind 75 LeetCode Challenges
-author: Mohsen Tabibian and Saeed Samadidana
-email: mohsentabibian@gmail.com
-description: Explore 75 essential problems frequently asked in tech interviews. Each solution includes detailed explanations and time/space complexity analysis. Boost your coding skills and interview preparedness!
-
-repository: 
-   url: https://github.com/mohsentabibian/LeetCode-Solutions
-   path: docs #LeetCode-Solutions/
-   branch: main
-
-# Execute notebooks when building (set to false if you don't want to run the code)
-execute:
-  execute_notebooks: auto
-
-# HTML settings
-html:
-  theme: jekyll-theme-Architect #jupyter
-  logo: logo.png
-  toc:
-    display: true
-  use_issues_button: true
-  use_repository_button: true
-
-#Build settings
-#repository_url: https://github.com/mohsentabibian/LeetCode-Solutions  # URL to the GitHub repository
-
-# Add GitHub buttons to the book
-# See https://jupyterbook.org/customize/config.html #add-a-link-to-your-repository
\ No newline at end of file
diff --git a/docs/_toc.yml b/docs/_toc.yml
deleted file mode 100644
index 0e4e1db..0000000
--- a/docs/_toc.yml
+++ /dev/null
@@ -1,88 +0,0 @@
-format: jb-book
-root: README
-chapters:
-  - file: 01. Array Hashing/README
-    sections:
-      - file: 01. Array Hashing/1. Two Sum
-      - file: 01. Array Hashing/128. Longest Consecutive Sequence
-      - file: 01. Array Hashing/217. Contains Duplicate
-      - file: 01. Array Hashing/238. Product of Array Except Self
-      - file: 01. Array Hashing/242. Valid Anagram
-      - file: 01. Array Hashing/347. Top K Frequent Elements
-      - file: 01. Array Hashing/49. Group Anagrams
-      - file: 01. Array Hashing/659. Encode and Decode Strings
-  - file: 02. Two Pointers/README
-    sections:
-      - file: 02. Two Pointers/11. Container With Most Water
-      - file: 02. Two Pointers/125. Valid Palindrome
-      - file: 02. Two Pointers/15. 3Sum
-  - file: 03. Sliding Window/README
-    sections:
-      - file: 03. Sliding Window/121. Best Time to Buy and Sell Stock
-      - file: 03. Sliding Window/3. Longest Substring Without Repeating Characters
-      - file: 03. Sliding Window/424. Longest Repeating Character Replacement
-      - file: 03. Sliding Window/76. Minimum Window Substring
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-    sections:
-      - file: 04. Stack/20. Valid Parentheses
-  - file: 05. Binary Search/README
-    sections:
-      - file: 05. Binary Search/153. Find Minimum in Rotated Sorted Array
-      - file: 05. Binary Search/33. Search in Rotated Sorted Array
-  - file: 06. Linked List/README
-    sections:
-      - file: 06. Linked List/141. Linked List Cycle
-      - file: 06. Linked List/143. Reorder List
-      - file: 06. Linked List/19. Remove Nth Node From End of List
-      - file: 06. Linked List/206. Reverse Linked List
-      - file: 06. Linked List/21. Merge Two Sorted Lists
-      - file: 06. Linked List/23. Merge k Sorted Lists
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-      - file: 07. Trees/100. Same Tree
-      - file: 07. Trees/102. Binary Tree Level Order Traversal
-      - file: 07. Trees/104. Maximum Depth of Binary Tree
-      - file: 07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal
-      - file: 07. Trees/124. Binary Tree Maximum Path Sum
-      - file: 07. Trees/199. Binary Tree Right Side View
-      - file: 07. Trees/226. Invert Binary Tree
-      - file: 07. Trees/230. Kth Smallest Element in a BST
-      - file: 07. Trees/235. Lowest Common Ancestor of a Binary Search Tree
-      - file: 07. Trees/297. Serialize and Deserialize Binary Tree
-      - file: 07. Trees/572. Subtree of Another Tree
-      - file: 07. Trees/98. Validate Binary Search Tree
-  - file: 08. Tries/README
-    sections:
-      - file: 08. Tries/208. Implement Trie (Prefix Tree)
-      - file: 08. Tries/211. Design Add and Search Words Data Structure
-      - file: 08. Tries/212. Word Search II
-  - file: 09. Heap - Priority Queue/README
-    sections:
-      - file: 09. Heap - Priority Queue/1046. Last Stone Weight
-      - file: 09. Heap - Priority Queue/295. Find Median from Data Stream
-      - file: 09. Heap - Priority Queue/703. Kth Largest Element in a Stream
-  - file: 10. Backtracking/README
-    sections:
-      - file: 10. Backtracking/39. Combination Sum
-      - file: 10. Backtracking/79. Word Search
-  - file: 11. Graphs/README
-    sections:
-      - file: 11. Graphs/133. Clone Graph
-      - file: 11. Graphs/178. Graph Valid Tree
-      - file: 11. Graphs/200. Number of Islands
-      - file: 11. Graphs/207. Course Schedule
-      - file: 11. Graphs/3651. Number of Connected Components in an Undirected Graph
-      - file: 11. Graphs/417. Pacific Atlantic Water Flow
-  - file: 12. Advanced Graphs/README
-    sections:
-      - file: 12. Advanced Graphs/892. Alien Dictionary
-  - file: 13. One-D Dynamic Programming/README
-    sections:
-      - file: 13. One-D Dynamic Programming/70. Climbing Stairs
-      - file: 13. One-D Dynamic Programming/198. House Robber
-      - file: 13. One-D Dynamic Programming/213. House Robber II
-  - file: 14. Two-D Dynamic Programming/README
-  - file: 15. Greedy/README
-  - file: 16. Intervals/README
-  - file: 17. Math Geometry/README
-  - file: 18. Bit Manipulation/README
\ No newline at end of file
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diff --git a/_build/html/objects.inv b/objects.inv
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diff --git a/_build/html/reports/Book.err.log b/reports/Book.err.log
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diff --git a/_build/html/search.html b/search.html
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diff --git a/_build/html/searchindex.js b/searchindex.js
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From ab62de199be76f7ee659d8e3174cec420ac09925 Mon Sep 17 00:00:00 2001
From: Mohsen Tabibian <36014419+mohsentabibian@users.noreply.github.com>
Date: Tue, 14 Nov 2023 16:57:05 +0000
Subject: [PATCH 2/7] jh

---
 .nojekyll                                     |   0
 .../1. Two Sum.ipynb                          |   0
 .../128. Longest Consecutive Sequence.ipynb   |   0
 .../217. Contains Duplicate.ipynb             |   0
 .../238. Product of Array Except Self.ipynb   |   0
 .../242. Valid Anagram.ipynb                  |   0
 .../347. Top K Frequent Elements.ipynb        |   0
 .../49. Group Anagrams.ipynb                  |   0
 .../659. Encode and Decode Strings.ipynb      |   0
 .../README.md                                 |   0
 .../11. Container With Most Water.ipynb       |   0
 .../125. Valid Palindrome.ipynb               |   0
 .../15. 3Sum.ipynb                            |   0
 .../README.md                                 |   0
 ...121. Best Time to Buy and Sell Stock.ipynb |   0
 ...bstring Without Repeating Characters.ipynb |   0
 ...gest Repeating Character Replacement.ipynb |   0
 .../76. Minimum Window Substring.ipynb        |   0
 .../README.md                                 |   0
 .../20. Valid Parentheses.ipynb               |   0
 {_sources/04. Stack => 04. Stack}/README.md   |   0
 ...Find Minimum in Rotated Sorted Array.ipynb |   0
 .../33. Search in Rotated Sorted Array.ipynb  |   0
 .../README.md                                 |   0
 .../141. Linked List Cycle.ipynb              |   0
 .../143. Reorder List.ipynb                   |   0
 ...19. Remove Nth Node From End of List.ipynb |   0
 .../206. Reverse Linked List.ipynb            |   0
 .../21. Merge Two Sorted Lists.ipynb          |   0
 .../23. Merge k Sorted Lists.ipynb            |   0
 .../README.md                                 |   0
 .../100. Same Tree.ipynb                      |   0
 ...2. Binary Tree Level Order Traversal.ipynb |   0
 .../104. Maximum Depth of Binary Tree.ipynb   |   0
 ... from Preorder and Inorder Traversal.ipynb |   0
 .../124. Binary Tree Maximum Path Sum.ipynb   |   0
 .../199. Binary Tree Right Side View.ipynb    |   0
 .../226. Invert Binary Tree.ipynb             |   0
 .../230. Kth Smallest Element in a BST.ipynb  |   0
 ...mon Ancestor of a Binary Search Tree.ipynb |   0
 ...erialize and Deserialize Binary Tree.ipynb |   0
 .../572. Subtree of Another Tree.ipynb        |   0
 .../98. Validate Binary Search Tree.ipynb     |   0
 {_sources/07. Trees => 07. Trees}/README.md   |   0
 .../208. Implement Trie (Prefix Tree).ipynb   |   0
 ... Add and Search Words Data Structure.ipynb |   0
 .../212. Word Search II.ipynb                 |   0
 {_sources/08. Tries => 08. Tries}/README.md   |   0
 .../1046. Last Stone Weight.ipynb             |   0
 .../295. Find Median from Data Stream.ipynb   |   0
 ...703. Kth Largest Element in a Stream.ipynb |   0
 .../README.md                                 |   0
 .../39. Combination Sum.ipynb                 |   0
 .../79. Word Search.ipynb                     |   0
 .../README.md                                 |   0
 .../133. Clone Graph.ipynb                    |   0
 .../178. Graph Valid Tree.ipynb               |   0
 .../200. Number of Islands.ipynb              |   0
 .../207. Course Schedule.ipynb                |   0
 ...ed Components in an Undirected Graph.ipynb |   0
 .../417. Pacific Atlantic Water Flow.ipynb    |   0
 {_sources/11. Graphs => 11. Graphs}/README.md |   0
 .../892. Alien Dictionary.ipynb               |   0
 .../README.md                                 |   0
 .../198. House Robber.ipynb                   |   0
 .../213. House Robber II.ipynb                |   0
 .../70. Climbing Stairs.ipynb                 |   0
 .../README.md                                 |   0
 .../README.md                                 |   0
 {_sources/15. Greedy => 15. Greedy}/README.md |   0
 .../16. Intervals => 16. Intervals}/README.md |   0
 .../README.md                                 |   0
 .../README.md                                 |   0
 Book.ipynb                                    | 369 +++++++
 .../CODE_OF_CONDUCT.md => CODE_OF_CONDUCT.md  |   0
 _sources/CONTRIBUTING.md => CONTRIBUTING.md   |   0
 LICENSE                                       |  21 +
 _sources/README.md => README.md               |   0
 _sources/SECURITY.md => SECURITY.md           |   0
 .../01. Array Hashing/1. Two Sum.doctree      | Bin 0 -> 23395 bytes
 .../128. Longest Consecutive Sequence.doctree | Bin 0 -> 27828 bytes
 .../217. Contains Duplicate.doctree           | Bin 0 -> 26460 bytes
 .../238. Product of Array Except Self.doctree | Bin 0 -> 29162 bytes
 .../242. Valid Anagram.doctree                | Bin 0 -> 31369 bytes
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+      "Requirement already satisfied: sphinx-thebe~=0.2.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.2.1)\n",
+      "Requirement already satisfied: sphinx-book-theme~=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (1.0.1)\n",
+      "Requirement already satisfied: sphinx_togglebutton in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.2)\n",
+      "Requirement already satisfied: sphinxcontrib-bibtex<=2.5.0,>=2.2.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (2.5.0)\n",
+      "Requirement already satisfied: sphinx-multitoc-numbering~=0.1.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.1.3)\n",
+      "Requirement already satisfied: attrs>=22.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (23.1.0)\n",
+      "Requirement already satisfied: jsonschema-specifications>=2023.03.6 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (2023.7.1)\n",
+      "Requirement already satisfied: referencing>=0.28.4 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (0.30.2)\n",
+      "Requirement already satisfied: rpds-py>=0.7.1 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (0.10.6)\n",
+      "Requirement already satisfied: uc-micro-py in /usr/local/python/3.10.8/lib/python3.10/site-packages (from linkify-it-py~=2.0.0->jupyter-book) (1.0.2)\n",
+      "Requirement already satisfied: importlib_metadata in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (6.8.0)\n",
+      "Requirement already satisfied: ipython in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (8.16.1)\n",
+      "Requirement already satisfied: jupyter-cache<0.7,>=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.6.1)\n",
+      "Requirement already satisfied: nbclient in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.7.4)\n",
+      "Requirement already satisfied: myst-parser~=0.18.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.18.1)\n",
+      "Requirement already satisfied: nbformat~=5.0 in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (5.9.2)\n",
+      "Requirement already satisfied: typing-extensions in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (4.8.0)\n",
+      "Requirement already satisfied: ipykernel in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (6.25.2)\n",
+      "Requirement already satisfied: sphinxcontrib-applehelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.0.7)\n",
+      "Requirement already satisfied: sphinxcontrib-devhelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.0.5)\n",
+      "Requirement already satisfied: sphinxcontrib-jsmath in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.0.1)\n",
+      "Requirement already satisfied: sphinxcontrib-htmlhelp>=2.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.0.4)\n",
+      "Requirement already satisfied: sphinxcontrib-serializinghtml>=1.1.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.1.9)\n",
+      "Requirement already satisfied: sphinxcontrib-qthelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.0.6)\n",
+      "Requirement already satisfied: Pygments>=2.0 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.16.1)\n",
+      "Requirement already satisfied: snowballstemmer>=1.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.2.0)\n",
+      "Requirement already satisfied: babel>=1.3 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.13.0)\n",
+      "Requirement already satisfied: alabaster<0.8,>=0.7 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (0.7.13)\n",
+      "Requirement already satisfied: imagesize in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.4.1)\n",
+      "Requirement already satisfied: requests>=2.5.0 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.31.0)\n",
+      "Requirement already satisfied: packaging in /home/codespace/.local/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (23.2)\n",
+      "Requirement already satisfied: MarkupSafe>=2.0 in /home/codespace/.local/lib/python3.10/site-packages (from Jinja2->jupyter-book) (2.1.3)\n",
+      "Requirement already satisfied: pydata-sphinx-theme>=0.13.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx-book-theme~=1.0.0->jupyter-book) (0.14.3)\n",
+      "Requirement already satisfied: pybtex>=0.24 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (0.24.0)\n",
+      "Requirement already satisfied: pybtex-docutils>=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (1.0.3)\n",
+      "Requirement already satisfied: setuptools in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx_togglebutton->jupyter-book) (68.0.0)\n",
+      "Requirement already satisfied: wheel in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx_togglebutton->jupyter-book) (0.41.3)\n",
+      "Requirement already satisfied: sqlalchemy<3,>=1.3.12 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (2.0.23)\n",
+      "Requirement already satisfied: tabulate in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (0.9.0)\n",
+      "Requirement already satisfied: markdown-it-py<3.0.0,>=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (2.2.0)\n",
+      "Requirement already satisfied: mdit-py-plugins~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.3.5)\n",
+      "Requirement already satisfied: jupyter-client>=6.1.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (8.4.0)\n",
+      "Requirement already satisfied: jupyter-core!=5.0.*,>=4.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (5.4.0)\n",
+      "Requirement already satisfied: traitlets>=5.3 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (5.11.2)\n",
+      "Requirement already satisfied: fastjsonschema in /home/codespace/.local/lib/python3.10/site-packages (from nbformat~=5.0->myst-nb~=0.17.1->jupyter-book) (2.18.1)\n",
+      "Requirement already satisfied: latexcodec>=1.0.4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pybtex>=0.24->sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (2.0.1)\n",
+      "Requirement already satisfied: six in /home/codespace/.local/lib/python3.10/site-packages (from pybtex>=0.24->sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (1.16.0)\n",
+      "Requirement already satisfied: beautifulsoup4 in /home/codespace/.local/lib/python3.10/site-packages (from pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (4.12.2)\n",
+      "Requirement already satisfied: accessible-pygments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (0.0.4)\n",
+      "Requirement already satisfied: charset-normalizer<4,>=2 in /home/codespace/.local/lib/python3.10/site-packages (from requests>=2.5.0->sphinx<6,>=4->jupyter-book) (3.3.0)\n",
+      "Requirement already satisfied: idna<4,>=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from requests>=2.5.0->sphinx<6,>=4->jupyter-book) (3.4)\n",
+      "Requirement already satisfied: urllib3<3,>=1.21.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from requests>=2.5.0->sphinx<6,>=4->jupyter-book) (1.26.18)\n",
+      "Requirement already satisfied: certifi>=2017.4.17 in /home/codespace/.local/lib/python3.10/site-packages (from requests>=2.5.0->sphinx<6,>=4->jupyter-book) (2023.7.22)\n",
+      "Requirement already satisfied: zipp>=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from importlib_metadata->myst-nb~=0.17.1->jupyter-book) (3.17.0)\n",
+      "Requirement already satisfied: comm>=0.1.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (0.1.4)\n",
+      "Requirement already satisfied: debugpy>=1.6.5 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (1.8.0)\n",
+      "Requirement already satisfied: matplotlib-inline>=0.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (0.1.6)\n",
+      "Requirement already satisfied: nest-asyncio in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (1.5.8)\n",
+      "Requirement already satisfied: psutil in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (5.9.6)\n",
+      "Requirement already satisfied: pyzmq>=20 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (25.1.1)\n",
+      "Requirement already satisfied: tornado>=6.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (6.3.3)\n",
+      "Requirement already satisfied: backcall in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.2.0)\n",
+      "Requirement already satisfied: decorator in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (5.1.1)\n",
+      "Requirement already satisfied: jedi>=0.16 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.19.1)\n",
+      "Requirement already satisfied: pickleshare in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.7.5)\n",
+      "Requirement already satisfied: prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (3.0.39)\n",
+      "Requirement already satisfied: stack-data in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.6.3)\n",
+      "Requirement already satisfied: exceptiongroup in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (1.1.3)\n",
+      "Requirement already satisfied: pexpect>4.3 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (4.8.0)\n",
+      "Requirement already satisfied: parso<0.9.0,>=0.8.3 in /home/codespace/.local/lib/python3.10/site-packages (from jedi>=0.16->ipython->myst-nb~=0.17.1->jupyter-book) (0.8.3)\n",
+      "Requirement already satisfied: python-dateutil>=2.8.2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-client>=6.1.12->nbclient->myst-nb~=0.17.1->jupyter-book) (2.8.2)\n",
+      "Requirement already satisfied: platformdirs>=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-core!=5.0.*,>=4.12->nbclient->myst-nb~=0.17.1->jupyter-book) (3.11.0)\n",
+      "Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py<3.0.0,>=1.0.0->myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.1.2)\n",
+      "Requirement already satisfied: ptyprocess>=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect>4.3->ipython->myst-nb~=0.17.1->jupyter-book) (0.7.0)\n",
+      "Requirement already satisfied: wcwidth in /home/codespace/.local/lib/python3.10/site-packages (from prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.8)\n",
+      "Requirement already satisfied: greenlet!=0.4.17 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sqlalchemy<3,>=1.3.12->jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (3.0.1)\n",
+      "Requirement already satisfied: soupsieve>1.2 in /home/codespace/.local/lib/python3.10/site-packages (from beautifulsoup4->pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (2.5)\n",
+      "Requirement already satisfied: executing>=1.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.0.0)\n",
+      "Requirement already satisfied: asttokens>=2.1.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.4.0)\n",
+      "Requirement already satisfied: pure-eval in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.2)\n",
+      "Note: you may need to restart the kernel to use updated packages.\n"
+     ]
+    }
+   ],
+   "source": [
+    "!python -m pip install --upgrade pip\n",
+    "%pip install jupyter-book"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 34,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions\n"
+     ]
+    }
+   ],
+   "source": [
+    "%cd /workspaces/LeetCode-Solutions"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "!jupyter-book toc migrate /workspaces/LeetCode-Solutions/_toc.yml -o /workspaces/LeetCode-Solutions/_toc.yml"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
+      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions\n",
+      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\n",
+      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\n",
+      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
+      "\u001b[01mmaking output directory... \u001b[39;49;00mdone\n",
+      "[etoc] Changing master_doc to 'README'\n",
+      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
+      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
+      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\n",
+      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
+      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 77 source files that are out of date\n",
+      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 77 added, 0 changed, 0 removed\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]  \n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.26 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.00 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]     \n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.92 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 1.28 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executed notebook in 0.91 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executing notebook using local CWD [mystnb]      \n",
+      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executed notebook in 0.91 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executed notebook in 1.17 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executed notebook in 1.16 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executing notebook using local CWD [mystnb] \n",
+      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executed notebook in 1.27 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executed notebook in 4.03 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executing notebook using local CWD [mystnb]   \n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/Book.ipynb: Executing notebook using local CWD [mystnb]                          \n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mSECURITY\u001b[39;49;00m                                              \n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb:30008: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Executing notebook failed: CellTimeoutError [mystnb.exec]\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Notebook exception traceback saved in: /workspaces/LeetCode-Solutions/_build/html/reports/Book.err.log [mystnb.exec]\u001b[39;49;00m\n",
+      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
+      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
+      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/CODE_OF_CONDUCT.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/SECURITY.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "done\n",
+      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
+      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mSECURITY\u001b[39;49;00m                                               raph\u001b[39;49;00m0m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:8: WARNING: 'myst' reference target not found: #ways-to-contribute\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:9: WARNING: 'myst' reference target not found: #getting-started\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:10: WARNING: 'myst' reference target not found: #contributing-process\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:11: WARNING: 'myst' reference target not found: #code-of-conduct\u001b[39;49;00m\n",
+      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
+      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
+      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
+      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
+      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
+      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
+      "[etoc] missing index.html written as redirect to 'README.html'\n",
+      "\u001b[01mbuild succeeded, 11 warnings.\u001b[39;49;00m\n",
+      "\n",
+      "The HTML pages are in _build/html.\n",
+      "\n",
+      "\u001b[92m===============================================================================\u001b[0m\n",
+      "\n",
+      "Finished generating HTML for book.\n",
+      "Your book's HTML pages are here:\n",
+      "    _build/html/\n",
+      "You can look at your book by opening this file in a browser:\n",
+      "    _build/html/index.html\n",
+      "Or paste this line directly into your browser bar:\n",
+      "    file:///workspaces/LeetCode-Solutions/_build/html/index.html            \n",
+      "\n",
+      "\u001b[92m===============================================================================\u001b[0m\n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "!jupyter-book build ."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: ghp-import in /usr/local/python/3.10.8/lib/python3.10/site-packages (2.1.0)\n",
+      "Requirement already satisfied: python-dateutil>=2.8.1 in /home/codespace/.local/lib/python3.10/site-packages (from ghp-import) (2.8.2)\n",
+      "Requirement already satisfied: six>=1.5 in /home/codespace/.local/lib/python3.10/site-packages (from python-dateutil>=2.8.1->ghp-import) (1.16.0)\n",
+      "Note: you may need to restart the kernel to use updated packages.\n"
+     ]
+    }
+   ],
+   "source": [
+    "%pip install ghp-import"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
diff --git a/_sources/CODE_OF_CONDUCT.md b/CODE_OF_CONDUCT.md
similarity index 100%
rename from _sources/CODE_OF_CONDUCT.md
rename to CODE_OF_CONDUCT.md
diff --git a/_sources/CONTRIBUTING.md b/CONTRIBUTING.md
similarity index 100%
rename from _sources/CONTRIBUTING.md
rename to CONTRIBUTING.md
diff --git a/LICENSE b/LICENSE
new file mode 100644
index 0000000..8818d1f
--- /dev/null
+++ b/LICENSE
@@ -0,0 +1,21 @@
+MIT License
+
+Copyright (c) 2023 Mohsen Tabibian
+
+Permission is hereby granted, free of charge, to any person obtaining a copy
+of this software and associated documentation files (the "Software"), to deal
+in the Software without restriction, including without limitation the rights
+to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
+copies of the Software, and to permit persons to whom the Software is
+furnished to do so, subject to the following conditions:
+
+The above copyright notice and this permission notice shall be included in all
+copies or substantial portions of the Software.
+
+THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
+IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
+FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
+AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
+LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
+OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
+SOFTWARE.
diff --git a/_sources/README.md b/README.md
similarity index 100%
rename from _sources/README.md
rename to README.md
diff --git a/_sources/SECURITY.md b/SECURITY.md
similarity index 100%
rename from _sources/SECURITY.md
rename to SECURITY.md
diff --git a/_build/.doctrees/01. Array Hashing/1. Two Sum.doctree b/_build/.doctrees/01. Array Hashing/1. Two Sum.doctree
new file mode 100644
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diff --git a/_build/.doctrees/01. Array Hashing/128. Longest Consecutive Sequence.doctree b/_build/.doctrees/01. Array Hashing/128. Longest Consecutive Sequence.doctree
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diff --git a/_build/.doctrees/01. Array Hashing/217. Contains Duplicate.doctree b/_build/.doctrees/01. Array Hashing/217. Contains Duplicate.doctree
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diff --git a/_build/.doctrees/01. Array Hashing/242. Valid Anagram.doctree b/_build/.doctrees/01. Array Hashing/242. Valid Anagram.doctree
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diff --git a/_build/.doctrees/01. Array Hashing/659. Encode and Decode Strings.doctree b/_build/.doctrees/01. Array Hashing/659. Encode and Decode Strings.doctree
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diff --git a/_build/.doctrees/02. Two Pointers/125. Valid Palindrome.doctree b/_build/.doctrees/02. Two Pointers/125. Valid Palindrome.doctree
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diff --git a/_build/.doctrees/03. Sliding Window/424. Longest Repeating Character Replacement.doctree b/_build/.doctrees/03. Sliding Window/424. Longest Repeating Character Replacement.doctree
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diff --git a/_build/.doctrees/05. Binary Search/33. Search in Rotated Sorted Array.doctree b/_build/.doctrees/05. Binary Search/33. Search in Rotated Sorted Array.doctree
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diff --git a/_build/.doctrees/06. Linked List/19. Remove Nth Node From End of List.doctree b/_build/.doctrees/06. Linked List/19. Remove Nth Node From End of List.doctree
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diff --git a/_build/.doctrees/06. Linked List/206. Reverse Linked List.doctree b/_build/.doctrees/06. Linked List/206. Reverse Linked List.doctree
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diff --git a/_build/.doctrees/06. Linked List/21. Merge Two Sorted Lists.doctree b/_build/.doctrees/06. Linked List/21. Merge Two Sorted Lists.doctree
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diff --git a/_build/.doctrees/07. Trees/230. Kth Smallest Element in a BST.doctree b/_build/.doctrees/07. Trees/230. Kth Smallest Element in a BST.doctree
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diff --git a/_build/.doctrees/07. Trees/297. Serialize and Deserialize Binary Tree.doctree b/_build/.doctrees/07. Trees/297. Serialize and Deserialize Binary Tree.doctree
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diff --git a/_build/.doctrees/08. Tries/211. Design Add and Search Words Data Structure.doctree b/_build/.doctrees/08. Tries/211. Design Add and Search Words Data Structure.doctree
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diff --git a/_build/.doctrees/09. Heap - Priority Queue/295. Find Median from Data Stream.doctree b/_build/.doctrees/09. Heap - Priority Queue/295. Find Median from Data Stream.doctree
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diff --git a/_build/.doctrees/11. Graphs/3651. Number of Connected Components in an Undirected Graph.doctree b/_build/.doctrees/11. Graphs/3651. Number of Connected Components in an Undirected Graph.doctree
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diff --git a/_build/.doctrees/12. Advanced Graphs/892. Alien Dictionary.doctree b/_build/.doctrees/12. Advanced Graphs/892. Alien Dictionary.doctree
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+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 100: Same Tree\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
+        "\n",
+        "---\n",
+        "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
+        "\n",
+        "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
+        "\n",
+        "**Constraints**\n",
+        "\n",
+        "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
+        "2. $-10^4$ <= `Node.val` <= $10^4$"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 5,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Definition for a binary tree node.\n",
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "def isSameTree(p, q):\n",
+        "    # Base case: If both p and q are None, the trees are the same.\n",
+        "    if not p and not q:\n",
+        "        return True\n",
+        "    \n",
+        "    # Base case: If either p or q is None (but not both), the trees are different.\n",
+        "    if not p or not q:\n",
+        "        return False\n",
+        "    \n",
+        "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
+        "    if p.val != q.val:\n",
+        "        return False\n",
+        "    \n",
+        "    # Recursively check the left and right subtrees of p and q.\n",
+        "    # If both subtrees are the same, the entire trees are the same.\n",
+        "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
+        "\n",
+        "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
+        "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
+        "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
+        "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
+        "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
+        "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 2,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 1\n",
+        "\n",
+        "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
+        "\n",
+        "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
+        "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
+        "print(isSameTree(p1, q1)) "
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 3,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 2\n",
+        "\n",
+        "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
+        "\n",
+        "p2 = TreeNode(1, TreeNode(2), None)\n",
+        "q2 = TreeNode(1, None, TreeNode(2))\n",
+        "print(isSameTree(p2, q2))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 4,
+      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 3\n",
+        "\n",
+        "#Input: p = [1,2,1], q = [1,1,2]\n",
+        "\n",
+        "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
+        "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
+        "print(isSameTree(p3, q3))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "**Time Complexity**\n",
+        "\n",
+        "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
+        "\n",
+        "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
+        "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
+        "\n",
+        "**Space Complexity**\n",
+        "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
+        "\n",
+        "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
+        "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
+        "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
+        "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
+        "\n",
+        "**In summary:**\n",
+        "\n",
+        "+ Time Complexity: $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
+        "+ Space Complexity: $O(n)$ due to the recursive call stack."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb b/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb
new file mode 100644
index 0000000..a77a03d
--- /dev/null
+++ b/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb	
@@ -0,0 +1,209 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 128. Longest Consecutive Sequence\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Longest Consecutive Sequence problem on LeetCode, click here!](https://leetcode.com/problems/longest-consecutive-sequence/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Given an unsorted array of integers `nums`, return *the length of the longest consecutive elements sequence.*\n",
+        "\n",
+        "You must write an algorithm that runs in `O(n)` time.\n",
+        "\n",
+        "*Constraints:**\n",
+        "- 0 <= `nums.length` <= $10^5$\n",
+        "- $-10^9$ <= `nums[i]` <= $10^9$\n",
+        "\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 58,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def longestConsecutive(nums):\n",
+        "    if not nums:\n",
+        "        return 0\n",
+        "\n",
+        "    num_set = set(nums)\n",
+        "    max_length = 0\n",
+        "\n",
+        "    for num in num_set:\n",
+        "        if num - 1 not in num_set:\n",
+        "            current_num = num\n",
+        "            current_length = 1\n",
+        "\n",
+        "            while current_num + 1 in num_set:\n",
+        "                current_num += 1\n",
+        "                current_length += 1\n",
+        "\n",
+        "            max_length = max(max_length, current_length)\n",
+        "\n",
+        "    return max_length"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We start by defining the `longestConsecutive` function that takes the `nums` array as input.\n",
+        "\n",
+        "2. We check if the `nums` array is empty. If it's empty, there are no consecutive elements, so we return 0.\n",
+        "\n",
+        "3. We create a Python set called `num_set` and insert all elements from the `nums` array into it. Using a set allows us to efficiently check for the existence of elements in O(1) time.\n",
+        "\n",
+        "4. We initialize a variable `max_length` to 0. This variable will keep track of the maximum length of consecutive elements sequence found.\n",
+        "\n",
+        "5. We iterate through the elements of the `num_set`. For each element `num`, we check if `num - 1` exists in the `num_set`. If it doesn't exist, it means `num` is the starting element of a potential consecutive sequence.\n",
+        "\n",
+        "6. Inside the loop, we initialize two variables: `current_num` to the current element `num` and `current_length` to 1. We start with a length of 1 because `num` itself is part of the sequence.\n",
+        "\n",
+        "7. We then enter a while loop that continues as long as `current_num + 1` exists in the `num_set`. This means we are incrementing the consecutive sequence.\n",
+        "\n",
+        "8. Inside the while loop, we increment `current_num` by 1 and also increment `current_length` by 1 to account for the next consecutive element.\n",
+        "\n",
+        "9. We compare `current_length` with the `max_length` and update `max_length` if the current sequence is longer.\n",
+        "\n",
+        "10. After the while loop, we move to the next element in the outer loop and repeat the process.\n",
+        "\n",
+        "11. Finally, we return the `max_length` as the result, which represents the length of the longest consecutive elements sequence in the `nums` array.\n",
+        "\n",
+        "The key idea here is to use a set to efficiently check for the existence of elements and to iterate through the elements, considering each element as the potential start of a consecutive sequence. By doing this, we can find the longest consecutive sequence in O(n) time complexity, where n is the number of elements in the array."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 59,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "4\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "nums1 = [100, 4, 200, 1, 3, 2]\n",
+        "print(longestConsecutive(nums1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 60,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "9\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "\n",
+        "nums2 = [0, 3, 7, 2, 5, 8, 4, 6, 0, 1]\n",
+        "print(longestConsecutive(nums2)) "
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "The code is designed to run in O(n) time complexity, where n is the number of elements in the `nums` array. Here's the breakdown:\n",
+        "\n",
+        "1. Constructing the `num_set` by inserting all elements from `nums` into it takes O(n) time because we perform an insertion operation for each element in `nums`.\n",
+        "\n",
+        "2. The main loop iterates through the elements in `num_set`. In the worst case, each element is visited only once, so the loop itself takes O(n) time.\n",
+        "\n",
+        "3. Within the loop, we have a while loop that may also take O(n) time in the worst case. However, this while loop is nested inside the main loop, so its overall time complexity remains O(n).\n",
+        "\n",
+        "Therefore, the overall time complexity of the code is O(n).\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "The space complexity of the code is determined by the space used by the `num_set` and a few additional variables. Here's the breakdown:\n",
+        "\n",
+        "1. `num_set` is a set that stores the unique elements from `nums`. In the worst case, it can store all n elements from `nums`, so the space complexity is O(n).\n",
+        "\n",
+        "2. The additional variables used, such as `max_length`, `current_num`, and `current_length`, have constant space requirements and do not depend on the size of the input array.\n",
+        "\n",
+        "Therefore, the overall space complexity of the code is O(n) due to the space used by the `num_set`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. Write an algorithm that not only returns the length of the longest consecutive elements sequence but also returns the actual consecutive sequence itself.\n",
+        "\n",
+        "2. Extend the problem to allow elements to be considered consecutive if they are within a certain absolute difference (e.g., less than or equal to k) instead of exactly 1. Write an algorithm that finds the longest such sequence and runs in O(n) time."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb b/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb
new file mode 100644
index 0000000..57bb55d
--- /dev/null
+++ b/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb	
@@ -0,0 +1,218 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 217. Contains Duplicate\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Contains Duplicate problem on LeetCode, click here!](https://leetcode.com/problems/contains-duplicate/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- 1 <= `nums.length` <= $10^5$\n",
+        "- $-10^9$ <= `nums[i]` <= $10^9$\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 19,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def containsDuplicate(nums):\n",
+        "    # Create an empty set to store unique elements\n",
+        "    unique_elements = set()\n",
+        "    \n",
+        "    # Iterate through the array\n",
+        "    for num in nums:\n",
+        "        # If the element is already in the set, return True\n",
+        "        if num in unique_elements:\n",
+        "            return True\n",
+        "        # Otherwise, add it to the set\n",
+        "        else:\n",
+        "            unique_elements.add(num)\n",
+        "    \n",
+        "    # If the loop completes without finding duplicates, return False\n",
+        "    return False"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. The `containsDuplicate` function takes a single argument, `nums`, which is the input integer array.\n",
+        "\n",
+        "2. Inside the function, an empty set called `unique_elements` is created. This set will be used to keep track of unique elements in the input array.\n",
+        "\n",
+        "3. The function then iterates through the input array `nums` using a `for` loop.\n",
+        "\n",
+        "4. For each element `num` in the array, it checks whether `num` is already in the `unique_elements` set using the `if num in unique_elements:` condition.\n",
+        "\n",
+        "5. If `num` is already in the set, it means there is a duplicate element in the array, and the function immediately returns `True`.\n",
+        "\n",
+        "6. If `num` is not in the set, it is added to the `unique_elements` set using `unique_elements.add(num)`.\n",
+        "\n",
+        "7. The loop continues to the next element, and the process repeats.\n",
+        "\n",
+        "8. If the loop completes without finding any duplicates, it means that all elements in the array are distinct, and the function returns `False`.\n",
+        "\n",
+        "The code efficiently utilizes a set data structure to keep track of unique elements while iterating through the array, allowing it to quickly detect duplicate elements. This code meets the problem's requirements and constraints, as explained earlier."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 20,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "nums1 = [1, 2, 3, 1]\n",
+        "print(containsDuplicate(nums1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 21,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "nums2 = [1, 2, 3, 4]\n",
+        "print(containsDuplicate(nums2))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 22,
+      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3\n",
+        "nums3 = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]\n",
+        "print(containsDuplicate(nums3))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code for the \"Contains Duplicate\" problem:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "The primary operation that affects the time complexity is the `for` loop that iterates through the input array `nums`. In the worst case, the loop will iterate through all `n` elements of the array, where `n` is the length of the input array.\n",
+        "\n",
+        "- Iterating through the array: O(n)\n",
+        "\n",
+        "Inside the loop, we perform two operations:\n",
+        "\n",
+        "1. Checking whether an element exists in the `unique_elements` set (`if num in unique_elements`). This operation has an average time complexity of O(1) for a set.\n",
+        "2. Adding an element to the `unique_elements` set (`unique_elements.add(num)`), which also has an average time complexity of O(1) for a set.\n",
+        "\n",
+        "Since these operations are performed for each element in the array, the overall time complexity remains O(n).\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "The space complexity is determined by the additional data structures used in the code, which are the `unique_elements` set.\n",
+        "\n",
+        "- `unique_elements` set: This set stores unique elements from the input array. In the worst case, if all elements in the input array are distinct, the set will store all `n` elements.\n",
+        "\n",
+        "Therefore, the space complexity is O(n) because, in the worst case, the set's size will grow linearly with the input array's size.\n",
+        "\n",
+        "**In summary**, the time complexity of the code is O(n), where `n` is the length of the input array, and the space complexity is also O(n) in the worst case. This code is efficient and meets the constraints of the problem."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **K Duplicates:** Modify the problem to return `true` if there are exactly K duplicate elements in the array. Write a function that takes an additional integer parameter K and returns `true` if there are exactly K duplicates.\n",
+        "2. **Frequency Count:** Write a function that returns a list of all the elements that appear more than once in the array, along with their frequencies."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb b/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb
new file mode 100644
index 0000000..1ff9a9c
--- /dev/null
+++ b/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb	
@@ -0,0 +1,197 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 238. Product of Array Except Self\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Product of Array Except Self problem on LeetCode, click here!](https://leetcode.com/problems/product-of-array-except-self/)\n",
+        "\n",
+        "---\n",
+        "Given an integer array `nums`, return *an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.*\n",
+        "\n",
+        "The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
+        "\n",
+        "You must write an algorithm that runs in O(n) time and without using the division operation.\n",
+        "\n",
+        "Constraints:\n",
+        "- 2 <= `nums.length` <= $10^5$\n",
+        "- `-30 <= nums[i] <= 30`\n",
+        "- The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
+        "\n",
+        "**Follow-up:** Can you solve the problem in O(1) extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 50,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def productExceptSelf(nums):\n",
+        "    n = len(nums)\n",
+        "    \n",
+        "    # Initialize two arrays to store products to the left and right of each element\n",
+        "    left_products = [1] * n\n",
+        "    right_products = [1] * n\n",
+        "    \n",
+        "    # Calculate products to the left of each element\n",
+        "    left_product = 1\n",
+        "    for i in range(n):\n",
+        "        left_products[i] = left_product\n",
+        "        left_product *= nums[i]\n",
+        "    \n",
+        "    # Calculate products to the right of each element\n",
+        "    right_product = 1\n",
+        "    for i in range(n - 1, -1, -1):\n",
+        "        right_products[i] = right_product\n",
+        "        right_product *= nums[i]\n",
+        "    \n",
+        "    # Calculate the final answer using left and right products\n",
+        "    answer = [left_products[i] * right_products[i] for i in range(n)]\n",
+        "    \n",
+        "    return answer"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. `n` is initialized as the length of the `nums` array.\n",
+        "2. Two arrays, `left_products` and `right_products`, are created, each with a length of `n` and filled with ones initially.\n",
+        "3. `left_product` is initialized to 1.\n",
+        "4. A loop iterates through the `nums` array, from left to right. For each element at index `i`, it stores the product of all elements to its left (including itself) in the `left_products` array and updates `left_product` accordingly.\n",
+        "5. `right_product` is initialized to 1.\n",
+        "6. Another loop iterates through the `nums` array in reverse order, from right to left. For each element at index `i`, it stores the product of all elements to its right (including itself) in the `right_products` array and updates `right_product`.\n",
+        "7. The final `answer` list is constructed by multiplying corresponding elements from the `left_products` and `right_products` arrays for each index `i`.\n",
+        "8. The function returns the `answer` list, which contains the desired results.\n",
+        "\n",
+        "Overall, this algorithm efficiently computes the product of all elements except the current element, as required. It runs in O(n) time complexity and uses O(n) extra space, which is within the problem's constraints."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 52,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[24, 12, 8, 6]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "nums = [1,2,3,4]\n",
+        "result1 = productExceptSelf(nums)\n",
+        "print(result1)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 53,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[0, 0, 9, 0, 0]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "nums = [-1,1,0,-3,3]\n",
+        "result2 = productExceptSelf(nums)\n",
+        "print(result2)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "1. The first loop iterates through the `nums` array to calculate the products to the left of each element. This loop runs in O(n) time, where n is the length of the `nums` array.\n",
+        "2. The second loop also iterates through the `nums` array in reverse order to calculate the products to the right of each element. This loop also runs in O(n) time.\n",
+        "3. The final loop constructs the `answer` list by multiplying elements from the `left_products` and `right_products` arrays. This loop runs in O(n) time.\n",
+        "\n",
+        "Since all the loops are independent and sequential, the overall time complexity is O(n).\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "1. Two additional arrays, `left_products` and `right_products`, are created with a length of `n`, where `n` is the length of the `nums` array. Therefore, these arrays consume O(n) extra space.\n",
+        "2. The `left_product` and `right_product` variables are used to keep track of the product to the left and right of each element, respectively. These variables occupy O(1) extra space.\n",
+        "3. The `answer` list, which contains the final results, also consumes O(n) space, but it's not counted toward the extra space complexity analysis as per the problem's constraints.\n",
+        "\n",
+        "Overall, the space complexity is O(n) for this solution."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Handle Zeros in the Input**: Modify the solution to handle cases where the input array contains zeros. For example, if `nums = [0, 1, 2, 3, 0]`, the output should be `[0, 0, 0, 0, 0]` because any element multiplied by zero results in zero.\n",
+        "\n",
+        "2. **In-Place Modification**: Attempt to solve the problem with in-place modification of the `nums` array, where the output is stored directly in the input array without using any additional space."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb b/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb
new file mode 100644
index 0000000..d165899
--- /dev/null
+++ b/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb	
@@ -0,0 +1,198 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 242. Valid Anagram\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Valid Anagram problem on LeetCode, click here!](https://leetcode.com/problems/valid-anagram/)\n",
+        "\n",
+        "---\n",
+        "Given two strings `s` and `t`, return `true` if `t` is an anagram of `s`, and `false` otherwise.\n",
+        "\n",
+        "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- 1 <= `s.length`, `t.length` <= $5 * 10^4$\n",
+        "- `s` and `t` consist of lowercase English letters.\n",
+        "\n",
+        "**Follow-up:** What if the inputs contain Unicode characters? How would you adapt your solution to such a case?\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 29,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def isAnagram(s: str, t: str) -> bool:\n",
+        "    # Check if the lengths of s and t are not equal\n",
+        "    if len(s) != len(t):\n",
+        "        return False\n",
+        "\n",
+        "    # Create dictionaries to store character frequencies for s and t\n",
+        "    countS, countT = {}, {}\n",
+        "\n",
+        "    # Count character frequencies in s\n",
+        "    for i in range(len(s)):\n",
+        "        countS[s[i]] = 1 + countS.get(s[i], 0)\n",
+        "\n",
+        "    # Count character frequencies in t\n",
+        "    for i in range(len(t)):\n",
+        "        countT[t[i]] = 1 + countT.get(t[i], 0)\n",
+        "\n",
+        "    # Check if the character frequencies in s and t are the same\n",
+        "    return countS == countT"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. The code defines a class called `Solution` with a method `isAnagram` that takes two input strings, `s` and `t`, and returns a boolean (`True` or `False`) based on whether `t` is an anagram of `s`.\n",
+        "\n",
+        "2. The first check performed in the method is whether the lengths of `s` and `t` are equal. If they are not equal, it immediately returns `False` because strings of different lengths cannot be anagrams of each other.\n",
+        "\n",
+        "3. Two dictionaries, `countS` and `countT`, are created to store the frequency of characters in `s` and `t`, respectively. These dictionaries will be used to count the occurrences of each character in the strings.\n",
+        "\n",
+        "4. The code then enters a loop to iterate through each character in string `s` using a `for` loop. Inside the loop, it updates the `countS` dictionary. The line `countS[s[i]] = 1 + countS.get(s[i], 0)` increments the count of character `s[i]` in `countS` by 1. If the character is not already in the dictionary, it initializes the count to 1.\n",
+        "\n",
+        "5. Similarly, the code iterates through each character in string `t` and updates the `countT` dictionary in the same way.\n",
+        "\n",
+        "6. After counting the character frequencies in both strings, the code compares the two dictionaries using `countS == countT`. If the dictionaries are equal, it means that both strings have the same character frequencies, and therefore, `t` is an anagram of `s`. In this case, the method returns `True`.\n",
+        "\n",
+        "7. If the dictionaries are not equal, the method returns `False`, indicating that `t` is not an anagram of `s`.\n",
+        "\n",
+        "In summary, this code efficiently determines whether two input strings are anagrams by counting the character frequencies in each string using dictionaries and then comparing these counts. If the character frequencies match, the strings are considered anagrams, and the method returns `True`; otherwise, it returns `False`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 30,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "s = \"anagram\"\n",
+        "t = \"nagaram\"\n",
+        "print(isAnagram(s, t))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 31,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "s = \"rat\"\n",
+        "t = \"car\"\n",
+        "print(isAnagram(s, t))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "1. The code first checks whether the lengths of the input strings `s` and `t` are equal, which takes constant time. This check has a time complexity of O(1).\n",
+        "\n",
+        "2. The code then iterates through both strings, `s` and `t`, once to count the character frequencies. Since both strings have a maximum length of 5 * 10^4, the worst-case scenario is that the code iterates through 5 * 10^4 characters in each string. This results in a linear time complexity of O(n), where n is the length of the longer of the two input strings (`s` or `t`).\n",
+        "\n",
+        "3. Finally, the code compares the two dictionaries `countS` and `countT` to check if the character frequencies match. This comparison takes O(n) time in the worst case, where n is the length of the longer string.\n",
+        "\n",
+        "Overall, the time complexity of the code is O(n), where n is the length of the longer input string.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "1. The code uses two dictionaries, `countS` and `countT`, to store the character frequencies of the input strings `s` and `t`. In the worst case, both dictionaries can contain all unique characters from the input strings. Since the input strings can have a maximum length of 5 * 10^4, the space complexity for these dictionaries is O(5 * 10^4) or simply O(n), where n is the length of the longer input string.\n",
+        "\n",
+        "2. The code uses a few additional variables for bookkeeping, but these variables have constant space requirements and do not depend on the input size. Therefore, they do not significantly impact the overall space complexity.\n",
+        "\n",
+        "**In summary**, the space complexity of the code is O(n), where n is the length of the longer input string."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Anagram Chains**: Given a list of words, find the longest chain of anagrams, where each word in the chain is an anagram of the previous word. For example, given [\"bat\", \"tab\", \"cat\", \"tac\", \"dog\"], the longest anagram chain is [\"bat\", \"tab\", \"cat\", \"tac\"].\n",
+        "\n",
+        "2. **Minimum Deletions to Make Anagrams**: Given two strings `s` and `t`, find the minimum number of character deletions required in both strings to make them anagrams of each other. For example, if `s = \"hello\"` and `t = \"billion\"`, you can remove the characters \"heo\" from `s` and \"billi\" from `t` to make them anagrams (\"lo\" and \"on\" are left in the two strings)."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb b/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb
new file mode 100644
index 0000000..ecfa494
--- /dev/null
+++ b/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb	
@@ -0,0 +1,215 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 347. Top K Frequent Elements\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Top K Frequent Elements problem on LeetCode, click here!](https://leetcode.com/problems/top-k-frequent-elements/)\n",
+        "\n",
+        "---\n",
+        "Given an integer array `nums` and an integer `k`, return the `k` most frequent elements. You may return the answer in **any order**.\n",
+        "\n",
+        "Constraints:\n",
+        "- 1 <= `nums.length` <= $10^5$\n",
+        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
+        "- `k` is in the range `[1, the number of unique elements in the array]`.\n",
+        "- It is **guaranteed** that the answer is **unique**.\n",
+        "\n",
+        "**Follow-up:** Your algorithm's time complexity must be better than $O(n\\ log\\ n)$, where n is the array's size.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 46,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def topKFrequent(nums, k):\n",
+        "    # Create a dictionary to store the frequency of each element\n",
+        "    count = {}\n",
+        "    # Create a list of lists to store elements with the same frequency\n",
+        "    frequency = [[] for i in range(len(nums) + 1)]\n",
+        "\n",
+        "    # Count the frequency of each element in nums\n",
+        "    for n in nums:\n",
+        "        count[n] = 1 + count.get(n, 0)\n",
+        "\n",
+        "    # Place elements in the freq list according to their frequency\n",
+        "    for n, c in count.items():\n",
+        "        frequency[c].append(n)\n",
+        "\n",
+        "    res = []\n",
+        "    # Traverse freq list from the end (higher frequencies)\n",
+        "    for i in range(len(frequency) - 1, 0, -1):\n",
+        "        for n in frequency[i]:\n",
+        "            res.append(n)\n",
+        "            if len(res) == k:\n",
+        "                return res"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. `def topKFrequent(nums, k):`: This is a function that takes two arguments: `nums`, which is the input array of integers, and `k`, which is the number of most frequent elements to return.\n",
+        "\n",
+        "2. `count = {}`: This dictionary `count` will be used to store the frequency of each unique element in the `nums` array. The keys are elements from the input array, and the values are their corresponding frequencies.\n",
+        "\n",
+        "3. `frequency = [[] for i in range(len(nums) + 1)]`: This creates a list of empty lists called `frequency`. It's used to store elements based on their frequencies, similar to the `freq` list in the previous code. The size of this list is set to be one greater than the length of the input `nums`.\n",
+        "\n",
+        "4. `for n in nums:`: This loop iterates through each element `n` in the input `nums` array.\n",
+        "\n",
+        "5. `count[n] = 1 + count.get(n, 0)`: This line counts the frequency of each element `n` in the `nums` array. It uses the `count.get(n, 0)` method to retrieve the current count of `n` from the `count` dictionary. If `n` is not in the dictionary, it defaults to 0. It then increments the count by 1.\n",
+        "\n",
+        "6. After the above loop, the `count` dictionary will contain counts of each unique element in the `nums` array.\n",
+        "\n",
+        "7. `for n, c in count.items():`: This loop iterates through the items (key-value pairs) of the `count` dictionary. `n` represents the element, and `c` represents its frequency.\n",
+        "\n",
+        "8. `frequency[c].append(n)`: This line places the element `n` into the bucket corresponding to its frequency `c`. Buckets are represented by the `frequency` list. For example, if an element `n` has a frequency of 3, it will be added to `frequency[3]`.\n",
+        "\n",
+        "9. After this loop, the `frequency` list will contain buckets of elements grouped by their frequencies.\n",
+        "\n",
+        "10. `res = []`: This list `res` will be used to store the k most frequent elements.\n",
+        "\n",
+        "11. `for i in range(len(freq) - 1, 0, -1):`: This loop iterates in reverse order through the `frequency` list, starting from the highest frequency and going down to 1. Note that there is a typo here; it should be `frequency` instead of `freq`.\n",
+        "\n",
+        "12. `for n in frequency[i]:`: For each element `n` in the current bucket (i.e., elements with frequency `i`), it appends `n` to the result list `res`.\n",
+        "\n",
+        "13. `if len(res) == k:`: Once `res` contains k elements, the code exits and returns the result.\n",
+        "\n",
+        "The code efficiently finds the k most frequent elements in the input array without using sorting, similar to the previous explanation, with the only difference being the variable names (e.g., `frequency` instead of `freq`)."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 48,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[1, 2]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "nums = [1,1,1,2,2,3]\n",
+        "k = 2\n",
+        "result1 = topKFrequent(nums, k)\n",
+        "print(result1)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 49,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[1]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "nums = [1]\n",
+        "k = 1\n",
+        "result2 = topKFrequent(nums, k)\n",
+        "print(result2)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "1. The first loop that counts the frequency of elements by iterating through `nums` has a time complexity of O(n), where n is the number of elements in the input array `nums`.\n",
+        "\n",
+        "2. The second loop iterates through the keys in the `count` dictionary, which has at most `k` unique elements (as per the constraints). Therefore, the second loop also has a time complexity of O(k).\n",
+        "\n",
+        "3. The third loop iterates through the `frequency` list, which has a length equal to the maximum frequency of elements in `nums`. In the worst case, this loop can have a time complexity of O(n), where n is the number of elements in `nums`.\n",
+        "\n",
+        "Overall, the dominant factor in terms of time complexity is the loop that iterates through the `count` dictionary. So, the total time complexity of the code is O(n + k).\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "1. The `count` dictionary stores the frequency of each unique element in `nums`. In the worst case, it can have at most `k` unique elements, so the space complexity for `count` is O(k).\n",
+        "\n",
+        "2. The `frequency` list is used to store elements grouped by their frequencies. It has a length of `len(nums) + 1`, which can be at most 105 based on the constraints. Therefore, the space complexity for `frequency` is O(105).\n",
+        "\n",
+        "3. The `res` list stores the k most frequent elements, which can be at most `k` elements, so the space complexity for `res` is O(k).\n",
+        "\n",
+        "**In summary**, the space complexity is dominated by the `count` dictionary and the `frequency` list, both of which have a space complexity of O(k + 105)."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Optimized k Most Frequent Elements:** Modify the code to find the k most frequent elements in an array while ensuring that the time complexity is O(n + klogk). You can use a priority queue (heap) to achieve this.\n",
+        "\n",
+        "2. **Handling Duplicate Frequencies:** Extend the code to handle cases where multiple elements have the same frequency and are among the k most frequent elements. Ensure that the output contains exactly k elements."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb b/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb
new file mode 100644
index 0000000..afe89ec
--- /dev/null
+++ b/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb	
@@ -0,0 +1,229 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 49. Group Anagrams\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Group Anagrams problem on LeetCode, click here!](https://leetcode.com/problems/group-anagrams/)\n",
+        "\n",
+        "---\n",
+        "Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.\n",
+        "\n",
+        "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- 1 <= `strs.length <= $10^4$\n",
+        "- `0 <= strs[i].length <= 100`\n",
+        "- `strs[i]` consists of lowercase English letters.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 36,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "from collections import defaultdict\n",
+        "\n",
+        "\n",
+        "def groupAnagrams(strs):\n",
+        "    ans = defaultdict(list)\n",
+        "\n",
+        "    # Iterate through the list of input strings\n",
+        "    for s in strs:\n",
+        "        # Initialize a list to represent character counts for each character (a-z)\n",
+        "        count = [0] * 26\n",
+        "\n",
+        "        # Count the occurrences of each character in the current word\n",
+        "        for c in s:\n",
+        "            count[ord(c) - ord(\"a\")] += 1\n",
+        "\n",
+        "        # Use a tuple of character counts as the key and append the word to the anagram group\n",
+        "        ans[tuple(count)].append(s)\n",
+        "\n",
+        "    # Convert the values (lists of anagrams) to a list of lists\n",
+        "    return list(ans.values())"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. Inside the method, a `defaultdict` called `ans` is created to store anagram groups. This dictionary will have a list as its default value, meaning that each key in the dictionary will be associated with an empty list by default.\n",
+        "\n",
+        "2. The code then iterates through the list of input strings, `strs`, using a `for` loop. For each word `s` in `strs`, it performs the following steps:\n",
+        "\n",
+        "   - It initializes a list `count` of length 26, where each element represents the count of a specific character (a-z) in the word `s`.\n",
+        "\n",
+        "   - The code then iterates through the characters of `s`. For each character `c`, it increments the corresponding count in the `count` list based on its ASCII value.\n",
+        "\n",
+        "   - After counting the characters in `s`, it converts the `count` list into a tuple `tuple(count)` to use as a key for the `ans` dictionary.\n",
+        "\n",
+        "   - It appends the word `s` to the list associated with the key (tuple) in the `ans` dictionary. This groups all anagrams of the same word together under the same key.\n",
+        "\n",
+        "3. After processing all words in `strs`, the code converts the values of the `ans` dictionary (which are lists of anagrams) to a list of lists using the `list()` constructor.\n",
+        "\n",
+        "4. Finally, the code returns the list of anagram groups, which is the result of grouping the anagrams in the input list.\n",
+        "\n",
+        "In summary, the code efficiently groups anagrams together by counting the characters in each word and using a dictionary to store them under the same key. The result is a list of lists, where each inner list represents a group of anagrams."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 37,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: Group anagrams from a list of words\n",
+        "strs1 = [\"eat\", \"tea\", \"tan\", \"ate\", \"nat\", \"bat\"]\n",
+        "result1 = groupAnagrams(strs1)\n",
+        "print(result1)\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 38,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[['']]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2: Group anagrams from a list with an empty string\n",
+        "strs2 = [\"\"]\n",
+        "result2 = groupAnagrams(strs2)\n",
+        "print(result2)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 40,
+      "id": "c8367e2a-e7d5-4e18-8235-2870e341a04b",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[['a']]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3: Group anagrams from a list with a single word\n",
+        "strs3 = [\"a\"]\n",
+        "result3 = groupAnagrams(strs3)\n",
+        "print(result3)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `groupAnagrams` method:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "1. First, we create an instance of the `Solution` class, which is an O(1) operation.\n",
+        "\n",
+        "2. Inside the `groupAnagrams` method, we iterate through the list of input strings `strs` once in a loop:\n",
+        "   - For each word in `strs`, we perform character counting, which is done in O(K) time, where K is the maximum length of a word in the list.\n",
+        "\n",
+        "3. The overall time complexity of the method is O(N * K), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "1. We use a `defaultdict` called `ans` to store the anagram groups. In the worst case, each word belongs to a distinct anagram group, so the space complexity for `ans` is O(N).\n",
+        "\n",
+        "2. Within the loop, we create a `count` list of length 26 to store character counts for each word. This is a fixed-size list and does not depend on the input size, so it has a constant space complexity of O(26), which is effectively O(1).\n",
+        "\n",
+        "3. The space used for other variables is also constant and does not depend on the input size.\n",
+        "\n",
+        "4. The final result, `result`, is a list of lists that contains the grouped anagrams. In the worst case, each word is a distinct anagram group, so the space complexity for `result` is O(N).\n",
+        "\n",
+        "5. The overall space complexity of the method is O(N) due to the space used by `ans` and `result`. The space used by the `count` list and other variables is constant and does not significantly affect the overall space complexity.\n",
+        "\n",
+        "**In summary,** the time complexity of the `groupAnagrams` method is O(N * K), and the space complexity is O(N), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Anagramic Palindromes**: Given a list of strings, find all the groups of anagrams where each group contains words that can be rearranged into a palindrome. For example, in the list [\"bat\", \"tab\", \"act\", \"tac\", \"dog\"], there are two groups: [\"bat\", \"tab\", \"act\", \"tac\"] and [\"dog\"].\n",
+        "\n",
+        "2. **Largest Anagram Group**: Given a list of strings, find the largest group of anagrams. Return the list of anagrams in that group. If there are multiple largest groups, return all of them."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb b/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb
new file mode 100644
index 0000000..037c476
--- /dev/null
+++ b/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb	
@@ -0,0 +1,204 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 659· Encode and Decode Strings\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Encode and Decode Strings problem on LintCode, click here!](https://www.lintcode.com/problem/659/)\n",
+        "\n",
+        "---\n",
+        "Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.\n",
+        "\n",
+        "Please implement `encode` and `decode`.\n",
+        "\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 54,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class Codec:\n",
+        "\n",
+        "    def encode(self, strs):\n",
+        "        \"\"\"Encodes a list of strings to a single string.\n",
+        "        \n",
+        "        :type strs: List[str]\n",
+        "        :rtype: str\n",
+        "        \"\"\"\n",
+        "        encoded = \"\"\n",
+        "        for s in strs:\n",
+        "            encoded += s.replace(':', '::') + ':;'\n",
+        "        return encoded\n",
+        "\n",
+        "    def decode(self, s):\n",
+        "        \"\"\"Decodes a single string to a list of strings.\n",
+        "        \n",
+        "        :type s: str\n",
+        "        :rtype: List[str]\n",
+        "        \"\"\"\n",
+        "        decoded = []\n",
+        "        i = 0\n",
+        "        while i < len(s):\n",
+        "            end = s.find(':;', i)\n",
+        "            if end == -1:\n",
+        "                end = len(s)\n",
+        "            decoded.append(s[i:end].replace('::', ':'))\n",
+        "            i = end + 2\n",
+        "        return decoded"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We define a class called `Codec` to encapsulate the encoding and decoding operations for a list of strings.\n",
+        "\n",
+        "2. In the `encode` method, we take a list of strings (`strs`) as input and return a single encoded string. The encoding process involves concatenating the strings together with a delimiter `:;`. We iterate through each string in the input list, replace any occurrence of `:` with `::` (to avoid conflicts with the delimiter), and then append `:;` to indicate the end of that string.\n",
+        "\n",
+        "3. In the `decode` method, we take an encoded string (`s`) as input and return a list of strings. The decoding process involves splitting the encoded string using the `:;` delimiter. We start from the beginning of the encoded string (`i = 0`) and repeatedly find the next occurrence of `:;`. We extract the substring between the current position (`i`) and the next delimiter (`end`). Before adding it to the result list, we replace any `::` with `:` to revert the encoding. We then update the current position `i` to be after the delimiter, so we can find the next substring in the next iteration.\n",
+        "\n",
+        "4. Finally, we create an instance of the `Codec` class and test it with two examples:\n",
+        "\n",
+        "By using the `:;` delimiter and handling the `::` escaping, this code can encode and decode lists of strings, preserving the original content even if it contains colons or the delimiter itself."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 55,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "['lint', 'code', 'love', 'you']\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "codec = Codec()\n",
+        "\n",
+        "input1 = [\"lint\", \"code\", \"love\", \"you\"]\n",
+        "encoded1 = codec.encode(input1)\n",
+        "decoded1 = codec.decode(encoded1)\n",
+        "print(decoded1)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 57,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "['we', 'say', ':', 'yes']\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "codec = Codec() \n",
+        "\n",
+        "input2 = [\"we\", \"say\", \":\", \"yes\"]\n",
+        "encoded2 = codec.encode(input2)\n",
+        "decoded2 = codec.decode(encoded2)\n",
+        "print(decoded2)  "
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `encode` and `decode` methods in the provided code.\n",
+        "\n",
+        "**Encode Method (`encode`):**\n",
+        "\n",
+        "- Time Complexity: O(n * m)\n",
+        "  - Here, `n` is the number of strings in the input list `strs`, and `m` is the average length of these strings.\n",
+        "  - In the worst case, for each string in the list, we iterate over its characters once to replace any colons (`:`) with double colons (`::`) and then append `:;`. This is done for each string in the list.\n",
+        "\n",
+        "- Space Complexity: O(n * m)\n",
+        "  - The space complexity is also O(n * m) because we build the encoded string by concatenating the modified strings. In the worst case, the encoded string can be of the same length as the original strings.\n",
+        "\n",
+        "**Decode Method (`decode`):**\n",
+        "\n",
+        "- Time Complexity: O(n * m)\n",
+        "  - Similar to the encode method, we iterate through the encoded string in the decode method. In the worst case, we may have to scan each character in the encoded string to find the `:;` delimiter.\n",
+        "\n",
+        "- Space Complexity: O(n * m)\n",
+        "  - The space complexity of the decode method is also O(n * m) because we build the list of decoded strings, which can be of the same length as the encoded string.\n",
+        "\n",
+        "Overall, both the `encode` and `decode` methods have time and space complexities of O(n * m), where `n` is the number of strings in the input list, and `m` is the average length of these strings. The space complexity arises from storing the encoded or decoded strings, and the time complexity arises from iterating through the characters in these strings."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Multiple Delimiters**: Modify the code to allow for multiple delimiters, not just one. Each string could have its own delimiter, and the code should be able to handle this complexity.\n",
+        "\n",
+        "2. **Optimize Encoding**: Modify the `encode` method to achieve a more efficient encoding algorithm. Try to minimize the length of the encoded string while ensuring that it can be correctly decoded."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/01. Array Hashing/README.md b/_build/html/_sources/01. Array Hashing/README.md
new file mode 100644
index 0000000..a331dab
--- /dev/null
+++ b/_build/html/_sources/01. Array Hashing/README.md	
@@ -0,0 +1,27 @@
+# Array & Hashing Solutions - Blind 75 LeetCode Problems
+![Array & Hashing Solutions](https://img.shields.io/badge/Array_&_Hashing_Solutions-blue)![Blind 75 LeetCode](https://img.shields.io/badge/Blind_75_LeetCode-Problems-blue?labelColor=red)
+
+This repository contains my solutions to Array & Hashing Problems from Blind 75 LeetCode problems. I've organized the solutions by categories for easier navigation and reference. Each problem solution is presented in Jupyter Notebook format (`.ipynb`).
+
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [217. Contains Duplicate](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/217.%20Contains%20Duplicate.ipynb)                            | Easy       |
+| [242. Valid Anagram](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/242.%20Valid%20Anagram.ipynb)                                      | Easy       |
+| [1. Two Sum](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/1.%20Two%20Sum.ipynb)                                                      | Easy       |
+| [49. Group Anagrams](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/49.%20Group%20Anagrams.ipynb)                                      | Medium     |
+| [347. Top K Frequent Elements](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.ipynb)              | Medium     |
+| [238. Product of Array Except Self](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.ipynb)  | Medium     |
+| [659. Encode and Decode Strings](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.ipynb)          | Medium     |
+| [128. Longest Consecutive Sequence](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.ipynb)      | Hard       |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
diff --git a/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb b/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb
new file mode 100644
index 0000000..284c832
--- /dev/null
+++ b/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb	
@@ -0,0 +1,220 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 11. Container With Most Water\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Container With Most Water problem on LeetCode, click here!](https://leetcode.com/problems/container-with-most-water/)\n",
+        "\n",
+        "---\n",
+        "You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the $i^{th}$ line are `(i, 0)` and `(i, height[i])`.\n",
+        "\n",
+        "Find two lines that together with the x-axis form a container, such that the container contains the most water.\n",
+        "\n",
+        "Return *the maximum amount of water a container can store.*\n",
+        "\n",
+        "**Notice** that you may not slant the container.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `n == height.length`\n",
+        "- 2 <= `n` <= $10^5$\n",
+        "- 0 <= `height[i]` <= $10^4$\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 2,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def maxArea(height):\n",
+        "    # Initialize the maximum area to 0.\n",
+        "    max_area = 0\n",
+        "    # Initialize two pointers, one at the beginning (left) and one at the end (right) of the array.\n",
+        "    left = 0\n",
+        "    right = len(height) - 1\n",
+        "\n",
+        "    # Iterate until the left pointer is less than the right pointer.\n",
+        "    while left < right:\n",
+        "        # Calculate the height of the container, which is the minimum height of the two lines.\n",
+        "        h = min(height[left], height[right])\n",
+        "        # Calculate the width of the container, which is the distance between the two pointers.\n",
+        "        width = right - left\n",
+        "        # Calculate the area of the container using height and width and update max_area if it's greater.\n",
+        "        max_area = max(max_area, h * width)\n",
+        "\n",
+        "        # Move the pointer with the shorter line inward.\n",
+        "        if height[left] < height[right]:\n",
+        "            left += 1\n",
+        "        else:\n",
+        "            right -= 1\n",
+        "\n",
+        "    # Continue this process until left < right, and then return the maximum area.\n",
+        "    return max_area"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We start with initializing `max_area` to 0, which will store the maximum area of water that can be contained by the lines.\n",
+        "\n",
+        "2. Two pointers, `left` and `right`, are initialized to the beginning and end of the input `height` array.\n",
+        "\n",
+        "3. The main loop runs while `left` is less than `right`. This ensures that we are checking all possible pairs of lines in a systematic way.\n",
+        "\n",
+        "4. Inside the loop:\n",
+        "   - We calculate the height of the container `h` as the minimum of the heights at the `left` and `right` pointers. This represents the height of the water the container can hold.\n",
+        "   - We calculate the width of the container as the difference between `right` and `left`. This represents the distance between the two lines.\n",
+        "   - We calculate the area of the container using `h * width` and update `max_area` if the calculated area is greater than the current `max_area`.\n",
+        "\n",
+        "5. After calculating the area and updating `max_area`, we move one of the pointers. We move the pointer pointing to the shorter line inward because moving the pointer pointing to the taller line won't increase the height and will only decrease the width, which will result in a smaller area.\n",
+        "\n",
+        "6. We continue this process until `left` is no longer less than `right`, indicating that we have checked all possible pairs of lines.\n",
+        "\n",
+        "7. Finally, we return the `max_area` which contains the maximum area of water that can be contained by two lines from the input array."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 7,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Output: 49\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "height1 = [1, 8, 6, 2, 5, 4, 8, 3, 7]\n",
+        "result1 = maxArea(height1)\n",
+        "print(\"Output:\", result1)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 8,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Output: 1\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "height2 = [1, 1]\n",
+        "result2 = maxArea(height2)\n",
+        "print(\"Output:\", result2)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 9,
+      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Output: 45\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Additional Example\n",
+        "height3 = [3, 9, 3, 4, 7, 2, 12, 6]\n",
+        "result3 = maxArea(height3)\n",
+        "print(\"Output:\", result3)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `maxArea` function:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "The function uses a two-pointer approach to find the maximum area of water. In the worst case, both the left and right pointers traverse the entire input array once. Therefore, the time complexity is O(n), where n is the number of elements in the `height` array.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "The function uses a constant amount of extra space for variables like `max_area`, `left`, `right`, `h`, and `width`. It does not use any data structures whose space consumption depends on the input size. As a result, the space complexity is O(1), which means it has constant space complexity.\n",
+        "\n",
+        "**In summary,** the `maxArea` function has a time complexity of O(n) and a space complexity of O(1), making it an efficient algorithm for finding the maximum area of water that can be contained by two lines in the input array."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Optimization Challenge:** Modify the maxArea function to not only return the maximum area but also the indices of the two lines that form the container with the maximum area.\n",
+        "2. **Variant Problem:** Instead of finding the maximum area of water, find the minimum amount of water required to fill all the gaps between the lines. You'll need to return the total water volume needed."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb b/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb
new file mode 100644
index 0000000..8520b3b
--- /dev/null
+++ b/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb	
@@ -0,0 +1,218 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 125. Valid Palindrome\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Valid Palindrome problem on LeetCode, click here!](https://leetcode.com/problems/valid-palindrome/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.\n",
+        "\n",
+        "Given a string `s`, *return `true` if it is a **palindrome**, or `false` otherwise.*\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- 1 <= `s.length` <= $2*10^5$\n",
+        "- `s` consists only of printable ASCII characters.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 14,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def isPalindrome(s):\n",
+        "    # Step 1: Convert the input string to lowercase\n",
+        "    s = s.lower()\n",
+        "    \n",
+        "    # Step 2: Remove all non-alphanumeric characters from the string\n",
+        "    cleaned_s = ''.join(c for c in s if c.isalnum())\n",
+        "    \n",
+        "    # Step 3: Initialize two pointers, one at the beginning and one at the end of the cleaned string\n",
+        "    left, right = 0, len(cleaned_s) - 1\n",
+        "    \n",
+        "    # Step 4: Compare characters using the two pointers while moving them towards each other\n",
+        "    while left < right:\n",
+        "        if cleaned_s[left] != cleaned_s[right]:\n",
+        "            return False  # If characters don't match, it's not a palindrome\n",
+        "        left += 1\n",
+        "        right -= 1\n",
+        "    \n",
+        "    # Step 5: If the loop completes without returning False, it's a palindrome\n",
+        "    return True"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. The `isPalindrome` function takes an input string `s`.\n",
+        "\n",
+        "2. It starts by converting the input string to lowercase using `s.lower()`. This step ensures that the function is case-insensitive when checking for palindromes.\n",
+        "\n",
+        "3. Next, it removes all non-alphanumeric characters from the string and stores the result in the `cleaned_s` variable. It does this by iterating through each character in the input string and only keeping characters that are alphanumeric (letters and digits). This step removes spaces, punctuation, and other non-alphanumeric characters.\n",
+        "\n",
+        "4. The function then initializes two pointers, `left` and `right`, which will be used to compare characters from the beginning and end of the cleaned string.\n",
+        "\n",
+        "5. In a while loop, it compares characters using the two pointers while moving them towards each other. The loop continues until the `left` pointer is less than the `right` pointer. This is done to check all characters in the cleaned string.\n",
+        "\n",
+        "6. Inside the loop, it checks if the characters at the `left` and `right` pointers don't match. If they don't match, it means the input is not a palindrome, so the function returns `False`.\n",
+        "\n",
+        "7. If the characters at the `left` and `right` pointers match, the pointers are moved closer to each other by incrementing `left` and decrementing `right`.\n",
+        "\n",
+        "8. The loop continues to compare characters until it either finds a mismatch (returning `False`) or until the `left` pointer is greater than or equal to the `right` pointer. If the loop completes without finding a mismatch, it means the input is a palindrome, so the function returns `True`.\n",
+        "\n",
+        "In summary, the code converts the input string to lowercase, removes non-alphanumeric characters, and then uses two pointers to compare characters from the cleaned string's beginning and end. If all characters match during the comparison, the function returns `True`, indicating that the input is a palindrome. If any characters do not match, it returns `False`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 16,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "input1 = \"A man, a plan, a canal: Panama\"\n",
+        "output1 = isPalindrome(input1)\n",
+        "print(output1)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 17,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "input2 = \"race a car\"\n",
+        "output2 = isPalindrome(input2)\n",
+        "print(output2)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 18,
+      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3\n",
+        "input3 = \"\"\n",
+        "output3 = isPalindrome(input3)\n",
+        "print(output3)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `isPalindrome` function that uses two pointers:\n",
+        "\n",
+        "1. **Time Complexity:**\n",
+        "   - Converting the input string to lowercase using `s.lower()` takes O(n) time, where 'n' is the length of the input string.\n",
+        "   - Removing non-alphanumeric characters using the list comprehension `''.join(c for c in s if c.isalnum())` also takes O(n) time because it iterates through each character in the string once.\n",
+        "   - The comparison of characters using the two pointers (`left` and `right`) is done in a loop that runs until `left` is less than `right`. In the worst case, the loop iterates through half of the string, so it takes O(n/2) time, which is still considered O(n).\n",
+        "\n",
+        "   Therefore, the overall time complexity of the `isPalindrome` function is O(n), where 'n' is the length of the input string.\n",
+        "\n",
+        "2. **Space Complexity:**\n",
+        "   - The space complexity is primarily determined by the additional string `cleaned_s` created to store the cleaned version of the input string. In the worst case, if all characters in the input string are alphanumeric, this cleaned string can be as large as the original input string.\n",
+        "   - Additionally, the function uses two pointers, `left` and `right`, which consume constant space and do not depend on the size of the input.\n",
+        "\n",
+        "**In summary,** the overall space complexity of the `isPalindrome` function is O(n) in the worst case, where 'n' is the length of the input string. The space complexity is dominated by the space required for `cleaned_s`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Longest Palindromic Substring:** Given a string `s`, find and return the longest palindromic substring within `s`. For example, if `s = \"babad\"`, the function should return `\"bab\"` or `\"aba\"`.\n",
+        "2. **Palindrome with Limited Characters:** Given a string `s` and a list of characters `allowed_chars`, check if `s` is a palindrome considering only the characters from `allowed_chars`. Ignore all other characters. For example, if `s = \"A man, a plan, a canal: Panama\"` and `allowed_chars = ['a', 'm', 'n', 'p']`, the function should return `True`."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb b/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb
new file mode 100644
index 0000000..06de549
--- /dev/null
+++ b/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb	
@@ -0,0 +1,250 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 15. 3Sum\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the 3Sum problem on LeetCode, click here!](https://leetcode.com/problems/3sum/)\n",
+        "\n",
+        "---\n",
+        "Given an integer array nums, *return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.*\n",
+        "\n",
+        "**Notice** that the solution set must not contain duplicate triplets.\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- `3 <= nums.length <= 3000`\n",
+        "- $-10^5$ <= `nums[i]` <= $10^5$\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 10,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def threeSum(nums):\n",
+        "    # Sort the input array in ascending order\n",
+        "    nums.sort()\n",
+        "    triplets = []\n",
+        "    \n",
+        "    for i in range(len(nums) - 2):\n",
+        "        # Skip duplicates to avoid duplicate triplets\n",
+        "        if i > 0 and nums[i] == nums[i - 1]:\n",
+        "            continue\n",
+        "        \n",
+        "        left, right = i + 1, len(nums) - 1\n",
+        "        \n",
+        "        while left < right:\n",
+        "            # Calculate the total sum of the current triplet\n",
+        "            total = nums[i] + nums[left] + nums[right]\n",
+        "            \n",
+        "            if total == 0:\n",
+        "                # If the total is zero, we found a valid triplet\n",
+        "                triplets.append([nums[i], nums[left], nums[right]])\n",
+        "                \n",
+        "                # Skip duplicates of left and right pointers\n",
+        "                while left < right and nums[left] == nums[left + 1]:\n",
+        "                    left += 1\n",
+        "                while left < right and nums[right] == nums[right - 1]:\n",
+        "                    right -= 1\n",
+        "                \n",
+        "                # Move the pointers to the next unique elements\n",
+        "                left += 1\n",
+        "                right -= 1\n",
+        "            elif total < 0:\n",
+        "                # If the total is negative, we need to increase the sum by moving the left pointer to the right\n",
+        "                left += 1\n",
+        "            else:\n",
+        "                # If the total is positive, we need to decrease the sum by moving the right pointer to the left\n",
+        "                right -= 1\n",
+        "    \n",
+        "    return triplets"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. It starts by sorting the input array `nums` in ascending order. Sorting helps in efficiently finding triplets that sum up to zero.\n",
+        "\n",
+        "2. It initializes an empty list `triplets` to store the unique triplets that meet the given conditions.\n",
+        "\n",
+        "3. The code then iterates through the `nums` array using a loop with index `i`. This loop will consider each element of the array as a potential starting point for a triplet.\n",
+        "\n",
+        "4. Inside the loop, it checks for duplicates and skips them. This is done to ensure that the solution set does not contain duplicate triplets. If `nums[i]` is the same as the previous element `nums[i-1]`, it continues to the next iteration of the loop.\n",
+        "\n",
+        "5. Two pointers, `left` and `right`, are initialized. `left` starts just after the current element `nums[i]`, and `right` starts at the end of the array.\n",
+        "\n",
+        "6. The code enters another loop with `left` and `right` pointers, trying to find a triplet that sums up to zero.\n",
+        "\n",
+        "7. It calculates the `total` sum of the current triplet as `nums[i] + nums[left] + nums[right]`.\n",
+        "\n",
+        "8. If `total` is zero, it means a valid triplet is found, so it appends `[nums[i], nums[left], nums[right]]` to the `triplets` list. Then, it moves both the `left` and `right` pointers to their next unique elements while skipping duplicates.\n",
+        "\n",
+        "9. If `total` is less than zero, it means the sum is negative, so it increments the `left` pointer to move towards a larger sum.\n",
+        "\n",
+        "10. If `total` is greater than zero, it means the sum is positive, so it decrements the `right` pointer to move towards a smaller sum.\n",
+        "\n",
+        "11. This process continues until all possible triplets have been considered.\n",
+        "\n",
+        "12. Finally, the function returns the `triplets` list, which contains all the unique triplets that sum up to zero in the sorted `nums` array."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 11,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[[-1, -1, 2], [-1, 0, 1]]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "nums = [-1, 0, 1, 2, -1, -4]\n",
+        "result = threeSum(nums)\n",
+        "print(result)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 12,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "nums = [0, 1, 1]\n",
+        "result = threeSum(nums)\n",
+        "print(result)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 13,
+      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[[0, 0, 0]]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3\n",
+        "nums = [0, 0, 0]\n",
+        "result = threeSum(nums)\n",
+        "print(result)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `threeSum` function:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "1. Sorting the input array `nums` takes O(n log n) time, where n is the length of the array.\n",
+        "\n",
+        "2. The outer loop runs for each element of the array, so it iterates O(n) times.\n",
+        "\n",
+        "3. Inside the outer loop, we have a while loop with two pointers (`left` and `right`). In the worst case, the while loop can iterate O(n) times (when all elements are unique).\n",
+        "\n",
+        "4. Inside the while loop, we have constant time operations (additions, comparisons, and list appends).\n",
+        "\n",
+        "Overall, the time complexity of the `threeSum` function is dominated by the sorting step, so it is O(n log n) due to the sorting. The other operations inside the loops contribute linearly but are dominated by the sorting step.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "1. The space used by the `triplets` list to store the output triplets is O(k), where k is the number of unique triplets that sum up to zero. In the worst case, this can be O(n^2/3) because there can be roughly O(n^2) possible triplets, and many of them may sum up to zero.\n",
+        "\n",
+        "2. Other than the `triplets` list, the function uses only a constant amount of additional space for variables like `i`, `left`, `right`, and `total`.\n",
+        "\n",
+        "Overall, the space complexity of the `threeSum` function is O(k), where k is the number of unique triplets that meet the criteria. It's important to note that the space complexity does not depend on the size of the input array but rather on the number of valid triplets found.\n",
+        "\n",
+        "**In summary:**\n",
+        "- Time Complexity: O(n log n) due to sorting (dominant factor) and O(n^2) in the worst case for the nested loops.\n",
+        "- Space Complexity: O(k) for storing the output triplets, where k is the number of unique triplets that meet the criteria."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **K-Sum Problem:** Generalize the problem to the K-Sum problem, where instead of finding triplets that sum to zero, you need to find K elements that sum to a target value. Implement a function for this generalization.\n",
+        "2. **Count the Number of Unique Triplets:** Modify the algorithm to count the number of unique triplets that satisfy the conditions instead of returning the triplets themselves. This will require some modifications to handle duplicates efficiently."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/02. Two Pointers/README.md b/_build/html/_sources/02. Two Pointers/README.md
new file mode 100644
index 0000000..0d4c7cf
--- /dev/null
+++ b/_build/html/_sources/02. Two Pointers/README.md	
@@ -0,0 +1,18 @@
+# Two Pointers Problems - Blind 75 LeetCode 
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)                            | Easy       |
+| [15. 3Sum](https://leetcode.com/problems/3sum/)                                                       | Medium     |
+| [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)           | Medium     |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb b/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb
new file mode 100644
index 0000000..554c940
--- /dev/null
+++ b/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb	
@@ -0,0 +1,191 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 121. Best Time to Buy and Sell Stock\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Best Time to Buy and Sell Stock problem on LeetCode, click here!](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "You are given an array `prices` where `prices[i]` is the price of a given stock on the $i^{th}$ day.\n",
+        "\n",
+        "You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.\n",
+        "\n",
+        "*Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.*\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- 1 <= `prices.length` <= $10^5$\n",
+        "- 0 <= `prices[i]` <= $10^4$\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 65,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def maxProfit(prices):\n",
+        "    if not prices:\n",
+        "        return 0\n",
+        "    \n",
+        "    min_price = prices[0]\n",
+        "    max_profit = 0\n",
+        "    \n",
+        "    for price in prices:\n",
+        "        # Update the minimum price if a lower price is encountered\n",
+        "        min_price = min(min_price, price)\n",
+        "        # Calculate the profit if selling at the current price\n",
+        "        max_profit = max(max_profit, price - min_price)\n",
+        "    \n",
+        "    return max_profit"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "- The `maxProfit` function takes a single argument, `prices`, which is a list of stock prices.\n",
+        "- The first line of the function checks if the `prices` list is empty. If it's empty, there are no prices to analyze, so the function returns 0 (no profit can be made).\n",
+        "\n",
+        "- Then, we initialize two variables:\n",
+        "  - `min_price`: This variable will keep track of the minimum price seen so far. We initialize it with the price of the first day in the list (`prices[0]`).\n",
+        "  - `max_profit`: This variable will keep track of the maximum profit we can achieve. We initialize it to 0 because initially, we haven't made any profit.\n",
+        "\n",
+        "- We start iterating through the `prices` list, where `price` represents the stock price for the current day.\n",
+        "\n",
+        "- In each iteration, we check if the current `price` is lower than the `min_price` we've seen so far. If it is, we update `min_price` to the current `price`. This is because we want to buy the stock at the lowest possible price.\n",
+        "\n",
+        "- Next, we calculate the profit we would make if we sold the stock at the current `price` (assuming we bought it at `min_price`). We do this by subtracting `min_price` from the current `price`. This represents the profit for the current day.\n",
+        "\n",
+        "- We also use the `max` function to keep track of the maximum profit seen so far. If the profit calculated for the current day is greater than the `max_profit` we've seen previously, we update `max_profit` with this higher value.\n",
+        "\n",
+        "- Finally, after iterating through all the days in the `prices` list, we return `max_profit`, which represents the maximum profit that can be achieved by buying and selling a single stock."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 66,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "5\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "prices1 = [7, 1, 5, 3, 6, 4]\n",
+        "print(maxProfit(prices1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 67,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "0\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "prices2 = [7, 6, 4, 3, 1]\n",
+        "print(maxProfit(prices2))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "- The code iterates through the `prices` array once, examining each price exactly once.\n",
+        "- Within the loop, there are constant-time operations such as comparisons, additions, and subtractions.\n",
+        "- Therefore, the time complexity of this code is O(n), where n is the length of the `prices` array. It performs a linear number of operations relative to the size of the input.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "- The space complexity of the code is constant, O(1).\n",
+        "- Regardless of the size of the input `prices` array, the code only uses a fixed amount of additional memory to store two variables (`min_price` and `max_profit`) and a few loop variables.\n",
+        "- The memory usage does not depend on the size of the input, so it is considered constant space complexity.\n",
+        "\n",
+        "**In summary:**\n",
+        "- Time Complexity: O(n) where n is the length of the `prices` array.\n",
+        "- Space Complexity: O(1) (constant space usage)."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Multiple Transactions Allowed:** Modify the problem to allow multiple transactions (buy and sell multiple times), but you must sell before buying again. Find the maximum profit in this scenario.\n",
+        "2. **K Transactions Allowed:** Extend the problem to allow at most k transactions. Find the maximum profit considering this constraint.\n",
+        "3. **With Transaction Fee:** Introduce a transaction fee for each buy/sell operation. Modify the code to maximize profit while considering the transaction fee."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb b/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb
new file mode 100644
index 0000000..4a516de
--- /dev/null
+++ b/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb	
@@ -0,0 +1,214 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 3. Longest Substring Without Repeating Characters\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Longest Substring Without Repeating Characters problem on LeetCode, click here!](https://leetcode.com/problems/longest-substring-without-repeating-characters/)\n",
+        "\n",
+        "---\n",
+        "Given a string `s`, find the length of the **longest substring** without repeating characters.\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- 0 <= `s.length` <= $5 * 10^4$\n",
+        "- `s` consists of English letters, digits, symbols and spaces.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 61,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def length_of_longest_substring(s):\n",
+        "    # Create a dictionary to store the index of each character's last occurrence.\n",
+        "    char_index = {}\n",
+        "    \n",
+        "    # Initialize variables to keep track of the start and end of the current substring.\n",
+        "    start = 0\n",
+        "    max_length = 0\n",
+        "    \n",
+        "    for end in range(len(s)):\n",
+        "        # If the character is in the dictionary and its last occurrence is after the start of the current substring,\n",
+        "        # update the start of the substring to the next character after its last occurrence.\n",
+        "        if s[end] in char_index and char_index[s[end]] >= start:\n",
+        "            start = char_index[s[end]] + 1\n",
+        "        \n",
+        "        # Update the last occurrence index of the current character.\n",
+        "        char_index[s[end]] = end\n",
+        "        \n",
+        "        # Calculate the length of the current substring and update the maximum length if necessary.\n",
+        "        current_length = end - start + 1\n",
+        "        max_length = max(max_length, current_length)\n",
+        "    \n",
+        "    return max_length"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We start by defining a function called `length_of_longest_substring` that takes a single argument `s`, which is the input string for which we want to find the length of the longest substring without repeating characters.\n",
+        "\n",
+        "2. We create a dictionary called `char_index` to store the index of the last occurrence of each character in the input string. This dictionary will help us keep track of where each character was last seen.\n",
+        "\n",
+        "3. We initialize two variables, `start` and `max_length`. \n",
+        "   - `start` represents the start index of the current substring without repeating characters. It starts at 0.\n",
+        "   - `max_length` is used to keep track of the maximum length found so far and is initially set to 0.\n",
+        "\n",
+        "4. We then iterate through the input string `s` using a for loop, where the loop variable `end` represents the current end index of the substring we are considering.\n",
+        "\n",
+        "5. Inside the loop:\n",
+        "   - We check if the character `s[end]` is already in the `char_index` dictionary and if its last occurrence is within or after the current substring. If so, it means that we've encountered a repeating character, and we need to update the `start` of the substring to the next character after the last occurrence of `s[end]`. This ensures that we have a new valid substring without repeating characters.\n",
+        "   - We then update the `char_index` dictionary by storing the current index `end` as the last occurrence index of the character `s[end]`.\n",
+        "\n",
+        "6. Next, we calculate the length of the current substring without repeating characters, which is `current_length = end - start + 1`. We add 1 to account for the fact that the indices are zero-based.\n",
+        "\n",
+        "7. We update the `max_length` with the maximum of its current value and the `current_length`. This step ensures that we keep track of the longest valid substring we have encountered so far.\n",
+        "\n",
+        "8. The loop continues to iterate through the input string, and at the end, we return the `max_length`, which represents the length of the longest substring without repeating characters."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 62,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "3\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "print(length_of_longest_substring(\"abcabcbb\"))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 63,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "1\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "print(length_of_longest_substring(\"bbbbb\"))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 64,
+      "id": "cf04eba0-0ead-4e8b-ba33-4316ea64a9d3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "3\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3:\n",
+        "print(length_of_longest_substring(\"pwwkew\"))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "Let's analyze the time and space complexity of the `length_of_longest_substring` function:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "- The function iterates through the input string `s` using a single for loop. The loop runs from the beginning of the string to the end once.\n",
+        "- Inside the loop, we perform constant-time operations such as dictionary lookups and updates, comparisons, and arithmetic operations.\n",
+        "- Therefore, the time complexity of the function is O(n), where n is the length of the input string `s`.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "- The primary data structure that consumes space in this function is the `char_index` dictionary.\n",
+        "- In the worst case, if there are no repeating characters in the input string, the dictionary can store all unique characters in the string.\n",
+        "- Therefore, the space complexity is O(min(n, m)), where n is the length of the input string `s`, and m is the number of unique characters in the string. In the worst case, when all characters are unique, m is equal to n, so the space complexity is O(n).\n",
+        "- Additionally, there are a few integer variables used for indices and lengths, which consume constant space.\n",
+        "- Overall, the space complexity of the function is O(min(n, m)) or simply O(n) in the worst case.\n",
+        "\n",
+        "**In summary,** the time complexity of the `length_of_longest_substring` function is O(n), and the space complexity is O(n) in the worst case due to the `char_index` dictionary. This algorithm provides an efficient way to find the length of the longest substring without repeating characters in linear time."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Longest Substring with K Distinct Characters:** Modify the problem to find the length of the longest substring with exactly K distinct characters. For example, given the input string \"abcabcbb\" and K = 2, the answer would be 4 because the longest substring with two distinct characters is \"abca.\"\n",
+        "2. **Longest Substring with Unique Characters:** Write a function to find the length of the longest substring in a given string where all characters are unique. For example, given the input string \"abcabcbb,\" the answer would be 4 because \"abcd\" is the longest substring with unique characters."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb b/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb
new file mode 100644
index 0000000..07e96a2
--- /dev/null
+++ b/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb	
@@ -0,0 +1,195 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 424. Longest Repeating Character Replacement\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Longest Repeating Character Replacement problem on LeetCode, click here!](https://leetcode.com/problems/longest-repeating-character-replacement/)\n",
+        "\n",
+        "---\n",
+        "You are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.\n",
+        "\n",
+        "*Return the length of the longest substring containing the same letter you can get after performing the above operations.*\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- 1 <= `s.length` <= $10^5$\n",
+        "- `s` consists of only uppercase English letters.\n",
+        "- `0 <= k <= s.length`"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 58,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def characterReplacement(s, k):\n",
+        "    max_length = 0  # Initialize the maximum length\n",
+        "    max_count = 0   # Initialize the maximum count of repeating characters\n",
+        "    start = 0       # Initialize the start of the sliding window\n",
+        "    char_count = {} # Dictionary to store the count of each character\n",
+        "\n",
+        "    for end in range(len(s)):\n",
+        "        # Update the count of the current character in the dictionary\n",
+        "        char_count[s[end]] = char_count.get(s[end], 0) + 1\n",
+        "        \n",
+        "        # Update the maximum count of repeating characters\n",
+        "        max_count = max(max_count, char_count[s[end]])\n",
+        "        \n",
+        "        # Check if the current window size is greater than k\n",
+        "        if (end - start + 1) - max_count > k:\n",
+        "            # Move the start of the window to the right\n",
+        "            char_count[s[start]] -= 1\n",
+        "            start += 1\n",
+        "        \n",
+        "        # Update the maximum length\n",
+        "        max_length = max(max_length, end - start + 1)\n",
+        "\n",
+        "    return max_length"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We initialize some variables:\n",
+        "   - `max_length` to keep track of the maximum length of the substring containing the same letter.\n",
+        "   - `max_count` to keep track of the maximum count of repeating characters within the current window.\n",
+        "   - `start` to represent the start index of the sliding window.\n",
+        "   - `char_count` is a dictionary that will store the count of each character within the current window.\n",
+        "\n",
+        "2. We use a `for` loop to iterate through the characters of the string `s`.\n",
+        "\n",
+        "3. Inside the loop, we do the following for each character at position `end`:\n",
+        "   - Update the count of the current character in the `char_count` dictionary.\n",
+        "   - Update the `max_count` to be the maximum of the current `max_count` and the count of the current character. This keeps track of the maximum count of repeating characters within the current window.\n",
+        "\n",
+        "4. We check if the current window size (the difference between `end` and `start` plus one) minus the `max_count` is greater than `k`. This condition checks whether we have exceeded the allowed number of replacements (`k`) within the current window.\n",
+        "\n",
+        "   - If we have exceeded the allowed replacements, it means we need to shrink the window from the left side. We do this by moving the `start` of the window to the right and decrementing the count of the character at `s[start]` in the `char_count` dictionary. This effectively removes characters from the left side of the window until we have a valid window again.\n",
+        "\n",
+        "   - By doing this, we ensure that the difference between the current window size and the `max_count` is always less than or equal to `k`.\n",
+        "\n",
+        "5. After each character, we update the `max_length` to be the maximum of the current `max_length` and the size of the current window (`end - start + 1`).\n",
+        "\n",
+        "6. Finally, we return `max_length`, which holds the length of the longest substring containing the same letter with at most `k` replacements."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 59,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "4\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "s1 = \"ABAB\"\n",
+        "k1 = 2\n",
+        "print(characterReplacement(s1, k1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 60,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "4\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "s2 = \"AABABBA\"\n",
+        "k2 = 1\n",
+        "print(characterReplacement(s2, k2))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "- The code uses a single `for` loop to iterate through the characters of the input string `s`. The loop runs from the beginning to the end of the string.\n",
+        "- Inside the loop, we perform constant-time operations such as updating the `char_count` dictionary and updating variables like `max_count` and `max_length`.\n",
+        "- The code's time complexity is primarily determined by the loop, which iterates through each character in the string once. Therefore, the time complexity is O(n), where n is the length of the input string `s`.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "- The code uses several variables to store information, but the space they consume is constant and does not depend on the size of the input string. These variables include `max_length`, `max_count`, `start`, and `char_count`. Therefore, the space complexity is O(1) or constant space.\n",
+        "\n",
+        "**In summary**, the time complexity of the code is O(n), where n is the length of the input string, and the space complexity is O(1), as it uses a constant amount of additional space regardless of the input size. This algorithm efficiently solves the problem with a linear time complexity."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Optimization Challenge:** Modify the code to find the longest substring containing the same letter with at most k replacements in O(n) time complexity and O(1) space complexity. Hint: You may need to update the approach to achieve this.\n",
+        "2. **Variation with Lowercase Letters:** Extend the problem to include both uppercase and lowercase English letters in the input string s. Write a function that can handle this extended input and still find the longest substring with at most k replacements efficiently."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb b/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb
new file mode 100644
index 0000000..33c2a44
--- /dev/null
+++ b/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb	
@@ -0,0 +1,273 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 76. Minimum Window Substring\n",
+        "\n",
+        "**Difficulty:** Hard\n",
+        "\n",
+        "**Link to Problem:** [To see the Minimum Window Substring problem on LeetCode, click here!](https://leetcode.com/problems/minimum-window-substring/)\n",
+        "\n",
+        "---\n",
+        "Given two strings `s` and `t` of lengths `m` and `n` respectively, return *the **minimum window substring** of `s` such that every character in `t` (**including duplicates**) is included in the window. If there is no such substring, return the empty string `\"\"`.*\n",
+        "\n",
+        "The testcases will be generated such that the answer is **unique**.\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- `m == s.length`\n",
+        "- `n == t.length`\n",
+        "- 1 <= `m, n` <= $10^5$\n",
+        "- `s` and `t` consist of uppercase and lowercase English letters.\n",
+        "\n",
+        "**Follow up**: Could you find an algorithm that runs in $O(m + n)$ time?"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 54,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def min_window_substring(s, t):\n",
+        "    if not s or not t:\n",
+        "        return \"\"\n",
+        "\n",
+        "    # Initialize dictionaries to keep track of character counts for t and the current window in s.\n",
+        "    t_dict = {}\n",
+        "    current_window_dict = {}\n",
+        "    \n",
+        "    # Populate t_dict with character counts for string t.\n",
+        "    for char in t:\n",
+        "        t_dict[char] = t_dict.get(char, 0) + 1\n",
+        "\n",
+        "    # Initialize pointers for the sliding window.\n",
+        "    left = 0\n",
+        "    min_len = float('inf')\n",
+        "    min_window = \"\"\n",
+        "    required_chars = len(t_dict)\n",
+        "\n",
+        "    # Initialize a variable to keep track of how many required characters have been found in the current window.\n",
+        "    found_chars = 0\n",
+        "\n",
+        "    # Iterate over the string s using the right pointer.\n",
+        "    for right in range(len(s)):\n",
+        "        char = s[right]\n",
+        "\n",
+        "        # Update the current_window_dict.\n",
+        "        current_window_dict[char] = current_window_dict.get(char, 0) + 1\n",
+        "\n",
+        "        # Check if the current character is a required character and if its count matches the required count.\n",
+        "        if char in t_dict and current_window_dict[char] == t_dict[char]:\n",
+        "            found_chars += 1\n",
+        "\n",
+        "        # Try to minimize the window by moving the left pointer.\n",
+        "        while left <= right and found_chars == required_chars:\n",
+        "            # Calculate the current window size.\n",
+        "            window_size = right - left + 1\n",
+        "\n",
+        "            # If the current window is smaller than the minimum found so far, update min_len and min_window.\n",
+        "            if window_size < min_len:\n",
+        "                min_len = window_size\n",
+        "                min_window = s[left:right+1]\n",
+        "\n",
+        "            # Move the left pointer to the right to shrink the window.\n",
+        "            left_char = s[left]\n",
+        "            current_window_dict[left_char] -= 1\n",
+        "\n",
+        "            # Check if the character removed from the window was a required character.\n",
+        "            if left_char in t_dict and current_window_dict[left_char] < t_dict[left_char]:\n",
+        "                found_chars -= 1\n",
+        "\n",
+        "            # Move the left pointer further to continue shrinking the window.\n",
+        "            left += 1\n",
+        "\n",
+        "    return min_window"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We start by checking if either of the input strings `s` or `t` is empty. If either of them is empty, we return an empty string since there can't be any valid substring in this case.\n",
+        "\n",
+        "2. We initialize two dictionaries: `t_dict` and `current_window_dict`. These dictionaries will be used to keep track of character counts in the string `t` and the current window in string `s`, respectively.\n",
+        "\n",
+        "3. We populate the `t_dict` dictionary by iterating through string `t`. For each character, we increment its count in the dictionary using `t_dict.get(char, 0) + 1`. This allows us to count the occurrences of each character in `t`.\n",
+        "\n",
+        "4. We initialize two pointers: `left` and `right`. The `left` pointer will represent the start of the current window, and the `right` pointer will represent the end of the current window. We also initialize `min_len` to store the length of the minimum window found so far, and `min_window` to store the actual minimum window substring.\n",
+        "\n",
+        "5. We determine the number of required characters in `t` (i.e., the number of distinct characters in `t`) and store it in the variable `required_chars`.\n",
+        "\n",
+        "6. We initialize a variable `found_chars` to keep track of how many required characters have been found in the current window. Initially, it is set to 0.\n",
+        "\n",
+        "7. We iterate over the string `s` using the `right` pointer. In each iteration, we do the following:\n",
+        "\n",
+        "   - Update the `current_window_dict` by incrementing the count of the current character.\n",
+        "   \n",
+        "   - Check if the current character is a required character (present in `t_dict`) and if its count in the `current_window_dict` matches the required count from `t_dict`. If so, we increment `found_chars`.\n",
+        "\n",
+        "8. After updating the window, we attempt to minimize the window size by moving the `left` pointer to the right. In this step, we:\n",
+        "\n",
+        "   - Calculate the size of the current window.\n",
+        "   \n",
+        "   - If the current window size is smaller than the minimum found so far (`min_len`), we update `min_len` and `min_window` to store the current window substring.\n",
+        "   \n",
+        "   - Move the `left` pointer to the right to shrink the window.\n",
+        "   \n",
+        "   - Check if the character being removed from the window was a required character. If it was, we decrement `found_chars`.\n",
+        "   \n",
+        "   - Continue moving the `left` pointer further to continue shrinking the window if the window still contains all required characters.\n",
+        "\n",
+        "9. Finally, we return `min_window`, which will contain the minimum window substring that contains all characters from `t`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 55,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "BANC\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "print(min_window_substring(\"ADOBECODEBANC\", \"ABC\"))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 56,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "a\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "print(min_window_substring(\"a\", \"a\")) "
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 57,
+      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": []
+        },
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3:\n",
+        "print(min_window_substring(\"a\", \"aa\")) "
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the `min_window_substring` function:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "- The main loop iterates through the string `s` from left to right using the `right` pointer. This loop runs in O(m) time, where 'm' is the length of string `s`.\n",
+        "\n",
+        "- Inside the loop, we have a while loop that moves the `left` pointer to the right to shrink the window as needed. In the worst case, this while loop can also run in O(m) time because in the worst case, we may have to move the `left` pointer all the way to the end of the string.\n",
+        "\n",
+        "- Within each iteration of the while loop, we perform constant time operations, such as updating dictionaries and comparing character counts.\n",
+        "\n",
+        "- The initialization of the `t_dict` dictionary takes O(n) time, where 'n' is the length of string `t`.\n",
+        "\n",
+        "Therefore, the overall time complexity of the function is O(m + n) because the dominant factor is the length of string `s`, and the length of string `t` has a smaller impact.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "- The space complexity is determined by the space used by the `t_dict` dictionary, the `current_window_dict` dictionary, and a few variables.\n",
+        "\n",
+        "- The `t_dict` dictionary stores character counts for string `t`. In the worst case, when all characters in `t` are distinct, this dictionary can have a maximum of 'n' key-value pairs. So, the space complexity for `t_dict` is O(n).\n",
+        "\n",
+        "- The `current_window_dict` dictionary stores character counts for the current window. In the worst case, it can have a maximum of 'm' key-value pairs. So, the space complexity for `current_window_dict` is also O(m).\n",
+        "\n",
+        "- Other variables used in the function, such as `left`, `right`, `min_len`, `min_window`, `required_chars`, and `found_chars`, are all of constant size and do not depend on the input sizes.\n",
+        "\n",
+        "**In summary**, the overall space complexity of the function is O(max(m, n)), which means it is determined by the larger of the two input strings' lengths. This is because the space used for `t_dict` and `current_window_dict` dominates the space complexity."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Smallest Distinct Substring:** Instead of finding the minimum window substring containing all characters, find the smallest distinct (unique) substring of s that contains all characters from t. This variation adds complexity because you must find a substring with distinct characters.\n",
+        "2. **No Extra Space:** Solve the problem without using any extra space, such as dictionaries or arrays, other than a constant amount of space. This is a significant optimization challenge."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/03. Sliding Window/README.md b/_build/html/_sources/03. Sliding Window/README.md
new file mode 100644
index 0000000..e1294ff
--- /dev/null
+++ b/_build/html/_sources/03. Sliding Window/README.md	
@@ -0,0 +1,19 @@
+# Sliding Window Problems - Blind 75 LeetCode 
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [121. Best Time to Buy And Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | Easy       |
+| [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) | Medium |
+| [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) | Medium |
+| [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)             | Hard       |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb b/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb
new file mode 100644
index 0000000..c47292f
--- /dev/null
+++ b/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb	
@@ -0,0 +1,220 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 20. Valid Parentheses\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Valid Parentheses problem on LeetCode, click here!](https://leetcode.com/problems/valid-parentheses/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Given a string s containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.\n",
+        "\n",
+        "An input string is valid if:\n",
+        "\n",
+        "1. Open brackets must be closed by the same type of brackets.\n",
+        "2. Open brackets must be closed in the correct order.\n",
+        "3. Every close bracket has a corresponding open bracket of the same type.\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- 1 <= `s.length` <= $10^4$\n",
+        "- `s` consists of parentheses only `'()[]{}'`."
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 49,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def isValid(s):\n",
+        "    # Create an empty stack to store opening brackets\n",
+        "    stack = []\n",
+        "    \n",
+        "    # Define a mapping for matching brackets\n",
+        "    bracket_mapping = {')': '(', '}': '{', ']': '['}\n",
+        "    \n",
+        "    # Iterate through the characters in the input string\n",
+        "    for char in s:\n",
+        "        # If the character is an opening bracket, push it onto the stack\n",
+        "        if char in bracket_mapping.values():\n",
+        "            stack.append(char)\n",
+        "        # If the character is a closing bracket\n",
+        "        elif char in bracket_mapping.keys():\n",
+        "            # Pop the top element from the stack if it exists, or use a dummy value '#'\n",
+        "            top_element = stack.pop() if stack else '#'\n",
+        "            # If the popped element does not match the corresponding opening bracket, return False\n",
+        "            if bracket_mapping[char] != top_element:\n",
+        "                return False\n",
+        "        # If the character is not a bracket, ignore it\n",
+        "        \n",
+        "    # After processing the entire string, if there are any unmatched opening brackets left in the stack, return False\n",
+        "    return len(stack) == 0"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We start by defining the `isValid` function that takes a single argument `s`, which is the input string containing parentheses and brackets.\n",
+        "\n",
+        "2. Inside the function, we create an empty stack, which is a list in Python, to store opening brackets as we encounter them in the input string. The stack will help us keep track of the brackets and their order.\n",
+        "\n",
+        "3. We define a `bracket_mapping` dictionary that maps each closing bracket to its corresponding opening bracket. This mapping will be used to check if a closing bracket matches the most recent opening bracket in the stack.\n",
+        "\n",
+        "4. We iterate through each character `char` in the input string `s` using a `for` loop.\n",
+        "\n",
+        "5. If the character `char` is an opening bracket (i.e., it exists in the `values()` of `bracket_mapping`), we push it onto the stack using the `append()` method.\n",
+        "\n",
+        "6. If the character `char` is a closing bracket (i.e., it exists in the `keys()` of `bracket_mapping`), we need to check if it matches the corresponding opening bracket. To do this, we pop the top element from the stack (if it exists) and store it in `top_element`. We use a dummy value `'#'` if the stack is empty to avoid errors.\n",
+        "\n",
+        "7. We then compare `top_element` with the corresponding opening bracket using `bracket_mapping[char]`. If they do not match, it means the string is not valid, and we return `False`.\n",
+        "\n",
+        "8. If the character `char` is not a bracket, we simply ignore it and continue the loop.\n",
+        "\n",
+        "9. After processing the entire string, we check if there are any unmatched opening brackets left in the stack. If the stack is empty, it means all opening brackets have been properly matched and closed, and we return `True`. Otherwise, we return `False`.\n",
+        "\n",
+        "10. Finally, we provide some test cases at the bottom of the code to demonstrate how the function works for different input strings."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 50,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "print(isValid(\"()\"))    "
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 51,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "print(isValid(\"()[]{}\")) "
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 52,
+      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3:\n",
+        "print(isValid(\"(]\"))     "
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the `isValid` function:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "- The function iterates through each character in the input string `s` once, performing constant-time operations for each character.\n",
+        "- Pushing and popping elements from the stack (list) also takes constant time in most cases.\n",
+        "- Therefore, the overall time complexity of the function is $O(n)$, where $n$ is the length of the input string `s`.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "- The space complexity of the function is determined by the space used by the stack and the `bracket_mapping` dictionary.\n",
+        "- In the worst case, if the input string `s` consists entirely of opening brackets, the stack can potentially contain all of these opening brackets, resulting in a space complexity of $O(n)$ in terms of the stack.\n",
+        "- The `bracket_mapping` dictionary has a constant number of key-value pairs (3 pairs in this case).\n",
+        "- Therefore, the overall space complexity of the function is $O(n)$, where $n$ is the length of the input string `s`, mainly due to the stack space.\n",
+        "\n",
+        "**In summary:**\n",
+        "- The time complexity is $O(n)$ because we iterate through the string once.\n",
+        "- The space complexity is $O(n)$ due to the stack used to keep track of opening brackets."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Valid Expressions:** Modify the problem to validate not only brackets but also arithmetic expressions containing parentheses, such as \"$2 * (3 + 5) / (4 - 2)$\".\n",
+        "2. **Valid Parentheses Combinations:**\n",
+        "Write a function to generate all valid combinations of n pairs of parentheses, where $n$ is a positive integer. For example, for $n = 3$, the valid combinations are [\"((()))\", \"(()())\", \"(())()\", \"()(())\", \"()()()\"]."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/04. Stack/README.md b/_build/html/_sources/04. Stack/README.md
new file mode 100644
index 0000000..2ed115b
--- /dev/null
+++ b/_build/html/_sources/04. Stack/README.md	
@@ -0,0 +1,16 @@
+# Stack Problems - Blind 75 LeetCode 
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)                            | Easy       |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb b/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb
new file mode 100644
index 0000000..98f2331
--- /dev/null
+++ b/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb	
@@ -0,0 +1,228 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 153. Find Minimum in Rotated Sorted Array\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Find Minimum in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:\n",
+        "\n",
+        "- `[4,5,6,7,0,1,2]` if it was rotated `4` times.\n",
+        "- `[0,1,2,4,5,6,7]` if it was rotated `7` times.\n",
+        "\n",
+        "Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.\n",
+        "\n",
+        "Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array.*\n",
+        "\n",
+        "You must write an algorithm that runs in $O(log\\ n)$ time.\n",
+        " \n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- `n == nums.length`\n",
+        "- `1 <= n <= 5000`\n",
+        "- `-5000 <= nums[i] <= 5000`\n",
+        "- All the integers of `nums` are **unique**.\n",
+        "- `nums` is sorted and rotated between `1` and `n` times."
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 39,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def findMin(nums):\n",
+        "    left, right = 0, len(nums) - 1\n",
+        "    \n",
+        "    while left < right:\n",
+        "        mid = left + (right - left) // 2\n",
+        "        \n",
+        "        # If the mid element is greater than the rightmost element,\n",
+        "        # it means the minimum element is in the right half.\n",
+        "        if nums[mid] > nums[right]:\n",
+        "            left = mid + 1\n",
+        "        # If the mid element is less than or equal to the rightmost element,\n",
+        "        # it means the minimum element is in the left half or at the mid itself.\n",
+        "        else:\n",
+        "            right = mid\n",
+        "    \n",
+        "    # The loop will break when left and right converge to the minimum element.\n",
+        "    # At this point, left (or right) will be pointing to the minimum element.\n",
+        "    return nums[left]"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. The function `findMin(nums)` takes an input list `nums`, which represents the rotated sorted array.\n",
+        "\n",
+        "2. Initialize two pointers `left` and `right` to track the current search range. Initially, `left` is set to 0 (the beginning of the array), and `right` is set to `len(nums) - 1` (the end of the array).\n",
+        "\n",
+        "3. Enter a `while` loop with the condition `left < right`. This loop continues until `left` and `right` converge to a single element, which will be the minimum element in the rotated array.\n",
+        "\n",
+        "4. Inside the loop, calculate the middle index `mid` using integer division. This helps in finding the middle element of the current search range.\n",
+        "\n",
+        "5. Check if the element at the middle index (`nums[mid]`) is greater than the element at the rightmost index (`nums[right]`). If this condition is true, it means that the minimum element must be in the right half of the current search range. So, update `left` to `mid + 1`, effectively eliminating the left half of the search range.\n",
+        "\n",
+        "6. If the condition from step 5 is not met, it implies that the minimum element can be in the left half of the current search range or could be the element at the `mid` index itself. In this case, update `right` to `mid`, effectively eliminating the right half of the search range.\n",
+        "\n",
+        "7. Repeat steps 4-6 until `left` and `right` converge to the minimum element.\n",
+        "\n",
+        "8. When the loop exits, it means that `left` (or `right`) points to the minimum element in the array.\n",
+        "\n",
+        "9. Return `nums[left]` (or `nums[right]`) as the result, which is the minimum element in the rotated sorted array."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 43,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "1\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 1:\n",
+        "nums1 = [3, 4, 5, 1, 2]\n",
+        "print(findMin(nums1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 41,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "0\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
+        "print(findMin(nums2))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 42,
+      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "11\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3:\n",
+        "nums3 = [11, 13, 15, 17]\n",
+        "print(findMin(nums3))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "The time complexity of this code is $O(log\\ n)$, where '$n$' is the length of the input array `nums`. This is because the binary search algorithm is employed, which continually divides the search range in half with each iteration.\n",
+        "\n",
+        "In each iteration of the `while` loop, the search range is halved, and this process continues until the `left` and `right` pointers converge to the minimum element. In the worst case, it will take logarithmic time to reduce the search range to a single element.\n",
+        "\n",
+        "Therefore, the binary search used in this code runs in $O(log\\ n)$ time complexity.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "The space complexity of this code is $O(1)$, which means it uses a constant amount of additional space that does not depend on the size of the input array.\n",
+        "\n",
+        "The algorithm uses a fixed number of variables (`left`, `right`, `mid`) to keep track of the search range and indices, but the number of these variables does not grow with the size of the input array. Hence, the space complexity is constant.\n",
+        "\n",
+        "**In summary:**\n",
+        "- Time Complexity: $O(log\\ n)$\n",
+        "- Space Complexity: $O(1)$"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. Solve the problem with a variation where the input array may contain duplicates.\n",
+        "\n",
+        "2. Modify the problem to return the index of the minimum element in the rotated sorted array, rather than the element itself. The algorithm should still run in $O(log\\ n)$ time.\n",
+        "\n",
+        "3. Implement a function that finds the maximum element in a rotated sorted array. How would you adapt the binary search algorithm to solve this problem efficiently in $O(log\\ n)$ time?\n",
+        "\n",
+        "4. Implement a function that can find the kth smallest element in a rotated sorted array. This is an extension of the original problem, and the algorithm should still run in $O(log\\ n)$ time."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb b/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb
new file mode 100644
index 0000000..d8d4202
--- /dev/null
+++ b/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb	
@@ -0,0 +1,245 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 33. Search in Rotated Sorted Array\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Search in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/search-in-rotated-sorted-array/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "There is an integer array `nums` sorted in ascending order (with **distinct** values).\n",
+        "\n",
+        "Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k (1 <= k < nums.length)` such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index 3 and become `[4,5,6,7,0,1,2]`.\n",
+        "\n",
+        "Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.*\n",
+        "\n",
+        "You must write an algorithm with $O(log\\ n)$ runtime complexity.\n",
+        " \n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- `1 <= nums.length <= 5000`\n",
+        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
+        "- All values of `nums` are **unique**.\n",
+        "- `nums` is an ascending array that is possibly rotated.\n",
+        "- $-10^4$ <= `target` <= $10^4$"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 44,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def search(nums, target):\n",
+        "    left, right = 0, len(nums) - 1\n",
+        "\n",
+        "    # Find the pivot index using binary search\n",
+        "    while left < right:\n",
+        "        mid = left + (right - left) // 2\n",
+        "\n",
+        "        if nums[mid] > nums[right]:\n",
+        "            left = mid + 1\n",
+        "        else:\n",
+        "            right = mid\n",
+        "\n",
+        "    pivot = left\n",
+        "    left, right = 0, len(nums) - 1\n",
+        "\n",
+        "    # Determine which part of the array to search in\n",
+        "    if target >= nums[pivot] and target <= nums[right]:\n",
+        "        left = pivot\n",
+        "    else:\n",
+        "        right = pivot\n",
+        "\n",
+        "    # Perform binary search to find the target\n",
+        "    while left <= right:\n",
+        "        mid = left + (right - left) // 2\n",
+        "\n",
+        "        if nums[mid] == target:\n",
+        "            return mid\n",
+        "        elif nums[mid] < target:\n",
+        "            left = mid + 1\n",
+        "        else:\n",
+        "            right = mid - 1\n",
+        "\n",
+        "    return -1"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. The `search` function takes two arguments: `nums`, which is the rotated sorted array, and `target`, which is the value we want to find in the array.\n",
+        "\n",
+        "2. We initialize two pointers, `left` and `right`, which represent the range in which we are going to perform binary search. Initially, `left` is set to 0, and `right` is set to the index of the last element in the array (`len(nums) - 1`).\n",
+        "\n",
+        "3. We start by finding the pivot index using binary search. The pivot index is the index at which the array is rotated. We use a `while` loop to continue the search until `left` is less than `right`.\n",
+        "\n",
+        "4. Inside the loop, we calculate the middle index `mid` using integer division. We then compare the value at `mid` with the value at `right`. If `nums[mid]` is greater than `nums[right]`, it means the pivot point lies to the right of `mid`, so we update `left` to `mid + 1`. Otherwise, the pivot point lies to the left of `mid`, so we update `right` to `mid`. This process continues until we find the pivot index.\n",
+        "\n",
+        "5. After finding the pivot index, we have divided the array into two parts: one part is sorted in ascending order, and the other part is also sorted in ascending order but rotated. We need to determine which part of the array contains the target value.\n",
+        "\n",
+        "6. We reset `left` and `right` pointers. If the target value is within the range `[nums[pivot], nums[right]]`, we set `left` to `pivot` (the start of the rotated part), indicating that we should search in the rotated part. Otherwise, we set `right` to `pivot` (the end of the sorted part), indicating that we should search in the sorted part.\n",
+        "\n",
+        "7. We then perform binary search again within the chosen range (`left` to `right`) to find the target element. The binary search continues until `left` is less than or equal to `right`.\n",
+        "\n",
+        "8. Inside the binary search loop, we calculate the middle index `mid` and compare `nums[mid]` with the target value. If they are equal, we return `mid` as the index of the target element. If `nums[mid]` is less than the target, we update `left` to `mid + 1` to search in the right half. If `nums[mid]` is greater than the target, we update `right` to `mid - 1` to search in the left half.\n",
+        "\n",
+        "9. If we exit the binary search loop without finding the target, we return -1 to indicate that the target is not present in the array."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 45,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "4\n"
+          ]
+        }
+      ],
+      "source": [
+        "nums1 = [4, 5, 6, 7, 0, 1, 2]\n",
+        "target1 = 0\n",
+        "result1 = search(nums1, target1)\n",
+        "print(result1) "
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 46,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "-1\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
+        "target2 = 3\n",
+        "result2 = search(nums2, target2)\n",
+        "print(result2) "
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 47,
+      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "-1\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3:\n",
+        "nums3 = [1]\n",
+        "target3 = 0\n",
+        "result3 = search(nums3, target3)\n",
+        "print(result3) "
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "1. Finding the pivot index using binary search takes $O(log\\ n)$ time, where '$n$' is the number of elements in the array.\n",
+        "2. After finding the pivot index, performing binary search to find the target element also takes $O(log\\ n)$ time in the worst case.\n",
+        "\n",
+        "The dominant factor in the time complexity is the binary search, and since we perform two binary searches sequentially, the overall time complexity of the code is $O(log\\ n)$.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "The space complexity of the code is very minimal:\n",
+        "\n",
+        "1. We use a constant amount of additional space for variables such as `left`, `right`, `pivot`, and `mid`. These variables do not depend on the input size, so they contribute $O(1)$ space complexity.\n",
+        "2. The function does not use any additional data structures that scale with the input size.\n",
+        "\n",
+        "Therefore, the space complexity of the code is $O(1)$, which means it has a constant space complexity and does not depend on the size of the input array.\n",
+        "\n",
+        "**In summary:**\n",
+        "- Time Complexity: $O(log\\ n)$\n",
+        "- Space Complexity: $O(1)$"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Find Minimum Element in Rotated Sorted Array**: Write an algorithm to find the minimum element in a rotated sorted array. This is a variation of the problem where you don't need to search for a target value but instead find the smallest element.\n",
+        "\n",
+        "2. **Search in a Circularly Sorted Array**: Consider an array that is sorted in a circular manner (e.g., [4, 5, 6, 7, 0, 1, 2, 3]). Adapt the search algorithm to work for circularly sorted arrays while maintaining $O(log\\ n)$ complexity."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/05. Binary Search/README.md b/_build/html/_sources/05. Binary Search/README.md
new file mode 100644
index 0000000..6086b9b
--- /dev/null
+++ b/_build/html/_sources/05. Binary Search/README.md	
@@ -0,0 +1,17 @@
+# Binary search Problems - Blind 75 LeetCode 
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [153. Find Minimum In Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) | Medium |
+| [33. Search In Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)   | Medium     |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
diff --git a/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb b/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb
new file mode 100644
index 0000000..0336bc8
--- /dev/null
+++ b/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb	
@@ -0,0 +1,271 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 141. Linked List Cycle\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Linked List Cycle problem on LeetCode, click here!](https://leetcode.com/problems/linked-list-cycle/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Given `head`, the head of a linked list, determine if the linked list has a cycle in it.\n",
+        "\n",
+        "There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is connected to. **Note that `pos` is not passed as a parameter.**\n",
+        "\n",
+        "Return `true` if there is a cycle in the linked list. Otherwise, return `false`.\n",
+        "\n",
+        " \n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- The number of the nodes in the list is in the range $[0, 10^4]$.\n",
+        "- $-10^5$ <= Node.val <= $10^5$\n",
+        "- `pos` is `-1` or a **valid index** in the linked-list.\n",
+        "\n",
+        "**Follow up:** Can you solve it using $O(1)$ (i.e. constant) memory?"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 14,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Definition for singly-linked list.\n",
+        "class ListNode:\n",
+        "    def __init__(self, val=0):\n",
+        "        self.val = val\n",
+        "        self.next = None\n",
+        "\n",
+        "class Solution:\n",
+        "    def hasCycle(self, head: ListNode) -> bool:\n",
+        "        if not head:\n",
+        "            return False\n",
+        "        \n",
+        "        slow = head\n",
+        "        fast = head\n",
+        "        \n",
+        "        while fast and fast.next:\n",
+        "            slow = slow.next\n",
+        "            fast = fast.next.next\n",
+        "            \n",
+        "            if slow == fast:\n",
+        "                return True\n",
+        "        \n",
+        "        return False"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The given code defines a class `Solution` with a method `hasCycle` for determining if a linked list has a cycle.\n",
+        "\n",
+        "The `ListNode` class represents nodes in the linked list, where each node has a `val` (a value associated with the node) and a `next` attribute (a reference to the next node in the list).\n",
+        "\n",
+        "The `hasCycle` method takes the `head` of a linked list as input and returns `True` if there is a cycle in the linked list, and `False` otherwise. It uses two pointers, `slow` and `fast`, to traverse the linked list. If there is a cycle, the `fast` pointer will eventually catch up to the `slow` pointer.\n",
+        "\n",
+        "The algorithm works as follows:\n",
+        "1. It checks if the input `head` is `None` (an empty list). If it is, the method returns `False` because an empty list can't have a cycle.\n",
+        "2. Two pointers, `slow` and `fast`, initially point to the `head` of the list.\n",
+        "3. The code enters a `while` loop where `fast` moves two steps at a time, and `slow` moves one step at a time.\n",
+        "4. If there is a cycle, `fast` will eventually catch up to `slow`, and the method returns `True`.\n",
+        "5. If the loop completes without finding a cycle, the method returns `False`."
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 15,
+      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Helper function to create a linked list with a cycle\n",
+        "def create_linked_list_with_cycle(values, pos):\n",
+        "    dummy = ListNode()\n",
+        "    current = dummy\n",
+        "    cycle_node = None\n",
+        "    \n",
+        "    for i, val in enumerate(values):\n",
+        "        current.next = ListNode(val)\n",
+        "        current = current.next\n",
+        "        \n",
+        "        if i == pos:\n",
+        "            cycle_node = current\n",
+        "    \n",
+        "    if cycle_node:\n",
+        "        current.next = cycle_node\n",
+        "    \n",
+        "    return dummy.next"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
+      "metadata": {},
+      "source": [
+        "This helper function is designed to create a linked list with a cycle for testing purposes. It takes two arguments:\n",
+        "\n",
+        "1. `values`: A list of values representing the nodes in the linked list.\n",
+        "2. `pos`: An integer representing the index of the node where the cycle begins.\n",
+        "\n",
+        "Here's how the function works:\n",
+        "\n",
+        "- It starts by creating an empty `ListNode` called `dummy`. This `dummy` node is used to simplify the creation of the linked list.\n",
+        "- It initializes a `current` pointer to the `dummy` node. This pointer will be used to traverse the linked list.\n",
+        "- It initializes a `cycle_node` variable to `None`. This variable will hold the reference to the node where the cycle begins, if any.\n",
+        "- It then iterates through the `values` list, creating a new `ListNode` for each value and appending it to the linked list.\n",
+        "- Inside the loop, it checks if the current index `i` is equal to the specified `pos`. If they match, it sets `cycle_node` to the current node. This simulates the creation of a cycle in the linked list.\n",
+        "- After the loop, if `cycle_node` is not `None` (indicating a cycle should be created), it connects the last node in the linked list to `cycle_node`, effectively creating a cycle.\n",
+        "- Finally, it returns the reference to the first node of the linked list (not the `dummy` node).\n",
+        "\n",
+        "This helper function allows you to easily create test cases where you can specify the values for the linked list and the position where the cycle starts. It's particularly useful for testing the code's ability to detect cycles in linked lists."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 20,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 1\n",
+        "head = [3,2,0,-4]\n",
+        "pos = 1\n",
+        "\n",
+        "head = create_linked_list_with_cycle(head, pos)\n",
+        "solution = Solution()\n",
+        "result = solution.hasCycle(head)\n",
+        "print(result)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 21,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 2\n",
+        "head = [1,2]\n",
+        "pos = 0\n",
+        "\n",
+        "head = create_linked_list_with_cycle(head, pos)\n",
+        "solution = Solution()\n",
+        "result = solution.hasCycle(head)\n",
+        "print(result)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 22,
+      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 3\n",
+        "head = [1]\n",
+        "pos = -1\n",
+        "\n",
+        "head = create_linked_list_with_cycle(head, pos)\n",
+        "solution = Solution()\n",
+        "result = solution.hasCycle(head)\n",
+        "print(result)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's discuss the time and space complexity of the code for detecting a cycle in a linked list:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "The time complexity of this code is $O(n)$, where \"$n$\" is the number of nodes in the linked list.\n",
+        "\n",
+        "The reason for this is that both the `slow` and `fast` pointers traverse the linked list, and they move at different speeds. In the worst case, when there is no cycle, the `fast` pointer will reach the end of the list after going through approximately $n/2$ nodes. In the case of a cycle, it may take some extra iterations for the `fast` pointer to catch up to the `slow` pointer. However, the total number of iterations is still proportional to the number of nodes in the linked list, making it $O(n)$.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "The space complexity of this code is $O(1)$, which means it uses constant space.\n",
+        "\n",
+        "The reason for this is that regardless of the size of the linked list, the code only uses two additional pointers (`slow` and `fast`) to traverse the list. These pointers do not depend on the size of the input linked list, so the space complexity remains constant.\n",
+        "\n",
+        "**In summary**, this code efficiently detects cycles in a linked list with a time complexity of $O(n)$ and a space complexity of $O(1)$. It is an example of an algorithm that solves a complex problem with minimal memory usage."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/06. Linked List/143. Reorder List.ipynb b/_build/html/_sources/06. Linked List/143. Reorder List.ipynb
new file mode 100644
index 0000000..a83545b
--- /dev/null
+++ b/_build/html/_sources/06. Linked List/143. Reorder List.ipynb	
@@ -0,0 +1,285 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 143. Reorder List\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Reorder List problem on LeetCode, click here!](https://leetcode.com/problems/reorder-list/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "You are given the head of a singly linked-list. The list can be represented as:\n",
+        "\n",
+        "$L_0 → L_1 → … → L_{n - 1} → L_n$\n",
+        "\n",
+        "Reorder the list to be on the following form:\n",
+        "\n",
+        "$L_0 → L_n → L_1 → L_{n - 1} → L_2 → L_{n - 2} → …$\n",
+        "\n",
+        "You may not modify the values in the list's nodes. Only nodes themselves may be changed.\n",
+        " \n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- The number of nodes in the list is in the range $[1, 5 * 10^4]$.\n",
+        "- `1 <= Node.val <= 1000`"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 28,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class ListNode:\n",
+        "    def __init__(self, val=0, next=None):\n",
+        "        self.val = val\n",
+        "        self.next = next\n",
+        "\n",
+        "def reorderList(head):\n",
+        "    if not head or not head.next:\n",
+        "        return head\n",
+        "\n",
+        "    # Step 1: Find the middle of the linked list\n",
+        "    slow, fast = head, head\n",
+        "    while fast.next and fast.next.next:\n",
+        "        slow = slow.next\n",
+        "        fast = fast.next.next\n",
+        "\n",
+        "    # Split the list into two halves\n",
+        "    first_half = head\n",
+        "    second_half = slow.next\n",
+        "    slow.next = None\n",
+        "\n",
+        "    # Step 2: Reverse the second half of the linked list\n",
+        "    prev = None\n",
+        "    current = second_half\n",
+        "    while current:\n",
+        "        next_node = current.next\n",
+        "        current.next = prev\n",
+        "        prev = current\n",
+        "        current = next_node\n",
+        "    second_half = prev\n",
+        "\n",
+        "    # Step 3: Merge the first and reversed second halves alternately\n",
+        "    p1, p2 = first_half, second_half\n",
+        "    while p2:\n",
+        "        next_p1 = p1.next\n",
+        "        next_p2 = p2.next\n",
+        "        p1.next = p2\n",
+        "        p2.next = next_p1\n",
+        "        p1 = next_p1\n",
+        "        p2 = next_p2\n",
+        "\n",
+        "    return head"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The `reorderList` function is designed to reorder a singly linked list following the specific pattern described in the problem statement. Here's a detailed explanation of how the function works:\n",
+        "\n",
+        "1. **Base Cases Handling**:\n",
+        "   - The function first checks if the input linked list is empty (`not head`) or contains only one element (`not head.next`). If either of these conditions is met, the list cannot be reordered, so the function returns the original list as-is.\n",
+        "\n",
+        "2. **Finding the Middle of the Linked List**:\n",
+        "   - To reorder the list, we first need to find the middle point so that we can split the list into two halves. This is done using two pointers, `slow` and `fast`, initialized to the head of the list.\n",
+        "   - `slow` moves one step at a time while `fast` moves two steps at a time. When `fast` reaches the end of the list or the second-to-last node, `slow` will be at the middle node.\n",
+        "\n",
+        "3. **Splitting the List**:\n",
+        "   - After finding the middle node, we split the list into two halves:\n",
+        "     - The first half, `first_half`, contains nodes from the beginning up to the middle.\n",
+        "     - The second half, `second_half`, contains nodes from the middle to the end.\n",
+        "   - To split the list, we set the `next` pointer of the node before the middle node to `None`.\n",
+        "\n",
+        "4. **Reversing the Second Half**:\n",
+        "   - We reverse the second half of the linked list using the `prev`, `current`, and `next_node` pointers.\n",
+        "   - `prev` is initially `None`, and we iterate through the second half. For each node, we:\n",
+        "     - Set the `next` of the current node to `prev`, effectively reversing the next pointer direction.\n",
+        "     - Update `prev` to the current node.\n",
+        "     - Move to the next node using the `next_node`.\n",
+        "\n",
+        "5. **Merging the Two Halves Alternately**:\n",
+        "   - We now have two linked lists: `first_half` and the reversed `second_half`.\n",
+        "   - We merge these two lists alternately by adjusting the `next` pointers of the nodes.\n",
+        "   - `p1` and `p2` are pointers to the current nodes in `first_half` and `second_half`, respectively.\n",
+        "   - We iterate through both lists while reordering nodes:\n",
+        "     - Set the `next` of `p1` to `p2` to link a node from the first half to a node from the reversed second half.\n",
+        "     - Update `p1` and `p2` to their respective next nodes.\n",
+        "     - Repeat this process until we have processed all nodes in both halves.\n",
+        "\n",
+        "6. **Returning the Reordered List**:\n",
+        "   - After the merging process is complete, the linked list is reordered as specified.\n",
+        "   - The function returns the `head` of the reordered list.\n",
+        "\n",
+        "The overall result is a singly linked list that has been reordered according to the given pattern. The time complexity of this algorithm is O(N), where N is the number of nodes in the linked list, as we traverse the list once to find the middle, once to reverse the second half, and once to merge the two halves."
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 29,
+      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Helper function to print the linked list\n",
+        "def printLinkedList(head):\n",
+        "    result = []\n",
+        "    while head:\n",
+        "        result.append(head.val)\n",
+        "        head = head.next\n",
+        "    return result"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
+      "metadata": {},
+      "source": [
+        "Let's explain the helper functions:\n",
+        "**printLinkedList Function**:\n",
+        "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
+        "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
+        "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
+        "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
+        "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
+        "\n",
+        "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 30,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[1, 4, 2, 3]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1\n",
+        "head1 = ListNode(1, ListNode(2, ListNode(3, ListNode(4))))\n",
+        "reorderList(head1)\n",
+        "print(printLinkedList(head1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 31,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[1, 5, 2, 4, 3]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "head2 = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))\n",
+        "reorderList(head2)\n",
+        "print(printLinkedList(head2))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the `reorderList` function:\n",
+        "\n",
+        "**Time Complexity**:\n",
+        "1. **Finding the Middle of the Linked List**: In this step, we use two pointers, one moving one step at a time (`slow`) and the other moving two steps at a time (`fast`). This process takes $O(N/2)$ time, where $N$ is the number of nodes in the linked list.\n",
+        "\n",
+        "2. **Splitting the List**: After finding the middle, we split the list into two halves by setting the `next` pointer of the node before the middle node to `None`. This step takes $O(1)$ time.\n",
+        "\n",
+        "3. **Reversing the Second Half**: We reverse the second half of the linked list using a while loop that iterates through the second half of the list once. Therefore, this step also takes $O(N/2)$ time.\n",
+        "\n",
+        "4. **Merging the Two Halves Alternately**: In this step, we merge the two halves alternately by adjusting the `next` pointers of the nodes. We iterate through both halves once, so this step takes $O(N/2)$ time.\n",
+        "\n",
+        "Overall, the time complexity of the `reorderList` function is dominated by the steps involving reversing and merging the two halves, both of which take $O(N/2)$ time. Therefore, the total time complexity is $O(N)$.\n",
+        "\n",
+        "**Space Complexity**:\n",
+        "The space complexity of the `reorderList` function is primarily determined by the variables and data structures used within the function. Let's break down the space complexity components:\n",
+        "\n",
+        "1. **Constant Space Variables**: The variables `slow`, `fast`, `prev`, `current`, `next_node`, `p1`, and `p2` are used to traverse and manipulate the linked list. These variables occupy constant space, regardless of the input size. Therefore, they contribute $O(1)$ to the space complexity.\n",
+        "\n",
+        "2. **Split Linked Lists**: In the splitting step, we create two new linked list segments (`first_half` and `second_half`) that store references to nodes from the original linked list. These segments occupy space proportional to half of the input list, $O(N/2)$. \n",
+        "\n",
+        "3. **Reversed Second Half**: During the reversal step, we reverse the second half of the linked list in-place without creating any additional data structures. Therefore, it doesn't contribute to additional space complexity.\n",
+        "\n",
+        "4. **Overall**: Combining the above components, the total space complexity is $O(N/2)$, which simplifies to $O(N)$ in terms of space complexity.\n",
+        "\n",
+        "**In summary**, the `reorderList` function has a time complexity of $O(N)$ and a space complexity of $O(N)$ due to the creation of two new linked list segments while splitting the list. The constant space variables used for traversal do not significantly impact the overall space complexity."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Reorder by K Elements:** Generalize the reorderList function to reorder the list in a different pattern, where you reorder the list by $K$ elements at a time. For example, for $K=3$, you would reorder it as ($L_0 → L_1 → L_2 → L_n → L_{n-1} → L_{n-2} → ...$).\n",
+        "You may need to handle cases where the number of nodes is not a multiple of $K$.\n",
+        "\n",
+        "2. **Reorder by Odd and Even Nodes:** Modify the reorderList function to reorder the list in a pattern where odd-indexed nodes (1-based index) come before even-indexed nodes. For example, ($L_0 → L_n → L_1 → L_{n-1} → L_2 → L_{n-2} → ...$)."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb b/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb
new file mode 100644
index 0000000..263124e
--- /dev/null
+++ b/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb	
@@ -0,0 +1,278 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 19. Remove Nth Node From End of List\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Remove Nth Node From End of List problem on LeetCode, click here!](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Given the head of a linked list, remove the $n^{th}$ node from the end of the list and return its head.\n",
+        " \n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- The number of nodes in the list is `sz`.\n",
+        "- `1 <= sz <= 30`\n",
+        "- `0 <= Node.val <= 100`\n",
+        "- `1 <= n <= sz`\n",
+        "\n",
+        "**Follow up:** Could you do this in one pass?"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 23,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class ListNode:\n",
+        "    def __init__(self, val=0, next=None):\n",
+        "        self.val = val\n",
+        "        self.next = next\n",
+        "\n",
+        "def removeNthFromEnd(head, n):\n",
+        "    # Create a dummy node to handle the case of removing the head node.\n",
+        "    dummy = ListNode(0)\n",
+        "    dummy.next = head\n",
+        "    first = dummy\n",
+        "    second = dummy\n",
+        "    \n",
+        "    # Advance the first pointer by n+1 nodes.\n",
+        "    for _ in range(n + 1):\n",
+        "        first = first.next\n",
+        "    \n",
+        "    # Move both pointers simultaneously until first reaches the end.\n",
+        "    while first is not None:\n",
+        "        first = first.next\n",
+        "        second = second.next\n",
+        "    \n",
+        "    # Remove the nth node by updating the next pointer of the previous node.\n",
+        "    second.next = second.next.next\n",
+        "    \n",
+        "    return dummy.next  # Return the modified head of the linked list"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We start by defining a `ListNode` class to represent individual nodes in a linked list. Each node has two attributes: `val` (the value of the node) and `next` (a reference to the next node in the list).\n",
+        "\n",
+        "2. The `removeNthFromEnd` function takes the head of a linked list (`head`) and the value `n`, which represents the position from the end of the list of the node to be removed. It returns the modified head of the linked list after the removal.\n",
+        "\n",
+        "3. We create a dummy node (`dummy`) at the beginning of the list. This dummy node simplifies the code by handling cases where the head of the list needs to be removed.\n",
+        "\n",
+        "4. We initialize two pointers, `first` and `second`, both initially pointing to the dummy node.\n",
+        "\n",
+        "5. To advance the `first` pointer by `n+1` nodes, we use a `for` loop. This positions the `first` pointer `n` nodes ahead of the `second` pointer.\n",
+        "\n",
+        "6. We then move both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This ensures that the `second` pointer ends up pointing to the node that needs to be removed.\n",
+        "\n",
+        "7. To remove the nth node from the end, we update the `next` pointer of the node pointed to by the `second` pointer to skip over the node to be removed.\n",
+        "\n",
+        "8. Finally, we return `dummy.next`, which is the modified head of the linked list without the removed node.\n",
+        "\n",
+        "9. Additionally, there are two helper functions:\n",
+        "   - `createLinkedList(values)` creates a linked list from a list of values.\n",
+        "   - `convertToList(head)` converts a linked list back into a list of values."
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 24,
+      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Helper function to create a linked list from a list of values.\n",
+        "def createLinkedList(values):\n",
+        "    if not values:\n",
+        "        return None\n",
+        "    head = ListNode(values[0])\n",
+        "    current = head\n",
+        "    for val in values[1:]:\n",
+        "        current.next = ListNode(val)\n",
+        "        current = current.next\n",
+        "    return head\n",
+        "\n",
+        "# Helper function to convert a linked list to a list of values.\n",
+        "def convertToList(head):\n",
+        "    result = []\n",
+        "    current = head\n",
+        "    while current:\n",
+        "        result.append(current.val)\n",
+        "        current = current.next\n",
+        "    return result"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
+      "metadata": {},
+      "source": [
+        "Let's explain the two helper functions:\n",
+        "\n",
+        "1. `createLinkedList(values)`: This function creates a linked list from a list of values.\n",
+        "\n",
+        "   - Input: `values` is a list containing values that you want to insert into the linked list.\n",
+        "   - Output: The function returns the head of the created linked list.\n",
+        "\n",
+        "   Explanation:\n",
+        "   - The function starts by checking if the `values` list is empty. If it is, it returns `None` to indicate an empty linked list.\n",
+        "   - If the `values` list is not empty, it creates the first node of the linked list with the value from the first element of `values`.\n",
+        "   - It then iterates through the remaining elements of `values` and creates new nodes for each value, connecting them together using the `next` attribute to form a linked list.\n",
+        "   - Finally, it returns the head of the linked list, which is the first node created.\n",
+        "   \n",
+        "\n",
+        "2. `convertToList(head)`: This function converts a linked list into a list of values.\n",
+        "\n",
+        "   - Input: `head` is the head node of the linked list that you want to convert.\n",
+        "   - Output: The function returns a list containing the values of the nodes in the linked list in order.\n",
+        "\n",
+        "   Explanation:\n",
+        "   - The function starts by initializing an empty list, `result`, to store the values of the linked list nodes.\n",
+        "   - It then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
+        "   - The iteration continues until it reaches the end of the linked list (i.e., the `next` attribute of the current node becomes `None`).\n",
+        "   - After the iteration is complete, the function returns the `result` list, which now contains the values of all the nodes in the linked list in the same order as they appear in the linked list.\n",
+        "\n",
+        "These helper functions are useful for creating linked lists from lists of values and converting linked lists back into lists of values, making it easier to work with linked list examples and test cases in a more familiar list format."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 25,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[1, 2, 3, 5]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1\n",
+        "head1 = createLinkedList([1, 2, 3, 4, 5])\n",
+        "n1 = 2\n",
+        "new_head1 = removeNthFromEnd(head1, n1)\n",
+        "print(convertToList(new_head1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 26,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "head2 = createLinkedList([1])\n",
+        "n2 = 1\n",
+        "new_head2 = removeNthFromEnd(head2, n2)\n",
+        "print(convertToList(new_head2))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 27,
+      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[1]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3\n",
+        "head3 = createLinkedList([1, 2])\n",
+        "n3 = 1\n",
+        "new_head3 = removeNthFromEnd(head3, n3)\n",
+        "print(convertToList(new_head3))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Certainly! Let's analyze the time and space complexity of the `removeNthFromEnd` function in isolation:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "- The `removeNthFromEnd` function performs two passes through the linked list. In the worst case, it needs to traverse the entire linked list of length $N$.\n",
+        "- The first pass positions the `first` pointer $n$ nodes ahead of the `second` pointer. This pass takes $O(N)$ time because it involves iterating through all N nodes.\n",
+        "- The second pass moves both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This also takes $O(N)$ time in the worst case.\n",
+        "- Therefore, the overall time complexity of the `removeNthFromEnd` function is $O(N)$.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "- The `removeNthFromEnd` function uses a constant amount of extra space for its variables (e.g., `first`, `second`, and `dummy`). This space usage does not depend on the size of the linked list.\n",
+        "- The space complexity is $O(1)$, indicating that the space used by the function is constant and independent of the size of the input linked list.\n",
+        "\n",
+        "**In summary**, the `removeNthFromEnd` function has a time complexity of $O(N)$ and a space complexity of $O(1)$. It efficiently removes the nth node from the end of the linked list with a single pass through the list and constant extra space usage."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb b/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb
new file mode 100644
index 0000000..c97de2d
--- /dev/null
+++ b/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb	
@@ -0,0 +1,316 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 206. Reverse Linked List\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Reverse Linked List problem on LeetCode, click here!](https://leetcode.com/problems/reverse-linked-list/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Given the `head` of a singly linked list, reverse the list, and return the reversed list.\n",
+        " \n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- The number of nodes in the list is the range `[0, 5000]`.\n",
+        "- `-5000 <= Node.val <= 5000`\n",
+        "\n",
+        "**Follow up:** A linked list can be reversed either iteratively or recursively. Could you implement both?"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 33,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Definition for singly-linked list.\n",
+        "class ListNode:\n",
+        "    def __init__(self, val=0, next=None):\n",
+        "        self.val = val\n",
+        "        self.next = next\n",
+        "\n",
+        "# Function to reverse a linked list iteratively\n",
+        "def reverseListIterative(head):\n",
+        "    prev = None\n",
+        "    current = head\n",
+        "    \n",
+        "    while current:\n",
+        "        next_node = current.next\n",
+        "        current.next = prev\n",
+        "        prev = current\n",
+        "        current = next_node\n",
+        "    \n",
+        "    return prev\n",
+        "\n",
+        "# Function to reverse a linked list recursively\n",
+        "def reverseListRecursive(head):\n",
+        "    if not head or not head.next:\n",
+        "        return head\n",
+        "\n",
+        "    new_head = reverseListRecursive(head.next)\n",
+        "    head.next.next = head\n",
+        "    head.next = None\n",
+        "\n",
+        "    return new_head"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. `reverseListIterative(head)`: This function reverses a linked list iteratively using a loop. Here's a step-by-step explanation:\n",
+        "\n",
+        "   - It takes the `head` of the input linked list as a parameter.\n",
+        "   - It initializes two pointers, `prev` and `current`, initially set to `None` and the `head` of the list, respectively.\n",
+        "   - It enters a `while` loop that continues until `current` reaches the end of the list (i.e., becomes `None`).\n",
+        "   - Inside the loop:\n",
+        "     - It stores the `next_node` which is the next node after `current`.\n",
+        "     - It updates the `next` pointer of `current` to point to the `prev` node. This effectively reverses the link direction.\n",
+        "     - It moves `prev` to `current` and `current` to `next_node`, advancing the pointers one step further in the list.\n",
+        "   - Once the loop completes, `prev` will be pointing to the new head of the reversed list (which was the last node of the original list), so it returns `prev`.\n",
+        "\n",
+        "2. `reverseListRecursive(head)`: This function reverses a linked list recursively. Here's how it works:\n",
+        "\n",
+        "   - It takes the `head` of the input linked list as a parameter.\n",
+        "   - The base case is checked: if `head` is `None` or `head.next` is `None`, meaning the list is empty or has only one node, it returns `head` as there's no need to reverse a list with zero or one element.\n",
+        "   - In the recursive case:\n",
+        "     - It calls itself with `head.next`, effectively moving down the list until it reaches the end.\n",
+        "     - Once it reaches the end, it starts reversing the links:\n",
+        "       - `head.next.next` is set to `head`, reversing the link between `head` and the next node.\n",
+        "       - `head.next` is set to `None` to avoid cycles.\n",
+        "   - Finally, it returns the new head of the reversed list, which will be the last node of the original list.\n",
+        "\n",
+        "**In summary,** `reverseListIterative` reverses the linked list by iterating through it and changing the next pointers, while `reverseListRecursive` reverses the linked list by recursively reaching the end and then reversing the links on the way back up the recursion stack. Both functions achieve the same result: reversing the linked list."
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 36,
+      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Helper function to create a linked list from a list of values\n",
+        "def createLinkedList(values):\n",
+        "    if not values:\n",
+        "        return None\n",
+        "    head = ListNode(values[0])\n",
+        "    current = head\n",
+        "    for val in values[1:]:\n",
+        "        current.next = ListNode(val)\n",
+        "        current = current.next\n",
+        "    return head\n",
+        "\n",
+        "# Helper function to convert a linked list to a list for testing\n",
+        "def linkedListToList(head):\n",
+        "    result = []\n",
+        "    current = head\n",
+        "    while current:\n",
+        "        result.append(current.val)\n",
+        "        current = current.next\n",
+        "    return result\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
+      "metadata": {},
+      "source": [
+        "Let's explain the two helper functions used in the provided code:\n",
+        "\n",
+        "1. `createLinkedList(values)`:\n",
+        "\n",
+        "   - Purpose: This function is used to create a linked list from a list of values. It takes a list of values as input and returns the head of the linked list.\n",
+        "\n",
+        "   - How it works:\n",
+        "     - It first checks if the input list `values` is empty. If it's empty, it returns `None` to indicate an empty linked list.\n",
+        "     - If `values` is not empty, it initializes a `head` node with the value of the first element in `values`.\n",
+        "     - It then iterates through the remaining values in `values`, creating new nodes for each value and linking them together to form a linked list.\n",
+        "     - Finally, it returns the `head` node of the newly created linked list.\n",
+        "\n",
+        "2. `linkedListToList(head)`:\n",
+        "\n",
+        "   - Purpose: This function is used to convert a linked list back into a Python list for testing and output purposes.\n",
+        "\n",
+        "   - How it works:\n",
+        "     - It takes the `head` of the linked list as input.\n",
+        "     - It initializes an empty list called `result`.\n",
+        "     - It then iterates through the linked list, starting from `head`, and appends each node's `val` (value) to the `result` list.\n",
+        "     - This process continues until the end of the linked list is reached.\n",
+        "     - Finally, it returns the `result` list, which contains the values from the linked list in the same order.\n",
+        "\n",
+        "These helper functions are used to simplify the process of creating linked lists from lists of values and converting linked lists back into lists, making it easier to work with linked lists in the provided examples and test cases."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 35,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "data": {
+            "text/plain": [
+              "[1, 1, 2, 3, 4, 4]"
+            ]
+          },
+          "execution_count": 35,
+          "metadata": {},
+          "output_type": "execute_result"
+        }
+      ],
+      "source": [
+        "# Example 1: Iterative\n",
+        "head1 = [1,2,3,4,5]\n",
+        "head1 = createLinkedList(head1)\n",
+        "reversed_head1 = reverseListIterative(head1)\n",
+        "print(linkedListToList(reversed_head1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 31,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "data": {
+            "text/plain": [
+              "[]"
+            ]
+          },
+          "execution_count": 37,
+          "metadata": {},
+          "output_type": "execute_result"
+        }
+      ],
+      "source": [
+        "# Example 2: Recursive\n",
+        "head2 = [1,2]\n",
+        "head2 = createLinkedList(head2)\n",
+        "reversed_head2 = reverseListRecursive(head2)\n",
+        "print(linkedListToList(reversed_head2))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 38,
+      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "data": {
+            "text/plain": [
+              "[0]"
+            ]
+          },
+          "execution_count": 38,
+          "metadata": {},
+          "output_type": "execute_result"
+        }
+      ],
+      "source": [
+        "# Example 3: Empty list\n",
+        "head3 = []\n",
+        "head3 = createLinkedList(head3)\n",
+        "reversed_head3 = reverseListIterative(head3)\n",
+        "print(linkedListToList(reversed_head3))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the two functions:\n",
+        "\n",
+        "1. `reverseListIterative(head)`:\n",
+        "\n",
+        "   - Time Complexity: $O(n)$\n",
+        "     - The iterative function visits each node in the linked list exactly once in a single pass, where \"$n$\" is the number of nodes in the list. Therefore, the time complexity is linear in the number of nodes.\n",
+        "\n",
+        "   - Space Complexity: $O(1)$\n",
+        "     - This function uses a constant amount of extra space for the `prev`, `current`, and `next_node` pointers. Regardless of the size of the input list, the amount of additional memory used remains the same, so the space complexity is constant.\n",
+        "\n",
+        "2. `reverseListRecursive(head)`:\n",
+        "\n",
+        "   - Time Complexity: $O(n)$\n",
+        "     - The recursive function also visits each node in the linked list exactly once. It recursively traverses the list from the head to the tail, reversing the links on the way back. Therefore, like the iterative approach, the time complexity is $O(n)$, where \"$n$\" is the number of nodes.\n",
+        "\n",
+        "   - Space Complexity: $O(n)$\n",
+        "     - The recursive function uses space on the call stack for each recursive call. In the worst case, when the list has \"$n$\" nodes, it will create \"$n$\" recursive function calls on the stack, resulting in a space complexity of $O(n)$. This is because each function call stores information about its state, including the `head` pointer, until it reaches the base case and starts returning.\n",
+        "\n",
+        "**In summary:**\n",
+        "\n",
+        "- The time complexity of both functions is $O(n)$ because they both visit each node once.\n",
+        "- The space complexity of `reverseListIterative` is $O(1)$ because it uses a constant amount of extra space.\n",
+        "- The space complexity of `reverseListRecursive` is $O(n)$ due to the recursive function calls and the associated call stack space.\n",
+        "\n",
+        "Both functions are efficient in terms of time complexity, but `reverseListIterative` is more memory-efficient because it uses a constant amount of additional space, while `reverseListRecursive` uses space on the call stack proportional to the number of nodes in the list."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **K-Group Reverse:** Modify the reverseListIterative function to reverse the linked list in groups of $K$ elements. For example, if the input linked list is [1, 2, 3, 4, 5, 6], and $K$ is 3, the output should be [3, 2, 1, 6, 5, 4].\n",
+        "\n",
+        "2. **Rotate Linked List:** Write a function to rotate a linked list to the right by $K$ places. For example, if the input is [1, 2, 3, 4, 5] and $K$ is 2, the output should be [4, 5, 1, 2, 3]."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb b/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb
new file mode 100644
index 0000000..876d633
--- /dev/null
+++ b/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb	
@@ -0,0 +1,249 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 21. Merge Two Sorted Lists\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Merge Two Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-two-sorted-lists/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "You are given the heads of two sorted linked lists `list1` and `list2`.\n",
+        "\n",
+        "Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.\n",
+        "\n",
+        "*Return the head of the merged linked list.*\n",
+        " \n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- The number of nodes in both lists is in the range `[0, 50]`.\n",
+        "- `-100 <= Node.val <= 100`\n",
+        "- Both `list1` and `list2` are sorted in non-decreasing order."
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class ListNode:\n",
+        "    def __init__(self, val=0, next=None):\n",
+        "        self.val = val\n",
+        "        self.next = next\n",
+        "\n",
+        "def mergeTwoLists(list1, list2):\n",
+        "    # Create a dummy node to simplify the code.\n",
+        "    dummy = ListNode()\n",
+        "    current = dummy\n",
+        "\n",
+        "    while list1 and list2:\n",
+        "        if list1.val < list2.val:\n",
+        "            current.next = list1\n",
+        "            list1 = list1.next\n",
+        "        else:\n",
+        "            current.next = list2\n",
+        "            list2 = list2.next\n",
+        "        current = current.next\n",
+        "\n",
+        "    # Append the remaining elements from either list.\n",
+        "    if list1:\n",
+        "        current.next = list1\n",
+        "    else:\n",
+        "        current.next = list2\n",
+        "\n",
+        "    return dummy.next"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We define a class `ListNode` to represent the nodes of a singly-linked list. Each node has a `val` attribute to store the value of the node and a `next` attribute to point to the next node in the list.\n",
+        "\n",
+        "2. We define the `mergeTwoLists` function that takes two linked list heads, `list1` and `list2`, as input.\n",
+        "\n",
+        "3. We create a `dummy` node at the beginning of the merged list. The `dummy` node simplifies the code by serving as a placeholder, and its `next` attribute will point to the actual merged list.\n",
+        "\n",
+        "4. We initialize a `current` pointer to point to the `dummy` node initially. This `current` pointer helps us traverse and build the merged list.\n",
+        "\n",
+        "5. We enter a `while` loop that continues until either `list1` or `list2` becomes empty. Inside the loop, we compare the values of the nodes at the current positions of `list1` and `list2`.\n",
+        "\n",
+        "6. If the `val` of the node in `list1` is smaller than the `val` of the node in `list2`, we attach the node from `list1` to the merged list by updating the `next` attribute of the `current` node to point to the node in `list1`. We then move the `list1` pointer to the next node.\n",
+        "\n",
+        "7. If the `val` of the node in `list2` is smaller than or equal to the `val` of the node in `list1`, we attach the node from `list2` to the merged list in a similar manner. We move the `list2` pointer to the next node.\n",
+        "\n",
+        "8. After attaching a node to the merged list, we move the `current` pointer to the newly added node. This step is essential for keeping track of the end of the merged list.\n",
+        "\n",
+        "9. The loop continues until either `list1` or `list2` becomes empty.\n",
+        "\n",
+        "10. Once the loop exits, we check if there are any remaining elements in `list1` or `list2`. If `list1` is not empty, we attach the remaining elements of `list1` to the merged list. If `list2` is not empty, we attach the remaining elements of `list2`.\n",
+        "\n",
+        "11. Finally, we return the `next` attribute of the `dummy` node as the head of the merged list, which represents the sorted merged list."
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Helper function to print the linked list\n",
+        "def printLinkedList(head):\n",
+        "    result = []\n",
+        "    while head:\n",
+        "        result.append(head.val)\n",
+        "        head = head.next\n",
+        "    return result"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
+      "metadata": {},
+      "source": [
+        "Let's explain the helper functions:\n",
+        "**printLinkedList Function**:\n",
+        "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
+        "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
+        "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
+        "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
+        "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
+        "\n",
+        "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Example 1:\n",
+        "# Input: list1 = [1,2,4], list2 = [1,3,4]\n",
+        "list1 = ListNode(1, ListNode(2, ListNode(4)))\n",
+        "list2 = ListNode(1, ListNode(3, ListNode(4)))\n",
+        "merged_list = mergeTwoLists(list1, list2)\n",
+        "printLinkedList(merged_list)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Example 2:\n",
+        "# Input: list1 = [], list2 = []\n",
+        "list1 = None\n",
+        "list2 = None\n",
+        "merged_list = mergeTwoLists(list1, list2)\n",
+        "printLinkedList(merged_list)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": null,
+      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Example 3:\n",
+        "# list1 = [], list2 = [0]\n",
+        "list1 = None\n",
+        "list2 = ListNode(0)\n",
+        "merged_list = mergeTwoLists(list1, list2)\n",
+        "printLinkedList(merged_list)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the `mergeTwoLists` function.\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "The time complexity of this function is $O(N + M$), where $N$ and $M$ are the lengths of `list1` and `list2`, respectively. Here's why:\n",
+        "\n",
+        "1. In the worst case, we need to traverse both `list1` and `list2` completely. This involves iterating through all the nodes in both lists once.\n",
+        "\n",
+        "2. The while loop runs until either `list1` or `list2` becomes empty. The number of iterations depends on the total number of nodes in both lists, which is N + M in the worst case.\n",
+        "\n",
+        "3. Inside the loop, we perform constant-time operations for each iteration, such as comparisons and updating pointers.\n",
+        "\n",
+        "Therefore, the dominant factor in the time complexity is the combined length of both input lists, resulting in a linear time complexity of $O(N + M)$.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "The space complexity of this function is $O(1)$, which means it uses a constant amount of extra space regardless of the input sizes. Here's why:\n",
+        "\n",
+        "1. We create a few extra pointers (`dummy`, `current`) and temporary variables (`val`) to manage the merging process. These consume a fixed amount of memory, regardless of the input sizes. The number of these extra variables is independent of the lengths of `list1` and `list2`.\n",
+        "\n",
+        "2. We do not create a new data structure or allocate memory for the merged list. Instead, we rearrange the existing nodes from `list1` and `list2` to form the merged list. This operation does not consume additional memory proportional to the input sizes.\n",
+        "\n",
+        "**In summary**, the `mergeTwoLists` function has a time complexity of $O(N + M)$ and a space complexity of $O(1)$, making it an efficient solution for merging two sorted linked lists."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Merge K Sorted Lists:** Extend the problem to merge $K$ sorted linked lists instead of just two. You need to efficiently merge $K$ lists into one sorted list.\n",
+        "\n",
+        "2. **Merge Lists in a Zigzag Pattern:** Merge two sorted lists in a zigzag pattern. For example, given [1, 2, 4] and [1, 3, 5], the merged list should be [1, 1, 2, 3, 4, 5]."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": "sql",
+      "file_extension": "",
+      "mimetype": "",
+      "name": "sql",
+      "version": "3.32.3"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb b/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb
new file mode 100644
index 0000000..a850734
--- /dev/null
+++ b/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb	
@@ -0,0 +1,285 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 23. Merge k Sorted Lists\n",
+        "\n",
+        "**Difficulty:** Hard\n",
+        "\n",
+        "**Link to Problem:** [To see the Merge k Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-k-sorted-lists/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "You are given an array of $k$ linked-lists lists, each linked-list is sorted in ascending order.\n",
+        "\n",
+        "Merge all the linked-lists into one sorted linked-list and return it.\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- `k == lists.length`\n",
+        "- 0 <= k <= $10^4$\n",
+        "- `0 <= lists[i].length <= 500`\n",
+        "- $-10^4$ <= lists[i][j] <= $10^4$\n",
+        "- `lists[i]` is sorted in ascending order.\n",
+        "- The sum of `lists[i].length` will not exceed $10^4$.\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 2,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Definition for singly-linked list.\n",
+        "class ListNode:\n",
+        "     def __init__(self, val=0, next=None):\n",
+        "         self.val = val\n",
+        "         self.next = next\n",
+        "\n",
+        "class Solution:\n",
+        "    def mergeKLists(self, lists: [[ListNode]]) -> ListNode:\n",
+        "        # Check if the input list of linked lists is empty\n",
+        "        if not lists or len(lists) == 0:\n",
+        "            return None\n",
+        "\n",
+        "        # Loop until there is only one merged list left\n",
+        "        while len(lists) > 1:\n",
+        "            mergedLists = []\n",
+        "\n",
+        "            # Merge pairs of lists\n",
+        "            for i in range(0, len(lists), 2):\n",
+        "                l1 = lists[i]\n",
+        "                l2 = lists[i + 1] if (i + 1) < len(lists) else None\n",
+        "                mergedLists.append(self.mergeList(l1, l2))\n",
+        "\n",
+        "            # Update the list of lists with merged results\n",
+        "            lists = mergedLists\n",
+        "        \n",
+        "        # Return the final merged list\n",
+        "        return lists[0]\n",
+        "\n",
+        "    def mergeList(self, l1, l2):\n",
+        "        # Create a dummy node to simplify merging\n",
+        "        dummy = ListNode()\n",
+        "        tail = dummy\n",
+        "\n",
+        "        # Merge the two sorted lists\n",
+        "        while l1 and l2:\n",
+        "            if l1.val < l2.val:\n",
+        "                tail.next = l1\n",
+        "                l1 = l1.next\n",
+        "            else:\n",
+        "                tail.next = l2\n",
+        "                l2 = l2.next\n",
+        "            tail = tail.next\n",
+        "\n",
+        "        # Append any remaining elements from l1 or l2 (if any)\n",
+        "        if l1:\n",
+        "            tail.next = l1\n",
+        "        if l2:\n",
+        "            tail.next = l2\n",
+        "\n",
+        "        # Return the merged result starting from the next of the dummy node\n",
+        "        return dummy.next\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The provided code defines a Python class `Solution` with two methods for merging k sorted linked lists:\n",
+        "\n",
+        "1. `mergeKLists(self, lists: List[ListNode]) -> ListNode`: This method takes a list of k sorted linked lists as input and returns a single merged sorted linked list. It uses a divide-and-conquer approach to repeatedly merge pairs of lists until only one merged list remains.\n",
+        "\n",
+        "2. `mergeList(self, l1, l2)`: This method takes two sorted linked lists, `l1` and `l2`, as input and merges them into a single sorted linked list. It uses a dummy node to simplify the merging process.\n",
+        "\n",
+        "Here's a high-level overview of how the code works:\n",
+        "\n",
+        "- The `mergeKLists` method checks if the input list of linked lists is empty or contains no lists. If there are no lists, it returns `None`.\n",
+        "\n",
+        "- Inside a `while` loop, the code repeatedly merges pairs of linked lists until only one merged list remains in the `lists` array. It does this by iterating through the input lists in pairs and calling the `mergeList` method to merge each pair.\n",
+        "\n",
+        "- The `mergeList` method takes two sorted linked lists, `l1` and `l2`, and merges them into a single sorted linked list. It uses a dummy node (`dummy`) and a `tail` pointer to keep track of the merged list while comparing and merging elements from `l1` and `l2`.\n",
+        "\n",
+        "- After merging all pairs of lists and updating the `lists` array with the merged results, the loop continues until only one merged list remains in the `lists` array.\n",
+        "\n",
+        "- Finally, the `mergeKLists` method returns the merged list.\n",
+        "\n",
+        "Overall, this code efficiently merges k sorted linked lists using a divide-and-conquer strategy, resulting in a single merged sorted linked list as the output."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 11,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> "
+          ]
+        }
+      ],
+      "source": [
+        "### Example 1\n",
+        "# Input: lists = [[1,4,5],[1,3,4],[2,6]]\n",
+        "\n",
+        "lists1 = [\n",
+        "    ListNode(1, ListNode(4, ListNode(5))),\n",
+        "    ListNode(1, ListNode(3, ListNode(4))),\n",
+        "    ListNode(2, ListNode(6))\n",
+        "]\n",
+        "solution = Solution()\n",
+        "result1 = solution.mergeKLists(lists1)\n",
+        "\n",
+        "# Print the result\n",
+        "if result1:\n",
+        "    current = result1\n",
+        "    while current:\n",
+        "        print(current.val, end=\" -> \")\n",
+        "        current = current.next\n",
+        "else:\n",
+        "    print(\"None\")  # Print \"None\" for input with a single None element"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 12,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "None\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 2\n",
+        "# Input: lists = []\n",
+        "\n",
+        "lists2 = []\n",
+        "solution = Solution()\n",
+        "result2 = solution.mergeKLists(lists2)\n",
+        "\n",
+        "# Print the result\n",
+        "if result2:\n",
+        "    current = result2\n",
+        "    while current:\n",
+        "        print(current.val, end=\" -> \")\n",
+        "        current = current.next\n",
+        "else:\n",
+        "    print(\"None\")  # Print \"None\" for input with a single None element"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 13,
+      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "None\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 3\n",
+        "# Input: lists = [[]]\n",
+        "\n",
+        "lists3 = [None]\n",
+        "solution = Solution()\n",
+        "result3 = solution.mergeKLists(lists3)\n",
+        "\n",
+        "# Print the result\n",
+        "if result3:\n",
+        "    current = result3\n",
+        "    while current:\n",
+        "        print(current.val, end=\" -> \")\n",
+        "        current = current.next\n",
+        "else:\n",
+        "    print(\"None\")  # Print \"None\" for input with a single None element"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "1. **Heap Initialization:** The code does not use a heap. Instead, it uses a divide-and-conquer approach. The initial check for empty input lists takes $O(1)$ time.\n",
+        "\n",
+        "2. **Merging:** The merging operation is performed in a divide-and-conquer fashion. In each iteration of the `while` loop, we merge pairs of linked lists, and the number of comparisons made is proportional to the total number of nodes across all the linked lists ($n$). In each merge step, we effectively process each node once. The number of iterations required to reduce k lists to 1 is $O(log\\ k)$.\n",
+        "\n",
+        "   Therefore, the overall time complexity of the code is $O(n * log\\ k)$, where n is the total number of nodes across all lists, and $k$ is the number of linked lists.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "1. **Heap Space:** The code doesn't use a heap data structure, so there's no additional space complexity due to a heap.\n",
+        "\n",
+        "2. **Merged Lists:** In the `mergeList` method, we create a new merged list. However, this list is not stored in memory for all lists; it's replaced with each merged pair. The space used for these merged lists is proportional to the size of the largest merged list, which is $O(n)$ in the worst case.\n",
+        "\n",
+        "3. **Additional Variables:** The code uses a few additional variables, such as `dummy` and `tail`, but these occupy a constant amount of space and don't depend on the input size.\n",
+        "\n",
+        "   Therefore, the overall space complexity of the code is $O(n)$, where n is the total number of nodes across all lists.\n",
+        "\n",
+        "In summary, the code's time complexity is $O(n * log(k))$, and its space complexity is $O(n)$. This code efficiently merges $k$ sorted linked lists using a divide-and-conquer approach with a relatively low space overhead."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/06. Linked List/README.md b/_build/html/_sources/06. Linked List/README.md
new file mode 100644
index 0000000..fb7b3ee
--- /dev/null
+++ b/_build/html/_sources/06. Linked List/README.md	
@@ -0,0 +1,21 @@
+# Linked List Problems - Blind 75 LeetCode 
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)                       | Easy       |
+| [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)                   | Easy       |
+| [143. Reorder List](https://leetcode.com/problems/reorder-list/)                                     | Medium     |
+| [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | Medium |
+| [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)                           | Medium     |
+| [23. Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)                       | Hard       |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
diff --git a/_build/html/_sources/07. Trees/100. Same Tree.ipynb b/_build/html/_sources/07. Trees/100. Same Tree.ipynb
new file mode 100644
index 0000000..a270fab
--- /dev/null
+++ b/_build/html/_sources/07. Trees/100. Same Tree.ipynb	
@@ -0,0 +1,211 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 100: Same Tree\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
+        "\n",
+        "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
+        "\n",
+        "**Constraints**\n",
+        "\n",
+        "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
+        "2. $-10^4$ <= Node.val <= $10^4$\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 5,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Definition for a binary tree node.\n",
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "def isSameTree(p, q):\n",
+        "    # Base case: If both p and q are None, the trees are the same.\n",
+        "    if not p and not q:\n",
+        "        return True\n",
+        "    \n",
+        "    # Base case: If either p or q is None (but not both), the trees are different.\n",
+        "    if not p or not q:\n",
+        "        return False\n",
+        "    \n",
+        "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
+        "    if p.val != q.val:\n",
+        "        return False\n",
+        "    \n",
+        "    # Recursively check the left and right subtrees of p and q.\n",
+        "    # If both subtrees are the same, the entire trees are the same.\n",
+        "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
+        "\n",
+        "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
+        "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
+        "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
+        "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
+        "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
+        "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 2,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 1\n",
+        "\n",
+        "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
+        "\n",
+        "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
+        "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
+        "print(isSameTree(p1, q1)) "
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 3,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 2\n",
+        "\n",
+        "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
+        "\n",
+        "p2 = TreeNode(1, TreeNode(2), None)\n",
+        "q2 = TreeNode(1, None, TreeNode(2))\n",
+        "print(isSameTree(p2, q2))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 4,
+      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 3\n",
+        "\n",
+        "#Input: p = [1,2,1], q = [1,1,2]\n",
+        "\n",
+        "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
+        "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
+        "print(isSameTree(p3, q3))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time Complexity\n",
+        "\n",
+        "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
+        "\n",
+        "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
+        "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
+        "\n",
+        "## Space Complexity\n",
+        "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
+        "\n",
+        "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
+        "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
+        "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
+        "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
+        "\n",
+        "## In summary:\n",
+        "\n",
+        "+ **Time Complexity:** $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
+        "+ **Space Complexity:** $O(n)$ due to the recursive call stack."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb b/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb
new file mode 100644
index 0000000..8cb08b7
--- /dev/null
+++ b/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb	
@@ -0,0 +1,271 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 102. Binary Tree Level Order Traversal\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Binary Tree Level Order Traversal problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-level-order-traversal/)\n",
+        "\n",
+        "---\n",
+        "Given the `root` of a binary tree, return *the level order traversal of its nodes' values*. (i.e., from left to right, level by level).\n",
+        "\n",
+        "**Constraints:**\n",
+        "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
+        "- `-1000 <= Node.val <= 1000`\n",
+        "\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 5,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "def levelOrder(root):\n",
+        "    # Check if the tree is empty, if so, return an empty list.\n",
+        "    if not root:\n",
+        "        return []\n",
+        "\n",
+        "    # Initialize an empty list to store the result.\n",
+        "    result = []\n",
+        "    \n",
+        "    # Initialize a queue with the root node to perform BFS.\n",
+        "    queue = [root]\n",
+        "\n",
+        "    while queue:\n",
+        "        # Initialize an empty list to store the values of nodes at the current level.\n",
+        "        level_values = []\n",
+        "        \n",
+        "        # Get the number of nodes at the current level.\n",
+        "        level_size = len(queue)\n",
+        "\n",
+        "        # Iterate through the nodes at the current level.\n",
+        "        for _ in range(level_size):\n",
+        "            # Dequeue the front node from the queue.\n",
+        "            node = queue.pop(0)\n",
+        "            \n",
+        "            # Append the value of the current node to the level_values list.\n",
+        "            level_values.append(node.val)\n",
+        "\n",
+        "            # Enqueue the left and right children of the current node if they exist.\n",
+        "            if node.left:\n",
+        "                queue.append(node.left)\n",
+        "            if node.right:\n",
+        "                queue.append(node.right)\n",
+        "\n",
+        "        # Append the level_values list (values at the current level) to the result list.\n",
+        "        result.append(level_values)\n",
+        "\n",
+        "    # Return the final result, which is a list of lists representing level order traversal.\n",
+        "    return result"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We define a `TreeNode` class that represents a node in the binary tree. Each `TreeNode` object has a value (`val`) and two child nodes: `left` and `right`.\n",
+        "\n",
+        "2. The `levelOrder` function takes the root of the binary tree as its input and returns the level order traversal of the tree as a list of lists.\n",
+        "\n",
+        "3. We start by checking if the input `root` is `None`, which indicates an empty tree. If the tree is empty, we return an empty list because there are no nodes to traverse.\n",
+        "\n",
+        "4. We initialize an empty list called `result` to store the final result, which will be a list of lists containing node values at each level.\n",
+        "\n",
+        "5. We initialize a queue called `queue` with the root node. This queue will be used for breadth-first traversal of the tree.\n",
+        "\n",
+        "6. We enter a while loop that continues until the `queue` is empty. Inside the loop, we perform the following steps:\n",
+        "\n",
+        "   - We initialize an empty list called `level_values` to store the values of nodes at the current level.\n",
+        "\n",
+        "   - We determine the number of nodes at the current level by getting the length of the `queue`. This is done to process nodes level by level.\n",
+        "\n",
+        "   - We iterate through the nodes at the current level using a for loop. For each node in the current level:\n",
+        "\n",
+        "     - We dequeue (remove) the front node from the `queue`.\n",
+        "\n",
+        "     - We append the value of the dequeued node to the `level_values` list, effectively collecting the values of nodes at the current level.\n",
+        "\n",
+        "     - If the dequeued node has a left child, we enqueue the left child to the `queue`.\n",
+        "\n",
+        "     - If the dequeued node has a right child, we enqueue the right child to the `queue`.\n",
+        "\n",
+        "   - After processing all nodes at the current level, we append the `level_values` list to the `result` list. This represents the values at the current level.\n",
+        "\n",
+        "7. The loop continues until all levels have been traversed, and the `queue` becomes empty.\n",
+        "\n",
+        "8. Finally, we return the `result` list, which contains lists of node values at each level, representing the level order traversal of the binary tree.\n",
+        "\n",
+        "The code effectively performs a breadth-first traversal of the binary tree, processing nodes level by level, and constructs the result list that represents the level order traversal."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 6,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[[3], [9, 20], [15, 7]]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Construct the tree\n",
+        "root = TreeNode(3)\n",
+        "root.left = TreeNode(9)\n",
+        "root.right = TreeNode(20)\n",
+        "root.right.left = TreeNode(15)\n",
+        "root.right.right = TreeNode(7)\n",
+        "\n",
+        "result = levelOrder(root)\n",
+        "print(result)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 7,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[[1]]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "\n",
+        "# Using the same tree as before\n",
+        "root = TreeNode(1)\n",
+        "result = levelOrder(root)\n",
+        "print(result)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 8,
+      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3:\n",
+        "# Creating a new tree for this example\n",
+        "root = None\n",
+        "result = levelOrder(root)\n",
+        "print(result)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `levelOrder` function:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "The time complexity of this function is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the breadth-first traversal.\n",
+        "\n",
+        "In the worst case, we have to enqueue and dequeue all nodes in the binary tree, which is proportional to the number of nodes (N). In each level, we process all nodes in that level, and since there are a total of log(N) levels in a balanced binary tree, the time complexity can also be approximated as O(N) for unbalanced trees.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "The space complexity of this function is O(N), where N is the number of nodes in the binary tree. Here's how the space complexity breaks down:\n",
+        "\n",
+        "1. The `result` list stores the level order traversal, and in the worst case, it contains N/2 levels (for a completely unbalanced binary tree). So, the space used by `result` is O(N).\n",
+        "\n",
+        "2. The `queue` data structure is used for BFS traversal. In the worst case, it can store all nodes at the last level of the tree. In a balanced binary tree, the maximum number of nodes at any level is 2^(log(N)), which is still O(N). In the case of an unbalanced tree, it can be even worse. So, the space used by `queue` is O(N).\n",
+        "\n",
+        "Overall, the dominant factor in terms of space complexity is the `queue`, and the space complexity is O(N).\n",
+        "\n",
+        "In summary, the function's time complexity is O(N), and its space complexity is also O(N). It is an efficient and optimal solution for performing a level order traversal of a binary tree."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Reverse Level Order Traversal:** Modify the `levelOrder` function to return the level order traversal in reverse order (from bottom to top). For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[15, 7], [9, 20], [3]]`.\n",
+        "\n",
+        "2. **Zigzag Level Order Traversal:** Write a function that performs a level order traversal of a binary tree in a zigzag pattern. In a zigzag traversal, the nodes at even levels are traversed from left to right, and nodes at odd levels are traversed from right to left. For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[3], [20, 9], [15, 7]]`."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb b/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb
new file mode 100644
index 0000000..213989d
--- /dev/null
+++ b/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb	
@@ -0,0 +1,209 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 104. Maximum Depth of Binary Tree\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Given the `root` of a binary tree, return its maximum depth.\n",
+        "\n",
+        "A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
+        "- `-100 <= Node.val <= 100`\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 19,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Definition for a binary tree node.\n",
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "def maxDepth(root):\n",
+        "    # Base case: If the root is None, the depth is 0.\n",
+        "    if root is None:\n",
+        "        return 0\n",
+        "    \n",
+        "    # Recursively calculate the maximum depth of the left and right subtrees.\n",
+        "    left_depth = maxDepth(root.left)\n",
+        "    right_depth = maxDepth(root.right)\n",
+        "    \n",
+        "    # Return the maximum depth of the tree by adding 1 to the maximum depth of the subtrees.\n",
+        "    return max(left_depth, right_depth) + 1"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "Let's go through the code step by step to understand how it works:\n",
+        "\n",
+        "1. We start by defining a `TreeNode` class to represent the nodes of the binary tree. Each node has three attributes:\n",
+        "   - `val`: the value stored in the node.\n",
+        "   - `left`: a reference to the left child node.\n",
+        "   - `right`: a reference to the right child node.\n",
+        "\n",
+        "2. The `maxDepth` function is the main function that calculates the maximum depth of the binary tree. It takes a single argument, `root`, which is the root node of the binary tree.\n",
+        "\n",
+        "3. In the `maxDepth` function, we have a base case:\n",
+        "   - If the `root` is `None`, it means we have reached the end of a branch of the tree (a leaf node or an empty subtree). In this case, the depth is 0 because there are no nodes to count.\n",
+        "\n",
+        "4. If the `root` is not `None`, we recursively calculate the maximum depth of the left and right subtrees:\n",
+        "   - We call `maxDepth` on the `root.left` to calculate the maximum depth of the left subtree and store it in the variable `left_depth`.\n",
+        "   - We call `maxDepth` on the `root.right` to calculate the maximum depth of the right subtree and store it in the variable `right_depth`.\n",
+        "\n",
+        "5. Finally, we return the maximum depth of the tree by taking the maximum of `left_depth` and `right_depth` and adding 1 to it. This is because we are counting the current level as well."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 20,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "3\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 1\n",
+        "# Input: root = [3,9,20,null,null,15,7]\n",
+        "\n",
+        "root = TreeNode(3)\n",
+        "root.left = TreeNode(9)\n",
+        "root.right = TreeNode(20)\n",
+        "root.right.left = TreeNode(15)\n",
+        "root.right.right = TreeNode(7)\n",
+        "\n",
+        "print(maxDepth(root))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a230be4f-1f47-4919-87e3-62fc3f072d97",
+      "metadata": {},
+      "source": [
+        "In this example, we create a binary tree based on the given input `[3,9,20,null,null,15,7]` and calculate its maximum depth, which is `3`. The tree structure is as follows:\n",
+        "   ```\n",
+        "       3\n",
+        "      / \\\n",
+        "     9  20\n",
+        "       /  \\\n",
+        "      15   7\n",
+        "   ```\n",
+        "\n",
+        "So, the maximum depth is the length of the longest path from the root to a leaf node, which is 3 in this case."
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 21,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "2\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 2\n",
+        "# Input: root = [1,null,2]\n",
+        "\n",
+        "root = TreeNode(1)\n",
+        "root.right = TreeNode(2)\n",
+        "\n",
+        "# Calculate the maximum depth of the tree and print the result\n",
+        "print(maxDepth(root))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "- The time complexity of the `maxDepth` function is $O(N)$, where $N$ is the number of nodes in the binary tree.\n",
+        "- This is because in the worst case, the function visits every node exactly once in a depth-first manner.\n",
+        "- The recursion explores all nodes of the tree, so the time complexity is linear with respect to the number of nodes.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "- The space complexity of the `maxDepth` function is $O(H)$, where H is the height of the binary tree.\n",
+        "- In the worst case, if the binary tree is completely unbalanced (skewed), the recursion stack can go as deep as the height of the tree.\n",
+        "- In the best case, if the binary tree is perfectly balanced, the height is $O(log\\ N)$, where $N$ is the number of nodes.\n",
+        "- Therefore, the space complexity can vary from $O(log\\ N)$ to $O(N)$ depending on the shape of the tree.\n",
+        "- In addition to the recursion stack, there is a small constant amount of space used for variables and function call overhead.\n",
+        "\n",
+        "**In summary:**\n",
+        "- The time complexity of the code is $O(N)$ as it visits each node once.\n",
+        "- The space complexity is $O(H)$, where H is the height of the tree, which can vary from $O(log\\ N)$ to $O(N)$ depending on the tree's shape."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb b/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb
new file mode 100644
index 0000000..31c9e69
--- /dev/null
+++ b/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb	
@@ -0,0 +1,313 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 105. Construct Binary Tree from Preorder and Inorder Traversal\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Construct Binary Tree from Preorder and Inorder Traversal problem on LeetCode, click here!](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)\n",
+        "\n",
+        "---\n",
+        "Given two integer arrays `preorder` and `inorder` where `preorder` is the preorder traversal of a binary tree and `inorder` is the inorder traversal of the same tree, construct and return *the binary tree.*\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `1 <= preorder.length <= 3000`\n",
+        "- `inorder.length == preorder.length`\n",
+        "- `-3000 <= preorder[i], inorder[i] <= 3000`\n",
+        "- `preorder` and `inorder` consist of **unique** values.\n",
+        "- Each value of `inorder` also appears in `preorder`.\n",
+        "- `preorder` is **guaranteed** to be the preorder traversal of the tree.\n",
+        "- `inorder` is **guaranteed** to be the inorder traversal of the tree.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 1,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "def buildTree(preorder, inorder):\n",
+        "    if not preorder:\n",
+        "        return None\n",
+        "\n",
+        "    # The first element in the preorder traversal is the root of the current subtree\n",
+        "    root_val = preorder[0]\n",
+        "    root = TreeNode(root_val)\n",
+        "\n",
+        "    # Find the index of the root value in the inorder traversal\n",
+        "    root_idx_inorder = inorder.index(root_val)\n",
+        "\n",
+        "    # Recursively build left and right subtrees\n",
+        "    root.left = buildTree(preorder[1:1 + root_idx_inorder], inorder[:root_idx_inorder])\n",
+        "    root.right = buildTree(preorder[1 + root_idx_inorder:], inorder[root_idx_inorder + 1:])\n",
+        "\n",
+        "    return root"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. `class TreeNode:`: This is a class definition for a binary tree node. It has three attributes:\n",
+        "   - `val`: The value of the node.\n",
+        "   - `left`: A reference to the left child node.\n",
+        "   - `right`: A reference to the right child node.\n",
+        "\n",
+        "   This class is used to create instances of binary tree nodes, which will be used to build the binary tree.\n",
+        "\n",
+        "2. `def buildTree(preorder, inorder):`: This is the main function that constructs a binary tree from its preorder and inorder traversals.\n",
+        "\n",
+        "   - `preorder`: A list representing the preorder traversal of the binary tree.\n",
+        "   - `inorder`: A list representing the inorder traversal of the binary tree.\n",
+        "\n",
+        "3. `if not preorder: return None`: This line checks if the `preorder` list is empty. If it is, it means there are no nodes to construct, so the function returns `None`.\n",
+        "\n",
+        "4. `root_val = preorder[0]`: The value of the root node is extracted from the first element of the `preorder` list.\n",
+        "\n",
+        "5. `root = TreeNode(root_val)`: A new `TreeNode` object is created with the `root_val` as its value. This represents the root of the current subtree.\n",
+        "\n",
+        "6. `root_idx_inorder = inorder.index(root_val)`: The index of the `root_val` in the `inorder` list is found. This index indicates the position of the root node in the inorder traversal.\n",
+        "\n",
+        "7. `root.left`: The left subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The left subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists up to the `root_idx_inorder`.\n",
+        "\n",
+        "8. `root.right`: The right subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The right subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists starting from `root_idx_inorder + 1`.\n",
+        "\n",
+        "9. Finally, the function returns the `root` of the subtree it just constructed.\n",
+        "\n",
+        "The recursive approach used here divides the problem of constructing the entire binary tree into smaller subproblems, starting with the root node and then recursively building the left and right subtrees until the entire tree is constructed."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "2d82857b-4fbe-474f-bf33-17b55c20147b",
+      "metadata": {},
+      "source": [
+        "## Helper Function \n",
+        "\n",
+        "Here is a helper function to represent the binary tree elements."
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 7,
+      "id": "13e6f3e1-659d-479d-9521-75629b8c782f",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def treeToList(root):\n",
+        "    # Initialize an empty list to store the elements of the binary tree\n",
+        "    result = []\n",
+        "    \n",
+        "    # Initialize a queue for level-order traversal starting with the root\n",
+        "    queue = [root]\n",
+        "\n",
+        "    while queue:\n",
+        "        # Pop the front node from the queue\n",
+        "        node = queue.pop(0)\n",
+        "        \n",
+        "        # If the node is not None (i.e., a valid node), add its value to the result list\n",
+        "        if node:\n",
+        "            result.append(node.val)\n",
+        "            \n",
+        "            # Add the left and right children of the node to the queue\n",
+        "            queue.append(node.left)\n",
+        "            queue.append(node.right)\n",
+        "        else:\n",
+        "            # If the node is None, add None to the result to represent an empty node\n",
+        "            result.append(None)\n",
+        "\n",
+        "    # Remove any trailing None values from the result list\n",
+        "    while result and result[-1] is None:\n",
+        "        result.pop()\n",
+        "\n",
+        "    # Return the resulting list representing the binary tree elements\n",
+        "    return result"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "86cd4c9d-a120-4dd1-a1bd-530436012574",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "Let's break down the `treeToList` helper function step by step:\n",
+        "\n",
+        "1. `def treeToList(root):`: This function takes the root of a binary tree as input and converts it into a list.\n",
+        "\n",
+        "2. `result = []`: Initialize an empty list called `result` to store the elements of the binary tree in list format.\n",
+        "\n",
+        "3. `queue = [root]`: Initialize a queue data structure for a level-order traversal, starting with the root node.\n",
+        "\n",
+        "4. `while queue:`: This starts a loop that continues until the queue is empty, indicating that all nodes have been processed.\n",
+        "\n",
+        "5. `node = queue.pop(0)`: Dequeue the first node from the queue, effectively processing it.\n",
+        "\n",
+        "6. `if node:`: Check if the `node` is not None, which means it's a valid node in the binary tree.\n",
+        "\n",
+        "    a. `result.append(node.val)`: If the node is valid, append its value (`node.val`) to the `result` list. This represents the value of the current node in the binary tree.\n",
+        "\n",
+        "    b. `queue.append(node.left)`: Enqueue the left child of the current node if it exists. This adds the left child to the queue for processing in the next iteration.\n",
+        "\n",
+        "    c. `queue.append(node.right)`: Enqueue the right child of the current node if it exists. This adds the right child to the queue for processing in the next iteration.\n",
+        "\n",
+        "7. `else:`: If the `node` is None, it represents an empty node in the binary tree.\n",
+        "\n",
+        "    a. `result.append(None)`: Append `None` to the `result` list to indicate an empty node.\n",
+        "\n",
+        "8. `while result and result[-1] is None:`: After the traversal is complete, there might be trailing `None` values in the `result` list. This loop removes any such trailing `None` values to ensure a clean representation of the tree's elements.\n",
+        "\n",
+        "9. `return result`: Return the `result` list, which now contains the elements of the binary tree in a format where `None` represents empty nodes and the order of elements reflects a level-order traversal of the tree.\n",
+        "\n",
+        "In summary, the `treeToList` function performs a level-order traversal of the binary tree using a queue, constructing a list where each element corresponds to a node's value or represents an empty node with `None`. This list represents the binary tree's elements in a structured format."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 16,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[3, 9, 20, None, None, 15, 7]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Example usage:\n",
+        "preorder = [3, 9, 20, 15, 7]\n",
+        "inorder = [9, 3, 15, 20, 7]\n",
+        "result = buildTree(preorder, inorder)\n",
+        "\n",
+        "print(treeToList(result))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 17,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[-1]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "\n",
+        "preorder = [-1]\n",
+        "inorder = [-1]\n",
+        "result = buildTree(preorder, inorder)\n",
+        "\n",
+        "print(treeToList(result))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `buildTree` function:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "The time complexity of the `buildTree` function can be analyzed in terms of the number of nodes in the binary tree. Let's assume there are `n` nodes in the binary tree.\n",
+        "\n",
+        "1. Finding the root value in the `preorder` traversal takes O(1) time as it's just extracting the first element from the list.\n",
+        "2. Finding the index of the root value in the `inorder` traversal using `inorder.index(root_val)` takes O(n) time in the worst case because in the worst case, it might have to search through all `n` elements in the `inorder` list to find the index.\n",
+        "3. The recursive calls to `buildTree` for the left and right subtrees are made once for each node in the tree. Therefore, the recursive calls have a combined time complexity of O(n) as they process each node once.\n",
+        "\n",
+        "The total time complexity of the `buildTree` function is O(n) due to the recursive calls and finding the root's index in the `inorder` list.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "The space complexity is determined by the space used by the function's call stack during recursion and any additional data structures used. \n",
+        "\n",
+        "1. The space used by the function call stack during recursion depends on the height of the binary tree. In the worst case, where the tree is highly unbalanced (e.g., a skewed tree), the space complexity for the call stack is O(n) as it can go as deep as the number of nodes in the tree.\n",
+        "\n",
+        "2. Additionally, the function creates TreeNode objects for each node in the binary tree. Therefore, the space complexity for these objects is also O(n).\n",
+        "\n",
+        "Overall, the space complexity of the `buildTree` function is O(n) due to the space used by the call stack and the TreeNode objects.\n",
+        "\n",
+        "In summary, the time complexity of `buildTree` is O(n), and the space complexity is O(n), where 'n' is the number of nodes in the binary tree."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Construct Binary Tree from Postorder and Inorder Traversal:**\n",
+        "   Modify the problem to construct a binary tree from its postorder and inorder traversals. Implement a function similar to `buildTree` but for postorder and inorder traversals.\n",
+        "\n",
+        "2. **Reconstruct Binary Tree with Duplicate Values:**\n",
+        "   Extend the problem to handle binary trees with duplicate values. Ensure that your solution correctly handles scenarios where there are duplicate values in the preorder and inorder traversals."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb b/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb
new file mode 100644
index 0000000..59d8e7d
--- /dev/null
+++ b/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb	
@@ -0,0 +1,249 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 124. Binary Tree Maximum Path Sum\n",
+        "\n",
+        "**Difficulty:** Hard\n",
+        "\n",
+        "**Link to Problem:** [To see the Binary Tree Maximum Path Sum problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-maximum-path-sum/)\n",
+        "\n",
+        "---\n",
+        "A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.\n",
+        "\n",
+        "The **path sum** of a path is the sum of the node's values in the path.\n",
+        "\n",
+        "Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- The number of nodes in the tree is in the range $[1, 3*10^4]$.\n",
+        "- `-1000 <= Node.val <= 1000`"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem is to find the maximum path sum in a binary tree. A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. The path sum of a path is the sum of the node values in that path.\n",
+        "\n",
+        "In this problem, you are given the root of a binary tree, and you need to find the maximum path sum among all possible paths in the tree. Note that the path does not need to start at the root node or end at a leaf node; it can start and end at any nodes in the tree.\n",
+        "\n",
+        "To find the maximum path sum, you need to consider both left and right subtrees of each node while traversing the tree. At each node, you have two choices:\n",
+        "\n",
+        "1. Include the current node in the path: In this case, you add the value of the current node to the sum and continue exploring both the left and right subtrees for possible extensions of the path.\n",
+        "\n",
+        "2. Start a new path from the current node: In this case, you do not include the current node in the path sum, and you choose either the left subtree or the right subtree to start a new path.\n",
+        "\n",
+        "To solve this problem, you can use a recursive approach to traverse the binary tree. For each node, you calculate the maximum path sum that can pass through that node (option 1) and also return the maximum path sum that can be extended from that node (option 2). You keep track of the global maximum path sum encountered during the traversal.\n",
+        "\n",
+        "Ultimately, the maximum path sum among all paths in the tree will be the maximum of the global maximum path sum and the maximum path sum calculated for each node.\n",
+        "\n",
+        "The constraints for this problem include:\n",
+        "\n",
+        "- The number of nodes in the tree is in the range $[1, 3 * 10^4]$.\n",
+        "- Node values are in the range [-1000, 1000].\n",
+        "\n",
+        "By considering all possible paths through the tree, the algorithm aims to find the most significant sum of node values in any path in the binary tree.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 18,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "class Solution:\n",
+        "    def maxPathSum(self, root):\n",
+        "        def max_path_sum_helper(node):\n",
+        "            if not node:\n",
+        "                return 0\n",
+        "            \n",
+        "            # Recursively calculate the maximum path sum for left and right subtrees\n",
+        "            left_max = max(0, max_path_sum_helper(node.left))\n",
+        "            right_max = max(0, max_path_sum_helper(node.right))\n",
+        "            \n",
+        "            # Update the global maximum path sum\n",
+        "            self.max_sum = max(self.max_sum, left_max + right_max + node.val)\n",
+        "            \n",
+        "            # Return the maximum path sum starting from the current node\n",
+        "            return max(left_max, right_max) + node.val\n",
+        "        \n",
+        "        self.max_sum = float('-inf')  # Initialize the global maximum to negative infinity\n",
+        "        max_path_sum_helper(root)  # Start the recursive traversal from the root\n",
+        "        return self.max_sum"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. The code starts by defining a `TreeNode` class to represent nodes in the binary tree. Each `TreeNode` has a `val` attribute representing its value, a `left` attribute pointing to the left child, and a `right` attribute pointing to the right child.\n",
+        "\n",
+        "2. Next, a `Solution` class is defined to contain the `maxPathSum` method, which will be used to find the maximum path sum in the binary tree.\n",
+        "\n",
+        "3. Within the `maxPathSum` method, a helper function called `max_path_sum_helper` is defined. This recursive function takes a single argument, `node`, which is the current node being considered during traversal.\n",
+        "\n",
+        "4. Inside the `max_path_sum_helper` function:\n",
+        "   - If `node` is `None` (i.e., the current node is a null node), it returns 0. This is the base case of the recursion.\n",
+        "   - It calculates the maximum path sum starting from the left subtree (`left_max`) and the right subtree (`right_max`). Importantly, it uses `max(0, ...)` to ensure that negative values (which would make the path sum smaller) are not considered. If a subtree's path sum is negative, it's better to not include it in the path.\n",
+        "   - The code then updates the global maximum path sum (`self.max_sum`) by checking if the sum of `left_max`, `right_max`, and the current node's value is greater than the current maximum.\n",
+        "   - Finally, it returns the maximum path sum starting from the current node, which is the maximum of `left_max` and `right_max` plus the current node's value.\n",
+        "\n",
+        "5. Before calling the `max_path_sum_helper` function, the `max_sum` attribute of the `Solution` class is initialized to negative infinity (`float('-inf')`). This is done to ensure that any valid path sum encountered during traversal will be greater than the initial value of `max_sum`.\n",
+        "\n",
+        "6. The `max_path_sum_helper` function is called with the root of the binary tree, effectively starting the traversal from the root node.\n",
+        "\n",
+        "7. Once the traversal is complete, the method returns the value of `self.max_sum`, which contains the maximum path sum found in the binary tree.\n",
+        "\n",
+        "8. Example usage at the bottom of the code demonstrates how to create binary trees and use the `maxPathSum` method to find the maximum path sum for two different tree structures.\n",
+        "\n",
+        "In summary, the code uses a recursive approach to explore all possible paths in the binary tree, ensuring that negative path sums are not considered. It keeps track of the maximum path sum encountered and returns that maximum value as the result."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 20,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "6\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Example usage:\n",
+        "root1 = TreeNode(1)\n",
+        "root1.left = TreeNode(2)\n",
+        "root1.right = TreeNode(3)\n",
+        "solution = Solution()\n",
+        "print(solution.maxPathSum(root1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 21,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "42\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "\n",
+        "root2 = TreeNode(-10)\n",
+        "root2.left = TreeNode(9)\n",
+        "root2.right = TreeNode(20)\n",
+        "root2.right.left = TreeNode(15)\n",
+        "root2.right.right = TreeNode(7)\n",
+        "print(solution.maxPathSum(root2))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the given code:\n",
+        "\n",
+        "Time Complexity:\n",
+        "- The code uses a recursive depth-first traversal to explore each node in the binary tree exactly once.\n",
+        "- During the traversal, for each node, we perform constant time operations, such as updating the `max_sum` variable and calculating the maximum path sum starting from that node.\n",
+        "- Therefore, the time complexity of the code is O(N), where N is the number of nodes in the binary tree.\n",
+        "\n",
+        "Space Complexity:\n",
+        "- The space complexity of the code is determined by the space used in the call stack during the recursive traversal.\n",
+        "- In the worst case, the depth of the call stack can be equal to the height of the binary tree. In an unbalanced tree, this could be O(N), but in a balanced binary tree, the height is O(log N).\n",
+        "- Additionally, the code uses a constant amount of space for variables like `left_max`, `right_max`, and `max_sum`.\n",
+        "- Therefore, the overall space complexity is O(H), where H is the height of the binary tree. In the worst case, it's O(N), and in the best case (a balanced binary tree), it's O(log N).\n",
+        "\n",
+        "In summary:\n",
+        "- Time Complexity: O(N)\n",
+        "- Space Complexity: O(H), where H is the height of the binary tree, ranging from O(log N) to O(N) depending on the tree's balance."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Print the Maximum Path:**\n",
+        "   Modify the code to not only find the maximum path sum but also print the nodes involved in the maximum path. You can print the nodes in the correct order from the root to the leaf.\n",
+        "\n",
+        "2. **Path Sum Count:**\n",
+        "   Instead of finding the maximum path sum, find the count of unique paths that sum to a given target value. Each path can start and end anywhere in the tree."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb b/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb
new file mode 100644
index 0000000..55dc944
--- /dev/null
+++ b/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb	
@@ -0,0 +1,257 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 199. Binary Tree Right Side View\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Binary Tree Right Side View problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-right-side-view/)\n",
+        "\n",
+        "---\n",
+        "Given the `root` of a binary tree, imagine yourself standing on the **right side** of it, return *the values of the nodes you can see ordered from top to bottom*.\n",
+        "\n",
+        " \n",
+        "\n",
+        "**Constraints:**\n",
+        "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
+        "- `-100 <= Node.val <= 100`\n",
+        "\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 9,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "def rightSideView(root):\n",
+        "    if not root:\n",
+        "        return []  # If the tree is empty, return an empty list.\n",
+        "\n",
+        "    result = []  # Initialize a list to store the right side view values.\n",
+        "    queue = [root]  # Initialize a queue for level-order traversal with the root node.\n",
+        "\n",
+        "    while queue:\n",
+        "        # Get the number of nodes at the current level.\n",
+        "        level_size = len(queue)\n",
+        "\n",
+        "        # Traverse all nodes at the current level and add the rightmost node to the result.\n",
+        "        for i in range(level_size):\n",
+        "            node = queue.pop(0)  # Dequeue the first node from the queue.\n",
+        "\n",
+        "            # If it's the rightmost node at this level, add its value to the result.\n",
+        "            if i == level_size - 1:\n",
+        "                result.append(node.val)\n",
+        "\n",
+        "            # Add the children of the current node to the queue.\n",
+        "            if node.left:\n",
+        "                queue.append(node.left)\n",
+        "            if node.right:\n",
+        "                queue.append(node.right)\n",
+        "\n",
+        "    return result  # Return the list of right side view values.\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We start with a binary tree represented using a TreeNode class. Each node has a value (`val`) and can have a left child (`left`) and a right child (`right`).\n",
+        "\n",
+        "2. The `rightSideView` function takes the root node of the binary tree as input and returns a list of values representing the nodes you can see when standing on the right side of the tree.\n",
+        "\n",
+        "3. We check if the `root` is `None`, which means the tree is empty. If it's empty, we return an empty list because there are no nodes to see.\n",
+        "\n",
+        "4. We initialize an empty list called `result` to store the right side view values.\n",
+        "\n",
+        "5. We initialize a queue called `queue` with the root node. This queue is used for a level-order traversal of the tree.\n",
+        "\n",
+        "6. We enter a while loop that continues until the `queue` is empty. Inside this loop, we perform the following steps for each level of the tree:\n",
+        "\n",
+        "   a. We determine the number of nodes at the current level by getting the length of the `queue`. This is important for processing nodes at the same level together.\n",
+        "\n",
+        "   b. We then iterate through the nodes at the current level using a for loop. For each node in the level, we dequeue it from the `queue` using `queue.pop(0)`.\n",
+        "\n",
+        "   c. If the node we dequeue is the rightmost node at the current level (determined by `i == level_size - 1`), we add its `val` to the `result` list. This is because, when standing on the right side of the tree, you can see the rightmost node at each level.\n",
+        "\n",
+        "   d. We enqueue the left and right children of the current node (if they exist) to the `queue`. This ensures that we process the next level in the subsequent iterations of the while loop.\n",
+        "\n",
+        "7. After processing all levels of the tree, the `result` list contains the values of the rightmost nodes in each level, ordered from top to bottom.\n",
+        "\n",
+        "8. Finally, we return the `result` list as the output of the function, which represents the right side view of the binary tree.\n",
+        "\n",
+        "The code effectively performs a level-order traversal of the binary tree while keeping track of the rightmost nodes at each level and adding them to the result list. This ensures that we obtain the correct order of nodes visible from the right side of the tree."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 10,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[1, 3, 4]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Construct the tree\n",
+        "root1 = TreeNode(1)\n",
+        "root1.left = TreeNode(2)\n",
+        "root1.right = TreeNode(3)\n",
+        "root1.left.right = TreeNode(5)\n",
+        "root1.right.right = TreeNode(4)\n",
+        "print(rightSideView(root1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 11,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[1, 3]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "\n",
+        "root2 = TreeNode(1)\n",
+        "root2.right = TreeNode(3)\n",
+        "print(rightSideView(root2))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 12,
+      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3:\n",
+        "# Creating a new tree for this example\n",
+        "root = None\n",
+        "result = rightSideView(root)\n",
+        "print(result)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the given code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "The time complexity of the code is O(N), where N is the number of nodes in the binary tree. Here's why:\n",
+        "\n",
+        "1. We perform a level-order traversal of the tree using a queue. In the worst case, we visit all nodes once, which takes O(N) time since we visit each node exactly once.\n",
+        "\n",
+        "2. For each node, we perform constant-time operations like dequeuing it from the queue, checking if it's the rightmost node at the current level, and enqueuing its children (if they exist). These operations do not depend on the size of the tree, so they do not contribute to the overall time complexity.\n",
+        "\n",
+        "Therefore, the dominant factor in the time complexity is the level-order traversal, making it O(N).\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "The space complexity of the code is also O(N). Here's why:\n",
+        "\n",
+        "1. We use a queue to perform level-order traversal. In the worst case, the queue can contain all nodes at the maximum width of the tree, which can be up to N/2 nodes for a completely unbalanced tree. Therefore, the space required for the queue is O(N).\n",
+        "\n",
+        "2. The `result` list stores the values of the rightmost nodes. In the worst case, when the binary tree is a complete binary tree, it can have roughly N/2 rightmost nodes. Therefore, the space required for the `result` list is also O(N).\n",
+        "\n",
+        "3. Other auxiliary variables like `level_size`, `i`, and `node` require only constant space and do not depend on the size of the tree.\n",
+        "\n",
+        "Combining the space used by the queue and the `result` list, we get a space complexity of O(N).\n",
+        "\n",
+        "In summary, the code has a time complexity of O(N) and a space complexity of O(N), where N is the number of nodes in the binary tree. These complexities are efficient and scale linearly with the size of the input tree."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Variant with Left Side View**: Modify the code to find and return the values of nodes visible from the left side of the binary tree instead of the right side.\n",
+        "\n",
+        "2. **Zigzag Right Side View**: Extend the code to return the right side view values in a zigzag order. In a zigzag order, you alternate between starting from the rightmost node at level 0, then the leftmost at level 1, rightmost at level 2, and so on."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb b/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb
new file mode 100644
index 0000000..8ea4609
--- /dev/null
+++ b/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb	
@@ -0,0 +1,223 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 226: Invert Binary Tree\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Given the root of a binary tree, invert the tree, and return its root.\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- The number of nodes in the tree is in the range `[0, 100]`.\n",
+        "- `-100 <= Node.val <= 100`\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 8,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Definition for a binary tree node.\n",
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "def invertTree(root):\n",
+        "    # Base case: If the root is None or the tree is empty, return None.\n",
+        "    if not root:\n",
+        "        return None\n",
+        "\n",
+        "    # Swap the left and right subtrees of the current node.\n",
+        "    root.left, root.right = root.right, root.left\n",
+        "\n",
+        "    # Recursively invert the left and right subtrees.\n",
+        "    invertTree(root.left)\n",
+        "    invertTree(root.right)\n",
+        "\n",
+        "    return root"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We start by defining a TreeNode class, which represents a node in a binary tree. Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
+        "\n",
+        "2. The `invertTree` function is defined to invert the binary tree. It takes the root node of the tree as an argument.\n",
+        "\n",
+        "3. In the `invertTree` function, we have a base case to handle the scenario when the root is `None` (empty tree). In such cases, we return `None` because there's nothing to invert.\n",
+        "\n",
+        "4. For non-empty trees, we swap the left and right subtrees of the current node. This effectively inverts the tree at the current node.\n",
+        "\n",
+        "5. We then recursively call `invertTree` on the left and right subtrees to invert them.\n",
+        "\n",
+        "6. Finally, we return the root of the inverted tree."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 12,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Inverted Tree (Example 1): <__main__.TreeNode object at 0x270c9c0>\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 1\n",
+        "# Input: root = [4,2,7,1,3,6,9]\n",
+        "\n",
+        "root1 = TreeNode(4)\n",
+        "root1.left = TreeNode(2)\n",
+        "root1.right = TreeNode(7)\n",
+        "root1.left.left = TreeNode(1)\n",
+        "root1.left.right = TreeNode(3)\n",
+        "root1.right.left = TreeNode(6)\n",
+        "root1.right.right = TreeNode(9)\n",
+        "\n",
+        "inverted_root1 = invertTree(root1)\n",
+        "print(\"Inverted Tree (Example 1):\", inverted_root1)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 15,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Inorder Traversal of Inverted Tree (Example 2): <__main__.TreeNode object at 0x2255fc0>\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 2\n",
+        "# Input: root = [2,1,3]\n",
+        "\n",
+        "root2 = TreeNode(2)\n",
+        "root2.left = TreeNode(1)\n",
+        "root2.right = TreeNode(3)\n",
+        "\n",
+        "inverted_root2 = invertTree(root2)\n",
+        "print(\"Inverted Tree (Example 2):\", inverted_root2)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 16,
+      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Inverted Tree (Example 3): None\n"
+          ]
+        }
+      ],
+      "source": [
+        "### Example 3\n",
+        "# Input: root = []\n",
+        "\n",
+        "root3 = None  # Empty tree\n",
+        "\n",
+        "inverted_root3 = invertTree(root3)\n",
+        "print(\"Inverted Tree (Example 3):\", inverted_root3)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "Let's discuss the time and space complexity of the provided Python code to invert a binary tree.\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "The time complexity of the `invertTree` function is $O(n)$, where $n$ is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal of the tree. In the worst case, we have to visit all nodes in the tree.\n",
+        "\n",
+        "The reason for this time complexity is the depth-first traversal of the tree, where we recursively process the left and right subtrees of each node.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "The space complexity of the code is determined by the function call stack during the recursion. In the worst case, the space complexity is $O(h)$, where $h$ is the height of the binary tree. \n",
+        "\n",
+        "In a completely unbalanced binary tree (essentially a linked list), the height $h$ is equal to the number of nodes $n$, resulting in a space complexity of $O(n)$. This occurs when the tree is skewed to one side.\n",
+        "\n",
+        "In a balanced binary tree, such as a full binary tree, the height $h$ is $O(log\\ n)$, and the space complexity is $O(log\\ n)$.\n",
+        "\n",
+        "The space complexity depends on how balanced the tree is. In practical scenarios, binary trees are often approximately balanced, so the space complexity is typically closer to $O(log\\ n)$.\n",
+        "\n",
+        "**In summary:**\n",
+        "\n",
+        "- Time Complexity: $O(n)$ where $n$ is the number of nodes.\n",
+        "- Space Complexity: $O(h)$, where $h$ is the height of the binary tree. In the worst case, it can be $O(n)$, and in a balanced tree, it is $O(log\\ n)$.\n",
+        "\n",
+        "Keep in mind that these complexities are based on the recursive implementation provided. Iterative solutions can achieve the same task with $O(1)$ space complexity, using auxiliary data structures like stacks or queues to mimic the recursion stack."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb b/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb
new file mode 100644
index 0000000..673ced2
--- /dev/null
+++ b/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb	
@@ -0,0 +1,221 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 230. Kth Smallest Element in a BST\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Kth Smallest Element in a BST problem on LeetCode, click here!](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)\n",
+        "\n",
+        "---\n",
+        "Given the `root` of a binary search tree, and an integer `k`, return *the $k^{th}$ smallest value (**1-indexed**) of all the values of the nodes in the tree.*\n",
+        "\n",
+        "**Constraints:**\n",
+        "- The number of nodes in the tree is `n`.\n",
+        "- 1 <= `k <= n` <= $10^4$\n",
+        "- 0 <= `Node.val` <= $10^4$\n",
+        "\n",
+        "**Follow up**: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 4,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "def kthSmallest(root, k):\n",
+        "    stack = []  # Initialize an empty stack to simulate the traversal\n",
+        "    while root or stack:\n",
+        "        while root:\n",
+        "            stack.append(root)  # Push the current node onto the stack\n",
+        "            root = root.left  # Move to the left child\n",
+        "        root = stack.pop()  # Pop a node from the stack\n",
+        "        k -= 1  # Decrement k since we've visited a node\n",
+        "        if k == 0:\n",
+        "            return root.val  # If k becomes 0, return the current node's value as the kth smallest\n",
+        "        root = root.right  # Move to the right child to continue the traversal"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. We start by defining a `TreeNode` class, which represents a node in the binary search tree (BST). Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
+        "\n",
+        "2. The `kthSmallest` function takes two parameters: `root`, which is the root of the BST, and `k`, which represents the kth smallest element we want to find.\n",
+        "\n",
+        "3. We initialize an empty stack (`stack`) to simulate the traversal of the BST. The stack will be used to keep track of nodes as we traverse the tree in an iterative manner.\n",
+        "\n",
+        "4. We enter a while loop that continues until either `root` becomes `None` (indicating we have traversed the entire tree) or the `stack` is empty.\n",
+        "\n",
+        "5. Within the loop, we start another while loop to traverse as far left as possible in the BST. We repeatedly push nodes onto the `stack` and move to their left children until we reach the leftmost leaf node. This is the smallest node in the BST.\n",
+        "\n",
+        "6. Once we have reached the leftmost leaf node (the smallest node), we pop nodes from the `stack` one by one. As we pop each node, we decrement `k` by 1 to keep track of the number of nodes we have visited.\n",
+        "\n",
+        "7. If `k` becomes 0 after decrementing, it means we have found the kth smallest element. In this case, we return the `val` of the current node as the result.\n",
+        "\n",
+        "8. If `k` is still greater than 0, it means we haven't found the kth smallest element yet. In this case, we move to the right child of the current node to continue the traversal, as the kth smallest element, if it exists, will be in the right subtree of the current node.\n",
+        "\n",
+        "9. The process continues until we find the kth smallest element or traverse the entire tree.\n",
+        "\n",
+        "10. Finally, we return the kth smallest element found.\n",
+        "\n",
+        "The key idea in this code is to perform an in-order traversal of the BST while keeping track of the kth smallest element. By visiting nodes in ascending order, we can efficiently find the kth smallest element in O(h + k) time, where h is the height of the BST and k is the desired kth element. This approach is both concise and efficient for this problem."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 5,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "1\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Example usage:\n",
+        "# Create the tree for the first example: root = [3,1,4,null,2], k = 1\n",
+        "root1 = TreeNode(3)\n",
+        "root1.left = TreeNode(1)\n",
+        "root1.right = TreeNode(4)\n",
+        "root1.left.right = TreeNode(2)\n",
+        "k1 = 1\n",
+        "print(kthSmallest(root1, k1)) "
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 6,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "3\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "\n",
+        "# Create the tree for the second example: root = [5,3,6,2,4,null,null,1], k = 3\n",
+        "root2 = TreeNode(5)\n",
+        "root2.left = TreeNode(3)\n",
+        "root2.right = TreeNode(6)\n",
+        "root2.left.left = TreeNode(2)\n",
+        "root2.left.right = TreeNode(4)\n",
+        "root2.left.left.left = TreeNode(1)\n",
+        "k2 = 3\n",
+        "print(kthSmallest(root2, k2))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the given code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "The time complexity of this code is O(h + k), where:\n",
+        "- h is the height of the binary search tree (BST).\n",
+        "- k is the desired kth smallest element we want to find.\n",
+        "\n",
+        "1. In the worst case, where the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the time complexity is O(n + k).\n",
+        "\n",
+        "2. In the best case, where the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the time complexity is O(log(n) + k).\n",
+        "\n",
+        "So, the time complexity can vary from O(log(n) + k) in the best-case scenario to O(n + k) in the worst-case scenario. Typically, for balanced BSTs, the time complexity is closer to O(log(n) + k).\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "The space complexity of this code is O(h) due to the stack used for the iterative in-order traversal, where h is the height of the BST.\n",
+        "\n",
+        "1. In the worst case, when the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the space complexity is O(n) because the stack can potentially store all n nodes.\n",
+        "\n",
+        "2. In the best case, when the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the space complexity is O(log(n)) because the stack will have at most log(n) nodes.\n",
+        "\n",
+        "So, the space complexity depends on the height of the BST and can vary from O(log(n)) in the best-case scenario to O(n) in the worst-case scenario.\n",
+        "\n",
+        "In practical terms, for balanced BSTs or moderately unbalanced BSTs, the space complexity is usually close to O(log(n)), and the code is efficient in terms of space usage."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Kth Largest Element**: Modify the code to find the kth largest element in the BST instead of the kth smallest element.\n",
+        "\n",
+        "2. **Kth Smallest Element in Two BSTs**: Given two BSTs, find the kth smallest element when considering both BSTs as a single sorted list. This involves merging the two BSTs while finding the kth element efficiently."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb b/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb
new file mode 100644
index 0000000..666bcd9
--- /dev/null
+++ b/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb	
@@ -0,0 +1,245 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 235. Lowest Common Ancestor of a Binary Search Tree\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Lowest Common Ancestor of a Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)\n",
+        "\n",
+        "---\n",
+        "Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.\n",
+        "\n",
+        "According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”\n",
+        "\n",
+        "**Constraints:**\n",
+        "- The number of nodes in the tree is in the range $[2, 10^5]$.\n",
+        "- $-10^9$ <= `Node.val` <= $10^9$\n",
+        "- All `Node.val` are **unique**.\n",
+        "- `p != q`\n",
+        "- `p` and `q` will exist in the BST.\n",
+        "\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 1,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "def lowestCommonAncestor(root, p, q):\n",
+        "    # Ensure p is less than q\n",
+        "    if p.val > q.val:\n",
+        "        p, q = q, p\n",
+        "\n",
+        "    while root:\n",
+        "        # If the current node value is greater than both p and q, move left\n",
+        "        if root.val > q.val:\n",
+        "            root = root.left\n",
+        "        # If the current node value is less than both p and q, move right\n",
+        "        elif root.val < p.val:\n",
+        "            root = root.right\n",
+        "        # If the current node value is between p and q (inclusive), or it matches either p or q, it's the LCA\n",
+        "        else:\n",
+        "            return root"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. `class TreeNode:` defines a simple class for representing nodes in a binary tree. Each `TreeNode` has three attributes:\n",
+        "   - `val`: The value of the node.\n",
+        "   - `left`: A reference to the left child node.\n",
+        "   - `right`: A reference to the right child node.\n",
+        "\n",
+        "2. `def lowestCommonAncestor(root, p, q):` is a function that takes three arguments:\n",
+        "   - `root`: The root node of the BST.\n",
+        "   - `p`: A TreeNode representing one of the nodes for which we want to find the LCA.\n",
+        "   - `q`: A TreeNode representing the other node for which we want to find the LCA.\n",
+        "\n",
+        "3. `if p.val > q.val:` checks if the value of `p` is greater than the value of `q`. In a BST, it's essential to ensure that `p` represents the smaller value, and `q` represents the larger value. If `p` is greater than `q`, the code swaps their values.\n",
+        "\n",
+        "4. The main logic is inside the `while` loop, which runs until `root` becomes `None`. It performs the following steps to find the LCA:\n",
+        "   - If the current `root` node's value is greater than the value of `q`, it means both `p` and `q` are on the left subtree of the current node. So, we move to the left child of the current node by setting `root = root.left`.\n",
+        "   - If the current `root` node's value is less than the value of `p`, it means both `p` and `q` are on the right subtree of the current node. So, we move to the right child of the current node by setting `root = root.right`.\n",
+        "   - If neither of the above conditions is met, it means the current `root` node's value is between `p` and `q`, or it matches either `p` or `q`. In this case, the current node is the lowest common ancestor (LCA), so we return `root`.\n",
+        "\n",
+        "The algorithm is efficient because it takes advantage of the properties of a BST. It eliminates subtrees that cannot contain the LCA by comparing the values of `p`, `q`, and the current `root` node. Eventually, it reaches the LCA node, and that node is returned as the result."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 2,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "6\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Construct the tree\n",
+        "root = TreeNode(6)\n",
+        "root.left = TreeNode(2)\n",
+        "root.right = TreeNode(8)\n",
+        "root.left.left = TreeNode(0)\n",
+        "root.left.right = TreeNode(4)\n",
+        "root.right.left = TreeNode(7)\n",
+        "root.right.right = TreeNode(9)\n",
+        "root.left.right.left = TreeNode(3)\n",
+        "root.left.right.right = TreeNode(5)\n",
+        "\n",
+        "p = root.left  # Node 2\n",
+        "q = root.right  # Node 8\n",
+        "\n",
+        "result = lowestCommonAncestor(root, p, q)\n",
+        "print(result.val)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 3,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "2\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "\n",
+        "# Using the same tree as before\n",
+        "p = root.left  # Node 2\n",
+        "q = root.left.right  # Node 4\n",
+        "\n",
+        "result = lowestCommonAncestor(root, p, q)\n",
+        "print(result.val) "
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 4,
+      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "2\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3:\n",
+        "# Creating a new tree for this example\n",
+        "root2 = TreeNode(2)\n",
+        "root2.left = TreeNode(1)\n",
+        "\n",
+        "p = root2  # Node 2\n",
+        "q = root2.left  # Node 1\n",
+        "\n",
+        "result = lowestCommonAncestor(root2, p, q)\n",
+        "print(result.val)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided algorithm for finding the lowest common ancestor (LCA) in a Binary Search Tree (BST).\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "The time complexity of this algorithm is O(h), where \"h\" is the height of the BST. In the worst case, where the tree is completely unbalanced (essentially a linked list), the height of the tree is O(n), where \"n\" is the number of nodes in the tree. However, in a well-balanced BST, the height is logarithmic, which is O(log n).\n",
+        "\n",
+        "The reason for this time complexity is that the algorithm efficiently narrows down the search space by traversing either left or right subtrees based on the values of the target nodes `p` and `q`. It eliminates entire subtrees that cannot contain the LCA, leading to a relatively quick search.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "The space complexity of the algorithm is O(1). This is because it uses a constant amount of extra space, regardless of the size of the input BST. The only variables used are `root`, `p`, and `q`, and there are no data structures like stacks or queues used for additional space. The algorithm performs a simple traversal without recursion, so it does not consume extra memory as the tree depth increases.\n",
+        "\n",
+        "**In summary**, the provided algorithm for finding the LCA in a BST is both time and space-efficient. Its time complexity is O(h), where \"h\" is the height of the tree, and its space complexity is O(1), making it suitable for practical use even in large BSTs."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "### Challenging Exercises:\n",
+        "\n",
+        "1. **Multiple LCAs:** Extend the algorithm to find all the lowest common ancestors of two given nodes `p` and `q` in a BST. In some cases, there can be multiple LCAs.\n",
+        "\n",
+        "2. **LCA with k Nodes:** Given a BST and k nodes, find the lowest common ancestor of these k nodes. This is an extension of the problem where you must find the LCA of more than two nodes."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb b/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb
new file mode 100644
index 0000000..75c48ea
--- /dev/null
+++ b/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb	
@@ -0,0 +1,297 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 297. Serialize and Deserialize Binary Tree\n",
+        "\n",
+        "**Difficulty:** Hard\n",
+        "\n",
+        "**Link to Problem:** [To see the Serialize and Deserialize Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)\n",
+        "\n",
+        "---\n",
+        "Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.\n",
+        "\n",
+        "Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.\n",
+        "\n",
+        "**Clarification:** The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
+        "- `-1000 <= Node.val <= 1000`"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "TThe problem is to design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of characters or bytes so that it can be easily stored in a file, transmitted over a network, or otherwise persisted. Deserialization is the reverse process of converting the serialized data back into the original data structure.\n",
+        "\n",
+        "In the context of this problem:\n",
+        "\n",
+        "1. **Serialization**: You are given a binary tree, and your task is to convert it into a string representation such that you can later recreate the same binary tree from this string. The format of serialization is flexible, but it should allow you to reconstruct the original binary tree accurately.\n",
+        "\n",
+        "2. **Deserialization**: Given a serialized string, you need to reconstruct the binary tree it represents, ensuring that it is identical to the original tree.\n",
+        "\n",
+        "Here are some key points to consider in solving this problem:\n",
+        "\n",
+        "- The input can include any valid binary tree, including trees with nodes having integer values within the range [-1000, 1000].\n",
+        "- You don't necessarily need to follow a specific format for serialization, but it should be designed in a way that allows unambiguous deserialization.\n",
+        "- The goal is to serialize the tree structure and its values and then deserialize it back into the same structure and values.\n",
+        "\n",
+        "A common approach for serialization is to use a traversal method like preorder traversal, where you visit nodes in the order: root, left subtree, right subtree. This way, you can serialize the tree into a string, and during deserialization, you can reconstruct the tree by parsing the string in the same order.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 12,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Definition for a binary tree node.\n",
+        "class TreeNode(object):\n",
+        "    def __init__(self, x):\n",
+        "        self.val = x\n",
+        "        self.left = None\n",
+        "        self.right = None\n",
+        "\n",
+        "class Codec:\n",
+        "    def serialize(self, root):\n",
+        "        serialized  = []\n",
+        "\n",
+        "        def dfs(node):\n",
+        "            if not node:\n",
+        "                # If the node is None, represent it as \"Null\" in the serialized string\n",
+        "                serialized .append(\"Null\")\n",
+        "                return\n",
+        "            # Convert the node's value to a string and add it to the serialized string\n",
+        "            serialized .append(str(node.val))\n",
+        "            dfs(node.left)\n",
+        "            dfs(node.right)\n",
+        "\n",
+        "        dfs(root)\n",
+        "        # Join the serialized values with \",\"\n",
+        "        return \",\".join(serialized )\n",
+        "\n",
+        "    def deserialize(self, data):\n",
+        "        vals = data.split(\",\")\n",
+        "        self.i = 0\n",
+        "\n",
+        "        def dfs():\n",
+        "            if vals[self.i] == \"Null\":\n",
+        "                self.i += 1\n",
+        "                return None\n",
+        "            # Convert the value to an integer to create a new node\n",
+        "            node = TreeNode(int(vals[self.i]))\n",
+        "            self.i += 1\n",
+        "            node.left = dfs()\n",
+        "            node.right = dfs()\n",
+        "            return node\n",
+        "\n",
+        "        return dfs()"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "Here's a step-by-step explanation of the code:\n",
+        "\n",
+        "1. **Definition of TreeNode**:\n",
+        "   - The code defines a simple `TreeNode` class that represents a node in a binary tree. Each `TreeNode` has three attributes:\n",
+        "     - `val`: The integer value stored in the node.\n",
+        "     - `left`: A reference to the left child node (or None if there is no left child).\n",
+        "     - `right`: A reference to the right child node (or None if there is no right child).\n",
+        "   - This class allows you to create binary tree nodes with integer values and left and right child references.\n",
+        "\n",
+        "2. **Serialization (serialize method)**:\n",
+        "   - The `serialize` method takes a `root` node as input, which is the root of the binary tree that needs to be serialized.\n",
+        "   - It initializes an empty list `serialized` to store the serialized elements.\n",
+        "   - The `dfs` (depth-first search) function is defined within the `serialize` method to perform a preorder traversal of the binary tree.\n",
+        "   - In the `dfs` function:\n",
+        "     - If the current `node` is `None` (i.e., a leaf node or a child of a leaf node), it appends the string \"Null\" to the `serialized` list to represent the absence of a node.\n",
+        "     - If the current `node` is not `None`, it appends the string representation of the `node.val` to the `serialized` list and then recursively calls `dfs` on the left and right children.\n",
+        "   - After the `dfs` traversal, the `serialized` list contains the serialized binary tree elements.\n",
+        "   - The method returns a string obtained by joining the elements in the `serialized` list with commas.\n",
+        "\n",
+        "3. **Deserialization (deserialize method)**:\n",
+        "   - The `deserialize` method takes a serialized string `data` as input, which represents a binary tree in the custom format.\n",
+        "   - It splits the `data` string by commas to obtain a list of elements called `vals`.\n",
+        "   - The method initializes an index `self.i` to 0. This index keeps track of the current position in the `vals` list during deserialization.\n",
+        "   - The `dfs` function is defined within the `deserialize` method to perform the deserialization process using a recursive approach.\n",
+        "   - In the `dfs` function:\n",
+        "     - If the current element in `vals` is \"Null,\" it means there is no node at this position in the binary tree, so it returns `None`.\n",
+        "     - If the current element is not \"Null,\" it converts the element to an integer to create a new `TreeNode` with that value.\n",
+        "     - It then recursively calls `dfs` to set the left and right children of the current node.\n",
+        "   - The method returns the root node of the reconstructed binary tree.\n",
+        "\n",
+        "Overall, this code demonstrates a way to serialize a binary tree into a string format and then deserialize it back into the original tree structure, allowing for the representation of both the tree structure and the values stored in the nodes."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 13,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "1,2,Null,Null,3,4,Null,Null,5,Null,Null\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Example usage:\n",
+        "# Serialize a binary tree\n",
+        "root = TreeNode(1)\n",
+        "root.left = TreeNode(2)\n",
+        "root.right = TreeNode(3)\n",
+        "root.right.left = TreeNode(4)\n",
+        "root.right.right = TreeNode(5)\n",
+        "\n",
+        "codec = Codec()\n",
+        "serialized_tree = codec.serialize(root)\n",
+        "print(serialized_tree)\n",
+        "\n",
+        "# Deserialize the serialized tree\n",
+        "new_root = codec.deserialize(serialized_tree)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 14,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Null\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "\n",
+        "# Example usage:\n",
+        "# Serialize a binary tree\n",
+        "root = None\n",
+        "\n",
+        "codec = Codec()\n",
+        "serialized_tree = codec.serialize(root)\n",
+        "print(serialized_tree)\n",
+        "\n",
+        "# Deserialize the serialized tree\n",
+        "new_root = codec.deserialize(serialized_tree)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code for serializing and deserializing a binary tree:\n",
+        "\n",
+        "1. **Serialization (serialize method)**:\n",
+        "\n",
+        "   - Time Complexity:\n",
+        "     - The `serialize` method uses a depth-first traversal (preorder) of the binary tree.\n",
+        "     - It visits each node exactly once.\n",
+        "     - At each node, it performs constant-time operations (string conversion and list append).\n",
+        "     - Therefore, the time complexity for serialization is O(N), where N is the number of nodes in the binary tree.\n",
+        "\n",
+        "   - Space Complexity:\n",
+        "     - The space complexity for serialization includes the space used by the `serialized` list and the recursive call stack.\n",
+        "     - The `serialized` list stores the serialized elements, and its space is proportional to the number of nodes.\n",
+        "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
+        "     - Therefore, the space complexity for serialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
+        "\n",
+        "2. **Deserialization (deserialize method)**:\n",
+        "\n",
+        "   - Time Complexity:\n",
+        "     - The `deserialize` method splits the serialized string into a list of elements, which takes O(N) time.\n",
+        "     - The `dfs` function is a recursive function that visits each element in the list exactly once.\n",
+        "     - At each element, it performs constant-time operations (string comparison, integer conversion, and recursive function calls).\n",
+        "     - Therefore, the time complexity for deserialization is O(N).\n",
+        "\n",
+        "   - Space Complexity:\n",
+        "     - The space complexity for deserialization includes the space used by the `vals` list and the recursive call stack.\n",
+        "     - The `vals` list stores the elements from the serialized string, and its space is proportional to the number of nodes.\n",
+        "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
+        "     - Therefore, the space complexity for deserialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
+        "\n",
+        "Overall, the provided code has a time complexity of O(N) for both serialization and deserialization and a space complexity of O(N) in the majority of cases (unless the tree is severely unbalanced, in which case the height H dominates the space complexity). It efficiently serializes and deserializes a binary tree while using a linear amount of memory."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Efficient Serialization**:\n",
+        "   Optimize the serialization process to minimize the length of the serialized string. Consider using techniques like binary encoding to represent node values more efficiently.\n",
+        "\n",
+        "2. **Custom Serialization Format**:\n",
+        "   Modify the serialization and deserialization methods to use a custom format different from the one provided. Ensure that the new format allows for accurate reconstruction of the original binary tree."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb b/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb
new file mode 100644
index 0000000..8a41c3b
--- /dev/null
+++ b/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb	
@@ -0,0 +1,265 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 572: Subtree of Another Tree\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Subtree of Another Tree problem on LeetCode, click here!](https://leetcode.com/problems/subtree-of-another-tree/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "**Problem Description:**\n",
+        "\n",
+        "Given the roots of two binary trees `root` and `subRoot`, return `True` if there is a subtree of `root` with the same structure and node values as `subRoot`, and `False` otherwise.\n",
+        "\n",
+        "A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "1. The number of nodes in the `root` tree is in the range $[1, 2000]$.\n",
+        "2. The number of nodes in the `subRoot` tree is in the range $[1, 1000]$.\n",
+        "3. $-10^4$ <= root.val <= $10^4$\n",
+        "4. $-10^4$ <= subRoot.val <= $10^4$\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 17,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Definition for a binary tree node.\n",
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "class Solution:\n",
+        "    def isSubtree(self, root, subRoot):\n",
+        "        if not root:\n",
+        "            return False\n",
+        "        \n",
+        "        # Check if the current subtree is equal to the subRoot\n",
+        "        if self.isSameTree(root, subRoot):\n",
+        "            return True\n",
+        "        \n",
+        "        # Recursively check the left and right subtrees\n",
+        "        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)\n",
+        "    \n",
+        "    def isSameTree(self, tree1, tree2):\n",
+        "        if not tree1 and not tree2:\n",
+        "            return True\n",
+        "        if not tree1 or not tree2:\n",
+        "            return False\n",
+        "        \n",
+        "        return (\n",
+        "            tree1.val == tree2.val and\n",
+        "            self.isSameTree(tree1.left, tree2.left) and\n",
+        "            self.isSameTree(tree1.right, tree2.right)\n",
+        "        )"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "This code defines a TreeNode class for the binary tree nodes and a Solution class with two methods:\n",
+        "\n",
+        "1. `isSubtree`: This method checks if there is a subtree of the `root` tree with the same structure and node values as the `subRoot` tree. It uses a helper function `isSameTree` to compare two trees for equality.\n",
+        "\n",
+        "2. `isSameTree`: This helper method recursively compares two trees to check if they are the same in structure and node values.\n",
+        "\n",
+        "### Here's a detailed explanation of the code:\n",
+        "\n",
+        "1. **TreeNode Class:**\n",
+        "\n",
+        "The `TreeNode` class is defined to represent nodes in a binary tree. Each node has a `val` (the node's value) and may have a `left` and `right` child.\n",
+        "\n",
+        "2. **Solution Class:**\n",
+        "\n",
+        "    + The `Solution` class contains the solution for the problem and defines two important methods:\n",
+        "\n",
+        "    + `isSubtree(self, root, subRoot)`:\n",
+        "\n",
+        "        - This method checks whether `subRoot` is a subtree of `root`. It takes two tree nodes, `root` and `subRoot`, as input arguments.\n",
+        "        - If `root` is `None`, it returns `False` because there is no subtree to search.\n",
+        "        - It then checks if the current subtree with `root` as its root is equal to `subRoot` using the `isSameTree` method. If they are the same, it returns `True`.\n",
+        "        - If the current subtree is not the same as `subRoot`, it recursively checks the left and right subtrees of `root` to see if `subRoot` is a subtree of any of them.\n",
+        "        - It returns `True` if `subRoot` is found in either the left or right subtree; otherwise, it returns `False`.\n",
+        "\n",
+        "    + `isSameTree(self, tree1, tree2)`:\n",
+        "\n",
+        "        - This method checks whether two trees, `tree1` and `tree2`, are the same.\n",
+        "        - If both `tree1` and `tree2` are `None`, they are considered the same tree, so it returns `True`.\n",
+        "        - If only one of them is `None` (but not both), they are different trees, so it returns `False`.\n",
+        "        - If both `tree1` and `tree2` have values, it checks if their values are equal and recursively checks if their left and right subtrees are the same.\n",
+        "        - It returns `True` if the trees are the same; otherwise, it returns `False`.\n",
+        "\n",
+        "The code effectively uses recursion to traverse the binary trees and check for subtree equality. The `isSubtree` method starts the recursive search, and the `isSameTree` method is used to compare individual subtrees. The approach is efficient and avoids unnecessary checks when possible.\n",
+        "\n",
+        "This solution allows you to determine if there exists a subtree within the `root` tree that matches the structure and node values of `subRoot`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 24,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Is subRoot a subtree of root? True\n"
+          ]
+        }
+      ],
+      "source": [
+        "#root = [3,4,5,1,2]\n",
+        "#subRoot = [4,1,2]\n",
+        "\n",
+        "# Example input\n",
+        "root = TreeNode(3)\n",
+        "root.left = TreeNode(4)\n",
+        "root.right = TreeNode(5)\n",
+        "root.left.left = TreeNode(1)\n",
+        "root.left.right = TreeNode(2)\n",
+        "\n",
+        "subRoot = TreeNode(4)\n",
+        "subRoot.left = TreeNode(1)\n",
+        "subRoot.right = TreeNode(2)\n",
+        "\n",
+        "# Create an instance of the Solution class\n",
+        "solution = Solution()\n",
+        "\n",
+        "# Test the isSubtree method\n",
+        "result = solution.isSubtree(root, subRoot)\n",
+        "print(\"Is subRoot a subtree of root?\", result)"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 25,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Is subRoot a subtree of root? False\n"
+          ]
+        }
+      ],
+      "source": [
+        "#root = [3,4,5,1,2,null,null,null,null,0] \n",
+        "#subRoot = [4,1,2]\n",
+        "\n",
+        "\n",
+        "# Example input\n",
+        "root = TreeNode(3)\n",
+        "root.left = TreeNode(4)\n",
+        "root.right = TreeNode(5)\n",
+        "root.left.left = TreeNode(1)\n",
+        "root.left.right = TreeNode(2)\n",
+        "root.left.right.left = TreeNode(0)\n",
+        "\n",
+        "subRoot = TreeNode(4)\n",
+        "subRoot.left = TreeNode(1)\n",
+        "subRoot.right = TreeNode(2)\n",
+        "\n",
+        "# Create an instance of the Solution class\n",
+        "solution = Solution()\n",
+        "\n",
+        "# Test the isSubtree method\n",
+        "result = solution.isSubtree(root, subRoot)\n",
+        "print(\"Is subRoot a subtree of root?\", result)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "86a34f0d-0789-498c-818c-2d30f84754d0",
+      "metadata": {},
+      "source": [
+        "Let's analyze the time and space complexity of the provided code for the \"Subtree of Another Tree\" problem:\n",
+        "\n",
+        "**Time Complexity**\n",
+        "\n",
+        "The time complexity of the code primarily depends on the recursive traversal of the binary tree. In both the `isSubtree` and `isSameTree` functions, we visit each node in the binary trees once. Let's break down the time complexity:\n",
+        "\n",
+        "1. The `isSubtree` function:\n",
+        "   - In the worst case, we need to visit every node in the `root` tree.\n",
+        "   - For each node in the `root` tree, we call the `isSameTree` function, which has its own traversal.\n",
+        "   - So, the total time complexity is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree.\n",
+        "\n",
+        "2. The `isSameTree` function:\n",
+        "   - In the worst case, we visit every node in both `tree1` and `tree2`.\n",
+        "   - The number of recursive calls made is proportional to the number of nodes in the trees.\n",
+        "   - So, the time complexity of this function is $O(max(n, m))$, where $n$ and $m$ are the numbers of nodes in `tree1` and `tree2`, respectively.\n",
+        "\n",
+        "Overall, the time complexity of the entire code is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree. In practice, it may be less than $O(n * m)$ if a subtree mismatch is detected early during the traversal.\n",
+        "\n",
+        "**Space Complexity**\n",
+        "\n",
+        "The space complexity of the code is determined by the function call stack during recursion and the space used by the recursive functions. Let's analyze the space complexity:\n",
+        "\n",
+        "1. The `isSubtree` function:\n",
+        "   - It uses the call stack for recursion.\n",
+        "   - The maximum depth of the recursion is equal to the height of the `root` tree, which can be $O(n)$ in the worst case (unbalanced tree).\n",
+        "   - Additionally, the function doesn't use any significant extra space other than the recursion stack.\n",
+        "\n",
+        "2. The `isSameTree` function:\n",
+        "   - It also uses the call stack for recursion.\n",
+        "   - The maximum depth of the recursion is equal to the height of the `tree1` or `tree2`, whichever is greater.\n",
+        "   - So, the maximum space used for the call stack is $O(max(n, m))$.\n",
+        "\n",
+        "In summary, the space complexity of the code is $O(max(n, m))$ due to the function call stack. It scales with the maximum height of the trees being compared.\n",
+        "\n",
+        "Overall, the code is efficient and works well for trees with moderate sizes. However, it's important to keep in mind that the worst-case time complexity is $O(n * m)$, so for very large trees, the performance may degrade."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb b/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb
new file mode 100644
index 0000000..ad86882
--- /dev/null
+++ b/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb	
@@ -0,0 +1,221 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 98. Validate Binary Search Tree\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Validate Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/validate-binary-search-tree/)\n",
+        "\n",
+        "---\n",
+        "Given the `root` of a binary tree, *determine if it is a valid binary search tree (BST)*.\n",
+        "\n",
+        "A **valid BST** is defined as follows:\n",
+        "\n",
+        "- The left subtree of a node contains only nodes with keys **less than** the node's key.\n",
+        "- The right subtree of a node contains only nodes with keys **greater than** the node's key.\n",
+        "- Both the left and right subtrees must also be binary search trees. \n",
+        "\n",
+        "**Constraints:**\n",
+        "- The number of nodes in the tree is in the range $[1, 10^4]$.\n",
+        "- $-2^{31}$ <= `Node.val` <= $2^{31} - 1$\n",
+        "\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to solve this problem:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 1,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class TreeNode:\n",
+        "    def __init__(self, val=0, left=None, right=None):\n",
+        "        self.val = val\n",
+        "        self.left = left\n",
+        "        self.right = right\n",
+        "\n",
+        "def is_valid_BST(root):\n",
+        "    def is_valid(node, min_val, max_val):\n",
+        "        # Base case: If the node is None, it's a valid BST.\n",
+        "        if node is None:\n",
+        "            return True\n",
+        "        \n",
+        "        # Check if the current node's value is within the valid range.\n",
+        "        if not (min_val < node.val < max_val):\n",
+        "            return False\n",
+        "        \n",
+        "        # Recursively check the left and right subtrees with updated ranges.\n",
+        "        return (is_valid(node.left, min_val, node.val) and\n",
+        "                is_valid(node.right, node.val, max_val))\n",
+        "    \n",
+        "    # Call the helper function starting with a wide range for root node.\n",
+        "    return is_valid(root, float('-inf'), float('inf'))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "- It defines a `TreeNode` class to represent nodes in the binary tree.\n",
+        "\n",
+        "- The `is_valid_BST` function takes the root node of the binary tree as input and returns `True` if the tree is a valid BST, and `False` otherwise.\n",
+        "\n",
+        "- Inside the `is_valid_BST` function, there is a helper function called `is_valid` that performs the actual validation using a recursive approach.\n",
+        "\n",
+        "- The `is_valid` function checks each node in the tree to ensure that it satisfies the properties of a BST:\n",
+        "  - The value of the current node must be within a valid range defined by `min_val` and `max_val`.\n",
+        "  - The left subtree of the current node should contain values less than the current node's value.\n",
+        "  - The right subtree of the current node should contain values greater than the current node's value.\n",
+        "\n",
+        "- If any of these conditions are violated, the function returns `False`, indicating that the tree is not a valid BST.\n",
+        "\n",
+        "- If all nodes satisfy these conditions, the function returns `True`, indicating that the tree is a valid BST.\n",
+        "\n",
+        "- The code calls the `is_valid` function with the root node and initial range values of negative infinity to positive infinity to start the validation.\n",
+        "\n",
+        "- The time complexity of the code is O(N), where N is the number of nodes in the tree, as it traverses each node once.\n",
+        "\n",
+        "- The space complexity depends on the height of the tree. In the average case for a balanced BST, the space complexity is O(log N), but in the worst case (skewed tree), it can be O(N) due to the recursive call stack.\n",
+        "\n",
+        "In summary, this code checks whether a given binary tree is a valid BST by recursively validating each node's value and its left and right subtrees while maintaining valid value ranges. If all nodes satisfy the BST properties, the tree is considered a valid BST."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 2,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Example usage:\n",
+        "# Create the tree for the first example: [2, 1, 3]\n",
+        "root1 = TreeNode(2)\n",
+        "root1.left = TreeNode(1)\n",
+        "root1.right = TreeNode(3)\n",
+        "print(is_valid_BST(root1))"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 3,
+      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "\n",
+        "# Create the tree for the second example: [5, 1, 4, None, None, 3, 6]\n",
+        "root2 = TreeNode(5)\n",
+        "root2.left = TreeNode(1)\n",
+        "root2.right = TreeNode(4)\n",
+        "root2.right.left = TreeNode(3)\n",
+        "root2.right.right = TreeNode(6)\n",
+        "print(is_valid_BST(root2))"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `is_valid_BST` function:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "- The time complexity of the function primarily depends on the number of nodes in the binary tree.\n",
+        "- In the worst case, we may have to visit every node in the tree once to validate whether it's part of a valid BST.\n",
+        "- Since we are performing a depth-first search (DFS) traversal of the tree, the time complexity is O(N), where N is the number of nodes in the tree.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "- The space complexity is determined by the space used by the recursive call stack during the DFS traversal.\n",
+        "- In the worst case, if the tree is completely unbalanced (a skewed tree), the maximum depth of the call stack would be equal to the height of the tree.\n",
+        "- In a balanced BST, the height is approximately log2(N), where N is the number of nodes.\n",
+        "- Therefore, the space complexity of the call stack is O(log N) for a balanced BST.\n",
+        "- In the worst case (skewed tree), the space complexity can be O(N) as the height of the tree can be equal to N.\n",
+        "\n",
+        "**Overall:**\n",
+        "- Time Complexity: O(N)\n",
+        "- Space Complexity: O(log N) on average for a balanced BST, and O(N) in the worst case for a skewed tree.\n",
+        "\n",
+        "The space complexity is typically dominated by the recursive call stack, and it varies depending on the shape of the binary tree. In practice, for balanced binary trees, the space complexity is often close to O(log N), which is quite efficient."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Complex Constraints**: Modify the problem to have more complex constraints on the tree. For example, consider a binary tree with constraints like \"The left subtree of a node contains only nodes with keys less than or equal to the node's key,\" and \"The right subtree of a node contains only nodes with keys greater than or equal to the node's key.\" How would you adapt the code to handle these constraints?\n",
+        "\n",
+        "2. **Handling Duplicates**: Modify the code to handle binary search trees that allow duplicate values. In a standard BST, each value is unique, but in this case, multiple nodes can have the same value. The tree should still be considered valid if it follows the BST property."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/07. Trees/README.md b/_build/html/_sources/07. Trees/README.md
new file mode 100644
index 0000000..1eba5d4
--- /dev/null
+++ b/_build/html/_sources/07. Trees/README.md	
@@ -0,0 +1,26 @@
+# Trees Problems - Blind 75 LeetCode 
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)                          | Easy       |
+| [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)     | Easy       |
+| [100. Same Tree](https://leetcode.com/problems/same-tree/)                                           | Easy       |
+| [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)               | Easy       |
+| [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/) | Easy |
+| [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/) | Medium |
+| [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)       | Medium     |
+| [230. Kth Smallest Element In a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)   | Medium     |
+| [105. Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) | Medium |
+| [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)     | Hard       |
+| [297. Serialize And Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/) | Hard |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
diff --git a/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb b/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb
new file mode 100644
index 0000000..9da2a17
--- /dev/null
+++ b/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb	
@@ -0,0 +1,249 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 208. Implement Trie (Prefix Tree)\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Implement Trie problem on LeetCode, click here!](https://leetcode.com/problems/implement-trie-prefix-tree/)\n",
+        "\n",
+        "---\n",
+        "A **trie** (pronounced as \"try\") or **prefix tree** is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.\n",
+        "\n",
+        "Implement the Trie class:\n",
+        "\n",
+        "- `Trie()` Initializes the trie object.\n",
+        "- `void insert(String word)` Inserts the string `word` into the trie.\n",
+        "- `boolean search(String word)` Returns `true` if the string `word` is in the trie (i.e., was inserted before), and `false` otherwise.\n",
+        "- `boolean startsWith(String prefix)` Returns `true` if there is a previously inserted string `word` that has the prefix `prefix`, and `false` otherwise.\n",
+        "\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `1 <= word.length, prefix.length <= 2000`\n",
+        "- `word` and `prefix` consist only of lowercase English letters.\n",
+        "- At most $3 * 10^4$ calls in total will be made to `insert`, `search`, and `startsWith`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem at hand is to implement a data structure known as a Trie (pronounced \"try\") or Prefix Tree. A Trie is a tree-like data structure that is used to efficiently store and retrieve a set of strings or words. It's particularly useful for tasks involving string manipulation, such as autocomplete and spell-checking. The problem statement defines the following operations for implementing the Trie class:\n",
+        "\n",
+        "1. `Trie()`: This is the constructor that initializes the Trie object.\n",
+        "\n",
+        "2. `void insert(String word)`: This method inserts the given string `word` into the Trie. It effectively adds the characters of the word to the Trie's structure, creating nodes for each character if they don't already exist. At the end of the word, a special flag is set to indicate that this node represents the end of a valid word.\n",
+        "\n",
+        "3. `boolean search(String word)`: This method checks if the given string `word` is present in the Trie. It starts from the root of the Trie and traverses the Trie by following the characters in the word. If it successfully traverses the Trie and reaches the end of the word, it returns `true`, indicating that the word exists in the Trie; otherwise, it returns `false`.\n",
+        "\n",
+        "4. `boolean startsWith(String prefix)`: This method checks if there is any previously inserted word in the Trie that has the given `prefix`. It's similar to the `search` method but does not require that the prefix be a complete word. If there is any word in the Trie that starts with the given prefix, it returns `true`; otherwise, it returns `false`.\n",
+        "\n",
+        "The problem also provides an example scenario where these operations are called, demonstrating the expected output for each operation.\n",
+        "\n",
+        "In summary, the goal is to create a data structure (Trie) that efficiently stores a set of strings and provides methods to insert new strings, search for complete words, and check if a given prefix exists in the stored set of strings.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 1,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class TrieNode:\n",
+        "    def __init__(self):\n",
+        "        self.children = {}           # A dictionary to store child nodes\n",
+        "        self.is_end_of_word = False  # Indicates if a word ends at this node\n",
+        "\n",
+        "class Trie:\n",
+        "    def __init__(self):\n",
+        "        self.root = TrieNode()       # Initialize the Trie with a root node\n",
+        "\n",
+        "    def insert(self, word):\n",
+        "        node = self.root             # Start from the root\n",
+        "        for char in word:\n",
+        "            if char not in node.children:\n",
+        "                node.children[char] = TrieNode()  # Create a new node if the character is not present\n",
+        "            node = node.children[char]           # Move to the child node\n",
+        "        node.is_end_of_word = True                # Mark the end of the inserted word\n",
+        "\n",
+        "    def search(self, word):\n",
+        "        node = self.root             # Start from the root\n",
+        "        for char in word:\n",
+        "            if char not in node.children:\n",
+        "                return False         # If the character is not found, the word is not in the Trie\n",
+        "            node = node.children[char]  # Move to the child node\n",
+        "        return node.is_end_of_word    # Check if the node represents the end of a valid word\n",
+        "\n",
+        "    def startsWith(self, prefix):\n",
+        "        node = self.root             # Start from the root\n",
+        "        for char in prefix:\n",
+        "            if char not in node.children:\n",
+        "                return False         # If the character is not found, no word starts with the prefix\n",
+        "            node = node.children[char]  # Move to the child node\n",
+        "        return True                  # Prefix found in the Trie"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "Let's explain how the code works step by step:\n",
+        "\n",
+        "1. `TrieNode` Class:\n",
+        "   - `TrieNode` is a class that represents nodes in the Trie. Each node has two attributes:\n",
+        "     - `children`: This is a dictionary that stores child nodes. Each key in the dictionary represents a character, and the corresponding value is the child node for that character.\n",
+        "     - `is_end_of_word`: This boolean flag indicates whether the node represents the end of a complete word. Initially, it's set to `False` for all nodes.\n",
+        "\n",
+        "2. `Trie` Class:\n",
+        "   - `Trie` is the main class that implements the Trie data structure. It has the following methods:\n",
+        "\n",
+        "   - `__init__(self)`: The constructor initializes the Trie object by creating a root node, which serves as the starting point for all Trie operations.\n",
+        "\n",
+        "   - `insert(self, word)`: This method inserts a string `word` into the Trie. It starts from the root node and iterates through each character in the word.\n",
+        "     - If a character is not present as a child node, it creates a new node for that character.\n",
+        "     - It then moves to the child node and continues the process until the entire word is inserted.\n",
+        "     - Finally, it sets the `is_end_of_word` flag to `True` for the last node to mark the end of the inserted word.\n",
+        "\n",
+        "   - `search(self, word)`: This method checks if a complete word (string) exists in the Trie. It starts from the root node and iterates through each character in the word.\n",
+        "     - If a character is not found as a child node, it immediately returns `False` because the word is not in the Trie.\n",
+        "     - It continues to move to the child node for each character.\n",
+        "     - After reaching the end of the word, it checks if the `is_end_of_word` flag is `True` for the last node to confirm the presence of the word.\n",
+        "\n",
+        "   - `startsWith(self, prefix)`: This method checks if there is any previously inserted word in the Trie that starts with a given `prefix`. It follows the same logic as the `search` method but does not require the entire word to be present.\n",
+        "     - If the prefix is found in the Trie, it returns `True`; otherwise, it returns `False`.\n",
+        "\n",
+        "3. Example Usage:\n",
+        "   - The code demonstrates how to create a Trie object, insert words (\"apple\" and \"app\"), and perform operations on the Trie:\n",
+        "     - It inserts the word \"apple\" into the Trie.\n",
+        "     - It checks if \"apple\" is in the Trie, which returns `True`.\n",
+        "     - It checks if \"app\" is in the Trie, which returns `False`.\n",
+        "     - It checks if there is any word in the Trie that starts with \"app,\" which returns `True`.\n",
+        "     - It inserts the word \"app\" into the Trie.\n",
+        "     - It checks if \"app\" is in the Trie again, which now returns `True`.\n",
+        "\n",
+        "In summary, the code efficiently implements a Trie data structure with the ability to insert words, search for complete words, and check for the existence of words with a given prefix. The Trie is organized as a tree of nodes, with each node representing a character in the words being stored."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 2,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n",
+            "False\n",
+            "True\n",
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Example usage:\n",
+        "\n",
+        "trie = Trie()\n",
+        "trie.insert(\"apple\")\n",
+        "print(trie.search(\"apple\"))   # Output: True\n",
+        "print(trie.search(\"app\"))     # Output: False\n",
+        "print(trie.startsWith(\"app\")) # Output: True\n",
+        "trie.insert(\"app\")\n",
+        "print(trie.search(\"app\"))     # Output: True"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the Trie implementation:\n",
+        "\n",
+        "1. Time Complexity:\n",
+        "   - Insertion (`insert` method): The time complexity of inserting a word into the Trie is O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to insert it into the Trie.\n",
+        "   - Search (`search` method): The time complexity of searching for a word in the Trie is also O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to determine if it exists in the Trie.\n",
+        "   - Starts with (`startsWith` method): The time complexity of checking if a prefix exists in the Trie is O(p), where p is the length of the prefix. In the worst case, you may need to traverse the entire prefix to determine its presence.\n",
+        "\n",
+        "   Therefore, for each operation (insertion, search, or starts with), the time complexity is O(m) or O(p), where m is the length of the word being operated on, and p is the length of the prefix.\n",
+        "\n",
+        "2. Space Complexity:\n",
+        "   - The space complexity of the Trie is determined by the number of nodes and characters stored in it. In the worst case, where none of the inserted words share common prefixes, the space complexity is O(N), where N is the total number of characters across all inserted words.\n",
+        "   - Each node in the Trie represents a character, and the number of nodes is directly related to the total number of characters.\n",
+        "\n",
+        "   In practice, the space complexity can be less than O(N) because common prefixes are shared among words in the Trie, leading to space optimization.\n",
+        "\n",
+        "In summary:\n",
+        "- Time complexity for each operation (insertion, search, starts with) is O(m) or O(p), where m is the length of the word and p is the length of the prefix.\n",
+        "- Space complexity is O(N) in the worst case, where N is the total number of characters across all inserted words. However, space optimization occurs due to common prefix sharing in practice, potentially reducing the actual space used."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Count Prefixes**:\n",
+        "   Implement a method to count the number of words in the Trie that have a specific prefix. This exercise requires you to maintain additional data in the Trie nodes to keep track of the count.\n",
+        "\n",
+        "2. **Auto-Complete Suggestions**:\n",
+        "   Implement an auto-complete feature using the Trie. Given a prefix, the program should return a list of suggested words based on the words previously inserted into the Trie."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb b/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb
new file mode 100644
index 0000000..2de472c
--- /dev/null
+++ b/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb	
@@ -0,0 +1,275 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 211. Design Add and Search Words Data Structure\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Design Add and Search Words Data Structure problem on LeetCode, click here!](https://leetcode.com/problems/design-add-and-search-words-data-structure/)\n",
+        "\n",
+        "---\n",
+        "Design a data structure that supports adding new words and finding if a string matches any previously added string.\n",
+        "\n",
+        "Implement the `WordDictionary` class:\n",
+        "\n",
+        "- `WordDictionary()` Initializes the object.\n",
+        "- `void addWord(word)` Adds `word` to the data structure, it can be matched later.\n",
+        "- `bool search(word)` Returns `true` if there is any string in the data structure that matches `word` or `false` otherwise. `word` may contain dots `'.'` where dots can be matched with any letter.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `1 <= word.length <= 25`\n",
+        "- `word` in `addWord` consists of lowercase English letters.\n",
+        "- `word` in `search` consist of `'.'` or lowercase English letters.\n",
+        "- There will be at most `2` dots in `word` for `search` queries.\n",
+        "- At most $10^4$ calls will be made to `addWord` and `search`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem is to design a data structure called `WordDictionary` that allows you to add words and search for words efficiently. The key feature of this data structure is that it should support wildcard search, where a dot ('.') can match any letter.\n",
+        "\n",
+        "Here are the specific requirements and explanations for the problem:\n",
+        "\n",
+        "1. Initialize the `WordDictionary` object using the constructor `WordDictionary()`.\n",
+        "\n",
+        "2. The `addWord(word)` method allows you to add a word to the data structure. Once added, you should be able to search for this word later. The words consist of lowercase English letters, and each word has a length between 1 and 25 characters.\n",
+        "\n",
+        "3. The `search(word)` method allows you to search for a word in the data structure. This method returns `True` if there is any string in the data structure that matches the given word or `False` otherwise. The word you are searching for may contain dots ('.'), where each dot can match any letter.\n",
+        "\n",
+        "4. The wildcard character ('.') in the `search` method allows for partial or fuzzy searching. For example, if the word dictionary contains the words \"bad,\" \"dad,\" and \"mad,\" then:\n",
+        "   - Searching for \"pad\" should return `False` because there is no exact match.\n",
+        "   - Searching for \"bad\" should return `True` because \"bad\" is in the dictionary.\n",
+        "   - Searching for \".ad\" should return `True` because it can match \"bad,\" \"dad,\" or \"mad.\"\n",
+        "   - Searching for \"b..\" should return `True` because it can match \"bad,\" \"bed,\" \"bet,\" etc.\n",
+        "\n",
+        "5. The problem specifies that there will be at most two dots ('.') in search queries, and there will be at most 10^4 calls to both `addWord` and `search`.\n",
+        "\n",
+        "In summary, you need to implement the `WordDictionary` class that efficiently supports adding words and searching for words, where the search can include wildcard characters ('.') that match any letter.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 3,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class TrieNode:\n",
+        "    def __init__(self):\n",
+        "        # Initialize a TrieNode with children and a flag to indicate the end of a word.\n",
+        "        self.children = {}\n",
+        "        self.is_end_of_word = False\n",
+        "\n",
+        "class WordDictionary:\n",
+        "\n",
+        "    def __init__(self):\n",
+        "        # Initialize the WordDictionary with a root node.\n",
+        "        self.root = TrieNode()\n",
+        "\n",
+        "    def addWord(self, word: str) -> None:\n",
+        "        node = self.root\n",
+        "        # Iterate through the characters in the word.\n",
+        "        for char in word:\n",
+        "            if char not in node.children:\n",
+        "                # Create a new node for the character if it doesn't exist.\n",
+        "                node.children[char] = TrieNode()\n",
+        "            # Move to the next node.\n",
+        "            node = node.children[char]\n",
+        "        # Mark the end of the word by setting the flag to True.\n",
+        "        node.is_end_of_word = True\n",
+        "\n",
+        "    def search_helper(self, node: TrieNode, word: str) -> bool:\n",
+        "        if not word:\n",
+        "            # If there are no characters left in the word, check if we reached the end of a word in the trie.\n",
+        "            return node.is_end_of_word\n",
+        "        \n",
+        "        char = word[0]\n",
+        "        if char == '.':\n",
+        "            # If the character is a dot, explore all possible child nodes.\n",
+        "            for child in node.children.values():\n",
+        "                if self.search_helper(child, word[1:]):\n",
+        "                    return True\n",
+        "        elif char in node.children:\n",
+        "            # If the character is in the children of the current node, move to the next node.\n",
+        "            return self.search_helper(node.children[char], word[1:])\n",
+        "        \n",
+        "        # If the character is not found and is not a dot, the word is not in the trie.\n",
+        "        return False\n",
+        "\n",
+        "    def search(self, word: str) -> bool:\n",
+        "        # Start the search from the root of the trie.\n",
+        "        return self.search_helper(self.root, word)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "This code defines a `WordDictionary` class that uses a Trie data structure to efficiently store and search for words. Here's a step-by-step explanation:\n",
+        "\n",
+        "1. `TrieNode` class: This class represents nodes in the Trie data structure. Each node has two attributes: `children` (a dictionary to store child nodes) and `is_end_of_word` (a flag to indicate whether the node marks the end of a word).\n",
+        "\n",
+        "2. `WordDictionary` class: This class initializes with an empty Trie by creating a root node.\n",
+        "\n",
+        "3. `addWord` method: This method adds a word to the Trie. It iterates through each character in the word and creates Trie nodes as necessary. It sets the `is_end_of_word` flag to `True` for the final character of the word to mark the end of the word.\n",
+        "\n",
+        "4. `search_helper` method: This is a recursive helper function for searching words in the Trie. It takes a Trie node (`node`) and a word to search (`word`) as input. If there are no characters left in the word (`not word`), it checks if the current node marks the end of a word. If it does, it returns `True`.\n",
+        "\n",
+        "5. If the first character of the word is a dot ('.'), the function explores all possible child nodes by iterating through `node.children.values()` and recursively calls `search_helper` for each child with the remaining part of the word (`word[1:]`).\n",
+        "\n",
+        "6. If the first character of the word is not a dot and is found in the children of the current node, the function recursively moves to the next node by calling `search_helper` on the child node and the remaining part of the word.\n",
+        "\n",
+        "7. If the character is not found in the children and is not a dot, the word is not in the Trie, so the function returns `False`.\n",
+        "\n",
+        "8. `search` method: This is the public method for searching words. It initiates the search process by calling `search_helper` starting from the root node of the Trie.\n",
+        "\n",
+        "Overall, this code efficiently implements a Trie-based data structure that allows adding and searching for words, including support for wildcard characters represented by dots ('.')."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 4,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n",
+            "True\n",
+            "True\n",
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Example usage:\n",
+        "\n",
+        "wordDictionary = WordDictionary()\n",
+        "wordDictionary.addWord(\"bad\")\n",
+        "wordDictionary.addWord(\"dad\")\n",
+        "wordDictionary.addWord(\"mad\")\n",
+        "print(wordDictionary.search(\"pad\"))  # Output: False\n",
+        "print(wordDictionary.search(\"bad\"))  # Output: True\n",
+        "print(wordDictionary.search(\".ad\"))  # Output: True\n",
+        "print(wordDictionary.search(\"b..\"))  # Output: True"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `WordDictionary` class methods:\n",
+        "\n",
+        "1. `addWord` Method:\n",
+        "\n",
+        "   - Time Complexity: O(L)\n",
+        "     - L is the length of the word being added.\n",
+        "     - In the worst case, we need to iterate through each character in the word and create nodes in the Trie. This takes O(L) time.\n",
+        "\n",
+        "   - Space Complexity: O(L)\n",
+        "     - The space complexity for storing the word in the Trie is also O(L) because we create nodes for each character in the word.\n",
+        "\n",
+        "2. `search_helper` Method (called by `search`):\n",
+        "\n",
+        "   - Time Complexity: O(2^M * L)\n",
+        "     - M is the maximum number of dots ('.') in the search word.\n",
+        "     - In the worst case, when there are M dots in the search word, we may explore all possible child nodes at each dot. This results in a branching factor of 26 (for lowercase letters) at each dot.\n",
+        "     - So, the time complexity is exponential in the number of dots and linear in the length of the word being searched (L).\n",
+        "\n",
+        "   - Space Complexity: O(L)\n",
+        "     - The space complexity for the recursive call stack is O(L) in the worst case because we may have to recurse to a depth equal to the length of the search word.\n",
+        "\n",
+        "3. `search` Method:\n",
+        "\n",
+        "   - Time Complexity: O(2^M * L)\n",
+        "     - The `search` method calls the `search_helper` method, which has the same time complexity explained above.\n",
+        "\n",
+        "   - Space Complexity: O(L)\n",
+        "     - The space complexity for the recursive call stack is O(L) in the worst case, as explained earlier.\n",
+        "\n",
+        "Overall, the time complexity of the `search` method is mainly affected by the number of dots ('.') in the search word. In the worst case, when there are multiple dots and many possibilities to explore at each dot, the time complexity can be exponential. However, for most practical cases, it performs reasonably well.\n",
+        "\n",
+        "The space complexity primarily depends on the space required to store the words in the Trie. It is proportional to the total number of characters in all added words and is linear with respect to the length of the words added.\n",
+        "\n",
+        "In summary:\n",
+        "\n",
+        "- Time Complexity for `addWord`: O(L)\n",
+        "- Time Complexity for `search`: O(2^M * L)\n",
+        "- Space Complexity: O(L) for storing words in the Trie, O(L) for the recursive call stack during search.\n",
+        "\n",
+        "Note: The space complexity analysis assumes that the dictionary contains a reasonable number of words and does not account for the overhead of Python objects or system-level memory allocation."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Dictionary Auto-Completion**: Extend the `WordDictionary` class to provide auto-completion suggestions based on the prefix of a word. Implement a method that returns a list of words that match the given prefix.\n",
+        "\n",
+        "2. **Support Word Deletion**: Extend the `WordDictionary` class to support word deletion. Add a method `deleteWord(word)` that removes a word from the data structure. Ensure that the search operation still works correctly after deletion."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/08. Tries/212. Word Search II.ipynb b/_build/html/_sources/08. Tries/212. Word Search II.ipynb
new file mode 100644
index 0000000..71597b8
--- /dev/null
+++ b/_build/html/_sources/08. Tries/212. Word Search II.ipynb	
@@ -0,0 +1,316 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 212. Word Search II\n",
+        "\n",
+        "**Difficulty:** Hard\n",
+        "\n",
+        "**Link to Problem:** [To see the Word Search II problem on LeetCode, click here!](https://leetcode.com/problems/word-search-ii/)\n",
+        "\n",
+        "---\n",
+        "Given an `m x n` `board` of characters and a list of strings `words`, the task is to return *all words on the board*.\n",
+        "\n",
+        "Each word must be constructed from letters of sequentially **adjacent cells**, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.\n",
+        "\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `m == board.length`\n",
+        "- `n == board[i].length`\n",
+        "- `1 <= m, n <= 12`\n",
+        "- `board[i][j]` is a lowercase English letter.\n",
+        "- 1 <= `words.length` <= $3 * 10^4$\n",
+        "- `1 <= words[i].length <= 10`\n",
+        "- `words[i]` consists of lowercase English letters.\n",
+        "- All the strings of `words` are unique."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "To solve the \"Word Search II\" problem, you can use a Trie data structure and perform a depth-first search (DFS) on the board. Here's a step-by-step approach to solve the problem:\n",
+        "\n",
+        "1. Build a Trie from the given list of words:\n",
+        "   - Create a Trie data structure to store all the words from the input list.\n",
+        "   - For each word in the input list, insert it into the Trie, character by character. Make sure to mark the end of each word in the Trie.\n",
+        "\n",
+        "2. Perform DFS on the board:\n",
+        "   - Iterate through each cell (i, j) on the board.\n",
+        "   - For each cell, start a DFS traversal from that cell to search for words.\n",
+        "   - During the DFS traversal, maintain a current path string.\n",
+        "   - At each cell, check if the current path string, concatenated with the character in the cell, matches any word prefix in the Trie. If it does, continue the DFS.\n",
+        "   - If the current path string matches a word in the Trie, add it to the result.\n",
+        "\n",
+        "3. Implement DFS with backtracking:\n",
+        "   - During the DFS traversal, you will explore neighboring cells (horizontally and vertically) if they are within bounds and haven't been visited before.\n",
+        "   - To prevent revisiting the same cell within the same word, mark the cell as visited (e.g., change its character to a special character like '#') before exploring it and restore its original character after the DFS traversal.\n",
+        "\n",
+        "4. Return the result:\n",
+        "   - After completing the DFS traversal for all cells on the board, you will have collected all the valid words in the result set.\n",
+        "   - Convert the result set to a list and return it as the final output.\n",
+        "\n",
+        "This approach efficiently finds all words on the board by utilizing the Trie data structure to prune unnecessary DFS paths and avoiding duplicate visits to the same cell within the same word. It ensures that each word is constructed from letters of sequentially adjacent cells without using the same letter cell more than once in a word.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 5,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Define a TrieNode class to represent nodes in the Trie.\n",
+        "class TrieNode:\n",
+        "    def __init__(self):\n",
+        "        self.children = {}  # Dictionary to store child nodes.\n",
+        "        self.is_end_of_word = False  # Flag to mark the end of a word.\n",
+        "\n",
+        "# Main function to find words on the board.\n",
+        "def findWords(board, words):\n",
+        "    # DFS function to search for words starting from a given cell.\n",
+        "    def dfs(node, i, j, path):\n",
+        "        char = board[i][j]  # Get the character at the current cell.\n",
+        "        curr_node = node.children.get(char)  # Find the corresponding Trie node.\n",
+        "\n",
+        "        if not curr_node:\n",
+        "            return  # If no matching Trie node, stop the search.\n",
+        "\n",
+        "        path += char  # Add the character to the current path.\n",
+        "\n",
+        "        if curr_node.is_end_of_word:\n",
+        "            result.add(path)  # If the path matches a word, add it to the result.\n",
+        "\n",
+        "        temp, board[i][j] = board[i][j], \"#\"  # Mark the cell as visited.\n",
+        "\n",
+        "        # Explore neighboring cells (up, down, left, right).\n",
+        "        if i > 0:\n",
+        "            dfs(curr_node, i - 1, j, path)\n",
+        "        if i < m - 1:\n",
+        "            dfs(curr_node, i + 1, j, path)\n",
+        "        if j > 0:\n",
+        "            dfs(curr_node, i, j - 1, path)\n",
+        "        if j < n - 1:\n",
+        "            dfs(curr_node, i, j + 1, path)\n",
+        "\n",
+        "        board[i][j] = temp  # Restore the cell's original character.\n",
+        "\n",
+        "    # Function to build a Trie from the list of words.\n",
+        "    def buildTrie():\n",
+        "        root = TrieNode()  # Create a root node.\n",
+        "        for word in words:\n",
+        "            node = root\n",
+        "            for char in word:\n",
+        "                if char not in node.children:\n",
+        "                    node.children[char] = TrieNode()  # Create a new node if needed.\n",
+        "                node = node.children[char]\n",
+        "            node.is_end_of_word = True  # Mark the end of the word.\n",
+        "        return root\n",
+        "\n",
+        "    m, n = len(board), len(board[0])  # Get the dimensions of the board.\n",
+        "    root = buildTrie()  # Build the Trie from the list of words.\n",
+        "    result = set()  # Use a set to store unique words found on the board.\n",
+        "\n",
+        "    # Iterate through each cell on the board and start DFS from there.\n",
+        "    for i in range(m):\n",
+        "        for j in range(n):\n",
+        "            dfs(root, i, j, \"\")  # Start DFS from each cell on the board.\n",
+        "\n",
+        "    return list(result)  # Convert the set to a list and return the result."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The code solves the \"Word Search II\" problem, where you have a grid of characters and a list of words. The goal is to find all words from the list that can be constructed by traversing adjacent cells (horizontally or vertically) in the grid, without using the same cell more than once for a single word.\n",
+        "\n",
+        "The code does the following:\n",
+        "\n",
+        "1. Defines a `TrieNode` class to represent nodes in a Trie data structure. Each node has children (other characters in the word) and a flag to mark the end of a word.\n",
+        "\n",
+        "2. Defines the `findWords` function, which takes the grid (`board`) and a list of words (`words`) as input.\n",
+        "\n",
+        "3. Inside the `findWords` function, there is a `dfs` (depth-first search) function. This function is used to explore the grid starting from a specific cell. It checks if the current path matches any word in the Trie and adds it to the result if found.\n",
+        "\n",
+        "4. The code also includes a `buildTrie` function that constructs a Trie data structure from the list of words. It iterates through each word, creating nodes for each character in the Trie and marking the end of words.\n",
+        "\n",
+        "5. The dimensions of the grid are obtained (m x n).\n",
+        "\n",
+        "6. A Trie is built using the `buildTrie` function.\n",
+        "\n",
+        "7. A set called `result` is used to store unique words found in the grid.\n",
+        "\n",
+        "8. The code iterates through each cell on the grid (using nested loops) and starts a DFS search from each cell.\n",
+        "\n",
+        "9. During the DFS, it explores neighboring cells (up, down, left, right) if they haven't been visited before and if the current path matches a prefix in the Trie.\n",
+        "\n",
+        "10. When a word is found during the DFS traversal, it is added to the `result` set.\n",
+        "\n",
+        "11. The grid cell is marked as visited during the DFS by changing its character to \"#\", and it is restored to its original character after the DFS traversal.\n",
+        "\n",
+        "12. Finally, the set of unique words found is converted to a list and returned as the output.\n",
+        "\n",
+        "The code effectively uses a Trie data structure to optimize the search and ensures that each word is constructed from adjacent cells without reusing the same cell for the same word."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 7,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "['eat', 'oath']\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "board1 = [[\"o\",\"a\",\"a\",\"n\"],[\"e\",\"t\",\"a\",\"e\"],[\"i\",\"h\",\"k\",\"r\"],[\"i\",\"f\",\"l\",\"v\"]]\n",
+        "words1 = [\"oath\",\"pea\",\"eat\",\"rain\"]\n",
+        "print(findWords(board1, words1))  # Output: [\"eat\", \"oath\"]"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 8,
+      "id": "c748b879-a2d8-4298-a42c-c7f61f9c7fc6",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "\n",
+        "board2 = [[\"a\",\"b\"],[\"c\",\"d\"]]\n",
+        "words2 = [\"abcb\"]\n",
+        "print(findWords(board2, words2))  # Output: []"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "1. **Building the Trie (`buildTrie` function):** \n",
+        "   - Suppose there are `N` words in the input list, and the average length of a word is `L`.\n",
+        "   - Building the Trie takes O(N * L) time, as we iterate through each word and insert each character into the Trie.\n",
+        "\n",
+        "2. **DFS Traversal (`dfs` function):** \n",
+        "   - In the worst case, the DFS function explores all possible paths in the grid.\n",
+        "   - The maximum number of recursive calls for each cell is 4 (up, down, left, right).\n",
+        "   - The maximum depth of the DFS is limited by the length of the longest word in the Trie, which is at most `L`.\n",
+        "   - Therefore, the DFS traversal for each cell takes $O(4^L)$ time.\n",
+        "\n",
+        "3. **Iterating Through the Grid (`findWords` function):**\n",
+        "   - In the worst case, we iterate through all `m` rows and `n` columns in the grid.\n",
+        "   - For each cell, we start a DFS traversal.\n",
+        "   - Therefore, iterating through the entire grid takes $O(m * n * 4^L)$ time.\n",
+        "\n",
+        "Overall, the time complexity of the code is $O(N * L + m * n * 4^L)$.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "1. **Trie Data Structure:**\n",
+        "   - The space required to store the Trie depends on the number of unique characters in all the words.\n",
+        "   - In the worst case, where all words are unique and have no common prefixes, the space complexity of the Trie is O(N * L).\n",
+        "   - In practice, it may be less if there are common prefixes among words.\n",
+        "\n",
+        "2. **DFS Stack (Recursive Calls):**\n",
+        "   - The depth of the recursive call stack during DFS is at most `L`, where `L` is the length of the longest word in the Trie.\n",
+        "   - Therefore, the space complexity for the recursive call stack is O(L).\n",
+        "\n",
+        "3. **Result Set (`result` set):**\n",
+        "   - The space used to store the result set depends on the number of valid words found in the grid.\n",
+        "   - In the worst case, when all words are found, the space complexity of the result set is O(N * L).\n",
+        "\n",
+        "Overall, the space complexity of the code is O(N * L) for the Trie and O(L) for the recursive call stack.\n",
+        "\n",
+        "In summary, the time complexity is dominated by the DFS traversal and is influenced by the number of words, their lengths, and the size of the grid. The space complexity is mainly determined by the Trie structure and the recursive call stack depth during DFS."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Word Search Paths:**\n",
+        "   Extend the basic word search exercise to find all possible paths (sequences of cells) that spell out a given word on the board. Return a list of paths if the word can be formed, otherwise an empty list.\n",
+        "\n",
+        "\n",
+        "2. **Reverse Word Search:**\n",
+        "   Modify the code to search for words in reverse order on the board. Given a 2D grid of characters and a list of words, find all words from the list that can be formed by traversing the board in reverse (right to left, bottom to top, etc.)."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/08. Tries/README.md b/_build/html/_sources/08. Tries/README.md
new file mode 100644
index 0000000..22c3bcd
--- /dev/null
+++ b/_build/html/_sources/08. Tries/README.md	
@@ -0,0 +1,18 @@
+# Trie Problems - Blind 75 LeetCode
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [208. Implement Trie Prefix Tree](https://leetcode.com/problems/implement-trie-prefix-tree/)         | Medium     |
+| [211. Design Add And Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/) | Medium |
+| [212. Word Search II](https://leetcode.com/problems/word-search-ii/)                                 | Hard       |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
diff --git a/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb b/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb
new file mode 100644
index 0000000..cb66b6a
--- /dev/null
+++ b/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb	
@@ -0,0 +1,254 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 1046. Last Stone Weight\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Last Stone Weight problem on LeetCode, click here!](https://leetcode.com/problems/last-stone-weight/)\n",
+        "\n",
+        "---\n",
+        "You are given an array of integers `stones` where `stones[i]` is the weight of the `i`th stone.\n",
+        "\n",
+        "We are playing a game with the stones. On each turn, you choose the **heaviest two stones** and smash them together. Suppose the heaviest two stones have weights `x` and `y` with `x <= y`. The result of this smash is:\n",
+        "\n",
+        "1. If `x == y`, both stones are destroyed.\n",
+        "2. If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - x`.\n",
+        "\n",
+        "At the end of the game, there is **at most one** stone left.\n",
+        "\n",
+        "Return *the weight of the last remaining stone*. If there are no stones left, return `0`.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `1 <= stones.length <= 30`\n",
+        "- `1 <= stones[i] <= 1000`"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The \"Last Stone Weight\" problem involves a game played with a collection of stones, each stone having a certain weight. The goal is to find the weight of the last remaining stone after applying a specific set of rules.\n",
+        "\n",
+        "Here's how the game works:\n",
+        "\n",
+        "1. You start with an array of stones, where each stone is represented by its weight. The weights of the stones are given in the input as an array.\n",
+        "\n",
+        "2. In each turn of the game, you select the two heaviest stones from the remaining stones. If there is only one stone left, you've found the answer.\n",
+        "\n",
+        "3. If there are two stones with weights `x` and `y`, where `x` is less than or equal to `y`, the following happens:\n",
+        "   - If `x` is equal to `y`, both stones are completely destroyed, and they are removed from the array of stones.\n",
+        "   - If `x` is not equal to `y`, the stone with weight `x` is destroyed, and the stone with weight `y` now has a new weight of `y - x`. The stone with weight `x` is removed from the array, and the modified stone with weight `y - x` remains in the array.\n",
+        "\n",
+        "4. The game continues with the modified array of stones until there is at most one stone left.\n",
+        "\n",
+        "5. The goal is to find the weight of the last remaining stone, or if there are no stones left, return `0`.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 14,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "import heapq\n",
+        "\n",
+        "def lastStoneWeight(stones):\n",
+        "    # Convert the input list to a max heap (negate the values to simulate a max heap)\n",
+        "    max_heap = [-stone for stone in stones]\n",
+        "    heapq.heapify(max_heap)\n",
+        "\n",
+        "    # Continue as long as there are more than 1 stone in the heap\n",
+        "    while len(max_heap) > 1:\n",
+        "        # Get the two heaviest stones\n",
+        "        y = -heapq.heappop(max_heap)  # Get the heaviest stone and negate it back to positive\n",
+        "        x = -heapq.heappop(max_heap)  # Get the second heaviest stone and negate it back to positive\n",
+        "\n",
+        "        # Calculate the new stone's weight after smashing\n",
+        "        if x != y:\n",
+        "            new_stone = y - x\n",
+        "            heapq.heappush(max_heap, -new_stone)  # Negate the value and push it back to the heap\n",
+        "\n",
+        "    # If there is one stone left, return its weight (negate it back to positive)\n",
+        "    if max_heap:\n",
+        "        return -max_heap[0]\n",
+        "    else:\n",
+        "        return 0"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The code defines a Python function called `lastStoneWeight` that takes a list of stone weights as input. It aims to find the weight of the last remaining stone after playing the stone smashing game described in the problem statement.\n",
+        "\n",
+        "Here's a step-by-step explanation of the code:\n",
+        "\n",
+        "1. We import the `heapq` library, which allows us to create and manipulate a heap (priority queue).\n",
+        "\n",
+        "2. Inside the `lastStoneWeight` function:\n",
+        "   - We create a max heap (negating stone weights to simulate a max heap) using the `heapq.heapify` function. This heap will keep track of the heaviest stones.\n",
+        "   \n",
+        "   - We enter a loop that continues as long as there are more than 1 stone in the heap.\n",
+        "   \n",
+        "   - Inside the loop, we:\n",
+        "     - Retrieve and negate the heaviest stone (`y`) from the heap.\n",
+        "     - Retrieve and negate the second heaviest stone (`x`) from the heap.\n",
+        "   \n",
+        "     - Calculate the new stone's weight after smashing (`y - x`), and negate it back to simulate a max heap.\n",
+        "     - Add the new stone back to the heap using `heapq.heappush`.\n",
+        "   \n",
+        "3. After the loop, if there is one stone left in the heap, we retrieve and negate it to get its actual weight, and return it as the result.\n",
+        "   \n",
+        "4. If there are no stones left (the heap is empty), we return `0`, indicating that no stones remain.\n",
+        "\n",
+        "5. Finally, we provide two test cases to demonstrate the function's usage and correctness.\n",
+        "\n",
+        "In summary, the code efficiently implements the stone smashing game by maintaining a max heap, selecting the heaviest stones in each turn, smashing them, and updating the heap until there is at most one stone left. It then returns the weight of the last remaining stone or 0 if there are no stones left."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 15,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "1\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "stones1 = [2, 7, 4, 1, 8, 1]\n",
+        "print(lastStoneWeight(stones1))  # Output: 1"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 16,
+      "id": "b7e79514-4821-48d0-bb9d-0f2de8150945",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "1\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2: \n",
+        "\n",
+        "stones2 = [1]\n",
+        "print(lastStoneWeight(stones2))  # Output: 1"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided Python code for the \"Last Stone Weight\" problem:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "1. The `heapq.heapify` operation to create the max heap from the input stone weights has a time complexity of O(n), where n is the number of stones.\n",
+        "\n",
+        "2. The loop that continues until there are more than 1 stone in the heap performs the following operations in each iteration:\n",
+        "   - Retrieving and negating the heaviest stone (`y`) from the heap, which takes O(log n) time.\n",
+        "   - Retrieving and negating the second heaviest stone (`x`) from the heap, also taking O(log n) time.\n",
+        "   - Calculating the new stone's weight and pushing it back into the heap using `heapq.heappush`, which takes O(log n) time.\n",
+        "\n",
+        "3. In the worst case, the loop runs until there is only one stone left, which requires approximately (n-1) iterations.\n",
+        "\n",
+        "Overall, the dominant time complexity is O(n) for the initial heap creation, and within the loop, each iteration takes O(log n) time. Therefore, the overall time complexity of the algorithm is O(n log n).\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "1. The space complexity is primarily determined by the space required for the max heap. In the worst case, this heap can contain all the stones, so the space complexity for the heap is O(n).\n",
+        "\n",
+        "2. The rest of the variables used in the function (e.g., `x`, `y`, `new_stone`) require only constant space, so they do not significantly contribute to the space complexity.\n",
+        "\n",
+        "Therefore, the overall space complexity of the algorithm is O(n) due to the space used for the max heap.\n",
+        "\n",
+        "In summary:\n",
+        "- Time Complexity: $O(n\\ log\\ n)$\n",
+        "- Space Complexity: $O(n)$\n",
+        "\n",
+        "The algorithm is efficient enough to handle the problem's constraints, as the worst-case time and space complexities are both linearithmic in terms of the number of stones."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Kth Last Stone Weight**: Modify the problem to find the weight of the Kth last remaining stone, where K is an integer input. Your function should work efficiently for large values of K.\n",
+        "\n",
+        "\n",
+        "2. **Multiple Stones Smash**: Modify the problem so that instead of smashing two stones at a time, you can smash up to K stones at each turn, where K is an integer input. Determine the weight of the last remaining stone in this variation."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb b/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb
new file mode 100644
index 0000000..847dd60
--- /dev/null
+++ b/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb	
@@ -0,0 +1,242 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 295. Find Median from Data Stream\n",
+        "\n",
+        "**Difficulty:** Hard\n",
+        "\n",
+        "**Link to Problem:** [To see the Find Median from Data Stream problem on LeetCode, click here!](https://leetcode.com/problems/find-median-from-data-stream/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.\n",
+        "\n",
+        "Implement the MedianFinder class:\n",
+        "\n",
+        "- `MedianFinder()` initializes the `MedianFinder` object.\n",
+        "- `void addNum(int num)` adds the integer `num` from the data stream to the data structure.\n",
+        "- `double findMedian()` returns the median of all elements so far. Answers within $10^{-5}$ of the actual answer will be accepted.\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- $-10^{5}$ <= `num` <= $10^{5}$\n",
+        "- There will be at least one element in the data structure before calling `findMedian`.\n",
+        "- At most $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
+        "\n",
+        "**Follow-up:**\n",
+        "\n",
+        "1. If all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?\n",
+        "2. If `99%` of all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem, \"Find Median from Data Stream,\" deals with efficiently computing the median of a sequence of numbers as they are incrementally provided. The median is the middle value in an ordered list of numbers. When the list has an even number of elements, the median is the average of the two middle values.\n",
+        "\n",
+        "Here's a breakdown of the problem:\n",
+        "\n",
+        "1. **Initialization**: You are asked to implement a `MedianFinder` class that supports the following operations:\n",
+        "\n",
+        "    - `MedianFinder()`: Initializes the MedianFinder object.\n",
+        "    - `addNum(int num)`: Adds an integer `num` from the data stream to the data structure.\n",
+        "    - `findMedian()`: Returns the median of all elements added to the data structure.\n",
+        "\n",
+        "2. **Median Calculation**: The `findMedian` operation should efficiently compute the median, and the result should have a precision of at least $10^{-5}$ (i.e., answers within this range will be considered correct).\n",
+        "\n",
+        "3. **Examples**:\n",
+        "   \n",
+        "   - If you add numbers [1, 2], the median should be 1.5 because (1 + 2) / 2 = 1.5.\n",
+        "   - If you add numbers [1, 2, 3], the median should be 2 because it's the middle value.\n",
+        "   \n",
+        "4. **Constraints**:\n",
+        "\n",
+        "   - The integers provided in the data stream are in the range from $-10^5$ to $10^5$.\n",
+        "   - There will be at least one element in the data structure before calling `findMedian`.\n",
+        "   - At most, $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
+        "\n",
+        "The problem can be solved by using two priority queues (heaps): one max-heap to store the smaller half of the numbers and one min-heap to store the larger half of the numbers. The max-heap ensures that the largest number in the smaller half is at the top, and the min-heap ensures that the smallest number in the larger half is at the top. These heaps are balanced to efficiently find the median when requested.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 1,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "import heapq\n",
+        "\n",
+        "class MedianFinder:\n",
+        "\n",
+        "    def __init__(self):\n",
+        "        # Initialize two heaps: a max-heap (small_half) for the smaller half of numbers,\n",
+        "        # and a min-heap (large_half) for the larger half of numbers.\n",
+        "        self.small_half = []  # Max-heap for the smaller half of the numbers\n",
+        "        self.large_half = []  # Min-heap for the larger half of the numbers\n",
+        "\n",
+        "    def addNum(self, num: int) -> None:\n",
+        "        # Add the number to the appropriate heap based on its value\n",
+        "        if not self.small_half or num <= -self.small_half[0]:\n",
+        "            # If the number is less than or equal to the current maximum in the smaller half,\n",
+        "            # add it to the smaller half (max-heap)\n",
+        "            heapq.heappush(self.small_half, -num)\n",
+        "        else:\n",
+        "            # Otherwise, add it to the larger half (min-heap)\n",
+        "            heapq.heappush(self.large_half, num)\n",
+        "\n",
+        "        # Balance the heaps if necessary to ensure the size difference is at most 1\n",
+        "        if len(self.small_half) > len(self.large_half) + 1:\n",
+        "            # If the size of the smaller half is more than one greater than the size of the larger half,\n",
+        "            # move the maximum from the smaller half to the larger half to balance them.\n",
+        "            heapq.heappush(self.large_half, -heapq.heappop(self.small_half))\n",
+        "        elif len(self.large_half) > len(self.small_half):\n",
+        "            # If the size of the larger half is greater than the size of the smaller half,\n",
+        "            # move the minimum from the larger half to the smaller half to balance them.\n",
+        "            heapq.heappush(self.small_half, -heapq.heappop(self.large_half))\n",
+        "\n",
+        "    def findMedian(self) -> float:\n",
+        "        if len(self.small_half) == len(self.large_half):\n",
+        "            # If both halves have the same size, there's an even number of elements,\n",
+        "            # so the median is the average of the top elements in both heaps.\n",
+        "            return (-self.small_half[0] + self.large_half[0]) / 2.0\n",
+        "        else:\n",
+        "            # If the smaller half has more elements, it contains the median (odd number of elements).\n",
+        "            return -self.small_half[0]"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The code defines a `MedianFinder` class for efficiently finding the median of a stream of numbers. It uses two heaps: a max-heap (`small_half`) to store the smaller half of the numbers and a min-heap (`large_half`) to store the larger half. Here's how the code works:\n",
+        "\n",
+        "1. The `__init__` method initializes the two heaps.\n",
+        "\n",
+        "2. The `addNum` method adds a number to the appropriate heap, ensuring that the heaps remain balanced. It does this by comparing the number to the current maximum in the smaller half and adding it to the appropriate heap. Then, it checks the sizes of the two heaps and rebalances them if necessary.\n",
+        "\n",
+        "3. The `findMedian` method calculates and returns the median. If both halves have the same size (even number of elements), it computes the average of the tops of both heaps. If the smaller half has more elements (odd number of elements), it returns the top of the max-heap, which is the median.\n",
+        "\n",
+        "In summary, this code efficiently maintains two heaps to divide the elements into smaller and larger halves, allowing for quick median retrieval. It offers a time complexity of O(log N) for adding elements and O(1) for finding the median, where N is the total number of elements processed. The space complexity is O(1) as the space used by the two heaps is independent of the input size."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 2,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "1.5\n",
+            "2\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Example usage:\n",
+        "medianFinder = MedianFinder()\n",
+        "medianFinder.addNum(1)\n",
+        "medianFinder.addNum(2)\n",
+        "print(medianFinder.findMedian())  # Output: 1.5\n",
+        "medianFinder.addNum(3)\n",
+        "print(medianFinder.findMedian())  # Output: 2.0\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `MedianFinder` class:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "1. `MedianFinder()` Initialization: This operation has a time complexity of O(1). It simply initializes the two heaps.\n",
+        "\n",
+        "2. `addNum(int num)`: The time complexity for adding a number is O(log N), where N is the total number of elements processed so far. This is because, in the worst case, you might need to perform logarithmic operations on the heaps to maintain their balance.\n",
+        "\n",
+        "3. `findMedian()`: Finding the median has a time complexity of O(1). It simply involves accessing the tops of the two heaps, which takes constant time.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "1. `MedianFinder()` Initialization: The space complexity for initialization is O(1). It involves creating two empty heaps, which do not depend on the size of the data stream.\n",
+        "\n",
+        "2. `addNum(int num)`: The space complexity for adding a number is also O(1). It involves adding the number to one of the two heaps, which do not significantly affect the overall space complexity.\n",
+        "\n",
+        "3. `findMedian()`: The space complexity for finding the median is O(1). It doesn't require any additional data structures that grow with the input size.\n",
+        "\n",
+        "In summary, the time complexity of the `addNum` operation is O(log N), and the time complexity of the `findMedian` operation is O(1). The space complexity is O(1) as well, as the space used by the two heaps is independent of the size of the data stream. This solution efficiently maintains and retrieves the median, making it suitable for the given problem constraints."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Generalized kth Element**: Extend the `MedianFinder` class to support finding the kth element in the data stream efficiently. This means finding the kth smallest or kth largest element in the stream.\n",
+        "\n",
+        "2. **Sliding Window Median**: Given an array and a sliding window size, find the median within the window as it slides over the array. This is an extension of the problem to a moving window scenario."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb b/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb
new file mode 100644
index 0000000..2e1522d
--- /dev/null
+++ b/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb	
@@ -0,0 +1,239 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 703. Kth Largest Element in a Stream\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Kth Largest Element in a Stream problem on LeetCode, click here!](https://leetcode.com/problems/kth-largest-element-in-a-stream/)\n",
+        "\n",
+        "---\n",
+        "Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.\n",
+        "\n",
+        "Implement the `KthLargest` class with the following methods:\n",
+        "\n",
+        "- `KthLargest(int k, int[] nums)`: Initializes the object with the integer `k` and the stream of integers `nums`.\n",
+        "- `int add(int val)`: Appends the integer `val` to the stream and returns the element representing the `kth` largest element in the stream.\n",
+        "\n",
+        "\n",
+        "**Constraints:**\n",
+        "\n",
+        "- 1 <= `k` <= $10^4$\n",
+        "- 0 <= `nums.length` <= $10^4$\n",
+        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
+        "- $-10^4$ <= `val` <= $10^4$\n",
+        "- At most $10^4$ calls will be made to `add`.\n",
+        "- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem you're trying to solve is to design a class called `KthLargest` that can efficiently find the kth largest element in a stream of integers. You need to support two main operations:\n",
+        "\n",
+        "To solve this problem efficiently, you can use a min-heap data structure. Here's how the approach works:\n",
+        "\n",
+        "1. Initialize a min-heap to store the k largest elements. Initially, it's empty.\n",
+        "\n",
+        "2. In the `__init__` method, populate the min-heap with the first `k` elements from `nums`. This ensures that you have the k largest elements seen so far.\n",
+        "\n",
+        "3. Whenever you add a new element to the stream using the `add` method, follow these steps:\n",
+        "   - Add the new element to the min-heap.\n",
+        "   - If the size of the min-heap exceeds `k`, remove the smallest element from the min-heap. This ensures that you always have the k largest elements in the min-heap.\n",
+        "   - The smallest element in the min-heap (the root) will always represent the kth largest element in the stream.\n",
+        "\n",
+        "Here's why this approach works efficiently:\n",
+        "\n",
+        "- By using a min-heap, you can quickly maintain the k largest elements, and finding the smallest element in the heap (the root) takes constant time.\n",
+        "\n",
+        "- When you add a new element, the min-heap's size is kept at most `k`, which ensures that you only track the k largest elements and discard the smaller ones.\n",
+        "\n",
+        "- The time complexity for adding an element is O(log k), which is very efficient compared to sorting the entire stream, which would be O(n log n).\n",
+        "\n",
+        "- This approach meets the constraints of the problem, including handling large streams and having a low time complexity for both initialization and adding elements.\n",
+        "\n",
+        "In summary, the min-heap approach efficiently tracks the kth largest element in the stream by maintaining a heap of the k largest elements seen so far, updating it as new elements are added. This approach provides a fast and scalable solution to the problem.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 12,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "import heapq\n",
+        "from typing import List  # Import the List type from the typing module\n",
+        "\n",
+        "class KthLargest:\n",
+        "\n",
+        "    def __init__(self, k: int, nums: List[int]):\n",
+        "        # Initialize the KthLargest object with k and nums.\n",
+        "        self.k = k\n",
+        "        # Create a min-heap to store the k largest elements.\n",
+        "        self.min_heap = []\n",
+        "        # Populate the min-heap with the first k elements from nums.\n",
+        "        for num in nums:\n",
+        "            self.add(num)\n",
+        "\n",
+        "    def add(self, val: int) -> int:\n",
+        "        # Add val to the min-heap.\n",
+        "        heapq.heappush(self.min_heap, val)\n",
+        "        # If the size of the min-heap exceeds k, remove the smallest element.\n",
+        "        if len(self.min_heap) > self.k:\n",
+        "            heapq.heappop(self.min_heap)\n",
+        "        # The root of the min-heap is the kth largest element.\n",
+        "        return self.min_heap[0]"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "- We start by importing the necessary modules:\n",
+        "  - `heapq`: This module provides functions to create and manipulate heaps.\n",
+        "  - `List` from the `typing` module: This is used to specify the type of the `nums` parameter.\n",
+        "\n",
+        "- We define the `KthLargest` class:\n",
+        "  - The `__init__` method initializes the object with the integer `k` and the list of integers `nums`.\n",
+        "  - It also creates an empty min-heap called `self.min_heap` to store the k largest elements.\n",
+        "  - It populates the min-heap with the first `k` elements from `nums` by calling the `add` method.\n",
+        "\n",
+        "- The `add` method:\n",
+        "  - Adds the new integer `val` to the min-heap using `heapq.heappush`. This maintains the min-heap property.\n",
+        "  - Checks if the size of the min-heap exceeds `k`. If it does, it removes the smallest element (the k+1th largest) using `heapq.heappop`.\n",
+        "  - Finally, it returns the smallest element in the min-heap, which is always the kth largest element.\n",
+        "\n",
+        "Overall, this code implements a class that efficiently finds the kth largest element in a stream of integers by maintaining a min-heap of the k largest elements seen so far.\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 13,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "4\n",
+            "5\n",
+            "5\n",
+            "8\n",
+            "8\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Example usage:\n",
+        "kthLargest = KthLargest(3, [4, 5, 8, 2])\n",
+        "print(kthLargest.add(3))  # Output: 4\n",
+        "print(kthLargest.add(5))  # Output: 5\n",
+        "print(kthLargest.add(10)) # Output: 5\n",
+        "print(kthLargest.add(9))  # Output: 8\n",
+        "print(kthLargest.add(4))  # Output: 8"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `KthLargest` class using the min-heap approach:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "1. `__init__` method:\n",
+        "   - In the `__init__` method, you iterate over the first `k` elements in `nums` and add them to the min-heap. Each `heapq.heappush` operation takes O(log k) time.\n",
+        "   - Therefore, the time complexity of the `__init__` method is O(k * log k).\n",
+        "\n",
+        "2. `add` method:\n",
+        "   - In the `add` method, you perform the following operations:\n",
+        "     - `heapq.heappush`: O(log k) to add the new element to the min-heap.\n",
+        "     - If the size of the min-heap exceeds `k`, you perform `heapq.heappop`, which is also O(log k).\n",
+        "     - Finally, you return the smallest element from the min-heap, which is O(1) because it's always the root.\n",
+        "   - Overall, the time complexity of the `add` method is O(log k).\n",
+        "\n",
+        "3. Overall, if you make `n` calls to the `add` method, the total time complexity is O(n * log k), where `n` is the total number of elements added to the stream.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "1. The space complexity is determined by the space used to store the min-heap and the instance variables.\n",
+        "2. The min-heap stores at most `k` elements at any given time, so its space complexity is O(k).\n",
+        "3. The instance variables (such as `self.k` and `self.min_heap`) have constant space requirements.\n",
+        "4. Therefore, the overall space complexity of the `KthLargest` class is O(k).\n",
+        "\n",
+        "In summary, the time complexity of the `KthLargest` class is O(n * log k) for `n` add operations, and the space complexity is O(k), where `k` is the parameter passed during initialization. This implementation efficiently maintains the kth largest element in the stream while meeting the problem's constraints."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Dynamic k:** Modify the `KthLargest` class to support dynamic changes in the value of `k`. Implement a method `update_k` that allows changing the value of `k` during the lifetime of the object. Ensure that the object can still correctly find the kth largest element based on the updated value of `k`.\n",
+        "\n",
+        "\n",
+        "2. **Implement kth Smallest:** Create a new class called `KthSmallest` that finds the kth smallest element in a stream instead of the kth largest. You may need to modify the data structure used in the implementation."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/09. Heap - Priority Queue/README.md b/_build/html/_sources/09. Heap - Priority Queue/README.md
new file mode 100644
index 0000000..2eac70c
--- /dev/null
+++ b/_build/html/_sources/09. Heap - Priority Queue/README.md	
@@ -0,0 +1,18 @@
+# Heap / Priority Queue Problems - Blind 75 LeetCode 
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [703. Kth Largest Element in a Stream](https://leetcode.com/problems/kth-largest-element-in-a-stream/)     | Easy       |
+| [1046. Last Stone Weight](https://leetcode.com/problems/last-stone-weight/)     | Easy       |
+| [295. Find Median From Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)     | Hard       |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
diff --git a/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb b/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb
new file mode 100644
index 0000000..228a62e
--- /dev/null
+++ b/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb	
@@ -0,0 +1,282 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 39. Combination Sum\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Combination Sum problem on LeetCode, click here!](https://leetcode.com/problems/combination-sum/)\n",
+        "\n",
+        "---\n",
+        "Given an array of **distinct** integers `candidates` and a target integer `target`, return *a list of all **unique combinations** of `candidates` where the chosen numbers sum to the `target`*. You may return the combinations in **any order**. \n",
+        "\n",
+        "The **same** number may be chosen from `candidates` an **unlimited number** of times. Two combinations are unique if the **frequency** of at least one of the chosen numbers is different.\n",
+        "\n",
+        "The test cases are generated such that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `1 <= candidates.length <= 30`\n",
+        "- `2 <= candidates[i] <= 40`\n",
+        "- All elements of `candidates` are **distinct**.\n",
+        "- `1 <= target <= 40`"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem is to find all unique combinations of numbers from a given array (candidates) such that their sum equals a target number. Here are the details of the problem:\n",
+        "\n",
+        "- You are given an array of distinct integers called \"candidates.\"\n",
+        "- You are also given a target integer called \"target.\"\n",
+        "\n",
+        "The goal is to find all unique combinations of numbers from the candidates array where the sum of the selected numbers is equal to the target. You can use the same number from the candidates array an unlimited number of times. A combination is considered unique if it has a different frequency (i.e., a different number of occurrences) of at least one chosen number compared to other combinations.\n",
+        "\n",
+        "For example:\n",
+        "\n",
+        "**Example 1:**\n",
+        "```python\n",
+        "Input: candidates = [2, 3, 6, 7], target = 7\n",
+        "Output: [[2, 2, 3], [7]]\n",
+        "```\n",
+        "In this example, there are two unique combinations that sum up to the target:\n",
+        "- 2 + 2 + 3 = 7\n",
+        "- 7 = 7\n",
+        "\n",
+        "**Example 2:**\n",
+        "```python\n",
+        "Input: candidates = [2, 3, 5], target = 8\n",
+        "Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n",
+        "```\n",
+        "Here, there are three unique combinations that sum up to the target.\n",
+        "\n",
+        "**Example 3:**\n",
+        "```python\n",
+        "Input: candidates = [2], target = 1\n",
+        "Output: []\n",
+        "```\n",
+        "In this case, there are no combinations that can be formed from the candidates to reach the target of 1, so the output is an empty list.\n",
+        "\n",
+        "The problem asks you to find and return these combinations in any order.\n",
+        "\n",
+        "The constraints for this problem include the length of the candidates array, the values of candidates, and the target value.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 6,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def combinationSum(candidates, target):\n",
+        "    # Define a recursive DFS function to find combinations\n",
+        "    def dfs(remaining, start, path):\n",
+        "        # If the remaining target is 0, we found a valid combination\n",
+        "        if remaining == 0:\n",
+        "            result.append(path)\n",
+        "            return\n",
+        "        # If the remaining target is negative, this path is invalid\n",
+        "        if remaining < 0:\n",
+        "            return\n",
+        "        # Iterate through candidates starting from 'start' to avoid duplicates\n",
+        "        for i in range(start, len(candidates)):\n",
+        "            # Explore the current candidate by subtracting it from the remaining target\n",
+        "            # Add the current candidate to the path\n",
+        "            dfs(remaining - candidates[i], i, path + [candidates[i]])\n",
+        "\n",
+        "    result = []  # Initialize an empty list to store the result\n",
+        "    candidates.sort()  # Sort the candidates for deduplication and early stopping\n",
+        "    dfs(target, 0, [])  # Start the DFS from the target value with an empty path\n",
+        "    return result  # Return the list of unique combinations"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The `combinationSum` function takes an array of distinct integers `candidates` and a target integer `target`. It returns a list of all unique combinations of `candidates` where the chosen numbers sum to the `target`. Each number in the `candidates` array can be used an unlimited number of times.\n",
+        "\n",
+        "Here's an overview of how the code works:\n",
+        "\n",
+        "1. The `combinationSum` function is defined, and it takes two arguments: `candidates` and `target`.\n",
+        "\n",
+        "2. Inside the function, there is a nested helper function called `backtrack`. This function is responsible for the actual combination generation using a recursive approach.\n",
+        "\n",
+        "3. The `backtrack` function is called with three arguments: `start`, `target`, and `path`. The `start` variable helps keep track of the current index in the `candidates` array, `target` keeps track of the remaining sum to be achieved, and `path` is a list that stores the current combination.\n",
+        "\n",
+        "4. Within the `backtrack` function, there are three main conditional statements:\n",
+        "   - If `target` becomes zero, it means we have found a combination that sums up to the target, so we add `path` to the `result` list.\n",
+        "   - If `target` becomes negative, it means the current combination doesn't work, so we return without doing anything.\n",
+        "   - Otherwise, we enter a loop that iterates over the `candidates` array, starting from the current index `start`.\n",
+        "\n",
+        "5. In the loop, the `backtrack` function is called recursively with the updated `target` and `path` after including the current candidate. This process explores different combinations by considering the current candidate or moving to the next candidate.\n",
+        "\n",
+        "6. After the loop completes, the `result` list contains all unique combinations that sum to the target.\n",
+        "\n",
+        "7. The `candidates` array is sorted to optimize the search. Sorting helps in avoiding unnecessary recursive branches and reduces the number of explored combinations.\n",
+        "\n",
+        "8. Finally, the `backtrack` function is initially called from the `combinationSum` function with `start` set to 0, `target` set to the original target value, and an empty `path`. The result is returned as a list of unique combinations.\n",
+        "\n",
+        "The code includes a few example cases to demonstrate how the function works with different inputs."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 7,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[[2, 2, 3], [7]]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Example 1\n",
+        "candidates1 = [2, 3, 6, 7]\n",
+        "target1 = 7\n",
+        "print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 8,
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "candidates2 = [2, 3, 5]\n",
+        "target2 = 8\n",
+        "print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 9,
+      "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3\n",
+        "candidates3 = [2]\n",
+        "target3 = 1\n",
+        "print(combinationSum(candidates3, target3))  # Output: []"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `combinationSum` function.\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "The time complexity of this function is influenced by the number of recursive calls made by the `backtrack` function and the amount of work done within each call. In the worst case, the function will explore all possible combinations.\n",
+        "\n",
+        "The worst-case scenario occurs when we have many combinations to reach the target. The time complexity can be expressed as $O(2^n)$, where 'n' is the maximum number of recursive calls. In this context, 'n' corresponds to the number of elements in the `candidates` list, and it may be up to 30 (as per the constraints). However, the actual number of combinations explored is generally much smaller because we eliminate branches when the target becomes negative, and we skip over candidates that cannot lead to a solution. The sorting step also helps to optimize the search.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "The space complexity is determined by the auxiliary space used for storing the results and the stack space for the recursive calls.\n",
+        "\n",
+        "1. The space required for the `result` list can be up to $O(2^n)$ in the worst case because each combination is stored.\n",
+        "\n",
+        "2. The stack space for the recursive calls can also be up to O(n) in the worst case, where 'n' is the number of elements in the `candidates` list.\n",
+        "\n",
+        "Therefore, the overall space complexity of the function is $O(2^n + n)$. However, in practice, the space used for the results list is often the dominant factor, and the actual space used may be much smaller than 2^n because not all combinations are explored.\n",
+        "\n",
+        "In summary, the time complexity is exponential but typically smaller in practice due to optimization, and the space complexity is influenced by the number of results and the depth of the recursion."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Find the Number of Unique Combinations:** Instead of listing all unique combinations, modify the function to return the count of unique combinations that sum to the target.\n",
+        "\n",
+        "2. **Combinations with a Maximum Value:** Add a maximum value constraint for each combination, so they cannot exceed a certain value. Modify the code to find combinations respecting this constraint."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/10. Backtracking/79. Word Search.ipynb b/_build/html/_sources/10. Backtracking/79. Word Search.ipynb
new file mode 100644
index 0000000..cee509b
--- /dev/null
+++ b/_build/html/_sources/10. Backtracking/79. Word Search.ipynb	
@@ -0,0 +1,298 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 79. Word Search\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Word Search problem on LeetCode, click here!](https://leetcode.com/problems/word-search/)\n",
+        "\n",
+        "---\n",
+        "Given an `m x n` grid of characters `board` and a string `word`, return *`true` if `word` exists in the grid.*\n",
+        "\n",
+        "The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `m == board.length`\n",
+        "- `n = board[i].length`\n",
+        "- `1 <= m, n <= 6`\n",
+        "- `1 <= word.length <= 15`\n",
+        "- `board` and `word` consists of only lowercase and uppercase English letters.\n",
+        "\n",
+        "**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem is to determine whether a given word can be constructed from characters in an `m x n` grid of characters. The word can be formed by moving sequentially through horizontally or vertically neighboring cells in the grid, and you cannot use the same cell more than once.\n",
+        "\n",
+        "Here are the specifics of the problem:\n",
+        "\n",
+        "- You are given an `m x n` grid, which is a 2D board of characters.\n",
+        "- You are also given a string `word` consisting of lowercase and uppercase English letters.\n",
+        "- Your task is to determine if the word can be found in the grid by moving from one cell to an adjacent cell (horizontally or vertically) and collecting the characters sequentially to form the word.\n",
+        "- You can't use the same cell more than once when forming the word.\n",
+        "- If it's possible to form the word in the grid, return `True`; otherwise, return `False`.\n",
+        "\n",
+        "For example, consider the following grid and word:\n",
+        "\n",
+        "```python\n",
+        "board = [[\"A\", \"B\", \"C\", \"E\"],\n",
+        "         [\"S\", \"F\", \"C\", \"S\"],\n",
+        "         [\"A\", \"D\", \"E\", \"E\"]]         \n",
+        "word = \"ABCCED\"\n",
+        "```\n",
+        "\n",
+        "In this case, the word \"ABCCED\" can be formed by moving through the cells (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) in the grid. Therefore, the function should return `True`.\n",
+        "\n",
+        "Conversely, if the word is impossible to construct, such as in the case of the word \"ABCB\" for the same grid, where there's no valid path that forms the word, the function should return `False`.\n",
+        "\n",
+        "The problem involves searching for a path through the grid that matches the characters in the word while respecting the constraints mentioned. It can be solved using depth-first search (DFS) or backtracking.\n",
+        "\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 3,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "from typing import List\n",
+        "\n",
+        "class Solution:\n",
+        "    def exist(self, board: List[List[str]], word: str) -> bool:\n",
+        "        def dfs(i, j, k):\n",
+        "            # Check if the current position is out of bounds or doesn't match the character in the word.\n",
+        "            if not (0 <= i < len(board)) or not (0 <= j < len(board[0])) or board[i][j] != word[k]:\n",
+        "                return False\n",
+        "            \n",
+        "            # If we have matched all characters in the word, return True.\n",
+        "            if k == len(word) - 1:\n",
+        "                return True\n",
+        "\n",
+        "            # Temporarily mark the current cell to prevent revisiting.\n",
+        "            tmp, board[i][j] = board[i][j], \"/\"\n",
+        "\n",
+        "            # Explore adjacent cells by making recursive calls in all four directions.\n",
+        "            found = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)\n",
+        "\n",
+        "            # Restore the original character in the cell.\n",
+        "            board[i][j] = tmp\n",
+        "\n",
+        "            return found\n",
+        "\n",
+        "        # Iterate through all cells in the board.\n",
+        "        for i in range(len(board)):\n",
+        "            for j in range(len(board[0])):\n",
+        "                # Check if the 'dfs' function starting from this cell returns True.\n",
+        "                if dfs(i, j, 0):\n",
+        "                    return True\n",
+        "\n",
+        "        # If no match is found after exploring all cells, return False.\n",
+        "        return False"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "1. Import the necessary module:\n",
+        "   - `from typing import List`: This line imports the `List` type from the `typing` module, which is used to specify the type of the `board` parameter in the method signature.\n",
+        "\n",
+        "2. Define the `Solution` class:\n",
+        "   - `class Solution:`: This line defines a class called `Solution` to encapsulate the solution for the word search problem.\n",
+        "\n",
+        "3. Define the `exist` method:\n",
+        "   - `def exist(self, board: List[List[str]], word: str) -> bool:`: This method is a member of the `Solution` class and takes two parameters: `board`, which is a 2D list of characters, and `word`, which is a string. It returns a boolean value, indicating whether the word exists in the board.\n",
+        "\n",
+        "4. Implement the inner `dfs` function:\n",
+        "   - `def dfs(i, j, k):`: This is a helper function used for depth-first search (DFS). It is called recursively to explore the board and search for the word. It takes three parameters: `i` and `j` representing the current position on the board, and `k` representing the current index in the `word` string.\n",
+        "\n",
+        "   - Inside `dfs`, it checks if the current position is out of bounds or if the character at the current position on the board does not match the character at the current index in the `word`. If either of these conditions is met, it returns `False`.\n",
+        "\n",
+        "   - If `k` is equal to the length of the `word` minus one, it means we have successfully matched all characters in the `word`, so it returns `True`.\n",
+        "\n",
+        "   - The function then temporarily replaces the character at the current position with a marker, \"/\", to prevent revisiting the same cell. It then explores adjacent cells by making recursive calls in all four directions: up, down, left, and right. If any of these recursive calls return `True`, it propagates the `True` result up the call stack. After the exploration is complete, it restores the original character in the cell.\n",
+        "\n",
+        "5. Iterate through the board:\n",
+        "   - The outer loop iterates through all cells in the `board` using two nested loops. For each cell, it checks if the `dfs` function starting from that cell (with `k` initially set to 0) returns `True`. If it does, it means the word exists, so the method returns `True`.\n",
+        "\n",
+        "6. Return `False`:\n",
+        "   - If the loop completes without finding the word in the board, the method returns `False`.\n",
+        "\n",
+        "To use this code, you can create an instance of the `Solution` class and call the `exist` method on that instance, passing the `board` and the `word` as arguments to check if the word exists in the board."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 4,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1: \n",
+        "\n",
+        "# Create an instance of the Solution class\n",
+        "solution = Solution()\n",
+        "\n",
+        "# Test case 1: Word \"ABCCED\" can be found in the grid\n",
+        "board1 = [[\"A\",\"B\",\"C\",\"E\"],\n",
+        "          [\"S\",\"F\",\"C\",\"S\"],\n",
+        "          [\"A\",\"D\",\"E\",\"E\"]]\n",
+        "word1 = \"ABCCED\"\n",
+        "print(solution.exist(board1, word1))  # Expected output: True"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 5,
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "\n",
+        "# Test case 2: Word \"SEE\" can be found in the grid\n",
+        "word2 = \"SEE\"\n",
+        "print(solution.exist(board1, word2))  # Expected output: True"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 6,
+      "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 3\n",
+        "\n",
+        "# Test case 3: Word \"ABCB\" cannot be found in the grid\n",
+        "word3 = \"ABCB\"\n",
+        "print(solution.exist(board1, word3))  # Expected output: False"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `exist` method in the given code.\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "The primary work in this code is done by the `dfs` function, which performs a depth-first search through the grid. The time complexity of the `dfs` function is $O(4^(n))$, where \"n\" is the length of the `word`. This is because, in the worst case, the function explores all possible directions (up, down, left, right) for each character in the `word`.\n",
+        "\n",
+        "The outer loops iterate through all cells in the `board`, which has dimensions `m x n`. So, the total number of iterations of `dfs` is O(m * n). However, since the `dfs` function has a higher time complexity, the overall time complexity is still dominated by the `dfs` function.\n",
+        "\n",
+        "Therefore, the overall time complexity of the `exist` method is $O(m * n * 4^n)$, where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "1. The primary space usage comes from the recursive calls in the `dfs` function. The depth of the recursion can be at most equal to the length of the `word`, which is O(n).\n",
+        "\n",
+        "2. Additionally, the `tmp` variable and the recursive calls on the call stack consume a small amount of additional memory.\n",
+        "\n",
+        "3. The marking of visited cells with the \"/\" character temporarily modifies the `board` in-place but doesn't significantly contribute to the space complexity.\n",
+        "\n",
+        "Therefore, the overall space complexity is O(n) due to the recursive call stack.\n",
+        "\n",
+        "In summary, the time complexity is $O(m * n * 4^n)$, and the space complexity is O(n), where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word. The time complexity is exponential in the worst case due to the depth-first search through all possible paths in the grid."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Optimization Challenge:** Modify the code to find all unique words that can be constructed from the grid. Instead of returning `True` or `False`, return a list of words found. Ensure that the same cell is not used more than once in constructing each word.\n",
+        "\n",
+        "2. **Board Generation:** Create a function to generate random grids of characters and then use the `exist` method to solve them. Test the efficiency and correctness of your code by generating larger boards and words."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/10. Backtracking/README.md b/_build/html/_sources/10. Backtracking/README.md
new file mode 100644
index 0000000..0d1ea13
--- /dev/null
+++ b/_build/html/_sources/10. Backtracking/README.md	
@@ -0,0 +1,17 @@
+# Backtracking Problems - Blind 75 LeetCode 
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [39. Combination Sum](https://leetcode.com/problems/combination-sum/)        | Medium     |
+| [79. Word Search](https://leetcode.com/problems/word-search/)                | Medium     |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb b/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb
new file mode 100644
index 0000000..07351b6
--- /dev/null
+++ b/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb	
@@ -0,0 +1,317 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 133. Clone Graph\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Clone Graph problem on LeetCode, click here!](https://leetcode.com/problems/clone-graph/)\n",
+        "\n",
+        "---\n",
+        "Given a reference of a node in a **connected** undirected graph.\n",
+        "\n",
+        "Return a **deep copy** (clone) of the graph.\n",
+        "\n",
+        "Each node in the graph contains a value (`int`) and a list (`List[Node]`) of its neighbors.\n",
+        "\n",
+        "```\n",
+        "class Node {\n",
+        "    public int val;\n",
+        "    public List<Node> neighbors;\n",
+        "}\n",
+        "```\n",
+        "\n",
+        "**Test case format:**\n",
+        "\n",
+        "For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with `val == 1`, the second node with `val == 2`, and so on. The graph is represented in the test case using an adjacency list.\n",
+        "\n",
+        "**An adjacency list** is a collection of unordered **lists** used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.\n",
+        "\n",
+        "The given node will always be the first node with `val = 1`. You must return the **copy of the given node** as a reference to the cloned graph.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- The number of nodes in the graph is in the range `[0, 100]`.\n",
+        "- `1 <= Node.val <= 100`\n",
+        "- `Node.val` is unique for each node.\n",
+        "- There are no repeated edges and no self-loops in the graph.\n",
+        "- The Graph is connected and all nodes can be visited starting from the given node."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem at hand is to clone a connected undirected graph, which means we need to create a deep copy of the original graph while preserving its structure and relationships between nodes. The graph is represented using adjacency lists, where each node has a value (an integer) and a list of its neighbors.\n",
+        "\n",
+        "Here are the key aspects of the problem:\n",
+        "\n",
+        "1. **Input**: You are given a reference to one of the nodes in the original graph. This reference node serves as the entry point to the graph. The entire graph can be explored starting from this node.\n",
+        "\n",
+        "2. **Output**: The goal is to create a deep copy of the graph and return a reference to the cloned graph. The cloned graph should have the same structure and relationships as the original one but should be a separate instance in memory.\n",
+        "\n",
+        "3. **Graph Structure**: The graph is composed of nodes (vertices) and edges. Nodes have values (integers) and are connected to other nodes through edges (undirected connections). Each node has a list of neighbors, which are other nodes it is connected to.\n",
+        "\n",
+        "4. **Connected Graph**: The problem assumes that the graph is connected, meaning that you can start from the reference node provided and traverse the entire graph by following the edges. This ensures that there are no isolated components in the graph.\n",
+        "\n",
+        "To solve this problem, you typically need to perform a graph traversal (e.g., depth-first search or breadth-first search) to visit all the nodes in the original graph, create corresponding nodes in the clone, and establish the same connections between nodes in the cloned graph as in the original one. To avoid duplication of nodes, you also need to keep track of visited nodes to ensure that the cloning process is efficient and that nodes are not duplicated in the clone.\n",
+        "\n",
+        "The problem can be challenging because you need to create a deep copy of the graph while handling cyclic dependencies between nodes and ensuring that the structure and relationships of the original graph are preserved in the clone. The solution requires recursion or a data structure to maintain the mapping between original nodes and their clones to achieve an efficient and accurate cloning process.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 5,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "# Definition for a Node.\n",
+        "class Node:\n",
+        "    def __init__(self, val=0, neighbors=None):\n",
+        "        self.val = val\n",
+        "        self.neighbors = neighbors if neighbors is not None else []\n",
+        "\n",
+        "class Solution:\n",
+        "    def cloneGraph(self, node: 'Node') -> 'Node':\n",
+        "        if not node:\n",
+        "            return None\n",
+        "\n",
+        "        visited = {}  # Dictionary to keep track of visited nodes\n",
+        "\n",
+        "        def dfs(original_node):\n",
+        "            if original_node in visited:\n",
+        "                return visited[original_node]\n",
+        "\n",
+        "            # Create a copy of the original node\n",
+        "            copy_node = Node(original_node.val)\n",
+        "            visited[original_node] = copy_node\n",
+        "\n",
+        "            # Recursively clone the neighbors of the original node\n",
+        "            for neighbor in original_node.neighbors:\n",
+        "                copy_node.neighbors.append(dfs(neighbor))\n",
+        "\n",
+        "            return copy_node\n",
+        "\n",
+        "        return dfs(node)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The given code defines a Python class, `Node`, and a solution class, `Solution`, to clone a connected undirected graph represented by nodes and their neighbors. Here's an explanation of the code:\n",
+        "\n",
+        "1. The `Node` class defines a node in the graph. Each node has a value (`val`) which is an integer and a list of neighbors (`neighbors`) that are other nodes in the graph.\n",
+        "\n",
+        "2. The `Solution` class contains the method `cloneGraph`, which takes a reference to the first node of the graph (`node`) as input and returns a deep copy of the entire graph, preserving the structure and relationships between nodes.\n",
+        "\n",
+        "3. Inside the `cloneGraph` method:\n",
+        "   - It first checks if the input `node` is `None`. If it's `None`, it means there is no graph to clone, and the method returns `None`.\n",
+        "\n",
+        "   - It initializes a `visited` dictionary to keep track of visited nodes. This dictionary is used to ensure that each node is cloned only once to avoid duplication.\n",
+        "\n",
+        "   - The `dfs` (depth-first search) function is defined within the `cloneGraph` method. It takes an original node as input and returns its corresponding clone. This function performs the actual cloning of the graph.\n",
+        "\n",
+        "   - Inside the `dfs` function:\n",
+        "     - It checks if the original node has been visited before by looking it up in the `visited` dictionary. If it has been visited, it returns the corresponding clone from the `visited` dictionary.\n",
+        "\n",
+        "     - If the original node is not visited, it creates a new node, `copy_node`, with the same value as the original node. This is the first step in creating the clone of the original node.\n",
+        "\n",
+        "     - It then recursively clones the neighbors of the original node by iterating through the `neighbors` list of the original node and appending the corresponding cloned neighbors to the `neighbors` list of the `copy_node`. This step ensures that the relationships between nodes are preserved in the clone.\n",
+        "\n",
+        "     - The `copy_node` is added to the `visited` dictionary with the original node as the key and the clone as the value.\n",
+        "\n",
+        "     - Finally, the `copy_node` is returned as the result of the DFS for the original node.\n",
+        "\n",
+        "4. The `cloneGraph` method returns the result of the DFS on the input `node`, which is the clone of the entire graph with its structure and relationships intact.\n",
+        "\n",
+        "In summary, the code uses depth-first search (DFS) to traverse the original graph, creating a clone for each node and its neighbors. It ensures that each node is cloned only once to avoid duplication and returns a reference to the cloned graph."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 7,
+      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[[1, 2, 4]]\n"
+          ]
+        }
+      ],
+      "source": [
+        "def get_adjacency_list(node):\n",
+        "    visited = {}  # Dictionary to keep track of visited nodes\n",
+        "\n",
+        "    def dfs(node):\n",
+        "        if not node:\n",
+        "            return []\n",
+        "\n",
+        "        if node in visited:\n",
+        "            return visited[node]\n",
+        "\n",
+        "        neighbors = [n.val for n in node.neighbors]\n",
+        "        visited[node] = neighbors\n",
+        "\n",
+        "        for neighbor in node.neighbors:\n",
+        "            dfs(neighbor)\n",
+        "\n",
+        "    dfs(node)\n",
+        "    result = []\n",
+        "\n",
+        "    def get_node_and_neighbors(node):\n",
+        "        result.append([node.val] + visited[node])\n",
+        "\n",
+        "    get_node_and_neighbors(node)\n",
+        "    return result\n",
+        "\n",
+        "\n",
+        "# Testing the code with the provided examples\n",
+        "adjList1 = [[2, 4], [1, 3], [2, 4], [1, 3]]\n",
+        "solution = Solution()\n",
+        "\n",
+        "# Create the graph from adjList\n",
+        "nodes = [Node(i) for i in range(1, len(adjList1) + 1)]\n",
+        "for i, adj in enumerate(adjList1):\n",
+        "    nodes[i].neighbors = [nodes[j - 1] for j in adj]\n",
+        "\n",
+        "# Clone the graph and get its adjacency list\n",
+        "cloned_node1 = solution.cloneGraph(nodes[0])\n",
+        "cloned_adjList1 = get_adjacency_list(cloned_node1)\n",
+        "\n",
+        "# Print the cloned graph as an adjacency list\n",
+        "print(cloned_adjList1)\n",
+        "\n",
+        "# Output should match adjList1\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 11,
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[[1]]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "# Input: adjList = [[]]\n",
+        "\n",
+        "adjList2 = [[]]\n",
+        "solution = Solution()\n",
+        "\n",
+        "# Create the graph from adjList\n",
+        "node1 = Node(1)\n",
+        "\n",
+        "# Clone the graph and get its adjacency list\n",
+        "cloned_node1 = solution.cloneGraph(node1)\n",
+        "cloned_adjList2 = get_adjacency_list(cloned_node1)\n",
+        "\n",
+        "# Print the cloned graph as an adjacency list\n",
+        "print(cloned_adjList2)\n",
+        "\n",
+        "# Output should be [[]], which represents an empty graph"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code for cloning a graph using depth-first search (DFS).\n",
+        "\n",
+        "Time Complexity:\n",
+        "1. Creating the graph: The time complexity to create the original graph from `adjList` is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. We iterate through each node and its neighbors in `adjList`.\n",
+        "\n",
+        "2. Cloning the graph (DFS): The depth-first search (DFS) is used to traverse the original graph and create a clone. The time complexity of DFS is O(V + E), as we visit each node and each edge once.\n",
+        "\n",
+        "Overall, the time complexity is O(V + E) for both creating the original graph and cloning it.\n",
+        "\n",
+        "Space Complexity:\n",
+        "1. Storage for the original graph: We need to store the original graph, which includes all nodes and their adjacency lists. In the worst case, this requires O(V + E) space, where V is the number of nodes, and E is the number of edges.\n",
+        "\n",
+        "2. Storage for the visited dictionary: The `visited` dictionary stores mappings from original nodes to their corresponding clones. In the worst case, this requires O(V) space because each node is visited once.\n",
+        "\n",
+        "3. Recursive stack for DFS: The space used for the recursive call stack during DFS can be up to O(V) in the worst case, where V is the number of nodes.\n",
+        "\n",
+        "Overall, the space complexity is O(V + E) for storing the original graph, and O(V) for auxiliary data structures, including the `visited` dictionary and the DFS call stack. Therefore, the overall space complexity is O(V + E).\n",
+        "\n",
+        "In summary, the time and space complexities of the provided code for cloning a graph are both O(V + E), where V is the number of nodes, and E is the number of edges in the graph."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Cyclic Graph**: Design a graph with cycles (nodes connected in a cycle). Test if the code can correctly clone graphs with cycles and doesn't fall into an infinite loop during traversal.\n",
+        "\n",
+        "\n",
+        "2. **Partial Graph Cloning**: Modify the code to clone only a subset of the graph starting from a given reference node, rather than the entire graph. Implement this as an extension to the problem."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb b/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb
new file mode 100644
index 0000000..e0235d6
--- /dev/null
+++ b/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb	
@@ -0,0 +1,248 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 178. Graph Valid Tree\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Graph Valid Tree problem on LintCode, click here!](https://www.lintcode.com/problem/178/)\n",
+        "\n",
+        "---\n",
+        "Given `n` nodes labeled from `0` to `n - 1` and a list of `undirected` edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- You can assume that no duplicate edges will appear in edges. Since all edges are `undirected`, `[0, 1]` is the same as `[1, 0]` and thus will not appear together in edges."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem is to determine whether a given set of undirected edges can form a valid tree. In this context, a \"tree\" is a specific type of graph with the following properties:\n",
+        "\n",
+        "1. It is a connected graph, meaning there is a path between any pair of nodes in the graph.\n",
+        "2. It has no cycles, which means there are no closed loops or circuits in the graph.\n",
+        "3. It has exactly `n - 1` edges, where `n` is the number of nodes in the tree.\n",
+        "\n",
+        "The problem is typically given with two pieces of information:\n",
+        "\n",
+        "1. The number of nodes (`n`), which are usually labeled from 0 to `n-1`.\n",
+        "2. A list of undirected edges, where each edge is represented as a pair of nodes.\n",
+        "\n",
+        "The task is to write a function or algorithm to determine whether the provided set of edges can be arranged to form a valid tree based on the properties mentioned above.\n",
+        "\n",
+        "To solve this problem, you need to check the following conditions:\n",
+        "\n",
+        "1. There should be exactly `n - 1` edges to ensure that the graph is connected and tree-like.\n",
+        "2. There should be no cycles in the graph. If there are any cycles, the set of edges cannot form a tree.\n",
+        "3. The graph should be connected, which means you can reach any node from any other node through a series of edges.\n",
+        "\n",
+        "The solution typically involves using graph traversal algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the graph and check these conditions. If all conditions are met, the given edges form a valid tree; otherwise, they do not.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 10,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def validTree(n, edges):\n",
+        "    if len(edges) != n - 1:\n",
+        "        return False  # A tree with n nodes must have n-1 edges.\n",
+        "\n",
+        "    # Create an adjacency list to represent the graph.\n",
+        "    adj_list = {i: [] for i in range(n)}\n",
+        "    for u, v in edges:\n",
+        "        adj_list[u].append(v)\n",
+        "        adj_list[v].append(u)\n",
+        "\n",
+        "    visited = set()\n",
+        "\n",
+        "    # Define a recursive DFS function.\n",
+        "    def dfs(node, parent):\n",
+        "        if node in visited:\n",
+        "            return False  # If we encounter a visited node, there's a cycle.\n",
+        "        visited.add(node)\n",
+        "\n",
+        "        for neighbor in adj_list[node]:\n",
+        "            if neighbor != parent:\n",
+        "                if not dfs(neighbor, node):\n",
+        "                    return False\n",
+        "\n",
+        "        return True\n",
+        "\n",
+        "    # Start DFS from any node (e.g., node 0).\n",
+        "    if not dfs(0, -1):\n",
+        "        return False  # If there's a cycle, it's not a valid tree.\n",
+        "\n",
+        "    return len(visited) == n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The `validTree` function checks if a given set of undirected edges forms a valid tree. A valid tree must satisfy these conditions:\n",
+        "\n",
+        "1. It has exactly `n - 1` edges, where `n` is the number of nodes.\n",
+        "2. There are no cycles in the graph.\n",
+        "3. The graph is connected, meaning all nodes are reachable from any starting node.\n",
+        "\n",
+        "The code accomplishes this as follows:\n",
+        "\n",
+        "- It first checks if the number of edges is equal to `n - 1`. If not, it immediately returns `False` since a tree with `n` nodes must have `n - 1` edges to be connected.\n",
+        "\n",
+        "- It then constructs an adjacency list (`adj_list`) to represent the graph efficiently. This data structure stores the neighbors of each node.\n",
+        "\n",
+        "- The code uses a `visited` set to keep track of nodes visited during depth-first search (DFS).\n",
+        "\n",
+        "- The `dfs` function is a recursive function that checks for cycles in the graph. It does this by visiting nodes and marking them as visited. If it encounters a node that has already been visited (indicating a cycle), it returns `False`.\n",
+        "\n",
+        "- The DFS function explores the neighbors of each node, avoiding revisiting the parent node to prevent cycles.\n",
+        "\n",
+        "- After the DFS is complete, the code checks if the number of visited nodes is equal to `n`, ensuring the graph is connected.\n",
+        "\n",
+        "- Finally, if all conditions are met, the code returns `True`, indicating that the given edges form a valid tree. Otherwise, it returns `False`.\n",
+        "\n",
+        "The code provides examples to demonstrate how to use the `validTree` function to check whether a set of edges forms a valid tree or not."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 11,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1\n",
+        "n1 = 5\n",
+        "edges1 = [[0, 1], [0, 2], [0, 3], [1, 4]]\n",
+        "print(validTree(n1, edges1))  # Output: true"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 12,
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "n2 = 5\n",
+        "edges2 = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]\n",
+        "print(validTree(n2, edges2))  # Output: false"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexities of the `validTree` function:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "1. Constructing the adjacency list (`adj_list`) takes O(E) time, where E is the number of edges in the input graph (|edges|).\n",
+        "\n",
+        "2. The DFS function, when considering all nodes, has a time complexity of O(V + E), where V is the number of nodes in the graph. This is because, in the worst case, the DFS function visits each node and each edge once.\n",
+        "\n",
+        "3. Checking if the number of visited nodes is equal to `n` takes O(1) time.\n",
+        "\n",
+        "Overall, the time complexity of the `validTree` function is O(E + V) in the worst case.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "1. The adjacency list (`adj_list`) consumes additional space to store the graph structure, taking O(V + E) space. This is because it stores information about each node and its adjacent nodes.\n",
+        "\n",
+        "2. The `visited` set keeps track of visited nodes, and its space complexity is O(V) in the worst case, as there can be at most V unique nodes in the set.\n",
+        "\n",
+        "3. The depth of the function call stack during DFS can be at most V, which contributes O(V) space to the space complexity.\n",
+        "\n",
+        "Overall, the space complexity of the `validTree` function is O(V + E) due to the adjacency list and O(V) due to the `visited` set and DFS call stack.\n",
+        "\n",
+        "In summary, the time complexity is O(E + V), and the space complexity is O(V + E). These complexities are based on the worst-case scenario, where all nodes and edges are considered. In practical cases, the actual complexities may be smaller, depending on the specific structure of the graph."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Minimum Spanning Tree (MST)**: Implement an algorithm to find the minimum spanning tree of a connected graph using either Kruskal's or Prim's algorithm. Compare the MST to the original graph to determine if the original graph is a tree. This exercise reinforces the concept of tree properties.\n",
+        "\n",
+        "2. **Detecting Cycles**: Modify the code to not only check if the input forms a tree but also to identify and print out any cycles present in the graph if it's not a tree. This exercise enhances your ability to detect and visualize cycles in a graph."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb b/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb
new file mode 100644
index 0000000..a8a7727
--- /dev/null
+++ b/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb	
@@ -0,0 +1,242 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 200. Number of Islands\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Number of Islands problem on LeetCode, click here!](https://leetcode.com/problems/number-of-islands/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "Given an `m x n` 2D binary grid `grid` which represents a map of `'1'`s (land) and `'0'`s (water), return *the number of islands*. \n",
+        "\n",
+        "An **island** is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `m == grid.length`\n",
+        "- `n == grid[i].length`\n",
+        "- `1 <= m, n <= 300`\n",
+        "- `grid[i][j]` is `'0'` or `'1'`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem is about counting the number of islands in a 2D binary grid. In this grid, '1' represents land, and '0' represents water. An island is defined as a group of '1' cells that are adjacent to each other either horizontally or vertically. It's important to note that diagonal connections do not count. Additionally, it's assumed that all four edges of the grid are surrounded by water, meaning the grid is finite and doesn't \"wrap around.\"\n",
+        "\n",
+        "The objective is to determine how many distinct islands exist in the grid. To clarify, an island consists of connected '1' cells, and these connections can only occur horizontally or vertically. If there are no '1' cells in the grid, the number of islands would be zero.\n",
+        "\n",
+        "The problem often involves using graph traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to identify and count these islands. You start at one '1' cell and explore its adjacent cells to determine the extent of the island. Once you've visited all the '1' cells in an island, you move on to the next unvisited '1' cell and repeat the process until you've counted all the islands in the grid.\n",
+        "\n",
+        "In essence, the problem asks you to analyze the grid to find groups of connected '1' cells and count them as individual islands. The goal is to return the total count of islands in the given grid.\n",
+        "\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 1,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "from typing import List\n",
+        "\n",
+        "class Solution:\n",
+        "    def numIslands(self, grid: List[List[str]]) -> int:\n",
+        "        if not grid:\n",
+        "            return 0\n",
+        "\n",
+        "        m, n = len(grid), len(grid[0])  # Get the dimensions of the grid\n",
+        "        num_islands = 0  # Initialize the count of islands\n",
+        "\n",
+        "        def dfs(row, col):\n",
+        "            if row < 0 or row >= m or col < 0 or col >= n or grid[row][col] == '0':\n",
+        "                return\n",
+        "\n",
+        "            grid[row][col] = '0'  # Mark the current cell as visited by changing it to '0'\n",
+        "            directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Define four possible directions to move\n",
+        "\n",
+        "            for dr, dc in directions:\n",
+        "                # Explore adjacent cells\n",
+        "                dfs(row + dr, col + dc)\n",
+        "\n",
+        "        for i in range(m):\n",
+        "            for j in range(n):\n",
+        "                if grid[i][j] == '1':\n",
+        "                    num_islands += 1  # Increment the island count\n",
+        "                    dfs(i, j)  # Start a DFS from the current '1' cell to explore the entire island\n",
+        "\n",
+        "        return num_islands\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The provided code defines a Python class called `Solution` with a method called `numIslands`. This class is designed to count the number of islands in a 2D binary grid.\n",
+        "\n",
+        "The `numIslands` method takes a 2D grid as input, where '1' represents land, and '0' represents water. It uses a depth-first search (DFS) approach to identify and count islands. The key steps in the code include:\n",
+        "\n",
+        "1. Checking if the input grid is empty, and if so, returning 0 (indicating there are no islands).\n",
+        "\n",
+        "2. Determining the dimensions of the grid (number of rows and columns).\n",
+        "\n",
+        "3. Initializing a variable `num_islands` to keep track of the count of islands.\n",
+        "\n",
+        "4. Defining a nested function `dfs` to perform depth-first search. This function recursively explores adjacent land cells connected to the current cell and marks them as visited by changing '1' to '0'.\n",
+        "\n",
+        "5. In the `dfs` function, it checks for boundary conditions and whether the current cell is '0'. If these conditions are met, it returns without further exploration.\n",
+        "\n",
+        "6. The `dfs` function explores neighboring cells in all four directions (up, down, left, right) using a loop.\n",
+        "\n",
+        "7. The main loop iterates through each cell of the grid. When a '1' cell is encountered, it increments the `num_islands` count and starts a DFS from that cell to explore the entire island.\n",
+        "\n",
+        "8. Finally, the method returns the total count of islands found in the grid.\n",
+        "\n",
+        "This code encapsulates the solution in a class, allowing you to create an instance of the `Solution` class and call the `numIslands` method on it to count the number of islands in a given grid."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 5,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Example 1 Output: 1\n"
+          ]
+        }
+      ],
+      "source": [
+        "grid1 = [\n",
+        "  [\"1\", \"1\", \"1\", \"1\", \"0\"],\n",
+        "  [\"1\", \"1\", \"0\", \"1\", \"0\"],\n",
+        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
+        "  [\"0\", \"0\", \"0\", \"0\", \"0\"]\n",
+        "]\n",
+        "\n",
+        "solution = Solution()\n",
+        "result1 = solution.numIslands(grid1)\n",
+        "print(\"Example 1 Output:\", result1)  # Expected output: 1\n"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 6,
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Example 2 Output: 3\n"
+          ]
+        }
+      ],
+      "source": [
+        "grid2 = [\n",
+        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
+        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
+        "  [\"0\", \"0\", \"1\", \"0\", \"0\"],\n",
+        "  [\"0\", \"0\", \"0\", \"1\", \"1\"]\n",
+        "]\n",
+        "\n",
+        "result2 = solution.numIslands(grid2)\n",
+        "print(\"Example 2 Output:\", result2)  # Expected output: 3\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code:\n",
+        "\n",
+        "Time Complexity:\n",
+        "- The code uses a depth-first search (DFS) to explore the grid and count the number of islands.\n",
+        "- In the worst case, it visits every cell in the grid exactly once.\n",
+        "- The DFS function explores neighboring cells in four directions (up, down, left, right), so it has a time complexity of $O(4^{max(m, n)})$, where m is the number of rows and n is the number of columns.\n",
+        "- Therefore, the overall time complexity of the code is O(m * n), where m is the number of rows, and n is the number of columns.\n",
+        "\n",
+        "Space Complexity:\n",
+        "- The space complexity is determined by the recursive calls and the depth of the DFS stack.\n",
+        "- In the worst case, when the entire grid consists of '1's, the depth of the DFS recursion can be as deep as max(m, n).\n",
+        "- Therefore, the space complexity of the code is O(max(m, n)), representing the maximum depth of the DFS stack.\n",
+        "\n",
+        "In practical terms, for reasonably sized grids, the time and space complexity is linear, and the code should perform well. However, it's important to keep in mind that the worst-case time complexity is O(m * n), and the space complexity can be O(max(m, n))."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Identify Largest Island:** In addition to counting the number of islands, modify the code to identify the largest island in the grid and return its size.\n",
+        "\n",
+        "2. **Optimize for Space:** Modify the solution to reduce its space complexity to O(1) without modifying the input grid. This is a more memory-efficient version of the problem."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb b/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb
new file mode 100644
index 0000000..37624f6
--- /dev/null
+++ b/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb	
@@ -0,0 +1,249 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 207. Course Schedule\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Course Schedule problem on LeetCode, click here!](https://leetcode.com/problems/course-schedule/)\n",
+        "\n",
+        "---\n",
+        "There are a total of `numCourses` courses to take, labeled from 0 to `numCourses - 1`. You are given an array `prerequisites` where `prerequisites[i] = [ai, bi]` indicates that you **must** take course `bi` first if you want to take course `ai`. \n",
+        "\n",
+        "- For example, the pair `[0, 1]`, indicates that to take course `0` you have to first take course `1`.\n",
+        "\n",
+        "Return `true` if you can finish all courses. Otherwise, return `false`.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- 1 <= `numCourses` <= 2000\n",
+        "- 0 <= `prerequisites.length` <= 5000\n",
+        "- `prerequisites[i].length` == 2\n",
+        "- 0 <= `ai`, `bi` < `numCourses`\n",
+        "- All the pairs `prerequisites[i]` are **unique**."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem is to determine whether it's possible to complete a set of courses given a list of prerequisites. You are given the following:\n",
+        "\n",
+        "- `numCourses`: The total number of courses to be taken, labeled from 0 to `numCourses - 1`.\n",
+        "- `prerequisites`: A list of prerequisite courses, where each element `prerequisites[i]` is a pair `[ai, bi]`, indicating that you must take course `bi` before you can take course `ai`.\n",
+        "\n",
+        "The goal is to check whether it's possible to complete all the courses while respecting the prerequisite requirements. In other words, you need to determine if there are any circular dependencies in the course prerequisites that would prevent you from taking all the courses.\n",
+        "\n",
+        "For example:\n",
+        "\n",
+        "- If `numCourses` is 2, and `prerequisites` is `[[1, 0]]`, it means you have two courses, and you must finish course 0 before you can take course 1. In this case, it's possible to complete all courses, so the function should return `True`.\n",
+        "\n",
+        "- If `numCourses` is 2, and `prerequisites` is `[[1, 0], [0, 1]]`, it means you have two courses, but there's a circular dependency where course 1 requires course 0 and course 0 requires course 1. In this case, it's impossible to complete all courses, so the function should return `False`.\n",
+        "\n",
+        "The problem is essentially about checking for the presence of cycles in a directed graph, where the courses are nodes, and the prerequisites represent directed edges between them. If there are no cycles, it's possible to complete all courses; otherwise, it's not possible.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 18,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "from collections import defaultdict, deque\n",
+        "from typing import List\n",
+        "\n",
+        "class Solution:\n",
+        "    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:\n",
+        "        # Create a graph using an adjacency list to represent the courses and their prerequisites.\n",
+        "        graph = defaultdict(list)  # Initialize an empty adjacency list.\n",
+        "        in_degree = [0] * numCourses  # Initialize an array to store in-degrees of courses.\n",
+        "\n",
+        "        # Populate the graph and calculate in-degrees.\n",
+        "        for course, prerequisite in prerequisites:\n",
+        "            graph[prerequisite].append(course)  # Add the course as a neighbor to its prerequisite.\n",
+        "            in_degree[course] += 1  # Increment the in-degree of the course.\n",
+        "\n",
+        "        # Initialize a queue with courses that have no prerequisites.\n",
+        "        queue = deque()\n",
+        "        for course in range(numCourses):\n",
+        "            if in_degree[course] == 0:\n",
+        "                queue.append(course)\n",
+        "\n",
+        "        # Perform topological sorting.\n",
+        "        while queue:\n",
+        "            course = queue.popleft()  # Take a course with no prerequisites.\n",
+        "            numCourses -= 1  # Decrement the count of remaining courses.\n",
+        "\n",
+        "            for neighbor in graph[course]:\n",
+        "                in_degree[neighbor] -= 1  # Remove the prerequisite relationship.\n",
+        "                if in_degree[neighbor] == 0:\n",
+        "                    queue.append(neighbor)  # If no more prerequisites, add to the queue.\n",
+        "\n",
+        "        # If all courses were successfully taken (numCourses becomes 0), return True.\n",
+        "        return numCourses == 0"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "This Python code is designed to determine whether it is possible to complete all the required courses given a set of prerequisites. It uses a topological sorting algorithm to accomplish this. Here's a brief explanation of the code:\n",
+        "\n",
+        "1. The code defines a `Solution` class with a `canFinish` method that takes two parameters: `numCourses` (the total number of courses) and `prerequisites` (a list of pairs indicating which course must be taken before another).\n",
+        "\n",
+        "2. It initializes an empty graph as an adjacency list using a `defaultdict` to represent the courses and their prerequisites. It also initializes a list called `in_degree` to keep track of the in-degrees of courses, which is used in the topological sorting process.\n",
+        "\n",
+        "3. The code then populates the graph by iterating through the `prerequisites` list. For each pair `(ai, bi)`, it adds course `ai` as a neighbor to its prerequisite course `bi` in the graph and increments the in-degree of course `ai`.\n",
+        "\n",
+        "4. Next, it initializes a queue with courses that have no prerequisites. It does this by iterating through all the courses and adding those with an in-degree of 0 to the queue.\n",
+        "\n",
+        "5. The code performs a topological sorting of the courses using a while loop. In each iteration of the loop, it dequeues a course from the front of the queue (a course with no prerequisites).\n",
+        "\n",
+        "6. For each neighbor (course that depends on the dequeued course), it decrements the in-degree of the neighbor. If the neighbor's in-degree becomes 0, it means all of its prerequisites have been taken, so the neighbor is added to the queue.\n",
+        "\n",
+        "7. The loop continues until there are no more courses with no prerequisites to dequeue, and the in-degrees are updated accordingly.\n",
+        "\n",
+        "8. After the loop, if all courses were successfully taken (i.e., `numCourses` becomes 0), it means that there were no circular dependencies, and it's possible to finish all courses. In this case, the method returns `True`. Otherwise, if there are remaining courses (indicating a circular dependency), it returns `False`."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 20,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "True\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1\n",
+        "numCourses1 = 2\n",
+        "prerequisites1 = [[1, 0]]\n",
+        "sol = Solution()\n",
+        "print(sol.canFinish(numCourses1, prerequisites1))  # Output: True"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 21,
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "False\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "numCourses2 = 2\n",
+        "prerequisites2 = [[1, 0], [0, 1]]\n",
+        "print(sol.canFinish(numCourses2, prerequisites2))  # Output: False"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the `canFinish` method:\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "\n",
+        "1. Constructing the graph: In the first loop where we iterate through the `prerequisites` list, we populate the graph with prerequisites. This loop has a time complexity of O(E), where E is the number of prerequisites (edges).\n",
+        "\n",
+        "2. Initializing in-degrees and finding courses with no prerequisites: In the worst case, we iterate through all `numCourses` courses to initialize in-degrees and find the courses with no prerequisites. This has a time complexity of O(V), where V is the number of courses (vertices).\n",
+        "\n",
+        "3. Topological Sorting: In the while loop, we perform topological sorting, processing each course once and its outgoing edges. In the worst case, each course is processed once. This has a time complexity of O(V + E).\n",
+        "\n",
+        "The overall time complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "\n",
+        "1. Graph Representation: The space complexity for storing the graph as an adjacency list is O(E), where E is the number of prerequisites.\n",
+        "\n",
+        "2. In-Degree Array: The space complexity for storing the in-degrees of courses is O(V), where V is the number of courses.\n",
+        "\n",
+        "3. Queue: The space complexity for the queue is O(V), as it may contain all the courses.\n",
+        "\n",
+        "The overall space complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
+        "\n",
+        "In most practical cases, the number of prerequisites (E) is much smaller than the number of courses (V), so the time and space complexity can often be considered as O(V)."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Optimizing Course Scheduling**: Given a list of courses with their durations and prerequisites, find the most efficient way to schedule these courses to minimize the time it takes to complete all of them.\n",
+        "\n",
+        "2. **Detecting Cycles in Course Dependencies**: Modify the original problem to not just determine if you can finish all courses but also to identify the specific courses that form a cycle in the prerequisite dependencies."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb b/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb
new file mode 100644
index 0000000..fb186f2
--- /dev/null
+++ b/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb	
@@ -0,0 +1,236 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 3651. Number of Connected Components in an Undirected Graph\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Number of Connected Components in an Undirected Graph problem on LintCode, click here!](https://www.lintcode.com/problem/3651/)\n",
+        "\n",
+        "---\n",
+        "In this problem, there is an undirected graph with `n` nodes. There is also an `edges` array. Where `edges[i] = [a, b]` means that there is an edge between *node a* and *node b* in the graph.\n",
+        "\n",
+        "You need to return *the number of connected components in that graph*."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem at hand is to determine the number of connected components in an undirected graph. Here's a more detailed explanation of the problem:\n",
+        "\n",
+        "- You are given an undirected graph, which consists of nodes and edges. The graph may have multiple distinct connected components.\n",
+        "\n",
+        "- A connected component is a group of nodes within the graph where each node can be reached from every other node in the same component. In other words, there is a path between any two nodes in the same connected component, but there are no paths connecting nodes from different components.\n",
+        "\n",
+        "- Your task is to write a function that takes as input the number of nodes `n` and a list of edges `edges`, where each edge is represented as a pair of nodes `[a, b]` indicating an edge between node `a` and node `b`.\n",
+        "\n",
+        "- The goal is to determine the number of distinct connected components in the graph.\n",
+        "\n",
+        "- You need to implement an algorithm that can identify these connected components and count how many there are in the given graph.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 13,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "def countConnectedComponents(n, edges):\n",
+        "    # Define a helper function for depth-first search (DFS).\n",
+        "    def dfs(node):\n",
+        "        if not visited[node]:  # If the node has not been visited yet:\n",
+        "            visited[node] = True  # Mark it as visited.\n",
+        "            for neighbor in graph[node]:  # Explore all neighbors of the current node.\n",
+        "                dfs(neighbor)  # Recursively call DFS on the neighbor.\n",
+        "\n",
+        "    # Create an empty adjacency list to represent the graph.\n",
+        "    graph = [[] for _ in range(n)]\n",
+        "\n",
+        "    # Initialize a boolean list to keep track of visited nodes.\n",
+        "    visited = [False] * n\n",
+        "\n",
+        "    # Populate the adjacency list with edges from the input.\n",
+        "    for a, b in edges:\n",
+        "        graph[a].append(b)\n",
+        "        graph[b].append(a)  # Since the graph is undirected, we add edges in both directions.\n",
+        "\n",
+        "    # Initialize a counter to keep track of the number of connected components.\n",
+        "    count = 0\n",
+        "\n",
+        "    # Iterate through all nodes in the graph.\n",
+        "    for i in range(n):\n",
+        "        if not visited[i]:  # If the node has not been visited:\n",
+        "            count += 1  # Increment the count since we've found a new connected component.\n",
+        "            dfs(i)  # Start a depth-first search from the unvisited node to explore the component.\n",
+        "\n",
+        "    return count"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The provided Python code calculates the number of connected components in an undirected graph. Here's a high-level explanation of how the code works:\n",
+        "\n",
+        "1. The `countConnectedComponents` function takes two arguments: `n` (the number of nodes in the graph) and `edges` (a list of edge pairs representing connections between nodes).\n",
+        "\n",
+        "2. Inside the function, there is a helper function called `dfs` (depth-first search), which is used to explore and mark nodes as visited.\n",
+        "\n",
+        "3. An empty adjacency list called `graph` is created to represent the graph. The `graph` is a list of lists where each element represents a node, and the list associated with each node contains its neighboring nodes.\n",
+        "\n",
+        "4. Another list called `visited` is created to keep track of visited nodes. Initially, all nodes are marked as not visited (`False`).\n",
+        "\n",
+        "5. The code populates the `graph` by iterating through the `edges` list and adding each edge to the adjacency list. Since the graph is undirected, both directions of each edge are added.\n",
+        "\n",
+        "6. A counter variable called `count` is initialized to keep track of the number of connected components.\n",
+        "\n",
+        "7. The code iterates through all nodes from `0` to `n-1`. For each node, it checks whether it has been visited yet. If it hasn't been visited, it means a new connected component is found. The counter `count` is incremented, and a depth-first search (DFS) is initiated from this unvisited node to explore the connected component.\n",
+        "\n",
+        "8. The `dfs` function recursively visits all nodes in the connected component, marking them as visited. It does this by checking each neighbor of the current node and recursively calling `dfs` on unvisited neighbors.\n",
+        "\n",
+        "9. After the loop through all nodes, the `count` variable contains the total number of connected components in the graph.\n",
+        "\n",
+        "10. The function returns the value of `count`, which represents the number of connected components.\n",
+        "\n",
+        "11. Two examples are provided at the end to demonstrate how to use the function with different inputs."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 14,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "1\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1\n",
+        "n1 = 3\n",
+        "edges1 = [[0, 1], [0, 2]]\n",
+        "print(countConnectedComponents(n1, edges1))  # Output: 1"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 15,
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "2\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "n2 = 6\n",
+        "edges2 = [[0, 1], [1, 2], [2, 3], [4, 5]]\n",
+        "print(countConnectedComponents(n2, edges2))  # Output: 2"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the code for counting the number of connected components in an undirected graph using depth-first search (DFS).\n",
+        "\n",
+        "**Time Complexity:**\n",
+        "- The code iterates through all nodes in the graph (from 0 to `n-1`).\n",
+        "- For each unvisited node, it initiates a depth-first search (DFS) from that node to explore the connected component.\n",
+        "- In the worst case, every node may belong to a separate connected component, and you perform a DFS for each node.\n",
+        "- The time complexity of the DFS for a single connected component is O(V + E), where V is the number of nodes in the component, and E is the number of edges.\n",
+        "- In the worst case, you perform O(n) DFS calls, each with its own O(V + E) complexity.\n",
+        "\n",
+        "So, the overall time complexity of the code is O(n * (V + E)). In the worst case, if the graph is connected (a single connected component), this simplifies to O(n * (n + m)), where m is the number of edges.\n",
+        "\n",
+        "**Space Complexity:**\n",
+        "- The main data structures used are the `graph` (adjacency list), `visited` (boolean array), and the recursive call stack for DFS.\n",
+        "- The `graph` has a space complexity of O(n + m), where n is the number of nodes, and m is the number of edges.\n",
+        "- The `visited` boolean array also has a space complexity of O(n) to keep track of whether nodes have been visited.\n",
+        "- The depth of the recursive call stack for DFS can go up to O(n) in the worst case, when all nodes are part of a single connected component.\n",
+        "\n",
+        "So, the overall space complexity of the code is O(n + m) for data structures and O(n) for the call stack. In the worst case, this simplifies to O(n + m).\n",
+        "\n",
+        "In summary, the time complexity is O(n * (V + E)), and the space complexity is O(n + m)."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Disconnected Components with Multiple Queries:** You are given an undirected graph with `n` nodes and `m` edges. Initially, the graph is disconnected, and you need to process `q` queries. Each query consists of adding an edge to connect two nodes. After each query, you need to output the number of connected components in the updated graph.\n",
+        "\n",
+        "2. **Maximum Size of Connected Components:** Given an undirected graph with `n` nodes and `m` edges, find the size of the largest connected component in the graph. Additionally, identify the nodes that belong to this largest component. You can assume there is at least one connected component in the graph.\n"
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb b/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb
new file mode 100644
index 0000000..eb34497
--- /dev/null
+++ b/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb	
@@ -0,0 +1,261 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 417. Pacific Atlantic Water Flow\n",
+        "\n",
+        "\n",
+        "**Difficulty:** Medium\n",
+        "\n",
+        "**Link to Problem:** [To see the Pacific Atlantic Water Flow problem on LeetCode, click here!](https://leetcode.com/problems/pacific-atlantic-water-flow/)\n",
+        "\n",
+        "---\n",
+        "There is an `m x n` rectangular island that borders both the **Pacific Ocean** and **Atlantic Ocean**. The **Pacific Ocea**n touches the island's left and top edges, and the **Atlantic Ocean** touches the island's right and bottom edges.\n",
+        "\n",
+        "The island is partitioned into a grid of square cells. You are given an `m x n` integer matrix `heights` where `heights[r][c]` represents the **height above sea level** of the cell at coordinate `(r, c)`.\n",
+        "\n",
+        "The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is **less than or equal to** the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.\n",
+        "\n",
+        "Return *a **2D list** of grid coordinates `result` where $result[i] = [r_i, c_i]$ denotes that rain water can flow from cell $(r_i, c_i)$ to **both** the Pacific and Atlantic oceans*.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `m == heights.length`\n",
+        "- `n == heights[r].length`\n",
+        "- `1 <= m, n <= 200`\n",
+        "- 0 <= `heights[r][c]` <= $10^5$"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem at hand involves a scenario with a rectangular island surrounded by both the Pacific Ocean and the Atlantic Ocean. The Pacific Ocean is in contact with the island's left and top edges, while the Atlantic Ocean is in contact with the island's right and bottom edges. The goal is to determine which cells on the island can allow rainwater to flow to both the Pacific and Atlantic Oceans.\n",
+        "\n",
+        "Here are the key details and constraints of the problem:\n",
+        "\n",
+        "1. The island is divided into a grid of square cells.\n",
+        "\n",
+        "2. Each cell is represented by an integer value in a matrix `heights`. The value at `heights[r][c]` represents the height above sea level of the cell at coordinates `(r, c)` on the island.\n",
+        "\n",
+        "3. Rainwater can flow from a cell to neighboring cells directly north, south, east, or west only if the neighboring cell's height is less than or equal to the current cell's height.\n",
+        "\n",
+        "4. Water can flow from any cell adjacent to the ocean into the ocean. This means water can flow from cells on the edge of the island to the Pacific Ocean and the Atlantic Ocean.\n",
+        "\n",
+        "The task is to find and return a 2D list of grid coordinates where rainwater can flow from a cell to both the Pacific and Atlantic Oceans. In other words, you need to identify the cells that can send water to both the left and top edges (Pacific) and the right and bottom edges (Atlantic) of the island.\n",
+        "\n",
+        "The problem is solved by performing a depth-first search (DFS) starting from the ocean edges and marking the cells that can flow to each ocean. Then, the algorithm identifies cells that are reachable from both oceans and returns their coordinates as the result.\n",
+        "\n",
+        "The problem involves both traversal of the island and backtracking to find the solution, which makes it a classic graph traversal problem.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 24,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "from typing import List\n",
+        "\n",
+        "class Solution:\n",
+        "    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:\n",
+        "        # Check if the input matrix is empty\n",
+        "        if not heights:\n",
+        "            return []\n",
+        "        \n",
+        "        m, n = len(heights), len(heights[0])\n",
+        "        \n",
+        "        # Initialize boolean matrices to keep track of cells reachable from each ocean\n",
+        "        pacific_reachable = [[False] * n for _ in range(m)]\n",
+        "        atlantic_reachable = [[False] * n for _ in range(m)]\n",
+        "        \n",
+        "        # Depth-First Search (DFS) function to mark cells that can be reached from an ocean\n",
+        "        def dfs(x, y, reachable):\n",
+        "            if reachable[x][y]:\n",
+        "                return\n",
+        "            reachable[x][y] = True\n",
+        "            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:\n",
+        "                new_x, new_y = x + dx, y + dy\n",
+        "                if 0 <= new_x < m and 0 <= new_y < n and heights[new_x][new_y] >= heights[x][y]:\n",
+        "                    dfs(new_x, new_y, reachable)\n",
+        "        \n",
+        "        # Start DFS from the Pacific Ocean (left and top edges)\n",
+        "        for i in range(m):\n",
+        "            dfs(i, 0, pacific_reachable)\n",
+        "            dfs(i, n - 1, atlantic_reachable)\n",
+        "        \n",
+        "        for j in range(n):\n",
+        "            dfs(0, j, pacific_reachable)\n",
+        "            dfs(m - 1, j, atlantic_reachable)\n",
+        "        \n",
+        "        result = []\n",
+        "        \n",
+        "        # Find cells that can flow to both the Pacific and Atlantic oceans\n",
+        "        for i in range(m):\n",
+        "            for j in range(n):\n",
+        "                if pacific_reachable[i][j] and atlantic_reachable[i][j]:\n",
+        "                    result.append([i, j])\n",
+        "        \n",
+        "        return result"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The code defines a class called `Solution` with a method named `pacificAtlantic`. This method takes a 2D matrix `heights`, where each element represents the height above sea level of a cell on a rectangular island. The goal of the method is to find and return a list of grid coordinates (cells) from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
+        "\n",
+        "The code uses a Depth-First Search (DFS) algorithm to traverse the island, starting from the ocean edges, and marks cells that are reachable from either the Pacific or Atlantic Ocean. It does this by creating two boolean matrices, `pacific_reachable` and `atlantic_reachable`, to keep track of cells that can be reached from each ocean.\n",
+        "\n",
+        "The code iterates through the entire island, initiating DFS searches from the edges of the island. If a cell is reachable from an ocean (either Pacific or Atlantic), it is marked as reachable in the corresponding boolean matrix. The DFS function ensures that the traversal follows the rule that water can flow to a neighboring cell if the neighboring cell's height is less than or equal to the current cell's height.\n",
+        "\n",
+        "After marking all reachable cells from both oceans, the code then looks for cells that are reachable from both the Pacific and Atlantic Oceans. If a cell satisfies this condition, it is added to the `result` list.\n",
+        "\n",
+        "Finally, the code returns the `result` list, which contains the grid coordinates of cells from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
+        "\n",
+        "The code provides a solution to the problem of finding cells on the island where water can reach both oceans, demonstrating a traversal algorithm that explores the connectivity of cells on the island."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 25,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1\n",
+        "heights1 = [\n",
+        "    [1,2,2,3,5],\n",
+        "    [3,2,3,4,4],\n",
+        "    [2,4,5,3,1],\n",
+        "    [6,7,1,4,5],\n",
+        "    [5,1,1,2,4]\n",
+        "]\n",
+        "\n",
+        "solution = Solution()\n",
+        "print(solution.pacificAtlantic(heights1)) # Expected Output: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 26,
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "[[0, 0]]\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "heights2 = [[1]]\n",
+        "print(solution.pacificAtlantic(heights2)) # Expected Output: [[0, 0]]"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Time Complexity:\n",
+        "The time complexity of this solution is O(m * n), where 'm' is the number of rows in the input matrix and 'n' is the number of columns. Here's a breakdown of the time complexity:\n",
+        "\n",
+        "1. Creating the `pacific_reachable` and `atlantic_reachable` matrices with dimensions m x n takes O(m * n) time.\n",
+        "\n",
+        "2. Initiating Depth-First Search (DFS) from the edges of the island takes O(m * n) time because it processes each cell once.\n",
+        "\n",
+        "3. The DFS function visits each cell at most once, and its time complexity is O(1) for each cell. The number of cells visited by the DFS in the worst case is m * n.\n",
+        "\n",
+        "4. The final step of identifying and appending cells to the result list takes O(m * n) time in the worst case.\n",
+        "\n",
+        "Overall, the time complexity is dominated by the DFS and is O(m * n).\n",
+        "\n",
+        "Space Complexity:\n",
+        "The space complexity of this solution is also O(m * n). Here's how the space is used:\n",
+        "\n",
+        "1. Two boolean matrices, `pacific_reachable` and `atlantic_reachable`, are created with dimensions m x n. Each matrix takes O(m * n) space, resulting in O(2 * m * n) space usage.\n",
+        "\n",
+        "2. The depth-first search (DFS) stack space during the recursive calls can go as deep as the diagonal of the grid, which is at most min(m, n). This additional space for the call stack is relatively small compared to the boolean matrices and can be considered O(min(m, n)).\n",
+        "\n",
+        "3. The `result` list stores the grid coordinates, and its size is determined by the number of cells that can flow to both oceans. In the worst case, this list can contain all m * n cells, so it takes O(m * n) space.\n",
+        "\n",
+        "The overall space complexity is the sum of these components, which is O(m * n) + O(min(m, n)) + O(m * n). In big O notation, we can simplify this to O(m * n)."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Optimization Challenge:** Modify the algorithm to find the maximum flow of water from any cell to both the Pacific and Atlantic Oceans. In other words, find the cell(s) that can send the most water to both oceans.\n",
+        "\n",
+        "2. **Efficiency Challenge:** Optimize the time and space complexity of the algorithm while maintaining correctness. Aim to reduce both time and space complexity as much as possible."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/11. Graphs/README.md b/_build/html/_sources/11. Graphs/README.md
new file mode 100644
index 0000000..aad262c
--- /dev/null
+++ b/_build/html/_sources/11. Graphs/README.md	
@@ -0,0 +1,23 @@
+# Graphs Problems - Blind 75 LeetCode
+
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)                           | Medium     |
+| [133. Clone Graph](https://leetcode.com/problems/clone-graph/)                                       | Medium     |
+| [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)       | Medium     |
+| [207. Course Schedule](https://leetcode.com/problems/course-schedule/)                                 | Medium     |
+| [323. Number of Connected Components In An Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/) | Medium |
+| [261. Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)                               | Medium     |
+
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb b/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb
new file mode 100644
index 0000000..81f0e5b
--- /dev/null
+++ b/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb	
@@ -0,0 +1,256 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {
+        "jp-MarkdownHeadingCollapsed": true
+      },
+      "source": [
+        "# 892. Alien Dictionary\n",
+        "\n",
+        "**Difficulty:** Hard\n",
+        "\n",
+        "**Link to Problem:** [To see the Alien Dictionary problem on LintCode, click here!](https://www.lintcode.com/problem/892/)\n",
+        "\n",
+        "---\n",
+        "\n",
+        "There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of **non-empty** words from the dictionary, where words are **sorted lexicographically by the rules of this new language**. Derive the order of letters in this language.\n",
+        "\n",
+        "**Constraints:**\n",
+        "- You may assume all letters are in lowercase.\n",
+        "- The dictionary is invalid, if string a is prefix of string b and b is appear before a.\n",
+        "- If the order is invalid, return an empty string.\n",
+        "- There may be multiple valid order of letters, return the smallest in normal lexicographical order.\n",
+        "- The letters in **one** string are of the same rank by default and are sorted in Human dictionary order."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The problem at hand involves determining the order of letters in a new alien language that uses the Latin alphabet. However, the order of letters in this alien language is unknown. To solve this problem, we are given a list of non-empty words from the dictionary, where the words are sorted lexicographically based on the rules of the alien language. The goal is to derive the correct order of letters in this new language.\n",
+        "\n",
+        "Here's a more detailed explanation of the problem:\n",
+        "\n",
+        "1. The alphabet: The new alien language uses the Latin alphabet, which consists of lowercase letters.\n",
+        "\n",
+        "2. Dictionary: We are provided with a list of non-empty words. These words are sorted in lexicographical order based on the rules of the alien language. The order of letters in the words reflects the correct order in the new language.\n",
+        "\n",
+        "3. Invalid order: The problem specifies that the dictionary is considered invalid if there exists a situation where one string 'a' is a prefix of another string 'b,' and 'b' appears before 'a' in the dictionary. This condition ensures that there are no inconsistencies in the order of letters.\n",
+        "\n",
+        "4. Multiple valid orders: It's possible that there are multiple valid orders of letters that satisfy the provided dictionary. However, we are instructed to return the smallest valid order in normal lexicographical order.\n",
+        "\n",
+        "To solve this problem, you need to analyze the given list of words and determine the correct order of letters in the alien language while adhering to the specified constraints. You should return the order as a string in lexicographical order if it's valid, and an empty string if the order is invalid.\n",
+        "\n",
+        "The key to solving this problem is to construct a directed graph of letter relationships based on the order information provided by the words in the dictionary. Then, perform a topological sort on this graph to derive the order of letters, ensuring that the order is consistent and valid according to the given constraints.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 7,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "from collections import defaultdict, deque\n",
+        "\n",
+        "def alienOrder(words):\n",
+        "    # Create an adjacency list to represent the graph\n",
+        "    graph = defaultdict(list)\n",
+        "    \n",
+        "    # Create a dictionary to store the in-degrees of each letter\n",
+        "    in_degree = {}\n",
+        "    \n",
+        "    # Initialize in-degrees to 0 for all letters\n",
+        "    for word in words:\n",
+        "        for char in word:\n",
+        "            if char not in in_degree:\n",
+        "                in_degree[char] = 0\n",
+        "    \n",
+        "    # Build the graph and update in-degrees\n",
+        "    for i in range(1, len(words)):\n",
+        "        word1, word2 = words[i-1], words[i]\n",
+        "        min_length = min(len(word1), len(word2))\n",
+        "        \n",
+        "        # Compare characters in the two words\n",
+        "        j = 0\n",
+        "        while j < min_length and word1[j] == word2[j]:\n",
+        "            j += 1\n",
+        "        \n",
+        "        if j < min_length:\n",
+        "            # If word1[j] is lexicographically before word2[j], add an edge\n",
+        "            graph[word1[j]].append(word2[j])\n",
+        "            in_degree[word2[j]] += 1\n",
+        "            \n",
+        "            # Break the loop to avoid further comparisons\n",
+        "            break\n",
+        "    \n",
+        "    # Perform topological sorting using Kahn's algorithm\n",
+        "    result = []\n",
+        "    queue = deque([char for char in in_degree if in_degree[char] == 0])\n",
+        "    \n",
+        "    while queue:\n",
+        "        char = queue.popleft()\n",
+        "        result.append(char)\n",
+        "        \n",
+        "        for neighbor in graph[char]:\n",
+        "            in_degree[neighbor] -= 1\n",
+        "            if in_degree[neighbor] == 0:\n",
+        "                queue.append(neighbor)\n",
+        "    \n",
+        "    # Check if there is a valid order\n",
+        "    if len(result) < len(in_degree):\n",
+        "        return \"\"\n",
+        "    \n",
+        "    return \"\".join(result)"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The code is designed to determine the order of letters in an unknown alien language that uses the Latin alphabet. It takes a list of non-empty words from a dictionary as input, where the words are sorted lexicographically based on the rules of this alien language.\n",
+        "\n",
+        "The code follows these main steps:\n",
+        "\n",
+        "1. Create a data structure to represent the graph of letters and their relationships in the alien language. It uses a defaultdict to store a list of letters that come after each letter. Additionally, it uses a dictionary called `in_degree` to keep track of the in-degrees (the number of letters that precede a letter) for each letter.\n",
+        "\n",
+        "2. Initialize the in-degrees of all letters to 0. This is done by iterating through all the words and characters in the words, adding each character to the `in_degree` dictionary with an initial in-degree of 0.\n",
+        "\n",
+        "3. Build the graph by comparing adjacent words in the dictionary. For each pair of adjacent words, the code finds the first differing character between the two words. If such a character exists, it means there's an order relationship between the letters represented by these characters. The code updates the graph and in-degrees accordingly.\n",
+        "\n",
+        "4. After constructing the graph, the code performs a topological sorting of the letters using Kahn's algorithm. It starts by initializing an empty result list and a queue containing letters with in-degrees of 0. The algorithm repeatedly removes a letter with an in-degree of 0 from the queue, adds it to the result list, and updates the in-degrees of its neighbors. This process continues until the queue is empty.\n",
+        "\n",
+        "5. Finally, the code checks if the topological sorting was successful by comparing the length of the result list with the number of unique letters in the input. If the lengths differ, it means there's a cycle in the graph, and the order is invalid. In such a case, an empty string is returned. Otherwise, the result list is joined together to form the smallest order of letters in normal lexicographical order, and it is returned as the output.\n",
+        "\n",
+        "The code provides a way to determine the order of letters in the alien language while handling various edge cases, including checking for invalid order conditions."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 8,
+      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "wrtef\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1\n",
+        "words1 = [\"wrt\", \"wrf\", \"er\", \"ett\", \"rftt\"]\n",
+        "print(alienOrder(words1))  # Output: \"wertf\""
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 9,
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "zx\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2\n",
+        "words2 = [\"z\", \"x\"]\n",
+        "print(alienOrder(words2))  # Output: \"zx\""
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Let's analyze the time and space complexity of the provided code for determining the order of letters in an alien language.\n",
+        "\n",
+        "Time Complexity:\n",
+        "1. Building the graph: The code iterates through the list of words and compares adjacent words. In the worst case, where all words have the same prefix, it takes O(N) time to build the graph, where N is the total number of characters in all the words.\n",
+        "2. Topological sorting: Performing a topological sort on the graph takes O(V + E) time, where V is the number of unique letters (vertices) and E is the number of relationships between letters (edges) in the graph.\n",
+        "\n",
+        "Overall, the time complexity is O(N + V + E), where N is the total number of characters in the words, V is the number of unique letters, and E is the number of order relationships between letters.\n",
+        "\n",
+        "Space Complexity:\n",
+        "1. Graph and In-degrees: The code uses data structures to represent the graph and in-degrees. The space complexity for these data structures is O(V + E), where V is the number of unique letters, and E is the number of order relationships between letters in the words.\n",
+        "2. Result List: The result list stores the order of letters, which can have a maximum size of V, where V is the number of unique letters.\n",
+        "3. Queue: The space used by the queue during topological sorting is also O(V).\n",
+        "\n",
+        "Overall, the space complexity is O(V + E) due to the data structures used for graph representation and the space needed for storing the result and the queue.\n",
+        "\n",
+        "In the worst case, where there are many unique letters and many order relationships, the space complexity is dominated by the number of unique letters and their relationships.\n",
+        "\n",
+        "In summary, the time complexity of the code is O(N + V + E), and the space complexity is O(V + E). The actual performance will depend on the specific input data, such as the number of unique letters and the relationships between them in the alien language."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Multiple Valid Orders**: Extend the problem to handle cases where there are multiple valid orders of letters in the alien language. Modify the code to return all valid orders instead of just one. Be mindful of performance.\n",
+        "\n",
+        "2. **Detect Invalid Dictionary**: Given a dictionary of words, write a function to detect whether the dictionary is valid or not based on the constraints mentioned in the problem statement."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/12. Advanced Graphs/README.md b/_build/html/_sources/12. Advanced Graphs/README.md
new file mode 100644
index 0000000..f362de7
--- /dev/null
+++ b/_build/html/_sources/12. Advanced Graphs/README.md	
@@ -0,0 +1,16 @@
+# Advanced Graphs Problems - Blind 75 LeetCode 
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [269. Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)                               | Hard       |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb b/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb
new file mode 100644
index 0000000..8d32e21
--- /dev/null
+++ b/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb	
@@ -0,0 +1,103 @@
+{
+  "metadata": {
+    "kernelspec": {
+      "name": "python",
+      "display_name": "Python (Pyodide)",
+      "language": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat_minor": 5,
+  "nbformat": 4,
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "source": "# House Robber\n**Difficulty:** Medium\n\n---\nYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and **it will automatically contact the police if two adjacent houses were broken into on the same night.**\n\nGiven an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight **without alerting the police**.*\n\n**Constraints:**\n- `1 <= nums.length <= 100`\n- `0 <= nums[i] <= 400`",
+      "metadata": {},
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49"
+    },
+    {
+      "cell_type": "markdown",
+      "source": "## Probelm Explanation:\n\nThe problem is a classic dynamic programming challenge known as the \"House Robber\" problem. The scenario is that you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, but the constraint is that adjacent houses have security systems connected. If two adjacent houses are broken into on the same night, the security systems will automatically contact the police.\n\nThe goal is to determine the maximum amount of money you can rob tonight without alerting the police. You need to find a strategy to maximize your earnings while avoiding robbing two adjacent houses.\n\nThe input to the problem is an array `nums`, where `nums[i]` represents the amount of money in the `i`-th house. The task is to return the maximum amount of money that can be robbed without triggering the security system.\n\nFor example:\n```\nInput: nums = [1,2,3,1]\nOutput: 4\nExplanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).\nTotal amount you can rob = 1 + 3 = 4.\n\nInput: nums = [2,7,9,3,1]\nOutput: 12\nExplanation: Rob house 1 (money = 2), rob house 3 (money = 9), and rob house 5 (money = 1).\nTotal amount you can rob = 2 + 9 + 1 = 12.\n```\n\nThe challenge involves designing an algorithm to efficiently solve this problem and find the optimal strategy for robbing houses to maximize the stolen money while adhering to the constraint of not robbing adjacent houses. The provided solution utilizes dynamic programming to achieve this.\n\n## Solution:\nHere's a Python function to implement this algorithm:",
+      "metadata": {},
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76"
+    },
+    {
+      "cell_type": "code",
+      "source": "from typing import List\n\nclass Solution:\n    def rob(self, nums: List[int]) -> int:\n        # Initialize variables to keep track of the maximum amount of money robbed at the previous and current houses\n        prev, curr = 0, 0\n\n        # Iterate through the houses\n        for n in nums:\n            # Calculate the maximum amount of money that can be robbed at the current house\n            # It is the maximum of either robbing the current house and the money robbed at the house before the previous one (prev),\n            # or skipping the current house and taking the maximum amount from the previous house (curr)\n            temp = max(n + prev, curr)\n\n            # Update the variables for the next iteration\n            prev, curr = curr, temp\n\n        # The final result is the maximum amount of money that can be robbed at the last house\n        return curr",
+      "metadata": {
+        "trusted": true
+      },
+      "execution_count": 4,
+      "outputs": [],
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5"
+    },
+    {
+      "cell_type": "markdown",
+      "source": "## Explanation:\n\nThe provided code is a Python implementation of the \"House Robber\" problem using a dynamic programming approach. It's part of a class `Solution` and includes a method `rob` that takes a list of integers `nums` as input and returns the maximum amount of money that can be robbed without alerting the police.\n\nThe key idea of the algorithm is to maintain two variables, `prev` and `curr`, which represent the maximum amount of money that can be robbed up to the previous house and the current house, respectively. The algorithm iterates through the houses, updating these variables based on the maximum amount that can be robbed at the current house, considering the constraint of not robbing adjacent houses.\n\nThe loop calculates a temporary variable `temp` that represents the maximum amount of money that can be robbed at the current house. It considers two scenarios: either robbing the current house and adding the money robbed at the house before the previous one (`n + prev`), or skipping the current house and taking the maximum amount from the previous house (`curr`). The maximum of these two scenarios is assigned to `temp`.\n\nThe `prev` and `curr` variables are then updated for the next iteration. `prev` is set to the previous value of `curr`, and `curr` is set to the calculated `temp`.\n\nThe final result is the value stored in the `curr` variable, representing the maximum amount of money that can be robbed at the last house. This value is returned as the result of the `rob` method.\n\nThe provided test cases at the end demonstrate the usage of the `rob` method with different input arrays.",
+      "metadata": {},
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9"
+    },
+    {
+      "cell_type": "markdown",
+      "source": "## Test cases:\n\nHere's how you can use this solution:",
+      "metadata": {},
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244"
+    },
+    {
+      "cell_type": "code",
+      "source": "# Example 1:\nsol = Solution()\n\nnums1 = [1, 2, 3, 1]\nprint(sol.rob(nums1))  # Output: 4",
+      "metadata": {
+        "trusted": true
+      },
+      "execution_count": 5,
+      "outputs": [
+        {
+          "name": "stdout",
+          "text": "4\n",
+          "output_type": "stream"
+        }
+      ],
+      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c"
+    },
+    {
+      "cell_type": "code",
+      "source": "# Example 2:\nnums2 = [2, 7, 9, 3, 1]\nprint(sol.rob(nums2))  # Output: 12",
+      "metadata": {
+        "trusted": true
+      },
+      "execution_count": 6,
+      "outputs": [
+        {
+          "name": "stdout",
+          "text": "12\n",
+          "output_type": "stream"
+        }
+      ],
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3"
+    },
+    {
+      "cell_type": "markdown",
+      "source": "## Time and Space Complexity Analysis\n\nLet's analyze the time and space complexity of the provided code.\n\n### Time Complexity:\n\nThe time complexity of the code is O(n), where n is the number of houses. The algorithm iterates through the array once, and in each iteration, it performs constant time operations. The loop runs for each house, and the number of operations inside the loop is independent of the input size. Therefore, the overall time complexity is linear, O(n).\n\n### Space Complexity:\n\nThe space complexity is O(1), constant space. The algorithm uses only a constant amount of extra space regardless of the input size. The variables `prev`, `curr`, and `temp` are the only additional variables used, and they do not depend on the size of the input array. Thus, the space complexity is constant, O(1).\n\nIn summary:\n- Time Complexity: O(n)\n- Space Complexity: O(1)",
+      "metadata": {},
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2"
+    },
+    {
+      "cell_type": "markdown",
+      "source": "## Challenging Exercises:\n\n1. **Variation with Maximum Number of Houses to Rob:** Modify the problem to include an additional constraint: you are allowed to rob at most 'k' houses without triggering the alarm. Find the maximum amount you can rob under this new constraint.\n\n2. **Robbing in a Circular Street:** Extend the problem to a circular street where the first and last houses are adjacent. You cannot rob both the first and last houses simultaneously. Find the maximum amount you can rob in this circular setting.\n\n\n**Link to Problem:** [To see the House Robber problem on LeetCode, click here!](https://leetcode.com/problems/house-robber/description/)\n",
+      "metadata": {},
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6"
+    }
+  ]
+}
\ No newline at end of file
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb b/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb
new file mode 100644
index 0000000..b1080e8
--- /dev/null
+++ b/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb	
@@ -0,0 +1,119 @@
+{
+  "metadata": {
+    "kernelspec": {
+      "name": "python",
+      "display_name": "Python (Pyodide)",
+      "language": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat_minor": 5,
+  "nbformat": 4,
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "source": "# House Robber II\n**Difficulty:** Medium\n\n---\nYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle**. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.\n\nGiven an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight without alerting the police*.\n\n**Constraints:**\n- `1 <= nums.length <= 100`\n- `0 <= nums[i] <= 1000`",
+      "metadata": {},
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49"
+    },
+    {
+      "cell_type": "markdown",
+      "source": "## Probelm Explanation:\n\nCertainly! The problem statement describes a scenario where you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. However, there is a security system in place: adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.\n\nThe twist in this problem is that the houses are arranged in a circle. That means the first house is the neighbor of the last one. So, robbing the first house has implications for the last house, and vice versa.\n\nThe goal is to find the maximum amount of money you can rob tonight without alerting the police. You need to implement a function that takes an integer array `nums` representing the amount of money in each house and returns the maximum amount of money that can be robbed without triggering the security system.\n\nHere are a few examples to illustrate the problem:\n\n1. **Example 1:**\n   ```python\n   Input: nums = [2, 3, 2]\n   Output: 3\n   Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2) because they are adjacent houses. The optimal strategy is to rob house 2 (money = 3).\n   ```\n\n2. **Example 2:**\n   ```python\n   Input: nums = [1, 2, 3, 1]\n   Output: 4\n   Explanation: Robbing house 1 (money = 1) and then robbing house 3 (money = 3) results in a total amount of 4. Robbing house 2 is skipped to avoid alerting the police.\n   ```\n\n3. **Example 3:**\n   ```python\n   Input: nums = [1, 2, 3]\n   Output: 3\n   Explanation: Since the houses are in a circle, you can choose either house 1, 2, or 3. The optimal strategy is to rob house 3 for a total amount of 3.\n   ```\n\nThe constraints for the problem are that the length of the `nums` array is between 1 and 100, and the amount of money in each house (`nums[i]`) is between 0 and 1000.\n\n## Solution:\nHere's a Python function to implement this algorithm:",
+      "metadata": {},
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76"
+    },
+    {
+      "cell_type": "code",
+      "source": "from typing import List\n\nclass Solution:\n    def rob(self, nums: List[int]) -> int:\n        def simple_rob(nums):\n            # Helper function for basic house robbery problem\n            prev, curr = 0, 0\n            for num in nums:\n                # Calculate the maximum amount of money that can be robbed\n                # without alerting the police for each house\n                prev, curr = curr, max(prev + num, curr)\n            return curr\n\n        # Check the base cases where the length of nums is 1 or 2\n        if len(nums) == 1:\n            return nums[0]\n        elif len(nums) == 2:\n            return max(nums[0], nums[1])\n\n        # Case 1: First house is robbed, last house is not\n        case1 = simple_rob(nums[:-1])\n\n        # Case 2: First house is not robbed, last house is\n        case2 = simple_rob(nums[1:])\n\n        # Maximum of the two cases\n        return max(case1, case2)",
+      "metadata": {
+        "trusted": true
+      },
+      "execution_count": 7,
+      "outputs": [],
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5"
+    },
+    {
+      "cell_type": "markdown",
+      "source": "## Explanation:\n\nThe provided code defines a class `Solution` with a method `rob`, which aims to determine the maximum amount of money a professional robber can steal from houses arranged in a circle without alerting the police. Each house contains a certain amount of money, and adjacent houses have a security system that triggers if both are robbed on the same night.\n\nThe code uses dynamic programming to solve the problem efficiently. It contains a helper function `simple_rob` that computes the maximum amount of money that can be robbed from a linear arrangement of houses. The main method `rob` handles the circular arrangement by considering two cases: robbing the first house and not robbing the last, and not robbing the first house and robbing the last. The result is the maximum of these two cases.\n\nThe class is instantiated as `solution`, and the `rob` method is applied to three different scenarios (test cases) to demonstrate its functionality. The final output represents the maximum amount of money the robber can steal without triggering the security system.",
+      "metadata": {},
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9"
+    },
+    {
+      "cell_type": "markdown",
+      "source": "## Test cases:\n\nHere's how you can use this solution:",
+      "metadata": {},
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244"
+    },
+    {
+      "cell_type": "code",
+      "source": "# Example 1:\n\n# Create an instance of the Solution class\nsolution = Solution()\n\nnums1 = [2, 3, 2]\nprint(solution.rob(nums1))  # Output: 3",
+      "metadata": {
+        "trusted": true
+      },
+      "execution_count": 8,
+      "outputs": [
+        {
+          "name": "stdout",
+          "text": "3\n",
+          "output_type": "stream"
+        }
+      ],
+      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c"
+    },
+    {
+      "cell_type": "code",
+      "source": "# Example 2:\nnums2 = [1, 2, 3, 1]\nprint(solution.rob(nums2))  # Output: 4",
+      "metadata": {
+        "trusted": true
+      },
+      "execution_count": 9,
+      "outputs": [
+        {
+          "name": "stdout",
+          "text": "4\n",
+          "output_type": "stream"
+        }
+      ],
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3"
+    },
+    {
+      "cell_type": "code",
+      "source": "# Example 2:\nnums3 = [1, 2, 3]\nprint(solution.rob(nums3))  # Output: 3",
+      "metadata": {
+        "trusted": true
+      },
+      "execution_count": 10,
+      "outputs": [
+        {
+          "name": "stdout",
+          "text": "3\n",
+          "output_type": "stream"
+        }
+      ],
+      "id": "e6e1554f-ae77-410c-8af2-e59e5428166c"
+    },
+    {
+      "cell_type": "markdown",
+      "source": "## Time and Space Complexity Analysis\n\nLet's analyze the time and space complexity of the provided code:\n\n### Time Complexity:\nThe time complexity is primarily determined by the `simple_rob` helper function, which has a time complexity of O(N), where N is the number of houses in the input array. The main `rob` function calls `simple_rob` twice, once for the case of robbing the first house and not robbing the last, and once for the case of not robbing the first house and robbing the last. Therefore, the overall time complexity is O(N).\n\n### Space Complexity:\nThe space complexity is O(1) because the space used by the algorithm is constant, regardless of the size of the input array. The only variables used are `prev` and `curr` in the `simple_rob` function, which represent the previous and current maximum amounts of money that can be robbed without alerting the police. These variables are updated iteratively, and no additional data structures are used that scale with the input size. Therefore, the space complexity is constant, or O(1).\n\nIn summary:\n- Time Complexity: O(N)\n- Space Complexity: O(1)",
+      "metadata": {},
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2"
+    },
+    {
+      "cell_type": "markdown",
+      "source": "## Challenging Exercises:\n\n1. **Optimization Challenge:** Optimize the space complexity of the solution. Can you solve the problem with O(1) space complexity without sacrificing the time complexity?\n\n2. **Circular Robbery Variations:** Consider variations of the circular house robbery problem, such as allowing the robber to skip a certain number of houses or restricting the number of consecutive houses that can be robbed. Modify the solution to accommodate these variations.\n\n**Link to Problem:** [To see the House Robber II problem on LeetCode, click here!](https://leetcode.com/problems/house-robber-ii/)",
+      "metadata": {},
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6"
+    }
+  ]
+}
\ No newline at end of file
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb b/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb
new file mode 100644
index 0000000..b727489
--- /dev/null
+++ b/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb	
@@ -0,0 +1,210 @@
+{
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+      "metadata": {},
+      "source": [
+        "# 70. Climbing Stairs\n",
+        "\n",
+        "**Difficulty:** Easy\n",
+        "\n",
+        "**Link to Problem:** [To see the Climbing Stairs problem on LeetCode, click here!](https://leetcode.com/problems/climbing-stairs/)\n",
+        "\n",
+        "---\n",
+        "You are climbing a staircase. It takes `n` steps to reach the top.\n",
+        "\n",
+        "Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?\n",
+        "\n",
+        "**Constraints:**\n",
+        "- `1 <= n <= 45`"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+      "metadata": {},
+      "source": [
+        "## Probelm Explanation:\n",
+        "\n",
+        "The \"Climbing Stairs\" problem is a classic algorithmic problem that asks you to find the number of distinct ways to climb a staircase with a given number of steps. Here's the problem statement:\n",
+        "\n",
+        "You are given a staircase with 'n' steps. Each time you can either climb 1 step or 2 steps. You need to find out how many distinct ways there are to reach the top of the staircase.\n",
+        "\n",
+        "For example, if 'n' is 2, there are two ways to climb to the top:\n",
+        "\n",
+        "1. Climbing two steps at once.\n",
+        "2. Climbing one step, and then another step.\n",
+        "\n",
+        "If 'n' is 3, there are three ways:\n",
+        "\n",
+        "1. Climbing one step, then one step, and finally one step.\n",
+        "2. Climbing one step, then two steps.\n",
+        "3. Climbing two steps, then one step.\n",
+        "\n",
+        "The problem essentially asks you to find a solution that counts the possible combinations of 1-step and 2-step climbs that can be taken to reach the top of the staircase. It's a classic example of a dynamic programming problem where you can break it down into smaller subproblems and build a solution incrementally, as shown in the code I provided earlier.\n",
+        "\n",
+        "## Solution:\n",
+        "Here's a Python function to implement this algorithm:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 1,
+      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [],
+      "source": [
+        "class Solution:\n",
+        "    def climbStairs(self, n: int) -> int:\n",
+        "        # If there are 1 or 2 steps, there are exactly n ways to reach the top.\n",
+        "        if n <= 2:\n",
+        "            return n\n",
+        "\n",
+        "        # Create an array to store the number of ways to reach each step.\n",
+        "        dp = [0] * (n + 1)\n",
+        "\n",
+        "        # There is one way to reach the first step (base case).\n",
+        "        dp[1] = 1\n",
+        "\n",
+        "        # There are two ways to reach the second step (base case).\n",
+        "        dp[2] = 2\n",
+        "\n",
+        "        # Calculate the number of ways to reach each step from the previous steps.\n",
+        "        for i in range(3, n + 1):\n",
+        "            # The number of ways to reach step 'i' is the sum of the ways to reach\n",
+        "            # the previous two steps, as you can take 1 or 2 steps at a time.\n",
+        "            dp[i] = dp[i - 1] + dp[i - 2]\n",
+        "\n",
+        "        # The result is the number of ways to reach the top, which is stored in dp[n].\n",
+        "        return dp[n]"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+      "metadata": {},
+      "source": [
+        "## Explanation:\n",
+        "\n",
+        "The code defines a class `Solution` with a method `climbStairs` that takes an integer `n` as input and returns the number of distinct ways to climb a staircase with `n` steps.\n",
+        "\n",
+        "1. If `n` is 1 or 2, the function returns `n` directly because there are only 1 or 2 distinct ways to reach the top in those cases.\n",
+        "\n",
+        "2. For `n` greater than 2, the code uses dynamic programming to calculate the number of ways to reach each step. It creates an array `dp` of length `n + 1` to store these values.\n",
+        "\n",
+        "3. The `dp[1]` and `dp[2]` are initialized to 1 and 2, respectively. This is because there is only one way to reach the first step (by taking 1 step), and there are two ways to reach the second step (by taking 2 steps or two 1-step climbs).\n",
+        "\n",
+        "4. The code then uses a loop to fill in the `dp` array for steps 3 and above. For each step `i` from 3 to `n`, it calculates the number of ways to reach that step by adding the number of ways to reach the previous two steps, as you can either take 1 step or 2 steps at a time.\n",
+        "\n",
+        "5. Finally, the function returns the value stored in `dp[n]`, which represents the total number of distinct ways to climb the staircase with `n` steps."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+      "metadata": {},
+      "source": [
+        "## Test cases:\n",
+        "\n",
+        "Here's how you can use this solution:"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 2,
+      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Example 1 - Input: n = 2, Output: 2\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 1:\n",
+        "n = 2\n",
+        "solution = Solution()\n",
+        "print(f\"Example 1 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 2"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "execution_count": 5,
+      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+      "metadata": {
+        "trusted": true
+      },
+      "outputs": [
+        {
+          "name": "stdout",
+          "output_type": "stream",
+          "text": [
+            "Example 2 - Input: n = 3, Output: 3\n"
+          ]
+        }
+      ],
+      "source": [
+        "# Example 2:\n",
+        "n = 3\n",
+        "print(f\"Example 2 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 3\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+      "metadata": {},
+      "source": [
+        "## Time and Space Complexity Analysis\n",
+        "\n",
+        "Time Complexity:\n",
+        "The time complexity of the solution is O(n) because we use a single for loop that iterates from 3 to n to fill the `dp` array. In each iteration, we perform constant time operations (addition and assignment). Therefore, the time complexity is linear with respect to the input value `n`.\n",
+        "\n",
+        "Space Complexity:\n",
+        "The space complexity of the solution is O(n) as well. We use an array `dp` of length `n + 1` to store the number of ways to reach each step. The size of this array is directly proportional to the input value `n`, which makes the space complexity linear in terms of `n`. The rest of the variables used in the function (e.g., `i`) occupy constant space and do not contribute significantly to the overall space complexity.\n",
+        "\n",
+        "In summary, the time complexity is O(n), and the space complexity is O(n) for this dynamic programming solution to the \"Climbing Stairs\" problem."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+      "metadata": {},
+      "source": [
+        "## Challenging Exercises:\n",
+        "\n",
+        "1. **Generalized Step Sizes**: Instead of being limited to 1 step or 2 steps, consider a scenario where you can take steps of sizes `[1, 2, 3]`. Implement a function that calculates the number of ways to climb the staircase with these different step sizes.\n",
+        "\n",
+        "2. **Minimize Steps**: Given an integer `n`, find the minimum number of steps required to reach the top of the staircase. Return both the minimum number of steps and the specific sequence of steps taken."
+      ]
+    }
+  ],
+  "metadata": {
+    "kernelspec": {
+      "display_name": "Python (Pyodide)",
+      "language": "python",
+      "name": "python"
+    },
+    "language_info": {
+      "codemirror_mode": {
+        "name": "python",
+        "version": 3
+      },
+      "file_extension": ".py",
+      "mimetype": "text/x-python",
+      "name": "python",
+      "nbconvert_exporter": "python",
+      "pygments_lexer": "ipython3",
+      "version": "3.8"
+    }
+  },
+  "nbformat": 4,
+  "nbformat_minor": 5
+}
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/README.md b/_build/html/_sources/13. One-D Dynamic Programming/README.md
new file mode 100644
index 0000000..73f4de3
--- /dev/null
+++ b/_build/html/_sources/13. One-D Dynamic Programming/README.md	
@@ -0,0 +1,25 @@
+# 1-D Dynamic Programming Problems - Blind 75 LeetCode 
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)                                   | Easy       |
+| [198. House Robber](https://leetcode.com/problems/house-robber/)                                       | Medium     |
+| [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)                                 | Medium     |
+| [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)       | Medium     |
+| [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)                   | Medium     |
+| [91. Decode Ways](https://leetcode.com/problems/decode-ways/)                                         | Medium     |
+| [322. Coin Change](https://leetcode.com/problems/coin-change/)                                         | Medium     |
+| [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)             | Medium     |
+| [139. Word Break](https://leetcode.com/problems/word-break/)                                           | Medium     |
+| [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | Medium     |
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/14. Two-D Dynamic Programming/README.md b/_build/html/_sources/14. Two-D Dynamic Programming/README.md
new file mode 100644
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--- /dev/null
+++ b/_build/html/_sources/14. Two-D Dynamic Programming/README.md	
@@ -0,0 +1,47 @@
+# Blind 75 LeetCode Problems
+
+This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
+
+## Problem List
+
+### 2-D Dynamic Programming
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [62. Unique Paths](https://leetcode.com/problems/unique-paths/)                                         | Medium     |
+| [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)       | Medium     |
+
+## How to Use
+
+1. Clone or download this repository to your local machine.
+
+```bash
+git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
+```
+
+2. Navigate to the directory of the specific problem you want to solve.
+
+```bash
+cd Blind-75-LeetCode/Problem-Directory
+```
+
+3. Open the problem's directory to find its description, code template, and solution(s).
+
+4. Solve the problem using your preferred programming language.
+
+5. Check the solution(s) provided in the directory for reference and to compare your solution.
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+## Disclaimer
+
+This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/15. Greedy/README.md b/_build/html/_sources/15. Greedy/README.md
new file mode 100644
index 0000000..82b6543
--- /dev/null
+++ b/_build/html/_sources/15. Greedy/README.md	
@@ -0,0 +1,47 @@
+# Blind 75 LeetCode Problems
+
+This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
+
+## Problem List
+
+### Greedy
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)                                   | Easy       |
+| [55. Jump Game](https://leetcode.com/problems/jump-game/)                                                 | Medium     |
+
+## How to Use
+
+1. Clone or download this repository to your local machine.
+
+```bash
+git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
+```
+
+2. Navigate to the directory of the specific problem you want to solve.
+
+```bash
+cd Blind-75-LeetCode/Problem-Directory
+```
+
+3. Open the problem's directory to find its description, code template, and solution(s).
+
+4. Solve the problem using your preferred programming language.
+
+5. Check the solution(s) provided in the directory for reference and to compare your solution.
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+## Disclaimer
+
+This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/16. Intervals/README.md b/_build/html/_sources/16. Intervals/README.md
new file mode 100644
index 0000000..a57713e
--- /dev/null
+++ b/_build/html/_sources/16. Intervals/README.md	
@@ -0,0 +1,50 @@
+# Blind 75 LeetCode Problems
+
+This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
+
+## Problem List
+
+### Intervals
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [57. Insert Interval](https://leetcode.com/problems/insert-interval/)                                     | Hard       |
+| [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)                                   | Medium     |
+| [435. Non Overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)             | Medium     |
+| [252. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)                                       | Easy       |
+| [253. Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)                                 | Medium     |
+
+## How to Use
+
+1. Clone or download this repository to your local machine.
+
+```bash
+git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
+```
+
+2. Navigate to the directory of the specific problem you want to solve.
+
+```bash
+cd Blind-75-LeetCode/Problem-Directory
+```
+
+3. Open the problem's directory to find its description, code template, and solution(s).
+
+4. Solve the problem using your preferred programming language.
+
+5. Check the solution(s) provided in the directory for reference and to compare your solution.
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+## Disclaimer
+
+This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/17. Math Geometry/README.md b/_build/html/_sources/17. Math Geometry/README.md
new file mode 100644
index 0000000..8b2f180
--- /dev/null
+++ b/_build/html/_sources/17. Math Geometry/README.md	
@@ -0,0 +1,48 @@
+# Blind 75 LeetCode Problems
+
+This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
+
+## Problem List
+
+### Math & Geometry
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [48. Rotate Image](https://leetcode.com/problems/rotate-image/)                                           | Medium     |
+| [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)                                         | Medium     |
+| [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)                               | Medium     |
+
+## How to Use
+
+1. Clone or download this repository to your local machine.
+
+```bash
+git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
+```
+
+2. Navigate to the directory of the specific problem you want to solve.
+
+```bash
+cd Blind-75-LeetCode/Problem-Directory
+```
+
+3. Open the problem's directory to find its description, code template, and solution(s).
+
+4. Solve the problem using your preferred programming language.
+
+5. Check the solution(s) provided in the directory for reference and to compare your solution.
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+## Disclaimer
+
+This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
+
+Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/18. Bit Manipulation/README.md b/_build/html/_sources/18. Bit Manipulation/README.md
new file mode 100644
index 0000000..05b2fef
--- /dev/null
+++ b/_build/html/_sources/18. Bit Manipulation/README.md	
@@ -0,0 +1,51 @@
+# Blind 75 LeetCode Problems
+
+This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
+
+## Problem List
+
+### Bit Manupulation
+
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)                                   | Easy       |
+| [338. Counting Bits](https://leetcode.com/problems/counting-bits/)                                         | Medium     |
+| [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)                                           | Easy       |
+| [268. Missing Number](https://leetcode.com/problems/missing-number/)                                       | Easy       |
+| [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)                             | Medium     |
+
+
+## How to Use
+
+1. Clone or download this repository to your local machine.
+
+```bash
+git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
+```
+
+2. Navigate to the directory of the specific problem you want to solve.
+
+```bash
+cd Blind-75-LeetCode/Problem-Directory
+```
+
+3. Open the problem's directory to find its description, code template, and solution(s).
+
+4. Solve the problem using your preferred programming language.
+
+5. Check the solution(s) provided in the directory for reference and to compare your solution.
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+## Disclaimer
+
+This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
+
+Happy coding!
\ No newline at end of file
diff --git a/_sources/Book.ipynb b/_build/html/_sources/Book.ipynb
similarity index 100%
rename from _sources/Book.ipynb
rename to _build/html/_sources/Book.ipynb
diff --git a/_build/html/_sources/CODE_OF_CONDUCT.md b/_build/html/_sources/CODE_OF_CONDUCT.md
new file mode 100644
index 0000000..5fe5041
--- /dev/null
+++ b/_build/html/_sources/CODE_OF_CONDUCT.md
@@ -0,0 +1,128 @@
+# Contributor Covenant Code of Conduct
+
+## Our Pledge
+
+We as members, contributors, and leaders pledge to make participation in our
+community a harassment-free experience for everyone, regardless of age, body
+size, visible or invisible disability, ethnicity, sex characteristics, gender
+identity and expression, level of experience, education, socio-economic status,
+nationality, personal appearance, race, religion, or sexual identity
+and orientation.
+
+We pledge to act and interact in ways that contribute to an open, welcoming,
+diverse, inclusive, and healthy community.
+
+## Our Standards
+
+Examples of behavior that contributes to a positive environment for our
+community include:
+
+* Demonstrating empathy and kindness toward other people
+* Being respectful of differing opinions, viewpoints, and experiences
+* Giving and gracefully accepting constructive feedback
+* Accepting responsibility and apologizing to those affected by our mistakes,
+  and learning from the experience
+* Focusing on what is best not just for us as individuals, but for the
+  overall community
+
+Examples of unacceptable behavior include:
+
+* The use of sexualized language or imagery, and sexual attention or
+  advances of any kind
+* Trolling, insulting or derogatory comments, and personal or political attacks
+* Public or private harassment
+* Publishing others' private information, such as a physical or email
+  address, without their explicit permission
+* Other conduct which could reasonably be considered inappropriate in a
+  professional setting
+
+## Enforcement Responsibilities
+
+Community leaders are responsible for clarifying and enforcing our standards of
+acceptable behavior and will take appropriate and fair corrective action in
+response to any behavior that they deem inappropriate, threatening, offensive,
+or harmful.
+
+Community leaders have the right and responsibility to remove, edit, or reject
+comments, commits, code, wiki edits, issues, and other contributions that are
+not aligned to this Code of Conduct, and will communicate reasons for moderation
+decisions when appropriate.
+
+## Scope
+
+This Code of Conduct applies within all community spaces, and also applies when
+an individual is officially representing the community in public spaces.
+Examples of representing our community include using an official e-mail address,
+posting via an official social media account, or acting as an appointed
+representative at an online or offline event.
+
+## Enforcement
+
+Instances of abusive, harassing, or otherwise unacceptable behavior may be
+reported to the community leaders responsible for enforcement at
+Email.
+All complaints will be reviewed and investigated promptly and fairly.
+
+All community leaders are obligated to respect the privacy and security of the
+reporter of any incident.
+
+## Enforcement Guidelines
+
+Community leaders will follow these Community Impact Guidelines in determining
+the consequences for any action they deem in violation of this Code of Conduct:
+
+### 1. Correction
+
+**Community Impact**: Use of inappropriate language or other behavior deemed
+unprofessional or unwelcome in the community.
+
+**Consequence**: A private, written warning from community leaders, providing
+clarity around the nature of the violation and an explanation of why the
+behavior was inappropriate. A public apology may be requested.
+
+### 2. Warning
+
+**Community Impact**: A violation through a single incident or series
+of actions.
+
+**Consequence**: A warning with consequences for continued behavior. No
+interaction with the people involved, including unsolicited interaction with
+those enforcing the Code of Conduct, for a specified period of time. This
+includes avoiding interactions in community spaces as well as external channels
+like social media. Violating these terms may lead to a temporary or
+permanent ban.
+
+### 3. Temporary Ban
+
+**Community Impact**: A serious violation of community standards, including
+sustained inappropriate behavior.
+
+**Consequence**: A temporary ban from any sort of interaction or public
+communication with the community for a specified period of time. No public or
+private interaction with the people involved, including unsolicited interaction
+with those enforcing the Code of Conduct, is allowed during this period.
+Violating these terms may lead to a permanent ban.
+
+### 4. Permanent Ban
+
+**Community Impact**: Demonstrating a pattern of violation of community
+standards, including sustained inappropriate behavior,  harassment of an
+individual, or aggression toward or disparagement of classes of individuals.
+
+**Consequence**: A permanent ban from any sort of public interaction within
+the community.
+
+## Attribution
+
+This Code of Conduct is adapted from the [Contributor Covenant][homepage],
+version 2.0, available at
+https://www.contributor-covenant.org/version/2/0/code_of_conduct.html.
+
+Community Impact Guidelines were inspired by [Mozilla's code of conduct
+enforcement ladder](https://github.com/mozilla/diversity).
+
+[homepage]: https://www.contributor-covenant.org
+
+For answers to common questions about this code of conduct, see the FAQ at
+https://www.contributor-covenant.org/faq. Translations are available at
+https://www.contributor-covenant.org/translations.
diff --git a/_build/html/_sources/CONTRIBUTING.md b/_build/html/_sources/CONTRIBUTING.md
new file mode 100644
index 0000000..ca794ef
--- /dev/null
+++ b/_build/html/_sources/CONTRIBUTING.md
@@ -0,0 +1,79 @@
+# Contributing Guidelines
+
+Thank you for considering contributing to the Blind 75 LeetCode repository! We welcome contributions from the community to help improve this resource and make it even more valuable.
+
+Please take a moment to review this document to understand how you can contribute.
+
+## Table of Contents
+- [Ways to Contribute](#ways-to-contribute)
+- [Getting Started](#getting-started)
+- [Contributing Process](#contributing-process)
+- [Code of Conduct](#code-of-conduct)
+
+## Ways to Contribute
+
+There are several ways you can contribute to this repository:
+
+1. **Suggest Improvements**: If you have ideas for making the Blind 75 LeetCode repository better, please open an issue to discuss your suggestions.
+
+2. **Report Issues**: If you encounter any issues with the existing content or code, please open an issue and provide as much detail as possible.
+
+3. **Contribute Solutions**: You can contribute solutions to the problems listed in the repository. Please follow the guidelines mentioned below when submitting your solutions.
+
+4. **Enhance Documentation**: Improve documentation, including explanations and time/space complexity analysis, for existing solutions.
+
+5. **Fix Bugs**: If you find and fix any bugs in the existing code or documentation, you are welcome to submit a pull request.
+
+## Getting Started
+
+Before you start contributing, please ensure you have the following prerequisites:
+
+- A GitHub account.
+- Familiarity with Git and GitHub.
+- Code editor or IDE for making changes.
+
+## Contributing Process
+
+Here's the general process for contributing:
+
+1. **Fork the Repository**: Click the "Fork" button at the top of this repository to create your copy.
+
+2. **Clone Your Fork**: Clone your fork to your local machine.
+
+    ```
+    git clone https://github.com/your-username/Blind-75-LeetCode.git
+    ```
+
+3. **Create a Branch**: Create a new branch for your contribution.
+
+    ```
+    git checkout -b feature/your-feature-name
+    ```
+
+4. **Make Changes**: Implement your changes, following the guidelines provided.
+
+5. **Test Your Changes**: Ensure that your changes do not introduce new issues and are consistent with the repository's style and structure.
+
+6. **Commit Changes**: Commit your changes with a meaningful commit message.
+
+    ```
+    git commit -m "Add solution to problem X"
+    ```
+
+7. **Push Changes**: Push your changes to your fork on GitHub.
+
+    ```
+    git push origin feature/your-feature-name
+    ```
+
+8. **Create a Pull Request**: Go to the original repository and create a pull request. Please provide a detailed description of your changes.
+
+9. **Review and Collaboration**: Your pull request will be reviewed, and you may need to make further changes. Be responsive to comments and feedback.
+
+10. **Merge**: Once your pull request is approved, it will be merged into the main branch.
+
+## Code of Conduct
+
+Please note that we have a [Code of Conduct](CODE_OF_CONDUCT.md) that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.
+
+We appreciate your contributions to the Blind 75 LeetCode repository and look forward to working together to make it an even better resource for the community!
diff --git a/_build/html/_sources/README.md b/_build/html/_sources/README.md
new file mode 100644
index 0000000..a9b8b2e
--- /dev/null
+++ b/_build/html/_sources/README.md
@@ -0,0 +1,216 @@
+# Blind 75 LeetCode Solutions
+
+![Blind 75 LeetCode](https://img.shields.io/badge/Blind_75_LeetCode-Solutions-blue?labelColor=red)
+
+This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
+
+## Problem List
+
+### Array & Hashing
+![Array & Hashing](https://img.shields.io/badge/Array_&_Hashing-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [217. Contains Duplicate](https://leetcode.com/problems/contains-duplicate/)                        | Easy       |
+| [242. Valid Anagram](https://leetcode.com/problems/valid-anagram/)                                  | Easy       |
+| [1. Two Sum](https://leetcode.com/problems/two-sum/)                                              | Easy       |
+| [49. Group Anagrams](https://leetcode.com/problems/group-anagrams/)                                | Medium     |
+| [347. Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/)              | Medium     |
+| [238. Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/)    | Medium     |
+| [659. Encode and Decode Strings](https://leetcode.com/problems/encode-and-decode-strings/)          | Medium     |
+| [128. Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/)    | Hard       |
+
+### Two Pointers
+![Array & Hashing](https://img.shields.io/badge/Two_Pointers-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)                            | Easy       |
+| [15. 3Sum](https://leetcode.com/problems/3sum/)                                                       | Medium     |
+| [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)           | Medium     |
+
+### Sliding Window
+![Array & Hashing](https://img.shields.io/badge/Sliding_Window-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [121. Best Time to Buy And Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | Easy       |
+| [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) | Medium |
+| [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) | Medium |
+| [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)             | Hard       |
+
+### Stack
+![Array & Hashing](https://img.shields.io/badge/Stack-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)                            | Easy       |
+
+### Binary Search
+![Array & Hashing](https://img.shields.io/badge/Binary_Search-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [153. Find Minimum In Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) | Medium |
+| [33. Search In Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)   | Medium     |
+
+### Linked List
+![Array & Hashing](https://img.shields.io/badge/Linked_List-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)                       | Easy       |
+| [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)                   | Easy       |
+| [143. Reorder List](https://leetcode.com/problems/reorder-list/)                                     | Medium     |
+| [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | Medium |
+| [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)                           | Medium     |
+| [23. Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)                       | Hard       |
+
+### Trees
+![Array & Hashing](https://img.shields.io/badge/Trees-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)                          | Easy       |
+| [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)     | Easy       |
+| [100. Same Tree](https://leetcode.com/problems/same-tree/)                                           | Easy       |
+| [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)               | Easy       |
+| [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/) | Easy |
+| [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/) | Medium |
+| [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)       | Medium     |
+| [230. Kth Smallest Element In a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)   | Medium     |
+| [105. Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) | Medium |
+| [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)     | Hard       |
+| [297. Serialize And Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/) | Hard |
+
+### Tries
+![Array & Hashing](https://img.shields.io/badge/Tries-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [208. Implement Trie Prefix Tree](https://leetcode.com/problems/implement-trie-prefix-tree/)         | Medium     |
+| [211. Design Add And Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/) | Medium |
+| [212. Word Search II](https://leetcode.com/problems/word-search-ii/)                                 | Hard       |
+
+### Heap / Priority Queue
+![Array & Hashing](https://img.shields.io/badge/Heap_/_Priority_Queue-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [295. Find Median From Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)     | Hard       |
+
+### Backtracking
+![Array & Hashing](https://img.shields.io/badge/Backtracking-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [39. Combination Sum](https://leetcode.com/problems/combination-sum/)                                 | Medium     |
+| [79. Word Search](https://leetcode.com/problems/word-search/)                                         | Medium     |
+
+### Graphs
+![Array & Hashing](https://img.shields.io/badge/Graphs-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)                           | Medium     |
+| [133. Clone Graph](https://leetcode.com/problems/clone-graph/)                                       | Medium     |
+| [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)       | Medium     |
+| [207. Course Schedule](https://leetcode.com/problems/course-schedule/)                                 | Medium     |
+| [323. Number of Connected Components In An Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/) | Medium |
+| [261. Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)                               | Medium     |
+
+### Advanced Graphs
+![Array & Hashing](https://img.shields.io/badge/Advanced_Graphs-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [269. Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)                               | Hard       |
+
+### 1-D Dynamic Programming
+![Array & Hashing](https://img.shields.io/badge/1_D_Dynamic_Programming-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)                                   | Easy       |
+| [198. House Robber](https://leetcode.com/problems/house-robber/)                                       | Easy       |
+| [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)                                 | Medium     |
+| [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)       | Medium     |
+| [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)                   | Medium     |
+| [91. Decode Ways](https://leetcode.com/problems/decode-ways/)                                         | Medium     |
+| [322. Coin Change](https://leetcode.com/problems/coin-change/)                                         | Medium     |
+| [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)             | Medium     |
+| [139. Word Break](https://leetcode.com/problems/word-break/)                                           | Medium     |
+| [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | Medium     |
+
+### 2-D Dynamic Programming
+![Array & Hashing](https://img.shields.io/badge/2_D_Dynamic_Programming-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [62. Unique Paths](https://leetcode.com/problems/unique-paths/)                                         | Medium     |
+| [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)       | Medium     |
+
+### Greedy
+![Array & Hashing](https://img.shields.io/badge/Greedy-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)                                   | Easy       |
+| [55. Jump Game](https://leetcode.com/problems/jump-game/)                                                 | Medium     |
+
+### Intervals
+![Array & Hashing](https://img.shields.io/badge/Intervals-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [57. Insert Interval](https://leetcode.com/problems/insert-interval/)                                     | Hard       |
+| [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)                                   | Medium     |
+| [435. Non Overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)             | Medium     |
+| [252. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)                                       | Easy       |
+| [253. Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)                                 | Medium     |
+
+### Math & Geometry
+![Array & Hashing](https://img.shields.io/badge/Math_&_Geometry-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [48. Rotate Image](https://leetcode.com/problems/rotate-image/)                                           | Medium     |
+| [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)                                         | Medium     |
+| [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)                               | Medium     |
+
+### Bit Manipulation
+![Array & Hashing](https://img.shields.io/badge/Bit_Manipulation-Solutions-blue?labelColor=red)
+| Problem Name                                    | Difficulty |
+| ----------------------------------------------- | ---------- |
+| [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)                                   | Easy       |
+| [338. Counting Bits](https://leetcode.com/problems/counting-bits/)                                         | Medium     |
+| [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)                                           | Easy       |
+| [268. Missing Number](https://leetcode.com/problems/missing-number/)                                       | Easy       |
+| [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)                             | Medium     |
+
+
+## How to Use
+
+1. Clone or download this repository to your local machine.
+
+```bash
+git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
+```
+
+2. Navigate to the directory of the specific problem you want to solve.
+
+```bash
+cd Blind-75-LeetCode/Problem-Directory
+```
+
+3. Open the problem's directory to find its description, code template, and solution(s).
+
+4. Solve the problem using your preferred programming language.
+
+5. Check the solution(s) provided in the directory for reference and to compare your solution.
+
+## Contribute
+
+If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
+
+## Resources
+
+- [LeetCode](https://leetcode.com/): The official LeetCode website.
+- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
+
+## Disclaimer
+
+This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
+
+## Code of Conduct
+
+We have a [Code of Conduct](CODE_OF_CONDUCT.md) that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.
+
+## Security Policy
+
+We take security seriously. Please review our [Security Policy](SECURITY.md) for information on reporting security vulnerabilities.
+
+Happy coding!
diff --git a/_build/html/_sources/SECURITY.md b/_build/html/_sources/SECURITY.md
new file mode 100644
index 0000000..7c19993
--- /dev/null
+++ b/_build/html/_sources/SECURITY.md
@@ -0,0 +1,42 @@
+# Security Policy
+
+## Reporting a Vulnerability
+
+We take the security of the Blind 75 LeetCode repository seriously. If you believe you have found a security vulnerability, please help us by following these steps:
+
+1. **Do Not Disclose Publicly**: Please do not disclose the issue publicly until it has been addressed by our team.
+
+2. **Report Privately**: Create a GitHub issue with the title "Security Vulnerability Report" and provide a detailed description of the vulnerability. You can use a sample report template as follows:
+
+    ```
+    **Description**
+    A clear and concise description of the security vulnerability.
+
+    **Steps to Reproduce**
+    Provide detailed steps on how to reproduce the vulnerability.
+
+    **Expected Behavior**
+    Describe what you expected to happen.
+
+    **Actual Behavior**
+    Describe what actually happened.
+
+    **Additional Information**
+    Any additional information that can help us understand and reproduce the issue, such as environment details, configuration, or screenshots.
+    ```
+
+3. **Secure Communication**: If the vulnerability involves sensitive data, please encrypt your communication with our PGP key. Contact us for our PGP key information.
+
+4. **Response**: We will acknowledge your report within 5 business days and provide an estimated timeline for the resolution.
+
+5. **Fix and Disclosure**: We will work to fix the issue promptly and privately. Once resolved, we will provide a security advisory and give credit to the reporter if desired.
+
+6. **Responsible Disclosure**: We encourage responsible disclosure. Please do not take advantage of the vulnerability or attempt to exploit it for malicious purposes.
+
+## Security Updates
+
+We are committed to promptly addressing and resolving any security issues. We recommend that you stay up-to-date with the latest releases of this repository, as they may include security updates and fixes.
+
+## Contact
+
+If you have any questions or need further assistance regarding security, you can contact us at mohsentabibian@gmail.com.
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diff --git a/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb b/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb
new file mode 100644
index 0000000..5d55ca7
--- /dev/null
+++ b/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb	
@@ -0,0 +1,203 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 100: Same Tree\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
+    "\n",
+    "---\n",
+    "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
+    "\n",
+    "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
+    "\n",
+    "**Constraints**\n",
+    "\n",
+    "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
+    "2. $-10^4$ <= `Node.val` <= $10^4$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Definition for a binary tree node.\n",
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "def isSameTree(p, q):\n",
+    "    # Base case: If both p and q are None, the trees are the same.\n",
+    "    if not p and not q:\n",
+    "        return True\n",
+    "    \n",
+    "    # Base case: If either p or q is None (but not both), the trees are different.\n",
+    "    if not p or not q:\n",
+    "        return False\n",
+    "    \n",
+    "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
+    "    if p.val != q.val:\n",
+    "        return False\n",
+    "    \n",
+    "    # Recursively check the left and right subtrees of p and q.\n",
+    "    # If both subtrees are the same, the entire trees are the same.\n",
+    "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
+    "\n",
+    "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
+    "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
+    "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
+    "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
+    "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
+    "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 1\n",
+    "\n",
+    "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
+    "\n",
+    "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
+    "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
+    "print(isSameTree(p1, q1)) "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 2\n",
+    "\n",
+    "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
+    "\n",
+    "p2 = TreeNode(1, TreeNode(2), None)\n",
+    "q2 = TreeNode(1, None, TreeNode(2))\n",
+    "print(isSameTree(p2, q2))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 3\n",
+    "\n",
+    "#Input: p = [1,2,1], q = [1,1,2]\n",
+    "\n",
+    "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
+    "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
+    "print(isSameTree(p3, q3))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "**Time Complexity**\n",
+    "\n",
+    "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
+    "\n",
+    "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
+    "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
+    "\n",
+    "**Space Complexity**\n",
+    "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
+    "\n",
+    "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
+    "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
+    "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
+    "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
+    "\n",
+    "**In summary:**\n",
+    "\n",
+    "+ Time Complexity: $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
+    "+ Space Complexity: $O(n)$ due to the recursive call stack."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb b/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb
new file mode 100644
index 0000000..9256d0e
--- /dev/null
+++ b/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb	
@@ -0,0 +1,203 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 128. Longest Consecutive Sequence\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Longest Consecutive Sequence problem on LeetCode, click here!](https://leetcode.com/problems/longest-consecutive-sequence/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Given an unsorted array of integers `nums`, return *the length of the longest consecutive elements sequence.*\n",
+    "\n",
+    "You must write an algorithm that runs in `O(n)` time.\n",
+    "\n",
+    "*Constraints:**\n",
+    "- 0 <= `nums.length` <= $10^5$\n",
+    "- $-10^9$ <= `nums[i]` <= $10^9$\n",
+    "\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def longestConsecutive(nums):\n",
+    "    if not nums:\n",
+    "        return 0\n",
+    "\n",
+    "    num_set = set(nums)\n",
+    "    max_length = 0\n",
+    "\n",
+    "    for num in num_set:\n",
+    "        if num - 1 not in num_set:\n",
+    "            current_num = num\n",
+    "            current_length = 1\n",
+    "\n",
+    "            while current_num + 1 in num_set:\n",
+    "                current_num += 1\n",
+    "                current_length += 1\n",
+    "\n",
+    "            max_length = max(max_length, current_length)\n",
+    "\n",
+    "    return max_length"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We start by defining the `longestConsecutive` function that takes the `nums` array as input.\n",
+    "\n",
+    "2. We check if the `nums` array is empty. If it's empty, there are no consecutive elements, so we return 0.\n",
+    "\n",
+    "3. We create a Python set called `num_set` and insert all elements from the `nums` array into it. Using a set allows us to efficiently check for the existence of elements in O(1) time.\n",
+    "\n",
+    "4. We initialize a variable `max_length` to 0. This variable will keep track of the maximum length of consecutive elements sequence found.\n",
+    "\n",
+    "5. We iterate through the elements of the `num_set`. For each element `num`, we check if `num - 1` exists in the `num_set`. If it doesn't exist, it means `num` is the starting element of a potential consecutive sequence.\n",
+    "\n",
+    "6. Inside the loop, we initialize two variables: `current_num` to the current element `num` and `current_length` to 1. We start with a length of 1 because `num` itself is part of the sequence.\n",
+    "\n",
+    "7. We then enter a while loop that continues as long as `current_num + 1` exists in the `num_set`. This means we are incrementing the consecutive sequence.\n",
+    "\n",
+    "8. Inside the while loop, we increment `current_num` by 1 and also increment `current_length` by 1 to account for the next consecutive element.\n",
+    "\n",
+    "9. We compare `current_length` with the `max_length` and update `max_length` if the current sequence is longer.\n",
+    "\n",
+    "10. After the while loop, we move to the next element in the outer loop and repeat the process.\n",
+    "\n",
+    "11. Finally, we return the `max_length` as the result, which represents the length of the longest consecutive elements sequence in the `nums` array.\n",
+    "\n",
+    "The key idea here is to use a set to efficiently check for the existence of elements and to iterate through the elements, considering each element as the potential start of a consecutive sequence. By doing this, we can find the longest consecutive sequence in O(n) time complexity, where n is the number of elements in the array."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "nums1 = [100, 4, 200, 1, 3, 2]\n",
+    "print(longestConsecutive(nums1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "9\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "\n",
+    "nums2 = [0, 3, 7, 2, 5, 8, 4, 6, 0, 1]\n",
+    "print(longestConsecutive(nums2)) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "The code is designed to run in O(n) time complexity, where n is the number of elements in the `nums` array. Here's the breakdown:\n",
+    "\n",
+    "1. Constructing the `num_set` by inserting all elements from `nums` into it takes O(n) time because we perform an insertion operation for each element in `nums`.\n",
+    "\n",
+    "2. The main loop iterates through the elements in `num_set`. In the worst case, each element is visited only once, so the loop itself takes O(n) time.\n",
+    "\n",
+    "3. Within the loop, we have a while loop that may also take O(n) time in the worst case. However, this while loop is nested inside the main loop, so its overall time complexity remains O(n).\n",
+    "\n",
+    "Therefore, the overall time complexity of the code is O(n).\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "The space complexity of the code is determined by the space used by the `num_set` and a few additional variables. Here's the breakdown:\n",
+    "\n",
+    "1. `num_set` is a set that stores the unique elements from `nums`. In the worst case, it can store all n elements from `nums`, so the space complexity is O(n).\n",
+    "\n",
+    "2. The additional variables used, such as `max_length`, `current_num`, and `current_length`, have constant space requirements and do not depend on the size of the input array.\n",
+    "\n",
+    "Therefore, the overall space complexity of the code is O(n) due to the space used by the `num_set`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. Write an algorithm that not only returns the length of the longest consecutive elements sequence but also returns the actual consecutive sequence itself.\n",
+    "\n",
+    "2. Extend the problem to allow elements to be considered consecutive if they are within a certain absolute difference (e.g., less than or equal to k) instead of exactly 1. Write an algorithm that finds the longest such sequence and runs in O(n) time."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb b/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb
new file mode 100644
index 0000000..f7dbbe9
--- /dev/null
+++ b/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb	
@@ -0,0 +1,210 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 217. Contains Duplicate\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Contains Duplicate problem on LeetCode, click here!](https://leetcode.com/problems/contains-duplicate/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- 1 <= `nums.length` <= $10^5$\n",
+    "- $-10^9$ <= `nums[i]` <= $10^9$\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def containsDuplicate(nums):\n",
+    "    # Create an empty set to store unique elements\n",
+    "    unique_elements = set()\n",
+    "    \n",
+    "    # Iterate through the array\n",
+    "    for num in nums:\n",
+    "        # If the element is already in the set, return True\n",
+    "        if num in unique_elements:\n",
+    "            return True\n",
+    "        # Otherwise, add it to the set\n",
+    "        else:\n",
+    "            unique_elements.add(num)\n",
+    "    \n",
+    "    # If the loop completes without finding duplicates, return False\n",
+    "    return False"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. The `containsDuplicate` function takes a single argument, `nums`, which is the input integer array.\n",
+    "\n",
+    "2. Inside the function, an empty set called `unique_elements` is created. This set will be used to keep track of unique elements in the input array.\n",
+    "\n",
+    "3. The function then iterates through the input array `nums` using a `for` loop.\n",
+    "\n",
+    "4. For each element `num` in the array, it checks whether `num` is already in the `unique_elements` set using the `if num in unique_elements:` condition.\n",
+    "\n",
+    "5. If `num` is already in the set, it means there is a duplicate element in the array, and the function immediately returns `True`.\n",
+    "\n",
+    "6. If `num` is not in the set, it is added to the `unique_elements` set using `unique_elements.add(num)`.\n",
+    "\n",
+    "7. The loop continues to the next element, and the process repeats.\n",
+    "\n",
+    "8. If the loop completes without finding any duplicates, it means that all elements in the array are distinct, and the function returns `False`.\n",
+    "\n",
+    "The code efficiently utilizes a set data structure to keep track of unique elements while iterating through the array, allowing it to quickly detect duplicate elements. This code meets the problem's requirements and constraints, as explained earlier."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "nums1 = [1, 2, 3, 1]\n",
+    "print(containsDuplicate(nums1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "nums2 = [1, 2, 3, 4]\n",
+    "print(containsDuplicate(nums2))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3\n",
+    "nums3 = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]\n",
+    "print(containsDuplicate(nums3))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code for the \"Contains Duplicate\" problem:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "The primary operation that affects the time complexity is the `for` loop that iterates through the input array `nums`. In the worst case, the loop will iterate through all `n` elements of the array, where `n` is the length of the input array.\n",
+    "\n",
+    "- Iterating through the array: O(n)\n",
+    "\n",
+    "Inside the loop, we perform two operations:\n",
+    "\n",
+    "1. Checking whether an element exists in the `unique_elements` set (`if num in unique_elements`). This operation has an average time complexity of O(1) for a set.\n",
+    "2. Adding an element to the `unique_elements` set (`unique_elements.add(num)`), which also has an average time complexity of O(1) for a set.\n",
+    "\n",
+    "Since these operations are performed for each element in the array, the overall time complexity remains O(n).\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "The space complexity is determined by the additional data structures used in the code, which are the `unique_elements` set.\n",
+    "\n",
+    "- `unique_elements` set: This set stores unique elements from the input array. In the worst case, if all elements in the input array are distinct, the set will store all `n` elements.\n",
+    "\n",
+    "Therefore, the space complexity is O(n) because, in the worst case, the set's size will grow linearly with the input array's size.\n",
+    "\n",
+    "**In summary**, the time complexity of the code is O(n), where `n` is the length of the input array, and the space complexity is also O(n) in the worst case. This code is efficient and meets the constraints of the problem."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **K Duplicates:** Modify the problem to return `true` if there are exactly K duplicate elements in the array. Write a function that takes an additional integer parameter K and returns `true` if there are exactly K duplicates.\n",
+    "2. **Frequency Count:** Write a function that returns a list of all the elements that appear more than once in the array, along with their frequencies."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb b/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb
new file mode 100644
index 0000000..c5a69c0
--- /dev/null
+++ b/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb	
@@ -0,0 +1,191 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 238. Product of Array Except Self\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Product of Array Except Self problem on LeetCode, click here!](https://leetcode.com/problems/product-of-array-except-self/)\n",
+    "\n",
+    "---\n",
+    "Given an integer array `nums`, return *an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.*\n",
+    "\n",
+    "The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
+    "\n",
+    "You must write an algorithm that runs in O(n) time and without using the division operation.\n",
+    "\n",
+    "Constraints:\n",
+    "- 2 <= `nums.length` <= $10^5$\n",
+    "- `-30 <= nums[i] <= 30`\n",
+    "- The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
+    "\n",
+    "**Follow-up:** Can you solve the problem in O(1) extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def productExceptSelf(nums):\n",
+    "    n = len(nums)\n",
+    "    \n",
+    "    # Initialize two arrays to store products to the left and right of each element\n",
+    "    left_products = [1] * n\n",
+    "    right_products = [1] * n\n",
+    "    \n",
+    "    # Calculate products to the left of each element\n",
+    "    left_product = 1\n",
+    "    for i in range(n):\n",
+    "        left_products[i] = left_product\n",
+    "        left_product *= nums[i]\n",
+    "    \n",
+    "    # Calculate products to the right of each element\n",
+    "    right_product = 1\n",
+    "    for i in range(n - 1, -1, -1):\n",
+    "        right_products[i] = right_product\n",
+    "        right_product *= nums[i]\n",
+    "    \n",
+    "    # Calculate the final answer using left and right products\n",
+    "    answer = [left_products[i] * right_products[i] for i in range(n)]\n",
+    "    \n",
+    "    return answer"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. `n` is initialized as the length of the `nums` array.\n",
+    "2. Two arrays, `left_products` and `right_products`, are created, each with a length of `n` and filled with ones initially.\n",
+    "3. `left_product` is initialized to 1.\n",
+    "4. A loop iterates through the `nums` array, from left to right. For each element at index `i`, it stores the product of all elements to its left (including itself) in the `left_products` array and updates `left_product` accordingly.\n",
+    "5. `right_product` is initialized to 1.\n",
+    "6. Another loop iterates through the `nums` array in reverse order, from right to left. For each element at index `i`, it stores the product of all elements to its right (including itself) in the `right_products` array and updates `right_product`.\n",
+    "7. The final `answer` list is constructed by multiplying corresponding elements from the `left_products` and `right_products` arrays for each index `i`.\n",
+    "8. The function returns the `answer` list, which contains the desired results.\n",
+    "\n",
+    "Overall, this algorithm efficiently computes the product of all elements except the current element, as required. It runs in O(n) time complexity and uses O(n) extra space, which is within the problem's constraints."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[24, 12, 8, 6]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "nums = [1,2,3,4]\n",
+    "result1 = productExceptSelf(nums)\n",
+    "print(result1)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[0, 0, 9, 0, 0]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "nums = [-1,1,0,-3,3]\n",
+    "result2 = productExceptSelf(nums)\n",
+    "print(result2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "1. The first loop iterates through the `nums` array to calculate the products to the left of each element. This loop runs in O(n) time, where n is the length of the `nums` array.\n",
+    "2. The second loop also iterates through the `nums` array in reverse order to calculate the products to the right of each element. This loop also runs in O(n) time.\n",
+    "3. The final loop constructs the `answer` list by multiplying elements from the `left_products` and `right_products` arrays. This loop runs in O(n) time.\n",
+    "\n",
+    "Since all the loops are independent and sequential, the overall time complexity is O(n).\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "1. Two additional arrays, `left_products` and `right_products`, are created with a length of `n`, where `n` is the length of the `nums` array. Therefore, these arrays consume O(n) extra space.\n",
+    "2. The `left_product` and `right_product` variables are used to keep track of the product to the left and right of each element, respectively. These variables occupy O(1) extra space.\n",
+    "3. The `answer` list, which contains the final results, also consumes O(n) space, but it's not counted toward the extra space complexity analysis as per the problem's constraints.\n",
+    "\n",
+    "Overall, the space complexity is O(n) for this solution."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Handle Zeros in the Input**: Modify the solution to handle cases where the input array contains zeros. For example, if `nums = [0, 1, 2, 3, 0]`, the output should be `[0, 0, 0, 0, 0]` because any element multiplied by zero results in zero.\n",
+    "\n",
+    "2. **In-Place Modification**: Attempt to solve the problem with in-place modification of the `nums` array, where the output is stored directly in the input array without using any additional space."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb b/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb
new file mode 100644
index 0000000..5649123
--- /dev/null
+++ b/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb	
@@ -0,0 +1,192 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 242. Valid Anagram\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Valid Anagram problem on LeetCode, click here!](https://leetcode.com/problems/valid-anagram/)\n",
+    "\n",
+    "---\n",
+    "Given two strings `s` and `t`, return `true` if `t` is an anagram of `s`, and `false` otherwise.\n",
+    "\n",
+    "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- 1 <= `s.length`, `t.length` <= $5 * 10^4$\n",
+    "- `s` and `t` consist of lowercase English letters.\n",
+    "\n",
+    "**Follow-up:** What if the inputs contain Unicode characters? How would you adapt your solution to such a case?\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def isAnagram(s: str, t: str) -> bool:\n",
+    "    # Check if the lengths of s and t are not equal\n",
+    "    if len(s) != len(t):\n",
+    "        return False\n",
+    "\n",
+    "    # Create dictionaries to store character frequencies for s and t\n",
+    "    countS, countT = {}, {}\n",
+    "\n",
+    "    # Count character frequencies in s\n",
+    "    for i in range(len(s)):\n",
+    "        countS[s[i]] = 1 + countS.get(s[i], 0)\n",
+    "\n",
+    "    # Count character frequencies in t\n",
+    "    for i in range(len(t)):\n",
+    "        countT[t[i]] = 1 + countT.get(t[i], 0)\n",
+    "\n",
+    "    # Check if the character frequencies in s and t are the same\n",
+    "    return countS == countT"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. The code defines a class called `Solution` with a method `isAnagram` that takes two input strings, `s` and `t`, and returns a boolean (`True` or `False`) based on whether `t` is an anagram of `s`.\n",
+    "\n",
+    "2. The first check performed in the method is whether the lengths of `s` and `t` are equal. If they are not equal, it immediately returns `False` because strings of different lengths cannot be anagrams of each other.\n",
+    "\n",
+    "3. Two dictionaries, `countS` and `countT`, are created to store the frequency of characters in `s` and `t`, respectively. These dictionaries will be used to count the occurrences of each character in the strings.\n",
+    "\n",
+    "4. The code then enters a loop to iterate through each character in string `s` using a `for` loop. Inside the loop, it updates the `countS` dictionary. The line `countS[s[i]] = 1 + countS.get(s[i], 0)` increments the count of character `s[i]` in `countS` by 1. If the character is not already in the dictionary, it initializes the count to 1.\n",
+    "\n",
+    "5. Similarly, the code iterates through each character in string `t` and updates the `countT` dictionary in the same way.\n",
+    "\n",
+    "6. After counting the character frequencies in both strings, the code compares the two dictionaries using `countS == countT`. If the dictionaries are equal, it means that both strings have the same character frequencies, and therefore, `t` is an anagram of `s`. In this case, the method returns `True`.\n",
+    "\n",
+    "7. If the dictionaries are not equal, the method returns `False`, indicating that `t` is not an anagram of `s`.\n",
+    "\n",
+    "In summary, this code efficiently determines whether two input strings are anagrams by counting the character frequencies in each string using dictionaries and then comparing these counts. If the character frequencies match, the strings are considered anagrams, and the method returns `True`; otherwise, it returns `False`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "s = \"anagram\"\n",
+    "t = \"nagaram\"\n",
+    "print(isAnagram(s, t))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "s = \"rat\"\n",
+    "t = \"car\"\n",
+    "print(isAnagram(s, t))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "1. The code first checks whether the lengths of the input strings `s` and `t` are equal, which takes constant time. This check has a time complexity of O(1).\n",
+    "\n",
+    "2. The code then iterates through both strings, `s` and `t`, once to count the character frequencies. Since both strings have a maximum length of 5 * 10^4, the worst-case scenario is that the code iterates through 5 * 10^4 characters in each string. This results in a linear time complexity of O(n), where n is the length of the longer of the two input strings (`s` or `t`).\n",
+    "\n",
+    "3. Finally, the code compares the two dictionaries `countS` and `countT` to check if the character frequencies match. This comparison takes O(n) time in the worst case, where n is the length of the longer string.\n",
+    "\n",
+    "Overall, the time complexity of the code is O(n), where n is the length of the longer input string.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "1. The code uses two dictionaries, `countS` and `countT`, to store the character frequencies of the input strings `s` and `t`. In the worst case, both dictionaries can contain all unique characters from the input strings. Since the input strings can have a maximum length of 5 * 10^4, the space complexity for these dictionaries is O(5 * 10^4) or simply O(n), where n is the length of the longer input string.\n",
+    "\n",
+    "2. The code uses a few additional variables for bookkeeping, but these variables have constant space requirements and do not depend on the input size. Therefore, they do not significantly impact the overall space complexity.\n",
+    "\n",
+    "**In summary**, the space complexity of the code is O(n), where n is the length of the longer input string."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Anagram Chains**: Given a list of words, find the longest chain of anagrams, where each word in the chain is an anagram of the previous word. For example, given [\"bat\", \"tab\", \"cat\", \"tac\", \"dog\"], the longest anagram chain is [\"bat\", \"tab\", \"cat\", \"tac\"].\n",
+    "\n",
+    "2. **Minimum Deletions to Make Anagrams**: Given two strings `s` and `t`, find the minimum number of character deletions required in both strings to make them anagrams of each other. For example, if `s = \"hello\"` and `t = \"billion\"`, you can remove the characters \"heo\" from `s` and \"billi\" from `t` to make them anagrams (\"lo\" and \"on\" are left in the two strings)."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb b/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb
new file mode 100644
index 0000000..1b11faf
--- /dev/null
+++ b/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb	
@@ -0,0 +1,209 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 347. Top K Frequent Elements\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Top K Frequent Elements problem on LeetCode, click here!](https://leetcode.com/problems/top-k-frequent-elements/)\n",
+    "\n",
+    "---\n",
+    "Given an integer array `nums` and an integer `k`, return the `k` most frequent elements. You may return the answer in **any order**.\n",
+    "\n",
+    "Constraints:\n",
+    "- 1 <= `nums.length` <= $10^5$\n",
+    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
+    "- `k` is in the range `[1, the number of unique elements in the array]`.\n",
+    "- It is **guaranteed** that the answer is **unique**.\n",
+    "\n",
+    "**Follow-up:** Your algorithm's time complexity must be better than $O(n\\ log\\ n)$, where n is the array's size.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def topKFrequent(nums, k):\n",
+    "    # Create a dictionary to store the frequency of each element\n",
+    "    count = {}\n",
+    "    # Create a list of lists to store elements with the same frequency\n",
+    "    frequency = [[] for i in range(len(nums) + 1)]\n",
+    "\n",
+    "    # Count the frequency of each element in nums\n",
+    "    for n in nums:\n",
+    "        count[n] = 1 + count.get(n, 0)\n",
+    "\n",
+    "    # Place elements in the freq list according to their frequency\n",
+    "    for n, c in count.items():\n",
+    "        frequency[c].append(n)\n",
+    "\n",
+    "    res = []\n",
+    "    # Traverse freq list from the end (higher frequencies)\n",
+    "    for i in range(len(frequency) - 1, 0, -1):\n",
+    "        for n in frequency[i]:\n",
+    "            res.append(n)\n",
+    "            if len(res) == k:\n",
+    "                return res"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. `def topKFrequent(nums, k):`: This is a function that takes two arguments: `nums`, which is the input array of integers, and `k`, which is the number of most frequent elements to return.\n",
+    "\n",
+    "2. `count = {}`: This dictionary `count` will be used to store the frequency of each unique element in the `nums` array. The keys are elements from the input array, and the values are their corresponding frequencies.\n",
+    "\n",
+    "3. `frequency = [[] for i in range(len(nums) + 1)]`: This creates a list of empty lists called `frequency`. It's used to store elements based on their frequencies, similar to the `freq` list in the previous code. The size of this list is set to be one greater than the length of the input `nums`.\n",
+    "\n",
+    "4. `for n in nums:`: This loop iterates through each element `n` in the input `nums` array.\n",
+    "\n",
+    "5. `count[n] = 1 + count.get(n, 0)`: This line counts the frequency of each element `n` in the `nums` array. It uses the `count.get(n, 0)` method to retrieve the current count of `n` from the `count` dictionary. If `n` is not in the dictionary, it defaults to 0. It then increments the count by 1.\n",
+    "\n",
+    "6. After the above loop, the `count` dictionary will contain counts of each unique element in the `nums` array.\n",
+    "\n",
+    "7. `for n, c in count.items():`: This loop iterates through the items (key-value pairs) of the `count` dictionary. `n` represents the element, and `c` represents its frequency.\n",
+    "\n",
+    "8. `frequency[c].append(n)`: This line places the element `n` into the bucket corresponding to its frequency `c`. Buckets are represented by the `frequency` list. For example, if an element `n` has a frequency of 3, it will be added to `frequency[3]`.\n",
+    "\n",
+    "9. After this loop, the `frequency` list will contain buckets of elements grouped by their frequencies.\n",
+    "\n",
+    "10. `res = []`: This list `res` will be used to store the k most frequent elements.\n",
+    "\n",
+    "11. `for i in range(len(freq) - 1, 0, -1):`: This loop iterates in reverse order through the `frequency` list, starting from the highest frequency and going down to 1. Note that there is a typo here; it should be `frequency` instead of `freq`.\n",
+    "\n",
+    "12. `for n in frequency[i]:`: For each element `n` in the current bucket (i.e., elements with frequency `i`), it appends `n` to the result list `res`.\n",
+    "\n",
+    "13. `if len(res) == k:`: Once `res` contains k elements, the code exits and returns the result.\n",
+    "\n",
+    "The code efficiently finds the k most frequent elements in the input array without using sorting, similar to the previous explanation, with the only difference being the variable names (e.g., `frequency` instead of `freq`)."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1, 2]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "nums = [1,1,1,2,2,3]\n",
+    "k = 2\n",
+    "result1 = topKFrequent(nums, k)\n",
+    "print(result1)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "nums = [1]\n",
+    "k = 1\n",
+    "result2 = topKFrequent(nums, k)\n",
+    "print(result2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "1. The first loop that counts the frequency of elements by iterating through `nums` has a time complexity of O(n), where n is the number of elements in the input array `nums`.\n",
+    "\n",
+    "2. The second loop iterates through the keys in the `count` dictionary, which has at most `k` unique elements (as per the constraints). Therefore, the second loop also has a time complexity of O(k).\n",
+    "\n",
+    "3. The third loop iterates through the `frequency` list, which has a length equal to the maximum frequency of elements in `nums`. In the worst case, this loop can have a time complexity of O(n), where n is the number of elements in `nums`.\n",
+    "\n",
+    "Overall, the dominant factor in terms of time complexity is the loop that iterates through the `count` dictionary. So, the total time complexity of the code is O(n + k).\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "1. The `count` dictionary stores the frequency of each unique element in `nums`. In the worst case, it can have at most `k` unique elements, so the space complexity for `count` is O(k).\n",
+    "\n",
+    "2. The `frequency` list is used to store elements grouped by their frequencies. It has a length of `len(nums) + 1`, which can be at most 105 based on the constraints. Therefore, the space complexity for `frequency` is O(105).\n",
+    "\n",
+    "3. The `res` list stores the k most frequent elements, which can be at most `k` elements, so the space complexity for `res` is O(k).\n",
+    "\n",
+    "**In summary**, the space complexity is dominated by the `count` dictionary and the `frequency` list, both of which have a space complexity of O(k + 105)."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Optimized k Most Frequent Elements:** Modify the code to find the k most frequent elements in an array while ensuring that the time complexity is O(n + klogk). You can use a priority queue (heap) to achieve this.\n",
+    "\n",
+    "2. **Handling Duplicate Frequencies:** Extend the code to handle cases where multiple elements have the same frequency and are among the k most frequent elements. Ensure that the output contains exactly k elements."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb b/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb
new file mode 100644
index 0000000..256e8f7
--- /dev/null
+++ b/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb	
@@ -0,0 +1,221 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 49. Group Anagrams\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Group Anagrams problem on LeetCode, click here!](https://leetcode.com/problems/group-anagrams/)\n",
+    "\n",
+    "---\n",
+    "Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.\n",
+    "\n",
+    "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- 1 <= `strs.length <= $10^4$\n",
+    "- `0 <= strs[i].length <= 100`\n",
+    "- `strs[i]` consists of lowercase English letters.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from collections import defaultdict\n",
+    "\n",
+    "\n",
+    "def groupAnagrams(strs):\n",
+    "    ans = defaultdict(list)\n",
+    "\n",
+    "    # Iterate through the list of input strings\n",
+    "    for s in strs:\n",
+    "        # Initialize a list to represent character counts for each character (a-z)\n",
+    "        count = [0] * 26\n",
+    "\n",
+    "        # Count the occurrences of each character in the current word\n",
+    "        for c in s:\n",
+    "            count[ord(c) - ord(\"a\")] += 1\n",
+    "\n",
+    "        # Use a tuple of character counts as the key and append the word to the anagram group\n",
+    "        ans[tuple(count)].append(s)\n",
+    "\n",
+    "    # Convert the values (lists of anagrams) to a list of lists\n",
+    "    return list(ans.values())"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. Inside the method, a `defaultdict` called `ans` is created to store anagram groups. This dictionary will have a list as its default value, meaning that each key in the dictionary will be associated with an empty list by default.\n",
+    "\n",
+    "2. The code then iterates through the list of input strings, `strs`, using a `for` loop. For each word `s` in `strs`, it performs the following steps:\n",
+    "\n",
+    "   - It initializes a list `count` of length 26, where each element represents the count of a specific character (a-z) in the word `s`.\n",
+    "\n",
+    "   - The code then iterates through the characters of `s`. For each character `c`, it increments the corresponding count in the `count` list based on its ASCII value.\n",
+    "\n",
+    "   - After counting the characters in `s`, it converts the `count` list into a tuple `tuple(count)` to use as a key for the `ans` dictionary.\n",
+    "\n",
+    "   - It appends the word `s` to the list associated with the key (tuple) in the `ans` dictionary. This groups all anagrams of the same word together under the same key.\n",
+    "\n",
+    "3. After processing all words in `strs`, the code converts the values of the `ans` dictionary (which are lists of anagrams) to a list of lists using the `list()` constructor.\n",
+    "\n",
+    "4. Finally, the code returns the list of anagram groups, which is the result of grouping the anagrams in the input list.\n",
+    "\n",
+    "In summary, the code efficiently groups anagrams together by counting the characters in each word and using a dictionary to store them under the same key. The result is a list of lists, where each inner list represents a group of anagrams."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: Group anagrams from a list of words\n",
+    "strs1 = [\"eat\", \"tea\", \"tan\", \"ate\", \"nat\", \"bat\"]\n",
+    "result1 = groupAnagrams(strs1)\n",
+    "print(result1)\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[['']]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2: Group anagrams from a list with an empty string\n",
+    "strs2 = [\"\"]\n",
+    "result2 = groupAnagrams(strs2)\n",
+    "print(result2)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "c8367e2a-e7d5-4e18-8235-2870e341a04b",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[['a']]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3: Group anagrams from a list with a single word\n",
+    "strs3 = [\"a\"]\n",
+    "result3 = groupAnagrams(strs3)\n",
+    "print(result3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `groupAnagrams` method:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "1. First, we create an instance of the `Solution` class, which is an O(1) operation.\n",
+    "\n",
+    "2. Inside the `groupAnagrams` method, we iterate through the list of input strings `strs` once in a loop:\n",
+    "   - For each word in `strs`, we perform character counting, which is done in O(K) time, where K is the maximum length of a word in the list.\n",
+    "\n",
+    "3. The overall time complexity of the method is O(N * K), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "1. We use a `defaultdict` called `ans` to store the anagram groups. In the worst case, each word belongs to a distinct anagram group, so the space complexity for `ans` is O(N).\n",
+    "\n",
+    "2. Within the loop, we create a `count` list of length 26 to store character counts for each word. This is a fixed-size list and does not depend on the input size, so it has a constant space complexity of O(26), which is effectively O(1).\n",
+    "\n",
+    "3. The space used for other variables is also constant and does not depend on the input size.\n",
+    "\n",
+    "4. The final result, `result`, is a list of lists that contains the grouped anagrams. In the worst case, each word is a distinct anagram group, so the space complexity for `result` is O(N).\n",
+    "\n",
+    "5. The overall space complexity of the method is O(N) due to the space used by `ans` and `result`. The space used by the `count` list and other variables is constant and does not significantly affect the overall space complexity.\n",
+    "\n",
+    "**In summary,** the time complexity of the `groupAnagrams` method is O(N * K), and the space complexity is O(N), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Anagramic Palindromes**: Given a list of strings, find all the groups of anagrams where each group contains words that can be rearranged into a palindrome. For example, in the list [\"bat\", \"tab\", \"act\", \"tac\", \"dog\"], there are two groups: [\"bat\", \"tab\", \"act\", \"tac\"] and [\"dog\"].\n",
+    "\n",
+    "2. **Largest Anagram Group**: Given a list of strings, find the largest group of anagrams. Return the list of anagrams in that group. If there are multiple largest groups, return all of them."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb b/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb
new file mode 100644
index 0000000..2fae2a5
--- /dev/null
+++ b/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb	
@@ -0,0 +1,198 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 659· Encode and Decode Strings\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Encode and Decode Strings problem on LintCode, click here!](https://www.lintcode.com/problem/659/)\n",
+    "\n",
+    "---\n",
+    "Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.\n",
+    "\n",
+    "Please implement `encode` and `decode`.\n",
+    "\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class Codec:\n",
+    "\n",
+    "    def encode(self, strs):\n",
+    "        \"\"\"Encodes a list of strings to a single string.\n",
+    "        \n",
+    "        :type strs: List[str]\n",
+    "        :rtype: str\n",
+    "        \"\"\"\n",
+    "        encoded = \"\"\n",
+    "        for s in strs:\n",
+    "            encoded += s.replace(':', '::') + ':;'\n",
+    "        return encoded\n",
+    "\n",
+    "    def decode(self, s):\n",
+    "        \"\"\"Decodes a single string to a list of strings.\n",
+    "        \n",
+    "        :type s: str\n",
+    "        :rtype: List[str]\n",
+    "        \"\"\"\n",
+    "        decoded = []\n",
+    "        i = 0\n",
+    "        while i < len(s):\n",
+    "            end = s.find(':;', i)\n",
+    "            if end == -1:\n",
+    "                end = len(s)\n",
+    "            decoded.append(s[i:end].replace('::', ':'))\n",
+    "            i = end + 2\n",
+    "        return decoded"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We define a class called `Codec` to encapsulate the encoding and decoding operations for a list of strings.\n",
+    "\n",
+    "2. In the `encode` method, we take a list of strings (`strs`) as input and return a single encoded string. The encoding process involves concatenating the strings together with a delimiter `:;`. We iterate through each string in the input list, replace any occurrence of `:` with `::` (to avoid conflicts with the delimiter), and then append `:;` to indicate the end of that string.\n",
+    "\n",
+    "3. In the `decode` method, we take an encoded string (`s`) as input and return a list of strings. The decoding process involves splitting the encoded string using the `:;` delimiter. We start from the beginning of the encoded string (`i = 0`) and repeatedly find the next occurrence of `:;`. We extract the substring between the current position (`i`) and the next delimiter (`end`). Before adding it to the result list, we replace any `::` with `:` to revert the encoding. We then update the current position `i` to be after the delimiter, so we can find the next substring in the next iteration.\n",
+    "\n",
+    "4. Finally, we create an instance of the `Codec` class and test it with two examples:\n",
+    "\n",
+    "By using the `:;` delimiter and handling the `::` escaping, this code can encode and decode lists of strings, preserving the original content even if it contains colons or the delimiter itself."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['lint', 'code', 'love', 'you']\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "codec = Codec()\n",
+    "\n",
+    "input1 = [\"lint\", \"code\", \"love\", \"you\"]\n",
+    "encoded1 = codec.encode(input1)\n",
+    "decoded1 = codec.decode(encoded1)\n",
+    "print(decoded1)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['we', 'say', ':', 'yes']\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "codec = Codec() \n",
+    "\n",
+    "input2 = [\"we\", \"say\", \":\", \"yes\"]\n",
+    "encoded2 = codec.encode(input2)\n",
+    "decoded2 = codec.decode(encoded2)\n",
+    "print(decoded2)  "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `encode` and `decode` methods in the provided code.\n",
+    "\n",
+    "**Encode Method (`encode`):**\n",
+    "\n",
+    "- Time Complexity: O(n * m)\n",
+    "  - Here, `n` is the number of strings in the input list `strs`, and `m` is the average length of these strings.\n",
+    "  - In the worst case, for each string in the list, we iterate over its characters once to replace any colons (`:`) with double colons (`::`) and then append `:;`. This is done for each string in the list.\n",
+    "\n",
+    "- Space Complexity: O(n * m)\n",
+    "  - The space complexity is also O(n * m) because we build the encoded string by concatenating the modified strings. In the worst case, the encoded string can be of the same length as the original strings.\n",
+    "\n",
+    "**Decode Method (`decode`):**\n",
+    "\n",
+    "- Time Complexity: O(n * m)\n",
+    "  - Similar to the encode method, we iterate through the encoded string in the decode method. In the worst case, we may have to scan each character in the encoded string to find the `:;` delimiter.\n",
+    "\n",
+    "- Space Complexity: O(n * m)\n",
+    "  - The space complexity of the decode method is also O(n * m) because we build the list of decoded strings, which can be of the same length as the encoded string.\n",
+    "\n",
+    "Overall, both the `encode` and `decode` methods have time and space complexities of O(n * m), where `n` is the number of strings in the input list, and `m` is the average length of these strings. The space complexity arises from storing the encoded or decoded strings, and the time complexity arises from iterating through the characters in these strings."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Multiple Delimiters**: Modify the code to allow for multiple delimiters, not just one. Each string could have its own delimiter, and the code should be able to handle this complexity.\n",
+    "\n",
+    "2. **Optimize Encoding**: Modify the `encode` method to achieve a more efficient encoding algorithm. Try to minimize the length of the encoded string while ensuring that it can be correctly decoded."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb b/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb
new file mode 100644
index 0000000..e330eb0
--- /dev/null
+++ b/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb	
@@ -0,0 +1,212 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 11. Container With Most Water\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Container With Most Water problem on LeetCode, click here!](https://leetcode.com/problems/container-with-most-water/)\n",
+    "\n",
+    "---\n",
+    "You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the $i^{th}$ line are `(i, 0)` and `(i, height[i])`.\n",
+    "\n",
+    "Find two lines that together with the x-axis form a container, such that the container contains the most water.\n",
+    "\n",
+    "Return *the maximum amount of water a container can store.*\n",
+    "\n",
+    "**Notice** that you may not slant the container.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `n == height.length`\n",
+    "- 2 <= `n` <= $10^5$\n",
+    "- 0 <= `height[i]` <= $10^4$\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def maxArea(height):\n",
+    "    # Initialize the maximum area to 0.\n",
+    "    max_area = 0\n",
+    "    # Initialize two pointers, one at the beginning (left) and one at the end (right) of the array.\n",
+    "    left = 0\n",
+    "    right = len(height) - 1\n",
+    "\n",
+    "    # Iterate until the left pointer is less than the right pointer.\n",
+    "    while left < right:\n",
+    "        # Calculate the height of the container, which is the minimum height of the two lines.\n",
+    "        h = min(height[left], height[right])\n",
+    "        # Calculate the width of the container, which is the distance between the two pointers.\n",
+    "        width = right - left\n",
+    "        # Calculate the area of the container using height and width and update max_area if it's greater.\n",
+    "        max_area = max(max_area, h * width)\n",
+    "\n",
+    "        # Move the pointer with the shorter line inward.\n",
+    "        if height[left] < height[right]:\n",
+    "            left += 1\n",
+    "        else:\n",
+    "            right -= 1\n",
+    "\n",
+    "    # Continue this process until left < right, and then return the maximum area.\n",
+    "    return max_area"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We start with initializing `max_area` to 0, which will store the maximum area of water that can be contained by the lines.\n",
+    "\n",
+    "2. Two pointers, `left` and `right`, are initialized to the beginning and end of the input `height` array.\n",
+    "\n",
+    "3. The main loop runs while `left` is less than `right`. This ensures that we are checking all possible pairs of lines in a systematic way.\n",
+    "\n",
+    "4. Inside the loop:\n",
+    "   - We calculate the height of the container `h` as the minimum of the heights at the `left` and `right` pointers. This represents the height of the water the container can hold.\n",
+    "   - We calculate the width of the container as the difference between `right` and `left`. This represents the distance between the two lines.\n",
+    "   - We calculate the area of the container using `h * width` and update `max_area` if the calculated area is greater than the current `max_area`.\n",
+    "\n",
+    "5. After calculating the area and updating `max_area`, we move one of the pointers. We move the pointer pointing to the shorter line inward because moving the pointer pointing to the taller line won't increase the height and will only decrease the width, which will result in a smaller area.\n",
+    "\n",
+    "6. We continue this process until `left` is no longer less than `right`, indicating that we have checked all possible pairs of lines.\n",
+    "\n",
+    "7. Finally, we return the `max_area` which contains the maximum area of water that can be contained by two lines from the input array."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Output: 49\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "height1 = [1, 8, 6, 2, 5, 4, 8, 3, 7]\n",
+    "result1 = maxArea(height1)\n",
+    "print(\"Output:\", result1)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Output: 1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "height2 = [1, 1]\n",
+    "result2 = maxArea(height2)\n",
+    "print(\"Output:\", result2)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Output: 45\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Additional Example\n",
+    "height3 = [3, 9, 3, 4, 7, 2, 12, 6]\n",
+    "result3 = maxArea(height3)\n",
+    "print(\"Output:\", result3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `maxArea` function:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "The function uses a two-pointer approach to find the maximum area of water. In the worst case, both the left and right pointers traverse the entire input array once. Therefore, the time complexity is O(n), where n is the number of elements in the `height` array.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "The function uses a constant amount of extra space for variables like `max_area`, `left`, `right`, `h`, and `width`. It does not use any data structures whose space consumption depends on the input size. As a result, the space complexity is O(1), which means it has constant space complexity.\n",
+    "\n",
+    "**In summary,** the `maxArea` function has a time complexity of O(n) and a space complexity of O(1), making it an efficient algorithm for finding the maximum area of water that can be contained by two lines in the input array."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Optimization Challenge:** Modify the maxArea function to not only return the maximum area but also the indices of the two lines that form the container with the maximum area.\n",
+    "2. **Variant Problem:** Instead of finding the maximum area of water, find the minimum amount of water required to fill all the gaps between the lines. You'll need to return the total water volume needed."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb b/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb
new file mode 100644
index 0000000..12c7daa
--- /dev/null
+++ b/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb	
@@ -0,0 +1,210 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 125. Valid Palindrome\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Valid Palindrome problem on LeetCode, click here!](https://leetcode.com/problems/valid-palindrome/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.\n",
+    "\n",
+    "Given a string `s`, *return `true` if it is a **palindrome**, or `false` otherwise.*\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- 1 <= `s.length` <= $2*10^5$\n",
+    "- `s` consists only of printable ASCII characters.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def isPalindrome(s):\n",
+    "    # Step 1: Convert the input string to lowercase\n",
+    "    s = s.lower()\n",
+    "    \n",
+    "    # Step 2: Remove all non-alphanumeric characters from the string\n",
+    "    cleaned_s = ''.join(c for c in s if c.isalnum())\n",
+    "    \n",
+    "    # Step 3: Initialize two pointers, one at the beginning and one at the end of the cleaned string\n",
+    "    left, right = 0, len(cleaned_s) - 1\n",
+    "    \n",
+    "    # Step 4: Compare characters using the two pointers while moving them towards each other\n",
+    "    while left < right:\n",
+    "        if cleaned_s[left] != cleaned_s[right]:\n",
+    "            return False  # If characters don't match, it's not a palindrome\n",
+    "        left += 1\n",
+    "        right -= 1\n",
+    "    \n",
+    "    # Step 5: If the loop completes without returning False, it's a palindrome\n",
+    "    return True"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. The `isPalindrome` function takes an input string `s`.\n",
+    "\n",
+    "2. It starts by converting the input string to lowercase using `s.lower()`. This step ensures that the function is case-insensitive when checking for palindromes.\n",
+    "\n",
+    "3. Next, it removes all non-alphanumeric characters from the string and stores the result in the `cleaned_s` variable. It does this by iterating through each character in the input string and only keeping characters that are alphanumeric (letters and digits). This step removes spaces, punctuation, and other non-alphanumeric characters.\n",
+    "\n",
+    "4. The function then initializes two pointers, `left` and `right`, which will be used to compare characters from the beginning and end of the cleaned string.\n",
+    "\n",
+    "5. In a while loop, it compares characters using the two pointers while moving them towards each other. The loop continues until the `left` pointer is less than the `right` pointer. This is done to check all characters in the cleaned string.\n",
+    "\n",
+    "6. Inside the loop, it checks if the characters at the `left` and `right` pointers don't match. If they don't match, it means the input is not a palindrome, so the function returns `False`.\n",
+    "\n",
+    "7. If the characters at the `left` and `right` pointers match, the pointers are moved closer to each other by incrementing `left` and decrementing `right`.\n",
+    "\n",
+    "8. The loop continues to compare characters until it either finds a mismatch (returning `False`) or until the `left` pointer is greater than or equal to the `right` pointer. If the loop completes without finding a mismatch, it means the input is a palindrome, so the function returns `True`.\n",
+    "\n",
+    "In summary, the code converts the input string to lowercase, removes non-alphanumeric characters, and then uses two pointers to compare characters from the cleaned string's beginning and end. If all characters match during the comparison, the function returns `True`, indicating that the input is a palindrome. If any characters do not match, it returns `False`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "input1 = \"A man, a plan, a canal: Panama\"\n",
+    "output1 = isPalindrome(input1)\n",
+    "print(output1)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "input2 = \"race a car\"\n",
+    "output2 = isPalindrome(input2)\n",
+    "print(output2)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3\n",
+    "input3 = \"\"\n",
+    "output3 = isPalindrome(input3)\n",
+    "print(output3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `isPalindrome` function that uses two pointers:\n",
+    "\n",
+    "1. **Time Complexity:**\n",
+    "   - Converting the input string to lowercase using `s.lower()` takes O(n) time, where 'n' is the length of the input string.\n",
+    "   - Removing non-alphanumeric characters using the list comprehension `''.join(c for c in s if c.isalnum())` also takes O(n) time because it iterates through each character in the string once.\n",
+    "   - The comparison of characters using the two pointers (`left` and `right`) is done in a loop that runs until `left` is less than `right`. In the worst case, the loop iterates through half of the string, so it takes O(n/2) time, which is still considered O(n).\n",
+    "\n",
+    "   Therefore, the overall time complexity of the `isPalindrome` function is O(n), where 'n' is the length of the input string.\n",
+    "\n",
+    "2. **Space Complexity:**\n",
+    "   - The space complexity is primarily determined by the additional string `cleaned_s` created to store the cleaned version of the input string. In the worst case, if all characters in the input string are alphanumeric, this cleaned string can be as large as the original input string.\n",
+    "   - Additionally, the function uses two pointers, `left` and `right`, which consume constant space and do not depend on the size of the input.\n",
+    "\n",
+    "**In summary,** the overall space complexity of the `isPalindrome` function is O(n) in the worst case, where 'n' is the length of the input string. The space complexity is dominated by the space required for `cleaned_s`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Longest Palindromic Substring:** Given a string `s`, find and return the longest palindromic substring within `s`. For example, if `s = \"babad\"`, the function should return `\"bab\"` or `\"aba\"`.\n",
+    "2. **Palindrome with Limited Characters:** Given a string `s` and a list of characters `allowed_chars`, check if `s` is a palindrome considering only the characters from `allowed_chars`. Ignore all other characters. For example, if `s = \"A man, a plan, a canal: Panama\"` and `allowed_chars = ['a', 'm', 'n', 'p']`, the function should return `True`."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb b/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb
new file mode 100644
index 0000000..15d189a
--- /dev/null
+++ b/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb	
@@ -0,0 +1,242 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 15. 3Sum\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the 3Sum problem on LeetCode, click here!](https://leetcode.com/problems/3sum/)\n",
+    "\n",
+    "---\n",
+    "Given an integer array nums, *return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.*\n",
+    "\n",
+    "**Notice** that the solution set must not contain duplicate triplets.\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- `3 <= nums.length <= 3000`\n",
+    "- $-10^5$ <= `nums[i]` <= $10^5$\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def threeSum(nums):\n",
+    "    # Sort the input array in ascending order\n",
+    "    nums.sort()\n",
+    "    triplets = []\n",
+    "    \n",
+    "    for i in range(len(nums) - 2):\n",
+    "        # Skip duplicates to avoid duplicate triplets\n",
+    "        if i > 0 and nums[i] == nums[i - 1]:\n",
+    "            continue\n",
+    "        \n",
+    "        left, right = i + 1, len(nums) - 1\n",
+    "        \n",
+    "        while left < right:\n",
+    "            # Calculate the total sum of the current triplet\n",
+    "            total = nums[i] + nums[left] + nums[right]\n",
+    "            \n",
+    "            if total == 0:\n",
+    "                # If the total is zero, we found a valid triplet\n",
+    "                triplets.append([nums[i], nums[left], nums[right]])\n",
+    "                \n",
+    "                # Skip duplicates of left and right pointers\n",
+    "                while left < right and nums[left] == nums[left + 1]:\n",
+    "                    left += 1\n",
+    "                while left < right and nums[right] == nums[right - 1]:\n",
+    "                    right -= 1\n",
+    "                \n",
+    "                # Move the pointers to the next unique elements\n",
+    "                left += 1\n",
+    "                right -= 1\n",
+    "            elif total < 0:\n",
+    "                # If the total is negative, we need to increase the sum by moving the left pointer to the right\n",
+    "                left += 1\n",
+    "            else:\n",
+    "                # If the total is positive, we need to decrease the sum by moving the right pointer to the left\n",
+    "                right -= 1\n",
+    "    \n",
+    "    return triplets"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. It starts by sorting the input array `nums` in ascending order. Sorting helps in efficiently finding triplets that sum up to zero.\n",
+    "\n",
+    "2. It initializes an empty list `triplets` to store the unique triplets that meet the given conditions.\n",
+    "\n",
+    "3. The code then iterates through the `nums` array using a loop with index `i`. This loop will consider each element of the array as a potential starting point for a triplet.\n",
+    "\n",
+    "4. Inside the loop, it checks for duplicates and skips them. This is done to ensure that the solution set does not contain duplicate triplets. If `nums[i]` is the same as the previous element `nums[i-1]`, it continues to the next iteration of the loop.\n",
+    "\n",
+    "5. Two pointers, `left` and `right`, are initialized. `left` starts just after the current element `nums[i]`, and `right` starts at the end of the array.\n",
+    "\n",
+    "6. The code enters another loop with `left` and `right` pointers, trying to find a triplet that sums up to zero.\n",
+    "\n",
+    "7. It calculates the `total` sum of the current triplet as `nums[i] + nums[left] + nums[right]`.\n",
+    "\n",
+    "8. If `total` is zero, it means a valid triplet is found, so it appends `[nums[i], nums[left], nums[right]]` to the `triplets` list. Then, it moves both the `left` and `right` pointers to their next unique elements while skipping duplicates.\n",
+    "\n",
+    "9. If `total` is less than zero, it means the sum is negative, so it increments the `left` pointer to move towards a larger sum.\n",
+    "\n",
+    "10. If `total` is greater than zero, it means the sum is positive, so it decrements the `right` pointer to move towards a smaller sum.\n",
+    "\n",
+    "11. This process continues until all possible triplets have been considered.\n",
+    "\n",
+    "12. Finally, the function returns the `triplets` list, which contains all the unique triplets that sum up to zero in the sorted `nums` array."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[-1, -1, 2], [-1, 0, 1]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "nums = [-1, 0, 1, 2, -1, -4]\n",
+    "result = threeSum(nums)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "nums = [0, 1, 1]\n",
+    "result = threeSum(nums)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[0, 0, 0]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3\n",
+    "nums = [0, 0, 0]\n",
+    "result = threeSum(nums)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `threeSum` function:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "1. Sorting the input array `nums` takes O(n log n) time, where n is the length of the array.\n",
+    "\n",
+    "2. The outer loop runs for each element of the array, so it iterates O(n) times.\n",
+    "\n",
+    "3. Inside the outer loop, we have a while loop with two pointers (`left` and `right`). In the worst case, the while loop can iterate O(n) times (when all elements are unique).\n",
+    "\n",
+    "4. Inside the while loop, we have constant time operations (additions, comparisons, and list appends).\n",
+    "\n",
+    "Overall, the time complexity of the `threeSum` function is dominated by the sorting step, so it is O(n log n) due to the sorting. The other operations inside the loops contribute linearly but are dominated by the sorting step.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "1. The space used by the `triplets` list to store the output triplets is O(k), where k is the number of unique triplets that sum up to zero. In the worst case, this can be O(n^2/3) because there can be roughly O(n^2) possible triplets, and many of them may sum up to zero.\n",
+    "\n",
+    "2. Other than the `triplets` list, the function uses only a constant amount of additional space for variables like `i`, `left`, `right`, and `total`.\n",
+    "\n",
+    "Overall, the space complexity of the `threeSum` function is O(k), where k is the number of unique triplets that meet the criteria. It's important to note that the space complexity does not depend on the size of the input array but rather on the number of valid triplets found.\n",
+    "\n",
+    "**In summary:**\n",
+    "- Time Complexity: O(n log n) due to sorting (dominant factor) and O(n^2) in the worst case for the nested loops.\n",
+    "- Space Complexity: O(k) for storing the output triplets, where k is the number of unique triplets that meet the criteria."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **K-Sum Problem:** Generalize the problem to the K-Sum problem, where instead of finding triplets that sum to zero, you need to find K elements that sum to a target value. Implement a function for this generalization.\n",
+    "2. **Count the Number of Unique Triplets:** Modify the algorithm to count the number of unique triplets that satisfy the conditions instead of returning the triplets themselves. This will require some modifications to handle duplicates efficiently."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb b/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb
new file mode 100644
index 0000000..451a21d
--- /dev/null
+++ b/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb	
@@ -0,0 +1,185 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 121. Best Time to Buy and Sell Stock\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Best Time to Buy and Sell Stock problem on LeetCode, click here!](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "You are given an array `prices` where `prices[i]` is the price of a given stock on the $i^{th}$ day.\n",
+    "\n",
+    "You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.\n",
+    "\n",
+    "*Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.*\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- 1 <= `prices.length` <= $10^5$\n",
+    "- 0 <= `prices[i]` <= $10^4$\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def maxProfit(prices):\n",
+    "    if not prices:\n",
+    "        return 0\n",
+    "    \n",
+    "    min_price = prices[0]\n",
+    "    max_profit = 0\n",
+    "    \n",
+    "    for price in prices:\n",
+    "        # Update the minimum price if a lower price is encountered\n",
+    "        min_price = min(min_price, price)\n",
+    "        # Calculate the profit if selling at the current price\n",
+    "        max_profit = max(max_profit, price - min_price)\n",
+    "    \n",
+    "    return max_profit"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "- The `maxProfit` function takes a single argument, `prices`, which is a list of stock prices.\n",
+    "- The first line of the function checks if the `prices` list is empty. If it's empty, there are no prices to analyze, so the function returns 0 (no profit can be made).\n",
+    "\n",
+    "- Then, we initialize two variables:\n",
+    "  - `min_price`: This variable will keep track of the minimum price seen so far. We initialize it with the price of the first day in the list (`prices[0]`).\n",
+    "  - `max_profit`: This variable will keep track of the maximum profit we can achieve. We initialize it to 0 because initially, we haven't made any profit.\n",
+    "\n",
+    "- We start iterating through the `prices` list, where `price` represents the stock price for the current day.\n",
+    "\n",
+    "- In each iteration, we check if the current `price` is lower than the `min_price` we've seen so far. If it is, we update `min_price` to the current `price`. This is because we want to buy the stock at the lowest possible price.\n",
+    "\n",
+    "- Next, we calculate the profit we would make if we sold the stock at the current `price` (assuming we bought it at `min_price`). We do this by subtracting `min_price` from the current `price`. This represents the profit for the current day.\n",
+    "\n",
+    "- We also use the `max` function to keep track of the maximum profit seen so far. If the profit calculated for the current day is greater than the `max_profit` we've seen previously, we update `max_profit` with this higher value.\n",
+    "\n",
+    "- Finally, after iterating through all the days in the `prices` list, we return `max_profit`, which represents the maximum profit that can be achieved by buying and selling a single stock."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "5\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "prices1 = [7, 1, 5, 3, 6, 4]\n",
+    "print(maxProfit(prices1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "prices2 = [7, 6, 4, 3, 1]\n",
+    "print(maxProfit(prices2))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "- The code iterates through the `prices` array once, examining each price exactly once.\n",
+    "- Within the loop, there are constant-time operations such as comparisons, additions, and subtractions.\n",
+    "- Therefore, the time complexity of this code is O(n), where n is the length of the `prices` array. It performs a linear number of operations relative to the size of the input.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "- The space complexity of the code is constant, O(1).\n",
+    "- Regardless of the size of the input `prices` array, the code only uses a fixed amount of additional memory to store two variables (`min_price` and `max_profit`) and a few loop variables.\n",
+    "- The memory usage does not depend on the size of the input, so it is considered constant space complexity.\n",
+    "\n",
+    "**In summary:**\n",
+    "- Time Complexity: O(n) where n is the length of the `prices` array.\n",
+    "- Space Complexity: O(1) (constant space usage)."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Multiple Transactions Allowed:** Modify the problem to allow multiple transactions (buy and sell multiple times), but you must sell before buying again. Find the maximum profit in this scenario.\n",
+    "2. **K Transactions Allowed:** Extend the problem to allow at most k transactions. Find the maximum profit considering this constraint.\n",
+    "3. **With Transaction Fee:** Introduce a transaction fee for each buy/sell operation. Modify the code to maximize profit while considering the transaction fee."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb b/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb
new file mode 100644
index 0000000..ded03f3
--- /dev/null
+++ b/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb	
@@ -0,0 +1,206 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 3. Longest Substring Without Repeating Characters\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Longest Substring Without Repeating Characters problem on LeetCode, click here!](https://leetcode.com/problems/longest-substring-without-repeating-characters/)\n",
+    "\n",
+    "---\n",
+    "Given a string `s`, find the length of the **longest substring** without repeating characters.\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- 0 <= `s.length` <= $5 * 10^4$\n",
+    "- `s` consists of English letters, digits, symbols and spaces.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def length_of_longest_substring(s):\n",
+    "    # Create a dictionary to store the index of each character's last occurrence.\n",
+    "    char_index = {}\n",
+    "    \n",
+    "    # Initialize variables to keep track of the start and end of the current substring.\n",
+    "    start = 0\n",
+    "    max_length = 0\n",
+    "    \n",
+    "    for end in range(len(s)):\n",
+    "        # If the character is in the dictionary and its last occurrence is after the start of the current substring,\n",
+    "        # update the start of the substring to the next character after its last occurrence.\n",
+    "        if s[end] in char_index and char_index[s[end]] >= start:\n",
+    "            start = char_index[s[end]] + 1\n",
+    "        \n",
+    "        # Update the last occurrence index of the current character.\n",
+    "        char_index[s[end]] = end\n",
+    "        \n",
+    "        # Calculate the length of the current substring and update the maximum length if necessary.\n",
+    "        current_length = end - start + 1\n",
+    "        max_length = max(max_length, current_length)\n",
+    "    \n",
+    "    return max_length"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We start by defining a function called `length_of_longest_substring` that takes a single argument `s`, which is the input string for which we want to find the length of the longest substring without repeating characters.\n",
+    "\n",
+    "2. We create a dictionary called `char_index` to store the index of the last occurrence of each character in the input string. This dictionary will help us keep track of where each character was last seen.\n",
+    "\n",
+    "3. We initialize two variables, `start` and `max_length`. \n",
+    "   - `start` represents the start index of the current substring without repeating characters. It starts at 0.\n",
+    "   - `max_length` is used to keep track of the maximum length found so far and is initially set to 0.\n",
+    "\n",
+    "4. We then iterate through the input string `s` using a for loop, where the loop variable `end` represents the current end index of the substring we are considering.\n",
+    "\n",
+    "5. Inside the loop:\n",
+    "   - We check if the character `s[end]` is already in the `char_index` dictionary and if its last occurrence is within or after the current substring. If so, it means that we've encountered a repeating character, and we need to update the `start` of the substring to the next character after the last occurrence of `s[end]`. This ensures that we have a new valid substring without repeating characters.\n",
+    "   - We then update the `char_index` dictionary by storing the current index `end` as the last occurrence index of the character `s[end]`.\n",
+    "\n",
+    "6. Next, we calculate the length of the current substring without repeating characters, which is `current_length = end - start + 1`. We add 1 to account for the fact that the indices are zero-based.\n",
+    "\n",
+    "7. We update the `max_length` with the maximum of its current value and the `current_length`. This step ensures that we keep track of the longest valid substring we have encountered so far.\n",
+    "\n",
+    "8. The loop continues to iterate through the input string, and at the end, we return the `max_length`, which represents the length of the longest substring without repeating characters."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "print(length_of_longest_substring(\"abcabcbb\"))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "print(length_of_longest_substring(\"bbbbb\"))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "cf04eba0-0ead-4e8b-ba33-4316ea64a9d3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3:\n",
+    "print(length_of_longest_substring(\"pwwkew\"))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "Let's analyze the time and space complexity of the `length_of_longest_substring` function:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "- The function iterates through the input string `s` using a single for loop. The loop runs from the beginning of the string to the end once.\n",
+    "- Inside the loop, we perform constant-time operations such as dictionary lookups and updates, comparisons, and arithmetic operations.\n",
+    "- Therefore, the time complexity of the function is O(n), where n is the length of the input string `s`.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "- The primary data structure that consumes space in this function is the `char_index` dictionary.\n",
+    "- In the worst case, if there are no repeating characters in the input string, the dictionary can store all unique characters in the string.\n",
+    "- Therefore, the space complexity is O(min(n, m)), where n is the length of the input string `s`, and m is the number of unique characters in the string. In the worst case, when all characters are unique, m is equal to n, so the space complexity is O(n).\n",
+    "- Additionally, there are a few integer variables used for indices and lengths, which consume constant space.\n",
+    "- Overall, the space complexity of the function is O(min(n, m)) or simply O(n) in the worst case.\n",
+    "\n",
+    "**In summary,** the time complexity of the `length_of_longest_substring` function is O(n), and the space complexity is O(n) in the worst case due to the `char_index` dictionary. This algorithm provides an efficient way to find the length of the longest substring without repeating characters in linear time."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Longest Substring with K Distinct Characters:** Modify the problem to find the length of the longest substring with exactly K distinct characters. For example, given the input string \"abcabcbb\" and K = 2, the answer would be 4 because the longest substring with two distinct characters is \"abca.\"\n",
+    "2. **Longest Substring with Unique Characters:** Write a function to find the length of the longest substring in a given string where all characters are unique. For example, given the input string \"abcabcbb,\" the answer would be 4 because \"abcd\" is the longest substring with unique characters."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb b/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb
new file mode 100644
index 0000000..df4b0d3
--- /dev/null
+++ b/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb	
@@ -0,0 +1,189 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 424. Longest Repeating Character Replacement\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Longest Repeating Character Replacement problem on LeetCode, click here!](https://leetcode.com/problems/longest-repeating-character-replacement/)\n",
+    "\n",
+    "---\n",
+    "You are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.\n",
+    "\n",
+    "*Return the length of the longest substring containing the same letter you can get after performing the above operations.*\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- 1 <= `s.length` <= $10^5$\n",
+    "- `s` consists of only uppercase English letters.\n",
+    "- `0 <= k <= s.length`"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def characterReplacement(s, k):\n",
+    "    max_length = 0  # Initialize the maximum length\n",
+    "    max_count = 0   # Initialize the maximum count of repeating characters\n",
+    "    start = 0       # Initialize the start of the sliding window\n",
+    "    char_count = {} # Dictionary to store the count of each character\n",
+    "\n",
+    "    for end in range(len(s)):\n",
+    "        # Update the count of the current character in the dictionary\n",
+    "        char_count[s[end]] = char_count.get(s[end], 0) + 1\n",
+    "        \n",
+    "        # Update the maximum count of repeating characters\n",
+    "        max_count = max(max_count, char_count[s[end]])\n",
+    "        \n",
+    "        # Check if the current window size is greater than k\n",
+    "        if (end - start + 1) - max_count > k:\n",
+    "            # Move the start of the window to the right\n",
+    "            char_count[s[start]] -= 1\n",
+    "            start += 1\n",
+    "        \n",
+    "        # Update the maximum length\n",
+    "        max_length = max(max_length, end - start + 1)\n",
+    "\n",
+    "    return max_length"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We initialize some variables:\n",
+    "   - `max_length` to keep track of the maximum length of the substring containing the same letter.\n",
+    "   - `max_count` to keep track of the maximum count of repeating characters within the current window.\n",
+    "   - `start` to represent the start index of the sliding window.\n",
+    "   - `char_count` is a dictionary that will store the count of each character within the current window.\n",
+    "\n",
+    "2. We use a `for` loop to iterate through the characters of the string `s`.\n",
+    "\n",
+    "3. Inside the loop, we do the following for each character at position `end`:\n",
+    "   - Update the count of the current character in the `char_count` dictionary.\n",
+    "   - Update the `max_count` to be the maximum of the current `max_count` and the count of the current character. This keeps track of the maximum count of repeating characters within the current window.\n",
+    "\n",
+    "4. We check if the current window size (the difference between `end` and `start` plus one) minus the `max_count` is greater than `k`. This condition checks whether we have exceeded the allowed number of replacements (`k`) within the current window.\n",
+    "\n",
+    "   - If we have exceeded the allowed replacements, it means we need to shrink the window from the left side. We do this by moving the `start` of the window to the right and decrementing the count of the character at `s[start]` in the `char_count` dictionary. This effectively removes characters from the left side of the window until we have a valid window again.\n",
+    "\n",
+    "   - By doing this, we ensure that the difference between the current window size and the `max_count` is always less than or equal to `k`.\n",
+    "\n",
+    "5. After each character, we update the `max_length` to be the maximum of the current `max_length` and the size of the current window (`end - start + 1`).\n",
+    "\n",
+    "6. Finally, we return `max_length`, which holds the length of the longest substring containing the same letter with at most `k` replacements."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "s1 = \"ABAB\"\n",
+    "k1 = 2\n",
+    "print(characterReplacement(s1, k1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "s2 = \"AABABBA\"\n",
+    "k2 = 1\n",
+    "print(characterReplacement(s2, k2))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "- The code uses a single `for` loop to iterate through the characters of the input string `s`. The loop runs from the beginning to the end of the string.\n",
+    "- Inside the loop, we perform constant-time operations such as updating the `char_count` dictionary and updating variables like `max_count` and `max_length`.\n",
+    "- The code's time complexity is primarily determined by the loop, which iterates through each character in the string once. Therefore, the time complexity is O(n), where n is the length of the input string `s`.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "- The code uses several variables to store information, but the space they consume is constant and does not depend on the size of the input string. These variables include `max_length`, `max_count`, `start`, and `char_count`. Therefore, the space complexity is O(1) or constant space.\n",
+    "\n",
+    "**In summary**, the time complexity of the code is O(n), where n is the length of the input string, and the space complexity is O(1), as it uses a constant amount of additional space regardless of the input size. This algorithm efficiently solves the problem with a linear time complexity."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Optimization Challenge:** Modify the code to find the longest substring containing the same letter with at most k replacements in O(n) time complexity and O(1) space complexity. Hint: You may need to update the approach to achieve this.\n",
+    "2. **Variation with Lowercase Letters:** Extend the problem to include both uppercase and lowercase English letters in the input string s. Write a function that can handle this extended input and still find the longest substring with at most k replacements efficiently."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb b/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb
new file mode 100644
index 0000000..cf22c86
--- /dev/null
+++ b/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb	
@@ -0,0 +1,260 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 76. Minimum Window Substring\n",
+    "\n",
+    "**Difficulty:** Hard\n",
+    "\n",
+    "**Link to Problem:** [To see the Minimum Window Substring problem on LeetCode, click here!](https://leetcode.com/problems/minimum-window-substring/)\n",
+    "\n",
+    "---\n",
+    "Given two strings `s` and `t` of lengths `m` and `n` respectively, return *the **minimum window substring** of `s` such that every character in `t` (**including duplicates**) is included in the window. If there is no such substring, return the empty string `\"\"`.*\n",
+    "\n",
+    "The testcases will be generated such that the answer is **unique**.\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- `m == s.length`\n",
+    "- `n == t.length`\n",
+    "- 1 <= `m, n` <= $10^5$\n",
+    "- `s` and `t` consist of uppercase and lowercase English letters.\n",
+    "\n",
+    "**Follow up**: Could you find an algorithm that runs in $O(m + n)$ time?"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def min_window_substring(s, t):\n",
+    "    if not s or not t:\n",
+    "        return \"\"\n",
+    "\n",
+    "    # Initialize dictionaries to keep track of character counts for t and the current window in s.\n",
+    "    t_dict = {}\n",
+    "    current_window_dict = {}\n",
+    "    \n",
+    "    # Populate t_dict with character counts for string t.\n",
+    "    for char in t:\n",
+    "        t_dict[char] = t_dict.get(char, 0) + 1\n",
+    "\n",
+    "    # Initialize pointers for the sliding window.\n",
+    "    left = 0\n",
+    "    min_len = float('inf')\n",
+    "    min_window = \"\"\n",
+    "    required_chars = len(t_dict)\n",
+    "\n",
+    "    # Initialize a variable to keep track of how many required characters have been found in the current window.\n",
+    "    found_chars = 0\n",
+    "\n",
+    "    # Iterate over the string s using the right pointer.\n",
+    "    for right in range(len(s)):\n",
+    "        char = s[right]\n",
+    "\n",
+    "        # Update the current_window_dict.\n",
+    "        current_window_dict[char] = current_window_dict.get(char, 0) + 1\n",
+    "\n",
+    "        # Check if the current character is a required character and if its count matches the required count.\n",
+    "        if char in t_dict and current_window_dict[char] == t_dict[char]:\n",
+    "            found_chars += 1\n",
+    "\n",
+    "        # Try to minimize the window by moving the left pointer.\n",
+    "        while left <= right and found_chars == required_chars:\n",
+    "            # Calculate the current window size.\n",
+    "            window_size = right - left + 1\n",
+    "\n",
+    "            # If the current window is smaller than the minimum found so far, update min_len and min_window.\n",
+    "            if window_size < min_len:\n",
+    "                min_len = window_size\n",
+    "                min_window = s[left:right+1]\n",
+    "\n",
+    "            # Move the left pointer to the right to shrink the window.\n",
+    "            left_char = s[left]\n",
+    "            current_window_dict[left_char] -= 1\n",
+    "\n",
+    "            # Check if the character removed from the window was a required character.\n",
+    "            if left_char in t_dict and current_window_dict[left_char] < t_dict[left_char]:\n",
+    "                found_chars -= 1\n",
+    "\n",
+    "            # Move the left pointer further to continue shrinking the window.\n",
+    "            left += 1\n",
+    "\n",
+    "    return min_window"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We start by checking if either of the input strings `s` or `t` is empty. If either of them is empty, we return an empty string since there can't be any valid substring in this case.\n",
+    "\n",
+    "2. We initialize two dictionaries: `t_dict` and `current_window_dict`. These dictionaries will be used to keep track of character counts in the string `t` and the current window in string `s`, respectively.\n",
+    "\n",
+    "3. We populate the `t_dict` dictionary by iterating through string `t`. For each character, we increment its count in the dictionary using `t_dict.get(char, 0) + 1`. This allows us to count the occurrences of each character in `t`.\n",
+    "\n",
+    "4. We initialize two pointers: `left` and `right`. The `left` pointer will represent the start of the current window, and the `right` pointer will represent the end of the current window. We also initialize `min_len` to store the length of the minimum window found so far, and `min_window` to store the actual minimum window substring.\n",
+    "\n",
+    "5. We determine the number of required characters in `t` (i.e., the number of distinct characters in `t`) and store it in the variable `required_chars`.\n",
+    "\n",
+    "6. We initialize a variable `found_chars` to keep track of how many required characters have been found in the current window. Initially, it is set to 0.\n",
+    "\n",
+    "7. We iterate over the string `s` using the `right` pointer. In each iteration, we do the following:\n",
+    "\n",
+    "   - Update the `current_window_dict` by incrementing the count of the current character.\n",
+    "   \n",
+    "   - Check if the current character is a required character (present in `t_dict`) and if its count in the `current_window_dict` matches the required count from `t_dict`. If so, we increment `found_chars`.\n",
+    "\n",
+    "8. After updating the window, we attempt to minimize the window size by moving the `left` pointer to the right. In this step, we:\n",
+    "\n",
+    "   - Calculate the size of the current window.\n",
+    "   \n",
+    "   - If the current window size is smaller than the minimum found so far (`min_len`), we update `min_len` and `min_window` to store the current window substring.\n",
+    "   \n",
+    "   - Move the `left` pointer to the right to shrink the window.\n",
+    "   \n",
+    "   - Check if the character being removed from the window was a required character. If it was, we decrement `found_chars`.\n",
+    "   \n",
+    "   - Continue moving the `left` pointer further to continue shrinking the window if the window still contains all required characters.\n",
+    "\n",
+    "9. Finally, we return `min_window`, which will contain the minimum window substring that contains all characters from `t`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "BANC\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "print(min_window_substring(\"ADOBECODEBANC\", \"ABC\"))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "print(min_window_substring(\"a\", \"a\")) "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3:\n",
+    "print(min_window_substring(\"a\", \"aa\")) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the `min_window_substring` function:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "- The main loop iterates through the string `s` from left to right using the `right` pointer. This loop runs in O(m) time, where 'm' is the length of string `s`.\n",
+    "\n",
+    "- Inside the loop, we have a while loop that moves the `left` pointer to the right to shrink the window as needed. In the worst case, this while loop can also run in O(m) time because in the worst case, we may have to move the `left` pointer all the way to the end of the string.\n",
+    "\n",
+    "- Within each iteration of the while loop, we perform constant time operations, such as updating dictionaries and comparing character counts.\n",
+    "\n",
+    "- The initialization of the `t_dict` dictionary takes O(n) time, where 'n' is the length of string `t`.\n",
+    "\n",
+    "Therefore, the overall time complexity of the function is O(m + n) because the dominant factor is the length of string `s`, and the length of string `t` has a smaller impact.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "- The space complexity is determined by the space used by the `t_dict` dictionary, the `current_window_dict` dictionary, and a few variables.\n",
+    "\n",
+    "- The `t_dict` dictionary stores character counts for string `t`. In the worst case, when all characters in `t` are distinct, this dictionary can have a maximum of 'n' key-value pairs. So, the space complexity for `t_dict` is O(n).\n",
+    "\n",
+    "- The `current_window_dict` dictionary stores character counts for the current window. In the worst case, it can have a maximum of 'm' key-value pairs. So, the space complexity for `current_window_dict` is also O(m).\n",
+    "\n",
+    "- Other variables used in the function, such as `left`, `right`, `min_len`, `min_window`, `required_chars`, and `found_chars`, are all of constant size and do not depend on the input sizes.\n",
+    "\n",
+    "**In summary**, the overall space complexity of the function is O(max(m, n)), which means it is determined by the larger of the two input strings' lengths. This is because the space used for `t_dict` and `current_window_dict` dominates the space complexity."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Smallest Distinct Substring:** Instead of finding the minimum window substring containing all characters, find the smallest distinct (unique) substring of s that contains all characters from t. This variation adds complexity because you must find a substring with distinct characters.\n",
+    "2. **No Extra Space:** Solve the problem without using any extra space, such as dictionaries or arrays, other than a constant amount of space. This is a significant optimization challenge."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb b/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb
new file mode 100644
index 0000000..04c0280
--- /dev/null
+++ b/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb	
@@ -0,0 +1,212 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 20. Valid Parentheses\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Valid Parentheses problem on LeetCode, click here!](https://leetcode.com/problems/valid-parentheses/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Given a string s containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.\n",
+    "\n",
+    "An input string is valid if:\n",
+    "\n",
+    "1. Open brackets must be closed by the same type of brackets.\n",
+    "2. Open brackets must be closed in the correct order.\n",
+    "3. Every close bracket has a corresponding open bracket of the same type.\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- 1 <= `s.length` <= $10^4$\n",
+    "- `s` consists of parentheses only `'()[]{}'`."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def isValid(s):\n",
+    "    # Create an empty stack to store opening brackets\n",
+    "    stack = []\n",
+    "    \n",
+    "    # Define a mapping for matching brackets\n",
+    "    bracket_mapping = {')': '(', '}': '{', ']': '['}\n",
+    "    \n",
+    "    # Iterate through the characters in the input string\n",
+    "    for char in s:\n",
+    "        # If the character is an opening bracket, push it onto the stack\n",
+    "        if char in bracket_mapping.values():\n",
+    "            stack.append(char)\n",
+    "        # If the character is a closing bracket\n",
+    "        elif char in bracket_mapping.keys():\n",
+    "            # Pop the top element from the stack if it exists, or use a dummy value '#'\n",
+    "            top_element = stack.pop() if stack else '#'\n",
+    "            # If the popped element does not match the corresponding opening bracket, return False\n",
+    "            if bracket_mapping[char] != top_element:\n",
+    "                return False\n",
+    "        # If the character is not a bracket, ignore it\n",
+    "        \n",
+    "    # After processing the entire string, if there are any unmatched opening brackets left in the stack, return False\n",
+    "    return len(stack) == 0"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We start by defining the `isValid` function that takes a single argument `s`, which is the input string containing parentheses and brackets.\n",
+    "\n",
+    "2. Inside the function, we create an empty stack, which is a list in Python, to store opening brackets as we encounter them in the input string. The stack will help us keep track of the brackets and their order.\n",
+    "\n",
+    "3. We define a `bracket_mapping` dictionary that maps each closing bracket to its corresponding opening bracket. This mapping will be used to check if a closing bracket matches the most recent opening bracket in the stack.\n",
+    "\n",
+    "4. We iterate through each character `char` in the input string `s` using a `for` loop.\n",
+    "\n",
+    "5. If the character `char` is an opening bracket (i.e., it exists in the `values()` of `bracket_mapping`), we push it onto the stack using the `append()` method.\n",
+    "\n",
+    "6. If the character `char` is a closing bracket (i.e., it exists in the `keys()` of `bracket_mapping`), we need to check if it matches the corresponding opening bracket. To do this, we pop the top element from the stack (if it exists) and store it in `top_element`. We use a dummy value `'#'` if the stack is empty to avoid errors.\n",
+    "\n",
+    "7. We then compare `top_element` with the corresponding opening bracket using `bracket_mapping[char]`. If they do not match, it means the string is not valid, and we return `False`.\n",
+    "\n",
+    "8. If the character `char` is not a bracket, we simply ignore it and continue the loop.\n",
+    "\n",
+    "9. After processing the entire string, we check if there are any unmatched opening brackets left in the stack. If the stack is empty, it means all opening brackets have been properly matched and closed, and we return `True`. Otherwise, we return `False`.\n",
+    "\n",
+    "10. Finally, we provide some test cases at the bottom of the code to demonstrate how the function works for different input strings."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "print(isValid(\"()\"))    "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "print(isValid(\"()[]{}\")) "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3:\n",
+    "print(isValid(\"(]\"))     "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the `isValid` function:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "- The function iterates through each character in the input string `s` once, performing constant-time operations for each character.\n",
+    "- Pushing and popping elements from the stack (list) also takes constant time in most cases.\n",
+    "- Therefore, the overall time complexity of the function is $O(n)$, where $n$ is the length of the input string `s`.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "- The space complexity of the function is determined by the space used by the stack and the `bracket_mapping` dictionary.\n",
+    "- In the worst case, if the input string `s` consists entirely of opening brackets, the stack can potentially contain all of these opening brackets, resulting in a space complexity of $O(n)$ in terms of the stack.\n",
+    "- The `bracket_mapping` dictionary has a constant number of key-value pairs (3 pairs in this case).\n",
+    "- Therefore, the overall space complexity of the function is $O(n)$, where $n$ is the length of the input string `s`, mainly due to the stack space.\n",
+    "\n",
+    "**In summary:**\n",
+    "- The time complexity is $O(n)$ because we iterate through the string once.\n",
+    "- The space complexity is $O(n)$ due to the stack used to keep track of opening brackets."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Valid Expressions:** Modify the problem to validate not only brackets but also arithmetic expressions containing parentheses, such as \"$2 * (3 + 5) / (4 - 2)$\".\n",
+    "2. **Valid Parentheses Combinations:**\n",
+    "Write a function to generate all valid combinations of n pairs of parentheses, where $n$ is a positive integer. For example, for $n = 3$, the valid combinations are [\"((()))\", \"(()())\", \"(())()\", \"()(())\", \"()()()\"]."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb b/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb
new file mode 100644
index 0000000..eb4858b
--- /dev/null
+++ b/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb	
@@ -0,0 +1,220 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 153. Find Minimum in Rotated Sorted Array\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Find Minimum in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:\n",
+    "\n",
+    "- `[4,5,6,7,0,1,2]` if it was rotated `4` times.\n",
+    "- `[0,1,2,4,5,6,7]` if it was rotated `7` times.\n",
+    "\n",
+    "Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.\n",
+    "\n",
+    "Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array.*\n",
+    "\n",
+    "You must write an algorithm that runs in $O(log\\ n)$ time.\n",
+    " \n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- `n == nums.length`\n",
+    "- `1 <= n <= 5000`\n",
+    "- `-5000 <= nums[i] <= 5000`\n",
+    "- All the integers of `nums` are **unique**.\n",
+    "- `nums` is sorted and rotated between `1` and `n` times."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def findMin(nums):\n",
+    "    left, right = 0, len(nums) - 1\n",
+    "    \n",
+    "    while left < right:\n",
+    "        mid = left + (right - left) // 2\n",
+    "        \n",
+    "        # If the mid element is greater than the rightmost element,\n",
+    "        # it means the minimum element is in the right half.\n",
+    "        if nums[mid] > nums[right]:\n",
+    "            left = mid + 1\n",
+    "        # If the mid element is less than or equal to the rightmost element,\n",
+    "        # it means the minimum element is in the left half or at the mid itself.\n",
+    "        else:\n",
+    "            right = mid\n",
+    "    \n",
+    "    # The loop will break when left and right converge to the minimum element.\n",
+    "    # At this point, left (or right) will be pointing to the minimum element.\n",
+    "    return nums[left]"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. The function `findMin(nums)` takes an input list `nums`, which represents the rotated sorted array.\n",
+    "\n",
+    "2. Initialize two pointers `left` and `right` to track the current search range. Initially, `left` is set to 0 (the beginning of the array), and `right` is set to `len(nums) - 1` (the end of the array).\n",
+    "\n",
+    "3. Enter a `while` loop with the condition `left < right`. This loop continues until `left` and `right` converge to a single element, which will be the minimum element in the rotated array.\n",
+    "\n",
+    "4. Inside the loop, calculate the middle index `mid` using integer division. This helps in finding the middle element of the current search range.\n",
+    "\n",
+    "5. Check if the element at the middle index (`nums[mid]`) is greater than the element at the rightmost index (`nums[right]`). If this condition is true, it means that the minimum element must be in the right half of the current search range. So, update `left` to `mid + 1`, effectively eliminating the left half of the search range.\n",
+    "\n",
+    "6. If the condition from step 5 is not met, it implies that the minimum element can be in the left half of the current search range or could be the element at the `mid` index itself. In this case, update `right` to `mid`, effectively eliminating the right half of the search range.\n",
+    "\n",
+    "7. Repeat steps 4-6 until `left` and `right` converge to the minimum element.\n",
+    "\n",
+    "8. When the loop exits, it means that `left` (or `right`) points to the minimum element in the array.\n",
+    "\n",
+    "9. Return `nums[left]` (or `nums[right]`) as the result, which is the minimum element in the rotated sorted array."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 1:\n",
+    "nums1 = [3, 4, 5, 1, 2]\n",
+    "print(findMin(nums1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
+    "print(findMin(nums2))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "11\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3:\n",
+    "nums3 = [11, 13, 15, 17]\n",
+    "print(findMin(nums3))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "The time complexity of this code is $O(log\\ n)$, where '$n$' is the length of the input array `nums`. This is because the binary search algorithm is employed, which continually divides the search range in half with each iteration.\n",
+    "\n",
+    "In each iteration of the `while` loop, the search range is halved, and this process continues until the `left` and `right` pointers converge to the minimum element. In the worst case, it will take logarithmic time to reduce the search range to a single element.\n",
+    "\n",
+    "Therefore, the binary search used in this code runs in $O(log\\ n)$ time complexity.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "The space complexity of this code is $O(1)$, which means it uses a constant amount of additional space that does not depend on the size of the input array.\n",
+    "\n",
+    "The algorithm uses a fixed number of variables (`left`, `right`, `mid`) to keep track of the search range and indices, but the number of these variables does not grow with the size of the input array. Hence, the space complexity is constant.\n",
+    "\n",
+    "**In summary:**\n",
+    "- Time Complexity: $O(log\\ n)$\n",
+    "- Space Complexity: $O(1)$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. Solve the problem with a variation where the input array may contain duplicates.\n",
+    "\n",
+    "2. Modify the problem to return the index of the minimum element in the rotated sorted array, rather than the element itself. The algorithm should still run in $O(log\\ n)$ time.\n",
+    "\n",
+    "3. Implement a function that finds the maximum element in a rotated sorted array. How would you adapt the binary search algorithm to solve this problem efficiently in $O(log\\ n)$ time?\n",
+    "\n",
+    "4. Implement a function that can find the kth smallest element in a rotated sorted array. This is an extension of the original problem, and the algorithm should still run in $O(log\\ n)$ time."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb b/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb
new file mode 100644
index 0000000..7b3139d
--- /dev/null
+++ b/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb	
@@ -0,0 +1,237 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 33. Search in Rotated Sorted Array\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Search in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/search-in-rotated-sorted-array/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "There is an integer array `nums` sorted in ascending order (with **distinct** values).\n",
+    "\n",
+    "Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k (1 <= k < nums.length)` such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index 3 and become `[4,5,6,7,0,1,2]`.\n",
+    "\n",
+    "Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.*\n",
+    "\n",
+    "You must write an algorithm with $O(log\\ n)$ runtime complexity.\n",
+    " \n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- `1 <= nums.length <= 5000`\n",
+    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
+    "- All values of `nums` are **unique**.\n",
+    "- `nums` is an ascending array that is possibly rotated.\n",
+    "- $-10^4$ <= `target` <= $10^4$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def search(nums, target):\n",
+    "    left, right = 0, len(nums) - 1\n",
+    "\n",
+    "    # Find the pivot index using binary search\n",
+    "    while left < right:\n",
+    "        mid = left + (right - left) // 2\n",
+    "\n",
+    "        if nums[mid] > nums[right]:\n",
+    "            left = mid + 1\n",
+    "        else:\n",
+    "            right = mid\n",
+    "\n",
+    "    pivot = left\n",
+    "    left, right = 0, len(nums) - 1\n",
+    "\n",
+    "    # Determine which part of the array to search in\n",
+    "    if target >= nums[pivot] and target <= nums[right]:\n",
+    "        left = pivot\n",
+    "    else:\n",
+    "        right = pivot\n",
+    "\n",
+    "    # Perform binary search to find the target\n",
+    "    while left <= right:\n",
+    "        mid = left + (right - left) // 2\n",
+    "\n",
+    "        if nums[mid] == target:\n",
+    "            return mid\n",
+    "        elif nums[mid] < target:\n",
+    "            left = mid + 1\n",
+    "        else:\n",
+    "            right = mid - 1\n",
+    "\n",
+    "    return -1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. The `search` function takes two arguments: `nums`, which is the rotated sorted array, and `target`, which is the value we want to find in the array.\n",
+    "\n",
+    "2. We initialize two pointers, `left` and `right`, which represent the range in which we are going to perform binary search. Initially, `left` is set to 0, and `right` is set to the index of the last element in the array (`len(nums) - 1`).\n",
+    "\n",
+    "3. We start by finding the pivot index using binary search. The pivot index is the index at which the array is rotated. We use a `while` loop to continue the search until `left` is less than `right`.\n",
+    "\n",
+    "4. Inside the loop, we calculate the middle index `mid` using integer division. We then compare the value at `mid` with the value at `right`. If `nums[mid]` is greater than `nums[right]`, it means the pivot point lies to the right of `mid`, so we update `left` to `mid + 1`. Otherwise, the pivot point lies to the left of `mid`, so we update `right` to `mid`. This process continues until we find the pivot index.\n",
+    "\n",
+    "5. After finding the pivot index, we have divided the array into two parts: one part is sorted in ascending order, and the other part is also sorted in ascending order but rotated. We need to determine which part of the array contains the target value.\n",
+    "\n",
+    "6. We reset `left` and `right` pointers. If the target value is within the range `[nums[pivot], nums[right]]`, we set `left` to `pivot` (the start of the rotated part), indicating that we should search in the rotated part. Otherwise, we set `right` to `pivot` (the end of the sorted part), indicating that we should search in the sorted part.\n",
+    "\n",
+    "7. We then perform binary search again within the chosen range (`left` to `right`) to find the target element. The binary search continues until `left` is less than or equal to `right`.\n",
+    "\n",
+    "8. Inside the binary search loop, we calculate the middle index `mid` and compare `nums[mid]` with the target value. If they are equal, we return `mid` as the index of the target element. If `nums[mid]` is less than the target, we update `left` to `mid + 1` to search in the right half. If `nums[mid]` is greater than the target, we update `right` to `mid - 1` to search in the left half.\n",
+    "\n",
+    "9. If we exit the binary search loop without finding the target, we return -1 to indicate that the target is not present in the array."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4\n"
+     ]
+    }
+   ],
+   "source": [
+    "nums1 = [4, 5, 6, 7, 0, 1, 2]\n",
+    "target1 = 0\n",
+    "result1 = search(nums1, target1)\n",
+    "print(result1) "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "-1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
+    "target2 = 3\n",
+    "result2 = search(nums2, target2)\n",
+    "print(result2) "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "-1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3:\n",
+    "nums3 = [1]\n",
+    "target3 = 0\n",
+    "result3 = search(nums3, target3)\n",
+    "print(result3) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "1. Finding the pivot index using binary search takes $O(log\\ n)$ time, where '$n$' is the number of elements in the array.\n",
+    "2. After finding the pivot index, performing binary search to find the target element also takes $O(log\\ n)$ time in the worst case.\n",
+    "\n",
+    "The dominant factor in the time complexity is the binary search, and since we perform two binary searches sequentially, the overall time complexity of the code is $O(log\\ n)$.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "The space complexity of the code is very minimal:\n",
+    "\n",
+    "1. We use a constant amount of additional space for variables such as `left`, `right`, `pivot`, and `mid`. These variables do not depend on the input size, so they contribute $O(1)$ space complexity.\n",
+    "2. The function does not use any additional data structures that scale with the input size.\n",
+    "\n",
+    "Therefore, the space complexity of the code is $O(1)$, which means it has a constant space complexity and does not depend on the size of the input array.\n",
+    "\n",
+    "**In summary:**\n",
+    "- Time Complexity: $O(log\\ n)$\n",
+    "- Space Complexity: $O(1)$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Find Minimum Element in Rotated Sorted Array**: Write an algorithm to find the minimum element in a rotated sorted array. This is a variation of the problem where you don't need to search for a target value but instead find the smallest element.\n",
+    "\n",
+    "2. **Search in a Circularly Sorted Array**: Consider an array that is sorted in a circular manner (e.g., [4, 5, 6, 7, 0, 1, 2, 3]). Adapt the search algorithm to work for circularly sorted arrays while maintaining $O(log\\ n)$ complexity."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb b/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb
new file mode 100644
index 0000000..c428c9f
--- /dev/null
+++ b/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb	
@@ -0,0 +1,261 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 141. Linked List Cycle\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Linked List Cycle problem on LeetCode, click here!](https://leetcode.com/problems/linked-list-cycle/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Given `head`, the head of a linked list, determine if the linked list has a cycle in it.\n",
+    "\n",
+    "There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is connected to. **Note that `pos` is not passed as a parameter.**\n",
+    "\n",
+    "Return `true` if there is a cycle in the linked list. Otherwise, return `false`.\n",
+    "\n",
+    " \n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- The number of the nodes in the list is in the range $[0, 10^4]$.\n",
+    "- $-10^5$ <= Node.val <= $10^5$\n",
+    "- `pos` is `-1` or a **valid index** in the linked-list.\n",
+    "\n",
+    "**Follow up:** Can you solve it using $O(1)$ (i.e. constant) memory?"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Definition for singly-linked list.\n",
+    "class ListNode:\n",
+    "    def __init__(self, val=0):\n",
+    "        self.val = val\n",
+    "        self.next = None\n",
+    "\n",
+    "class Solution:\n",
+    "    def hasCycle(self, head: ListNode) -> bool:\n",
+    "        if not head:\n",
+    "            return False\n",
+    "        \n",
+    "        slow = head\n",
+    "        fast = head\n",
+    "        \n",
+    "        while fast and fast.next:\n",
+    "            slow = slow.next\n",
+    "            fast = fast.next.next\n",
+    "            \n",
+    "            if slow == fast:\n",
+    "                return True\n",
+    "        \n",
+    "        return False"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The given code defines a class `Solution` with a method `hasCycle` for determining if a linked list has a cycle.\n",
+    "\n",
+    "The `ListNode` class represents nodes in the linked list, where each node has a `val` (a value associated with the node) and a `next` attribute (a reference to the next node in the list).\n",
+    "\n",
+    "The `hasCycle` method takes the `head` of a linked list as input and returns `True` if there is a cycle in the linked list, and `False` otherwise. It uses two pointers, `slow` and `fast`, to traverse the linked list. If there is a cycle, the `fast` pointer will eventually catch up to the `slow` pointer.\n",
+    "\n",
+    "The algorithm works as follows:\n",
+    "1. It checks if the input `head` is `None` (an empty list). If it is, the method returns `False` because an empty list can't have a cycle.\n",
+    "2. Two pointers, `slow` and `fast`, initially point to the `head` of the list.\n",
+    "3. The code enters a `while` loop where `fast` moves two steps at a time, and `slow` moves one step at a time.\n",
+    "4. If there is a cycle, `fast` will eventually catch up to `slow`, and the method returns `True`.\n",
+    "5. If the loop completes without finding a cycle, the method returns `False`."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Helper function to create a linked list with a cycle\n",
+    "def create_linked_list_with_cycle(values, pos):\n",
+    "    dummy = ListNode()\n",
+    "    current = dummy\n",
+    "    cycle_node = None\n",
+    "    \n",
+    "    for i, val in enumerate(values):\n",
+    "        current.next = ListNode(val)\n",
+    "        current = current.next\n",
+    "        \n",
+    "        if i == pos:\n",
+    "            cycle_node = current\n",
+    "    \n",
+    "    if cycle_node:\n",
+    "        current.next = cycle_node\n",
+    "    \n",
+    "    return dummy.next"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
+   "metadata": {},
+   "source": [
+    "This helper function is designed to create a linked list with a cycle for testing purposes. It takes two arguments:\n",
+    "\n",
+    "1. `values`: A list of values representing the nodes in the linked list.\n",
+    "2. `pos`: An integer representing the index of the node where the cycle begins.\n",
+    "\n",
+    "Here's how the function works:\n",
+    "\n",
+    "- It starts by creating an empty `ListNode` called `dummy`. This `dummy` node is used to simplify the creation of the linked list.\n",
+    "- It initializes a `current` pointer to the `dummy` node. This pointer will be used to traverse the linked list.\n",
+    "- It initializes a `cycle_node` variable to `None`. This variable will hold the reference to the node where the cycle begins, if any.\n",
+    "- It then iterates through the `values` list, creating a new `ListNode` for each value and appending it to the linked list.\n",
+    "- Inside the loop, it checks if the current index `i` is equal to the specified `pos`. If they match, it sets `cycle_node` to the current node. This simulates the creation of a cycle in the linked list.\n",
+    "- After the loop, if `cycle_node` is not `None` (indicating a cycle should be created), it connects the last node in the linked list to `cycle_node`, effectively creating a cycle.\n",
+    "- Finally, it returns the reference to the first node of the linked list (not the `dummy` node).\n",
+    "\n",
+    "This helper function allows you to easily create test cases where you can specify the values for the linked list and the position where the cycle starts. It's particularly useful for testing the code's ability to detect cycles in linked lists."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 1\n",
+    "head = [3,2,0,-4]\n",
+    "pos = 1\n",
+    "\n",
+    "head = create_linked_list_with_cycle(head, pos)\n",
+    "solution = Solution()\n",
+    "result = solution.hasCycle(head)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 2\n",
+    "head = [1,2]\n",
+    "pos = 0\n",
+    "\n",
+    "head = create_linked_list_with_cycle(head, pos)\n",
+    "solution = Solution()\n",
+    "result = solution.hasCycle(head)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 3\n",
+    "head = [1]\n",
+    "pos = -1\n",
+    "\n",
+    "head = create_linked_list_with_cycle(head, pos)\n",
+    "solution = Solution()\n",
+    "result = solution.hasCycle(head)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's discuss the time and space complexity of the code for detecting a cycle in a linked list:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "The time complexity of this code is $O(n)$, where \"$n$\" is the number of nodes in the linked list.\n",
+    "\n",
+    "The reason for this is that both the `slow` and `fast` pointers traverse the linked list, and they move at different speeds. In the worst case, when there is no cycle, the `fast` pointer will reach the end of the list after going through approximately $n/2$ nodes. In the case of a cycle, it may take some extra iterations for the `fast` pointer to catch up to the `slow` pointer. However, the total number of iterations is still proportional to the number of nodes in the linked list, making it $O(n)$.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "The space complexity of this code is $O(1)$, which means it uses constant space.\n",
+    "\n",
+    "The reason for this is that regardless of the size of the linked list, the code only uses two additional pointers (`slow` and `fast`) to traverse the list. These pointers do not depend on the size of the input linked list, so the space complexity remains constant.\n",
+    "\n",
+    "**In summary**, this code efficiently detects cycles in a linked list with a time complexity of $O(n)$ and a space complexity of $O(1)$. It is an example of an algorithm that solves a complex problem with minimal memory usage."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb b/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb
new file mode 100644
index 0000000..4ec3f3b
--- /dev/null
+++ b/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb	
@@ -0,0 +1,277 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 143. Reorder List\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Reorder List problem on LeetCode, click here!](https://leetcode.com/problems/reorder-list/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "You are given the head of a singly linked-list. The list can be represented as:\n",
+    "\n",
+    "$L_0 → L_1 → … → L_{n - 1} → L_n$\n",
+    "\n",
+    "Reorder the list to be on the following form:\n",
+    "\n",
+    "$L_0 → L_n → L_1 → L_{n - 1} → L_2 → L_{n - 2} → …$\n",
+    "\n",
+    "You may not modify the values in the list's nodes. Only nodes themselves may be changed.\n",
+    " \n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- The number of nodes in the list is in the range $[1, 5 * 10^4]$.\n",
+    "- `1 <= Node.val <= 1000`"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class ListNode:\n",
+    "    def __init__(self, val=0, next=None):\n",
+    "        self.val = val\n",
+    "        self.next = next\n",
+    "\n",
+    "def reorderList(head):\n",
+    "    if not head or not head.next:\n",
+    "        return head\n",
+    "\n",
+    "    # Step 1: Find the middle of the linked list\n",
+    "    slow, fast = head, head\n",
+    "    while fast.next and fast.next.next:\n",
+    "        slow = slow.next\n",
+    "        fast = fast.next.next\n",
+    "\n",
+    "    # Split the list into two halves\n",
+    "    first_half = head\n",
+    "    second_half = slow.next\n",
+    "    slow.next = None\n",
+    "\n",
+    "    # Step 2: Reverse the second half of the linked list\n",
+    "    prev = None\n",
+    "    current = second_half\n",
+    "    while current:\n",
+    "        next_node = current.next\n",
+    "        current.next = prev\n",
+    "        prev = current\n",
+    "        current = next_node\n",
+    "    second_half = prev\n",
+    "\n",
+    "    # Step 3: Merge the first and reversed second halves alternately\n",
+    "    p1, p2 = first_half, second_half\n",
+    "    while p2:\n",
+    "        next_p1 = p1.next\n",
+    "        next_p2 = p2.next\n",
+    "        p1.next = p2\n",
+    "        p2.next = next_p1\n",
+    "        p1 = next_p1\n",
+    "        p2 = next_p2\n",
+    "\n",
+    "    return head"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The `reorderList` function is designed to reorder a singly linked list following the specific pattern described in the problem statement. Here's a detailed explanation of how the function works:\n",
+    "\n",
+    "1. **Base Cases Handling**:\n",
+    "   - The function first checks if the input linked list is empty (`not head`) or contains only one element (`not head.next`). If either of these conditions is met, the list cannot be reordered, so the function returns the original list as-is.\n",
+    "\n",
+    "2. **Finding the Middle of the Linked List**:\n",
+    "   - To reorder the list, we first need to find the middle point so that we can split the list into two halves. This is done using two pointers, `slow` and `fast`, initialized to the head of the list.\n",
+    "   - `slow` moves one step at a time while `fast` moves two steps at a time. When `fast` reaches the end of the list or the second-to-last node, `slow` will be at the middle node.\n",
+    "\n",
+    "3. **Splitting the List**:\n",
+    "   - After finding the middle node, we split the list into two halves:\n",
+    "     - The first half, `first_half`, contains nodes from the beginning up to the middle.\n",
+    "     - The second half, `second_half`, contains nodes from the middle to the end.\n",
+    "   - To split the list, we set the `next` pointer of the node before the middle node to `None`.\n",
+    "\n",
+    "4. **Reversing the Second Half**:\n",
+    "   - We reverse the second half of the linked list using the `prev`, `current`, and `next_node` pointers.\n",
+    "   - `prev` is initially `None`, and we iterate through the second half. For each node, we:\n",
+    "     - Set the `next` of the current node to `prev`, effectively reversing the next pointer direction.\n",
+    "     - Update `prev` to the current node.\n",
+    "     - Move to the next node using the `next_node`.\n",
+    "\n",
+    "5. **Merging the Two Halves Alternately**:\n",
+    "   - We now have two linked lists: `first_half` and the reversed `second_half`.\n",
+    "   - We merge these two lists alternately by adjusting the `next` pointers of the nodes.\n",
+    "   - `p1` and `p2` are pointers to the current nodes in `first_half` and `second_half`, respectively.\n",
+    "   - We iterate through both lists while reordering nodes:\n",
+    "     - Set the `next` of `p1` to `p2` to link a node from the first half to a node from the reversed second half.\n",
+    "     - Update `p1` and `p2` to their respective next nodes.\n",
+    "     - Repeat this process until we have processed all nodes in both halves.\n",
+    "\n",
+    "6. **Returning the Reordered List**:\n",
+    "   - After the merging process is complete, the linked list is reordered as specified.\n",
+    "   - The function returns the `head` of the reordered list.\n",
+    "\n",
+    "The overall result is a singly linked list that has been reordered according to the given pattern. The time complexity of this algorithm is O(N), where N is the number of nodes in the linked list, as we traverse the list once to find the middle, once to reverse the second half, and once to merge the two halves."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Helper function to print the linked list\n",
+    "def printLinkedList(head):\n",
+    "    result = []\n",
+    "    while head:\n",
+    "        result.append(head.val)\n",
+    "        head = head.next\n",
+    "    return result"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
+   "metadata": {},
+   "source": [
+    "Let's explain the helper functions:\n",
+    "**printLinkedList Function**:\n",
+    "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
+    "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
+    "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
+    "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
+    "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
+    "\n",
+    "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1, 4, 2, 3]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1\n",
+    "head1 = ListNode(1, ListNode(2, ListNode(3, ListNode(4))))\n",
+    "reorderList(head1)\n",
+    "print(printLinkedList(head1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1, 5, 2, 4, 3]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "head2 = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))\n",
+    "reorderList(head2)\n",
+    "print(printLinkedList(head2))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the `reorderList` function:\n",
+    "\n",
+    "**Time Complexity**:\n",
+    "1. **Finding the Middle of the Linked List**: In this step, we use two pointers, one moving one step at a time (`slow`) and the other moving two steps at a time (`fast`). This process takes $O(N/2)$ time, where $N$ is the number of nodes in the linked list.\n",
+    "\n",
+    "2. **Splitting the List**: After finding the middle, we split the list into two halves by setting the `next` pointer of the node before the middle node to `None`. This step takes $O(1)$ time.\n",
+    "\n",
+    "3. **Reversing the Second Half**: We reverse the second half of the linked list using a while loop that iterates through the second half of the list once. Therefore, this step also takes $O(N/2)$ time.\n",
+    "\n",
+    "4. **Merging the Two Halves Alternately**: In this step, we merge the two halves alternately by adjusting the `next` pointers of the nodes. We iterate through both halves once, so this step takes $O(N/2)$ time.\n",
+    "\n",
+    "Overall, the time complexity of the `reorderList` function is dominated by the steps involving reversing and merging the two halves, both of which take $O(N/2)$ time. Therefore, the total time complexity is $O(N)$.\n",
+    "\n",
+    "**Space Complexity**:\n",
+    "The space complexity of the `reorderList` function is primarily determined by the variables and data structures used within the function. Let's break down the space complexity components:\n",
+    "\n",
+    "1. **Constant Space Variables**: The variables `slow`, `fast`, `prev`, `current`, `next_node`, `p1`, and `p2` are used to traverse and manipulate the linked list. These variables occupy constant space, regardless of the input size. Therefore, they contribute $O(1)$ to the space complexity.\n",
+    "\n",
+    "2. **Split Linked Lists**: In the splitting step, we create two new linked list segments (`first_half` and `second_half`) that store references to nodes from the original linked list. These segments occupy space proportional to half of the input list, $O(N/2)$. \n",
+    "\n",
+    "3. **Reversed Second Half**: During the reversal step, we reverse the second half of the linked list in-place without creating any additional data structures. Therefore, it doesn't contribute to additional space complexity.\n",
+    "\n",
+    "4. **Overall**: Combining the above components, the total space complexity is $O(N/2)$, which simplifies to $O(N)$ in terms of space complexity.\n",
+    "\n",
+    "**In summary**, the `reorderList` function has a time complexity of $O(N)$ and a space complexity of $O(N)$ due to the creation of two new linked list segments while splitting the list. The constant space variables used for traversal do not significantly impact the overall space complexity."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Reorder by K Elements:** Generalize the reorderList function to reorder the list in a different pattern, where you reorder the list by $K$ elements at a time. For example, for $K=3$, you would reorder it as ($L_0 → L_1 → L_2 → L_n → L_{n-1} → L_{n-2} → ...$).\n",
+    "You may need to handle cases where the number of nodes is not a multiple of $K$.\n",
+    "\n",
+    "2. **Reorder by Odd and Even Nodes:** Modify the reorderList function to reorder the list in a pattern where odd-indexed nodes (1-based index) come before even-indexed nodes. For example, ($L_0 → L_n → L_1 → L_{n-1} → L_2 → L_{n-2} → ...$)."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb b/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb
new file mode 100644
index 0000000..73dbf66
--- /dev/null
+++ b/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb	
@@ -0,0 +1,268 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 19. Remove Nth Node From End of List\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Remove Nth Node From End of List problem on LeetCode, click here!](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Given the head of a linked list, remove the $n^{th}$ node from the end of the list and return its head.\n",
+    " \n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- The number of nodes in the list is `sz`.\n",
+    "- `1 <= sz <= 30`\n",
+    "- `0 <= Node.val <= 100`\n",
+    "- `1 <= n <= sz`\n",
+    "\n",
+    "**Follow up:** Could you do this in one pass?"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class ListNode:\n",
+    "    def __init__(self, val=0, next=None):\n",
+    "        self.val = val\n",
+    "        self.next = next\n",
+    "\n",
+    "def removeNthFromEnd(head, n):\n",
+    "    # Create a dummy node to handle the case of removing the head node.\n",
+    "    dummy = ListNode(0)\n",
+    "    dummy.next = head\n",
+    "    first = dummy\n",
+    "    second = dummy\n",
+    "    \n",
+    "    # Advance the first pointer by n+1 nodes.\n",
+    "    for _ in range(n + 1):\n",
+    "        first = first.next\n",
+    "    \n",
+    "    # Move both pointers simultaneously until first reaches the end.\n",
+    "    while first is not None:\n",
+    "        first = first.next\n",
+    "        second = second.next\n",
+    "    \n",
+    "    # Remove the nth node by updating the next pointer of the previous node.\n",
+    "    second.next = second.next.next\n",
+    "    \n",
+    "    return dummy.next  # Return the modified head of the linked list"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We start by defining a `ListNode` class to represent individual nodes in a linked list. Each node has two attributes: `val` (the value of the node) and `next` (a reference to the next node in the list).\n",
+    "\n",
+    "2. The `removeNthFromEnd` function takes the head of a linked list (`head`) and the value `n`, which represents the position from the end of the list of the node to be removed. It returns the modified head of the linked list after the removal.\n",
+    "\n",
+    "3. We create a dummy node (`dummy`) at the beginning of the list. This dummy node simplifies the code by handling cases where the head of the list needs to be removed.\n",
+    "\n",
+    "4. We initialize two pointers, `first` and `second`, both initially pointing to the dummy node.\n",
+    "\n",
+    "5. To advance the `first` pointer by `n+1` nodes, we use a `for` loop. This positions the `first` pointer `n` nodes ahead of the `second` pointer.\n",
+    "\n",
+    "6. We then move both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This ensures that the `second` pointer ends up pointing to the node that needs to be removed.\n",
+    "\n",
+    "7. To remove the nth node from the end, we update the `next` pointer of the node pointed to by the `second` pointer to skip over the node to be removed.\n",
+    "\n",
+    "8. Finally, we return `dummy.next`, which is the modified head of the linked list without the removed node.\n",
+    "\n",
+    "9. Additionally, there are two helper functions:\n",
+    "   - `createLinkedList(values)` creates a linked list from a list of values.\n",
+    "   - `convertToList(head)` converts a linked list back into a list of values."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Helper function to create a linked list from a list of values.\n",
+    "def createLinkedList(values):\n",
+    "    if not values:\n",
+    "        return None\n",
+    "    head = ListNode(values[0])\n",
+    "    current = head\n",
+    "    for val in values[1:]:\n",
+    "        current.next = ListNode(val)\n",
+    "        current = current.next\n",
+    "    return head\n",
+    "\n",
+    "# Helper function to convert a linked list to a list of values.\n",
+    "def convertToList(head):\n",
+    "    result = []\n",
+    "    current = head\n",
+    "    while current:\n",
+    "        result.append(current.val)\n",
+    "        current = current.next\n",
+    "    return result"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
+   "metadata": {},
+   "source": [
+    "Let's explain the two helper functions:\n",
+    "\n",
+    "1. `createLinkedList(values)`: This function creates a linked list from a list of values.\n",
+    "\n",
+    "   - Input: `values` is a list containing values that you want to insert into the linked list.\n",
+    "   - Output: The function returns the head of the created linked list.\n",
+    "\n",
+    "   Explanation:\n",
+    "   - The function starts by checking if the `values` list is empty. If it is, it returns `None` to indicate an empty linked list.\n",
+    "   - If the `values` list is not empty, it creates the first node of the linked list with the value from the first element of `values`.\n",
+    "   - It then iterates through the remaining elements of `values` and creates new nodes for each value, connecting them together using the `next` attribute to form a linked list.\n",
+    "   - Finally, it returns the head of the linked list, which is the first node created.\n",
+    "   \n",
+    "\n",
+    "2. `convertToList(head)`: This function converts a linked list into a list of values.\n",
+    "\n",
+    "   - Input: `head` is the head node of the linked list that you want to convert.\n",
+    "   - Output: The function returns a list containing the values of the nodes in the linked list in order.\n",
+    "\n",
+    "   Explanation:\n",
+    "   - The function starts by initializing an empty list, `result`, to store the values of the linked list nodes.\n",
+    "   - It then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
+    "   - The iteration continues until it reaches the end of the linked list (i.e., the `next` attribute of the current node becomes `None`).\n",
+    "   - After the iteration is complete, the function returns the `result` list, which now contains the values of all the nodes in the linked list in the same order as they appear in the linked list.\n",
+    "\n",
+    "These helper functions are useful for creating linked lists from lists of values and converting linked lists back into lists of values, making it easier to work with linked list examples and test cases in a more familiar list format."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1, 2, 3, 5]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1\n",
+    "head1 = createLinkedList([1, 2, 3, 4, 5])\n",
+    "n1 = 2\n",
+    "new_head1 = removeNthFromEnd(head1, n1)\n",
+    "print(convertToList(new_head1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "head2 = createLinkedList([1])\n",
+    "n2 = 1\n",
+    "new_head2 = removeNthFromEnd(head2, n2)\n",
+    "print(convertToList(new_head2))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3\n",
+    "head3 = createLinkedList([1, 2])\n",
+    "n3 = 1\n",
+    "new_head3 = removeNthFromEnd(head3, n3)\n",
+    "print(convertToList(new_head3))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Certainly! Let's analyze the time and space complexity of the `removeNthFromEnd` function in isolation:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "- The `removeNthFromEnd` function performs two passes through the linked list. In the worst case, it needs to traverse the entire linked list of length $N$.\n",
+    "- The first pass positions the `first` pointer $n$ nodes ahead of the `second` pointer. This pass takes $O(N)$ time because it involves iterating through all N nodes.\n",
+    "- The second pass moves both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This also takes $O(N)$ time in the worst case.\n",
+    "- Therefore, the overall time complexity of the `removeNthFromEnd` function is $O(N)$.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "- The `removeNthFromEnd` function uses a constant amount of extra space for its variables (e.g., `first`, `second`, and `dummy`). This space usage does not depend on the size of the linked list.\n",
+    "- The space complexity is $O(1)$, indicating that the space used by the function is constant and independent of the size of the input linked list.\n",
+    "\n",
+    "**In summary**, the `removeNthFromEnd` function has a time complexity of $O(N)$ and a space complexity of $O(1)$. It efficiently removes the nth node from the end of the linked list with a single pass through the list and constant extra space usage."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb b/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb
new file mode 100644
index 0000000..09de0a3
--- /dev/null
+++ b/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb	
@@ -0,0 +1,297 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 206. Reverse Linked List\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Reverse Linked List problem on LeetCode, click here!](https://leetcode.com/problems/reverse-linked-list/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Given the `head` of a singly linked list, reverse the list, and return the reversed list.\n",
+    " \n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- The number of nodes in the list is the range `[0, 5000]`.\n",
+    "- `-5000 <= Node.val <= 5000`\n",
+    "\n",
+    "**Follow up:** A linked list can be reversed either iteratively or recursively. Could you implement both?"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Definition for singly-linked list.\n",
+    "class ListNode:\n",
+    "    def __init__(self, val=0, next=None):\n",
+    "        self.val = val\n",
+    "        self.next = next\n",
+    "\n",
+    "# Function to reverse a linked list iteratively\n",
+    "def reverseListIterative(head):\n",
+    "    prev = None\n",
+    "    current = head\n",
+    "    \n",
+    "    while current:\n",
+    "        next_node = current.next\n",
+    "        current.next = prev\n",
+    "        prev = current\n",
+    "        current = next_node\n",
+    "    \n",
+    "    return prev\n",
+    "\n",
+    "# Function to reverse a linked list recursively\n",
+    "def reverseListRecursive(head):\n",
+    "    if not head or not head.next:\n",
+    "        return head\n",
+    "\n",
+    "    new_head = reverseListRecursive(head.next)\n",
+    "    head.next.next = head\n",
+    "    head.next = None\n",
+    "\n",
+    "    return new_head"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. `reverseListIterative(head)`: This function reverses a linked list iteratively using a loop. Here's a step-by-step explanation:\n",
+    "\n",
+    "   - It takes the `head` of the input linked list as a parameter.\n",
+    "   - It initializes two pointers, `prev` and `current`, initially set to `None` and the `head` of the list, respectively.\n",
+    "   - It enters a `while` loop that continues until `current` reaches the end of the list (i.e., becomes `None`).\n",
+    "   - Inside the loop:\n",
+    "     - It stores the `next_node` which is the next node after `current`.\n",
+    "     - It updates the `next` pointer of `current` to point to the `prev` node. This effectively reverses the link direction.\n",
+    "     - It moves `prev` to `current` and `current` to `next_node`, advancing the pointers one step further in the list.\n",
+    "   - Once the loop completes, `prev` will be pointing to the new head of the reversed list (which was the last node of the original list), so it returns `prev`.\n",
+    "\n",
+    "2. `reverseListRecursive(head)`: This function reverses a linked list recursively. Here's how it works:\n",
+    "\n",
+    "   - It takes the `head` of the input linked list as a parameter.\n",
+    "   - The base case is checked: if `head` is `None` or `head.next` is `None`, meaning the list is empty or has only one node, it returns `head` as there's no need to reverse a list with zero or one element.\n",
+    "   - In the recursive case:\n",
+    "     - It calls itself with `head.next`, effectively moving down the list until it reaches the end.\n",
+    "     - Once it reaches the end, it starts reversing the links:\n",
+    "       - `head.next.next` is set to `head`, reversing the link between `head` and the next node.\n",
+    "       - `head.next` is set to `None` to avoid cycles.\n",
+    "   - Finally, it returns the new head of the reversed list, which will be the last node of the original list.\n",
+    "\n",
+    "**In summary,** `reverseListIterative` reverses the linked list by iterating through it and changing the next pointers, while `reverseListRecursive` reverses the linked list by recursively reaching the end and then reversing the links on the way back up the recursion stack. Both functions achieve the same result: reversing the linked list."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Helper function to create a linked list from a list of values\n",
+    "def createLinkedList(values):\n",
+    "    if not values:\n",
+    "        return None\n",
+    "    head = ListNode(values[0])\n",
+    "    current = head\n",
+    "    for val in values[1:]:\n",
+    "        current.next = ListNode(val)\n",
+    "        current = current.next\n",
+    "    return head\n",
+    "\n",
+    "# Helper function to convert a linked list to a list for testing\n",
+    "def linkedListToList(head):\n",
+    "    result = []\n",
+    "    current = head\n",
+    "    while current:\n",
+    "        result.append(current.val)\n",
+    "        current = current.next\n",
+    "    return result\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
+   "metadata": {},
+   "source": [
+    "Let's explain the two helper functions used in the provided code:\n",
+    "\n",
+    "1. `createLinkedList(values)`:\n",
+    "\n",
+    "   - Purpose: This function is used to create a linked list from a list of values. It takes a list of values as input and returns the head of the linked list.\n",
+    "\n",
+    "   - How it works:\n",
+    "     - It first checks if the input list `values` is empty. If it's empty, it returns `None` to indicate an empty linked list.\n",
+    "     - If `values` is not empty, it initializes a `head` node with the value of the first element in `values`.\n",
+    "     - It then iterates through the remaining values in `values`, creating new nodes for each value and linking them together to form a linked list.\n",
+    "     - Finally, it returns the `head` node of the newly created linked list.\n",
+    "\n",
+    "2. `linkedListToList(head)`:\n",
+    "\n",
+    "   - Purpose: This function is used to convert a linked list back into a Python list for testing and output purposes.\n",
+    "\n",
+    "   - How it works:\n",
+    "     - It takes the `head` of the linked list as input.\n",
+    "     - It initializes an empty list called `result`.\n",
+    "     - It then iterates through the linked list, starting from `head`, and appends each node's `val` (value) to the `result` list.\n",
+    "     - This process continues until the end of the linked list is reached.\n",
+    "     - Finally, it returns the `result` list, which contains the values from the linked list in the same order.\n",
+    "\n",
+    "These helper functions are used to simplify the process of creating linked lists from lists of values and converting linked lists back into lists, making it easier to work with linked lists in the provided examples and test cases."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[5, 4, 3, 2, 1]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: Iterative\n",
+    "head1 = [1,2,3,4,5]\n",
+    "head1 = createLinkedList(head1)\n",
+    "reversed_head1 = reverseListIterative(head1)\n",
+    "print(linkedListToList(reversed_head1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[2, 1]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2: Recursive\n",
+    "head2 = [1,2]\n",
+    "head2 = createLinkedList(head2)\n",
+    "reversed_head2 = reverseListRecursive(head2)\n",
+    "print(linkedListToList(reversed_head2))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3: Empty list\n",
+    "head3 = []\n",
+    "head3 = createLinkedList(head3)\n",
+    "reversed_head3 = reverseListIterative(head3)\n",
+    "print(linkedListToList(reversed_head3))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the two functions:\n",
+    "\n",
+    "1. `reverseListIterative(head)`:\n",
+    "\n",
+    "   - Time Complexity: $O(n)$\n",
+    "     - The iterative function visits each node in the linked list exactly once in a single pass, where \"$n$\" is the number of nodes in the list. Therefore, the time complexity is linear in the number of nodes.\n",
+    "\n",
+    "   - Space Complexity: $O(1)$\n",
+    "     - This function uses a constant amount of extra space for the `prev`, `current`, and `next_node` pointers. Regardless of the size of the input list, the amount of additional memory used remains the same, so the space complexity is constant.\n",
+    "\n",
+    "2. `reverseListRecursive(head)`:\n",
+    "\n",
+    "   - Time Complexity: $O(n)$\n",
+    "     - The recursive function also visits each node in the linked list exactly once. It recursively traverses the list from the head to the tail, reversing the links on the way back. Therefore, like the iterative approach, the time complexity is $O(n)$, where \"$n$\" is the number of nodes.\n",
+    "\n",
+    "   - Space Complexity: $O(n)$\n",
+    "     - The recursive function uses space on the call stack for each recursive call. In the worst case, when the list has \"$n$\" nodes, it will create \"$n$\" recursive function calls on the stack, resulting in a space complexity of $O(n)$. This is because each function call stores information about its state, including the `head` pointer, until it reaches the base case and starts returning.\n",
+    "\n",
+    "**In summary:**\n",
+    "\n",
+    "- The time complexity of both functions is $O(n)$ because they both visit each node once.\n",
+    "- The space complexity of `reverseListIterative` is $O(1)$ because it uses a constant amount of extra space.\n",
+    "- The space complexity of `reverseListRecursive` is $O(n)$ due to the recursive function calls and the associated call stack space.\n",
+    "\n",
+    "Both functions are efficient in terms of time complexity, but `reverseListIterative` is more memory-efficient because it uses a constant amount of additional space, while `reverseListRecursive` uses space on the call stack proportional to the number of nodes in the list."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **K-Group Reverse:** Modify the reverseListIterative function to reverse the linked list in groups of $K$ elements. For example, if the input linked list is [1, 2, 3, 4, 5, 6], and $K$ is 3, the output should be [3, 2, 1, 6, 5, 4].\n",
+    "\n",
+    "2. **Rotate Linked List:** Write a function to rotate a linked list to the right by $K$ places. For example, if the input is [1, 2, 3, 4, 5] and $K$ is 2, the output should be [4, 5, 1, 2, 3]."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb b/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb
new file mode 100644
index 0000000..b433cf3
--- /dev/null
+++ b/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb	
@@ -0,0 +1,277 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 21. Merge Two Sorted Lists\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Merge Two Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-two-sorted-lists/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "You are given the heads of two sorted linked lists `list1` and `list2`.\n",
+    "\n",
+    "Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.\n",
+    "\n",
+    "*Return the head of the merged linked list.*\n",
+    " \n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- The number of nodes in both lists is in the range `[0, 50]`.\n",
+    "- `-100 <= Node.val <= 100`\n",
+    "- Both `list1` and `list2` are sorted in non-decreasing order."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class ListNode:\n",
+    "    def __init__(self, val=0, next=None):\n",
+    "        self.val = val\n",
+    "        self.next = next\n",
+    "\n",
+    "def mergeTwoLists(list1, list2):\n",
+    "    # Create a dummy node to simplify the code.\n",
+    "    dummy = ListNode()\n",
+    "    current = dummy\n",
+    "\n",
+    "    while list1 and list2:\n",
+    "        if list1.val < list2.val:\n",
+    "            current.next = list1\n",
+    "            list1 = list1.next\n",
+    "        else:\n",
+    "            current.next = list2\n",
+    "            list2 = list2.next\n",
+    "        current = current.next\n",
+    "\n",
+    "    # Append the remaining elements from either list.\n",
+    "    if list1:\n",
+    "        current.next = list1\n",
+    "    else:\n",
+    "        current.next = list2\n",
+    "\n",
+    "    return dummy.next"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We define a class `ListNode` to represent the nodes of a singly-linked list. Each node has a `val` attribute to store the value of the node and a `next` attribute to point to the next node in the list.\n",
+    "\n",
+    "2. We define the `mergeTwoLists` function that takes two linked list heads, `list1` and `list2`, as input.\n",
+    "\n",
+    "3. We create a `dummy` node at the beginning of the merged list. The `dummy` node simplifies the code by serving as a placeholder, and its `next` attribute will point to the actual merged list.\n",
+    "\n",
+    "4. We initialize a `current` pointer to point to the `dummy` node initially. This `current` pointer helps us traverse and build the merged list.\n",
+    "\n",
+    "5. We enter a `while` loop that continues until either `list1` or `list2` becomes empty. Inside the loop, we compare the values of the nodes at the current positions of `list1` and `list2`.\n",
+    "\n",
+    "6. If the `val` of the node in `list1` is smaller than the `val` of the node in `list2`, we attach the node from `list1` to the merged list by updating the `next` attribute of the `current` node to point to the node in `list1`. We then move the `list1` pointer to the next node.\n",
+    "\n",
+    "7. If the `val` of the node in `list2` is smaller than or equal to the `val` of the node in `list1`, we attach the node from `list2` to the merged list in a similar manner. We move the `list2` pointer to the next node.\n",
+    "\n",
+    "8. After attaching a node to the merged list, we move the `current` pointer to the newly added node. This step is essential for keeping track of the end of the merged list.\n",
+    "\n",
+    "9. The loop continues until either `list1` or `list2` becomes empty.\n",
+    "\n",
+    "10. Once the loop exits, we check if there are any remaining elements in `list1` or `list2`. If `list1` is not empty, we attach the remaining elements of `list1` to the merged list. If `list2` is not empty, we attach the remaining elements of `list2`.\n",
+    "\n",
+    "11. Finally, we return the `next` attribute of the `dummy` node as the head of the merged list, which represents the sorted merged list."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Helper function to print the linked list\n",
+    "def printLinkedList(head):\n",
+    "    result = []\n",
+    "    while head:\n",
+    "        result.append(head.val)\n",
+    "        head = head.next\n",
+    "    return result"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
+   "metadata": {},
+   "source": [
+    "Let's explain the helper functions:\n",
+    "**printLinkedList Function**:\n",
+    "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
+    "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
+    "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
+    "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
+    "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
+    "\n",
+    "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "[1, 1, 2, 3, 4, 4]"
+      ]
+     },
+     "execution_count": 3,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "# Input: list1 = [1,2,4], list2 = [1,3,4]\n",
+    "list1 = ListNode(1, ListNode(2, ListNode(4)))\n",
+    "list2 = ListNode(1, ListNode(3, ListNode(4)))\n",
+    "merged_list = mergeTwoLists(list1, list2)\n",
+    "printLinkedList(merged_list)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "[]"
+      ]
+     },
+     "execution_count": 4,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "# Input: list1 = [], list2 = []\n",
+    "list1 = None\n",
+    "list2 = None\n",
+    "merged_list = mergeTwoLists(list1, list2)\n",
+    "printLinkedList(merged_list)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "[0]"
+      ]
+     },
+     "execution_count": 5,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "# Example 3:\n",
+    "# list1 = [], list2 = [0]\n",
+    "list1 = None\n",
+    "list2 = ListNode(0)\n",
+    "merged_list = mergeTwoLists(list1, list2)\n",
+    "printLinkedList(merged_list)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the `mergeTwoLists` function.\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "The time complexity of this function is $O(N + M$), where $N$ and $M$ are the lengths of `list1` and `list2`, respectively. Here's why:\n",
+    "\n",
+    "1. In the worst case, we need to traverse both `list1` and `list2` completely. This involves iterating through all the nodes in both lists once.\n",
+    "\n",
+    "2. The while loop runs until either `list1` or `list2` becomes empty. The number of iterations depends on the total number of nodes in both lists, which is N + M in the worst case.\n",
+    "\n",
+    "3. Inside the loop, we perform constant-time operations for each iteration, such as comparisons and updating pointers.\n",
+    "\n",
+    "Therefore, the dominant factor in the time complexity is the combined length of both input lists, resulting in a linear time complexity of $O(N + M)$.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "The space complexity of this function is $O(1)$, which means it uses a constant amount of extra space regardless of the input sizes. Here's why:\n",
+    "\n",
+    "1. We create a few extra pointers (`dummy`, `current`) and temporary variables (`val`) to manage the merging process. These consume a fixed amount of memory, regardless of the input sizes. The number of these extra variables is independent of the lengths of `list1` and `list2`.\n",
+    "\n",
+    "2. We do not create a new data structure or allocate memory for the merged list. Instead, we rearrange the existing nodes from `list1` and `list2` to form the merged list. This operation does not consume additional memory proportional to the input sizes.\n",
+    "\n",
+    "**In summary**, the `mergeTwoLists` function has a time complexity of $O(N + M)$ and a space complexity of $O(1)$, making it an efficient solution for merging two sorted linked lists."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Merge K Sorted Lists:** Extend the problem to merge $K$ sorted linked lists instead of just two. You need to efficiently merge $K$ lists into one sorted list.\n",
+    "\n",
+    "2. **Merge Lists in a Zigzag Pattern:** Merge two sorted lists in a zigzag pattern. For example, given [1, 2, 4] and [1, 3, 5], the merged list should be [1, 1, 2, 3, 4, 5]."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb b/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb
new file mode 100644
index 0000000..0568867
--- /dev/null
+++ b/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb	
@@ -0,0 +1,277 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 23. Merge k Sorted Lists\n",
+    "\n",
+    "**Difficulty:** Hard\n",
+    "\n",
+    "**Link to Problem:** [To see the Merge k Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-k-sorted-lists/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "You are given an array of $k$ linked-lists lists, each linked-list is sorted in ascending order.\n",
+    "\n",
+    "Merge all the linked-lists into one sorted linked-list and return it.\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- `k == lists.length`\n",
+    "- 0 <= k <= $10^4$\n",
+    "- `0 <= lists[i].length <= 500`\n",
+    "- $-10^4$ <= lists[i][j] <= $10^4$\n",
+    "- `lists[i]` is sorted in ascending order.\n",
+    "- The sum of `lists[i].length` will not exceed $10^4$.\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Definition for singly-linked list.\n",
+    "class ListNode:\n",
+    "     def __init__(self, val=0, next=None):\n",
+    "         self.val = val\n",
+    "         self.next = next\n",
+    "\n",
+    "class Solution:\n",
+    "    def mergeKLists(self, lists: [[ListNode]]) -> ListNode:\n",
+    "        # Check if the input list of linked lists is empty\n",
+    "        if not lists or len(lists) == 0:\n",
+    "            return None\n",
+    "\n",
+    "        # Loop until there is only one merged list left\n",
+    "        while len(lists) > 1:\n",
+    "            mergedLists = []\n",
+    "\n",
+    "            # Merge pairs of lists\n",
+    "            for i in range(0, len(lists), 2):\n",
+    "                l1 = lists[i]\n",
+    "                l2 = lists[i + 1] if (i + 1) < len(lists) else None\n",
+    "                mergedLists.append(self.mergeList(l1, l2))\n",
+    "\n",
+    "            # Update the list of lists with merged results\n",
+    "            lists = mergedLists\n",
+    "        \n",
+    "        # Return the final merged list\n",
+    "        return lists[0]\n",
+    "\n",
+    "    def mergeList(self, l1, l2):\n",
+    "        # Create a dummy node to simplify merging\n",
+    "        dummy = ListNode()\n",
+    "        tail = dummy\n",
+    "\n",
+    "        # Merge the two sorted lists\n",
+    "        while l1 and l2:\n",
+    "            if l1.val < l2.val:\n",
+    "                tail.next = l1\n",
+    "                l1 = l1.next\n",
+    "            else:\n",
+    "                tail.next = l2\n",
+    "                l2 = l2.next\n",
+    "            tail = tail.next\n",
+    "\n",
+    "        # Append any remaining elements from l1 or l2 (if any)\n",
+    "        if l1:\n",
+    "            tail.next = l1\n",
+    "        if l2:\n",
+    "            tail.next = l2\n",
+    "\n",
+    "        # Return the merged result starting from the next of the dummy node\n",
+    "        return dummy.next\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The provided code defines a Python class `Solution` with two methods for merging k sorted linked lists:\n",
+    "\n",
+    "1. `mergeKLists(self, lists: List[ListNode]) -> ListNode`: This method takes a list of k sorted linked lists as input and returns a single merged sorted linked list. It uses a divide-and-conquer approach to repeatedly merge pairs of lists until only one merged list remains.\n",
+    "\n",
+    "2. `mergeList(self, l1, l2)`: This method takes two sorted linked lists, `l1` and `l2`, as input and merges them into a single sorted linked list. It uses a dummy node to simplify the merging process.\n",
+    "\n",
+    "Here's a high-level overview of how the code works:\n",
+    "\n",
+    "- The `mergeKLists` method checks if the input list of linked lists is empty or contains no lists. If there are no lists, it returns `None`.\n",
+    "\n",
+    "- Inside a `while` loop, the code repeatedly merges pairs of linked lists until only one merged list remains in the `lists` array. It does this by iterating through the input lists in pairs and calling the `mergeList` method to merge each pair.\n",
+    "\n",
+    "- The `mergeList` method takes two sorted linked lists, `l1` and `l2`, and merges them into a single sorted linked list. It uses a dummy node (`dummy`) and a `tail` pointer to keep track of the merged list while comparing and merging elements from `l1` and `l2`.\n",
+    "\n",
+    "- After merging all pairs of lists and updating the `lists` array with the merged results, the loop continues until only one merged list remains in the `lists` array.\n",
+    "\n",
+    "- Finally, the `mergeKLists` method returns the merged list.\n",
+    "\n",
+    "Overall, this code efficiently merges k sorted linked lists using a divide-and-conquer strategy, resulting in a single merged sorted linked list as the output."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> "
+     ]
+    }
+   ],
+   "source": [
+    "### Example 1\n",
+    "# Input: lists = [[1,4,5],[1,3,4],[2,6]]\n",
+    "\n",
+    "lists1 = [\n",
+    "    ListNode(1, ListNode(4, ListNode(5))),\n",
+    "    ListNode(1, ListNode(3, ListNode(4))),\n",
+    "    ListNode(2, ListNode(6))\n",
+    "]\n",
+    "solution = Solution()\n",
+    "result1 = solution.mergeKLists(lists1)\n",
+    "\n",
+    "# Print the result\n",
+    "if result1:\n",
+    "    current = result1\n",
+    "    while current:\n",
+    "        print(current.val, end=\" -> \")\n",
+    "        current = current.next\n",
+    "else:\n",
+    "    print(\"None\")  # Print \"None\" for input with a single None element"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "None\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 2\n",
+    "# Input: lists = []\n",
+    "\n",
+    "lists2 = []\n",
+    "solution = Solution()\n",
+    "result2 = solution.mergeKLists(lists2)\n",
+    "\n",
+    "# Print the result\n",
+    "if result2:\n",
+    "    current = result2\n",
+    "    while current:\n",
+    "        print(current.val, end=\" -> \")\n",
+    "        current = current.next\n",
+    "else:\n",
+    "    print(\"None\")  # Print \"None\" for input with a single None element"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "None\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 3\n",
+    "# Input: lists = [[]]\n",
+    "\n",
+    "lists3 = [None]\n",
+    "solution = Solution()\n",
+    "result3 = solution.mergeKLists(lists3)\n",
+    "\n",
+    "# Print the result\n",
+    "if result3:\n",
+    "    current = result3\n",
+    "    while current:\n",
+    "        print(current.val, end=\" -> \")\n",
+    "        current = current.next\n",
+    "else:\n",
+    "    print(\"None\")  # Print \"None\" for input with a single None element"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "1. **Heap Initialization:** The code does not use a heap. Instead, it uses a divide-and-conquer approach. The initial check for empty input lists takes $O(1)$ time.\n",
+    "\n",
+    "2. **Merging:** The merging operation is performed in a divide-and-conquer fashion. In each iteration of the `while` loop, we merge pairs of linked lists, and the number of comparisons made is proportional to the total number of nodes across all the linked lists ($n$). In each merge step, we effectively process each node once. The number of iterations required to reduce k lists to 1 is $O(log\\ k)$.\n",
+    "\n",
+    "   Therefore, the overall time complexity of the code is $O(n * log\\ k)$, where n is the total number of nodes across all lists, and $k$ is the number of linked lists.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "1. **Heap Space:** The code doesn't use a heap data structure, so there's no additional space complexity due to a heap.\n",
+    "\n",
+    "2. **Merged Lists:** In the `mergeList` method, we create a new merged list. However, this list is not stored in memory for all lists; it's replaced with each merged pair. The space used for these merged lists is proportional to the size of the largest merged list, which is $O(n)$ in the worst case.\n",
+    "\n",
+    "3. **Additional Variables:** The code uses a few additional variables, such as `dummy` and `tail`, but these occupy a constant amount of space and don't depend on the input size.\n",
+    "\n",
+    "   Therefore, the overall space complexity of the code is $O(n)$, where n is the total number of nodes across all lists.\n",
+    "\n",
+    "In summary, the code's time complexity is $O(n * log(k))$, and its space complexity is $O(n)$. This code efficiently merges $k$ sorted linked lists using a divide-and-conquer approach with a relatively low space overhead."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb b/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb
new file mode 100644
index 0000000..2238b48
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb	
@@ -0,0 +1,203 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 100: Same Tree\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
+    "\n",
+    "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
+    "\n",
+    "**Constraints**\n",
+    "\n",
+    "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
+    "2. $-10^4$ <= Node.val <= $10^4$\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Definition for a binary tree node.\n",
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "def isSameTree(p, q):\n",
+    "    # Base case: If both p and q are None, the trees are the same.\n",
+    "    if not p and not q:\n",
+    "        return True\n",
+    "    \n",
+    "    # Base case: If either p or q is None (but not both), the trees are different.\n",
+    "    if not p or not q:\n",
+    "        return False\n",
+    "    \n",
+    "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
+    "    if p.val != q.val:\n",
+    "        return False\n",
+    "    \n",
+    "    # Recursively check the left and right subtrees of p and q.\n",
+    "    # If both subtrees are the same, the entire trees are the same.\n",
+    "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
+    "\n",
+    "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
+    "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
+    "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
+    "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
+    "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
+    "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 1\n",
+    "\n",
+    "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
+    "\n",
+    "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
+    "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
+    "print(isSameTree(p1, q1)) "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 2\n",
+    "\n",
+    "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
+    "\n",
+    "p2 = TreeNode(1, TreeNode(2), None)\n",
+    "q2 = TreeNode(1, None, TreeNode(2))\n",
+    "print(isSameTree(p2, q2))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 3\n",
+    "\n",
+    "#Input: p = [1,2,1], q = [1,1,2]\n",
+    "\n",
+    "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
+    "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
+    "print(isSameTree(p3, q3))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time Complexity\n",
+    "\n",
+    "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
+    "\n",
+    "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
+    "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
+    "\n",
+    "## Space Complexity\n",
+    "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
+    "\n",
+    "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
+    "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
+    "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
+    "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
+    "\n",
+    "## In summary:\n",
+    "\n",
+    "+ **Time Complexity:** $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
+    "+ **Space Complexity:** $O(n)$ due to the recursive call stack."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb b/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb
new file mode 100644
index 0000000..9d2cc7f
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb	
@@ -0,0 +1,263 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 102. Binary Tree Level Order Traversal\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Binary Tree Level Order Traversal problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-level-order-traversal/)\n",
+    "\n",
+    "---\n",
+    "Given the `root` of a binary tree, return *the level order traversal of its nodes' values*. (i.e., from left to right, level by level).\n",
+    "\n",
+    "**Constraints:**\n",
+    "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
+    "- `-1000 <= Node.val <= 1000`\n",
+    "\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "def levelOrder(root):\n",
+    "    # Check if the tree is empty, if so, return an empty list.\n",
+    "    if not root:\n",
+    "        return []\n",
+    "\n",
+    "    # Initialize an empty list to store the result.\n",
+    "    result = []\n",
+    "    \n",
+    "    # Initialize a queue with the root node to perform BFS.\n",
+    "    queue = [root]\n",
+    "\n",
+    "    while queue:\n",
+    "        # Initialize an empty list to store the values of nodes at the current level.\n",
+    "        level_values = []\n",
+    "        \n",
+    "        # Get the number of nodes at the current level.\n",
+    "        level_size = len(queue)\n",
+    "\n",
+    "        # Iterate through the nodes at the current level.\n",
+    "        for _ in range(level_size):\n",
+    "            # Dequeue the front node from the queue.\n",
+    "            node = queue.pop(0)\n",
+    "            \n",
+    "            # Append the value of the current node to the level_values list.\n",
+    "            level_values.append(node.val)\n",
+    "\n",
+    "            # Enqueue the left and right children of the current node if they exist.\n",
+    "            if node.left:\n",
+    "                queue.append(node.left)\n",
+    "            if node.right:\n",
+    "                queue.append(node.right)\n",
+    "\n",
+    "        # Append the level_values list (values at the current level) to the result list.\n",
+    "        result.append(level_values)\n",
+    "\n",
+    "    # Return the final result, which is a list of lists representing level order traversal.\n",
+    "    return result"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We define a `TreeNode` class that represents a node in the binary tree. Each `TreeNode` object has a value (`val`) and two child nodes: `left` and `right`.\n",
+    "\n",
+    "2. The `levelOrder` function takes the root of the binary tree as its input and returns the level order traversal of the tree as a list of lists.\n",
+    "\n",
+    "3. We start by checking if the input `root` is `None`, which indicates an empty tree. If the tree is empty, we return an empty list because there are no nodes to traverse.\n",
+    "\n",
+    "4. We initialize an empty list called `result` to store the final result, which will be a list of lists containing node values at each level.\n",
+    "\n",
+    "5. We initialize a queue called `queue` with the root node. This queue will be used for breadth-first traversal of the tree.\n",
+    "\n",
+    "6. We enter a while loop that continues until the `queue` is empty. Inside the loop, we perform the following steps:\n",
+    "\n",
+    "   - We initialize an empty list called `level_values` to store the values of nodes at the current level.\n",
+    "\n",
+    "   - We determine the number of nodes at the current level by getting the length of the `queue`. This is done to process nodes level by level.\n",
+    "\n",
+    "   - We iterate through the nodes at the current level using a for loop. For each node in the current level:\n",
+    "\n",
+    "     - We dequeue (remove) the front node from the `queue`.\n",
+    "\n",
+    "     - We append the value of the dequeued node to the `level_values` list, effectively collecting the values of nodes at the current level.\n",
+    "\n",
+    "     - If the dequeued node has a left child, we enqueue the left child to the `queue`.\n",
+    "\n",
+    "     - If the dequeued node has a right child, we enqueue the right child to the `queue`.\n",
+    "\n",
+    "   - After processing all nodes at the current level, we append the `level_values` list to the `result` list. This represents the values at the current level.\n",
+    "\n",
+    "7. The loop continues until all levels have been traversed, and the `queue` becomes empty.\n",
+    "\n",
+    "8. Finally, we return the `result` list, which contains lists of node values at each level, representing the level order traversal of the binary tree.\n",
+    "\n",
+    "The code effectively performs a breadth-first traversal of the binary tree, processing nodes level by level, and constructs the result list that represents the level order traversal."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[3], [9, 20], [15, 7]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Construct the tree\n",
+    "root = TreeNode(3)\n",
+    "root.left = TreeNode(9)\n",
+    "root.right = TreeNode(20)\n",
+    "root.right.left = TreeNode(15)\n",
+    "root.right.right = TreeNode(7)\n",
+    "\n",
+    "result = levelOrder(root)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[1]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "\n",
+    "# Using the same tree as before\n",
+    "root = TreeNode(1)\n",
+    "result = levelOrder(root)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3:\n",
+    "# Creating a new tree for this example\n",
+    "root = None\n",
+    "result = levelOrder(root)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `levelOrder` function:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "The time complexity of this function is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the breadth-first traversal.\n",
+    "\n",
+    "In the worst case, we have to enqueue and dequeue all nodes in the binary tree, which is proportional to the number of nodes (N). In each level, we process all nodes in that level, and since there are a total of log(N) levels in a balanced binary tree, the time complexity can also be approximated as O(N) for unbalanced trees.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "The space complexity of this function is O(N), where N is the number of nodes in the binary tree. Here's how the space complexity breaks down:\n",
+    "\n",
+    "1. The `result` list stores the level order traversal, and in the worst case, it contains N/2 levels (for a completely unbalanced binary tree). So, the space used by `result` is O(N).\n",
+    "\n",
+    "2. The `queue` data structure is used for BFS traversal. In the worst case, it can store all nodes at the last level of the tree. In a balanced binary tree, the maximum number of nodes at any level is 2^(log(N)), which is still O(N). In the case of an unbalanced tree, it can be even worse. So, the space used by `queue` is O(N).\n",
+    "\n",
+    "Overall, the dominant factor in terms of space complexity is the `queue`, and the space complexity is O(N).\n",
+    "\n",
+    "In summary, the function's time complexity is O(N), and its space complexity is also O(N). It is an efficient and optimal solution for performing a level order traversal of a binary tree."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Reverse Level Order Traversal:** Modify the `levelOrder` function to return the level order traversal in reverse order (from bottom to top). For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[15, 7], [9, 20], [3]]`.\n",
+    "\n",
+    "2. **Zigzag Level Order Traversal:** Write a function that performs a level order traversal of a binary tree in a zigzag pattern. In a zigzag traversal, the nodes at even levels are traversed from left to right, and nodes at odd levels are traversed from right to left. For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[3], [20, 9], [15, 7]]`."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb b/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb
new file mode 100644
index 0000000..87c5eaa
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb	
@@ -0,0 +1,203 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 104. Maximum Depth of Binary Tree\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Given the `root` of a binary tree, return its maximum depth.\n",
+    "\n",
+    "A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
+    "- `-100 <= Node.val <= 100`\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Definition for a binary tree node.\n",
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "def maxDepth(root):\n",
+    "    # Base case: If the root is None, the depth is 0.\n",
+    "    if root is None:\n",
+    "        return 0\n",
+    "    \n",
+    "    # Recursively calculate the maximum depth of the left and right subtrees.\n",
+    "    left_depth = maxDepth(root.left)\n",
+    "    right_depth = maxDepth(root.right)\n",
+    "    \n",
+    "    # Return the maximum depth of the tree by adding 1 to the maximum depth of the subtrees.\n",
+    "    return max(left_depth, right_depth) + 1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "Let's go through the code step by step to understand how it works:\n",
+    "\n",
+    "1. We start by defining a `TreeNode` class to represent the nodes of the binary tree. Each node has three attributes:\n",
+    "   - `val`: the value stored in the node.\n",
+    "   - `left`: a reference to the left child node.\n",
+    "   - `right`: a reference to the right child node.\n",
+    "\n",
+    "2. The `maxDepth` function is the main function that calculates the maximum depth of the binary tree. It takes a single argument, `root`, which is the root node of the binary tree.\n",
+    "\n",
+    "3. In the `maxDepth` function, we have a base case:\n",
+    "   - If the `root` is `None`, it means we have reached the end of a branch of the tree (a leaf node or an empty subtree). In this case, the depth is 0 because there are no nodes to count.\n",
+    "\n",
+    "4. If the `root` is not `None`, we recursively calculate the maximum depth of the left and right subtrees:\n",
+    "   - We call `maxDepth` on the `root.left` to calculate the maximum depth of the left subtree and store it in the variable `left_depth`.\n",
+    "   - We call `maxDepth` on the `root.right` to calculate the maximum depth of the right subtree and store it in the variable `right_depth`.\n",
+    "\n",
+    "5. Finally, we return the maximum depth of the tree by taking the maximum of `left_depth` and `right_depth` and adding 1 to it. This is because we are counting the current level as well."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 1\n",
+    "# Input: root = [3,9,20,null,null,15,7]\n",
+    "\n",
+    "root = TreeNode(3)\n",
+    "root.left = TreeNode(9)\n",
+    "root.right = TreeNode(20)\n",
+    "root.right.left = TreeNode(15)\n",
+    "root.right.right = TreeNode(7)\n",
+    "\n",
+    "print(maxDepth(root))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a230be4f-1f47-4919-87e3-62fc3f072d97",
+   "metadata": {},
+   "source": [
+    "In this example, we create a binary tree based on the given input `[3,9,20,null,null,15,7]` and calculate its maximum depth, which is `3`. The tree structure is as follows:\n",
+    "   ```\n",
+    "       3\n",
+    "      / \\\n",
+    "     9  20\n",
+    "       /  \\\n",
+    "      15   7\n",
+    "   ```\n",
+    "\n",
+    "So, the maximum depth is the length of the longest path from the root to a leaf node, which is 3 in this case."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 2\n",
+    "# Input: root = [1,null,2]\n",
+    "\n",
+    "root = TreeNode(1)\n",
+    "root.right = TreeNode(2)\n",
+    "\n",
+    "# Calculate the maximum depth of the tree and print the result\n",
+    "print(maxDepth(root))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "- The time complexity of the `maxDepth` function is $O(N)$, where $N$ is the number of nodes in the binary tree.\n",
+    "- This is because in the worst case, the function visits every node exactly once in a depth-first manner.\n",
+    "- The recursion explores all nodes of the tree, so the time complexity is linear with respect to the number of nodes.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "- The space complexity of the `maxDepth` function is $O(H)$, where H is the height of the binary tree.\n",
+    "- In the worst case, if the binary tree is completely unbalanced (skewed), the recursion stack can go as deep as the height of the tree.\n",
+    "- In the best case, if the binary tree is perfectly balanced, the height is $O(log\\ N)$, where $N$ is the number of nodes.\n",
+    "- Therefore, the space complexity can vary from $O(log\\ N)$ to $O(N)$ depending on the shape of the tree.\n",
+    "- In addition to the recursion stack, there is a small constant amount of space used for variables and function call overhead.\n",
+    "\n",
+    "**In summary:**\n",
+    "- The time complexity of the code is $O(N)$ as it visits each node once.\n",
+    "- The space complexity is $O(H)$, where H is the height of the tree, which can vary from $O(log\\ N)$ to $O(N)$ depending on the tree's shape."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb b/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb
new file mode 100644
index 0000000..0d5bbd6
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb	
@@ -0,0 +1,305 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 105. Construct Binary Tree from Preorder and Inorder Traversal\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Construct Binary Tree from Preorder and Inorder Traversal problem on LeetCode, click here!](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)\n",
+    "\n",
+    "---\n",
+    "Given two integer arrays `preorder` and `inorder` where `preorder` is the preorder traversal of a binary tree and `inorder` is the inorder traversal of the same tree, construct and return *the binary tree.*\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `1 <= preorder.length <= 3000`\n",
+    "- `inorder.length == preorder.length`\n",
+    "- `-3000 <= preorder[i], inorder[i] <= 3000`\n",
+    "- `preorder` and `inorder` consist of **unique** values.\n",
+    "- Each value of `inorder` also appears in `preorder`.\n",
+    "- `preorder` is **guaranteed** to be the preorder traversal of the tree.\n",
+    "- `inorder` is **guaranteed** to be the inorder traversal of the tree.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "def buildTree(preorder, inorder):\n",
+    "    if not preorder:\n",
+    "        return None\n",
+    "\n",
+    "    # The first element in the preorder traversal is the root of the current subtree\n",
+    "    root_val = preorder[0]\n",
+    "    root = TreeNode(root_val)\n",
+    "\n",
+    "    # Find the index of the root value in the inorder traversal\n",
+    "    root_idx_inorder = inorder.index(root_val)\n",
+    "\n",
+    "    # Recursively build left and right subtrees\n",
+    "    root.left = buildTree(preorder[1:1 + root_idx_inorder], inorder[:root_idx_inorder])\n",
+    "    root.right = buildTree(preorder[1 + root_idx_inorder:], inorder[root_idx_inorder + 1:])\n",
+    "\n",
+    "    return root"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. `class TreeNode:`: This is a class definition for a binary tree node. It has three attributes:\n",
+    "   - `val`: The value of the node.\n",
+    "   - `left`: A reference to the left child node.\n",
+    "   - `right`: A reference to the right child node.\n",
+    "\n",
+    "   This class is used to create instances of binary tree nodes, which will be used to build the binary tree.\n",
+    "\n",
+    "2. `def buildTree(preorder, inorder):`: This is the main function that constructs a binary tree from its preorder and inorder traversals.\n",
+    "\n",
+    "   - `preorder`: A list representing the preorder traversal of the binary tree.\n",
+    "   - `inorder`: A list representing the inorder traversal of the binary tree.\n",
+    "\n",
+    "3. `if not preorder: return None`: This line checks if the `preorder` list is empty. If it is, it means there are no nodes to construct, so the function returns `None`.\n",
+    "\n",
+    "4. `root_val = preorder[0]`: The value of the root node is extracted from the first element of the `preorder` list.\n",
+    "\n",
+    "5. `root = TreeNode(root_val)`: A new `TreeNode` object is created with the `root_val` as its value. This represents the root of the current subtree.\n",
+    "\n",
+    "6. `root_idx_inorder = inorder.index(root_val)`: The index of the `root_val` in the `inorder` list is found. This index indicates the position of the root node in the inorder traversal.\n",
+    "\n",
+    "7. `root.left`: The left subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The left subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists up to the `root_idx_inorder`.\n",
+    "\n",
+    "8. `root.right`: The right subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The right subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists starting from `root_idx_inorder + 1`.\n",
+    "\n",
+    "9. Finally, the function returns the `root` of the subtree it just constructed.\n",
+    "\n",
+    "The recursive approach used here divides the problem of constructing the entire binary tree into smaller subproblems, starting with the root node and then recursively building the left and right subtrees until the entire tree is constructed."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "2d82857b-4fbe-474f-bf33-17b55c20147b",
+   "metadata": {},
+   "source": [
+    "## Helper Function \n",
+    "\n",
+    "Here is a helper function to represent the binary tree elements."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "13e6f3e1-659d-479d-9521-75629b8c782f",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def treeToList(root):\n",
+    "    # Initialize an empty list to store the elements of the binary tree\n",
+    "    result = []\n",
+    "    \n",
+    "    # Initialize a queue for level-order traversal starting with the root\n",
+    "    queue = [root]\n",
+    "\n",
+    "    while queue:\n",
+    "        # Pop the front node from the queue\n",
+    "        node = queue.pop(0)\n",
+    "        \n",
+    "        # If the node is not None (i.e., a valid node), add its value to the result list\n",
+    "        if node:\n",
+    "            result.append(node.val)\n",
+    "            \n",
+    "            # Add the left and right children of the node to the queue\n",
+    "            queue.append(node.left)\n",
+    "            queue.append(node.right)\n",
+    "        else:\n",
+    "            # If the node is None, add None to the result to represent an empty node\n",
+    "            result.append(None)\n",
+    "\n",
+    "    # Remove any trailing None values from the result list\n",
+    "    while result and result[-1] is None:\n",
+    "        result.pop()\n",
+    "\n",
+    "    # Return the resulting list representing the binary tree elements\n",
+    "    return result"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "86cd4c9d-a120-4dd1-a1bd-530436012574",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "Let's break down the `treeToList` helper function step by step:\n",
+    "\n",
+    "1. `def treeToList(root):`: This function takes the root of a binary tree as input and converts it into a list.\n",
+    "\n",
+    "2. `result = []`: Initialize an empty list called `result` to store the elements of the binary tree in list format.\n",
+    "\n",
+    "3. `queue = [root]`: Initialize a queue data structure for a level-order traversal, starting with the root node.\n",
+    "\n",
+    "4. `while queue:`: This starts a loop that continues until the queue is empty, indicating that all nodes have been processed.\n",
+    "\n",
+    "5. `node = queue.pop(0)`: Dequeue the first node from the queue, effectively processing it.\n",
+    "\n",
+    "6. `if node:`: Check if the `node` is not None, which means it's a valid node in the binary tree.\n",
+    "\n",
+    "    a. `result.append(node.val)`: If the node is valid, append its value (`node.val`) to the `result` list. This represents the value of the current node in the binary tree.\n",
+    "\n",
+    "    b. `queue.append(node.left)`: Enqueue the left child of the current node if it exists. This adds the left child to the queue for processing in the next iteration.\n",
+    "\n",
+    "    c. `queue.append(node.right)`: Enqueue the right child of the current node if it exists. This adds the right child to the queue for processing in the next iteration.\n",
+    "\n",
+    "7. `else:`: If the `node` is None, it represents an empty node in the binary tree.\n",
+    "\n",
+    "    a. `result.append(None)`: Append `None` to the `result` list to indicate an empty node.\n",
+    "\n",
+    "8. `while result and result[-1] is None:`: After the traversal is complete, there might be trailing `None` values in the `result` list. This loop removes any such trailing `None` values to ensure a clean representation of the tree's elements.\n",
+    "\n",
+    "9. `return result`: Return the `result` list, which now contains the elements of the binary tree in a format where `None` represents empty nodes and the order of elements reflects a level-order traversal of the tree.\n",
+    "\n",
+    "In summary, the `treeToList` function performs a level-order traversal of the binary tree using a queue, constructing a list where each element corresponds to a node's value or represents an empty node with `None`. This list represents the binary tree's elements in a structured format."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[3, 9, 20, None, None, 15, 7]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Example usage:\n",
+    "preorder = [3, 9, 20, 15, 7]\n",
+    "inorder = [9, 3, 15, 20, 7]\n",
+    "result = buildTree(preorder, inorder)\n",
+    "\n",
+    "print(treeToList(result))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[-1]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "\n",
+    "preorder = [-1]\n",
+    "inorder = [-1]\n",
+    "result = buildTree(preorder, inorder)\n",
+    "\n",
+    "print(treeToList(result))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `buildTree` function:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "The time complexity of the `buildTree` function can be analyzed in terms of the number of nodes in the binary tree. Let's assume there are `n` nodes in the binary tree.\n",
+    "\n",
+    "1. Finding the root value in the `preorder` traversal takes O(1) time as it's just extracting the first element from the list.\n",
+    "2. Finding the index of the root value in the `inorder` traversal using `inorder.index(root_val)` takes O(n) time in the worst case because in the worst case, it might have to search through all `n` elements in the `inorder` list to find the index.\n",
+    "3. The recursive calls to `buildTree` for the left and right subtrees are made once for each node in the tree. Therefore, the recursive calls have a combined time complexity of O(n) as they process each node once.\n",
+    "\n",
+    "The total time complexity of the `buildTree` function is O(n) due to the recursive calls and finding the root's index in the `inorder` list.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "The space complexity is determined by the space used by the function's call stack during recursion and any additional data structures used. \n",
+    "\n",
+    "1. The space used by the function call stack during recursion depends on the height of the binary tree. In the worst case, where the tree is highly unbalanced (e.g., a skewed tree), the space complexity for the call stack is O(n) as it can go as deep as the number of nodes in the tree.\n",
+    "\n",
+    "2. Additionally, the function creates TreeNode objects for each node in the binary tree. Therefore, the space complexity for these objects is also O(n).\n",
+    "\n",
+    "Overall, the space complexity of the `buildTree` function is O(n) due to the space used by the call stack and the TreeNode objects.\n",
+    "\n",
+    "In summary, the time complexity of `buildTree` is O(n), and the space complexity is O(n), where 'n' is the number of nodes in the binary tree."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Construct Binary Tree from Postorder and Inorder Traversal:**\n",
+    "   Modify the problem to construct a binary tree from its postorder and inorder traversals. Implement a function similar to `buildTree` but for postorder and inorder traversals.\n",
+    "\n",
+    "2. **Reconstruct Binary Tree with Duplicate Values:**\n",
+    "   Extend the problem to handle binary trees with duplicate values. Ensure that your solution correctly handles scenarios where there are duplicate values in the preorder and inorder traversals."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb b/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb
new file mode 100644
index 0000000..b661fa1
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb	
@@ -0,0 +1,243 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 124. Binary Tree Maximum Path Sum\n",
+    "\n",
+    "**Difficulty:** Hard\n",
+    "\n",
+    "**Link to Problem:** [To see the Binary Tree Maximum Path Sum problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-maximum-path-sum/)\n",
+    "\n",
+    "---\n",
+    "A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.\n",
+    "\n",
+    "The **path sum** of a path is the sum of the node's values in the path.\n",
+    "\n",
+    "Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- The number of nodes in the tree is in the range $[1, 3*10^4]$.\n",
+    "- `-1000 <= Node.val <= 1000`"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem is to find the maximum path sum in a binary tree. A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. The path sum of a path is the sum of the node values in that path.\n",
+    "\n",
+    "In this problem, you are given the root of a binary tree, and you need to find the maximum path sum among all possible paths in the tree. Note that the path does not need to start at the root node or end at a leaf node; it can start and end at any nodes in the tree.\n",
+    "\n",
+    "To find the maximum path sum, you need to consider both left and right subtrees of each node while traversing the tree. At each node, you have two choices:\n",
+    "\n",
+    "1. Include the current node in the path: In this case, you add the value of the current node to the sum and continue exploring both the left and right subtrees for possible extensions of the path.\n",
+    "\n",
+    "2. Start a new path from the current node: In this case, you do not include the current node in the path sum, and you choose either the left subtree or the right subtree to start a new path.\n",
+    "\n",
+    "To solve this problem, you can use a recursive approach to traverse the binary tree. For each node, you calculate the maximum path sum that can pass through that node (option 1) and also return the maximum path sum that can be extended from that node (option 2). You keep track of the global maximum path sum encountered during the traversal.\n",
+    "\n",
+    "Ultimately, the maximum path sum among all paths in the tree will be the maximum of the global maximum path sum and the maximum path sum calculated for each node.\n",
+    "\n",
+    "The constraints for this problem include:\n",
+    "\n",
+    "- The number of nodes in the tree is in the range $[1, 3 * 10^4]$.\n",
+    "- Node values are in the range [-1000, 1000].\n",
+    "\n",
+    "By considering all possible paths through the tree, the algorithm aims to find the most significant sum of node values in any path in the binary tree.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "class Solution:\n",
+    "    def maxPathSum(self, root):\n",
+    "        def max_path_sum_helper(node):\n",
+    "            if not node:\n",
+    "                return 0\n",
+    "            \n",
+    "            # Recursively calculate the maximum path sum for left and right subtrees\n",
+    "            left_max = max(0, max_path_sum_helper(node.left))\n",
+    "            right_max = max(0, max_path_sum_helper(node.right))\n",
+    "            \n",
+    "            # Update the global maximum path sum\n",
+    "            self.max_sum = max(self.max_sum, left_max + right_max + node.val)\n",
+    "            \n",
+    "            # Return the maximum path sum starting from the current node\n",
+    "            return max(left_max, right_max) + node.val\n",
+    "        \n",
+    "        self.max_sum = float('-inf')  # Initialize the global maximum to negative infinity\n",
+    "        max_path_sum_helper(root)  # Start the recursive traversal from the root\n",
+    "        return self.max_sum"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. The code starts by defining a `TreeNode` class to represent nodes in the binary tree. Each `TreeNode` has a `val` attribute representing its value, a `left` attribute pointing to the left child, and a `right` attribute pointing to the right child.\n",
+    "\n",
+    "2. Next, a `Solution` class is defined to contain the `maxPathSum` method, which will be used to find the maximum path sum in the binary tree.\n",
+    "\n",
+    "3. Within the `maxPathSum` method, a helper function called `max_path_sum_helper` is defined. This recursive function takes a single argument, `node`, which is the current node being considered during traversal.\n",
+    "\n",
+    "4. Inside the `max_path_sum_helper` function:\n",
+    "   - If `node` is `None` (i.e., the current node is a null node), it returns 0. This is the base case of the recursion.\n",
+    "   - It calculates the maximum path sum starting from the left subtree (`left_max`) and the right subtree (`right_max`). Importantly, it uses `max(0, ...)` to ensure that negative values (which would make the path sum smaller) are not considered. If a subtree's path sum is negative, it's better to not include it in the path.\n",
+    "   - The code then updates the global maximum path sum (`self.max_sum`) by checking if the sum of `left_max`, `right_max`, and the current node's value is greater than the current maximum.\n",
+    "   - Finally, it returns the maximum path sum starting from the current node, which is the maximum of `left_max` and `right_max` plus the current node's value.\n",
+    "\n",
+    "5. Before calling the `max_path_sum_helper` function, the `max_sum` attribute of the `Solution` class is initialized to negative infinity (`float('-inf')`). This is done to ensure that any valid path sum encountered during traversal will be greater than the initial value of `max_sum`.\n",
+    "\n",
+    "6. The `max_path_sum_helper` function is called with the root of the binary tree, effectively starting the traversal from the root node.\n",
+    "\n",
+    "7. Once the traversal is complete, the method returns the value of `self.max_sum`, which contains the maximum path sum found in the binary tree.\n",
+    "\n",
+    "8. Example usage at the bottom of the code demonstrates how to create binary trees and use the `maxPathSum` method to find the maximum path sum for two different tree structures.\n",
+    "\n",
+    "In summary, the code uses a recursive approach to explore all possible paths in the binary tree, ensuring that negative path sums are not considered. It keeps track of the maximum path sum encountered and returns that maximum value as the result."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "6\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Example usage:\n",
+    "root1 = TreeNode(1)\n",
+    "root1.left = TreeNode(2)\n",
+    "root1.right = TreeNode(3)\n",
+    "solution = Solution()\n",
+    "print(solution.maxPathSum(root1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "42\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "\n",
+    "root2 = TreeNode(-10)\n",
+    "root2.left = TreeNode(9)\n",
+    "root2.right = TreeNode(20)\n",
+    "root2.right.left = TreeNode(15)\n",
+    "root2.right.right = TreeNode(7)\n",
+    "print(solution.maxPathSum(root2))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the given code:\n",
+    "\n",
+    "Time Complexity:\n",
+    "- The code uses a recursive depth-first traversal to explore each node in the binary tree exactly once.\n",
+    "- During the traversal, for each node, we perform constant time operations, such as updating the `max_sum` variable and calculating the maximum path sum starting from that node.\n",
+    "- Therefore, the time complexity of the code is O(N), where N is the number of nodes in the binary tree.\n",
+    "\n",
+    "Space Complexity:\n",
+    "- The space complexity of the code is determined by the space used in the call stack during the recursive traversal.\n",
+    "- In the worst case, the depth of the call stack can be equal to the height of the binary tree. In an unbalanced tree, this could be O(N), but in a balanced binary tree, the height is O(log N).\n",
+    "- Additionally, the code uses a constant amount of space for variables like `left_max`, `right_max`, and `max_sum`.\n",
+    "- Therefore, the overall space complexity is O(H), where H is the height of the binary tree. In the worst case, it's O(N), and in the best case (a balanced binary tree), it's O(log N).\n",
+    "\n",
+    "In summary:\n",
+    "- Time Complexity: O(N)\n",
+    "- Space Complexity: O(H), where H is the height of the binary tree, ranging from O(log N) to O(N) depending on the tree's balance."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Print the Maximum Path:**\n",
+    "   Modify the code to not only find the maximum path sum but also print the nodes involved in the maximum path. You can print the nodes in the correct order from the root to the leaf.\n",
+    "\n",
+    "2. **Path Sum Count:**\n",
+    "   Instead of finding the maximum path sum, find the count of unique paths that sum to a given target value. Each path can start and end anywhere in the tree."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb b/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb
new file mode 100644
index 0000000..5bbdf42
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb	
@@ -0,0 +1,249 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 199. Binary Tree Right Side View\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Binary Tree Right Side View problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-right-side-view/)\n",
+    "\n",
+    "---\n",
+    "Given the `root` of a binary tree, imagine yourself standing on the **right side** of it, return *the values of the nodes you can see ordered from top to bottom*.\n",
+    "\n",
+    " \n",
+    "\n",
+    "**Constraints:**\n",
+    "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
+    "- `-100 <= Node.val <= 100`\n",
+    "\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "def rightSideView(root):\n",
+    "    if not root:\n",
+    "        return []  # If the tree is empty, return an empty list.\n",
+    "\n",
+    "    result = []  # Initialize a list to store the right side view values.\n",
+    "    queue = [root]  # Initialize a queue for level-order traversal with the root node.\n",
+    "\n",
+    "    while queue:\n",
+    "        # Get the number of nodes at the current level.\n",
+    "        level_size = len(queue)\n",
+    "\n",
+    "        # Traverse all nodes at the current level and add the rightmost node to the result.\n",
+    "        for i in range(level_size):\n",
+    "            node = queue.pop(0)  # Dequeue the first node from the queue.\n",
+    "\n",
+    "            # If it's the rightmost node at this level, add its value to the result.\n",
+    "            if i == level_size - 1:\n",
+    "                result.append(node.val)\n",
+    "\n",
+    "            # Add the children of the current node to the queue.\n",
+    "            if node.left:\n",
+    "                queue.append(node.left)\n",
+    "            if node.right:\n",
+    "                queue.append(node.right)\n",
+    "\n",
+    "    return result  # Return the list of right side view values.\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We start with a binary tree represented using a TreeNode class. Each node has a value (`val`) and can have a left child (`left`) and a right child (`right`).\n",
+    "\n",
+    "2. The `rightSideView` function takes the root node of the binary tree as input and returns a list of values representing the nodes you can see when standing on the right side of the tree.\n",
+    "\n",
+    "3. We check if the `root` is `None`, which means the tree is empty. If it's empty, we return an empty list because there are no nodes to see.\n",
+    "\n",
+    "4. We initialize an empty list called `result` to store the right side view values.\n",
+    "\n",
+    "5. We initialize a queue called `queue` with the root node. This queue is used for a level-order traversal of the tree.\n",
+    "\n",
+    "6. We enter a while loop that continues until the `queue` is empty. Inside this loop, we perform the following steps for each level of the tree:\n",
+    "\n",
+    "   a. We determine the number of nodes at the current level by getting the length of the `queue`. This is important for processing nodes at the same level together.\n",
+    "\n",
+    "   b. We then iterate through the nodes at the current level using a for loop. For each node in the level, we dequeue it from the `queue` using `queue.pop(0)`.\n",
+    "\n",
+    "   c. If the node we dequeue is the rightmost node at the current level (determined by `i == level_size - 1`), we add its `val` to the `result` list. This is because, when standing on the right side of the tree, you can see the rightmost node at each level.\n",
+    "\n",
+    "   d. We enqueue the left and right children of the current node (if they exist) to the `queue`. This ensures that we process the next level in the subsequent iterations of the while loop.\n",
+    "\n",
+    "7. After processing all levels of the tree, the `result` list contains the values of the rightmost nodes in each level, ordered from top to bottom.\n",
+    "\n",
+    "8. Finally, we return the `result` list as the output of the function, which represents the right side view of the binary tree.\n",
+    "\n",
+    "The code effectively performs a level-order traversal of the binary tree while keeping track of the rightmost nodes at each level and adding them to the result list. This ensures that we obtain the correct order of nodes visible from the right side of the tree."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1, 3, 4]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Construct the tree\n",
+    "root1 = TreeNode(1)\n",
+    "root1.left = TreeNode(2)\n",
+    "root1.right = TreeNode(3)\n",
+    "root1.left.right = TreeNode(5)\n",
+    "root1.right.right = TreeNode(4)\n",
+    "print(rightSideView(root1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1, 3]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "\n",
+    "root2 = TreeNode(1)\n",
+    "root2.right = TreeNode(3)\n",
+    "print(rightSideView(root2))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3:\n",
+    "# Creating a new tree for this example\n",
+    "root = None\n",
+    "result = rightSideView(root)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the given code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "The time complexity of the code is O(N), where N is the number of nodes in the binary tree. Here's why:\n",
+    "\n",
+    "1. We perform a level-order traversal of the tree using a queue. In the worst case, we visit all nodes once, which takes O(N) time since we visit each node exactly once.\n",
+    "\n",
+    "2. For each node, we perform constant-time operations like dequeuing it from the queue, checking if it's the rightmost node at the current level, and enqueuing its children (if they exist). These operations do not depend on the size of the tree, so they do not contribute to the overall time complexity.\n",
+    "\n",
+    "Therefore, the dominant factor in the time complexity is the level-order traversal, making it O(N).\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "The space complexity of the code is also O(N). Here's why:\n",
+    "\n",
+    "1. We use a queue to perform level-order traversal. In the worst case, the queue can contain all nodes at the maximum width of the tree, which can be up to N/2 nodes for a completely unbalanced tree. Therefore, the space required for the queue is O(N).\n",
+    "\n",
+    "2. The `result` list stores the values of the rightmost nodes. In the worst case, when the binary tree is a complete binary tree, it can have roughly N/2 rightmost nodes. Therefore, the space required for the `result` list is also O(N).\n",
+    "\n",
+    "3. Other auxiliary variables like `level_size`, `i`, and `node` require only constant space and do not depend on the size of the tree.\n",
+    "\n",
+    "Combining the space used by the queue and the `result` list, we get a space complexity of O(N).\n",
+    "\n",
+    "In summary, the code has a time complexity of O(N) and a space complexity of O(N), where N is the number of nodes in the binary tree. These complexities are efficient and scale linearly with the size of the input tree."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Variant with Left Side View**: Modify the code to find and return the values of nodes visible from the left side of the binary tree instead of the right side.\n",
+    "\n",
+    "2. **Zigzag Right Side View**: Extend the code to return the right side view values in a zigzag order. In a zigzag order, you alternate between starting from the rightmost node at level 0, then the leftmost at level 1, rightmost at level 2, and so on."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb b/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb
new file mode 100644
index 0000000..3e542c9
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb	
@@ -0,0 +1,215 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 226: Invert Binary Tree\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Given the root of a binary tree, invert the tree, and return its root.\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- The number of nodes in the tree is in the range `[0, 100]`.\n",
+    "- `-100 <= Node.val <= 100`\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Definition for a binary tree node.\n",
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "def invertTree(root):\n",
+    "    # Base case: If the root is None or the tree is empty, return None.\n",
+    "    if not root:\n",
+    "        return None\n",
+    "\n",
+    "    # Swap the left and right subtrees of the current node.\n",
+    "    root.left, root.right = root.right, root.left\n",
+    "\n",
+    "    # Recursively invert the left and right subtrees.\n",
+    "    invertTree(root.left)\n",
+    "    invertTree(root.right)\n",
+    "\n",
+    "    return root"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We start by defining a TreeNode class, which represents a node in a binary tree. Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
+    "\n",
+    "2. The `invertTree` function is defined to invert the binary tree. It takes the root node of the tree as an argument.\n",
+    "\n",
+    "3. In the `invertTree` function, we have a base case to handle the scenario when the root is `None` (empty tree). In such cases, we return `None` because there's nothing to invert.\n",
+    "\n",
+    "4. For non-empty trees, we swap the left and right subtrees of the current node. This effectively inverts the tree at the current node.\n",
+    "\n",
+    "5. We then recursively call `invertTree` on the left and right subtrees to invert them.\n",
+    "\n",
+    "6. Finally, we return the root of the inverted tree."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Inverted Tree (Example 1): <__main__.TreeNode object at 0x7faa5c445ab0>\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 1\n",
+    "# Input: root = [4,2,7,1,3,6,9]\n",
+    "\n",
+    "root1 = TreeNode(4)\n",
+    "root1.left = TreeNode(2)\n",
+    "root1.right = TreeNode(7)\n",
+    "root1.left.left = TreeNode(1)\n",
+    "root1.left.right = TreeNode(3)\n",
+    "root1.right.left = TreeNode(6)\n",
+    "root1.right.right = TreeNode(9)\n",
+    "\n",
+    "inverted_root1 = invertTree(root1)\n",
+    "print(\"Inverted Tree (Example 1):\", inverted_root1)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Inverted Tree (Example 2): <__main__.TreeNode object at 0x7faa5c444310>\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 2\n",
+    "# Input: root = [2,1,3]\n",
+    "\n",
+    "root2 = TreeNode(2)\n",
+    "root2.left = TreeNode(1)\n",
+    "root2.right = TreeNode(3)\n",
+    "\n",
+    "inverted_root2 = invertTree(root2)\n",
+    "print(\"Inverted Tree (Example 2):\", inverted_root2)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Inverted Tree (Example 3): None\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Example 3\n",
+    "# Input: root = []\n",
+    "\n",
+    "root3 = None  # Empty tree\n",
+    "\n",
+    "inverted_root3 = invertTree(root3)\n",
+    "print(\"Inverted Tree (Example 3):\", inverted_root3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "Let's discuss the time and space complexity of the provided Python code to invert a binary tree.\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "The time complexity of the `invertTree` function is $O(n)$, where $n$ is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal of the tree. In the worst case, we have to visit all nodes in the tree.\n",
+    "\n",
+    "The reason for this time complexity is the depth-first traversal of the tree, where we recursively process the left and right subtrees of each node.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "The space complexity of the code is determined by the function call stack during the recursion. In the worst case, the space complexity is $O(h)$, where $h$ is the height of the binary tree. \n",
+    "\n",
+    "In a completely unbalanced binary tree (essentially a linked list), the height $h$ is equal to the number of nodes $n$, resulting in a space complexity of $O(n)$. This occurs when the tree is skewed to one side.\n",
+    "\n",
+    "In a balanced binary tree, such as a full binary tree, the height $h$ is $O(log\\ n)$, and the space complexity is $O(log\\ n)$.\n",
+    "\n",
+    "The space complexity depends on how balanced the tree is. In practical scenarios, binary trees are often approximately balanced, so the space complexity is typically closer to $O(log\\ n)$.\n",
+    "\n",
+    "**In summary:**\n",
+    "\n",
+    "- Time Complexity: $O(n)$ where $n$ is the number of nodes.\n",
+    "- Space Complexity: $O(h)$, where $h$ is the height of the binary tree. In the worst case, it can be $O(n)$, and in a balanced tree, it is $O(log\\ n)$.\n",
+    "\n",
+    "Keep in mind that these complexities are based on the recursive implementation provided. Iterative solutions can achieve the same task with $O(1)$ space complexity, using auxiliary data structures like stacks or queues to mimic the recursion stack."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb b/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb
new file mode 100644
index 0000000..fdad9ba
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb	
@@ -0,0 +1,215 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 230. Kth Smallest Element in a BST\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Kth Smallest Element in a BST problem on LeetCode, click here!](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)\n",
+    "\n",
+    "---\n",
+    "Given the `root` of a binary search tree, and an integer `k`, return *the $k^{th}$ smallest value (**1-indexed**) of all the values of the nodes in the tree.*\n",
+    "\n",
+    "**Constraints:**\n",
+    "- The number of nodes in the tree is `n`.\n",
+    "- 1 <= `k <= n` <= $10^4$\n",
+    "- 0 <= `Node.val` <= $10^4$\n",
+    "\n",
+    "**Follow up**: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "def kthSmallest(root, k):\n",
+    "    stack = []  # Initialize an empty stack to simulate the traversal\n",
+    "    while root or stack:\n",
+    "        while root:\n",
+    "            stack.append(root)  # Push the current node onto the stack\n",
+    "            root = root.left  # Move to the left child\n",
+    "        root = stack.pop()  # Pop a node from the stack\n",
+    "        k -= 1  # Decrement k since we've visited a node\n",
+    "        if k == 0:\n",
+    "            return root.val  # If k becomes 0, return the current node's value as the kth smallest\n",
+    "        root = root.right  # Move to the right child to continue the traversal"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. We start by defining a `TreeNode` class, which represents a node in the binary search tree (BST). Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
+    "\n",
+    "2. The `kthSmallest` function takes two parameters: `root`, which is the root of the BST, and `k`, which represents the kth smallest element we want to find.\n",
+    "\n",
+    "3. We initialize an empty stack (`stack`) to simulate the traversal of the BST. The stack will be used to keep track of nodes as we traverse the tree in an iterative manner.\n",
+    "\n",
+    "4. We enter a while loop that continues until either `root` becomes `None` (indicating we have traversed the entire tree) or the `stack` is empty.\n",
+    "\n",
+    "5. Within the loop, we start another while loop to traverse as far left as possible in the BST. We repeatedly push nodes onto the `stack` and move to their left children until we reach the leftmost leaf node. This is the smallest node in the BST.\n",
+    "\n",
+    "6. Once we have reached the leftmost leaf node (the smallest node), we pop nodes from the `stack` one by one. As we pop each node, we decrement `k` by 1 to keep track of the number of nodes we have visited.\n",
+    "\n",
+    "7. If `k` becomes 0 after decrementing, it means we have found the kth smallest element. In this case, we return the `val` of the current node as the result.\n",
+    "\n",
+    "8. If `k` is still greater than 0, it means we haven't found the kth smallest element yet. In this case, we move to the right child of the current node to continue the traversal, as the kth smallest element, if it exists, will be in the right subtree of the current node.\n",
+    "\n",
+    "9. The process continues until we find the kth smallest element or traverse the entire tree.\n",
+    "\n",
+    "10. Finally, we return the kth smallest element found.\n",
+    "\n",
+    "The key idea in this code is to perform an in-order traversal of the BST while keeping track of the kth smallest element. By visiting nodes in ascending order, we can efficiently find the kth smallest element in O(h + k) time, where h is the height of the BST and k is the desired kth element. This approach is both concise and efficient for this problem."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Example usage:\n",
+    "# Create the tree for the first example: root = [3,1,4,null,2], k = 1\n",
+    "root1 = TreeNode(3)\n",
+    "root1.left = TreeNode(1)\n",
+    "root1.right = TreeNode(4)\n",
+    "root1.left.right = TreeNode(2)\n",
+    "k1 = 1\n",
+    "print(kthSmallest(root1, k1)) "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "\n",
+    "# Create the tree for the second example: root = [5,3,6,2,4,null,null,1], k = 3\n",
+    "root2 = TreeNode(5)\n",
+    "root2.left = TreeNode(3)\n",
+    "root2.right = TreeNode(6)\n",
+    "root2.left.left = TreeNode(2)\n",
+    "root2.left.right = TreeNode(4)\n",
+    "root2.left.left.left = TreeNode(1)\n",
+    "k2 = 3\n",
+    "print(kthSmallest(root2, k2))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the given code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "The time complexity of this code is O(h + k), where:\n",
+    "- h is the height of the binary search tree (BST).\n",
+    "- k is the desired kth smallest element we want to find.\n",
+    "\n",
+    "1. In the worst case, where the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the time complexity is O(n + k).\n",
+    "\n",
+    "2. In the best case, where the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the time complexity is O(log(n) + k).\n",
+    "\n",
+    "So, the time complexity can vary from O(log(n) + k) in the best-case scenario to O(n + k) in the worst-case scenario. Typically, for balanced BSTs, the time complexity is closer to O(log(n) + k).\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "The space complexity of this code is O(h) due to the stack used for the iterative in-order traversal, where h is the height of the BST.\n",
+    "\n",
+    "1. In the worst case, when the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the space complexity is O(n) because the stack can potentially store all n nodes.\n",
+    "\n",
+    "2. In the best case, when the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the space complexity is O(log(n)) because the stack will have at most log(n) nodes.\n",
+    "\n",
+    "So, the space complexity depends on the height of the BST and can vary from O(log(n)) in the best-case scenario to O(n) in the worst-case scenario.\n",
+    "\n",
+    "In practical terms, for balanced BSTs or moderately unbalanced BSTs, the space complexity is usually close to O(log(n)), and the code is efficient in terms of space usage."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Kth Largest Element**: Modify the code to find the kth largest element in the BST instead of the kth smallest element.\n",
+    "\n",
+    "2. **Kth Smallest Element in Two BSTs**: Given two BSTs, find the kth smallest element when considering both BSTs as a single sorted list. This involves merging the two BSTs while finding the kth element efficiently."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb b/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb
new file mode 100644
index 0000000..0179ca4
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb	
@@ -0,0 +1,237 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 235. Lowest Common Ancestor of a Binary Search Tree\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Lowest Common Ancestor of a Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)\n",
+    "\n",
+    "---\n",
+    "Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.\n",
+    "\n",
+    "According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”\n",
+    "\n",
+    "**Constraints:**\n",
+    "- The number of nodes in the tree is in the range $[2, 10^5]$.\n",
+    "- $-10^9$ <= `Node.val` <= $10^9$\n",
+    "- All `Node.val` are **unique**.\n",
+    "- `p != q`\n",
+    "- `p` and `q` will exist in the BST.\n",
+    "\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "def lowestCommonAncestor(root, p, q):\n",
+    "    # Ensure p is less than q\n",
+    "    if p.val > q.val:\n",
+    "        p, q = q, p\n",
+    "\n",
+    "    while root:\n",
+    "        # If the current node value is greater than both p and q, move left\n",
+    "        if root.val > q.val:\n",
+    "            root = root.left\n",
+    "        # If the current node value is less than both p and q, move right\n",
+    "        elif root.val < p.val:\n",
+    "            root = root.right\n",
+    "        # If the current node value is between p and q (inclusive), or it matches either p or q, it's the LCA\n",
+    "        else:\n",
+    "            return root"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. `class TreeNode:` defines a simple class for representing nodes in a binary tree. Each `TreeNode` has three attributes:\n",
+    "   - `val`: The value of the node.\n",
+    "   - `left`: A reference to the left child node.\n",
+    "   - `right`: A reference to the right child node.\n",
+    "\n",
+    "2. `def lowestCommonAncestor(root, p, q):` is a function that takes three arguments:\n",
+    "   - `root`: The root node of the BST.\n",
+    "   - `p`: A TreeNode representing one of the nodes for which we want to find the LCA.\n",
+    "   - `q`: A TreeNode representing the other node for which we want to find the LCA.\n",
+    "\n",
+    "3. `if p.val > q.val:` checks if the value of `p` is greater than the value of `q`. In a BST, it's essential to ensure that `p` represents the smaller value, and `q` represents the larger value. If `p` is greater than `q`, the code swaps their values.\n",
+    "\n",
+    "4. The main logic is inside the `while` loop, which runs until `root` becomes `None`. It performs the following steps to find the LCA:\n",
+    "   - If the current `root` node's value is greater than the value of `q`, it means both `p` and `q` are on the left subtree of the current node. So, we move to the left child of the current node by setting `root = root.left`.\n",
+    "   - If the current `root` node's value is less than the value of `p`, it means both `p` and `q` are on the right subtree of the current node. So, we move to the right child of the current node by setting `root = root.right`.\n",
+    "   - If neither of the above conditions is met, it means the current `root` node's value is between `p` and `q`, or it matches either `p` or `q`. In this case, the current node is the lowest common ancestor (LCA), so we return `root`.\n",
+    "\n",
+    "The algorithm is efficient because it takes advantage of the properties of a BST. It eliminates subtrees that cannot contain the LCA by comparing the values of `p`, `q`, and the current `root` node. Eventually, it reaches the LCA node, and that node is returned as the result."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "6\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Construct the tree\n",
+    "root = TreeNode(6)\n",
+    "root.left = TreeNode(2)\n",
+    "root.right = TreeNode(8)\n",
+    "root.left.left = TreeNode(0)\n",
+    "root.left.right = TreeNode(4)\n",
+    "root.right.left = TreeNode(7)\n",
+    "root.right.right = TreeNode(9)\n",
+    "root.left.right.left = TreeNode(3)\n",
+    "root.left.right.right = TreeNode(5)\n",
+    "\n",
+    "p = root.left  # Node 2\n",
+    "q = root.right  # Node 8\n",
+    "\n",
+    "result = lowestCommonAncestor(root, p, q)\n",
+    "print(result.val)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "\n",
+    "# Using the same tree as before\n",
+    "p = root.left  # Node 2\n",
+    "q = root.left.right  # Node 4\n",
+    "\n",
+    "result = lowestCommonAncestor(root, p, q)\n",
+    "print(result.val) "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3:\n",
+    "# Creating a new tree for this example\n",
+    "root2 = TreeNode(2)\n",
+    "root2.left = TreeNode(1)\n",
+    "\n",
+    "p = root2  # Node 2\n",
+    "q = root2.left  # Node 1\n",
+    "\n",
+    "result = lowestCommonAncestor(root2, p, q)\n",
+    "print(result.val)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided algorithm for finding the lowest common ancestor (LCA) in a Binary Search Tree (BST).\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "The time complexity of this algorithm is O(h), where \"h\" is the height of the BST. In the worst case, where the tree is completely unbalanced (essentially a linked list), the height of the tree is O(n), where \"n\" is the number of nodes in the tree. However, in a well-balanced BST, the height is logarithmic, which is O(log n).\n",
+    "\n",
+    "The reason for this time complexity is that the algorithm efficiently narrows down the search space by traversing either left or right subtrees based on the values of the target nodes `p` and `q`. It eliminates entire subtrees that cannot contain the LCA, leading to a relatively quick search.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "The space complexity of the algorithm is O(1). This is because it uses a constant amount of extra space, regardless of the size of the input BST. The only variables used are `root`, `p`, and `q`, and there are no data structures like stacks or queues used for additional space. The algorithm performs a simple traversal without recursion, so it does not consume extra memory as the tree depth increases.\n",
+    "\n",
+    "**In summary**, the provided algorithm for finding the LCA in a BST is both time and space-efficient. Its time complexity is O(h), where \"h\" is the height of the tree, and its space complexity is O(1), making it suitable for practical use even in large BSTs."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "### Challenging Exercises:\n",
+    "\n",
+    "1. **Multiple LCAs:** Extend the algorithm to find all the lowest common ancestors of two given nodes `p` and `q` in a BST. In some cases, there can be multiple LCAs.\n",
+    "\n",
+    "2. **LCA with k Nodes:** Given a BST and k nodes, find the lowest common ancestor of these k nodes. This is an extension of the problem where you must find the LCA of more than two nodes."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb b/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb
new file mode 100644
index 0000000..78a78b5
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb	
@@ -0,0 +1,291 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 297. Serialize and Deserialize Binary Tree\n",
+    "\n",
+    "**Difficulty:** Hard\n",
+    "\n",
+    "**Link to Problem:** [To see the Serialize and Deserialize Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)\n",
+    "\n",
+    "---\n",
+    "Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.\n",
+    "\n",
+    "Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.\n",
+    "\n",
+    "**Clarification:** The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
+    "- `-1000 <= Node.val <= 1000`"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "TThe problem is to design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of characters or bytes so that it can be easily stored in a file, transmitted over a network, or otherwise persisted. Deserialization is the reverse process of converting the serialized data back into the original data structure.\n",
+    "\n",
+    "In the context of this problem:\n",
+    "\n",
+    "1. **Serialization**: You are given a binary tree, and your task is to convert it into a string representation such that you can later recreate the same binary tree from this string. The format of serialization is flexible, but it should allow you to reconstruct the original binary tree accurately.\n",
+    "\n",
+    "2. **Deserialization**: Given a serialized string, you need to reconstruct the binary tree it represents, ensuring that it is identical to the original tree.\n",
+    "\n",
+    "Here are some key points to consider in solving this problem:\n",
+    "\n",
+    "- The input can include any valid binary tree, including trees with nodes having integer values within the range [-1000, 1000].\n",
+    "- You don't necessarily need to follow a specific format for serialization, but it should be designed in a way that allows unambiguous deserialization.\n",
+    "- The goal is to serialize the tree structure and its values and then deserialize it back into the same structure and values.\n",
+    "\n",
+    "A common approach for serialization is to use a traversal method like preorder traversal, where you visit nodes in the order: root, left subtree, right subtree. This way, you can serialize the tree into a string, and during deserialization, you can reconstruct the tree by parsing the string in the same order.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Definition for a binary tree node.\n",
+    "class TreeNode(object):\n",
+    "    def __init__(self, x):\n",
+    "        self.val = x\n",
+    "        self.left = None\n",
+    "        self.right = None\n",
+    "\n",
+    "class Codec:\n",
+    "    def serialize(self, root):\n",
+    "        serialized  = []\n",
+    "\n",
+    "        def dfs(node):\n",
+    "            if not node:\n",
+    "                # If the node is None, represent it as \"Null\" in the serialized string\n",
+    "                serialized .append(\"Null\")\n",
+    "                return\n",
+    "            # Convert the node's value to a string and add it to the serialized string\n",
+    "            serialized .append(str(node.val))\n",
+    "            dfs(node.left)\n",
+    "            dfs(node.right)\n",
+    "\n",
+    "        dfs(root)\n",
+    "        # Join the serialized values with \",\"\n",
+    "        return \",\".join(serialized )\n",
+    "\n",
+    "    def deserialize(self, data):\n",
+    "        vals = data.split(\",\")\n",
+    "        self.i = 0\n",
+    "\n",
+    "        def dfs():\n",
+    "            if vals[self.i] == \"Null\":\n",
+    "                self.i += 1\n",
+    "                return None\n",
+    "            # Convert the value to an integer to create a new node\n",
+    "            node = TreeNode(int(vals[self.i]))\n",
+    "            self.i += 1\n",
+    "            node.left = dfs()\n",
+    "            node.right = dfs()\n",
+    "            return node\n",
+    "\n",
+    "        return dfs()"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "Here's a step-by-step explanation of the code:\n",
+    "\n",
+    "1. **Definition of TreeNode**:\n",
+    "   - The code defines a simple `TreeNode` class that represents a node in a binary tree. Each `TreeNode` has three attributes:\n",
+    "     - `val`: The integer value stored in the node.\n",
+    "     - `left`: A reference to the left child node (or None if there is no left child).\n",
+    "     - `right`: A reference to the right child node (or None if there is no right child).\n",
+    "   - This class allows you to create binary tree nodes with integer values and left and right child references.\n",
+    "\n",
+    "2. **Serialization (serialize method)**:\n",
+    "   - The `serialize` method takes a `root` node as input, which is the root of the binary tree that needs to be serialized.\n",
+    "   - It initializes an empty list `serialized` to store the serialized elements.\n",
+    "   - The `dfs` (depth-first search) function is defined within the `serialize` method to perform a preorder traversal of the binary tree.\n",
+    "   - In the `dfs` function:\n",
+    "     - If the current `node` is `None` (i.e., a leaf node or a child of a leaf node), it appends the string \"Null\" to the `serialized` list to represent the absence of a node.\n",
+    "     - If the current `node` is not `None`, it appends the string representation of the `node.val` to the `serialized` list and then recursively calls `dfs` on the left and right children.\n",
+    "   - After the `dfs` traversal, the `serialized` list contains the serialized binary tree elements.\n",
+    "   - The method returns a string obtained by joining the elements in the `serialized` list with commas.\n",
+    "\n",
+    "3. **Deserialization (deserialize method)**:\n",
+    "   - The `deserialize` method takes a serialized string `data` as input, which represents a binary tree in the custom format.\n",
+    "   - It splits the `data` string by commas to obtain a list of elements called `vals`.\n",
+    "   - The method initializes an index `self.i` to 0. This index keeps track of the current position in the `vals` list during deserialization.\n",
+    "   - The `dfs` function is defined within the `deserialize` method to perform the deserialization process using a recursive approach.\n",
+    "   - In the `dfs` function:\n",
+    "     - If the current element in `vals` is \"Null,\" it means there is no node at this position in the binary tree, so it returns `None`.\n",
+    "     - If the current element is not \"Null,\" it converts the element to an integer to create a new `TreeNode` with that value.\n",
+    "     - It then recursively calls `dfs` to set the left and right children of the current node.\n",
+    "   - The method returns the root node of the reconstructed binary tree.\n",
+    "\n",
+    "Overall, this code demonstrates a way to serialize a binary tree into a string format and then deserialize it back into the original tree structure, allowing for the representation of both the tree structure and the values stored in the nodes."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1,2,Null,Null,3,4,Null,Null,5,Null,Null\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Example usage:\n",
+    "# Serialize a binary tree\n",
+    "root = TreeNode(1)\n",
+    "root.left = TreeNode(2)\n",
+    "root.right = TreeNode(3)\n",
+    "root.right.left = TreeNode(4)\n",
+    "root.right.right = TreeNode(5)\n",
+    "\n",
+    "codec = Codec()\n",
+    "serialized_tree = codec.serialize(root)\n",
+    "print(serialized_tree)\n",
+    "\n",
+    "# Deserialize the serialized tree\n",
+    "new_root = codec.deserialize(serialized_tree)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Null\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "\n",
+    "# Example usage:\n",
+    "# Serialize a binary tree\n",
+    "root = None\n",
+    "\n",
+    "codec = Codec()\n",
+    "serialized_tree = codec.serialize(root)\n",
+    "print(serialized_tree)\n",
+    "\n",
+    "# Deserialize the serialized tree\n",
+    "new_root = codec.deserialize(serialized_tree)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code for serializing and deserializing a binary tree:\n",
+    "\n",
+    "1. **Serialization (serialize method)**:\n",
+    "\n",
+    "   - Time Complexity:\n",
+    "     - The `serialize` method uses a depth-first traversal (preorder) of the binary tree.\n",
+    "     - It visits each node exactly once.\n",
+    "     - At each node, it performs constant-time operations (string conversion and list append).\n",
+    "     - Therefore, the time complexity for serialization is O(N), where N is the number of nodes in the binary tree.\n",
+    "\n",
+    "   - Space Complexity:\n",
+    "     - The space complexity for serialization includes the space used by the `serialized` list and the recursive call stack.\n",
+    "     - The `serialized` list stores the serialized elements, and its space is proportional to the number of nodes.\n",
+    "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
+    "     - Therefore, the space complexity for serialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
+    "\n",
+    "2. **Deserialization (deserialize method)**:\n",
+    "\n",
+    "   - Time Complexity:\n",
+    "     - The `deserialize` method splits the serialized string into a list of elements, which takes O(N) time.\n",
+    "     - The `dfs` function is a recursive function that visits each element in the list exactly once.\n",
+    "     - At each element, it performs constant-time operations (string comparison, integer conversion, and recursive function calls).\n",
+    "     - Therefore, the time complexity for deserialization is O(N).\n",
+    "\n",
+    "   - Space Complexity:\n",
+    "     - The space complexity for deserialization includes the space used by the `vals` list and the recursive call stack.\n",
+    "     - The `vals` list stores the elements from the serialized string, and its space is proportional to the number of nodes.\n",
+    "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
+    "     - Therefore, the space complexity for deserialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
+    "\n",
+    "Overall, the provided code has a time complexity of O(N) for both serialization and deserialization and a space complexity of O(N) in the majority of cases (unless the tree is severely unbalanced, in which case the height H dominates the space complexity). It efficiently serializes and deserializes a binary tree while using a linear amount of memory."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Efficient Serialization**:\n",
+    "   Optimize the serialization process to minimize the length of the serialized string. Consider using techniques like binary encoding to represent node values more efficiently.\n",
+    "\n",
+    "2. **Custom Serialization Format**:\n",
+    "   Modify the serialization and deserialization methods to use a custom format different from the one provided. Ensure that the new format allows for accurate reconstruction of the original binary tree."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
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+++ b/_build/jupyter_execute/07. Trees/572. Subtree of Another Tree.ipynb	
@@ -0,0 +1,259 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 572: Subtree of Another Tree\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Subtree of Another Tree problem on LeetCode, click here!](https://leetcode.com/problems/subtree-of-another-tree/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "**Problem Description:**\n",
+    "\n",
+    "Given the roots of two binary trees `root` and `subRoot`, return `True` if there is a subtree of `root` with the same structure and node values as `subRoot`, and `False` otherwise.\n",
+    "\n",
+    "A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "1. The number of nodes in the `root` tree is in the range $[1, 2000]$.\n",
+    "2. The number of nodes in the `subRoot` tree is in the range $[1, 1000]$.\n",
+    "3. $-10^4$ <= root.val <= $10^4$\n",
+    "4. $-10^4$ <= subRoot.val <= $10^4$\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Definition for a binary tree node.\n",
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "class Solution:\n",
+    "    def isSubtree(self, root, subRoot):\n",
+    "        if not root:\n",
+    "            return False\n",
+    "        \n",
+    "        # Check if the current subtree is equal to the subRoot\n",
+    "        if self.isSameTree(root, subRoot):\n",
+    "            return True\n",
+    "        \n",
+    "        # Recursively check the left and right subtrees\n",
+    "        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)\n",
+    "    \n",
+    "    def isSameTree(self, tree1, tree2):\n",
+    "        if not tree1 and not tree2:\n",
+    "            return True\n",
+    "        if not tree1 or not tree2:\n",
+    "            return False\n",
+    "        \n",
+    "        return (\n",
+    "            tree1.val == tree2.val and\n",
+    "            self.isSameTree(tree1.left, tree2.left) and\n",
+    "            self.isSameTree(tree1.right, tree2.right)\n",
+    "        )"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "This code defines a TreeNode class for the binary tree nodes and a Solution class with two methods:\n",
+    "\n",
+    "1. `isSubtree`: This method checks if there is a subtree of the `root` tree with the same structure and node values as the `subRoot` tree. It uses a helper function `isSameTree` to compare two trees for equality.\n",
+    "\n",
+    "2. `isSameTree`: This helper method recursively compares two trees to check if they are the same in structure and node values.\n",
+    "\n",
+    "### Here's a detailed explanation of the code:\n",
+    "\n",
+    "1. **TreeNode Class:**\n",
+    "\n",
+    "The `TreeNode` class is defined to represent nodes in a binary tree. Each node has a `val` (the node's value) and may have a `left` and `right` child.\n",
+    "\n",
+    "2. **Solution Class:**\n",
+    "\n",
+    "    + The `Solution` class contains the solution for the problem and defines two important methods:\n",
+    "\n",
+    "    + `isSubtree(self, root, subRoot)`:\n",
+    "\n",
+    "        - This method checks whether `subRoot` is a subtree of `root`. It takes two tree nodes, `root` and `subRoot`, as input arguments.\n",
+    "        - If `root` is `None`, it returns `False` because there is no subtree to search.\n",
+    "        - It then checks if the current subtree with `root` as its root is equal to `subRoot` using the `isSameTree` method. If they are the same, it returns `True`.\n",
+    "        - If the current subtree is not the same as `subRoot`, it recursively checks the left and right subtrees of `root` to see if `subRoot` is a subtree of any of them.\n",
+    "        - It returns `True` if `subRoot` is found in either the left or right subtree; otherwise, it returns `False`.\n",
+    "\n",
+    "    + `isSameTree(self, tree1, tree2)`:\n",
+    "\n",
+    "        - This method checks whether two trees, `tree1` and `tree2`, are the same.\n",
+    "        - If both `tree1` and `tree2` are `None`, they are considered the same tree, so it returns `True`.\n",
+    "        - If only one of them is `None` (but not both), they are different trees, so it returns `False`.\n",
+    "        - If both `tree1` and `tree2` have values, it checks if their values are equal and recursively checks if their left and right subtrees are the same.\n",
+    "        - It returns `True` if the trees are the same; otherwise, it returns `False`.\n",
+    "\n",
+    "The code effectively uses recursion to traverse the binary trees and check for subtree equality. The `isSubtree` method starts the recursive search, and the `isSameTree` method is used to compare individual subtrees. The approach is efficient and avoids unnecessary checks when possible.\n",
+    "\n",
+    "This solution allows you to determine if there exists a subtree within the `root` tree that matches the structure and node values of `subRoot`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Is subRoot a subtree of root? True\n"
+     ]
+    }
+   ],
+   "source": [
+    "#root = [3,4,5,1,2]\n",
+    "#subRoot = [4,1,2]\n",
+    "\n",
+    "# Example input\n",
+    "root = TreeNode(3)\n",
+    "root.left = TreeNode(4)\n",
+    "root.right = TreeNode(5)\n",
+    "root.left.left = TreeNode(1)\n",
+    "root.left.right = TreeNode(2)\n",
+    "\n",
+    "subRoot = TreeNode(4)\n",
+    "subRoot.left = TreeNode(1)\n",
+    "subRoot.right = TreeNode(2)\n",
+    "\n",
+    "# Create an instance of the Solution class\n",
+    "solution = Solution()\n",
+    "\n",
+    "# Test the isSubtree method\n",
+    "result = solution.isSubtree(root, subRoot)\n",
+    "print(\"Is subRoot a subtree of root?\", result)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Is subRoot a subtree of root? False\n"
+     ]
+    }
+   ],
+   "source": [
+    "#root = [3,4,5,1,2,null,null,null,null,0] \n",
+    "#subRoot = [4,1,2]\n",
+    "\n",
+    "\n",
+    "# Example input\n",
+    "root = TreeNode(3)\n",
+    "root.left = TreeNode(4)\n",
+    "root.right = TreeNode(5)\n",
+    "root.left.left = TreeNode(1)\n",
+    "root.left.right = TreeNode(2)\n",
+    "root.left.right.left = TreeNode(0)\n",
+    "\n",
+    "subRoot = TreeNode(4)\n",
+    "subRoot.left = TreeNode(1)\n",
+    "subRoot.right = TreeNode(2)\n",
+    "\n",
+    "# Create an instance of the Solution class\n",
+    "solution = Solution()\n",
+    "\n",
+    "# Test the isSubtree method\n",
+    "result = solution.isSubtree(root, subRoot)\n",
+    "print(\"Is subRoot a subtree of root?\", result)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "86a34f0d-0789-498c-818c-2d30f84754d0",
+   "metadata": {},
+   "source": [
+    "Let's analyze the time and space complexity of the provided code for the \"Subtree of Another Tree\" problem:\n",
+    "\n",
+    "**Time Complexity**\n",
+    "\n",
+    "The time complexity of the code primarily depends on the recursive traversal of the binary tree. In both the `isSubtree` and `isSameTree` functions, we visit each node in the binary trees once. Let's break down the time complexity:\n",
+    "\n",
+    "1. The `isSubtree` function:\n",
+    "   - In the worst case, we need to visit every node in the `root` tree.\n",
+    "   - For each node in the `root` tree, we call the `isSameTree` function, which has its own traversal.\n",
+    "   - So, the total time complexity is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree.\n",
+    "\n",
+    "2. The `isSameTree` function:\n",
+    "   - In the worst case, we visit every node in both `tree1` and `tree2`.\n",
+    "   - The number of recursive calls made is proportional to the number of nodes in the trees.\n",
+    "   - So, the time complexity of this function is $O(max(n, m))$, where $n$ and $m$ are the numbers of nodes in `tree1` and `tree2`, respectively.\n",
+    "\n",
+    "Overall, the time complexity of the entire code is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree. In practice, it may be less than $O(n * m)$ if a subtree mismatch is detected early during the traversal.\n",
+    "\n",
+    "**Space Complexity**\n",
+    "\n",
+    "The space complexity of the code is determined by the function call stack during recursion and the space used by the recursive functions. Let's analyze the space complexity:\n",
+    "\n",
+    "1. The `isSubtree` function:\n",
+    "   - It uses the call stack for recursion.\n",
+    "   - The maximum depth of the recursion is equal to the height of the `root` tree, which can be $O(n)$ in the worst case (unbalanced tree).\n",
+    "   - Additionally, the function doesn't use any significant extra space other than the recursion stack.\n",
+    "\n",
+    "2. The `isSameTree` function:\n",
+    "   - It also uses the call stack for recursion.\n",
+    "   - The maximum depth of the recursion is equal to the height of the `tree1` or `tree2`, whichever is greater.\n",
+    "   - So, the maximum space used for the call stack is $O(max(n, m))$.\n",
+    "\n",
+    "In summary, the space complexity of the code is $O(max(n, m))$ due to the function call stack. It scales with the maximum height of the trees being compared.\n",
+    "\n",
+    "Overall, the code is efficient and works well for trees with moderate sizes. However, it's important to keep in mind that the worst-case time complexity is $O(n * m)$, so for very large trees, the performance may degrade."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb b/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb
new file mode 100644
index 0000000..0cbe2fc
--- /dev/null
+++ b/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb	
@@ -0,0 +1,215 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 98. Validate Binary Search Tree\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Validate Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/validate-binary-search-tree/)\n",
+    "\n",
+    "---\n",
+    "Given the `root` of a binary tree, *determine if it is a valid binary search tree (BST)*.\n",
+    "\n",
+    "A **valid BST** is defined as follows:\n",
+    "\n",
+    "- The left subtree of a node contains only nodes with keys **less than** the node's key.\n",
+    "- The right subtree of a node contains only nodes with keys **greater than** the node's key.\n",
+    "- Both the left and right subtrees must also be binary search trees. \n",
+    "\n",
+    "**Constraints:**\n",
+    "- The number of nodes in the tree is in the range $[1, 10^4]$.\n",
+    "- $-2^{31}$ <= `Node.val` <= $2^{31} - 1$\n",
+    "\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to solve this problem:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class TreeNode:\n",
+    "    def __init__(self, val=0, left=None, right=None):\n",
+    "        self.val = val\n",
+    "        self.left = left\n",
+    "        self.right = right\n",
+    "\n",
+    "def is_valid_BST(root):\n",
+    "    def is_valid(node, min_val, max_val):\n",
+    "        # Base case: If the node is None, it's a valid BST.\n",
+    "        if node is None:\n",
+    "            return True\n",
+    "        \n",
+    "        # Check if the current node's value is within the valid range.\n",
+    "        if not (min_val < node.val < max_val):\n",
+    "            return False\n",
+    "        \n",
+    "        # Recursively check the left and right subtrees with updated ranges.\n",
+    "        return (is_valid(node.left, min_val, node.val) and\n",
+    "                is_valid(node.right, node.val, max_val))\n",
+    "    \n",
+    "    # Call the helper function starting with a wide range for root node.\n",
+    "    return is_valid(root, float('-inf'), float('inf'))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "- It defines a `TreeNode` class to represent nodes in the binary tree.\n",
+    "\n",
+    "- The `is_valid_BST` function takes the root node of the binary tree as input and returns `True` if the tree is a valid BST, and `False` otherwise.\n",
+    "\n",
+    "- Inside the `is_valid_BST` function, there is a helper function called `is_valid` that performs the actual validation using a recursive approach.\n",
+    "\n",
+    "- The `is_valid` function checks each node in the tree to ensure that it satisfies the properties of a BST:\n",
+    "  - The value of the current node must be within a valid range defined by `min_val` and `max_val`.\n",
+    "  - The left subtree of the current node should contain values less than the current node's value.\n",
+    "  - The right subtree of the current node should contain values greater than the current node's value.\n",
+    "\n",
+    "- If any of these conditions are violated, the function returns `False`, indicating that the tree is not a valid BST.\n",
+    "\n",
+    "- If all nodes satisfy these conditions, the function returns `True`, indicating that the tree is a valid BST.\n",
+    "\n",
+    "- The code calls the `is_valid` function with the root node and initial range values of negative infinity to positive infinity to start the validation.\n",
+    "\n",
+    "- The time complexity of the code is O(N), where N is the number of nodes in the tree, as it traverses each node once.\n",
+    "\n",
+    "- The space complexity depends on the height of the tree. In the average case for a balanced BST, the space complexity is O(log N), but in the worst case (skewed tree), it can be O(N) due to the recursive call stack.\n",
+    "\n",
+    "In summary, this code checks whether a given binary tree is a valid BST by recursively validating each node's value and its left and right subtrees while maintaining valid value ranges. If all nodes satisfy the BST properties, the tree is considered a valid BST."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Example usage:\n",
+    "# Create the tree for the first example: [2, 1, 3]\n",
+    "root1 = TreeNode(2)\n",
+    "root1.left = TreeNode(1)\n",
+    "root1.right = TreeNode(3)\n",
+    "print(is_valid_BST(root1))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "\n",
+    "# Create the tree for the second example: [5, 1, 4, None, None, 3, 6]\n",
+    "root2 = TreeNode(5)\n",
+    "root2.left = TreeNode(1)\n",
+    "root2.right = TreeNode(4)\n",
+    "root2.right.left = TreeNode(3)\n",
+    "root2.right.right = TreeNode(6)\n",
+    "print(is_valid_BST(root2))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `is_valid_BST` function:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "- The time complexity of the function primarily depends on the number of nodes in the binary tree.\n",
+    "- In the worst case, we may have to visit every node in the tree once to validate whether it's part of a valid BST.\n",
+    "- Since we are performing a depth-first search (DFS) traversal of the tree, the time complexity is O(N), where N is the number of nodes in the tree.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "- The space complexity is determined by the space used by the recursive call stack during the DFS traversal.\n",
+    "- In the worst case, if the tree is completely unbalanced (a skewed tree), the maximum depth of the call stack would be equal to the height of the tree.\n",
+    "- In a balanced BST, the height is approximately log2(N), where N is the number of nodes.\n",
+    "- Therefore, the space complexity of the call stack is O(log N) for a balanced BST.\n",
+    "- In the worst case (skewed tree), the space complexity can be O(N) as the height of the tree can be equal to N.\n",
+    "\n",
+    "**Overall:**\n",
+    "- Time Complexity: O(N)\n",
+    "- Space Complexity: O(log N) on average for a balanced BST, and O(N) in the worst case for a skewed tree.\n",
+    "\n",
+    "The space complexity is typically dominated by the recursive call stack, and it varies depending on the shape of the binary tree. In practice, for balanced binary trees, the space complexity is often close to O(log N), which is quite efficient."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Complex Constraints**: Modify the problem to have more complex constraints on the tree. For example, consider a binary tree with constraints like \"The left subtree of a node contains only nodes with keys less than or equal to the node's key,\" and \"The right subtree of a node contains only nodes with keys greater than or equal to the node's key.\" How would you adapt the code to handle these constraints?\n",
+    "\n",
+    "2. **Handling Duplicates**: Modify the code to handle binary search trees that allow duplicate values. In a standard BST, each value is unique, but in this case, multiple nodes can have the same value. The tree should still be considered valid if it follows the BST property."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb b/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb
new file mode 100644
index 0000000..5c54c25
--- /dev/null
+++ b/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb	
@@ -0,0 +1,245 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 208. Implement Trie (Prefix Tree)\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Implement Trie problem on LeetCode, click here!](https://leetcode.com/problems/implement-trie-prefix-tree/)\n",
+    "\n",
+    "---\n",
+    "A **trie** (pronounced as \"try\") or **prefix tree** is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.\n",
+    "\n",
+    "Implement the Trie class:\n",
+    "\n",
+    "- `Trie()` Initializes the trie object.\n",
+    "- `void insert(String word)` Inserts the string `word` into the trie.\n",
+    "- `boolean search(String word)` Returns `true` if the string `word` is in the trie (i.e., was inserted before), and `false` otherwise.\n",
+    "- `boolean startsWith(String prefix)` Returns `true` if there is a previously inserted string `word` that has the prefix `prefix`, and `false` otherwise.\n",
+    "\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `1 <= word.length, prefix.length <= 2000`\n",
+    "- `word` and `prefix` consist only of lowercase English letters.\n",
+    "- At most $3 * 10^4$ calls in total will be made to `insert`, `search`, and `startsWith`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem at hand is to implement a data structure known as a Trie (pronounced \"try\") or Prefix Tree. A Trie is a tree-like data structure that is used to efficiently store and retrieve a set of strings or words. It's particularly useful for tasks involving string manipulation, such as autocomplete and spell-checking. The problem statement defines the following operations for implementing the Trie class:\n",
+    "\n",
+    "1. `Trie()`: This is the constructor that initializes the Trie object.\n",
+    "\n",
+    "2. `void insert(String word)`: This method inserts the given string `word` into the Trie. It effectively adds the characters of the word to the Trie's structure, creating nodes for each character if they don't already exist. At the end of the word, a special flag is set to indicate that this node represents the end of a valid word.\n",
+    "\n",
+    "3. `boolean search(String word)`: This method checks if the given string `word` is present in the Trie. It starts from the root of the Trie and traverses the Trie by following the characters in the word. If it successfully traverses the Trie and reaches the end of the word, it returns `true`, indicating that the word exists in the Trie; otherwise, it returns `false`.\n",
+    "\n",
+    "4. `boolean startsWith(String prefix)`: This method checks if there is any previously inserted word in the Trie that has the given `prefix`. It's similar to the `search` method but does not require that the prefix be a complete word. If there is any word in the Trie that starts with the given prefix, it returns `true`; otherwise, it returns `false`.\n",
+    "\n",
+    "The problem also provides an example scenario where these operations are called, demonstrating the expected output for each operation.\n",
+    "\n",
+    "In summary, the goal is to create a data structure (Trie) that efficiently stores a set of strings and provides methods to insert new strings, search for complete words, and check if a given prefix exists in the stored set of strings.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class TrieNode:\n",
+    "    def __init__(self):\n",
+    "        self.children = {}           # A dictionary to store child nodes\n",
+    "        self.is_end_of_word = False  # Indicates if a word ends at this node\n",
+    "\n",
+    "class Trie:\n",
+    "    def __init__(self):\n",
+    "        self.root = TrieNode()       # Initialize the Trie with a root node\n",
+    "\n",
+    "    def insert(self, word):\n",
+    "        node = self.root             # Start from the root\n",
+    "        for char in word:\n",
+    "            if char not in node.children:\n",
+    "                node.children[char] = TrieNode()  # Create a new node if the character is not present\n",
+    "            node = node.children[char]           # Move to the child node\n",
+    "        node.is_end_of_word = True                # Mark the end of the inserted word\n",
+    "\n",
+    "    def search(self, word):\n",
+    "        node = self.root             # Start from the root\n",
+    "        for char in word:\n",
+    "            if char not in node.children:\n",
+    "                return False         # If the character is not found, the word is not in the Trie\n",
+    "            node = node.children[char]  # Move to the child node\n",
+    "        return node.is_end_of_word    # Check if the node represents the end of a valid word\n",
+    "\n",
+    "    def startsWith(self, prefix):\n",
+    "        node = self.root             # Start from the root\n",
+    "        for char in prefix:\n",
+    "            if char not in node.children:\n",
+    "                return False         # If the character is not found, no word starts with the prefix\n",
+    "            node = node.children[char]  # Move to the child node\n",
+    "        return True                  # Prefix found in the Trie"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "Let's explain how the code works step by step:\n",
+    "\n",
+    "1. `TrieNode` Class:\n",
+    "   - `TrieNode` is a class that represents nodes in the Trie. Each node has two attributes:\n",
+    "     - `children`: This is a dictionary that stores child nodes. Each key in the dictionary represents a character, and the corresponding value is the child node for that character.\n",
+    "     - `is_end_of_word`: This boolean flag indicates whether the node represents the end of a complete word. Initially, it's set to `False` for all nodes.\n",
+    "\n",
+    "2. `Trie` Class:\n",
+    "   - `Trie` is the main class that implements the Trie data structure. It has the following methods:\n",
+    "\n",
+    "   - `__init__(self)`: The constructor initializes the Trie object by creating a root node, which serves as the starting point for all Trie operations.\n",
+    "\n",
+    "   - `insert(self, word)`: This method inserts a string `word` into the Trie. It starts from the root node and iterates through each character in the word.\n",
+    "     - If a character is not present as a child node, it creates a new node for that character.\n",
+    "     - It then moves to the child node and continues the process until the entire word is inserted.\n",
+    "     - Finally, it sets the `is_end_of_word` flag to `True` for the last node to mark the end of the inserted word.\n",
+    "\n",
+    "   - `search(self, word)`: This method checks if a complete word (string) exists in the Trie. It starts from the root node and iterates through each character in the word.\n",
+    "     - If a character is not found as a child node, it immediately returns `False` because the word is not in the Trie.\n",
+    "     - It continues to move to the child node for each character.\n",
+    "     - After reaching the end of the word, it checks if the `is_end_of_word` flag is `True` for the last node to confirm the presence of the word.\n",
+    "\n",
+    "   - `startsWith(self, prefix)`: This method checks if there is any previously inserted word in the Trie that starts with a given `prefix`. It follows the same logic as the `search` method but does not require the entire word to be present.\n",
+    "     - If the prefix is found in the Trie, it returns `True`; otherwise, it returns `False`.\n",
+    "\n",
+    "3. Example Usage:\n",
+    "   - The code demonstrates how to create a Trie object, insert words (\"apple\" and \"app\"), and perform operations on the Trie:\n",
+    "     - It inserts the word \"apple\" into the Trie.\n",
+    "     - It checks if \"apple\" is in the Trie, which returns `True`.\n",
+    "     - It checks if \"app\" is in the Trie, which returns `False`.\n",
+    "     - It checks if there is any word in the Trie that starts with \"app,\" which returns `True`.\n",
+    "     - It inserts the word \"app\" into the Trie.\n",
+    "     - It checks if \"app\" is in the Trie again, which now returns `True`.\n",
+    "\n",
+    "In summary, the code efficiently implements a Trie data structure with the ability to insert words, search for complete words, and check for the existence of words with a given prefix. The Trie is organized as a tree of nodes, with each node representing a character in the words being stored."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n",
+      "False\n",
+      "True\n",
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Example usage:\n",
+    "\n",
+    "trie = Trie()\n",
+    "trie.insert(\"apple\")\n",
+    "print(trie.search(\"apple\"))   # Output: True\n",
+    "print(trie.search(\"app\"))     # Output: False\n",
+    "print(trie.startsWith(\"app\")) # Output: True\n",
+    "trie.insert(\"app\")\n",
+    "print(trie.search(\"app\"))     # Output: True"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the Trie implementation:\n",
+    "\n",
+    "1. Time Complexity:\n",
+    "   - Insertion (`insert` method): The time complexity of inserting a word into the Trie is O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to insert it into the Trie.\n",
+    "   - Search (`search` method): The time complexity of searching for a word in the Trie is also O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to determine if it exists in the Trie.\n",
+    "   - Starts with (`startsWith` method): The time complexity of checking if a prefix exists in the Trie is O(p), where p is the length of the prefix. In the worst case, you may need to traverse the entire prefix to determine its presence.\n",
+    "\n",
+    "   Therefore, for each operation (insertion, search, or starts with), the time complexity is O(m) or O(p), where m is the length of the word being operated on, and p is the length of the prefix.\n",
+    "\n",
+    "2. Space Complexity:\n",
+    "   - The space complexity of the Trie is determined by the number of nodes and characters stored in it. In the worst case, where none of the inserted words share common prefixes, the space complexity is O(N), where N is the total number of characters across all inserted words.\n",
+    "   - Each node in the Trie represents a character, and the number of nodes is directly related to the total number of characters.\n",
+    "\n",
+    "   In practice, the space complexity can be less than O(N) because common prefixes are shared among words in the Trie, leading to space optimization.\n",
+    "\n",
+    "In summary:\n",
+    "- Time complexity for each operation (insertion, search, starts with) is O(m) or O(p), where m is the length of the word and p is the length of the prefix.\n",
+    "- Space complexity is O(N) in the worst case, where N is the total number of characters across all inserted words. However, space optimization occurs due to common prefix sharing in practice, potentially reducing the actual space used."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Count Prefixes**:\n",
+    "   Implement a method to count the number of words in the Trie that have a specific prefix. This exercise requires you to maintain additional data in the Trie nodes to keep track of the count.\n",
+    "\n",
+    "2. **Auto-Complete Suggestions**:\n",
+    "   Implement an auto-complete feature using the Trie. Given a prefix, the program should return a list of suggested words based on the words previously inserted into the Trie."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/08. Tries/211. Design Add and Search Words Data Structure.ipynb b/_build/jupyter_execute/08. Tries/211. Design Add and Search Words Data Structure.ipynb
new file mode 100644
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--- /dev/null
+++ b/_build/jupyter_execute/08. Tries/211. Design Add and Search Words Data Structure.ipynb	
@@ -0,0 +1,271 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 211. Design Add and Search Words Data Structure\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Design Add and Search Words Data Structure problem on LeetCode, click here!](https://leetcode.com/problems/design-add-and-search-words-data-structure/)\n",
+    "\n",
+    "---\n",
+    "Design a data structure that supports adding new words and finding if a string matches any previously added string.\n",
+    "\n",
+    "Implement the `WordDictionary` class:\n",
+    "\n",
+    "- `WordDictionary()` Initializes the object.\n",
+    "- `void addWord(word)` Adds `word` to the data structure, it can be matched later.\n",
+    "- `bool search(word)` Returns `true` if there is any string in the data structure that matches `word` or `false` otherwise. `word` may contain dots `'.'` where dots can be matched with any letter.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `1 <= word.length <= 25`\n",
+    "- `word` in `addWord` consists of lowercase English letters.\n",
+    "- `word` in `search` consist of `'.'` or lowercase English letters.\n",
+    "- There will be at most `2` dots in `word` for `search` queries.\n",
+    "- At most $10^4$ calls will be made to `addWord` and `search`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem is to design a data structure called `WordDictionary` that allows you to add words and search for words efficiently. The key feature of this data structure is that it should support wildcard search, where a dot ('.') can match any letter.\n",
+    "\n",
+    "Here are the specific requirements and explanations for the problem:\n",
+    "\n",
+    "1. Initialize the `WordDictionary` object using the constructor `WordDictionary()`.\n",
+    "\n",
+    "2. The `addWord(word)` method allows you to add a word to the data structure. Once added, you should be able to search for this word later. The words consist of lowercase English letters, and each word has a length between 1 and 25 characters.\n",
+    "\n",
+    "3. The `search(word)` method allows you to search for a word in the data structure. This method returns `True` if there is any string in the data structure that matches the given word or `False` otherwise. The word you are searching for may contain dots ('.'), where each dot can match any letter.\n",
+    "\n",
+    "4. The wildcard character ('.') in the `search` method allows for partial or fuzzy searching. For example, if the word dictionary contains the words \"bad,\" \"dad,\" and \"mad,\" then:\n",
+    "   - Searching for \"pad\" should return `False` because there is no exact match.\n",
+    "   - Searching for \"bad\" should return `True` because \"bad\" is in the dictionary.\n",
+    "   - Searching for \".ad\" should return `True` because it can match \"bad,\" \"dad,\" or \"mad.\"\n",
+    "   - Searching for \"b..\" should return `True` because it can match \"bad,\" \"bed,\" \"bet,\" etc.\n",
+    "\n",
+    "5. The problem specifies that there will be at most two dots ('.') in search queries, and there will be at most 10^4 calls to both `addWord` and `search`.\n",
+    "\n",
+    "In summary, you need to implement the `WordDictionary` class that efficiently supports adding words and searching for words, where the search can include wildcard characters ('.') that match any letter.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class TrieNode:\n",
+    "    def __init__(self):\n",
+    "        # Initialize a TrieNode with children and a flag to indicate the end of a word.\n",
+    "        self.children = {}\n",
+    "        self.is_end_of_word = False\n",
+    "\n",
+    "class WordDictionary:\n",
+    "\n",
+    "    def __init__(self):\n",
+    "        # Initialize the WordDictionary with a root node.\n",
+    "        self.root = TrieNode()\n",
+    "\n",
+    "    def addWord(self, word: str) -> None:\n",
+    "        node = self.root\n",
+    "        # Iterate through the characters in the word.\n",
+    "        for char in word:\n",
+    "            if char not in node.children:\n",
+    "                # Create a new node for the character if it doesn't exist.\n",
+    "                node.children[char] = TrieNode()\n",
+    "            # Move to the next node.\n",
+    "            node = node.children[char]\n",
+    "        # Mark the end of the word by setting the flag to True.\n",
+    "        node.is_end_of_word = True\n",
+    "\n",
+    "    def search_helper(self, node: TrieNode, word: str) -> bool:\n",
+    "        if not word:\n",
+    "            # If there are no characters left in the word, check if we reached the end of a word in the trie.\n",
+    "            return node.is_end_of_word\n",
+    "        \n",
+    "        char = word[0]\n",
+    "        if char == '.':\n",
+    "            # If the character is a dot, explore all possible child nodes.\n",
+    "            for child in node.children.values():\n",
+    "                if self.search_helper(child, word[1:]):\n",
+    "                    return True\n",
+    "        elif char in node.children:\n",
+    "            # If the character is in the children of the current node, move to the next node.\n",
+    "            return self.search_helper(node.children[char], word[1:])\n",
+    "        \n",
+    "        # If the character is not found and is not a dot, the word is not in the trie.\n",
+    "        return False\n",
+    "\n",
+    "    def search(self, word: str) -> bool:\n",
+    "        # Start the search from the root of the trie.\n",
+    "        return self.search_helper(self.root, word)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "This code defines a `WordDictionary` class that uses a Trie data structure to efficiently store and search for words. Here's a step-by-step explanation:\n",
+    "\n",
+    "1. `TrieNode` class: This class represents nodes in the Trie data structure. Each node has two attributes: `children` (a dictionary to store child nodes) and `is_end_of_word` (a flag to indicate whether the node marks the end of a word).\n",
+    "\n",
+    "2. `WordDictionary` class: This class initializes with an empty Trie by creating a root node.\n",
+    "\n",
+    "3. `addWord` method: This method adds a word to the Trie. It iterates through each character in the word and creates Trie nodes as necessary. It sets the `is_end_of_word` flag to `True` for the final character of the word to mark the end of the word.\n",
+    "\n",
+    "4. `search_helper` method: This is a recursive helper function for searching words in the Trie. It takes a Trie node (`node`) and a word to search (`word`) as input. If there are no characters left in the word (`not word`), it checks if the current node marks the end of a word. If it does, it returns `True`.\n",
+    "\n",
+    "5. If the first character of the word is a dot ('.'), the function explores all possible child nodes by iterating through `node.children.values()` and recursively calls `search_helper` for each child with the remaining part of the word (`word[1:]`).\n",
+    "\n",
+    "6. If the first character of the word is not a dot and is found in the children of the current node, the function recursively moves to the next node by calling `search_helper` on the child node and the remaining part of the word.\n",
+    "\n",
+    "7. If the character is not found in the children and is not a dot, the word is not in the Trie, so the function returns `False`.\n",
+    "\n",
+    "8. `search` method: This is the public method for searching words. It initiates the search process by calling `search_helper` starting from the root node of the Trie.\n",
+    "\n",
+    "Overall, this code efficiently implements a Trie-based data structure that allows adding and searching for words, including support for wildcard characters represented by dots ('.')."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n",
+      "True\n",
+      "True\n",
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Example usage:\n",
+    "\n",
+    "wordDictionary = WordDictionary()\n",
+    "wordDictionary.addWord(\"bad\")\n",
+    "wordDictionary.addWord(\"dad\")\n",
+    "wordDictionary.addWord(\"mad\")\n",
+    "print(wordDictionary.search(\"pad\"))  # Output: False\n",
+    "print(wordDictionary.search(\"bad\"))  # Output: True\n",
+    "print(wordDictionary.search(\".ad\"))  # Output: True\n",
+    "print(wordDictionary.search(\"b..\"))  # Output: True"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `WordDictionary` class methods:\n",
+    "\n",
+    "1. `addWord` Method:\n",
+    "\n",
+    "   - Time Complexity: O(L)\n",
+    "     - L is the length of the word being added.\n",
+    "     - In the worst case, we need to iterate through each character in the word and create nodes in the Trie. This takes O(L) time.\n",
+    "\n",
+    "   - Space Complexity: O(L)\n",
+    "     - The space complexity for storing the word in the Trie is also O(L) because we create nodes for each character in the word.\n",
+    "\n",
+    "2. `search_helper` Method (called by `search`):\n",
+    "\n",
+    "   - Time Complexity: O(2^M * L)\n",
+    "     - M is the maximum number of dots ('.') in the search word.\n",
+    "     - In the worst case, when there are M dots in the search word, we may explore all possible child nodes at each dot. This results in a branching factor of 26 (for lowercase letters) at each dot.\n",
+    "     - So, the time complexity is exponential in the number of dots and linear in the length of the word being searched (L).\n",
+    "\n",
+    "   - Space Complexity: O(L)\n",
+    "     - The space complexity for the recursive call stack is O(L) in the worst case because we may have to recurse to a depth equal to the length of the search word.\n",
+    "\n",
+    "3. `search` Method:\n",
+    "\n",
+    "   - Time Complexity: O(2^M * L)\n",
+    "     - The `search` method calls the `search_helper` method, which has the same time complexity explained above.\n",
+    "\n",
+    "   - Space Complexity: O(L)\n",
+    "     - The space complexity for the recursive call stack is O(L) in the worst case, as explained earlier.\n",
+    "\n",
+    "Overall, the time complexity of the `search` method is mainly affected by the number of dots ('.') in the search word. In the worst case, when there are multiple dots and many possibilities to explore at each dot, the time complexity can be exponential. However, for most practical cases, it performs reasonably well.\n",
+    "\n",
+    "The space complexity primarily depends on the space required to store the words in the Trie. It is proportional to the total number of characters in all added words and is linear with respect to the length of the words added.\n",
+    "\n",
+    "In summary:\n",
+    "\n",
+    "- Time Complexity for `addWord`: O(L)\n",
+    "- Time Complexity for `search`: O(2^M * L)\n",
+    "- Space Complexity: O(L) for storing words in the Trie, O(L) for the recursive call stack during search.\n",
+    "\n",
+    "Note: The space complexity analysis assumes that the dictionary contains a reasonable number of words and does not account for the overhead of Python objects or system-level memory allocation."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Dictionary Auto-Completion**: Extend the `WordDictionary` class to provide auto-completion suggestions based on the prefix of a word. Implement a method that returns a list of words that match the given prefix.\n",
+    "\n",
+    "2. **Support Word Deletion**: Extend the `WordDictionary` class to support word deletion. Add a method `deleteWord(word)` that removes a word from the data structure. Ensure that the search operation still works correctly after deletion."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb b/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb
new file mode 100644
index 0000000..e582b9d
--- /dev/null
+++ b/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb	
@@ -0,0 +1,310 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 212. Word Search II\n",
+    "\n",
+    "**Difficulty:** Hard\n",
+    "\n",
+    "**Link to Problem:** [To see the Word Search II problem on LeetCode, click here!](https://leetcode.com/problems/word-search-ii/)\n",
+    "\n",
+    "---\n",
+    "Given an `m x n` `board` of characters and a list of strings `words`, the task is to return *all words on the board*.\n",
+    "\n",
+    "Each word must be constructed from letters of sequentially **adjacent cells**, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.\n",
+    "\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `m == board.length`\n",
+    "- `n == board[i].length`\n",
+    "- `1 <= m, n <= 12`\n",
+    "- `board[i][j]` is a lowercase English letter.\n",
+    "- 1 <= `words.length` <= $3 * 10^4$\n",
+    "- `1 <= words[i].length <= 10`\n",
+    "- `words[i]` consists of lowercase English letters.\n",
+    "- All the strings of `words` are unique."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "To solve the \"Word Search II\" problem, you can use a Trie data structure and perform a depth-first search (DFS) on the board. Here's a step-by-step approach to solve the problem:\n",
+    "\n",
+    "1. Build a Trie from the given list of words:\n",
+    "   - Create a Trie data structure to store all the words from the input list.\n",
+    "   - For each word in the input list, insert it into the Trie, character by character. Make sure to mark the end of each word in the Trie.\n",
+    "\n",
+    "2. Perform DFS on the board:\n",
+    "   - Iterate through each cell (i, j) on the board.\n",
+    "   - For each cell, start a DFS traversal from that cell to search for words.\n",
+    "   - During the DFS traversal, maintain a current path string.\n",
+    "   - At each cell, check if the current path string, concatenated with the character in the cell, matches any word prefix in the Trie. If it does, continue the DFS.\n",
+    "   - If the current path string matches a word in the Trie, add it to the result.\n",
+    "\n",
+    "3. Implement DFS with backtracking:\n",
+    "   - During the DFS traversal, you will explore neighboring cells (horizontally and vertically) if they are within bounds and haven't been visited before.\n",
+    "   - To prevent revisiting the same cell within the same word, mark the cell as visited (e.g., change its character to a special character like '#') before exploring it and restore its original character after the DFS traversal.\n",
+    "\n",
+    "4. Return the result:\n",
+    "   - After completing the DFS traversal for all cells on the board, you will have collected all the valid words in the result set.\n",
+    "   - Convert the result set to a list and return it as the final output.\n",
+    "\n",
+    "This approach efficiently finds all words on the board by utilizing the Trie data structure to prune unnecessary DFS paths and avoiding duplicate visits to the same cell within the same word. It ensures that each word is constructed from letters of sequentially adjacent cells without using the same letter cell more than once in a word.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Define a TrieNode class to represent nodes in the Trie.\n",
+    "class TrieNode:\n",
+    "    def __init__(self):\n",
+    "        self.children = {}  # Dictionary to store child nodes.\n",
+    "        self.is_end_of_word = False  # Flag to mark the end of a word.\n",
+    "\n",
+    "# Main function to find words on the board.\n",
+    "def findWords(board, words):\n",
+    "    # DFS function to search for words starting from a given cell.\n",
+    "    def dfs(node, i, j, path):\n",
+    "        char = board[i][j]  # Get the character at the current cell.\n",
+    "        curr_node = node.children.get(char)  # Find the corresponding Trie node.\n",
+    "\n",
+    "        if not curr_node:\n",
+    "            return  # If no matching Trie node, stop the search.\n",
+    "\n",
+    "        path += char  # Add the character to the current path.\n",
+    "\n",
+    "        if curr_node.is_end_of_word:\n",
+    "            result.add(path)  # If the path matches a word, add it to the result.\n",
+    "\n",
+    "        temp, board[i][j] = board[i][j], \"#\"  # Mark the cell as visited.\n",
+    "\n",
+    "        # Explore neighboring cells (up, down, left, right).\n",
+    "        if i > 0:\n",
+    "            dfs(curr_node, i - 1, j, path)\n",
+    "        if i < m - 1:\n",
+    "            dfs(curr_node, i + 1, j, path)\n",
+    "        if j > 0:\n",
+    "            dfs(curr_node, i, j - 1, path)\n",
+    "        if j < n - 1:\n",
+    "            dfs(curr_node, i, j + 1, path)\n",
+    "\n",
+    "        board[i][j] = temp  # Restore the cell's original character.\n",
+    "\n",
+    "    # Function to build a Trie from the list of words.\n",
+    "    def buildTrie():\n",
+    "        root = TrieNode()  # Create a root node.\n",
+    "        for word in words:\n",
+    "            node = root\n",
+    "            for char in word:\n",
+    "                if char not in node.children:\n",
+    "                    node.children[char] = TrieNode()  # Create a new node if needed.\n",
+    "                node = node.children[char]\n",
+    "            node.is_end_of_word = True  # Mark the end of the word.\n",
+    "        return root\n",
+    "\n",
+    "    m, n = len(board), len(board[0])  # Get the dimensions of the board.\n",
+    "    root = buildTrie()  # Build the Trie from the list of words.\n",
+    "    result = set()  # Use a set to store unique words found on the board.\n",
+    "\n",
+    "    # Iterate through each cell on the board and start DFS from there.\n",
+    "    for i in range(m):\n",
+    "        for j in range(n):\n",
+    "            dfs(root, i, j, \"\")  # Start DFS from each cell on the board.\n",
+    "\n",
+    "    return list(result)  # Convert the set to a list and return the result."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The code solves the \"Word Search II\" problem, where you have a grid of characters and a list of words. The goal is to find all words from the list that can be constructed by traversing adjacent cells (horizontally or vertically) in the grid, without using the same cell more than once for a single word.\n",
+    "\n",
+    "The code does the following:\n",
+    "\n",
+    "1. Defines a `TrieNode` class to represent nodes in a Trie data structure. Each node has children (other characters in the word) and a flag to mark the end of a word.\n",
+    "\n",
+    "2. Defines the `findWords` function, which takes the grid (`board`) and a list of words (`words`) as input.\n",
+    "\n",
+    "3. Inside the `findWords` function, there is a `dfs` (depth-first search) function. This function is used to explore the grid starting from a specific cell. It checks if the current path matches any word in the Trie and adds it to the result if found.\n",
+    "\n",
+    "4. The code also includes a `buildTrie` function that constructs a Trie data structure from the list of words. It iterates through each word, creating nodes for each character in the Trie and marking the end of words.\n",
+    "\n",
+    "5. The dimensions of the grid are obtained (m x n).\n",
+    "\n",
+    "6. A Trie is built using the `buildTrie` function.\n",
+    "\n",
+    "7. A set called `result` is used to store unique words found in the grid.\n",
+    "\n",
+    "8. The code iterates through each cell on the grid (using nested loops) and starts a DFS search from each cell.\n",
+    "\n",
+    "9. During the DFS, it explores neighboring cells (up, down, left, right) if they haven't been visited before and if the current path matches a prefix in the Trie.\n",
+    "\n",
+    "10. When a word is found during the DFS traversal, it is added to the `result` set.\n",
+    "\n",
+    "11. The grid cell is marked as visited during the DFS by changing its character to \"#\", and it is restored to its original character after the DFS traversal.\n",
+    "\n",
+    "12. Finally, the set of unique words found is converted to a list and returned as the output.\n",
+    "\n",
+    "The code effectively uses a Trie data structure to optimize the search and ensures that each word is constructed from adjacent cells without reusing the same cell for the same word."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['eat', 'oath']\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "board1 = [[\"o\",\"a\",\"a\",\"n\"],[\"e\",\"t\",\"a\",\"e\"],[\"i\",\"h\",\"k\",\"r\"],[\"i\",\"f\",\"l\",\"v\"]]\n",
+    "words1 = [\"oath\",\"pea\",\"eat\",\"rain\"]\n",
+    "print(findWords(board1, words1))  # Output: [\"eat\", \"oath\"]"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "c748b879-a2d8-4298-a42c-c7f61f9c7fc6",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "\n",
+    "board2 = [[\"a\",\"b\"],[\"c\",\"d\"]]\n",
+    "words2 = [\"abcb\"]\n",
+    "print(findWords(board2, words2))  # Output: []"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "1. **Building the Trie (`buildTrie` function):** \n",
+    "   - Suppose there are `N` words in the input list, and the average length of a word is `L`.\n",
+    "   - Building the Trie takes O(N * L) time, as we iterate through each word and insert each character into the Trie.\n",
+    "\n",
+    "2. **DFS Traversal (`dfs` function):** \n",
+    "   - In the worst case, the DFS function explores all possible paths in the grid.\n",
+    "   - The maximum number of recursive calls for each cell is 4 (up, down, left, right).\n",
+    "   - The maximum depth of the DFS is limited by the length of the longest word in the Trie, which is at most `L`.\n",
+    "   - Therefore, the DFS traversal for each cell takes $O(4^L)$ time.\n",
+    "\n",
+    "3. **Iterating Through the Grid (`findWords` function):**\n",
+    "   - In the worst case, we iterate through all `m` rows and `n` columns in the grid.\n",
+    "   - For each cell, we start a DFS traversal.\n",
+    "   - Therefore, iterating through the entire grid takes $O(m * n * 4^L)$ time.\n",
+    "\n",
+    "Overall, the time complexity of the code is $O(N * L + m * n * 4^L)$.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "1. **Trie Data Structure:**\n",
+    "   - The space required to store the Trie depends on the number of unique characters in all the words.\n",
+    "   - In the worst case, where all words are unique and have no common prefixes, the space complexity of the Trie is O(N * L).\n",
+    "   - In practice, it may be less if there are common prefixes among words.\n",
+    "\n",
+    "2. **DFS Stack (Recursive Calls):**\n",
+    "   - The depth of the recursive call stack during DFS is at most `L`, where `L` is the length of the longest word in the Trie.\n",
+    "   - Therefore, the space complexity for the recursive call stack is O(L).\n",
+    "\n",
+    "3. **Result Set (`result` set):**\n",
+    "   - The space used to store the result set depends on the number of valid words found in the grid.\n",
+    "   - In the worst case, when all words are found, the space complexity of the result set is O(N * L).\n",
+    "\n",
+    "Overall, the space complexity of the code is O(N * L) for the Trie and O(L) for the recursive call stack.\n",
+    "\n",
+    "In summary, the time complexity is dominated by the DFS traversal and is influenced by the number of words, their lengths, and the size of the grid. The space complexity is mainly determined by the Trie structure and the recursive call stack depth during DFS."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Word Search Paths:**\n",
+    "   Extend the basic word search exercise to find all possible paths (sequences of cells) that spell out a given word on the board. Return a list of paths if the word can be formed, otherwise an empty list.\n",
+    "\n",
+    "\n",
+    "2. **Reverse Word Search:**\n",
+    "   Modify the code to search for words in reverse order on the board. Given a 2D grid of characters and a list of words, find all words from the list that can be formed by traversing the board in reverse (right to left, bottom to top, etc.)."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb b/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb
new file mode 100644
index 0000000..f82ad98
--- /dev/null
+++ b/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb	
@@ -0,0 +1,248 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 1046. Last Stone Weight\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Last Stone Weight problem on LeetCode, click here!](https://leetcode.com/problems/last-stone-weight/)\n",
+    "\n",
+    "---\n",
+    "You are given an array of integers `stones` where `stones[i]` is the weight of the `i`th stone.\n",
+    "\n",
+    "We are playing a game with the stones. On each turn, you choose the **heaviest two stones** and smash them together. Suppose the heaviest two stones have weights `x` and `y` with `x <= y`. The result of this smash is:\n",
+    "\n",
+    "1. If `x == y`, both stones are destroyed.\n",
+    "2. If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - x`.\n",
+    "\n",
+    "At the end of the game, there is **at most one** stone left.\n",
+    "\n",
+    "Return *the weight of the last remaining stone*. If there are no stones left, return `0`.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `1 <= stones.length <= 30`\n",
+    "- `1 <= stones[i] <= 1000`"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The \"Last Stone Weight\" problem involves a game played with a collection of stones, each stone having a certain weight. The goal is to find the weight of the last remaining stone after applying a specific set of rules.\n",
+    "\n",
+    "Here's how the game works:\n",
+    "\n",
+    "1. You start with an array of stones, where each stone is represented by its weight. The weights of the stones are given in the input as an array.\n",
+    "\n",
+    "2. In each turn of the game, you select the two heaviest stones from the remaining stones. If there is only one stone left, you've found the answer.\n",
+    "\n",
+    "3. If there are two stones with weights `x` and `y`, where `x` is less than or equal to `y`, the following happens:\n",
+    "   - If `x` is equal to `y`, both stones are completely destroyed, and they are removed from the array of stones.\n",
+    "   - If `x` is not equal to `y`, the stone with weight `x` is destroyed, and the stone with weight `y` now has a new weight of `y - x`. The stone with weight `x` is removed from the array, and the modified stone with weight `y - x` remains in the array.\n",
+    "\n",
+    "4. The game continues with the modified array of stones until there is at most one stone left.\n",
+    "\n",
+    "5. The goal is to find the weight of the last remaining stone, or if there are no stones left, return `0`.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import heapq\n",
+    "\n",
+    "def lastStoneWeight(stones):\n",
+    "    # Convert the input list to a max heap (negate the values to simulate a max heap)\n",
+    "    max_heap = [-stone for stone in stones]\n",
+    "    heapq.heapify(max_heap)\n",
+    "\n",
+    "    # Continue as long as there are more than 1 stone in the heap\n",
+    "    while len(max_heap) > 1:\n",
+    "        # Get the two heaviest stones\n",
+    "        y = -heapq.heappop(max_heap)  # Get the heaviest stone and negate it back to positive\n",
+    "        x = -heapq.heappop(max_heap)  # Get the second heaviest stone and negate it back to positive\n",
+    "\n",
+    "        # Calculate the new stone's weight after smashing\n",
+    "        if x != y:\n",
+    "            new_stone = y - x\n",
+    "            heapq.heappush(max_heap, -new_stone)  # Negate the value and push it back to the heap\n",
+    "\n",
+    "    # If there is one stone left, return its weight (negate it back to positive)\n",
+    "    if max_heap:\n",
+    "        return -max_heap[0]\n",
+    "    else:\n",
+    "        return 0"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The code defines a Python function called `lastStoneWeight` that takes a list of stone weights as input. It aims to find the weight of the last remaining stone after playing the stone smashing game described in the problem statement.\n",
+    "\n",
+    "Here's a step-by-step explanation of the code:\n",
+    "\n",
+    "1. We import the `heapq` library, which allows us to create and manipulate a heap (priority queue).\n",
+    "\n",
+    "2. Inside the `lastStoneWeight` function:\n",
+    "   - We create a max heap (negating stone weights to simulate a max heap) using the `heapq.heapify` function. This heap will keep track of the heaviest stones.\n",
+    "   \n",
+    "   - We enter a loop that continues as long as there are more than 1 stone in the heap.\n",
+    "   \n",
+    "   - Inside the loop, we:\n",
+    "     - Retrieve and negate the heaviest stone (`y`) from the heap.\n",
+    "     - Retrieve and negate the second heaviest stone (`x`) from the heap.\n",
+    "   \n",
+    "     - Calculate the new stone's weight after smashing (`y - x`), and negate it back to simulate a max heap.\n",
+    "     - Add the new stone back to the heap using `heapq.heappush`.\n",
+    "   \n",
+    "3. After the loop, if there is one stone left in the heap, we retrieve and negate it to get its actual weight, and return it as the result.\n",
+    "   \n",
+    "4. If there are no stones left (the heap is empty), we return `0`, indicating that no stones remain.\n",
+    "\n",
+    "5. Finally, we provide two test cases to demonstrate the function's usage and correctness.\n",
+    "\n",
+    "In summary, the code efficiently implements the stone smashing game by maintaining a max heap, selecting the heaviest stones in each turn, smashing them, and updating the heap until there is at most one stone left. It then returns the weight of the last remaining stone or 0 if there are no stones left."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "stones1 = [2, 7, 4, 1, 8, 1]\n",
+    "print(lastStoneWeight(stones1))  # Output: 1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "b7e79514-4821-48d0-bb9d-0f2de8150945",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2: \n",
+    "\n",
+    "stones2 = [1]\n",
+    "print(lastStoneWeight(stones2))  # Output: 1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided Python code for the \"Last Stone Weight\" problem:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "1. The `heapq.heapify` operation to create the max heap from the input stone weights has a time complexity of O(n), where n is the number of stones.\n",
+    "\n",
+    "2. The loop that continues until there are more than 1 stone in the heap performs the following operations in each iteration:\n",
+    "   - Retrieving and negating the heaviest stone (`y`) from the heap, which takes O(log n) time.\n",
+    "   - Retrieving and negating the second heaviest stone (`x`) from the heap, also taking O(log n) time.\n",
+    "   - Calculating the new stone's weight and pushing it back into the heap using `heapq.heappush`, which takes O(log n) time.\n",
+    "\n",
+    "3. In the worst case, the loop runs until there is only one stone left, which requires approximately (n-1) iterations.\n",
+    "\n",
+    "Overall, the dominant time complexity is O(n) for the initial heap creation, and within the loop, each iteration takes O(log n) time. Therefore, the overall time complexity of the algorithm is O(n log n).\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "1. The space complexity is primarily determined by the space required for the max heap. In the worst case, this heap can contain all the stones, so the space complexity for the heap is O(n).\n",
+    "\n",
+    "2. The rest of the variables used in the function (e.g., `x`, `y`, `new_stone`) require only constant space, so they do not significantly contribute to the space complexity.\n",
+    "\n",
+    "Therefore, the overall space complexity of the algorithm is O(n) due to the space used for the max heap.\n",
+    "\n",
+    "In summary:\n",
+    "- Time Complexity: $O(n\\ log\\ n)$\n",
+    "- Space Complexity: $O(n)$\n",
+    "\n",
+    "The algorithm is efficient enough to handle the problem's constraints, as the worst-case time and space complexities are both linearithmic in terms of the number of stones."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Kth Last Stone Weight**: Modify the problem to find the weight of the Kth last remaining stone, where K is an integer input. Your function should work efficiently for large values of K.\n",
+    "\n",
+    "\n",
+    "2. **Multiple Stones Smash**: Modify the problem so that instead of smashing two stones at a time, you can smash up to K stones at each turn, where K is an integer input. Determine the weight of the last remaining stone in this variation."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb b/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb
new file mode 100644
index 0000000..6c43a8d
--- /dev/null
+++ b/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb	
@@ -0,0 +1,238 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 295. Find Median from Data Stream\n",
+    "\n",
+    "**Difficulty:** Hard\n",
+    "\n",
+    "**Link to Problem:** [To see the Find Median from Data Stream problem on LeetCode, click here!](https://leetcode.com/problems/find-median-from-data-stream/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.\n",
+    "\n",
+    "Implement the MedianFinder class:\n",
+    "\n",
+    "- `MedianFinder()` initializes the `MedianFinder` object.\n",
+    "- `void addNum(int num)` adds the integer `num` from the data stream to the data structure.\n",
+    "- `double findMedian()` returns the median of all elements so far. Answers within $10^{-5}$ of the actual answer will be accepted.\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- $-10^{5}$ <= `num` <= $10^{5}$\n",
+    "- There will be at least one element in the data structure before calling `findMedian`.\n",
+    "- At most $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
+    "\n",
+    "**Follow-up:**\n",
+    "\n",
+    "1. If all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?\n",
+    "2. If `99%` of all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem, \"Find Median from Data Stream,\" deals with efficiently computing the median of a sequence of numbers as they are incrementally provided. The median is the middle value in an ordered list of numbers. When the list has an even number of elements, the median is the average of the two middle values.\n",
+    "\n",
+    "Here's a breakdown of the problem:\n",
+    "\n",
+    "1. **Initialization**: You are asked to implement a `MedianFinder` class that supports the following operations:\n",
+    "\n",
+    "    - `MedianFinder()`: Initializes the MedianFinder object.\n",
+    "    - `addNum(int num)`: Adds an integer `num` from the data stream to the data structure.\n",
+    "    - `findMedian()`: Returns the median of all elements added to the data structure.\n",
+    "\n",
+    "2. **Median Calculation**: The `findMedian` operation should efficiently compute the median, and the result should have a precision of at least $10^{-5}$ (i.e., answers within this range will be considered correct).\n",
+    "\n",
+    "3. **Examples**:\n",
+    "   \n",
+    "   - If you add numbers [1, 2], the median should be 1.5 because (1 + 2) / 2 = 1.5.\n",
+    "   - If you add numbers [1, 2, 3], the median should be 2 because it's the middle value.\n",
+    "   \n",
+    "4. **Constraints**:\n",
+    "\n",
+    "   - The integers provided in the data stream are in the range from $-10^5$ to $10^5$.\n",
+    "   - There will be at least one element in the data structure before calling `findMedian`.\n",
+    "   - At most, $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
+    "\n",
+    "The problem can be solved by using two priority queues (heaps): one max-heap to store the smaller half of the numbers and one min-heap to store the larger half of the numbers. The max-heap ensures that the largest number in the smaller half is at the top, and the min-heap ensures that the smallest number in the larger half is at the top. These heaps are balanced to efficiently find the median when requested.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import heapq\n",
+    "\n",
+    "class MedianFinder:\n",
+    "\n",
+    "    def __init__(self):\n",
+    "        # Initialize two heaps: a max-heap (small_half) for the smaller half of numbers,\n",
+    "        # and a min-heap (large_half) for the larger half of numbers.\n",
+    "        self.small_half = []  # Max-heap for the smaller half of the numbers\n",
+    "        self.large_half = []  # Min-heap for the larger half of the numbers\n",
+    "\n",
+    "    def addNum(self, num: int) -> None:\n",
+    "        # Add the number to the appropriate heap based on its value\n",
+    "        if not self.small_half or num <= -self.small_half[0]:\n",
+    "            # If the number is less than or equal to the current maximum in the smaller half,\n",
+    "            # add it to the smaller half (max-heap)\n",
+    "            heapq.heappush(self.small_half, -num)\n",
+    "        else:\n",
+    "            # Otherwise, add it to the larger half (min-heap)\n",
+    "            heapq.heappush(self.large_half, num)\n",
+    "\n",
+    "        # Balance the heaps if necessary to ensure the size difference is at most 1\n",
+    "        if len(self.small_half) > len(self.large_half) + 1:\n",
+    "            # If the size of the smaller half is more than one greater than the size of the larger half,\n",
+    "            # move the maximum from the smaller half to the larger half to balance them.\n",
+    "            heapq.heappush(self.large_half, -heapq.heappop(self.small_half))\n",
+    "        elif len(self.large_half) > len(self.small_half):\n",
+    "            # If the size of the larger half is greater than the size of the smaller half,\n",
+    "            # move the minimum from the larger half to the smaller half to balance them.\n",
+    "            heapq.heappush(self.small_half, -heapq.heappop(self.large_half))\n",
+    "\n",
+    "    def findMedian(self) -> float:\n",
+    "        if len(self.small_half) == len(self.large_half):\n",
+    "            # If both halves have the same size, there's an even number of elements,\n",
+    "            # so the median is the average of the top elements in both heaps.\n",
+    "            return (-self.small_half[0] + self.large_half[0]) / 2.0\n",
+    "        else:\n",
+    "            # If the smaller half has more elements, it contains the median (odd number of elements).\n",
+    "            return -self.small_half[0]"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The code defines a `MedianFinder` class for efficiently finding the median of a stream of numbers. It uses two heaps: a max-heap (`small_half`) to store the smaller half of the numbers and a min-heap (`large_half`) to store the larger half. Here's how the code works:\n",
+    "\n",
+    "1. The `__init__` method initializes the two heaps.\n",
+    "\n",
+    "2. The `addNum` method adds a number to the appropriate heap, ensuring that the heaps remain balanced. It does this by comparing the number to the current maximum in the smaller half and adding it to the appropriate heap. Then, it checks the sizes of the two heaps and rebalances them if necessary.\n",
+    "\n",
+    "3. The `findMedian` method calculates and returns the median. If both halves have the same size (even number of elements), it computes the average of the tops of both heaps. If the smaller half has more elements (odd number of elements), it returns the top of the max-heap, which is the median.\n",
+    "\n",
+    "In summary, this code efficiently maintains two heaps to divide the elements into smaller and larger halves, allowing for quick median retrieval. It offers a time complexity of O(log N) for adding elements and O(1) for finding the median, where N is the total number of elements processed. The space complexity is O(1) as the space used by the two heaps is independent of the input size."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1.5\n",
+      "2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Example usage:\n",
+    "medianFinder = MedianFinder()\n",
+    "medianFinder.addNum(1)\n",
+    "medianFinder.addNum(2)\n",
+    "print(medianFinder.findMedian())  # Output: 1.5\n",
+    "medianFinder.addNum(3)\n",
+    "print(medianFinder.findMedian())  # Output: 2.0\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `MedianFinder` class:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "1. `MedianFinder()` Initialization: This operation has a time complexity of O(1). It simply initializes the two heaps.\n",
+    "\n",
+    "2. `addNum(int num)`: The time complexity for adding a number is O(log N), where N is the total number of elements processed so far. This is because, in the worst case, you might need to perform logarithmic operations on the heaps to maintain their balance.\n",
+    "\n",
+    "3. `findMedian()`: Finding the median has a time complexity of O(1). It simply involves accessing the tops of the two heaps, which takes constant time.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "1. `MedianFinder()` Initialization: The space complexity for initialization is O(1). It involves creating two empty heaps, which do not depend on the size of the data stream.\n",
+    "\n",
+    "2. `addNum(int num)`: The space complexity for adding a number is also O(1). It involves adding the number to one of the two heaps, which do not significantly affect the overall space complexity.\n",
+    "\n",
+    "3. `findMedian()`: The space complexity for finding the median is O(1). It doesn't require any additional data structures that grow with the input size.\n",
+    "\n",
+    "In summary, the time complexity of the `addNum` operation is O(log N), and the time complexity of the `findMedian` operation is O(1). The space complexity is O(1) as well, as the space used by the two heaps is independent of the size of the data stream. This solution efficiently maintains and retrieves the median, making it suitable for the given problem constraints."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Generalized kth Element**: Extend the `MedianFinder` class to support finding the kth element in the data stream efficiently. This means finding the kth smallest or kth largest element in the stream.\n",
+    "\n",
+    "2. **Sliding Window Median**: Given an array and a sliding window size, find the median within the window as it slides over the array. This is an extension of the problem to a moving window scenario."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb b/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb
new file mode 100644
index 0000000..e6ab9e3
--- /dev/null
+++ b/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb	
@@ -0,0 +1,235 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 703. Kth Largest Element in a Stream\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Kth Largest Element in a Stream problem on LeetCode, click here!](https://leetcode.com/problems/kth-largest-element-in-a-stream/)\n",
+    "\n",
+    "---\n",
+    "Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.\n",
+    "\n",
+    "Implement the `KthLargest` class with the following methods:\n",
+    "\n",
+    "- `KthLargest(int k, int[] nums)`: Initializes the object with the integer `k` and the stream of integers `nums`.\n",
+    "- `int add(int val)`: Appends the integer `val` to the stream and returns the element representing the `kth` largest element in the stream.\n",
+    "\n",
+    "\n",
+    "**Constraints:**\n",
+    "\n",
+    "- 1 <= `k` <= $10^4$\n",
+    "- 0 <= `nums.length` <= $10^4$\n",
+    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
+    "- $-10^4$ <= `val` <= $10^4$\n",
+    "- At most $10^4$ calls will be made to `add`.\n",
+    "- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem you're trying to solve is to design a class called `KthLargest` that can efficiently find the kth largest element in a stream of integers. You need to support two main operations:\n",
+    "\n",
+    "To solve this problem efficiently, you can use a min-heap data structure. Here's how the approach works:\n",
+    "\n",
+    "1. Initialize a min-heap to store the k largest elements. Initially, it's empty.\n",
+    "\n",
+    "2. In the `__init__` method, populate the min-heap with the first `k` elements from `nums`. This ensures that you have the k largest elements seen so far.\n",
+    "\n",
+    "3. Whenever you add a new element to the stream using the `add` method, follow these steps:\n",
+    "   - Add the new element to the min-heap.\n",
+    "   - If the size of the min-heap exceeds `k`, remove the smallest element from the min-heap. This ensures that you always have the k largest elements in the min-heap.\n",
+    "   - The smallest element in the min-heap (the root) will always represent the kth largest element in the stream.\n",
+    "\n",
+    "Here's why this approach works efficiently:\n",
+    "\n",
+    "- By using a min-heap, you can quickly maintain the k largest elements, and finding the smallest element in the heap (the root) takes constant time.\n",
+    "\n",
+    "- When you add a new element, the min-heap's size is kept at most `k`, which ensures that you only track the k largest elements and discard the smaller ones.\n",
+    "\n",
+    "- The time complexity for adding an element is O(log k), which is very efficient compared to sorting the entire stream, which would be O(n log n).\n",
+    "\n",
+    "- This approach meets the constraints of the problem, including handling large streams and having a low time complexity for both initialization and adding elements.\n",
+    "\n",
+    "In summary, the min-heap approach efficiently tracks the kth largest element in the stream by maintaining a heap of the k largest elements seen so far, updating it as new elements are added. This approach provides a fast and scalable solution to the problem.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import heapq\n",
+    "from typing import List  # Import the List type from the typing module\n",
+    "\n",
+    "class KthLargest:\n",
+    "\n",
+    "    def __init__(self, k: int, nums: List[int]):\n",
+    "        # Initialize the KthLargest object with k and nums.\n",
+    "        self.k = k\n",
+    "        # Create a min-heap to store the k largest elements.\n",
+    "        self.min_heap = []\n",
+    "        # Populate the min-heap with the first k elements from nums.\n",
+    "        for num in nums:\n",
+    "            self.add(num)\n",
+    "\n",
+    "    def add(self, val: int) -> int:\n",
+    "        # Add val to the min-heap.\n",
+    "        heapq.heappush(self.min_heap, val)\n",
+    "        # If the size of the min-heap exceeds k, remove the smallest element.\n",
+    "        if len(self.min_heap) > self.k:\n",
+    "            heapq.heappop(self.min_heap)\n",
+    "        # The root of the min-heap is the kth largest element.\n",
+    "        return self.min_heap[0]"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "- We start by importing the necessary modules:\n",
+    "  - `heapq`: This module provides functions to create and manipulate heaps.\n",
+    "  - `List` from the `typing` module: This is used to specify the type of the `nums` parameter.\n",
+    "\n",
+    "- We define the `KthLargest` class:\n",
+    "  - The `__init__` method initializes the object with the integer `k` and the list of integers `nums`.\n",
+    "  - It also creates an empty min-heap called `self.min_heap` to store the k largest elements.\n",
+    "  - It populates the min-heap with the first `k` elements from `nums` by calling the `add` method.\n",
+    "\n",
+    "- The `add` method:\n",
+    "  - Adds the new integer `val` to the min-heap using `heapq.heappush`. This maintains the min-heap property.\n",
+    "  - Checks if the size of the min-heap exceeds `k`. If it does, it removes the smallest element (the k+1th largest) using `heapq.heappop`.\n",
+    "  - Finally, it returns the smallest element in the min-heap, which is always the kth largest element.\n",
+    "\n",
+    "Overall, this code implements a class that efficiently finds the kth largest element in a stream of integers by maintaining a min-heap of the k largest elements seen so far.\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4\n",
+      "5\n",
+      "5\n",
+      "8\n",
+      "8\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Example usage:\n",
+    "kthLargest = KthLargest(3, [4, 5, 8, 2])\n",
+    "print(kthLargest.add(3))  # Output: 4\n",
+    "print(kthLargest.add(5))  # Output: 5\n",
+    "print(kthLargest.add(10)) # Output: 5\n",
+    "print(kthLargest.add(9))  # Output: 8\n",
+    "print(kthLargest.add(4))  # Output: 8"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `KthLargest` class using the min-heap approach:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "1. `__init__` method:\n",
+    "   - In the `__init__` method, you iterate over the first `k` elements in `nums` and add them to the min-heap. Each `heapq.heappush` operation takes O(log k) time.\n",
+    "   - Therefore, the time complexity of the `__init__` method is O(k * log k).\n",
+    "\n",
+    "2. `add` method:\n",
+    "   - In the `add` method, you perform the following operations:\n",
+    "     - `heapq.heappush`: O(log k) to add the new element to the min-heap.\n",
+    "     - If the size of the min-heap exceeds `k`, you perform `heapq.heappop`, which is also O(log k).\n",
+    "     - Finally, you return the smallest element from the min-heap, which is O(1) because it's always the root.\n",
+    "   - Overall, the time complexity of the `add` method is O(log k).\n",
+    "\n",
+    "3. Overall, if you make `n` calls to the `add` method, the total time complexity is O(n * log k), where `n` is the total number of elements added to the stream.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "1. The space complexity is determined by the space used to store the min-heap and the instance variables.\n",
+    "2. The min-heap stores at most `k` elements at any given time, so its space complexity is O(k).\n",
+    "3. The instance variables (such as `self.k` and `self.min_heap`) have constant space requirements.\n",
+    "4. Therefore, the overall space complexity of the `KthLargest` class is O(k).\n",
+    "\n",
+    "In summary, the time complexity of the `KthLargest` class is O(n * log k) for `n` add operations, and the space complexity is O(k), where `k` is the parameter passed during initialization. This implementation efficiently maintains the kth largest element in the stream while meeting the problem's constraints."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Dynamic k:** Modify the `KthLargest` class to support dynamic changes in the value of `k`. Implement a method `update_k` that allows changing the value of `k` during the lifetime of the object. Ensure that the object can still correctly find the kth largest element based on the updated value of `k`.\n",
+    "\n",
+    "\n",
+    "2. **Implement kth Smallest:** Create a new class called `KthSmallest` that finds the kth smallest element in a stream instead of the kth largest. You may need to modify the data structure used in the implementation."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb b/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb
new file mode 100644
index 0000000..85d59ac
--- /dev/null
+++ b/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb	
@@ -0,0 +1,274 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 39. Combination Sum\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Combination Sum problem on LeetCode, click here!](https://leetcode.com/problems/combination-sum/)\n",
+    "\n",
+    "---\n",
+    "Given an array of **distinct** integers `candidates` and a target integer `target`, return *a list of all **unique combinations** of `candidates` where the chosen numbers sum to the `target`*. You may return the combinations in **any order**. \n",
+    "\n",
+    "The **same** number may be chosen from `candidates` an **unlimited number** of times. Two combinations are unique if the **frequency** of at least one of the chosen numbers is different.\n",
+    "\n",
+    "The test cases are generated such that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `1 <= candidates.length <= 30`\n",
+    "- `2 <= candidates[i] <= 40`\n",
+    "- All elements of `candidates` are **distinct**.\n",
+    "- `1 <= target <= 40`"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem is to find all unique combinations of numbers from a given array (candidates) such that their sum equals a target number. Here are the details of the problem:\n",
+    "\n",
+    "- You are given an array of distinct integers called \"candidates.\"\n",
+    "- You are also given a target integer called \"target.\"\n",
+    "\n",
+    "The goal is to find all unique combinations of numbers from the candidates array where the sum of the selected numbers is equal to the target. You can use the same number from the candidates array an unlimited number of times. A combination is considered unique if it has a different frequency (i.e., a different number of occurrences) of at least one chosen number compared to other combinations.\n",
+    "\n",
+    "For example:\n",
+    "\n",
+    "**Example 1:**\n",
+    "```python\n",
+    "Input: candidates = [2, 3, 6, 7], target = 7\n",
+    "Output: [[2, 2, 3], [7]]\n",
+    "```\n",
+    "In this example, there are two unique combinations that sum up to the target:\n",
+    "- 2 + 2 + 3 = 7\n",
+    "- 7 = 7\n",
+    "\n",
+    "**Example 2:**\n",
+    "```python\n",
+    "Input: candidates = [2, 3, 5], target = 8\n",
+    "Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n",
+    "```\n",
+    "Here, there are three unique combinations that sum up to the target.\n",
+    "\n",
+    "**Example 3:**\n",
+    "```python\n",
+    "Input: candidates = [2], target = 1\n",
+    "Output: []\n",
+    "```\n",
+    "In this case, there are no combinations that can be formed from the candidates to reach the target of 1, so the output is an empty list.\n",
+    "\n",
+    "The problem asks you to find and return these combinations in any order.\n",
+    "\n",
+    "The constraints for this problem include the length of the candidates array, the values of candidates, and the target value.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def combinationSum(candidates, target):\n",
+    "    # Define a recursive DFS function to find combinations\n",
+    "    def dfs(remaining, start, path):\n",
+    "        # If the remaining target is 0, we found a valid combination\n",
+    "        if remaining == 0:\n",
+    "            result.append(path)\n",
+    "            return\n",
+    "        # If the remaining target is negative, this path is invalid\n",
+    "        if remaining < 0:\n",
+    "            return\n",
+    "        # Iterate through candidates starting from 'start' to avoid duplicates\n",
+    "        for i in range(start, len(candidates)):\n",
+    "            # Explore the current candidate by subtracting it from the remaining target\n",
+    "            # Add the current candidate to the path\n",
+    "            dfs(remaining - candidates[i], i, path + [candidates[i]])\n",
+    "\n",
+    "    result = []  # Initialize an empty list to store the result\n",
+    "    candidates.sort()  # Sort the candidates for deduplication and early stopping\n",
+    "    dfs(target, 0, [])  # Start the DFS from the target value with an empty path\n",
+    "    return result  # Return the list of unique combinations"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The `combinationSum` function takes an array of distinct integers `candidates` and a target integer `target`. It returns a list of all unique combinations of `candidates` where the chosen numbers sum to the `target`. Each number in the `candidates` array can be used an unlimited number of times.\n",
+    "\n",
+    "Here's an overview of how the code works:\n",
+    "\n",
+    "1. The `combinationSum` function is defined, and it takes two arguments: `candidates` and `target`.\n",
+    "\n",
+    "2. Inside the function, there is a nested helper function called `backtrack`. This function is responsible for the actual combination generation using a recursive approach.\n",
+    "\n",
+    "3. The `backtrack` function is called with three arguments: `start`, `target`, and `path`. The `start` variable helps keep track of the current index in the `candidates` array, `target` keeps track of the remaining sum to be achieved, and `path` is a list that stores the current combination.\n",
+    "\n",
+    "4. Within the `backtrack` function, there are three main conditional statements:\n",
+    "   - If `target` becomes zero, it means we have found a combination that sums up to the target, so we add `path` to the `result` list.\n",
+    "   - If `target` becomes negative, it means the current combination doesn't work, so we return without doing anything.\n",
+    "   - Otherwise, we enter a loop that iterates over the `candidates` array, starting from the current index `start`.\n",
+    "\n",
+    "5. In the loop, the `backtrack` function is called recursively with the updated `target` and `path` after including the current candidate. This process explores different combinations by considering the current candidate or moving to the next candidate.\n",
+    "\n",
+    "6. After the loop completes, the `result` list contains all unique combinations that sum to the target.\n",
+    "\n",
+    "7. The `candidates` array is sorted to optimize the search. Sorting helps in avoiding unnecessary recursive branches and reduces the number of explored combinations.\n",
+    "\n",
+    "8. Finally, the `backtrack` function is initially called from the `combinationSum` function with `start` set to 0, `target` set to the original target value, and an empty `path`. The result is returned as a list of unique combinations.\n",
+    "\n",
+    "The code includes a few example cases to demonstrate how the function works with different inputs."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[2, 2, 3], [7]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Example 1\n",
+    "candidates1 = [2, 3, 6, 7]\n",
+    "target1 = 7\n",
+    "print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "candidates2 = [2, 3, 5]\n",
+    "target2 = 8\n",
+    "print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3\n",
+    "candidates3 = [2]\n",
+    "target3 = 1\n",
+    "print(combinationSum(candidates3, target3))  # Output: []"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `combinationSum` function.\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "The time complexity of this function is influenced by the number of recursive calls made by the `backtrack` function and the amount of work done within each call. In the worst case, the function will explore all possible combinations.\n",
+    "\n",
+    "The worst-case scenario occurs when we have many combinations to reach the target. The time complexity can be expressed as $O(2^n)$, where 'n' is the maximum number of recursive calls. In this context, 'n' corresponds to the number of elements in the `candidates` list, and it may be up to 30 (as per the constraints). However, the actual number of combinations explored is generally much smaller because we eliminate branches when the target becomes negative, and we skip over candidates that cannot lead to a solution. The sorting step also helps to optimize the search.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "The space complexity is determined by the auxiliary space used for storing the results and the stack space for the recursive calls.\n",
+    "\n",
+    "1. The space required for the `result` list can be up to $O(2^n)$ in the worst case because each combination is stored.\n",
+    "\n",
+    "2. The stack space for the recursive calls can also be up to O(n) in the worst case, where 'n' is the number of elements in the `candidates` list.\n",
+    "\n",
+    "Therefore, the overall space complexity of the function is $O(2^n + n)$. However, in practice, the space used for the results list is often the dominant factor, and the actual space used may be much smaller than 2^n because not all combinations are explored.\n",
+    "\n",
+    "In summary, the time complexity is exponential but typically smaller in practice due to optimization, and the space complexity is influenced by the number of results and the depth of the recursion."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Find the Number of Unique Combinations:** Instead of listing all unique combinations, modify the function to return the count of unique combinations that sum to the target.\n",
+    "\n",
+    "2. **Combinations with a Maximum Value:** Add a maximum value constraint for each combination, so they cannot exceed a certain value. Modify the code to find combinations respecting this constraint."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb b/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb
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+++ b/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb	
@@ -0,0 +1,290 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 79. Word Search\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Word Search problem on LeetCode, click here!](https://leetcode.com/problems/word-search/)\n",
+    "\n",
+    "---\n",
+    "Given an `m x n` grid of characters `board` and a string `word`, return *`true` if `word` exists in the grid.*\n",
+    "\n",
+    "The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `m == board.length`\n",
+    "- `n = board[i].length`\n",
+    "- `1 <= m, n <= 6`\n",
+    "- `1 <= word.length <= 15`\n",
+    "- `board` and `word` consists of only lowercase and uppercase English letters.\n",
+    "\n",
+    "**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem is to determine whether a given word can be constructed from characters in an `m x n` grid of characters. The word can be formed by moving sequentially through horizontally or vertically neighboring cells in the grid, and you cannot use the same cell more than once.\n",
+    "\n",
+    "Here are the specifics of the problem:\n",
+    "\n",
+    "- You are given an `m x n` grid, which is a 2D board of characters.\n",
+    "- You are also given a string `word` consisting of lowercase and uppercase English letters.\n",
+    "- Your task is to determine if the word can be found in the grid by moving from one cell to an adjacent cell (horizontally or vertically) and collecting the characters sequentially to form the word.\n",
+    "- You can't use the same cell more than once when forming the word.\n",
+    "- If it's possible to form the word in the grid, return `True`; otherwise, return `False`.\n",
+    "\n",
+    "For example, consider the following grid and word:\n",
+    "\n",
+    "```python\n",
+    "board = [[\"A\", \"B\", \"C\", \"E\"],\n",
+    "         [\"S\", \"F\", \"C\", \"S\"],\n",
+    "         [\"A\", \"D\", \"E\", \"E\"]]         \n",
+    "word = \"ABCCED\"\n",
+    "```\n",
+    "\n",
+    "In this case, the word \"ABCCED\" can be formed by moving through the cells (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) in the grid. Therefore, the function should return `True`.\n",
+    "\n",
+    "Conversely, if the word is impossible to construct, such as in the case of the word \"ABCB\" for the same grid, where there's no valid path that forms the word, the function should return `False`.\n",
+    "\n",
+    "The problem involves searching for a path through the grid that matches the characters in the word while respecting the constraints mentioned. It can be solved using depth-first search (DFS) or backtracking.\n",
+    "\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from typing import List\n",
+    "\n",
+    "class Solution:\n",
+    "    def exist(self, board: List[List[str]], word: str) -> bool:\n",
+    "        def dfs(i, j, k):\n",
+    "            # Check if the current position is out of bounds or doesn't match the character in the word.\n",
+    "            if not (0 <= i < len(board)) or not (0 <= j < len(board[0])) or board[i][j] != word[k]:\n",
+    "                return False\n",
+    "            \n",
+    "            # If we have matched all characters in the word, return True.\n",
+    "            if k == len(word) - 1:\n",
+    "                return True\n",
+    "\n",
+    "            # Temporarily mark the current cell to prevent revisiting.\n",
+    "            tmp, board[i][j] = board[i][j], \"/\"\n",
+    "\n",
+    "            # Explore adjacent cells by making recursive calls in all four directions.\n",
+    "            found = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)\n",
+    "\n",
+    "            # Restore the original character in the cell.\n",
+    "            board[i][j] = tmp\n",
+    "\n",
+    "            return found\n",
+    "\n",
+    "        # Iterate through all cells in the board.\n",
+    "        for i in range(len(board)):\n",
+    "            for j in range(len(board[0])):\n",
+    "                # Check if the 'dfs' function starting from this cell returns True.\n",
+    "                if dfs(i, j, 0):\n",
+    "                    return True\n",
+    "\n",
+    "        # If no match is found after exploring all cells, return False.\n",
+    "        return False"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "1. Import the necessary module:\n",
+    "   - `from typing import List`: This line imports the `List` type from the `typing` module, which is used to specify the type of the `board` parameter in the method signature.\n",
+    "\n",
+    "2. Define the `Solution` class:\n",
+    "   - `class Solution:`: This line defines a class called `Solution` to encapsulate the solution for the word search problem.\n",
+    "\n",
+    "3. Define the `exist` method:\n",
+    "   - `def exist(self, board: List[List[str]], word: str) -> bool:`: This method is a member of the `Solution` class and takes two parameters: `board`, which is a 2D list of characters, and `word`, which is a string. It returns a boolean value, indicating whether the word exists in the board.\n",
+    "\n",
+    "4. Implement the inner `dfs` function:\n",
+    "   - `def dfs(i, j, k):`: This is a helper function used for depth-first search (DFS). It is called recursively to explore the board and search for the word. It takes three parameters: `i` and `j` representing the current position on the board, and `k` representing the current index in the `word` string.\n",
+    "\n",
+    "   - Inside `dfs`, it checks if the current position is out of bounds or if the character at the current position on the board does not match the character at the current index in the `word`. If either of these conditions is met, it returns `False`.\n",
+    "\n",
+    "   - If `k` is equal to the length of the `word` minus one, it means we have successfully matched all characters in the `word`, so it returns `True`.\n",
+    "\n",
+    "   - The function then temporarily replaces the character at the current position with a marker, \"/\", to prevent revisiting the same cell. It then explores adjacent cells by making recursive calls in all four directions: up, down, left, and right. If any of these recursive calls return `True`, it propagates the `True` result up the call stack. After the exploration is complete, it restores the original character in the cell.\n",
+    "\n",
+    "5. Iterate through the board:\n",
+    "   - The outer loop iterates through all cells in the `board` using two nested loops. For each cell, it checks if the `dfs` function starting from that cell (with `k` initially set to 0) returns `True`. If it does, it means the word exists, so the method returns `True`.\n",
+    "\n",
+    "6. Return `False`:\n",
+    "   - If the loop completes without finding the word in the board, the method returns `False`.\n",
+    "\n",
+    "To use this code, you can create an instance of the `Solution` class and call the `exist` method on that instance, passing the `board` and the `word` as arguments to check if the word exists in the board."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1: \n",
+    "\n",
+    "# Create an instance of the Solution class\n",
+    "solution = Solution()\n",
+    "\n",
+    "# Test case 1: Word \"ABCCED\" can be found in the grid\n",
+    "board1 = [[\"A\",\"B\",\"C\",\"E\"],\n",
+    "          [\"S\",\"F\",\"C\",\"S\"],\n",
+    "          [\"A\",\"D\",\"E\",\"E\"]]\n",
+    "word1 = \"ABCCED\"\n",
+    "print(solution.exist(board1, word1))  # Expected output: True"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "\n",
+    "# Test case 2: Word \"SEE\" can be found in the grid\n",
+    "word2 = \"SEE\"\n",
+    "print(solution.exist(board1, word2))  # Expected output: True"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 3\n",
+    "\n",
+    "# Test case 3: Word \"ABCB\" cannot be found in the grid\n",
+    "word3 = \"ABCB\"\n",
+    "print(solution.exist(board1, word3))  # Expected output: False"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `exist` method in the given code.\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "The primary work in this code is done by the `dfs` function, which performs a depth-first search through the grid. The time complexity of the `dfs` function is $O(4^(n))$, where \"n\" is the length of the `word`. This is because, in the worst case, the function explores all possible directions (up, down, left, right) for each character in the `word`.\n",
+    "\n",
+    "The outer loops iterate through all cells in the `board`, which has dimensions `m x n`. So, the total number of iterations of `dfs` is O(m * n). However, since the `dfs` function has a higher time complexity, the overall time complexity is still dominated by the `dfs` function.\n",
+    "\n",
+    "Therefore, the overall time complexity of the `exist` method is $O(m * n * 4^n)$, where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "1. The primary space usage comes from the recursive calls in the `dfs` function. The depth of the recursion can be at most equal to the length of the `word`, which is O(n).\n",
+    "\n",
+    "2. Additionally, the `tmp` variable and the recursive calls on the call stack consume a small amount of additional memory.\n",
+    "\n",
+    "3. The marking of visited cells with the \"/\" character temporarily modifies the `board` in-place but doesn't significantly contribute to the space complexity.\n",
+    "\n",
+    "Therefore, the overall space complexity is O(n) due to the recursive call stack.\n",
+    "\n",
+    "In summary, the time complexity is $O(m * n * 4^n)$, and the space complexity is O(n), where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word. The time complexity is exponential in the worst case due to the depth-first search through all possible paths in the grid."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Optimization Challenge:** Modify the code to find all unique words that can be constructed from the grid. Instead of returning `True` or `False`, return a list of words found. Ensure that the same cell is not used more than once in constructing each word.\n",
+    "\n",
+    "2. **Board Generation:** Create a function to generate random grids of characters and then use the `exist` method to solve them. Test the efficiency and correctness of your code by generating larger boards and words."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb b/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb
new file mode 100644
index 0000000..56b7647
--- /dev/null
+++ b/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb	
@@ -0,0 +1,311 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 133. Clone Graph\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Clone Graph problem on LeetCode, click here!](https://leetcode.com/problems/clone-graph/)\n",
+    "\n",
+    "---\n",
+    "Given a reference of a node in a **connected** undirected graph.\n",
+    "\n",
+    "Return a **deep copy** (clone) of the graph.\n",
+    "\n",
+    "Each node in the graph contains a value (`int`) and a list (`List[Node]`) of its neighbors.\n",
+    "\n",
+    "```\n",
+    "class Node {\n",
+    "    public int val;\n",
+    "    public List<Node> neighbors;\n",
+    "}\n",
+    "```\n",
+    "\n",
+    "**Test case format:**\n",
+    "\n",
+    "For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with `val == 1`, the second node with `val == 2`, and so on. The graph is represented in the test case using an adjacency list.\n",
+    "\n",
+    "**An adjacency list** is a collection of unordered **lists** used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.\n",
+    "\n",
+    "The given node will always be the first node with `val = 1`. You must return the **copy of the given node** as a reference to the cloned graph.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- The number of nodes in the graph is in the range `[0, 100]`.\n",
+    "- `1 <= Node.val <= 100`\n",
+    "- `Node.val` is unique for each node.\n",
+    "- There are no repeated edges and no self-loops in the graph.\n",
+    "- The Graph is connected and all nodes can be visited starting from the given node."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem at hand is to clone a connected undirected graph, which means we need to create a deep copy of the original graph while preserving its structure and relationships between nodes. The graph is represented using adjacency lists, where each node has a value (an integer) and a list of its neighbors.\n",
+    "\n",
+    "Here are the key aspects of the problem:\n",
+    "\n",
+    "1. **Input**: You are given a reference to one of the nodes in the original graph. This reference node serves as the entry point to the graph. The entire graph can be explored starting from this node.\n",
+    "\n",
+    "2. **Output**: The goal is to create a deep copy of the graph and return a reference to the cloned graph. The cloned graph should have the same structure and relationships as the original one but should be a separate instance in memory.\n",
+    "\n",
+    "3. **Graph Structure**: The graph is composed of nodes (vertices) and edges. Nodes have values (integers) and are connected to other nodes through edges (undirected connections). Each node has a list of neighbors, which are other nodes it is connected to.\n",
+    "\n",
+    "4. **Connected Graph**: The problem assumes that the graph is connected, meaning that you can start from the reference node provided and traverse the entire graph by following the edges. This ensures that there are no isolated components in the graph.\n",
+    "\n",
+    "To solve this problem, you typically need to perform a graph traversal (e.g., depth-first search or breadth-first search) to visit all the nodes in the original graph, create corresponding nodes in the clone, and establish the same connections between nodes in the cloned graph as in the original one. To avoid duplication of nodes, you also need to keep track of visited nodes to ensure that the cloning process is efficient and that nodes are not duplicated in the clone.\n",
+    "\n",
+    "The problem can be challenging because you need to create a deep copy of the graph while handling cyclic dependencies between nodes and ensuring that the structure and relationships of the original graph are preserved in the clone. The solution requires recursion or a data structure to maintain the mapping between original nodes and their clones to achieve an efficient and accurate cloning process.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "# Definition for a Node.\n",
+    "class Node:\n",
+    "    def __init__(self, val=0, neighbors=None):\n",
+    "        self.val = val\n",
+    "        self.neighbors = neighbors if neighbors is not None else []\n",
+    "\n",
+    "class Solution:\n",
+    "    def cloneGraph(self, node: 'Node') -> 'Node':\n",
+    "        if not node:\n",
+    "            return None\n",
+    "\n",
+    "        visited = {}  # Dictionary to keep track of visited nodes\n",
+    "\n",
+    "        def dfs(original_node):\n",
+    "            if original_node in visited:\n",
+    "                return visited[original_node]\n",
+    "\n",
+    "            # Create a copy of the original node\n",
+    "            copy_node = Node(original_node.val)\n",
+    "            visited[original_node] = copy_node\n",
+    "\n",
+    "            # Recursively clone the neighbors of the original node\n",
+    "            for neighbor in original_node.neighbors:\n",
+    "                copy_node.neighbors.append(dfs(neighbor))\n",
+    "\n",
+    "            return copy_node\n",
+    "\n",
+    "        return dfs(node)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The given code defines a Python class, `Node`, and a solution class, `Solution`, to clone a connected undirected graph represented by nodes and their neighbors. Here's an explanation of the code:\n",
+    "\n",
+    "1. The `Node` class defines a node in the graph. Each node has a value (`val`) which is an integer and a list of neighbors (`neighbors`) that are other nodes in the graph.\n",
+    "\n",
+    "2. The `Solution` class contains the method `cloneGraph`, which takes a reference to the first node of the graph (`node`) as input and returns a deep copy of the entire graph, preserving the structure and relationships between nodes.\n",
+    "\n",
+    "3. Inside the `cloneGraph` method:\n",
+    "   - It first checks if the input `node` is `None`. If it's `None`, it means there is no graph to clone, and the method returns `None`.\n",
+    "\n",
+    "   - It initializes a `visited` dictionary to keep track of visited nodes. This dictionary is used to ensure that each node is cloned only once to avoid duplication.\n",
+    "\n",
+    "   - The `dfs` (depth-first search) function is defined within the `cloneGraph` method. It takes an original node as input and returns its corresponding clone. This function performs the actual cloning of the graph.\n",
+    "\n",
+    "   - Inside the `dfs` function:\n",
+    "     - It checks if the original node has been visited before by looking it up in the `visited` dictionary. If it has been visited, it returns the corresponding clone from the `visited` dictionary.\n",
+    "\n",
+    "     - If the original node is not visited, it creates a new node, `copy_node`, with the same value as the original node. This is the first step in creating the clone of the original node.\n",
+    "\n",
+    "     - It then recursively clones the neighbors of the original node by iterating through the `neighbors` list of the original node and appending the corresponding cloned neighbors to the `neighbors` list of the `copy_node`. This step ensures that the relationships between nodes are preserved in the clone.\n",
+    "\n",
+    "     - The `copy_node` is added to the `visited` dictionary with the original node as the key and the clone as the value.\n",
+    "\n",
+    "     - Finally, the `copy_node` is returned as the result of the DFS for the original node.\n",
+    "\n",
+    "4. The `cloneGraph` method returns the result of the DFS on the input `node`, which is the clone of the entire graph with its structure and relationships intact.\n",
+    "\n",
+    "In summary, the code uses depth-first search (DFS) to traverse the original graph, creating a clone for each node and its neighbors. It ensures that each node is cloned only once to avoid duplication and returns a reference to the cloned graph."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[1, 2, 4]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "def get_adjacency_list(node):\n",
+    "    visited = {}  # Dictionary to keep track of visited nodes\n",
+    "\n",
+    "    def dfs(node):\n",
+    "        if not node:\n",
+    "            return []\n",
+    "\n",
+    "        if node in visited:\n",
+    "            return visited[node]\n",
+    "\n",
+    "        neighbors = [n.val for n in node.neighbors]\n",
+    "        visited[node] = neighbors\n",
+    "\n",
+    "        for neighbor in node.neighbors:\n",
+    "            dfs(neighbor)\n",
+    "\n",
+    "    dfs(node)\n",
+    "    result = []\n",
+    "\n",
+    "    def get_node_and_neighbors(node):\n",
+    "        result.append([node.val] + visited[node])\n",
+    "\n",
+    "    get_node_and_neighbors(node)\n",
+    "    return result\n",
+    "\n",
+    "\n",
+    "# Testing the code with the provided examples\n",
+    "adjList1 = [[2, 4], [1, 3], [2, 4], [1, 3]]\n",
+    "solution = Solution()\n",
+    "\n",
+    "# Create the graph from adjList\n",
+    "nodes = [Node(i) for i in range(1, len(adjList1) + 1)]\n",
+    "for i, adj in enumerate(adjList1):\n",
+    "    nodes[i].neighbors = [nodes[j - 1] for j in adj]\n",
+    "\n",
+    "# Clone the graph and get its adjacency list\n",
+    "cloned_node1 = solution.cloneGraph(nodes[0])\n",
+    "cloned_adjList1 = get_adjacency_list(cloned_node1)\n",
+    "\n",
+    "# Print the cloned graph as an adjacency list\n",
+    "print(cloned_adjList1)\n",
+    "\n",
+    "# Output should match adjList1\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[1]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "# Input: adjList = [[]]\n",
+    "\n",
+    "adjList2 = [[]]\n",
+    "solution = Solution()\n",
+    "\n",
+    "# Create the graph from adjList\n",
+    "node1 = Node(1)\n",
+    "\n",
+    "# Clone the graph and get its adjacency list\n",
+    "cloned_node1 = solution.cloneGraph(node1)\n",
+    "cloned_adjList2 = get_adjacency_list(cloned_node1)\n",
+    "\n",
+    "# Print the cloned graph as an adjacency list\n",
+    "print(cloned_adjList2)\n",
+    "\n",
+    "# Output should be [[]], which represents an empty graph"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code for cloning a graph using depth-first search (DFS).\n",
+    "\n",
+    "Time Complexity:\n",
+    "1. Creating the graph: The time complexity to create the original graph from `adjList` is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. We iterate through each node and its neighbors in `adjList`.\n",
+    "\n",
+    "2. Cloning the graph (DFS): The depth-first search (DFS) is used to traverse the original graph and create a clone. The time complexity of DFS is O(V + E), as we visit each node and each edge once.\n",
+    "\n",
+    "Overall, the time complexity is O(V + E) for both creating the original graph and cloning it.\n",
+    "\n",
+    "Space Complexity:\n",
+    "1. Storage for the original graph: We need to store the original graph, which includes all nodes and their adjacency lists. In the worst case, this requires O(V + E) space, where V is the number of nodes, and E is the number of edges.\n",
+    "\n",
+    "2. Storage for the visited dictionary: The `visited` dictionary stores mappings from original nodes to their corresponding clones. In the worst case, this requires O(V) space because each node is visited once.\n",
+    "\n",
+    "3. Recursive stack for DFS: The space used for the recursive call stack during DFS can be up to O(V) in the worst case, where V is the number of nodes.\n",
+    "\n",
+    "Overall, the space complexity is O(V + E) for storing the original graph, and O(V) for auxiliary data structures, including the `visited` dictionary and the DFS call stack. Therefore, the overall space complexity is O(V + E).\n",
+    "\n",
+    "In summary, the time and space complexities of the provided code for cloning a graph are both O(V + E), where V is the number of nodes, and E is the number of edges in the graph."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Cyclic Graph**: Design a graph with cycles (nodes connected in a cycle). Test if the code can correctly clone graphs with cycles and doesn't fall into an infinite loop during traversal.\n",
+    "\n",
+    "\n",
+    "2. **Partial Graph Cloning**: Modify the code to clone only a subset of the graph starting from a given reference node, rather than the entire graph. Implement this as an extension to the problem."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb b/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb
new file mode 100644
index 0000000..0dd53e9
--- /dev/null
+++ b/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb	
@@ -0,0 +1,242 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 178. Graph Valid Tree\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Graph Valid Tree problem on LintCode, click here!](https://www.lintcode.com/problem/178/)\n",
+    "\n",
+    "---\n",
+    "Given `n` nodes labeled from `0` to `n - 1` and a list of `undirected` edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- You can assume that no duplicate edges will appear in edges. Since all edges are `undirected`, `[0, 1]` is the same as `[1, 0]` and thus will not appear together in edges."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem is to determine whether a given set of undirected edges can form a valid tree. In this context, a \"tree\" is a specific type of graph with the following properties:\n",
+    "\n",
+    "1. It is a connected graph, meaning there is a path between any pair of nodes in the graph.\n",
+    "2. It has no cycles, which means there are no closed loops or circuits in the graph.\n",
+    "3. It has exactly `n - 1` edges, where `n` is the number of nodes in the tree.\n",
+    "\n",
+    "The problem is typically given with two pieces of information:\n",
+    "\n",
+    "1. The number of nodes (`n`), which are usually labeled from 0 to `n-1`.\n",
+    "2. A list of undirected edges, where each edge is represented as a pair of nodes.\n",
+    "\n",
+    "The task is to write a function or algorithm to determine whether the provided set of edges can be arranged to form a valid tree based on the properties mentioned above.\n",
+    "\n",
+    "To solve this problem, you need to check the following conditions:\n",
+    "\n",
+    "1. There should be exactly `n - 1` edges to ensure that the graph is connected and tree-like.\n",
+    "2. There should be no cycles in the graph. If there are any cycles, the set of edges cannot form a tree.\n",
+    "3. The graph should be connected, which means you can reach any node from any other node through a series of edges.\n",
+    "\n",
+    "The solution typically involves using graph traversal algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the graph and check these conditions. If all conditions are met, the given edges form a valid tree; otherwise, they do not.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def validTree(n, edges):\n",
+    "    if len(edges) != n - 1:\n",
+    "        return False  # A tree with n nodes must have n-1 edges.\n",
+    "\n",
+    "    # Create an adjacency list to represent the graph.\n",
+    "    adj_list = {i: [] for i in range(n)}\n",
+    "    for u, v in edges:\n",
+    "        adj_list[u].append(v)\n",
+    "        adj_list[v].append(u)\n",
+    "\n",
+    "    visited = set()\n",
+    "\n",
+    "    # Define a recursive DFS function.\n",
+    "    def dfs(node, parent):\n",
+    "        if node in visited:\n",
+    "            return False  # If we encounter a visited node, there's a cycle.\n",
+    "        visited.add(node)\n",
+    "\n",
+    "        for neighbor in adj_list[node]:\n",
+    "            if neighbor != parent:\n",
+    "                if not dfs(neighbor, node):\n",
+    "                    return False\n",
+    "\n",
+    "        return True\n",
+    "\n",
+    "    # Start DFS from any node (e.g., node 0).\n",
+    "    if not dfs(0, -1):\n",
+    "        return False  # If there's a cycle, it's not a valid tree.\n",
+    "\n",
+    "    return len(visited) == n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The `validTree` function checks if a given set of undirected edges forms a valid tree. A valid tree must satisfy these conditions:\n",
+    "\n",
+    "1. It has exactly `n - 1` edges, where `n` is the number of nodes.\n",
+    "2. There are no cycles in the graph.\n",
+    "3. The graph is connected, meaning all nodes are reachable from any starting node.\n",
+    "\n",
+    "The code accomplishes this as follows:\n",
+    "\n",
+    "- It first checks if the number of edges is equal to `n - 1`. If not, it immediately returns `False` since a tree with `n` nodes must have `n - 1` edges to be connected.\n",
+    "\n",
+    "- It then constructs an adjacency list (`adj_list`) to represent the graph efficiently. This data structure stores the neighbors of each node.\n",
+    "\n",
+    "- The code uses a `visited` set to keep track of nodes visited during depth-first search (DFS).\n",
+    "\n",
+    "- The `dfs` function is a recursive function that checks for cycles in the graph. It does this by visiting nodes and marking them as visited. If it encounters a node that has already been visited (indicating a cycle), it returns `False`.\n",
+    "\n",
+    "- The DFS function explores the neighbors of each node, avoiding revisiting the parent node to prevent cycles.\n",
+    "\n",
+    "- After the DFS is complete, the code checks if the number of visited nodes is equal to `n`, ensuring the graph is connected.\n",
+    "\n",
+    "- Finally, if all conditions are met, the code returns `True`, indicating that the given edges form a valid tree. Otherwise, it returns `False`.\n",
+    "\n",
+    "The code provides examples to demonstrate how to use the `validTree` function to check whether a set of edges forms a valid tree or not."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1\n",
+    "n1 = 5\n",
+    "edges1 = [[0, 1], [0, 2], [0, 3], [1, 4]]\n",
+    "print(validTree(n1, edges1))  # Output: true"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "n2 = 5\n",
+    "edges2 = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]\n",
+    "print(validTree(n2, edges2))  # Output: false"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexities of the `validTree` function:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "1. Constructing the adjacency list (`adj_list`) takes O(E) time, where E is the number of edges in the input graph (|edges|).\n",
+    "\n",
+    "2. The DFS function, when considering all nodes, has a time complexity of O(V + E), where V is the number of nodes in the graph. This is because, in the worst case, the DFS function visits each node and each edge once.\n",
+    "\n",
+    "3. Checking if the number of visited nodes is equal to `n` takes O(1) time.\n",
+    "\n",
+    "Overall, the time complexity of the `validTree` function is O(E + V) in the worst case.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "1. The adjacency list (`adj_list`) consumes additional space to store the graph structure, taking O(V + E) space. This is because it stores information about each node and its adjacent nodes.\n",
+    "\n",
+    "2. The `visited` set keeps track of visited nodes, and its space complexity is O(V) in the worst case, as there can be at most V unique nodes in the set.\n",
+    "\n",
+    "3. The depth of the function call stack during DFS can be at most V, which contributes O(V) space to the space complexity.\n",
+    "\n",
+    "Overall, the space complexity of the `validTree` function is O(V + E) due to the adjacency list and O(V) due to the `visited` set and DFS call stack.\n",
+    "\n",
+    "In summary, the time complexity is O(E + V), and the space complexity is O(V + E). These complexities are based on the worst-case scenario, where all nodes and edges are considered. In practical cases, the actual complexities may be smaller, depending on the specific structure of the graph."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Minimum Spanning Tree (MST)**: Implement an algorithm to find the minimum spanning tree of a connected graph using either Kruskal's or Prim's algorithm. Compare the MST to the original graph to determine if the original graph is a tree. This exercise reinforces the concept of tree properties.\n",
+    "\n",
+    "2. **Detecting Cycles**: Modify the code to not only check if the input forms a tree but also to identify and print out any cycles present in the graph if it's not a tree. This exercise enhances your ability to detect and visualize cycles in a graph."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb b/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb
new file mode 100644
index 0000000..5a878a4
--- /dev/null
+++ b/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb	
@@ -0,0 +1,236 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 200. Number of Islands\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Number of Islands problem on LeetCode, click here!](https://leetcode.com/problems/number-of-islands/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "Given an `m x n` 2D binary grid `grid` which represents a map of `'1'`s (land) and `'0'`s (water), return *the number of islands*. \n",
+    "\n",
+    "An **island** is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `m == grid.length`\n",
+    "- `n == grid[i].length`\n",
+    "- `1 <= m, n <= 300`\n",
+    "- `grid[i][j]` is `'0'` or `'1'`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem is about counting the number of islands in a 2D binary grid. In this grid, '1' represents land, and '0' represents water. An island is defined as a group of '1' cells that are adjacent to each other either horizontally or vertically. It's important to note that diagonal connections do not count. Additionally, it's assumed that all four edges of the grid are surrounded by water, meaning the grid is finite and doesn't \"wrap around.\"\n",
+    "\n",
+    "The objective is to determine how many distinct islands exist in the grid. To clarify, an island consists of connected '1' cells, and these connections can only occur horizontally or vertically. If there are no '1' cells in the grid, the number of islands would be zero.\n",
+    "\n",
+    "The problem often involves using graph traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to identify and count these islands. You start at one '1' cell and explore its adjacent cells to determine the extent of the island. Once you've visited all the '1' cells in an island, you move on to the next unvisited '1' cell and repeat the process until you've counted all the islands in the grid.\n",
+    "\n",
+    "In essence, the problem asks you to analyze the grid to find groups of connected '1' cells and count them as individual islands. The goal is to return the total count of islands in the given grid.\n",
+    "\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from typing import List\n",
+    "\n",
+    "class Solution:\n",
+    "    def numIslands(self, grid: List[List[str]]) -> int:\n",
+    "        if not grid:\n",
+    "            return 0\n",
+    "\n",
+    "        m, n = len(grid), len(grid[0])  # Get the dimensions of the grid\n",
+    "        num_islands = 0  # Initialize the count of islands\n",
+    "\n",
+    "        def dfs(row, col):\n",
+    "            if row < 0 or row >= m or col < 0 or col >= n or grid[row][col] == '0':\n",
+    "                return\n",
+    "\n",
+    "            grid[row][col] = '0'  # Mark the current cell as visited by changing it to '0'\n",
+    "            directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Define four possible directions to move\n",
+    "\n",
+    "            for dr, dc in directions:\n",
+    "                # Explore adjacent cells\n",
+    "                dfs(row + dr, col + dc)\n",
+    "\n",
+    "        for i in range(m):\n",
+    "            for j in range(n):\n",
+    "                if grid[i][j] == '1':\n",
+    "                    num_islands += 1  # Increment the island count\n",
+    "                    dfs(i, j)  # Start a DFS from the current '1' cell to explore the entire island\n",
+    "\n",
+    "        return num_islands\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The provided code defines a Python class called `Solution` with a method called `numIslands`. This class is designed to count the number of islands in a 2D binary grid.\n",
+    "\n",
+    "The `numIslands` method takes a 2D grid as input, where '1' represents land, and '0' represents water. It uses a depth-first search (DFS) approach to identify and count islands. The key steps in the code include:\n",
+    "\n",
+    "1. Checking if the input grid is empty, and if so, returning 0 (indicating there are no islands).\n",
+    "\n",
+    "2. Determining the dimensions of the grid (number of rows and columns).\n",
+    "\n",
+    "3. Initializing a variable `num_islands` to keep track of the count of islands.\n",
+    "\n",
+    "4. Defining a nested function `dfs` to perform depth-first search. This function recursively explores adjacent land cells connected to the current cell and marks them as visited by changing '1' to '0'.\n",
+    "\n",
+    "5. In the `dfs` function, it checks for boundary conditions and whether the current cell is '0'. If these conditions are met, it returns without further exploration.\n",
+    "\n",
+    "6. The `dfs` function explores neighboring cells in all four directions (up, down, left, right) using a loop.\n",
+    "\n",
+    "7. The main loop iterates through each cell of the grid. When a '1' cell is encountered, it increments the `num_islands` count and starts a DFS from that cell to explore the entire island.\n",
+    "\n",
+    "8. Finally, the method returns the total count of islands found in the grid.\n",
+    "\n",
+    "This code encapsulates the solution in a class, allowing you to create an instance of the `Solution` class and call the `numIslands` method on it to count the number of islands in a given grid."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1 Output: 1\n"
+     ]
+    }
+   ],
+   "source": [
+    "grid1 = [\n",
+    "  [\"1\", \"1\", \"1\", \"1\", \"0\"],\n",
+    "  [\"1\", \"1\", \"0\", \"1\", \"0\"],\n",
+    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
+    "  [\"0\", \"0\", \"0\", \"0\", \"0\"]\n",
+    "]\n",
+    "\n",
+    "solution = Solution()\n",
+    "result1 = solution.numIslands(grid1)\n",
+    "print(\"Example 1 Output:\", result1)  # Expected output: 1\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 2 Output: 3\n"
+     ]
+    }
+   ],
+   "source": [
+    "grid2 = [\n",
+    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
+    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
+    "  [\"0\", \"0\", \"1\", \"0\", \"0\"],\n",
+    "  [\"0\", \"0\", \"0\", \"1\", \"1\"]\n",
+    "]\n",
+    "\n",
+    "result2 = solution.numIslands(grid2)\n",
+    "print(\"Example 2 Output:\", result2)  # Expected output: 3\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "Time Complexity:\n",
+    "- The code uses a depth-first search (DFS) to explore the grid and count the number of islands.\n",
+    "- In the worst case, it visits every cell in the grid exactly once.\n",
+    "- The DFS function explores neighboring cells in four directions (up, down, left, right), so it has a time complexity of $O(4^{max(m, n)})$, where m is the number of rows and n is the number of columns.\n",
+    "- Therefore, the overall time complexity of the code is O(m * n), where m is the number of rows, and n is the number of columns.\n",
+    "\n",
+    "Space Complexity:\n",
+    "- The space complexity is determined by the recursive calls and the depth of the DFS stack.\n",
+    "- In the worst case, when the entire grid consists of '1's, the depth of the DFS recursion can be as deep as max(m, n).\n",
+    "- Therefore, the space complexity of the code is O(max(m, n)), representing the maximum depth of the DFS stack.\n",
+    "\n",
+    "In practical terms, for reasonably sized grids, the time and space complexity is linear, and the code should perform well. However, it's important to keep in mind that the worst-case time complexity is O(m * n), and the space complexity can be O(max(m, n))."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Identify Largest Island:** In addition to counting the number of islands, modify the code to identify the largest island in the grid and return its size.\n",
+    "\n",
+    "2. **Optimize for Space:** Modify the solution to reduce its space complexity to O(1) without modifying the input grid. This is a more memory-efficient version of the problem."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb b/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb
new file mode 100644
index 0000000..4176cab
--- /dev/null
+++ b/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb	
@@ -0,0 +1,243 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 207. Course Schedule\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Course Schedule problem on LeetCode, click here!](https://leetcode.com/problems/course-schedule/)\n",
+    "\n",
+    "---\n",
+    "There are a total of `numCourses` courses to take, labeled from 0 to `numCourses - 1`. You are given an array `prerequisites` where `prerequisites[i] = [ai, bi]` indicates that you **must** take course `bi` first if you want to take course `ai`. \n",
+    "\n",
+    "- For example, the pair `[0, 1]`, indicates that to take course `0` you have to first take course `1`.\n",
+    "\n",
+    "Return `true` if you can finish all courses. Otherwise, return `false`.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- 1 <= `numCourses` <= 2000\n",
+    "- 0 <= `prerequisites.length` <= 5000\n",
+    "- `prerequisites[i].length` == 2\n",
+    "- 0 <= `ai`, `bi` < `numCourses`\n",
+    "- All the pairs `prerequisites[i]` are **unique**."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem is to determine whether it's possible to complete a set of courses given a list of prerequisites. You are given the following:\n",
+    "\n",
+    "- `numCourses`: The total number of courses to be taken, labeled from 0 to `numCourses - 1`.\n",
+    "- `prerequisites`: A list of prerequisite courses, where each element `prerequisites[i]` is a pair `[ai, bi]`, indicating that you must take course `bi` before you can take course `ai`.\n",
+    "\n",
+    "The goal is to check whether it's possible to complete all the courses while respecting the prerequisite requirements. In other words, you need to determine if there are any circular dependencies in the course prerequisites that would prevent you from taking all the courses.\n",
+    "\n",
+    "For example:\n",
+    "\n",
+    "- If `numCourses` is 2, and `prerequisites` is `[[1, 0]]`, it means you have two courses, and you must finish course 0 before you can take course 1. In this case, it's possible to complete all courses, so the function should return `True`.\n",
+    "\n",
+    "- If `numCourses` is 2, and `prerequisites` is `[[1, 0], [0, 1]]`, it means you have two courses, but there's a circular dependency where course 1 requires course 0 and course 0 requires course 1. In this case, it's impossible to complete all courses, so the function should return `False`.\n",
+    "\n",
+    "The problem is essentially about checking for the presence of cycles in a directed graph, where the courses are nodes, and the prerequisites represent directed edges between them. If there are no cycles, it's possible to complete all courses; otherwise, it's not possible.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from collections import defaultdict, deque\n",
+    "from typing import List\n",
+    "\n",
+    "class Solution:\n",
+    "    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:\n",
+    "        # Create a graph using an adjacency list to represent the courses and their prerequisites.\n",
+    "        graph = defaultdict(list)  # Initialize an empty adjacency list.\n",
+    "        in_degree = [0] * numCourses  # Initialize an array to store in-degrees of courses.\n",
+    "\n",
+    "        # Populate the graph and calculate in-degrees.\n",
+    "        for course, prerequisite in prerequisites:\n",
+    "            graph[prerequisite].append(course)  # Add the course as a neighbor to its prerequisite.\n",
+    "            in_degree[course] += 1  # Increment the in-degree of the course.\n",
+    "\n",
+    "        # Initialize a queue with courses that have no prerequisites.\n",
+    "        queue = deque()\n",
+    "        for course in range(numCourses):\n",
+    "            if in_degree[course] == 0:\n",
+    "                queue.append(course)\n",
+    "\n",
+    "        # Perform topological sorting.\n",
+    "        while queue:\n",
+    "            course = queue.popleft()  # Take a course with no prerequisites.\n",
+    "            numCourses -= 1  # Decrement the count of remaining courses.\n",
+    "\n",
+    "            for neighbor in graph[course]:\n",
+    "                in_degree[neighbor] -= 1  # Remove the prerequisite relationship.\n",
+    "                if in_degree[neighbor] == 0:\n",
+    "                    queue.append(neighbor)  # If no more prerequisites, add to the queue.\n",
+    "\n",
+    "        # If all courses were successfully taken (numCourses becomes 0), return True.\n",
+    "        return numCourses == 0"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "This Python code is designed to determine whether it is possible to complete all the required courses given a set of prerequisites. It uses a topological sorting algorithm to accomplish this. Here's a brief explanation of the code:\n",
+    "\n",
+    "1. The code defines a `Solution` class with a `canFinish` method that takes two parameters: `numCourses` (the total number of courses) and `prerequisites` (a list of pairs indicating which course must be taken before another).\n",
+    "\n",
+    "2. It initializes an empty graph as an adjacency list using a `defaultdict` to represent the courses and their prerequisites. It also initializes a list called `in_degree` to keep track of the in-degrees of courses, which is used in the topological sorting process.\n",
+    "\n",
+    "3. The code then populates the graph by iterating through the `prerequisites` list. For each pair `(ai, bi)`, it adds course `ai` as a neighbor to its prerequisite course `bi` in the graph and increments the in-degree of course `ai`.\n",
+    "\n",
+    "4. Next, it initializes a queue with courses that have no prerequisites. It does this by iterating through all the courses and adding those with an in-degree of 0 to the queue.\n",
+    "\n",
+    "5. The code performs a topological sorting of the courses using a while loop. In each iteration of the loop, it dequeues a course from the front of the queue (a course with no prerequisites).\n",
+    "\n",
+    "6. For each neighbor (course that depends on the dequeued course), it decrements the in-degree of the neighbor. If the neighbor's in-degree becomes 0, it means all of its prerequisites have been taken, so the neighbor is added to the queue.\n",
+    "\n",
+    "7. The loop continues until there are no more courses with no prerequisites to dequeue, and the in-degrees are updated accordingly.\n",
+    "\n",
+    "8. After the loop, if all courses were successfully taken (i.e., `numCourses` becomes 0), it means that there were no circular dependencies, and it's possible to finish all courses. In this case, the method returns `True`. Otherwise, if there are remaining courses (indicating a circular dependency), it returns `False`."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "True\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1\n",
+    "numCourses1 = 2\n",
+    "prerequisites1 = [[1, 0]]\n",
+    "sol = Solution()\n",
+    "print(sol.canFinish(numCourses1, prerequisites1))  # Output: True"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "numCourses2 = 2\n",
+    "prerequisites2 = [[1, 0], [0, 1]]\n",
+    "print(sol.canFinish(numCourses2, prerequisites2))  # Output: False"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the `canFinish` method:\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "\n",
+    "1. Constructing the graph: In the first loop where we iterate through the `prerequisites` list, we populate the graph with prerequisites. This loop has a time complexity of O(E), where E is the number of prerequisites (edges).\n",
+    "\n",
+    "2. Initializing in-degrees and finding courses with no prerequisites: In the worst case, we iterate through all `numCourses` courses to initialize in-degrees and find the courses with no prerequisites. This has a time complexity of O(V), where V is the number of courses (vertices).\n",
+    "\n",
+    "3. Topological Sorting: In the while loop, we perform topological sorting, processing each course once and its outgoing edges. In the worst case, each course is processed once. This has a time complexity of O(V + E).\n",
+    "\n",
+    "The overall time complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "\n",
+    "1. Graph Representation: The space complexity for storing the graph as an adjacency list is O(E), where E is the number of prerequisites.\n",
+    "\n",
+    "2. In-Degree Array: The space complexity for storing the in-degrees of courses is O(V), where V is the number of courses.\n",
+    "\n",
+    "3. Queue: The space complexity for the queue is O(V), as it may contain all the courses.\n",
+    "\n",
+    "The overall space complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
+    "\n",
+    "In most practical cases, the number of prerequisites (E) is much smaller than the number of courses (V), so the time and space complexity can often be considered as O(V)."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Optimizing Course Scheduling**: Given a list of courses with their durations and prerequisites, find the most efficient way to schedule these courses to minimize the time it takes to complete all of them.\n",
+    "\n",
+    "2. **Detecting Cycles in Course Dependencies**: Modify the original problem to not just determine if you can finish all courses but also to identify the specific courses that form a cycle in the prerequisite dependencies."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb b/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb
new file mode 100644
index 0000000..a2789c4
--- /dev/null
+++ b/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb	
@@ -0,0 +1,230 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 3651. Number of Connected Components in an Undirected Graph\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Number of Connected Components in an Undirected Graph problem on LintCode, click here!](https://www.lintcode.com/problem/3651/)\n",
+    "\n",
+    "---\n",
+    "In this problem, there is an undirected graph with `n` nodes. There is also an `edges` array. Where `edges[i] = [a, b]` means that there is an edge between *node a* and *node b* in the graph.\n",
+    "\n",
+    "You need to return *the number of connected components in that graph*."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem at hand is to determine the number of connected components in an undirected graph. Here's a more detailed explanation of the problem:\n",
+    "\n",
+    "- You are given an undirected graph, which consists of nodes and edges. The graph may have multiple distinct connected components.\n",
+    "\n",
+    "- A connected component is a group of nodes within the graph where each node can be reached from every other node in the same component. In other words, there is a path between any two nodes in the same connected component, but there are no paths connecting nodes from different components.\n",
+    "\n",
+    "- Your task is to write a function that takes as input the number of nodes `n` and a list of edges `edges`, where each edge is represented as a pair of nodes `[a, b]` indicating an edge between node `a` and node `b`.\n",
+    "\n",
+    "- The goal is to determine the number of distinct connected components in the graph.\n",
+    "\n",
+    "- You need to implement an algorithm that can identify these connected components and count how many there are in the given graph.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def countConnectedComponents(n, edges):\n",
+    "    # Define a helper function for depth-first search (DFS).\n",
+    "    def dfs(node):\n",
+    "        if not visited[node]:  # If the node has not been visited yet:\n",
+    "            visited[node] = True  # Mark it as visited.\n",
+    "            for neighbor in graph[node]:  # Explore all neighbors of the current node.\n",
+    "                dfs(neighbor)  # Recursively call DFS on the neighbor.\n",
+    "\n",
+    "    # Create an empty adjacency list to represent the graph.\n",
+    "    graph = [[] for _ in range(n)]\n",
+    "\n",
+    "    # Initialize a boolean list to keep track of visited nodes.\n",
+    "    visited = [False] * n\n",
+    "\n",
+    "    # Populate the adjacency list with edges from the input.\n",
+    "    for a, b in edges:\n",
+    "        graph[a].append(b)\n",
+    "        graph[b].append(a)  # Since the graph is undirected, we add edges in both directions.\n",
+    "\n",
+    "    # Initialize a counter to keep track of the number of connected components.\n",
+    "    count = 0\n",
+    "\n",
+    "    # Iterate through all nodes in the graph.\n",
+    "    for i in range(n):\n",
+    "        if not visited[i]:  # If the node has not been visited:\n",
+    "            count += 1  # Increment the count since we've found a new connected component.\n",
+    "            dfs(i)  # Start a depth-first search from the unvisited node to explore the component.\n",
+    "\n",
+    "    return count"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The provided Python code calculates the number of connected components in an undirected graph. Here's a high-level explanation of how the code works:\n",
+    "\n",
+    "1. The `countConnectedComponents` function takes two arguments: `n` (the number of nodes in the graph) and `edges` (a list of edge pairs representing connections between nodes).\n",
+    "\n",
+    "2. Inside the function, there is a helper function called `dfs` (depth-first search), which is used to explore and mark nodes as visited.\n",
+    "\n",
+    "3. An empty adjacency list called `graph` is created to represent the graph. The `graph` is a list of lists where each element represents a node, and the list associated with each node contains its neighboring nodes.\n",
+    "\n",
+    "4. Another list called `visited` is created to keep track of visited nodes. Initially, all nodes are marked as not visited (`False`).\n",
+    "\n",
+    "5. The code populates the `graph` by iterating through the `edges` list and adding each edge to the adjacency list. Since the graph is undirected, both directions of each edge are added.\n",
+    "\n",
+    "6. A counter variable called `count` is initialized to keep track of the number of connected components.\n",
+    "\n",
+    "7. The code iterates through all nodes from `0` to `n-1`. For each node, it checks whether it has been visited yet. If it hasn't been visited, it means a new connected component is found. The counter `count` is incremented, and a depth-first search (DFS) is initiated from this unvisited node to explore the connected component.\n",
+    "\n",
+    "8. The `dfs` function recursively visits all nodes in the connected component, marking them as visited. It does this by checking each neighbor of the current node and recursively calling `dfs` on unvisited neighbors.\n",
+    "\n",
+    "9. After the loop through all nodes, the `count` variable contains the total number of connected components in the graph.\n",
+    "\n",
+    "10. The function returns the value of `count`, which represents the number of connected components.\n",
+    "\n",
+    "11. Two examples are provided at the end to demonstrate how to use the function with different inputs."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1\n",
+    "n1 = 3\n",
+    "edges1 = [[0, 1], [0, 2]]\n",
+    "print(countConnectedComponents(n1, edges1))  # Output: 1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "n2 = 6\n",
+    "edges2 = [[0, 1], [1, 2], [2, 3], [4, 5]]\n",
+    "print(countConnectedComponents(n2, edges2))  # Output: 2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the code for counting the number of connected components in an undirected graph using depth-first search (DFS).\n",
+    "\n",
+    "**Time Complexity:**\n",
+    "- The code iterates through all nodes in the graph (from 0 to `n-1`).\n",
+    "- For each unvisited node, it initiates a depth-first search (DFS) from that node to explore the connected component.\n",
+    "- In the worst case, every node may belong to a separate connected component, and you perform a DFS for each node.\n",
+    "- The time complexity of the DFS for a single connected component is O(V + E), where V is the number of nodes in the component, and E is the number of edges.\n",
+    "- In the worst case, you perform O(n) DFS calls, each with its own O(V + E) complexity.\n",
+    "\n",
+    "So, the overall time complexity of the code is O(n * (V + E)). In the worst case, if the graph is connected (a single connected component), this simplifies to O(n * (n + m)), where m is the number of edges.\n",
+    "\n",
+    "**Space Complexity:**\n",
+    "- The main data structures used are the `graph` (adjacency list), `visited` (boolean array), and the recursive call stack for DFS.\n",
+    "- The `graph` has a space complexity of O(n + m), where n is the number of nodes, and m is the number of edges.\n",
+    "- The `visited` boolean array also has a space complexity of O(n) to keep track of whether nodes have been visited.\n",
+    "- The depth of the recursive call stack for DFS can go up to O(n) in the worst case, when all nodes are part of a single connected component.\n",
+    "\n",
+    "So, the overall space complexity of the code is O(n + m) for data structures and O(n) for the call stack. In the worst case, this simplifies to O(n + m).\n",
+    "\n",
+    "In summary, the time complexity is O(n * (V + E)), and the space complexity is O(n + m)."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Disconnected Components with Multiple Queries:** You are given an undirected graph with `n` nodes and `m` edges. Initially, the graph is disconnected, and you need to process `q` queries. Each query consists of adding an edge to connect two nodes. After each query, you need to output the number of connected components in the updated graph.\n",
+    "\n",
+    "2. **Maximum Size of Connected Components:** Given an undirected graph with `n` nodes and `m` edges, find the size of the largest connected component in the graph. Additionally, identify the nodes that belong to this largest component. You can assume there is at least one connected component in the graph.\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb b/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb
new file mode 100644
index 0000000..c685b23
--- /dev/null
+++ b/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb	
@@ -0,0 +1,255 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 417. Pacific Atlantic Water Flow\n",
+    "\n",
+    "\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "**Link to Problem:** [To see the Pacific Atlantic Water Flow problem on LeetCode, click here!](https://leetcode.com/problems/pacific-atlantic-water-flow/)\n",
+    "\n",
+    "---\n",
+    "There is an `m x n` rectangular island that borders both the **Pacific Ocean** and **Atlantic Ocean**. The **Pacific Ocea**n touches the island's left and top edges, and the **Atlantic Ocean** touches the island's right and bottom edges.\n",
+    "\n",
+    "The island is partitioned into a grid of square cells. You are given an `m x n` integer matrix `heights` where `heights[r][c]` represents the **height above sea level** of the cell at coordinate `(r, c)`.\n",
+    "\n",
+    "The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is **less than or equal to** the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.\n",
+    "\n",
+    "Return *a **2D list** of grid coordinates `result` where $result[i] = [r_i, c_i]$ denotes that rain water can flow from cell $(r_i, c_i)$ to **both** the Pacific and Atlantic oceans*.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `m == heights.length`\n",
+    "- `n == heights[r].length`\n",
+    "- `1 <= m, n <= 200`\n",
+    "- 0 <= `heights[r][c]` <= $10^5$"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem at hand involves a scenario with a rectangular island surrounded by both the Pacific Ocean and the Atlantic Ocean. The Pacific Ocean is in contact with the island's left and top edges, while the Atlantic Ocean is in contact with the island's right and bottom edges. The goal is to determine which cells on the island can allow rainwater to flow to both the Pacific and Atlantic Oceans.\n",
+    "\n",
+    "Here are the key details and constraints of the problem:\n",
+    "\n",
+    "1. The island is divided into a grid of square cells.\n",
+    "\n",
+    "2. Each cell is represented by an integer value in a matrix `heights`. The value at `heights[r][c]` represents the height above sea level of the cell at coordinates `(r, c)` on the island.\n",
+    "\n",
+    "3. Rainwater can flow from a cell to neighboring cells directly north, south, east, or west only if the neighboring cell's height is less than or equal to the current cell's height.\n",
+    "\n",
+    "4. Water can flow from any cell adjacent to the ocean into the ocean. This means water can flow from cells on the edge of the island to the Pacific Ocean and the Atlantic Ocean.\n",
+    "\n",
+    "The task is to find and return a 2D list of grid coordinates where rainwater can flow from a cell to both the Pacific and Atlantic Oceans. In other words, you need to identify the cells that can send water to both the left and top edges (Pacific) and the right and bottom edges (Atlantic) of the island.\n",
+    "\n",
+    "The problem is solved by performing a depth-first search (DFS) starting from the ocean edges and marking the cells that can flow to each ocean. Then, the algorithm identifies cells that are reachable from both oceans and returns their coordinates as the result.\n",
+    "\n",
+    "The problem involves both traversal of the island and backtracking to find the solution, which makes it a classic graph traversal problem.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from typing import List\n",
+    "\n",
+    "class Solution:\n",
+    "    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:\n",
+    "        # Check if the input matrix is empty\n",
+    "        if not heights:\n",
+    "            return []\n",
+    "        \n",
+    "        m, n = len(heights), len(heights[0])\n",
+    "        \n",
+    "        # Initialize boolean matrices to keep track of cells reachable from each ocean\n",
+    "        pacific_reachable = [[False] * n for _ in range(m)]\n",
+    "        atlantic_reachable = [[False] * n for _ in range(m)]\n",
+    "        \n",
+    "        # Depth-First Search (DFS) function to mark cells that can be reached from an ocean\n",
+    "        def dfs(x, y, reachable):\n",
+    "            if reachable[x][y]:\n",
+    "                return\n",
+    "            reachable[x][y] = True\n",
+    "            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:\n",
+    "                new_x, new_y = x + dx, y + dy\n",
+    "                if 0 <= new_x < m and 0 <= new_y < n and heights[new_x][new_y] >= heights[x][y]:\n",
+    "                    dfs(new_x, new_y, reachable)\n",
+    "        \n",
+    "        # Start DFS from the Pacific Ocean (left and top edges)\n",
+    "        for i in range(m):\n",
+    "            dfs(i, 0, pacific_reachable)\n",
+    "            dfs(i, n - 1, atlantic_reachable)\n",
+    "        \n",
+    "        for j in range(n):\n",
+    "            dfs(0, j, pacific_reachable)\n",
+    "            dfs(m - 1, j, atlantic_reachable)\n",
+    "        \n",
+    "        result = []\n",
+    "        \n",
+    "        # Find cells that can flow to both the Pacific and Atlantic oceans\n",
+    "        for i in range(m):\n",
+    "            for j in range(n):\n",
+    "                if pacific_reachable[i][j] and atlantic_reachable[i][j]:\n",
+    "                    result.append([i, j])\n",
+    "        \n",
+    "        return result"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The code defines a class called `Solution` with a method named `pacificAtlantic`. This method takes a 2D matrix `heights`, where each element represents the height above sea level of a cell on a rectangular island. The goal of the method is to find and return a list of grid coordinates (cells) from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
+    "\n",
+    "The code uses a Depth-First Search (DFS) algorithm to traverse the island, starting from the ocean edges, and marks cells that are reachable from either the Pacific or Atlantic Ocean. It does this by creating two boolean matrices, `pacific_reachable` and `atlantic_reachable`, to keep track of cells that can be reached from each ocean.\n",
+    "\n",
+    "The code iterates through the entire island, initiating DFS searches from the edges of the island. If a cell is reachable from an ocean (either Pacific or Atlantic), it is marked as reachable in the corresponding boolean matrix. The DFS function ensures that the traversal follows the rule that water can flow to a neighboring cell if the neighboring cell's height is less than or equal to the current cell's height.\n",
+    "\n",
+    "After marking all reachable cells from both oceans, the code then looks for cells that are reachable from both the Pacific and Atlantic Oceans. If a cell satisfies this condition, it is added to the `result` list.\n",
+    "\n",
+    "Finally, the code returns the `result` list, which contains the grid coordinates of cells from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
+    "\n",
+    "The code provides a solution to the problem of finding cells on the island where water can reach both oceans, demonstrating a traversal algorithm that explores the connectivity of cells on the island."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1\n",
+    "heights1 = [\n",
+    "    [1,2,2,3,5],\n",
+    "    [3,2,3,4,4],\n",
+    "    [2,4,5,3,1],\n",
+    "    [6,7,1,4,5],\n",
+    "    [5,1,1,2,4]\n",
+    "]\n",
+    "\n",
+    "solution = Solution()\n",
+    "print(solution.pacificAtlantic(heights1)) # Expected Output: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[0, 0]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "heights2 = [[1]]\n",
+    "print(solution.pacificAtlantic(heights2)) # Expected Output: [[0, 0]]"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Time Complexity:\n",
+    "The time complexity of this solution is O(m * n), where 'm' is the number of rows in the input matrix and 'n' is the number of columns. Here's a breakdown of the time complexity:\n",
+    "\n",
+    "1. Creating the `pacific_reachable` and `atlantic_reachable` matrices with dimensions m x n takes O(m * n) time.\n",
+    "\n",
+    "2. Initiating Depth-First Search (DFS) from the edges of the island takes O(m * n) time because it processes each cell once.\n",
+    "\n",
+    "3. The DFS function visits each cell at most once, and its time complexity is O(1) for each cell. The number of cells visited by the DFS in the worst case is m * n.\n",
+    "\n",
+    "4. The final step of identifying and appending cells to the result list takes O(m * n) time in the worst case.\n",
+    "\n",
+    "Overall, the time complexity is dominated by the DFS and is O(m * n).\n",
+    "\n",
+    "Space Complexity:\n",
+    "The space complexity of this solution is also O(m * n). Here's how the space is used:\n",
+    "\n",
+    "1. Two boolean matrices, `pacific_reachable` and `atlantic_reachable`, are created with dimensions m x n. Each matrix takes O(m * n) space, resulting in O(2 * m * n) space usage.\n",
+    "\n",
+    "2. The depth-first search (DFS) stack space during the recursive calls can go as deep as the diagonal of the grid, which is at most min(m, n). This additional space for the call stack is relatively small compared to the boolean matrices and can be considered O(min(m, n)).\n",
+    "\n",
+    "3. The `result` list stores the grid coordinates, and its size is determined by the number of cells that can flow to both oceans. In the worst case, this list can contain all m * n cells, so it takes O(m * n) space.\n",
+    "\n",
+    "The overall space complexity is the sum of these components, which is O(m * n) + O(min(m, n)) + O(m * n). In big O notation, we can simplify this to O(m * n)."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Optimization Challenge:** Modify the algorithm to find the maximum flow of water from any cell to both the Pacific and Atlantic Oceans. In other words, find the cell(s) that can send the most water to both oceans.\n",
+    "\n",
+    "2. **Efficiency Challenge:** Optimize the time and space complexity of the algorithm while maintaining correctness. Aim to reduce both time and space complexity as much as possible."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb b/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb
new file mode 100644
index 0000000..19db846
--- /dev/null
+++ b/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb	
@@ -0,0 +1,250 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {
+    "jp-MarkdownHeadingCollapsed": true
+   },
+   "source": [
+    "# 892. Alien Dictionary\n",
+    "\n",
+    "**Difficulty:** Hard\n",
+    "\n",
+    "**Link to Problem:** [To see the Alien Dictionary problem on LintCode, click here!](https://www.lintcode.com/problem/892/)\n",
+    "\n",
+    "---\n",
+    "\n",
+    "There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of **non-empty** words from the dictionary, where words are **sorted lexicographically by the rules of this new language**. Derive the order of letters in this language.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- You may assume all letters are in lowercase.\n",
+    "- The dictionary is invalid, if string a is prefix of string b and b is appear before a.\n",
+    "- If the order is invalid, return an empty string.\n",
+    "- There may be multiple valid order of letters, return the smallest in normal lexicographical order.\n",
+    "- The letters in **one** string are of the same rank by default and are sorted in Human dictionary order."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem at hand involves determining the order of letters in a new alien language that uses the Latin alphabet. However, the order of letters in this alien language is unknown. To solve this problem, we are given a list of non-empty words from the dictionary, where the words are sorted lexicographically based on the rules of the alien language. The goal is to derive the correct order of letters in this new language.\n",
+    "\n",
+    "Here's a more detailed explanation of the problem:\n",
+    "\n",
+    "1. The alphabet: The new alien language uses the Latin alphabet, which consists of lowercase letters.\n",
+    "\n",
+    "2. Dictionary: We are provided with a list of non-empty words. These words are sorted in lexicographical order based on the rules of the alien language. The order of letters in the words reflects the correct order in the new language.\n",
+    "\n",
+    "3. Invalid order: The problem specifies that the dictionary is considered invalid if there exists a situation where one string 'a' is a prefix of another string 'b,' and 'b' appears before 'a' in the dictionary. This condition ensures that there are no inconsistencies in the order of letters.\n",
+    "\n",
+    "4. Multiple valid orders: It's possible that there are multiple valid orders of letters that satisfy the provided dictionary. However, we are instructed to return the smallest valid order in normal lexicographical order.\n",
+    "\n",
+    "To solve this problem, you need to analyze the given list of words and determine the correct order of letters in the alien language while adhering to the specified constraints. You should return the order as a string in lexicographical order if it's valid, and an empty string if the order is invalid.\n",
+    "\n",
+    "The key to solving this problem is to construct a directed graph of letter relationships based on the order information provided by the words in the dictionary. Then, perform a topological sort on this graph to derive the order of letters, ensuring that the order is consistent and valid according to the given constraints.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from collections import defaultdict, deque\n",
+    "\n",
+    "def alienOrder(words):\n",
+    "    # Create an adjacency list to represent the graph\n",
+    "    graph = defaultdict(list)\n",
+    "    \n",
+    "    # Create a dictionary to store the in-degrees of each letter\n",
+    "    in_degree = {}\n",
+    "    \n",
+    "    # Initialize in-degrees to 0 for all letters\n",
+    "    for word in words:\n",
+    "        for char in word:\n",
+    "            if char not in in_degree:\n",
+    "                in_degree[char] = 0\n",
+    "    \n",
+    "    # Build the graph and update in-degrees\n",
+    "    for i in range(1, len(words)):\n",
+    "        word1, word2 = words[i-1], words[i]\n",
+    "        min_length = min(len(word1), len(word2))\n",
+    "        \n",
+    "        # Compare characters in the two words\n",
+    "        j = 0\n",
+    "        while j < min_length and word1[j] == word2[j]:\n",
+    "            j += 1\n",
+    "        \n",
+    "        if j < min_length:\n",
+    "            # If word1[j] is lexicographically before word2[j], add an edge\n",
+    "            graph[word1[j]].append(word2[j])\n",
+    "            in_degree[word2[j]] += 1\n",
+    "            \n",
+    "            # Break the loop to avoid further comparisons\n",
+    "            break\n",
+    "    \n",
+    "    # Perform topological sorting using Kahn's algorithm\n",
+    "    result = []\n",
+    "    queue = deque([char for char in in_degree if in_degree[char] == 0])\n",
+    "    \n",
+    "    while queue:\n",
+    "        char = queue.popleft()\n",
+    "        result.append(char)\n",
+    "        \n",
+    "        for neighbor in graph[char]:\n",
+    "            in_degree[neighbor] -= 1\n",
+    "            if in_degree[neighbor] == 0:\n",
+    "                queue.append(neighbor)\n",
+    "    \n",
+    "    # Check if there is a valid order\n",
+    "    if len(result) < len(in_degree):\n",
+    "        return \"\"\n",
+    "    \n",
+    "    return \"\".join(result)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The code is designed to determine the order of letters in an unknown alien language that uses the Latin alphabet. It takes a list of non-empty words from a dictionary as input, where the words are sorted lexicographically based on the rules of this alien language.\n",
+    "\n",
+    "The code follows these main steps:\n",
+    "\n",
+    "1. Create a data structure to represent the graph of letters and their relationships in the alien language. It uses a defaultdict to store a list of letters that come after each letter. Additionally, it uses a dictionary called `in_degree` to keep track of the in-degrees (the number of letters that precede a letter) for each letter.\n",
+    "\n",
+    "2. Initialize the in-degrees of all letters to 0. This is done by iterating through all the words and characters in the words, adding each character to the `in_degree` dictionary with an initial in-degree of 0.\n",
+    "\n",
+    "3. Build the graph by comparing adjacent words in the dictionary. For each pair of adjacent words, the code finds the first differing character between the two words. If such a character exists, it means there's an order relationship between the letters represented by these characters. The code updates the graph and in-degrees accordingly.\n",
+    "\n",
+    "4. After constructing the graph, the code performs a topological sorting of the letters using Kahn's algorithm. It starts by initializing an empty result list and a queue containing letters with in-degrees of 0. The algorithm repeatedly removes a letter with an in-degree of 0 from the queue, adds it to the result list, and updates the in-degrees of its neighbors. This process continues until the queue is empty.\n",
+    "\n",
+    "5. Finally, the code checks if the topological sorting was successful by comparing the length of the result list with the number of unique letters in the input. If the lengths differ, it means there's a cycle in the graph, and the order is invalid. In such a case, an empty string is returned. Otherwise, the result list is joined together to form the smallest order of letters in normal lexicographical order, and it is returned as the output.\n",
+    "\n",
+    "The code provides a way to determine the order of letters in the alien language while handling various edge cases, including checking for invalid order conditions."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "wrtef\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1\n",
+    "words1 = [\"wrt\", \"wrf\", \"er\", \"ett\", \"rftt\"]\n",
+    "print(alienOrder(words1))  # Output: \"wertf\""
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "zx\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2\n",
+    "words2 = [\"z\", \"x\"]\n",
+    "print(alienOrder(words2))  # Output: \"zx\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code for determining the order of letters in an alien language.\n",
+    "\n",
+    "Time Complexity:\n",
+    "1. Building the graph: The code iterates through the list of words and compares adjacent words. In the worst case, where all words have the same prefix, it takes O(N) time to build the graph, where N is the total number of characters in all the words.\n",
+    "2. Topological sorting: Performing a topological sort on the graph takes O(V + E) time, where V is the number of unique letters (vertices) and E is the number of relationships between letters (edges) in the graph.\n",
+    "\n",
+    "Overall, the time complexity is O(N + V + E), where N is the total number of characters in the words, V is the number of unique letters, and E is the number of order relationships between letters.\n",
+    "\n",
+    "Space Complexity:\n",
+    "1. Graph and In-degrees: The code uses data structures to represent the graph and in-degrees. The space complexity for these data structures is O(V + E), where V is the number of unique letters, and E is the number of order relationships between letters in the words.\n",
+    "2. Result List: The result list stores the order of letters, which can have a maximum size of V, where V is the number of unique letters.\n",
+    "3. Queue: The space used by the queue during topological sorting is also O(V).\n",
+    "\n",
+    "Overall, the space complexity is O(V + E) due to the data structures used for graph representation and the space needed for storing the result and the queue.\n",
+    "\n",
+    "In the worst case, where there are many unique letters and many order relationships, the space complexity is dominated by the number of unique letters and their relationships.\n",
+    "\n",
+    "In summary, the time complexity of the code is O(N + V + E), and the space complexity is O(V + E). The actual performance will depend on the specific input data, such as the number of unique letters and the relationships between them in the alien language."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Multiple Valid Orders**: Extend the problem to handle cases where there are multiple valid orders of letters in the alien language. Modify the code to return all valid orders instead of just one. Be mindful of performance.\n",
+    "\n",
+    "2. **Detect Invalid Dictionary**: Given a dictionary of words, write a function to detect whether the dictionary is valid or not based on the constraints mentioned in the problem statement."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb b/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb
new file mode 100644
index 0000000..9539363
--- /dev/null
+++ b/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb	
@@ -0,0 +1,212 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# House Robber\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "---\n",
+    "You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and **it will automatically contact the police if two adjacent houses were broken into on the same night.**\n",
+    "\n",
+    "Given an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight **without alerting the police**.*\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `1 <= nums.length <= 100`\n",
+    "- `0 <= nums[i] <= 400`"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The problem is a classic dynamic programming challenge known as the \"House Robber\" problem. The scenario is that you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, but the constraint is that adjacent houses have security systems connected. If two adjacent houses are broken into on the same night, the security systems will automatically contact the police.\n",
+    "\n",
+    "The goal is to determine the maximum amount of money you can rob tonight without alerting the police. You need to find a strategy to maximize your earnings while avoiding robbing two adjacent houses.\n",
+    "\n",
+    "The input to the problem is an array `nums`, where `nums[i]` represents the amount of money in the `i`-th house. The task is to return the maximum amount of money that can be robbed without triggering the security system.\n",
+    "\n",
+    "For example:\n",
+    "```\n",
+    "Input: nums = [1,2,3,1]\n",
+    "Output: 4\n",
+    "Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).\n",
+    "Total amount you can rob = 1 + 3 = 4.\n",
+    "\n",
+    "Input: nums = [2,7,9,3,1]\n",
+    "Output: 12\n",
+    "Explanation: Rob house 1 (money = 2), rob house 3 (money = 9), and rob house 5 (money = 1).\n",
+    "Total amount you can rob = 2 + 9 + 1 = 12.\n",
+    "```\n",
+    "\n",
+    "The challenge involves designing an algorithm to efficiently solve this problem and find the optimal strategy for robbing houses to maximize the stolen money while adhering to the constraint of not robbing adjacent houses. The provided solution utilizes dynamic programming to achieve this.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from typing import List\n",
+    "\n",
+    "class Solution:\n",
+    "    def rob(self, nums: List[int]) -> int:\n",
+    "        # Initialize variables to keep track of the maximum amount of money robbed at the previous and current houses\n",
+    "        prev, curr = 0, 0\n",
+    "\n",
+    "        # Iterate through the houses\n",
+    "        for n in nums:\n",
+    "            # Calculate the maximum amount of money that can be robbed at the current house\n",
+    "            # It is the maximum of either robbing the current house and the money robbed at the house before the previous one (prev),\n",
+    "            # or skipping the current house and taking the maximum amount from the previous house (curr)\n",
+    "            temp = max(n + prev, curr)\n",
+    "\n",
+    "            # Update the variables for the next iteration\n",
+    "            prev, curr = curr, temp\n",
+    "\n",
+    "        # The final result is the maximum amount of money that can be robbed at the last house\n",
+    "        return curr"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The provided code is a Python implementation of the \"House Robber\" problem using a dynamic programming approach. It's part of a class `Solution` and includes a method `rob` that takes a list of integers `nums` as input and returns the maximum amount of money that can be robbed without alerting the police.\n",
+    "\n",
+    "The key idea of the algorithm is to maintain two variables, `prev` and `curr`, which represent the maximum amount of money that can be robbed up to the previous house and the current house, respectively. The algorithm iterates through the houses, updating these variables based on the maximum amount that can be robbed at the current house, considering the constraint of not robbing adjacent houses.\n",
+    "\n",
+    "The loop calculates a temporary variable `temp` that represents the maximum amount of money that can be robbed at the current house. It considers two scenarios: either robbing the current house and adding the money robbed at the house before the previous one (`n + prev`), or skipping the current house and taking the maximum amount from the previous house (`curr`). The maximum of these two scenarios is assigned to `temp`.\n",
+    "\n",
+    "The `prev` and `curr` variables are then updated for the next iteration. `prev` is set to the previous value of `curr`, and `curr` is set to the calculated `temp`.\n",
+    "\n",
+    "The final result is the value stored in the `curr` variable, representing the maximum amount of money that can be robbed at the last house. This value is returned as the result of the `rob` method.\n",
+    "\n",
+    "The provided test cases at the end demonstrate the usage of the `rob` method with different input arrays."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "sol = Solution()\n",
+    "\n",
+    "nums1 = [1, 2, 3, 1]\n",
+    "print(sol.rob(nums1))  # Output: 4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "12\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "nums2 = [2, 7, 9, 3, 1]\n",
+    "print(sol.rob(nums2))  # Output: 12"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code.\n",
+    "\n",
+    "### Time Complexity:\n",
+    "\n",
+    "The time complexity of the code is O(n), where n is the number of houses. The algorithm iterates through the array once, and in each iteration, it performs constant time operations. The loop runs for each house, and the number of operations inside the loop is independent of the input size. Therefore, the overall time complexity is linear, O(n).\n",
+    "\n",
+    "### Space Complexity:\n",
+    "\n",
+    "The space complexity is O(1), constant space. The algorithm uses only a constant amount of extra space regardless of the input size. The variables `prev`, `curr`, and `temp` are the only additional variables used, and they do not depend on the size of the input array. Thus, the space complexity is constant, O(1).\n",
+    "\n",
+    "In summary:\n",
+    "- Time Complexity: O(n)\n",
+    "- Space Complexity: O(1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Variation with Maximum Number of Houses to Rob:** Modify the problem to include an additional constraint: you are allowed to rob at most 'k' houses without triggering the alarm. Find the maximum amount you can rob under this new constraint.\n",
+    "\n",
+    "2. **Robbing in a Circular Street:** Extend the problem to a circular street where the first and last houses are adjacent. You cannot rob both the first and last houses simultaneously. Find the maximum amount you can rob in this circular setting.\n",
+    "\n",
+    "\n",
+    "**Link to Problem:** [To see the House Robber problem on LeetCode, click here!](https://leetcode.com/problems/house-robber/description/)\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb b/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb
new file mode 100644
index 0000000..d210017
--- /dev/null
+++ b/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb	
@@ -0,0 +1,243 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# House Robber II\n",
+    "**Difficulty:** Medium\n",
+    "\n",
+    "---\n",
+    "You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle**. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.\n",
+    "\n",
+    "Given an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight without alerting the police*.\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `1 <= nums.length <= 100`\n",
+    "- `0 <= nums[i] <= 1000`"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "Certainly! The problem statement describes a scenario where you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. However, there is a security system in place: adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.\n",
+    "\n",
+    "The twist in this problem is that the houses are arranged in a circle. That means the first house is the neighbor of the last one. So, robbing the first house has implications for the last house, and vice versa.\n",
+    "\n",
+    "The goal is to find the maximum amount of money you can rob tonight without alerting the police. You need to implement a function that takes an integer array `nums` representing the amount of money in each house and returns the maximum amount of money that can be robbed without triggering the security system.\n",
+    "\n",
+    "Here are a few examples to illustrate the problem:\n",
+    "\n",
+    "1. **Example 1:**\n",
+    "   ```python\n",
+    "   Input: nums = [2, 3, 2]\n",
+    "   Output: 3\n",
+    "   Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2) because they are adjacent houses. The optimal strategy is to rob house 2 (money = 3).\n",
+    "   ```\n",
+    "\n",
+    "2. **Example 2:**\n",
+    "   ```python\n",
+    "   Input: nums = [1, 2, 3, 1]\n",
+    "   Output: 4\n",
+    "   Explanation: Robbing house 1 (money = 1) and then robbing house 3 (money = 3) results in a total amount of 4. Robbing house 2 is skipped to avoid alerting the police.\n",
+    "   ```\n",
+    "\n",
+    "3. **Example 3:**\n",
+    "   ```python\n",
+    "   Input: nums = [1, 2, 3]\n",
+    "   Output: 3\n",
+    "   Explanation: Since the houses are in a circle, you can choose either house 1, 2, or 3. The optimal strategy is to rob house 3 for a total amount of 3.\n",
+    "   ```\n",
+    "\n",
+    "The constraints for the problem are that the length of the `nums` array is between 1 and 100, and the amount of money in each house (`nums[i]`) is between 0 and 1000.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from typing import List\n",
+    "\n",
+    "class Solution:\n",
+    "    def rob(self, nums: List[int]) -> int:\n",
+    "        def simple_rob(nums):\n",
+    "            # Helper function for basic house robbery problem\n",
+    "            prev, curr = 0, 0\n",
+    "            for num in nums:\n",
+    "                # Calculate the maximum amount of money that can be robbed\n",
+    "                # without alerting the police for each house\n",
+    "                prev, curr = curr, max(prev + num, curr)\n",
+    "            return curr\n",
+    "\n",
+    "        # Check the base cases where the length of nums is 1 or 2\n",
+    "        if len(nums) == 1:\n",
+    "            return nums[0]\n",
+    "        elif len(nums) == 2:\n",
+    "            return max(nums[0], nums[1])\n",
+    "\n",
+    "        # Case 1: First house is robbed, last house is not\n",
+    "        case1 = simple_rob(nums[:-1])\n",
+    "\n",
+    "        # Case 2: First house is not robbed, last house is\n",
+    "        case2 = simple_rob(nums[1:])\n",
+    "\n",
+    "        # Maximum of the two cases\n",
+    "        return max(case1, case2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The provided code defines a class `Solution` with a method `rob`, which aims to determine the maximum amount of money a professional robber can steal from houses arranged in a circle without alerting the police. Each house contains a certain amount of money, and adjacent houses have a security system that triggers if both are robbed on the same night.\n",
+    "\n",
+    "The code uses dynamic programming to solve the problem efficiently. It contains a helper function `simple_rob` that computes the maximum amount of money that can be robbed from a linear arrangement of houses. The main method `rob` handles the circular arrangement by considering two cases: robbing the first house and not robbing the last, and not robbing the first house and robbing the last. The result is the maximum of these two cases.\n",
+    "\n",
+    "The class is instantiated as `solution`, and the `rob` method is applied to three different scenarios (test cases) to demonstrate its functionality. The final output represents the maximum amount of money the robber can steal without triggering the security system."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "\n",
+    "# Create an instance of the Solution class\n",
+    "solution = Solution()\n",
+    "\n",
+    "nums1 = [2, 3, 2]\n",
+    "print(solution.rob(nums1))  # Output: 3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "4\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "nums2 = [1, 2, 3, 1]\n",
+    "print(solution.rob(nums2))  # Output: 4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "e6e1554f-ae77-410c-8af2-e59e5428166c",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "nums3 = [1, 2, 3]\n",
+    "print(solution.rob(nums3))  # Output: 3"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Let's analyze the time and space complexity of the provided code:\n",
+    "\n",
+    "### Time Complexity:\n",
+    "The time complexity is primarily determined by the `simple_rob` helper function, which has a time complexity of O(N), where N is the number of houses in the input array. The main `rob` function calls `simple_rob` twice, once for the case of robbing the first house and not robbing the last, and once for the case of not robbing the first house and robbing the last. Therefore, the overall time complexity is O(N).\n",
+    "\n",
+    "### Space Complexity:\n",
+    "The space complexity is O(1) because the space used by the algorithm is constant, regardless of the size of the input array. The only variables used are `prev` and `curr` in the `simple_rob` function, which represent the previous and current maximum amounts of money that can be robbed without alerting the police. These variables are updated iteratively, and no additional data structures are used that scale with the input size. Therefore, the space complexity is constant, or O(1).\n",
+    "\n",
+    "In summary:\n",
+    "- Time Complexity: O(N)\n",
+    "- Space Complexity: O(1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Optimization Challenge:** Optimize the space complexity of the solution. Can you solve the problem with O(1) space complexity without sacrificing the time complexity?\n",
+    "\n",
+    "2. **Circular Robbery Variations:** Consider variations of the circular house robbery problem, such as allowing the robber to skip a certain number of houses or restricting the number of consecutive houses that can be robbed. Modify the solution to accommodate these variations.\n",
+    "\n",
+    "**Link to Problem:** [To see the House Robber II problem on LeetCode, click here!](https://leetcode.com/problems/house-robber-ii/)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb b/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb
new file mode 100644
index 0000000..32c5da3
--- /dev/null
+++ b/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb	
@@ -0,0 +1,204 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
+   "metadata": {},
+   "source": [
+    "# 70. Climbing Stairs\n",
+    "\n",
+    "**Difficulty:** Easy\n",
+    "\n",
+    "**Link to Problem:** [To see the Climbing Stairs problem on LeetCode, click here!](https://leetcode.com/problems/climbing-stairs/)\n",
+    "\n",
+    "---\n",
+    "You are climbing a staircase. It takes `n` steps to reach the top.\n",
+    "\n",
+    "Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?\n",
+    "\n",
+    "**Constraints:**\n",
+    "- `1 <= n <= 45`"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
+   "metadata": {},
+   "source": [
+    "## Probelm Explanation:\n",
+    "\n",
+    "The \"Climbing Stairs\" problem is a classic algorithmic problem that asks you to find the number of distinct ways to climb a staircase with a given number of steps. Here's the problem statement:\n",
+    "\n",
+    "You are given a staircase with 'n' steps. Each time you can either climb 1 step or 2 steps. You need to find out how many distinct ways there are to reach the top of the staircase.\n",
+    "\n",
+    "For example, if 'n' is 2, there are two ways to climb to the top:\n",
+    "\n",
+    "1. Climbing two steps at once.\n",
+    "2. Climbing one step, and then another step.\n",
+    "\n",
+    "If 'n' is 3, there are three ways:\n",
+    "\n",
+    "1. Climbing one step, then one step, and finally one step.\n",
+    "2. Climbing one step, then two steps.\n",
+    "3. Climbing two steps, then one step.\n",
+    "\n",
+    "The problem essentially asks you to find a solution that counts the possible combinations of 1-step and 2-step climbs that can be taken to reach the top of the staircase. It's a classic example of a dynamic programming problem where you can break it down into smaller subproblems and build a solution incrementally, as shown in the code I provided earlier.\n",
+    "\n",
+    "## Solution:\n",
+    "Here's a Python function to implement this algorithm:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "class Solution:\n",
+    "    def climbStairs(self, n: int) -> int:\n",
+    "        # If there are 1 or 2 steps, there are exactly n ways to reach the top.\n",
+    "        if n <= 2:\n",
+    "            return n\n",
+    "\n",
+    "        # Create an array to store the number of ways to reach each step.\n",
+    "        dp = [0] * (n + 1)\n",
+    "\n",
+    "        # There is one way to reach the first step (base case).\n",
+    "        dp[1] = 1\n",
+    "\n",
+    "        # There are two ways to reach the second step (base case).\n",
+    "        dp[2] = 2\n",
+    "\n",
+    "        # Calculate the number of ways to reach each step from the previous steps.\n",
+    "        for i in range(3, n + 1):\n",
+    "            # The number of ways to reach step 'i' is the sum of the ways to reach\n",
+    "            # the previous two steps, as you can take 1 or 2 steps at a time.\n",
+    "            dp[i] = dp[i - 1] + dp[i - 2]\n",
+    "\n",
+    "        # The result is the number of ways to reach the top, which is stored in dp[n].\n",
+    "        return dp[n]"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
+   "metadata": {},
+   "source": [
+    "## Explanation:\n",
+    "\n",
+    "The code defines a class `Solution` with a method `climbStairs` that takes an integer `n` as input and returns the number of distinct ways to climb a staircase with `n` steps.\n",
+    "\n",
+    "1. If `n` is 1 or 2, the function returns `n` directly because there are only 1 or 2 distinct ways to reach the top in those cases.\n",
+    "\n",
+    "2. For `n` greater than 2, the code uses dynamic programming to calculate the number of ways to reach each step. It creates an array `dp` of length `n + 1` to store these values.\n",
+    "\n",
+    "3. The `dp[1]` and `dp[2]` are initialized to 1 and 2, respectively. This is because there is only one way to reach the first step (by taking 1 step), and there are two ways to reach the second step (by taking 2 steps or two 1-step climbs).\n",
+    "\n",
+    "4. The code then uses a loop to fill in the `dp` array for steps 3 and above. For each step `i` from 3 to `n`, it calculates the number of ways to reach that step by adding the number of ways to reach the previous two steps, as you can either take 1 step or 2 steps at a time.\n",
+    "\n",
+    "5. Finally, the function returns the value stored in `dp[n]`, which represents the total number of distinct ways to climb the staircase with `n` steps."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
+   "metadata": {},
+   "source": [
+    "## Test cases:\n",
+    "\n",
+    "Here's how you can use this solution:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1 - Input: n = 2, Output: 2\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 1:\n",
+    "n = 2\n",
+    "solution = Solution()\n",
+    "print(f\"Example 1 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 2 - Input: n = 3, Output: 3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Example 2:\n",
+    "n = 3\n",
+    "print(f\"Example 2 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 3\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
+   "metadata": {},
+   "source": [
+    "## Time and Space Complexity Analysis\n",
+    "\n",
+    "Time Complexity:\n",
+    "The time complexity of the solution is O(n) because we use a single for loop that iterates from 3 to n to fill the `dp` array. In each iteration, we perform constant time operations (addition and assignment). Therefore, the time complexity is linear with respect to the input value `n`.\n",
+    "\n",
+    "Space Complexity:\n",
+    "The space complexity of the solution is O(n) as well. We use an array `dp` of length `n + 1` to store the number of ways to reach each step. The size of this array is directly proportional to the input value `n`, which makes the space complexity linear in terms of `n`. The rest of the variables used in the function (e.g., `i`) occupy constant space and do not contribute significantly to the overall space complexity.\n",
+    "\n",
+    "In summary, the time complexity is O(n), and the space complexity is O(n) for this dynamic programming solution to the \"Climbing Stairs\" problem."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
+   "metadata": {},
+   "source": [
+    "## Challenging Exercises:\n",
+    "\n",
+    "1. **Generalized Step Sizes**: Instead of being limited to 1 step or 2 steps, consider a scenario where you can take steps of sizes `[1, 2, 3]`. Implement a function that calculates the number of ways to climb the staircase with these different step sizes.\n",
+    "\n",
+    "2. **Minimize Steps**: Given an integer `n`, find the minimum number of steps required to reach the top of the staircase. Return both the minimum number of steps and the specific sequence of steps taken."
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python (Pyodide)",
+   "language": "python",
+   "name": "python"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
\ No newline at end of file
diff --git a/_build/jupyter_execute/Book.ipynb b/_build/jupyter_execute/Book.ipynb
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@@ -0,0 +1,914 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: pip in /usr/local/python/3.10.8/lib/python3.10/site-packages (23.3.1)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: jupyter-book in /usr/local/python/3.10.8/lib/python3.10/site-packages (0.15.1)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: click<9,>=7.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (8.1.7)\r\n",
+      "Requirement already satisfied: docutils<0.19,>=0.15 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.18.1)\r\n",
+      "Requirement already satisfied: Jinja2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (3.1.2)\r\n",
+      "Requirement already satisfied: jsonschema<5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (4.19.1)\r\n",
+      "Requirement already satisfied: linkify-it-py~=2.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (2.0.2)\r\n",
+      "Requirement already satisfied: myst-nb~=0.17.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.17.2)\r\n",
+      "Requirement already satisfied: pyyaml in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (6.0.1)\r\n",
+      "Requirement already satisfied: sphinx<6,>=4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (5.0.2)\r\n",
+      "Requirement already satisfied: sphinx-comments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.0.3)\r\n",
+      "Requirement already satisfied: sphinx-copybutton in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.5.2)\r\n",
+      "Requirement already satisfied: sphinx-external-toc~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.1)\r\n",
+      "Requirement already satisfied: sphinx-jupyterbook-latex~=0.5.2 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.5.2)\r\n",
+      "Requirement already satisfied: sphinx-design~=0.3.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.0)\r\n",
+      "Requirement already satisfied: sphinx-thebe~=0.2.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.2.1)\r\n",
+      "Requirement already satisfied: sphinx-book-theme~=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (1.0.1)\r\n",
+      "Requirement already satisfied: sphinx_togglebutton in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.2)\r\n",
+      "Requirement already satisfied: sphinxcontrib-bibtex<=2.5.0,>=2.2.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (2.5.0)\r\n",
+      "Requirement already satisfied: sphinx-multitoc-numbering~=0.1.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.1.3)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: attrs>=22.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (23.1.0)\r\n",
+      "Requirement already satisfied: jsonschema-specifications>=2023.03.6 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (2023.7.1)\r\n",
+      "Requirement already satisfied: referencing>=0.28.4 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (0.30.2)\r\n",
+      "Requirement already satisfied: rpds-py>=0.7.1 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (0.10.6)\r\n",
+      "Requirement already satisfied: uc-micro-py in /usr/local/python/3.10.8/lib/python3.10/site-packages (from linkify-it-py~=2.0.0->jupyter-book) (1.0.2)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: importlib_metadata in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (6.8.0)\r\n",
+      "Requirement already satisfied: ipython in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (8.16.1)\r\n",
+      "Requirement already satisfied: jupyter-cache<0.7,>=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.6.1)\r\n",
+      "Requirement already satisfied: nbclient in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.7.4)\r\n",
+      "Requirement already satisfied: myst-parser~=0.18.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.18.1)\r\n",
+      "Requirement already satisfied: nbformat~=5.0 in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (5.9.2)\r\n",
+      "Requirement already satisfied: typing-extensions in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (4.8.0)\r\n",
+      "Requirement already satisfied: ipykernel in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (6.25.2)\r\n",
+      "Requirement already satisfied: sphinxcontrib-applehelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.0.7)\r\n",
+      "Requirement already satisfied: sphinxcontrib-devhelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.0.5)\r\n",
+      "Requirement already satisfied: sphinxcontrib-jsmath in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.0.1)\r\n",
+      "Requirement already satisfied: sphinxcontrib-htmlhelp>=2.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.0.4)\r\n",
+      "Requirement already satisfied: sphinxcontrib-serializinghtml>=1.1.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.1.9)\r\n",
+      "Requirement already satisfied: sphinxcontrib-qthelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.0.6)\r\n",
+      "Requirement already satisfied: Pygments>=2.0 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.16.1)\r\n",
+      "Requirement already satisfied: snowballstemmer>=1.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.2.0)\r\n",
+      "Requirement already satisfied: babel>=1.3 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.13.0)\r\n",
+      "Requirement already satisfied: alabaster<0.8,>=0.7 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (0.7.13)\r\n",
+      "Requirement already satisfied: imagesize in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.4.1)\r\n",
+      "Requirement already satisfied: requests>=2.5.0 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.31.0)\r\n",
+      "Requirement already satisfied: packaging in /home/codespace/.local/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (23.2)\r\n",
+      "Requirement already satisfied: MarkupSafe>=2.0 in /home/codespace/.local/lib/python3.10/site-packages (from Jinja2->jupyter-book) (2.1.3)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: pydata-sphinx-theme>=0.13.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx-book-theme~=1.0.0->jupyter-book) (0.14.3)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: pybtex>=0.24 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (0.24.0)\r\n",
+      "Requirement already satisfied: pybtex-docutils>=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (1.0.3)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: setuptools in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx_togglebutton->jupyter-book) (68.0.0)\r\n",
+      "Requirement already satisfied: wheel in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx_togglebutton->jupyter-book) (0.41.3)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: sqlalchemy<3,>=1.3.12 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (2.0.23)\r\n",
+      "Requirement already satisfied: tabulate in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (0.9.0)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: markdown-it-py<3.0.0,>=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (2.2.0)\r\n",
+      "Requirement already satisfied: mdit-py-plugins~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.3.5)\r\n",
+      "Requirement already satisfied: jupyter-client>=6.1.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (8.4.0)\r\n",
+      "Requirement already satisfied: jupyter-core!=5.0.*,>=4.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (5.4.0)\r\n",
+      "Requirement already satisfied: traitlets>=5.3 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (5.11.2)\r\n",
+      "Requirement already satisfied: fastjsonschema in /home/codespace/.local/lib/python3.10/site-packages (from nbformat~=5.0->myst-nb~=0.17.1->jupyter-book) (2.18.1)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: latexcodec>=1.0.4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pybtex>=0.24->sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (2.0.1)\r\n",
+      "Requirement already satisfied: six in /home/codespace/.local/lib/python3.10/site-packages (from pybtex>=0.24->sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (1.16.0)\r\n",
+      "Requirement already satisfied: beautifulsoup4 in /home/codespace/.local/lib/python3.10/site-packages (from pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (4.12.2)\r\n",
+      "Requirement already satisfied: accessible-pygments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (0.0.4)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: charset-normalizer<4,>=2 in /home/codespace/.local/lib/python3.10/site-packages (from requests>=2.5.0->sphinx<6,>=4->jupyter-book) (3.3.0)\r\n",
+      "Requirement already satisfied: idna<4,>=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from requests>=2.5.0->sphinx<6,>=4->jupyter-book) (3.4)\r\n",
+      "Requirement already satisfied: urllib3<3,>=1.21.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from requests>=2.5.0->sphinx<6,>=4->jupyter-book) (1.26.18)\r\n",
+      "Requirement already satisfied: certifi>=2017.4.17 in /home/codespace/.local/lib/python3.10/site-packages (from requests>=2.5.0->sphinx<6,>=4->jupyter-book) (2023.7.22)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: zipp>=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from importlib_metadata->myst-nb~=0.17.1->jupyter-book) (3.17.0)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: comm>=0.1.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (0.1.4)\r\n",
+      "Requirement already satisfied: debugpy>=1.6.5 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (1.8.0)\r\n",
+      "Requirement already satisfied: matplotlib-inline>=0.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (0.1.6)\r\n",
+      "Requirement already satisfied: nest-asyncio in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (1.5.8)\r\n",
+      "Requirement already satisfied: psutil in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (5.9.6)\r\n",
+      "Requirement already satisfied: pyzmq>=20 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (25.1.1)\r\n",
+      "Requirement already satisfied: tornado>=6.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (6.3.3)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: backcall in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.2.0)\r\n",
+      "Requirement already satisfied: decorator in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (5.1.1)\r\n",
+      "Requirement already satisfied: jedi>=0.16 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.19.1)\r\n",
+      "Requirement already satisfied: pickleshare in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.7.5)\r\n",
+      "Requirement already satisfied: prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (3.0.39)\r\n",
+      "Requirement already satisfied: stack-data in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.6.3)\r\n",
+      "Requirement already satisfied: exceptiongroup in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (1.1.3)\r\n",
+      "Requirement already satisfied: pexpect>4.3 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (4.8.0)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: parso<0.9.0,>=0.8.3 in /home/codespace/.local/lib/python3.10/site-packages (from jedi>=0.16->ipython->myst-nb~=0.17.1->jupyter-book) (0.8.3)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: python-dateutil>=2.8.2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-client>=6.1.12->nbclient->myst-nb~=0.17.1->jupyter-book) (2.8.2)\r\n",
+      "Requirement already satisfied: platformdirs>=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-core!=5.0.*,>=4.12->nbclient->myst-nb~=0.17.1->jupyter-book) (3.11.0)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py<3.0.0,>=1.0.0->myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.1.2)\r\n",
+      "Requirement already satisfied: ptyprocess>=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect>4.3->ipython->myst-nb~=0.17.1->jupyter-book) (0.7.0)\r\n",
+      "Requirement already satisfied: wcwidth in /home/codespace/.local/lib/python3.10/site-packages (from prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.8)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: greenlet!=0.4.17 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sqlalchemy<3,>=1.3.12->jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (3.0.1)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: soupsieve>1.2 in /home/codespace/.local/lib/python3.10/site-packages (from beautifulsoup4->pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (2.5)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: executing>=1.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.0.0)\r\n",
+      "Requirement already satisfied: asttokens>=2.1.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.4.0)\r\n",
+      "Requirement already satisfied: pure-eval in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.2)\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Note: you may need to restart the kernel to use updated packages.\n"
+     ]
+    }
+   ],
+   "source": [
+    "!python -m pip install --upgrade pip\n",
+    "%pip install jupyter-book"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions\n"
+     ]
+    }
+   ],
+   "source": [
+    "%cd /workspaces/LeetCode-Solutions"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Traceback (most recent call last):\r\n",
+      "  File \"/home/codespace/.python/current/bin/jupyter-book\", line 8, in <module>\r\n",
+      "    sys.exit(main())\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1157, in __call__\r\n",
+      "    return self.main(*args, **kwargs)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1078, in main\r\n",
+      "    rv = self.invoke(ctx)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
+      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
+      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1434, in invoke\r\n",
+      "    return ctx.invoke(self.callback, **ctx.params)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 783, in invoke\r\n",
+      "    return __callback(*args, **kwargs)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx_external_toc/cli.py\", line 139, in migrate_toc\r\n",
+      "    toc = migrate_jupyter_book(Path(toc_file))\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx_external_toc/tools.py\", line 327, in migrate_jupyter_book\r\n",
+      "    raise MalformedError(\"no top-level 'file' key found\")\r\n",
+      "sphinx_external_toc.parsing.MalformedError: no top-level 'file' key found\r\n"
+     ]
+    }
+   ],
+   "source": [
+    "!jupyter-book toc migrate /workspaces/LeetCode-Solutions/_toc.yml -o /workspaces/LeetCode-Solutions/_toc.yml"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions\r\n",
+      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\r\n",
+      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\r\n",
+      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[etoc] Changing master_doc to 'README'\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\r\n",
+      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\r\n",
+      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\r\n",
+      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 77 source files that are out of date\r\n",
+      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 77 added, 0 changed, 0 removed\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[  1%] \u001b[35m01. Array Hashing/1. Two Sum\u001b[39;49;00m                          \r"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.07 seconds [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mreading sources... \u001b[39;49;00m[  2%] \u001b[35m01. Array Hashing/128. Longest Consecutive Sequence\u001b[39;49;00m   \r",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.17 seconds [mystnb]\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[  3%] \u001b[35m01. Array Hashing/217. Contains Duplicate\u001b[39;49;00m             \r",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.19 seconds [mystnb]\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[  5%] \u001b[35m01. Array Hashing/238. Product of Array Except Self\u001b[39;49;00m   \r",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 1.26 seconds [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mreading sources... \u001b[39;49;00m[  6%] \u001b[35m01. Array Hashing/242. Valid Anagram\u001b[39;49;00m                  \r",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 1.18 seconds [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mreading sources... \u001b[39;49;00m[  7%] \u001b[35m01. Array Hashing/347. Top K Frequent Elements\u001b[39;49;00m        \r",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.13 seconds [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mreading sources... \u001b[39;49;00m[  9%] \u001b[35m01. Array Hashing/49. Group Anagrams\u001b[39;49;00m                  \r",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 4.02 seconds [mystnb]\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 10%] \u001b[35m01. Array Hashing/659. Encode and Decode Strings\u001b[39;49;00m      \r"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 11%] \u001b[35m01. Array Hashing/README\u001b[39;49;00m                              \r",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 12%] \u001b[35m02. Two Pointers/11. Container With Most Water\u001b[39;49;00m        \r",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.26 seconds [mystnb]\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 14%] \u001b[35m02. Two Pointers/125. Valid Palindrome\u001b[39;49;00m                \r",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 1.07 seconds [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 15%] \u001b[35m02. Two Pointers/15. 3Sum\u001b[39;49;00m                             \r",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.00 seconds [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 16%] \u001b[35m02. Two Pointers/README\u001b[39;49;00m                               \r",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 18%] \u001b[35m03. Sliding Window/121. Best Time to Buy and Sell Stock\u001b[39;49;00m\r"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.89 seconds [mystnb]\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 19%] \u001b[35m03. Sliding Window/3. Longest Substring Without Repeating Characters\u001b[39;49;00m\r"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 1.16 seconds [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 20%] \u001b[35m03. Sliding Window/424. Longest Repeating Character Replacement\u001b[39;49;00m\r",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 22%] \u001b[35m03. Sliding Window/76. Minimum Window Substring\u001b[39;49;00m       \r",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 0.87 seconds [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 23%] \u001b[35m03. Sliding Window/README\u001b[39;49;00m                             \r",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 24%] \u001b[35m04. Stack/20. Valid Parentheses\u001b[39;49;00m                       \r",
+      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 4.02 seconds [mystnb]\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 25%] \u001b[35m04. Stack/README\u001b[39;49;00m                                      \r",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 27%] \u001b[35m05. Binary Search/153. Find Minimum in Rotated Sorted Array\u001b[39;49;00m\r"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 0.98 seconds [mystnb]\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 28%] \u001b[35m05. Binary Search/33. Search in Rotated Sorted Array\u001b[39;49;00m  \r"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 29%] \u001b[35m05. Binary Search/README\u001b[39;49;00m                              \r"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 31%] \u001b[35m06. Linked List/141. Linked List Cycle\u001b[39;49;00m                \r",
+      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 0.96 seconds [mystnb]\r\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[ 32%] \u001b[35m06. Linked List/143. Reorder List\u001b[39;49;00m                     \r"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]\r\n"
+     ]
+    },
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "^C\r\n",
+      "\r\n",
+      "\u001b[91mException occurred:\u001b[39;49;00m\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 512, in _async_cleanup_kernel\r\n",
+      "    await ensure_async(self.km.cleanup_resources())\r\n",
+      "AttributeError: 'NoneType' object has no attribute 'cleanup_resources'\r\n",
+      "\u001b[91mThe full traceback has been saved in /tmp/sphinx-err-tl19fjln.log, if you want to report the issue to the developers.\u001b[39;49;00m\r\n",
+      "Please also report this if it was a user error, so that a better error message can be provided next time.\r\n",
+      "A bug report can be filed in the tracker at <https://github.com/sphinx-doc/sphinx/issues>. Thanks!\r\n",
+      "Traceback (most recent call last):\r\n",
+      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py\", line 89, in wrapper\r\n",
+      "    self._ready.set_result(None)\r\n",
+      "asyncio.exceptions.InvalidStateError: invalid state\r\n",
+      "\r\n",
+      "During handling of the above exception, another exception occurred:\r\n",
+      "\r\n",
+      "Traceback (most recent call last):\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 505, in _async_cleanup_kernel\r\n",
+      "    await ensure_async(self.km.shutdown_kernel(now=now))\r\n",
+      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py\", line 189, in ensure_async\r\n",
+      "    result = await obj\r\n",
+      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py\", line 92, in wrapper\r\n",
+      "    self._ready.set_exception(e)\r\n",
+      "asyncio.exceptions.InvalidStateError: invalid state\r\n",
+      "\r\n",
+      "During handling of the above exception, another exception occurred:\r\n",
+      "\r\n",
+      "Traceback (most recent call last):\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/sphinx.py\", line 171, in build_sphinx\r\n",
+      "    app.build(force_all, filenames)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/application.py\", line 329, in build\r\n",
+      "    self.builder.build_update()\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 288, in build_update\r\n",
+      "    self.build(to_build,\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 302, in build\r\n",
+      "    updated_docnames = set(self.read())\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 409, in read\r\n",
+      "    self._read_serial(docnames)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 430, in _read_serial\r\n",
+      "    self.read_doc(docname)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 483, in read_doc\r\n",
+      "    publisher.publish()\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/core.py\", line 217, in publish\r\n",
+      "    self.document = self.reader.read(self.source, self.parser,\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/io.py\", line 103, in read\r\n",
+      "    self.parse()\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/readers/__init__.py\", line 78, in parse\r\n",
+      "    self.parser.parse(self.input, document)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/sphinx_.py\", line 150, in parse\r\n",
+      "    with create_client(\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/base.py\", line 83, in __enter__\r\n",
+      "    self.start_client()\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/direct.py\", line 40, in start_client\r\n",
+      "    result = single_nb_execution(\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_cache/executors/utils.py\", line 58, in single_nb_execution\r\n",
+      "    executenb(\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 1305, in execute\r\n",
+      "    return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute()\r\n",
+      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py\", line 173, in wrapped\r\n",
+      "    return loop.run_until_complete(inner)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/asyncio/base_events.py\", line 649, in run_until_complete\r\n",
+      "    return future.result()\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 689, in async_execute\r\n",
+      "    async with self.async_setup_kernel(**kwargs):\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/contextlib.py\", line 206, in __aexit__\r\n",
+      "    await anext(self.gen)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 656, in async_setup_kernel\r\n",
+      "    await self._async_cleanup_kernel()\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 512, in _async_cleanup_kernel\r\n",
+      "    await ensure_async(self.km.cleanup_resources())\r\n",
+      "AttributeError: 'NoneType' object has no attribute 'cleanup_resources'\r\n",
+      "\r\n",
+      "The above exception was the direct cause of the following exception:\r\n",
+      "\r\n",
+      "Traceback (most recent call last):\r\n",
+      "  File \"/home/codespace/.python/current/bin/jupyter-book\", line 8, in <module>\r\n",
+      "    sys.exit(main())\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1157, in __call__\r\n",
+      "    return self.main(*args, **kwargs)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1078, in main\r\n",
+      "    rv = self.invoke(ctx)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
+      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1434, in invoke\r\n",
+      "    return ctx.invoke(self.callback, **ctx.params)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 783, in invoke\r\n",
+      "    return __callback(*args, **kwargs)\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py\", line 317, in build\r\n",
+      "    builder_specific_actions(\r\n",
+      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py\", line 525, in builder_specific_actions\r\n",
+      "    raise RuntimeError(_message_box(msg, color=\"red\", doprint=False)) from result\r\n",
+      "RuntimeError: \r\n",
+      "\u001b[91m===============================================================================\u001b[0m\r\n",
+      "\r\n",
+      "There was an error in building your book. Look above for the cause.\r\n",
+      "\r\n",
+      "\u001b[91m===============================================================================\u001b[0m\r\n",
+      "\r\n"
+     ]
+    }
+   ],
+   "source": [
+    "!jupyter-book build ."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "!jupyter-book build /workspaces/LeetCode-Solutions/ --builder pdfhtml"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/html": [
+       "<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>"
+      ],
+      "text/plain": [
+       "<IPython.core.display.HTML object>"
+      ]
+     },
+     "execution_count": 5,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "from IPython.display import HTML\n",
+    "HTML('<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>')\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 40,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "\u001b[92m===============================================================================\u001b[0m\n",
+      "\n",
+      "Your book template can be found at\n",
+      "\n",
+      "    mybook/\n",
+      "\n",
+      "\u001b[92m===============================================================================\u001b[0m\n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "!jupyter-book create mybook"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 45,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
+      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions/mybook\n",
+      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_config.yml\n",
+      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_build/html\n",
+      "[sphinxcontrib-bibtex] Beware that docutils versions 0.18 and 0.19 (you are running 0.18.1) are known to generate invalid html for citations. If this issue affects you, please use docutils<0.18 (or >=0.20 once released) instead. For more details, see https://sourceforge.net/p/docutils/patches/195/\n",
+      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
+      "[etoc] Changing master_doc to 'intro'\n",
+      "checking bibtex cache... out of date\n",
+      "parsing bibtex file /workspaces/LeetCode-Solutions/mybook/references.bib... parsed 5 entries\n",
+      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
+      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='force', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
+      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/mybook/_build/.jupyter_cache\n",
+      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
+      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 14 source files that are out of date\n",
+      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 14 added, 0 changed, 0 removed\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.35 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.97 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 0.83 seconds [mystnb]\n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mintro\u001b[39;49;00m                                                 \n",
+      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
+      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
+      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "done\n",
+      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
+      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mintro\u001b[39;49;00m                                                  \n",
+      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
+      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
+      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
+      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
+      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
+      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
+      "[etoc] missing index.html written as redirect to 'intro.html'\n",
+      "\u001b[01mbuild succeeded, 2 warnings.\u001b[39;49;00m\n",
+      "\n",
+      "The HTML pages are in mybook/_build/html.\n",
+      "\n",
+      "\u001b[92m===============================================================================\u001b[0m\n",
+      "\n",
+      "Finished generating HTML for book.\n",
+      "Your book's HTML pages are here:\n",
+      "    mybook/_build/html/\n",
+      "You can look at your book by opening this file in a browser:\n",
+      "    mybook/_build/html/index.html\n",
+      "Or paste this line directly into your browser bar:\n",
+      "    file:///workspaces/LeetCode-Solutions/mybook/_build/html/index.html            \n",
+      "\n",
+      "\u001b[92m===============================================================================\u001b[0m\n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "!jupyter-book build mybook"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 31,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "format: jb-book\n",
+      "root: intro\n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "!jupyter-book toc from-project /workspaces/LeetCode-Solutions/mybook/ -f jb-book"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.10.8"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
\ No newline at end of file
diff --git a/_config.yml b/_config.yml
new file mode 100644
index 0000000..3def3c7
--- /dev/null
+++ b/_config.yml
@@ -0,0 +1,31 @@
+# Jupyter Book configuration
+
+# Project information
+title: "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges" # Python Problem-Solving Pro: Conquering the Blind 75 LeetCode Challenges
+author: Mohsen Tabibian and Saeed Samadidana
+email: mohsentabibian@gmail.com
+description: Explore 75 essential problems frequently asked in tech interviews. Each solution includes detailed explanations and time/space complexity analysis. Boost your coding skills and interview preparedness!
+
+repository: 
+   url: https://github.com/mohsentabibian/LeetCode-Solutions
+   path: docs #LeetCode-Solutions/
+   branch: main
+
+# Execute notebooks when building (set to false if you don't want to run the code)
+execute:
+  execute_notebooks: auto
+
+# HTML settings
+html:
+  theme: jekyll-theme-Architect #jupyter
+  logo: logo.png
+  toc:
+    display: true
+  use_issues_button: true
+  use_repository_button: true
+
+#Build settings
+#repository_url: https://github.com/mohsentabibian/LeetCode-Solutions  # URL to the GitHub repository
+
+# Add GitHub buttons to the book
+# See https://jupyterbook.org/customize/config.html #add-a-link-to-your-repository
\ No newline at end of file
diff --git a/_toc.yml b/_toc.yml
new file mode 100644
index 0000000..0e4e1db
--- /dev/null
+++ b/_toc.yml
@@ -0,0 +1,88 @@
+format: jb-book
+root: README
+chapters:
+  - file: 01. Array Hashing/README
+    sections:
+      - file: 01. Array Hashing/1. Two Sum
+      - file: 01. Array Hashing/128. Longest Consecutive Sequence
+      - file: 01. Array Hashing/217. Contains Duplicate
+      - file: 01. Array Hashing/238. Product of Array Except Self
+      - file: 01. Array Hashing/242. Valid Anagram
+      - file: 01. Array Hashing/347. Top K Frequent Elements
+      - file: 01. Array Hashing/49. Group Anagrams
+      - file: 01. Array Hashing/659. Encode and Decode Strings
+  - file: 02. Two Pointers/README
+    sections:
+      - file: 02. Two Pointers/11. Container With Most Water
+      - file: 02. Two Pointers/125. Valid Palindrome
+      - file: 02. Two Pointers/15. 3Sum
+  - file: 03. Sliding Window/README
+    sections:
+      - file: 03. Sliding Window/121. Best Time to Buy and Sell Stock
+      - file: 03. Sliding Window/3. Longest Substring Without Repeating Characters
+      - file: 03. Sliding Window/424. Longest Repeating Character Replacement
+      - file: 03. Sliding Window/76. Minimum Window Substring
+  - file: 04. Stack/README
+    sections:
+      - file: 04. Stack/20. Valid Parentheses
+  - file: 05. Binary Search/README
+    sections:
+      - file: 05. Binary Search/153. Find Minimum in Rotated Sorted Array
+      - file: 05. Binary Search/33. Search in Rotated Sorted Array
+  - file: 06. Linked List/README
+    sections:
+      - file: 06. Linked List/141. Linked List Cycle
+      - file: 06. Linked List/143. Reorder List
+      - file: 06. Linked List/19. Remove Nth Node From End of List
+      - file: 06. Linked List/206. Reverse Linked List
+      - file: 06. Linked List/21. Merge Two Sorted Lists
+      - file: 06. Linked List/23. Merge k Sorted Lists
+  - file: 07. Trees/README
+    sections:
+      - file: 07. Trees/100. Same Tree
+      - file: 07. Trees/102. Binary Tree Level Order Traversal
+      - file: 07. Trees/104. Maximum Depth of Binary Tree
+      - file: 07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal
+      - file: 07. Trees/124. Binary Tree Maximum Path Sum
+      - file: 07. Trees/199. Binary Tree Right Side View
+      - file: 07. Trees/226. Invert Binary Tree
+      - file: 07. Trees/230. Kth Smallest Element in a BST
+      - file: 07. Trees/235. Lowest Common Ancestor of a Binary Search Tree
+      - file: 07. Trees/297. Serialize and Deserialize Binary Tree
+      - file: 07. Trees/572. Subtree of Another Tree
+      - file: 07. Trees/98. Validate Binary Search Tree
+  - file: 08. Tries/README
+    sections:
+      - file: 08. Tries/208. Implement Trie (Prefix Tree)
+      - file: 08. Tries/211. Design Add and Search Words Data Structure
+      - file: 08. Tries/212. Word Search II
+  - file: 09. Heap - Priority Queue/README
+    sections:
+      - file: 09. Heap - Priority Queue/1046. Last Stone Weight
+      - file: 09. Heap - Priority Queue/295. Find Median from Data Stream
+      - file: 09. Heap - Priority Queue/703. Kth Largest Element in a Stream
+  - file: 10. Backtracking/README
+    sections:
+      - file: 10. Backtracking/39. Combination Sum
+      - file: 10. Backtracking/79. Word Search
+  - file: 11. Graphs/README
+    sections:
+      - file: 11. Graphs/133. Clone Graph
+      - file: 11. Graphs/178. Graph Valid Tree
+      - file: 11. Graphs/200. Number of Islands
+      - file: 11. Graphs/207. Course Schedule
+      - file: 11. Graphs/3651. Number of Connected Components in an Undirected Graph
+      - file: 11. Graphs/417. Pacific Atlantic Water Flow
+  - file: 12. Advanced Graphs/README
+    sections:
+      - file: 12. Advanced Graphs/892. Alien Dictionary
+  - file: 13. One-D Dynamic Programming/README
+    sections:
+      - file: 13. One-D Dynamic Programming/70. Climbing Stairs
+      - file: 13. One-D Dynamic Programming/198. House Robber
+      - file: 13. One-D Dynamic Programming/213. House Robber II
+  - file: 14. Two-D Dynamic Programming/README
+  - file: 15. Greedy/README
+  - file: 16. Intervals/README
+  - file: 17. Math Geometry/README
+  - file: 18. Bit Manipulation/README
\ No newline at end of file
diff --git a/logo.png b/logo.png
new file mode 100644
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GIT binary patch
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HcmV?d00001


From 9c27f747ff0761250b58a667bd21c204d6472a06 Mon Sep 17 00:00:00 2001
From: Mohsen Tabibian <36014419+mohsentabibian@users.noreply.github.com>
Date: Tue, 14 Nov 2023 16:57:21 +0000
Subject: [PATCH 3/7] u

---
 Book.ipynb | 24 +++++++++++++++++++++---
 1 file changed, 21 insertions(+), 3 deletions(-)

diff --git a/Book.ipynb b/Book.ipynb
index c1d7507..0ed4a15 100644
--- a/Book.ipynb
+++ b/Book.ipynb
@@ -339,10 +339,28 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 2,
    "metadata": {},
-   "outputs": [],
-   "source": []
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Enumerating objects: 376, done.\n",
+      "Counting objects: 100% (376/376), done.\n",
+      "Delta compression using up to 2 threads\n",
+      "Compressing objects: 100% (155/155), done.\n",
+      "Writing objects: 100% (358/358), 2.00 MiB | 12.46 MiB/s, done.\n",
+      "Total 358 (delta 149), reused 354 (delta 148), pack-reused 0\n",
+      "remote: Resolving deltas: 100% (149/149), completed with 1 local object.\u001b[K\n",
+      "To https://github.com/mohsentabibian/LeetCode-Solutions\n",
+      "   39aff16..9bcf446  gh-pages -> gh-pages\n"
+     ]
+    }
+   ],
+   "source": [
+    "!ghp-import -n -p -f _build/html"
+   ]
   }
  ],
  "metadata": {

From 9b1f6d21fbc1ef5cbc509adb92e1611ca77fe16a Mon Sep 17 00:00:00 2001
From: Mohsen Tabibian <36014419+mohsentabibian@users.noreply.github.com>
Date: Tue, 14 Nov 2023 17:13:41 +0000
Subject: [PATCH 4/7] Update documentation

---
 _build/html/.buildinfo => .buildinfo          |   0
 .nojekyll                                     |   0
 .../1. Two Sum.html                           |   0
 .../128. Longest Consecutive Sequence.html    |   0
 .../217. Contains Duplicate.html              |   0
 .../238. Product of Array Except Self.html    |   0
 .../242. Valid Anagram.html                   |   0
 .../347. Top K Frequent Elements.html         |   0
 .../49. Group Anagrams.html                   |   0
 .../659. Encode and Decode Strings.html       |   0
 .../README.html                               |   0
 .../11. Container With Most Water.html        |   0
 .../125. Valid Palindrome.html                |   0
 .../15. 3Sum.html                             |   0
 .../README.html                               |   0
 .../121. Best Time to Buy and Sell Stock.html |   0
 ...ubstring Without Repeating Characters.html |   0
 ...ngest Repeating Character Replacement.html |   0
 .../76. Minimum Window Substring.html         |   0
 .../README.html                               |   0
 .../20. Valid Parentheses.html                |   0
 .../html/04. Stack => 04. Stack}/README.html  |   0
 ... Find Minimum in Rotated Sorted Array.html |   0
 .../33. Search in Rotated Sorted Array.html   |   0
 .../README.html                               |   0
 .../141. Linked List Cycle.html               |   0
 .../143. Reorder List.html                    |   0
 .../19. Remove Nth Node From End of List.html |   0
 .../206. Reverse Linked List.html             |   0
 .../21. Merge Two Sorted Lists.html           |   0
 .../23. Merge k Sorted Lists.html             |   0
 .../README.html                               |   0
 .../100. Same Tree.html                       |   0
 ...02. Binary Tree Level Order Traversal.html |   0
 .../104. Maximum Depth of Binary Tree.html    |   0
 ...e from Preorder and Inorder Traversal.html |   0
 .../124. Binary Tree Maximum Path Sum.html    |   0
 .../199. Binary Tree Right Side View.html     |   0
 .../226. Invert Binary Tree.html              |   0
 .../230. Kth Smallest Element in a BST.html   |   0
 ...mmon Ancestor of a Binary Search Tree.html |   0
 ...Serialize and Deserialize Binary Tree.html |   0
 .../572. Subtree of Another Tree.html         |   0
 .../98. Validate Binary Search Tree.html      |   0
 .../html/07. Trees => 07. Trees}/README.html  |   0
 .../208. Implement Trie (Prefix Tree).html    |   0
 ...n Add and Search Words Data Structure.html |   0
 .../212. Word Search II.html                  |   0
 .../html/08. Tries => 08. Tries}/README.html  |   0
 .../1046. Last Stone Weight.html              |   0
 .../295. Find Median from Data Stream.html    |   0
 .../703. Kth Largest Element in a Stream.html |   0
 .../README.html                               |   0
 .../39. Combination Sum.html                  |   0
 .../79. Word Search.html                      |   0
 .../README.html                               |   0
 .../133. Clone Graph.html                     |   0
 .../178. Graph Valid Tree.html                |   0
 .../200. Number of Islands.html               |   0
 .../207. Course Schedule.html                 |   0
 ...ted Components in an Undirected Graph.html |   0
 .../417. Pacific Atlantic Water Flow.html     |   0
 .../11. Graphs => 11. Graphs}/README.html     |   0
 .../892. Alien Dictionary.html                |   0
 .../README.html                               |   0
 .../198. House Robber.html                    |   0
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-{
- "cells": [
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: pip in /usr/local/python/3.10.8/lib/python3.10/site-packages (23.3.1)\n",
-      "Requirement already satisfied: jupyter-book in /usr/local/python/3.10.8/lib/python3.10/site-packages (0.15.1)\n",
-      "Requirement already satisfied: click<9,>=7.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (8.1.7)\n",
-      "Requirement already satisfied: docutils<0.19,>=0.15 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.18.1)\n",
-      "Requirement already satisfied: Jinja2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (3.1.2)\n",
-      "Requirement already satisfied: jsonschema<5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (4.19.1)\n",
-      "Requirement already satisfied: linkify-it-py~=2.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (2.0.2)\n",
-      "Requirement already satisfied: myst-nb~=0.17.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.17.2)\n",
-      "Requirement already satisfied: pyyaml in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (6.0.1)\n",
-      "Requirement already satisfied: sphinx<6,>=4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (5.0.2)\n",
-      "Requirement already satisfied: sphinx-comments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.0.3)\n",
-      "Requirement already satisfied: sphinx-copybutton in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.5.2)\n",
-      "Requirement already satisfied: sphinx-external-toc~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.1)\n",
-      "Requirement already satisfied: sphinx-jupyterbook-latex~=0.5.2 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.5.2)\n",
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-      "Requirement already satisfied: sphinx-multitoc-numbering~=0.1.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.1.3)\n",
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-      "Requirement already satisfied: referencing>=0.28.4 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (0.30.2)\n",
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-      "Requirement already satisfied: uc-micro-py in /usr/local/python/3.10.8/lib/python3.10/site-packages (from linkify-it-py~=2.0.0->jupyter-book) (1.0.2)\n",
-      "Requirement already satisfied: importlib_metadata in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (6.8.0)\n",
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-      "Requirement already satisfied: jupyter-cache<0.7,>=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.6.1)\n",
-      "Requirement already satisfied: nbclient in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.7.4)\n",
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-      "Requirement already satisfied: sphinxcontrib-htmlhelp>=2.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.0.4)\n",
-      "Requirement already satisfied: sphinxcontrib-serializinghtml>=1.1.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.1.9)\n",
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-      "Requirement already satisfied: pybtex>=0.24 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (0.24.0)\n",
-      "Requirement already satisfied: pybtex-docutils>=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (1.0.3)\n",
-      "Requirement already satisfied: setuptools in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx_togglebutton->jupyter-book) (68.0.0)\n",
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-      "Requirement already satisfied: mdit-py-plugins~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.3.5)\n",
-      "Requirement already satisfied: jupyter-client>=6.1.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (8.4.0)\n",
-      "Requirement already satisfied: jupyter-core!=5.0.*,>=4.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (5.4.0)\n",
-      "Requirement already satisfied: traitlets>=5.3 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (5.11.2)\n",
-      "Requirement already satisfied: fastjsonschema in /home/codespace/.local/lib/python3.10/site-packages (from nbformat~=5.0->myst-nb~=0.17.1->jupyter-book) (2.18.1)\n",
-      "Requirement already satisfied: latexcodec>=1.0.4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pybtex>=0.24->sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (2.0.1)\n",
-      "Requirement already satisfied: six in /home/codespace/.local/lib/python3.10/site-packages (from pybtex>=0.24->sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (1.16.0)\n",
-      "Requirement already satisfied: beautifulsoup4 in /home/codespace/.local/lib/python3.10/site-packages (from pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (4.12.2)\n",
-      "Requirement already satisfied: accessible-pygments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (0.0.4)\n",
-      "Requirement already satisfied: charset-normalizer<4,>=2 in /home/codespace/.local/lib/python3.10/site-packages (from requests>=2.5.0->sphinx<6,>=4->jupyter-book) (3.3.0)\n",
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-      "Requirement already satisfied: comm>=0.1.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (0.1.4)\n",
-      "Requirement already satisfied: debugpy>=1.6.5 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (1.8.0)\n",
-      "Requirement already satisfied: matplotlib-inline>=0.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (0.1.6)\n",
-      "Requirement already satisfied: nest-asyncio in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (1.5.8)\n",
-      "Requirement already satisfied: psutil in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (5.9.6)\n",
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-      "Requirement already satisfied: tornado>=6.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel->myst-nb~=0.17.1->jupyter-book) (6.3.3)\n",
-      "Requirement already satisfied: backcall in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.2.0)\n",
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-      "Requirement already satisfied: pickleshare in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.7.5)\n",
-      "Requirement already satisfied: prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (3.0.39)\n",
-      "Requirement already satisfied: stack-data in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.6.3)\n",
-      "Requirement already satisfied: exceptiongroup in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (1.1.3)\n",
-      "Requirement already satisfied: pexpect>4.3 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (4.8.0)\n",
-      "Requirement already satisfied: parso<0.9.0,>=0.8.3 in /home/codespace/.local/lib/python3.10/site-packages (from jedi>=0.16->ipython->myst-nb~=0.17.1->jupyter-book) (0.8.3)\n",
-      "Requirement already satisfied: python-dateutil>=2.8.2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-client>=6.1.12->nbclient->myst-nb~=0.17.1->jupyter-book) (2.8.2)\n",
-      "Requirement already satisfied: platformdirs>=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-core!=5.0.*,>=4.12->nbclient->myst-nb~=0.17.1->jupyter-book) (3.11.0)\n",
-      "Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py<3.0.0,>=1.0.0->myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.1.2)\n",
-      "Requirement already satisfied: ptyprocess>=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect>4.3->ipython->myst-nb~=0.17.1->jupyter-book) (0.7.0)\n",
-      "Requirement already satisfied: wcwidth in /home/codespace/.local/lib/python3.10/site-packages (from prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.8)\n",
-      "Requirement already satisfied: greenlet!=0.4.17 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sqlalchemy<3,>=1.3.12->jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (3.0.1)\n",
-      "Requirement already satisfied: soupsieve>1.2 in /home/codespace/.local/lib/python3.10/site-packages (from beautifulsoup4->pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (2.5)\n",
-      "Requirement already satisfied: executing>=1.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.0.0)\n",
-      "Requirement already satisfied: asttokens>=2.1.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.4.0)\n",
-      "Requirement already satisfied: pure-eval in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.2)\n",
-      "Note: you may need to restart the kernel to use updated packages.\n"
-     ]
-    }
-   ],
-   "source": [
-    "!python -m pip install --upgrade pip\n",
-    "%pip install jupyter-book"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 34,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions\n"
-     ]
-    }
-   ],
-   "source": [
-    "%cd /workspaces/LeetCode-Solutions"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book toc migrate /workspaces/LeetCode-Solutions/_toc.yml -o /workspaces/LeetCode-Solutions/_toc.yml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "\u001b[01mmaking output directory... \u001b[39;49;00mdone\n",
-      "[etoc] Changing master_doc to 'README'\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 77 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 77 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]  \n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.26 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.00 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]     \n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.92 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 1.28 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executed notebook in 0.91 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executing notebook using local CWD [mystnb]      \n",
-      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executed notebook in 0.91 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executed notebook in 1.17 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executed notebook in 1.16 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executing notebook using local CWD [mystnb] \n",
-      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executed notebook in 1.27 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executed notebook in 4.03 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executing notebook using local CWD [mystnb]   \n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/Book.ipynb: Executing notebook using local CWD [mystnb]                          \n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mSECURITY\u001b[39;49;00m                                              \n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb:30008: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Executing notebook failed: CellTimeoutError [mystnb.exec]\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Notebook exception traceback saved in: /workspaces/LeetCode-Solutions/_build/html/reports/Book.err.log [mystnb.exec]\u001b[39;49;00m\n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CODE_OF_CONDUCT.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/SECURITY.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mSECURITY\u001b[39;49;00m                                               raph\u001b[39;49;00m0m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:8: WARNING: 'myst' reference target not found: #ways-to-contribute\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:9: WARNING: 'myst' reference target not found: #getting-started\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:10: WARNING: 'myst' reference target not found: #contributing-process\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md:11: WARNING: 'myst' reference target not found: #code-of-conduct\u001b[39;49;00m\n",
-      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
-      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
-      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
-      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
-      "[etoc] missing index.html written as redirect to 'README.html'\n",
-      "\u001b[01mbuild succeeded, 11 warnings.\u001b[39;49;00m\n",
-      "\n",
-      "The HTML pages are in _build/html.\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Finished generating HTML for book.\n",
-      "Your book's HTML pages are here:\n",
-      "    _build/html/\n",
-      "You can look at your book by opening this file in a browser:\n",
-      "    _build/html/index.html\n",
-      "Or paste this line directly into your browser bar:\n",
-      "    file:///workspaces/LeetCode-Solutions/_build/html/index.html            \n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build ."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: ghp-import in /usr/local/python/3.10.8/lib/python3.10/site-packages (2.1.0)\n",
-      "Requirement already satisfied: python-dateutil>=2.8.1 in /home/codespace/.local/lib/python3.10/site-packages (from ghp-import) (2.8.2)\n",
-      "Requirement already satisfied: six>=1.5 in /home/codespace/.local/lib/python3.10/site-packages (from python-dateutil>=2.8.1->ghp-import) (1.16.0)\n",
-      "Note: you may need to restart the kernel to use updated packages.\n"
-     ]
-    }
-   ],
-   "source": [
-    "%pip install ghp-import"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Enumerating objects: 376, done.\n",
-      "Counting objects: 100% (376/376), done.\n",
-      "Delta compression using up to 2 threads\n",
-      "Compressing objects: 100% (155/155), done.\n",
-      "Writing objects: 100% (358/358), 2.00 MiB | 12.46 MiB/s, done.\n",
-      "Total 358 (delta 149), reused 354 (delta 148), pack-reused 0\n",
-      "remote: Resolving deltas: 100% (149/149), completed with 1 local object.\u001b[K\n",
-      "To https://github.com/mohsentabibian/LeetCode-Solutions\n",
-      "   39aff16..9bcf446  gh-pages -> gh-pages\n"
-     ]
-    }
-   ],
-   "source": [
-    "!ghp-import -n -p -f _build/html"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 2
-}
diff --git a/_build/html/CODE_OF_CONDUCT.html b/CODE_OF_CONDUCT.html
similarity index 100%
rename from _build/html/CODE_OF_CONDUCT.html
rename to CODE_OF_CONDUCT.html
diff --git a/_build/html/CONTRIBUTING.html b/CONTRIBUTING.html
similarity index 100%
rename from _build/html/CONTRIBUTING.html
rename to CONTRIBUTING.html
diff --git a/LICENSE b/LICENSE
deleted file mode 100644
index 8818d1f..0000000
--- a/LICENSE
+++ /dev/null
@@ -1,21 +0,0 @@
-MIT License
-
-Copyright (c) 2023 Mohsen Tabibian
-
-Permission is hereby granted, free of charge, to any person obtaining a copy
-of this software and associated documentation files (the "Software"), to deal
-in the Software without restriction, including without limitation the rights
-to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
-copies of the Software, and to permit persons to whom the Software is
-furnished to do so, subject to the following conditions:
-
-The above copyright notice and this permission notice shall be included in all
-copies or substantial portions of the Software.
-
-THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
-IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
-FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
-AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
-LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
-OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
-SOFTWARE.
diff --git a/_build/html/README.html b/README.html
similarity index 100%
rename from _build/html/README.html
rename to README.html
diff --git a/_build/html/SECURITY.html b/SECURITY.html
similarity index 100%
rename from _build/html/SECURITY.html
rename to SECURITY.html
diff --git a/_build/.doctrees/01. Array Hashing/1. Two Sum.doctree b/_build/.doctrees/01. Array Hashing/1. Two Sum.doctree
deleted file mode 100644
index 5390e0939c900e91a25d8cd0cb7d0f9dcb08d699..0000000000000000000000000000000000000000
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diff --git a/_build/.doctrees/03. Sliding Window/424. Longest Repeating Character Replacement.doctree b/_build/.doctrees/03. Sliding Window/424. Longest Repeating Character Replacement.doctree
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diff --git a/_build/.doctrees/07. Trees/297. Serialize and Deserialize Binary Tree.doctree b/_build/.doctrees/07. Trees/297. Serialize and Deserialize Binary Tree.doctree
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diff --git a/_build/.doctrees/08. Tries/211. Design Add and Search Words Data Structure.doctree b/_build/.doctrees/08. Tries/211. Design Add and Search Words Data Structure.doctree
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diff --git a/_build/.doctrees/11. Graphs/3651. Number of Connected Components in an Undirected Graph.doctree b/_build/.doctrees/11. Graphs/3651. Number of Connected Components in an Undirected Graph.doctree
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diff --git a/_build/html/_sources/01. Array Hashing/1. Two Sum.ipynb b/_build/html/_sources/01. Array Hashing/1. Two Sum.ipynb
deleted file mode 100644
index b3b517f..0000000
--- a/_build/html/_sources/01. Array Hashing/1. Two Sum.ipynb	
+++ /dev/null
@@ -1,211 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 100: Same Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-        "\n",
-        "---\n",
-        "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-        "\n",
-        "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-        "\n",
-        "**Constraints**\n",
-        "\n",
-        "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-        "2. $-10^4$ <= `Node.val` <= $10^4$"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def isSameTree(p, q):\n",
-        "    # Base case: If both p and q are None, the trees are the same.\n",
-        "    if not p and not q:\n",
-        "        return True\n",
-        "    \n",
-        "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-        "    if not p or not q:\n",
-        "        return False\n",
-        "    \n",
-        "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-        "    if p.val != q.val:\n",
-        "        return False\n",
-        "    \n",
-        "    # Recursively check the left and right subtrees of p and q.\n",
-        "    # If both subtrees are the same, the entire trees are the same.\n",
-        "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-        "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-        "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-        "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-        "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-        "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "\n",
-        "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-        "\n",
-        "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "print(isSameTree(p1, q1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "\n",
-        "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-        "\n",
-        "p2 = TreeNode(1, TreeNode(2), None)\n",
-        "q2 = TreeNode(1, None, TreeNode(2))\n",
-        "print(isSameTree(p2, q2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "\n",
-        "#Input: p = [1,2,1], q = [1,1,2]\n",
-        "\n",
-        "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-        "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-        "print(isSameTree(p3, q3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "**Time Complexity**\n",
-        "\n",
-        "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-        "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-        "\n",
-        "**Space Complexity**\n",
-        "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-        "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-        "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-        "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "+ Time Complexity: $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-        "+ Space Complexity: $O(n)$ due to the recursive call stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb b/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb
deleted file mode 100644
index a77a03d..0000000
--- a/_build/html/_sources/01. Array Hashing/128. Longest Consecutive Sequence.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 128. Longest Consecutive Sequence\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Consecutive Sequence problem on LeetCode, click here!](https://leetcode.com/problems/longest-consecutive-sequence/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an unsorted array of integers `nums`, return *the length of the longest consecutive elements sequence.*\n",
-        "\n",
-        "You must write an algorithm that runs in `O(n)` time.\n",
-        "\n",
-        "*Constraints:**\n",
-        "- 0 <= `nums.length` <= $10^5$\n",
-        "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 58,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def longestConsecutive(nums):\n",
-        "    if not nums:\n",
-        "        return 0\n",
-        "\n",
-        "    num_set = set(nums)\n",
-        "    max_length = 0\n",
-        "\n",
-        "    for num in num_set:\n",
-        "        if num - 1 not in num_set:\n",
-        "            current_num = num\n",
-        "            current_length = 1\n",
-        "\n",
-        "            while current_num + 1 in num_set:\n",
-        "                current_num += 1\n",
-        "                current_length += 1\n",
-        "\n",
-        "            max_length = max(max_length, current_length)\n",
-        "\n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining the `longestConsecutive` function that takes the `nums` array as input.\n",
-        "\n",
-        "2. We check if the `nums` array is empty. If it's empty, there are no consecutive elements, so we return 0.\n",
-        "\n",
-        "3. We create a Python set called `num_set` and insert all elements from the `nums` array into it. Using a set allows us to efficiently check for the existence of elements in O(1) time.\n",
-        "\n",
-        "4. We initialize a variable `max_length` to 0. This variable will keep track of the maximum length of consecutive elements sequence found.\n",
-        "\n",
-        "5. We iterate through the elements of the `num_set`. For each element `num`, we check if `num - 1` exists in the `num_set`. If it doesn't exist, it means `num` is the starting element of a potential consecutive sequence.\n",
-        "\n",
-        "6. Inside the loop, we initialize two variables: `current_num` to the current element `num` and `current_length` to 1. We start with a length of 1 because `num` itself is part of the sequence.\n",
-        "\n",
-        "7. We then enter a while loop that continues as long as `current_num + 1` exists in the `num_set`. This means we are incrementing the consecutive sequence.\n",
-        "\n",
-        "8. Inside the while loop, we increment `current_num` by 1 and also increment `current_length` by 1 to account for the next consecutive element.\n",
-        "\n",
-        "9. We compare `current_length` with the `max_length` and update `max_length` if the current sequence is longer.\n",
-        "\n",
-        "10. After the while loop, we move to the next element in the outer loop and repeat the process.\n",
-        "\n",
-        "11. Finally, we return the `max_length` as the result, which represents the length of the longest consecutive elements sequence in the `nums` array.\n",
-        "\n",
-        "The key idea here is to use a set to efficiently check for the existence of elements and to iterate through the elements, considering each element as the potential start of a consecutive sequence. By doing this, we can find the longest consecutive sequence in O(n) time complexity, where n is the number of elements in the array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 59,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "nums1 = [100, 4, 200, 1, 3, 2]\n",
-        "print(longestConsecutive(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 60,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "9\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "nums2 = [0, 3, 7, 2, 5, 8, 4, 6, 0, 1]\n",
-        "print(longestConsecutive(nums2)) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The code is designed to run in O(n) time complexity, where n is the number of elements in the `nums` array. Here's the breakdown:\n",
-        "\n",
-        "1. Constructing the `num_set` by inserting all elements from `nums` into it takes O(n) time because we perform an insertion operation for each element in `nums`.\n",
-        "\n",
-        "2. The main loop iterates through the elements in `num_set`. In the worst case, each element is visited only once, so the loop itself takes O(n) time.\n",
-        "\n",
-        "3. Within the loop, we have a while loop that may also take O(n) time in the worst case. However, this while loop is nested inside the main loop, so its overall time complexity remains O(n).\n",
-        "\n",
-        "Therefore, the overall time complexity of the code is O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of the code is determined by the space used by the `num_set` and a few additional variables. Here's the breakdown:\n",
-        "\n",
-        "1. `num_set` is a set that stores the unique elements from `nums`. In the worst case, it can store all n elements from `nums`, so the space complexity is O(n).\n",
-        "\n",
-        "2. The additional variables used, such as `max_length`, `current_num`, and `current_length`, have constant space requirements and do not depend on the size of the input array.\n",
-        "\n",
-        "Therefore, the overall space complexity of the code is O(n) due to the space used by the `num_set`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. Write an algorithm that not only returns the length of the longest consecutive elements sequence but also returns the actual consecutive sequence itself.\n",
-        "\n",
-        "2. Extend the problem to allow elements to be considered consecutive if they are within a certain absolute difference (e.g., less than or equal to k) instead of exactly 1. Write an algorithm that finds the longest such sequence and runs in O(n) time."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb b/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb
deleted file mode 100644
index 57bb55d..0000000
--- a/_build/html/_sources/01. Array Hashing/217. Contains Duplicate.ipynb	
+++ /dev/null
@@ -1,218 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 217. Contains Duplicate\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Contains Duplicate problem on LeetCode, click here!](https://leetcode.com/problems/contains-duplicate/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `nums.length` <= $10^5$\n",
-        "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 19,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def containsDuplicate(nums):\n",
-        "    # Create an empty set to store unique elements\n",
-        "    unique_elements = set()\n",
-        "    \n",
-        "    # Iterate through the array\n",
-        "    for num in nums:\n",
-        "        # If the element is already in the set, return True\n",
-        "        if num in unique_elements:\n",
-        "            return True\n",
-        "        # Otherwise, add it to the set\n",
-        "        else:\n",
-        "            unique_elements.add(num)\n",
-        "    \n",
-        "    # If the loop completes without finding duplicates, return False\n",
-        "    return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `containsDuplicate` function takes a single argument, `nums`, which is the input integer array.\n",
-        "\n",
-        "2. Inside the function, an empty set called `unique_elements` is created. This set will be used to keep track of unique elements in the input array.\n",
-        "\n",
-        "3. The function then iterates through the input array `nums` using a `for` loop.\n",
-        "\n",
-        "4. For each element `num` in the array, it checks whether `num` is already in the `unique_elements` set using the `if num in unique_elements:` condition.\n",
-        "\n",
-        "5. If `num` is already in the set, it means there is a duplicate element in the array, and the function immediately returns `True`.\n",
-        "\n",
-        "6. If `num` is not in the set, it is added to the `unique_elements` set using `unique_elements.add(num)`.\n",
-        "\n",
-        "7. The loop continues to the next element, and the process repeats.\n",
-        "\n",
-        "8. If the loop completes without finding any duplicates, it means that all elements in the array are distinct, and the function returns `False`.\n",
-        "\n",
-        "The code efficiently utilizes a set data structure to keep track of unique elements while iterating through the array, allowing it to quickly detect duplicate elements. This code meets the problem's requirements and constraints, as explained earlier."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "nums1 = [1, 2, 3, 1]\n",
-        "print(containsDuplicate(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "nums2 = [1, 2, 3, 4]\n",
-        "print(containsDuplicate(nums2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 22,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "nums3 = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]\n",
-        "print(containsDuplicate(nums3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for the \"Contains Duplicate\" problem:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The primary operation that affects the time complexity is the `for` loop that iterates through the input array `nums`. In the worst case, the loop will iterate through all `n` elements of the array, where `n` is the length of the input array.\n",
-        "\n",
-        "- Iterating through the array: O(n)\n",
-        "\n",
-        "Inside the loop, we perform two operations:\n",
-        "\n",
-        "1. Checking whether an element exists in the `unique_elements` set (`if num in unique_elements`). This operation has an average time complexity of O(1) for a set.\n",
-        "2. Adding an element to the `unique_elements` set (`unique_elements.add(num)`), which also has an average time complexity of O(1) for a set.\n",
-        "\n",
-        "Since these operations are performed for each element in the array, the overall time complexity remains O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity is determined by the additional data structures used in the code, which are the `unique_elements` set.\n",
-        "\n",
-        "- `unique_elements` set: This set stores unique elements from the input array. In the worst case, if all elements in the input array are distinct, the set will store all `n` elements.\n",
-        "\n",
-        "Therefore, the space complexity is O(n) because, in the worst case, the set's size will grow linearly with the input array's size.\n",
-        "\n",
-        "**In summary**, the time complexity of the code is O(n), where `n` is the length of the input array, and the space complexity is also O(n) in the worst case. This code is efficient and meets the constraints of the problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K Duplicates:** Modify the problem to return `true` if there are exactly K duplicate elements in the array. Write a function that takes an additional integer parameter K and returns `true` if there are exactly K duplicates.\n",
-        "2. **Frequency Count:** Write a function that returns a list of all the elements that appear more than once in the array, along with their frequencies."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb b/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb
deleted file mode 100644
index 1ff9a9c..0000000
--- a/_build/html/_sources/01. Array Hashing/238. Product of Array Except Self.ipynb	
+++ /dev/null
@@ -1,197 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 238. Product of Array Except Self\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Product of Array Except Self problem on LeetCode, click here!](https://leetcode.com/problems/product-of-array-except-self/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array `nums`, return *an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.*\n",
-        "\n",
-        "The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-        "\n",
-        "You must write an algorithm that runs in O(n) time and without using the division operation.\n",
-        "\n",
-        "Constraints:\n",
-        "- 2 <= `nums.length` <= $10^5$\n",
-        "- `-30 <= nums[i] <= 30`\n",
-        "- The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-        "\n",
-        "**Follow-up:** Can you solve the problem in O(1) extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 50,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def productExceptSelf(nums):\n",
-        "    n = len(nums)\n",
-        "    \n",
-        "    # Initialize two arrays to store products to the left and right of each element\n",
-        "    left_products = [1] * n\n",
-        "    right_products = [1] * n\n",
-        "    \n",
-        "    # Calculate products to the left of each element\n",
-        "    left_product = 1\n",
-        "    for i in range(n):\n",
-        "        left_products[i] = left_product\n",
-        "        left_product *= nums[i]\n",
-        "    \n",
-        "    # Calculate products to the right of each element\n",
-        "    right_product = 1\n",
-        "    for i in range(n - 1, -1, -1):\n",
-        "        right_products[i] = right_product\n",
-        "        right_product *= nums[i]\n",
-        "    \n",
-        "    # Calculate the final answer using left and right products\n",
-        "    answer = [left_products[i] * right_products[i] for i in range(n)]\n",
-        "    \n",
-        "    return answer"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `n` is initialized as the length of the `nums` array.\n",
-        "2. Two arrays, `left_products` and `right_products`, are created, each with a length of `n` and filled with ones initially.\n",
-        "3. `left_product` is initialized to 1.\n",
-        "4. A loop iterates through the `nums` array, from left to right. For each element at index `i`, it stores the product of all elements to its left (including itself) in the `left_products` array and updates `left_product` accordingly.\n",
-        "5. `right_product` is initialized to 1.\n",
-        "6. Another loop iterates through the `nums` array in reverse order, from right to left. For each element at index `i`, it stores the product of all elements to its right (including itself) in the `right_products` array and updates `right_product`.\n",
-        "7. The final `answer` list is constructed by multiplying corresponding elements from the `left_products` and `right_products` arrays for each index `i`.\n",
-        "8. The function returns the `answer` list, which contains the desired results.\n",
-        "\n",
-        "Overall, this algorithm efficiently computes the product of all elements except the current element, as required. It runs in O(n) time complexity and uses O(n) extra space, which is within the problem's constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 52,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[24, 12, 8, 6]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "nums = [1,2,3,4]\n",
-        "result1 = productExceptSelf(nums)\n",
-        "print(result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 53,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[0, 0, 9, 0, 0]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums = [-1,1,0,-3,3]\n",
-        "result2 = productExceptSelf(nums)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The first loop iterates through the `nums` array to calculate the products to the left of each element. This loop runs in O(n) time, where n is the length of the `nums` array.\n",
-        "2. The second loop also iterates through the `nums` array in reverse order to calculate the products to the right of each element. This loop also runs in O(n) time.\n",
-        "3. The final loop constructs the `answer` list by multiplying elements from the `left_products` and `right_products` arrays. This loop runs in O(n) time.\n",
-        "\n",
-        "Since all the loops are independent and sequential, the overall time complexity is O(n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. Two additional arrays, `left_products` and `right_products`, are created with a length of `n`, where `n` is the length of the `nums` array. Therefore, these arrays consume O(n) extra space.\n",
-        "2. The `left_product` and `right_product` variables are used to keep track of the product to the left and right of each element, respectively. These variables occupy O(1) extra space.\n",
-        "3. The `answer` list, which contains the final results, also consumes O(n) space, but it's not counted toward the extra space complexity analysis as per the problem's constraints.\n",
-        "\n",
-        "Overall, the space complexity is O(n) for this solution."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Handle Zeros in the Input**: Modify the solution to handle cases where the input array contains zeros. For example, if `nums = [0, 1, 2, 3, 0]`, the output should be `[0, 0, 0, 0, 0]` because any element multiplied by zero results in zero.\n",
-        "\n",
-        "2. **In-Place Modification**: Attempt to solve the problem with in-place modification of the `nums` array, where the output is stored directly in the input array without using any additional space."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb b/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb
deleted file mode 100644
index d165899..0000000
--- a/_build/html/_sources/01. Array Hashing/242. Valid Anagram.ipynb	
+++ /dev/null
@@ -1,198 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 242. Valid Anagram\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Anagram problem on LeetCode, click here!](https://leetcode.com/problems/valid-anagram/)\n",
-        "\n",
-        "---\n",
-        "Given two strings `s` and `t`, return `true` if `t` is an anagram of `s`, and `false` otherwise.\n",
-        "\n",
-        "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `s.length`, `t.length` <= $5 * 10^4$\n",
-        "- `s` and `t` consist of lowercase English letters.\n",
-        "\n",
-        "**Follow-up:** What if the inputs contain Unicode characters? How would you adapt your solution to such a case?\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 29,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isAnagram(s: str, t: str) -> bool:\n",
-        "    # Check if the lengths of s and t are not equal\n",
-        "    if len(s) != len(t):\n",
-        "        return False\n",
-        "\n",
-        "    # Create dictionaries to store character frequencies for s and t\n",
-        "    countS, countT = {}, {}\n",
-        "\n",
-        "    # Count character frequencies in s\n",
-        "    for i in range(len(s)):\n",
-        "        countS[s[i]] = 1 + countS.get(s[i], 0)\n",
-        "\n",
-        "    # Count character frequencies in t\n",
-        "    for i in range(len(t)):\n",
-        "        countT[t[i]] = 1 + countT.get(t[i], 0)\n",
-        "\n",
-        "    # Check if the character frequencies in s and t are the same\n",
-        "    return countS == countT"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The code defines a class called `Solution` with a method `isAnagram` that takes two input strings, `s` and `t`, and returns a boolean (`True` or `False`) based on whether `t` is an anagram of `s`.\n",
-        "\n",
-        "2. The first check performed in the method is whether the lengths of `s` and `t` are equal. If they are not equal, it immediately returns `False` because strings of different lengths cannot be anagrams of each other.\n",
-        "\n",
-        "3. Two dictionaries, `countS` and `countT`, are created to store the frequency of characters in `s` and `t`, respectively. These dictionaries will be used to count the occurrences of each character in the strings.\n",
-        "\n",
-        "4. The code then enters a loop to iterate through each character in string `s` using a `for` loop. Inside the loop, it updates the `countS` dictionary. The line `countS[s[i]] = 1 + countS.get(s[i], 0)` increments the count of character `s[i]` in `countS` by 1. If the character is not already in the dictionary, it initializes the count to 1.\n",
-        "\n",
-        "5. Similarly, the code iterates through each character in string `t` and updates the `countT` dictionary in the same way.\n",
-        "\n",
-        "6. After counting the character frequencies in both strings, the code compares the two dictionaries using `countS == countT`. If the dictionaries are equal, it means that both strings have the same character frequencies, and therefore, `t` is an anagram of `s`. In this case, the method returns `True`.\n",
-        "\n",
-        "7. If the dictionaries are not equal, the method returns `False`, indicating that `t` is not an anagram of `s`.\n",
-        "\n",
-        "In summary, this code efficiently determines whether two input strings are anagrams by counting the character frequencies in each string using dictionaries and then comparing these counts. If the character frequencies match, the strings are considered anagrams, and the method returns `True`; otherwise, it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 30,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "s = \"anagram\"\n",
-        "t = \"nagaram\"\n",
-        "print(isAnagram(s, t))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "s = \"rat\"\n",
-        "t = \"car\"\n",
-        "print(isAnagram(s, t))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. The code first checks whether the lengths of the input strings `s` and `t` are equal, which takes constant time. This check has a time complexity of O(1).\n",
-        "\n",
-        "2. The code then iterates through both strings, `s` and `t`, once to count the character frequencies. Since both strings have a maximum length of 5 * 10^4, the worst-case scenario is that the code iterates through 5 * 10^4 characters in each string. This results in a linear time complexity of O(n), where n is the length of the longer of the two input strings (`s` or `t`).\n",
-        "\n",
-        "3. Finally, the code compares the two dictionaries `countS` and `countT` to check if the character frequencies match. This comparison takes O(n) time in the worst case, where n is the length of the longer string.\n",
-        "\n",
-        "Overall, the time complexity of the code is O(n), where n is the length of the longer input string.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The code uses two dictionaries, `countS` and `countT`, to store the character frequencies of the input strings `s` and `t`. In the worst case, both dictionaries can contain all unique characters from the input strings. Since the input strings can have a maximum length of 5 * 10^4, the space complexity for these dictionaries is O(5 * 10^4) or simply O(n), where n is the length of the longer input string.\n",
-        "\n",
-        "2. The code uses a few additional variables for bookkeeping, but these variables have constant space requirements and do not depend on the input size. Therefore, they do not significantly impact the overall space complexity.\n",
-        "\n",
-        "**In summary**, the space complexity of the code is O(n), where n is the length of the longer input string."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Anagram Chains**: Given a list of words, find the longest chain of anagrams, where each word in the chain is an anagram of the previous word. For example, given [\"bat\", \"tab\", \"cat\", \"tac\", \"dog\"], the longest anagram chain is [\"bat\", \"tab\", \"cat\", \"tac\"].\n",
-        "\n",
-        "2. **Minimum Deletions to Make Anagrams**: Given two strings `s` and `t`, find the minimum number of character deletions required in both strings to make them anagrams of each other. For example, if `s = \"hello\"` and `t = \"billion\"`, you can remove the characters \"heo\" from `s` and \"billi\" from `t` to make them anagrams (\"lo\" and \"on\" are left in the two strings)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb b/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb
deleted file mode 100644
index ecfa494..0000000
--- a/_build/html/_sources/01. Array Hashing/347. Top K Frequent Elements.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 347. Top K Frequent Elements\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Top K Frequent Elements problem on LeetCode, click here!](https://leetcode.com/problems/top-k-frequent-elements/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array `nums` and an integer `k`, return the `k` most frequent elements. You may return the answer in **any order**.\n",
-        "\n",
-        "Constraints:\n",
-        "- 1 <= `nums.length` <= $10^5$\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- `k` is in the range `[1, the number of unique elements in the array]`.\n",
-        "- It is **guaranteed** that the answer is **unique**.\n",
-        "\n",
-        "**Follow-up:** Your algorithm's time complexity must be better than $O(n\\ log\\ n)$, where n is the array's size.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 46,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def topKFrequent(nums, k):\n",
-        "    # Create a dictionary to store the frequency of each element\n",
-        "    count = {}\n",
-        "    # Create a list of lists to store elements with the same frequency\n",
-        "    frequency = [[] for i in range(len(nums) + 1)]\n",
-        "\n",
-        "    # Count the frequency of each element in nums\n",
-        "    for n in nums:\n",
-        "        count[n] = 1 + count.get(n, 0)\n",
-        "\n",
-        "    # Place elements in the freq list according to their frequency\n",
-        "    for n, c in count.items():\n",
-        "        frequency[c].append(n)\n",
-        "\n",
-        "    res = []\n",
-        "    # Traverse freq list from the end (higher frequencies)\n",
-        "    for i in range(len(frequency) - 1, 0, -1):\n",
-        "        for n in frequency[i]:\n",
-        "            res.append(n)\n",
-        "            if len(res) == k:\n",
-        "                return res"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `def topKFrequent(nums, k):`: This is a function that takes two arguments: `nums`, which is the input array of integers, and `k`, which is the number of most frequent elements to return.\n",
-        "\n",
-        "2. `count = {}`: This dictionary `count` will be used to store the frequency of each unique element in the `nums` array. The keys are elements from the input array, and the values are their corresponding frequencies.\n",
-        "\n",
-        "3. `frequency = [[] for i in range(len(nums) + 1)]`: This creates a list of empty lists called `frequency`. It's used to store elements based on their frequencies, similar to the `freq` list in the previous code. The size of this list is set to be one greater than the length of the input `nums`.\n",
-        "\n",
-        "4. `for n in nums:`: This loop iterates through each element `n` in the input `nums` array.\n",
-        "\n",
-        "5. `count[n] = 1 + count.get(n, 0)`: This line counts the frequency of each element `n` in the `nums` array. It uses the `count.get(n, 0)` method to retrieve the current count of `n` from the `count` dictionary. If `n` is not in the dictionary, it defaults to 0. It then increments the count by 1.\n",
-        "\n",
-        "6. After the above loop, the `count` dictionary will contain counts of each unique element in the `nums` array.\n",
-        "\n",
-        "7. `for n, c in count.items():`: This loop iterates through the items (key-value pairs) of the `count` dictionary. `n` represents the element, and `c` represents its frequency.\n",
-        "\n",
-        "8. `frequency[c].append(n)`: This line places the element `n` into the bucket corresponding to its frequency `c`. Buckets are represented by the `frequency` list. For example, if an element `n` has a frequency of 3, it will be added to `frequency[3]`.\n",
-        "\n",
-        "9. After this loop, the `frequency` list will contain buckets of elements grouped by their frequencies.\n",
-        "\n",
-        "10. `res = []`: This list `res` will be used to store the k most frequent elements.\n",
-        "\n",
-        "11. `for i in range(len(freq) - 1, 0, -1):`: This loop iterates in reverse order through the `frequency` list, starting from the highest frequency and going down to 1. Note that there is a typo here; it should be `frequency` instead of `freq`.\n",
-        "\n",
-        "12. `for n in frequency[i]:`: For each element `n` in the current bucket (i.e., elements with frequency `i`), it appends `n` to the result list `res`.\n",
-        "\n",
-        "13. `if len(res) == k:`: Once `res` contains k elements, the code exits and returns the result.\n",
-        "\n",
-        "The code efficiently finds the k most frequent elements in the input array without using sorting, similar to the previous explanation, with the only difference being the variable names (e.g., `frequency` instead of `freq`)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 48,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 2]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "nums = [1,1,1,2,2,3]\n",
-        "k = 2\n",
-        "result1 = topKFrequent(nums, k)\n",
-        "print(result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 49,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums = [1]\n",
-        "k = 1\n",
-        "result2 = topKFrequent(nums, k)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The first loop that counts the frequency of elements by iterating through `nums` has a time complexity of O(n), where n is the number of elements in the input array `nums`.\n",
-        "\n",
-        "2. The second loop iterates through the keys in the `count` dictionary, which has at most `k` unique elements (as per the constraints). Therefore, the second loop also has a time complexity of O(k).\n",
-        "\n",
-        "3. The third loop iterates through the `frequency` list, which has a length equal to the maximum frequency of elements in `nums`. In the worst case, this loop can have a time complexity of O(n), where n is the number of elements in `nums`.\n",
-        "\n",
-        "Overall, the dominant factor in terms of time complexity is the loop that iterates through the `count` dictionary. So, the total time complexity of the code is O(n + k).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The `count` dictionary stores the frequency of each unique element in `nums`. In the worst case, it can have at most `k` unique elements, so the space complexity for `count` is O(k).\n",
-        "\n",
-        "2. The `frequency` list is used to store elements grouped by their frequencies. It has a length of `len(nums) + 1`, which can be at most 105 based on the constraints. Therefore, the space complexity for `frequency` is O(105).\n",
-        "\n",
-        "3. The `res` list stores the k most frequent elements, which can be at most `k` elements, so the space complexity for `res` is O(k).\n",
-        "\n",
-        "**In summary**, the space complexity is dominated by the `count` dictionary and the `frequency` list, both of which have a space complexity of O(k + 105)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimized k Most Frequent Elements:** Modify the code to find the k most frequent elements in an array while ensuring that the time complexity is O(n + klogk). You can use a priority queue (heap) to achieve this.\n",
-        "\n",
-        "2. **Handling Duplicate Frequencies:** Extend the code to handle cases where multiple elements have the same frequency and are among the k most frequent elements. Ensure that the output contains exactly k elements."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb b/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb
deleted file mode 100644
index afe89ec..0000000
--- a/_build/html/_sources/01. Array Hashing/49. Group Anagrams.ipynb	
+++ /dev/null
@@ -1,229 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 49. Group Anagrams\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Group Anagrams problem on LeetCode, click here!](https://leetcode.com/problems/group-anagrams/)\n",
-        "\n",
-        "---\n",
-        "Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.\n",
-        "\n",
-        "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `strs.length <= $10^4$\n",
-        "- `0 <= strs[i].length <= 100`\n",
-        "- `strs[i]` consists of lowercase English letters.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 36,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict\n",
-        "\n",
-        "\n",
-        "def groupAnagrams(strs):\n",
-        "    ans = defaultdict(list)\n",
-        "\n",
-        "    # Iterate through the list of input strings\n",
-        "    for s in strs:\n",
-        "        # Initialize a list to represent character counts for each character (a-z)\n",
-        "        count = [0] * 26\n",
-        "\n",
-        "        # Count the occurrences of each character in the current word\n",
-        "        for c in s:\n",
-        "            count[ord(c) - ord(\"a\")] += 1\n",
-        "\n",
-        "        # Use a tuple of character counts as the key and append the word to the anagram group\n",
-        "        ans[tuple(count)].append(s)\n",
-        "\n",
-        "    # Convert the values (lists of anagrams) to a list of lists\n",
-        "    return list(ans.values())"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. Inside the method, a `defaultdict` called `ans` is created to store anagram groups. This dictionary will have a list as its default value, meaning that each key in the dictionary will be associated with an empty list by default.\n",
-        "\n",
-        "2. The code then iterates through the list of input strings, `strs`, using a `for` loop. For each word `s` in `strs`, it performs the following steps:\n",
-        "\n",
-        "   - It initializes a list `count` of length 26, where each element represents the count of a specific character (a-z) in the word `s`.\n",
-        "\n",
-        "   - The code then iterates through the characters of `s`. For each character `c`, it increments the corresponding count in the `count` list based on its ASCII value.\n",
-        "\n",
-        "   - After counting the characters in `s`, it converts the `count` list into a tuple `tuple(count)` to use as a key for the `ans` dictionary.\n",
-        "\n",
-        "   - It appends the word `s` to the list associated with the key (tuple) in the `ans` dictionary. This groups all anagrams of the same word together under the same key.\n",
-        "\n",
-        "3. After processing all words in `strs`, the code converts the values of the `ans` dictionary (which are lists of anagrams) to a list of lists using the `list()` constructor.\n",
-        "\n",
-        "4. Finally, the code returns the list of anagram groups, which is the result of grouping the anagrams in the input list.\n",
-        "\n",
-        "In summary, the code efficiently groups anagrams together by counting the characters in each word and using a dictionary to store them under the same key. The result is a list of lists, where each inner list represents a group of anagrams."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 37,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: Group anagrams from a list of words\n",
-        "strs1 = [\"eat\", \"tea\", \"tan\", \"ate\", \"nat\", \"bat\"]\n",
-        "result1 = groupAnagrams(strs1)\n",
-        "print(result1)\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 38,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2: Group anagrams from a list with an empty string\n",
-        "strs2 = [\"\"]\n",
-        "result2 = groupAnagrams(strs2)\n",
-        "print(result2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 40,
-      "id": "c8367e2a-e7d5-4e18-8235-2870e341a04b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[['a']]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3: Group anagrams from a list with a single word\n",
-        "strs3 = [\"a\"]\n",
-        "result3 = groupAnagrams(strs3)\n",
-        "print(result3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `groupAnagrams` method:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. First, we create an instance of the `Solution` class, which is an O(1) operation.\n",
-        "\n",
-        "2. Inside the `groupAnagrams` method, we iterate through the list of input strings `strs` once in a loop:\n",
-        "   - For each word in `strs`, we perform character counting, which is done in O(K) time, where K is the maximum length of a word in the list.\n",
-        "\n",
-        "3. The overall time complexity of the method is O(N * K), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. We use a `defaultdict` called `ans` to store the anagram groups. In the worst case, each word belongs to a distinct anagram group, so the space complexity for `ans` is O(N).\n",
-        "\n",
-        "2. Within the loop, we create a `count` list of length 26 to store character counts for each word. This is a fixed-size list and does not depend on the input size, so it has a constant space complexity of O(26), which is effectively O(1).\n",
-        "\n",
-        "3. The space used for other variables is also constant and does not depend on the input size.\n",
-        "\n",
-        "4. The final result, `result`, is a list of lists that contains the grouped anagrams. In the worst case, each word is a distinct anagram group, so the space complexity for `result` is O(N).\n",
-        "\n",
-        "5. The overall space complexity of the method is O(N) due to the space used by `ans` and `result`. The space used by the `count` list and other variables is constant and does not significantly affect the overall space complexity.\n",
-        "\n",
-        "**In summary,** the time complexity of the `groupAnagrams` method is O(N * K), and the space complexity is O(N), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Anagramic Palindromes**: Given a list of strings, find all the groups of anagrams where each group contains words that can be rearranged into a palindrome. For example, in the list [\"bat\", \"tab\", \"act\", \"tac\", \"dog\"], there are two groups: [\"bat\", \"tab\", \"act\", \"tac\"] and [\"dog\"].\n",
-        "\n",
-        "2. **Largest Anagram Group**: Given a list of strings, find the largest group of anagrams. Return the list of anagrams in that group. If there are multiple largest groups, return all of them."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb b/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb
deleted file mode 100644
index 037c476..0000000
--- a/_build/html/_sources/01. Array Hashing/659. Encode and Decode Strings.ipynb	
+++ /dev/null
@@ -1,204 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 659· Encode and Decode Strings\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Encode and Decode Strings problem on LintCode, click here!](https://www.lintcode.com/problem/659/)\n",
-        "\n",
-        "---\n",
-        "Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.\n",
-        "\n",
-        "Please implement `encode` and `decode`.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 54,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class Codec:\n",
-        "\n",
-        "    def encode(self, strs):\n",
-        "        \"\"\"Encodes a list of strings to a single string.\n",
-        "        \n",
-        "        :type strs: List[str]\n",
-        "        :rtype: str\n",
-        "        \"\"\"\n",
-        "        encoded = \"\"\n",
-        "        for s in strs:\n",
-        "            encoded += s.replace(':', '::') + ':;'\n",
-        "        return encoded\n",
-        "\n",
-        "    def decode(self, s):\n",
-        "        \"\"\"Decodes a single string to a list of strings.\n",
-        "        \n",
-        "        :type s: str\n",
-        "        :rtype: List[str]\n",
-        "        \"\"\"\n",
-        "        decoded = []\n",
-        "        i = 0\n",
-        "        while i < len(s):\n",
-        "            end = s.find(':;', i)\n",
-        "            if end == -1:\n",
-        "                end = len(s)\n",
-        "            decoded.append(s[i:end].replace('::', ':'))\n",
-        "            i = end + 2\n",
-        "        return decoded"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a class called `Codec` to encapsulate the encoding and decoding operations for a list of strings.\n",
-        "\n",
-        "2. In the `encode` method, we take a list of strings (`strs`) as input and return a single encoded string. The encoding process involves concatenating the strings together with a delimiter `:;`. We iterate through each string in the input list, replace any occurrence of `:` with `::` (to avoid conflicts with the delimiter), and then append `:;` to indicate the end of that string.\n",
-        "\n",
-        "3. In the `decode` method, we take an encoded string (`s`) as input and return a list of strings. The decoding process involves splitting the encoded string using the `:;` delimiter. We start from the beginning of the encoded string (`i = 0`) and repeatedly find the next occurrence of `:;`. We extract the substring between the current position (`i`) and the next delimiter (`end`). Before adding it to the result list, we replace any `::` with `:` to revert the encoding. We then update the current position `i` to be after the delimiter, so we can find the next substring in the next iteration.\n",
-        "\n",
-        "4. Finally, we create an instance of the `Codec` class and test it with two examples:\n",
-        "\n",
-        "By using the `:;` delimiter and handling the `::` escaping, this code can encode and decode lists of strings, preserving the original content even if it contains colons or the delimiter itself."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 55,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['lint', 'code', 'love', 'you']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "codec = Codec()\n",
-        "\n",
-        "input1 = [\"lint\", \"code\", \"love\", \"you\"]\n",
-        "encoded1 = codec.encode(input1)\n",
-        "decoded1 = codec.decode(encoded1)\n",
-        "print(decoded1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 57,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['we', 'say', ':', 'yes']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "codec = Codec() \n",
-        "\n",
-        "input2 = [\"we\", \"say\", \":\", \"yes\"]\n",
-        "encoded2 = codec.encode(input2)\n",
-        "decoded2 = codec.decode(encoded2)\n",
-        "print(decoded2)  "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `encode` and `decode` methods in the provided code.\n",
-        "\n",
-        "**Encode Method (`encode`):**\n",
-        "\n",
-        "- Time Complexity: O(n * m)\n",
-        "  - Here, `n` is the number of strings in the input list `strs`, and `m` is the average length of these strings.\n",
-        "  - In the worst case, for each string in the list, we iterate over its characters once to replace any colons (`:`) with double colons (`::`) and then append `:;`. This is done for each string in the list.\n",
-        "\n",
-        "- Space Complexity: O(n * m)\n",
-        "  - The space complexity is also O(n * m) because we build the encoded string by concatenating the modified strings. In the worst case, the encoded string can be of the same length as the original strings.\n",
-        "\n",
-        "**Decode Method (`decode`):**\n",
-        "\n",
-        "- Time Complexity: O(n * m)\n",
-        "  - Similar to the encode method, we iterate through the encoded string in the decode method. In the worst case, we may have to scan each character in the encoded string to find the `:;` delimiter.\n",
-        "\n",
-        "- Space Complexity: O(n * m)\n",
-        "  - The space complexity of the decode method is also O(n * m) because we build the list of decoded strings, which can be of the same length as the encoded string.\n",
-        "\n",
-        "Overall, both the `encode` and `decode` methods have time and space complexities of O(n * m), where `n` is the number of strings in the input list, and `m` is the average length of these strings. The space complexity arises from storing the encoded or decoded strings, and the time complexity arises from iterating through the characters in these strings."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Delimiters**: Modify the code to allow for multiple delimiters, not just one. Each string could have its own delimiter, and the code should be able to handle this complexity.\n",
-        "\n",
-        "2. **Optimize Encoding**: Modify the `encode` method to achieve a more efficient encoding algorithm. Try to minimize the length of the encoded string while ensuring that it can be correctly decoded."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/01. Array Hashing/README.md b/_build/html/_sources/01. Array Hashing/README.md
deleted file mode 100644
index a331dab..0000000
--- a/_build/html/_sources/01. Array Hashing/README.md	
+++ /dev/null
@@ -1,27 +0,0 @@
-# Array & Hashing Solutions - Blind 75 LeetCode Problems
-![Array & Hashing Solutions](https://img.shields.io/badge/Array_&_Hashing_Solutions-blue)![Blind 75 LeetCode](https://img.shields.io/badge/Blind_75_LeetCode-Problems-blue?labelColor=red)
-
-This repository contains my solutions to Array & Hashing Problems from Blind 75 LeetCode problems. I've organized the solutions by categories for easier navigation and reference. Each problem solution is presented in Jupyter Notebook format (`.ipynb`).
-
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [217. Contains Duplicate](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/217.%20Contains%20Duplicate.ipynb)                            | Easy       |
-| [242. Valid Anagram](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/242.%20Valid%20Anagram.ipynb)                                      | Easy       |
-| [1. Two Sum](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/1.%20Two%20Sum.ipynb)                                                      | Easy       |
-| [49. Group Anagrams](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/49.%20Group%20Anagrams.ipynb)                                      | Medium     |
-| [347. Top K Frequent Elements](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.ipynb)              | Medium     |
-| [238. Product of Array Except Self](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.ipynb)  | Medium     |
-| [659. Encode and Decode Strings](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.ipynb)          | Medium     |
-| [128. Longest Consecutive Sequence](https://github.com/mohsentabibian/LeetCode-Solutions/blob/main/01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.ipynb)      | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb b/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb
deleted file mode 100644
index 284c832..0000000
--- a/_build/html/_sources/02. Two Pointers/11. Container With Most Water.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 11. Container With Most Water\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Container With Most Water problem on LeetCode, click here!](https://leetcode.com/problems/container-with-most-water/)\n",
-        "\n",
-        "---\n",
-        "You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the $i^{th}$ line are `(i, 0)` and `(i, height[i])`.\n",
-        "\n",
-        "Find two lines that together with the x-axis form a container, such that the container contains the most water.\n",
-        "\n",
-        "Return *the maximum amount of water a container can store.*\n",
-        "\n",
-        "**Notice** that you may not slant the container.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `n == height.length`\n",
-        "- 2 <= `n` <= $10^5$\n",
-        "- 0 <= `height[i]` <= $10^4$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def maxArea(height):\n",
-        "    # Initialize the maximum area to 0.\n",
-        "    max_area = 0\n",
-        "    # Initialize two pointers, one at the beginning (left) and one at the end (right) of the array.\n",
-        "    left = 0\n",
-        "    right = len(height) - 1\n",
-        "\n",
-        "    # Iterate until the left pointer is less than the right pointer.\n",
-        "    while left < right:\n",
-        "        # Calculate the height of the container, which is the minimum height of the two lines.\n",
-        "        h = min(height[left], height[right])\n",
-        "        # Calculate the width of the container, which is the distance between the two pointers.\n",
-        "        width = right - left\n",
-        "        # Calculate the area of the container using height and width and update max_area if it's greater.\n",
-        "        max_area = max(max_area, h * width)\n",
-        "\n",
-        "        # Move the pointer with the shorter line inward.\n",
-        "        if height[left] < height[right]:\n",
-        "            left += 1\n",
-        "        else:\n",
-        "            right -= 1\n",
-        "\n",
-        "    # Continue this process until left < right, and then return the maximum area.\n",
-        "    return max_area"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start with initializing `max_area` to 0, which will store the maximum area of water that can be contained by the lines.\n",
-        "\n",
-        "2. Two pointers, `left` and `right`, are initialized to the beginning and end of the input `height` array.\n",
-        "\n",
-        "3. The main loop runs while `left` is less than `right`. This ensures that we are checking all possible pairs of lines in a systematic way.\n",
-        "\n",
-        "4. Inside the loop:\n",
-        "   - We calculate the height of the container `h` as the minimum of the heights at the `left` and `right` pointers. This represents the height of the water the container can hold.\n",
-        "   - We calculate the width of the container as the difference between `right` and `left`. This represents the distance between the two lines.\n",
-        "   - We calculate the area of the container using `h * width` and update `max_area` if the calculated area is greater than the current `max_area`.\n",
-        "\n",
-        "5. After calculating the area and updating `max_area`, we move one of the pointers. We move the pointer pointing to the shorter line inward because moving the pointer pointing to the taller line won't increase the height and will only decrease the width, which will result in a smaller area.\n",
-        "\n",
-        "6. We continue this process until `left` is no longer less than `right`, indicating that we have checked all possible pairs of lines.\n",
-        "\n",
-        "7. Finally, we return the `max_area` which contains the maximum area of water that can be contained by two lines from the input array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 49\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "height1 = [1, 8, 6, 2, 5, 4, 8, 3, 7]\n",
-        "result1 = maxArea(height1)\n",
-        "print(\"Output:\", result1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "height2 = [1, 1]\n",
-        "result2 = maxArea(height2)\n",
-        "print(\"Output:\", result2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Output: 45\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Additional Example\n",
-        "height3 = [3, 9, 3, 4, 7, 2, 12, 6]\n",
-        "result3 = maxArea(height3)\n",
-        "print(\"Output:\", result3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `maxArea` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The function uses a two-pointer approach to find the maximum area of water. In the worst case, both the left and right pointers traverse the entire input array once. Therefore, the time complexity is O(n), where n is the number of elements in the `height` array.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The function uses a constant amount of extra space for variables like `max_area`, `left`, `right`, `h`, and `width`. It does not use any data structures whose space consumption depends on the input size. As a result, the space complexity is O(1), which means it has constant space complexity.\n",
-        "\n",
-        "**In summary,** the `maxArea` function has a time complexity of O(n) and a space complexity of O(1), making it an efficient algorithm for finding the maximum area of water that can be contained by two lines in the input array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the maxArea function to not only return the maximum area but also the indices of the two lines that form the container with the maximum area.\n",
-        "2. **Variant Problem:** Instead of finding the maximum area of water, find the minimum amount of water required to fill all the gaps between the lines. You'll need to return the total water volume needed."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb b/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb
deleted file mode 100644
index 8520b3b..0000000
--- a/_build/html/_sources/02. Two Pointers/125. Valid Palindrome.ipynb	
+++ /dev/null
@@ -1,218 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 125. Valid Palindrome\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Palindrome problem on LeetCode, click here!](https://leetcode.com/problems/valid-palindrome/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.\n",
-        "\n",
-        "Given a string `s`, *return `true` if it is a **palindrome**, or `false` otherwise.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $2*10^5$\n",
-        "- `s` consists only of printable ASCII characters.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isPalindrome(s):\n",
-        "    # Step 1: Convert the input string to lowercase\n",
-        "    s = s.lower()\n",
-        "    \n",
-        "    # Step 2: Remove all non-alphanumeric characters from the string\n",
-        "    cleaned_s = ''.join(c for c in s if c.isalnum())\n",
-        "    \n",
-        "    # Step 3: Initialize two pointers, one at the beginning and one at the end of the cleaned string\n",
-        "    left, right = 0, len(cleaned_s) - 1\n",
-        "    \n",
-        "    # Step 4: Compare characters using the two pointers while moving them towards each other\n",
-        "    while left < right:\n",
-        "        if cleaned_s[left] != cleaned_s[right]:\n",
-        "            return False  # If characters don't match, it's not a palindrome\n",
-        "        left += 1\n",
-        "        right -= 1\n",
-        "    \n",
-        "    # Step 5: If the loop completes without returning False, it's a palindrome\n",
-        "    return True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `isPalindrome` function takes an input string `s`.\n",
-        "\n",
-        "2. It starts by converting the input string to lowercase using `s.lower()`. This step ensures that the function is case-insensitive when checking for palindromes.\n",
-        "\n",
-        "3. Next, it removes all non-alphanumeric characters from the string and stores the result in the `cleaned_s` variable. It does this by iterating through each character in the input string and only keeping characters that are alphanumeric (letters and digits). This step removes spaces, punctuation, and other non-alphanumeric characters.\n",
-        "\n",
-        "4. The function then initializes two pointers, `left` and `right`, which will be used to compare characters from the beginning and end of the cleaned string.\n",
-        "\n",
-        "5. In a while loop, it compares characters using the two pointers while moving them towards each other. The loop continues until the `left` pointer is less than the `right` pointer. This is done to check all characters in the cleaned string.\n",
-        "\n",
-        "6. Inside the loop, it checks if the characters at the `left` and `right` pointers don't match. If they don't match, it means the input is not a palindrome, so the function returns `False`.\n",
-        "\n",
-        "7. If the characters at the `left` and `right` pointers match, the pointers are moved closer to each other by incrementing `left` and decrementing `right`.\n",
-        "\n",
-        "8. The loop continues to compare characters until it either finds a mismatch (returning `False`) or until the `left` pointer is greater than or equal to the `right` pointer. If the loop completes without finding a mismatch, it means the input is a palindrome, so the function returns `True`.\n",
-        "\n",
-        "In summary, the code converts the input string to lowercase, removes non-alphanumeric characters, and then uses two pointers to compare characters from the cleaned string's beginning and end. If all characters match during the comparison, the function returns `True`, indicating that the input is a palindrome. If any characters do not match, it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "input1 = \"A man, a plan, a canal: Panama\"\n",
-        "output1 = isPalindrome(input1)\n",
-        "print(output1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "input2 = \"race a car\"\n",
-        "output2 = isPalindrome(input2)\n",
-        "print(output2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "input3 = \"\"\n",
-        "output3 = isPalindrome(input3)\n",
-        "print(output3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `isPalindrome` function that uses two pointers:\n",
-        "\n",
-        "1. **Time Complexity:**\n",
-        "   - Converting the input string to lowercase using `s.lower()` takes O(n) time, where 'n' is the length of the input string.\n",
-        "   - Removing non-alphanumeric characters using the list comprehension `''.join(c for c in s if c.isalnum())` also takes O(n) time because it iterates through each character in the string once.\n",
-        "   - The comparison of characters using the two pointers (`left` and `right`) is done in a loop that runs until `left` is less than `right`. In the worst case, the loop iterates through half of the string, so it takes O(n/2) time, which is still considered O(n).\n",
-        "\n",
-        "   Therefore, the overall time complexity of the `isPalindrome` function is O(n), where 'n' is the length of the input string.\n",
-        "\n",
-        "2. **Space Complexity:**\n",
-        "   - The space complexity is primarily determined by the additional string `cleaned_s` created to store the cleaned version of the input string. In the worst case, if all characters in the input string are alphanumeric, this cleaned string can be as large as the original input string.\n",
-        "   - Additionally, the function uses two pointers, `left` and `right`, which consume constant space and do not depend on the size of the input.\n",
-        "\n",
-        "**In summary,** the overall space complexity of the `isPalindrome` function is O(n) in the worst case, where 'n' is the length of the input string. The space complexity is dominated by the space required for `cleaned_s`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Longest Palindromic Substring:** Given a string `s`, find and return the longest palindromic substring within `s`. For example, if `s = \"babad\"`, the function should return `\"bab\"` or `\"aba\"`.\n",
-        "2. **Palindrome with Limited Characters:** Given a string `s` and a list of characters `allowed_chars`, check if `s` is a palindrome considering only the characters from `allowed_chars`. Ignore all other characters. For example, if `s = \"A man, a plan, a canal: Panama\"` and `allowed_chars = ['a', 'm', 'n', 'p']`, the function should return `True`."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb b/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb
deleted file mode 100644
index 06de549..0000000
--- a/_build/html/_sources/02. Two Pointers/15. 3Sum.ipynb	
+++ /dev/null
@@ -1,250 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 15. 3Sum\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the 3Sum problem on LeetCode, click here!](https://leetcode.com/problems/3sum/)\n",
-        "\n",
-        "---\n",
-        "Given an integer array nums, *return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.*\n",
-        "\n",
-        "**Notice** that the solution set must not contain duplicate triplets.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `3 <= nums.length <= 3000`\n",
-        "- $-10^5$ <= `nums[i]` <= $10^5$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def threeSum(nums):\n",
-        "    # Sort the input array in ascending order\n",
-        "    nums.sort()\n",
-        "    triplets = []\n",
-        "    \n",
-        "    for i in range(len(nums) - 2):\n",
-        "        # Skip duplicates to avoid duplicate triplets\n",
-        "        if i > 0 and nums[i] == nums[i - 1]:\n",
-        "            continue\n",
-        "        \n",
-        "        left, right = i + 1, len(nums) - 1\n",
-        "        \n",
-        "        while left < right:\n",
-        "            # Calculate the total sum of the current triplet\n",
-        "            total = nums[i] + nums[left] + nums[right]\n",
-        "            \n",
-        "            if total == 0:\n",
-        "                # If the total is zero, we found a valid triplet\n",
-        "                triplets.append([nums[i], nums[left], nums[right]])\n",
-        "                \n",
-        "                # Skip duplicates of left and right pointers\n",
-        "                while left < right and nums[left] == nums[left + 1]:\n",
-        "                    left += 1\n",
-        "                while left < right and nums[right] == nums[right - 1]:\n",
-        "                    right -= 1\n",
-        "                \n",
-        "                # Move the pointers to the next unique elements\n",
-        "                left += 1\n",
-        "                right -= 1\n",
-        "            elif total < 0:\n",
-        "                # If the total is negative, we need to increase the sum by moving the left pointer to the right\n",
-        "                left += 1\n",
-        "            else:\n",
-        "                # If the total is positive, we need to decrease the sum by moving the right pointer to the left\n",
-        "                right -= 1\n",
-        "    \n",
-        "    return triplets"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. It starts by sorting the input array `nums` in ascending order. Sorting helps in efficiently finding triplets that sum up to zero.\n",
-        "\n",
-        "2. It initializes an empty list `triplets` to store the unique triplets that meet the given conditions.\n",
-        "\n",
-        "3. The code then iterates through the `nums` array using a loop with index `i`. This loop will consider each element of the array as a potential starting point for a triplet.\n",
-        "\n",
-        "4. Inside the loop, it checks for duplicates and skips them. This is done to ensure that the solution set does not contain duplicate triplets. If `nums[i]` is the same as the previous element `nums[i-1]`, it continues to the next iteration of the loop.\n",
-        "\n",
-        "5. Two pointers, `left` and `right`, are initialized. `left` starts just after the current element `nums[i]`, and `right` starts at the end of the array.\n",
-        "\n",
-        "6. The code enters another loop with `left` and `right` pointers, trying to find a triplet that sums up to zero.\n",
-        "\n",
-        "7. It calculates the `total` sum of the current triplet as `nums[i] + nums[left] + nums[right]`.\n",
-        "\n",
-        "8. If `total` is zero, it means a valid triplet is found, so it appends `[nums[i], nums[left], nums[right]]` to the `triplets` list. Then, it moves both the `left` and `right` pointers to their next unique elements while skipping duplicates.\n",
-        "\n",
-        "9. If `total` is less than zero, it means the sum is negative, so it increments the `left` pointer to move towards a larger sum.\n",
-        "\n",
-        "10. If `total` is greater than zero, it means the sum is positive, so it decrements the `right` pointer to move towards a smaller sum.\n",
-        "\n",
-        "11. This process continues until all possible triplets have been considered.\n",
-        "\n",
-        "12. Finally, the function returns the `triplets` list, which contains all the unique triplets that sum up to zero in the sorted `nums` array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[-1, -1, 2], [-1, 0, 1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "nums = [-1, 0, 1, 2, -1, -4]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "nums = [0, 1, 1]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 0, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "nums = [0, 0, 0]\n",
-        "result = threeSum(nums)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `threeSum` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. Sorting the input array `nums` takes O(n log n) time, where n is the length of the array.\n",
-        "\n",
-        "2. The outer loop runs for each element of the array, so it iterates O(n) times.\n",
-        "\n",
-        "3. Inside the outer loop, we have a while loop with two pointers (`left` and `right`). In the worst case, the while loop can iterate O(n) times (when all elements are unique).\n",
-        "\n",
-        "4. Inside the while loop, we have constant time operations (additions, comparisons, and list appends).\n",
-        "\n",
-        "Overall, the time complexity of the `threeSum` function is dominated by the sorting step, so it is O(n log n) due to the sorting. The other operations inside the loops contribute linearly but are dominated by the sorting step.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The space used by the `triplets` list to store the output triplets is O(k), where k is the number of unique triplets that sum up to zero. In the worst case, this can be O(n^2/3) because there can be roughly O(n^2) possible triplets, and many of them may sum up to zero.\n",
-        "\n",
-        "2. Other than the `triplets` list, the function uses only a constant amount of additional space for variables like `i`, `left`, `right`, and `total`.\n",
-        "\n",
-        "Overall, the space complexity of the `threeSum` function is O(k), where k is the number of unique triplets that meet the criteria. It's important to note that the space complexity does not depend on the size of the input array but rather on the number of valid triplets found.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: O(n log n) due to sorting (dominant factor) and O(n^2) in the worst case for the nested loops.\n",
-        "- Space Complexity: O(k) for storing the output triplets, where k is the number of unique triplets that meet the criteria."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K-Sum Problem:** Generalize the problem to the K-Sum problem, where instead of finding triplets that sum to zero, you need to find K elements that sum to a target value. Implement a function for this generalization.\n",
-        "2. **Count the Number of Unique Triplets:** Modify the algorithm to count the number of unique triplets that satisfy the conditions instead of returning the triplets themselves. This will require some modifications to handle duplicates efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/02. Two Pointers/README.md b/_build/html/_sources/02. Two Pointers/README.md
deleted file mode 100644
index 0d4c7cf..0000000
--- a/_build/html/_sources/02. Two Pointers/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Two Pointers Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)                            | Easy       |
-| [15. 3Sum](https://leetcode.com/problems/3sum/)                                                       | Medium     |
-| [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)           | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb b/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb
deleted file mode 100644
index 554c940..0000000
--- a/_build/html/_sources/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb	
+++ /dev/null
@@ -1,191 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 121. Best Time to Buy and Sell Stock\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Best Time to Buy and Sell Stock problem on LeetCode, click here!](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given an array `prices` where `prices[i]` is the price of a given stock on the $i^{th}$ day.\n",
-        "\n",
-        "You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.\n",
-        "\n",
-        "*Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `prices.length` <= $10^5$\n",
-        "- 0 <= `prices[i]` <= $10^4$\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 65,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def maxProfit(prices):\n",
-        "    if not prices:\n",
-        "        return 0\n",
-        "    \n",
-        "    min_price = prices[0]\n",
-        "    max_profit = 0\n",
-        "    \n",
-        "    for price in prices:\n",
-        "        # Update the minimum price if a lower price is encountered\n",
-        "        min_price = min(min_price, price)\n",
-        "        # Calculate the profit if selling at the current price\n",
-        "        max_profit = max(max_profit, price - min_price)\n",
-        "    \n",
-        "    return max_profit"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- The `maxProfit` function takes a single argument, `prices`, which is a list of stock prices.\n",
-        "- The first line of the function checks if the `prices` list is empty. If it's empty, there are no prices to analyze, so the function returns 0 (no profit can be made).\n",
-        "\n",
-        "- Then, we initialize two variables:\n",
-        "  - `min_price`: This variable will keep track of the minimum price seen so far. We initialize it with the price of the first day in the list (`prices[0]`).\n",
-        "  - `max_profit`: This variable will keep track of the maximum profit we can achieve. We initialize it to 0 because initially, we haven't made any profit.\n",
-        "\n",
-        "- We start iterating through the `prices` list, where `price` represents the stock price for the current day.\n",
-        "\n",
-        "- In each iteration, we check if the current `price` is lower than the `min_price` we've seen so far. If it is, we update `min_price` to the current `price`. This is because we want to buy the stock at the lowest possible price.\n",
-        "\n",
-        "- Next, we calculate the profit we would make if we sold the stock at the current `price` (assuming we bought it at `min_price`). We do this by subtracting `min_price` from the current `price`. This represents the profit for the current day.\n",
-        "\n",
-        "- We also use the `max` function to keep track of the maximum profit seen so far. If the profit calculated for the current day is greater than the `max_profit` we've seen previously, we update `max_profit` with this higher value.\n",
-        "\n",
-        "- Finally, after iterating through all the days in the `prices` list, we return `max_profit`, which represents the maximum profit that can be achieved by buying and selling a single stock."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 66,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "5\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "prices1 = [7, 1, 5, 3, 6, 4]\n",
-        "print(maxProfit(prices1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 67,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "0\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "prices2 = [7, 6, 4, 3, 1]\n",
-        "print(maxProfit(prices2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code iterates through the `prices` array once, examining each price exactly once.\n",
-        "- Within the loop, there are constant-time operations such as comparisons, additions, and subtractions.\n",
-        "- Therefore, the time complexity of this code is O(n), where n is the length of the `prices` array. It performs a linear number of operations relative to the size of the input.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the code is constant, O(1).\n",
-        "- Regardless of the size of the input `prices` array, the code only uses a fixed amount of additional memory to store two variables (`min_price` and `max_profit`) and a few loop variables.\n",
-        "- The memory usage does not depend on the size of the input, so it is considered constant space complexity.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: O(n) where n is the length of the `prices` array.\n",
-        "- Space Complexity: O(1) (constant space usage)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Transactions Allowed:** Modify the problem to allow multiple transactions (buy and sell multiple times), but you must sell before buying again. Find the maximum profit in this scenario.\n",
-        "2. **K Transactions Allowed:** Extend the problem to allow at most k transactions. Find the maximum profit considering this constraint.\n",
-        "3. **With Transaction Fee:** Introduce a transaction fee for each buy/sell operation. Modify the code to maximize profit while considering the transaction fee."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb b/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb
deleted file mode 100644
index 4a516de..0000000
--- a/_build/html/_sources/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb	
+++ /dev/null
@@ -1,214 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 3. Longest Substring Without Repeating Characters\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Substring Without Repeating Characters problem on LeetCode, click here!](https://leetcode.com/problems/longest-substring-without-repeating-characters/)\n",
-        "\n",
-        "---\n",
-        "Given a string `s`, find the length of the **longest substring** without repeating characters.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 0 <= `s.length` <= $5 * 10^4$\n",
-        "- `s` consists of English letters, digits, symbols and spaces.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 61,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def length_of_longest_substring(s):\n",
-        "    # Create a dictionary to store the index of each character's last occurrence.\n",
-        "    char_index = {}\n",
-        "    \n",
-        "    # Initialize variables to keep track of the start and end of the current substring.\n",
-        "    start = 0\n",
-        "    max_length = 0\n",
-        "    \n",
-        "    for end in range(len(s)):\n",
-        "        # If the character is in the dictionary and its last occurrence is after the start of the current substring,\n",
-        "        # update the start of the substring to the next character after its last occurrence.\n",
-        "        if s[end] in char_index and char_index[s[end]] >= start:\n",
-        "            start = char_index[s[end]] + 1\n",
-        "        \n",
-        "        # Update the last occurrence index of the current character.\n",
-        "        char_index[s[end]] = end\n",
-        "        \n",
-        "        # Calculate the length of the current substring and update the maximum length if necessary.\n",
-        "        current_length = end - start + 1\n",
-        "        max_length = max(max_length, current_length)\n",
-        "    \n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a function called `length_of_longest_substring` that takes a single argument `s`, which is the input string for which we want to find the length of the longest substring without repeating characters.\n",
-        "\n",
-        "2. We create a dictionary called `char_index` to store the index of the last occurrence of each character in the input string. This dictionary will help us keep track of where each character was last seen.\n",
-        "\n",
-        "3. We initialize two variables, `start` and `max_length`. \n",
-        "   - `start` represents the start index of the current substring without repeating characters. It starts at 0.\n",
-        "   - `max_length` is used to keep track of the maximum length found so far and is initially set to 0.\n",
-        "\n",
-        "4. We then iterate through the input string `s` using a for loop, where the loop variable `end` represents the current end index of the substring we are considering.\n",
-        "\n",
-        "5. Inside the loop:\n",
-        "   - We check if the character `s[end]` is already in the `char_index` dictionary and if its last occurrence is within or after the current substring. If so, it means that we've encountered a repeating character, and we need to update the `start` of the substring to the next character after the last occurrence of `s[end]`. This ensures that we have a new valid substring without repeating characters.\n",
-        "   - We then update the `char_index` dictionary by storing the current index `end` as the last occurrence index of the character `s[end]`.\n",
-        "\n",
-        "6. Next, we calculate the length of the current substring without repeating characters, which is `current_length = end - start + 1`. We add 1 to account for the fact that the indices are zero-based.\n",
-        "\n",
-        "7. We update the `max_length` with the maximum of its current value and the `current_length`. This step ensures that we keep track of the longest valid substring we have encountered so far.\n",
-        "\n",
-        "8. The loop continues to iterate through the input string, and at the end, we return the `max_length`, which represents the length of the longest substring without repeating characters."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 62,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(length_of_longest_substring(\"abcabcbb\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 63,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(length_of_longest_substring(\"bbbbb\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 64,
-      "id": "cf04eba0-0ead-4e8b-ba33-4316ea64a9d3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(length_of_longest_substring(\"pwwkew\"))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "Let's analyze the time and space complexity of the `length_of_longest_substring` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The function iterates through the input string `s` using a single for loop. The loop runs from the beginning of the string to the end once.\n",
-        "- Inside the loop, we perform constant-time operations such as dictionary lookups and updates, comparisons, and arithmetic operations.\n",
-        "- Therefore, the time complexity of the function is O(n), where n is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The primary data structure that consumes space in this function is the `char_index` dictionary.\n",
-        "- In the worst case, if there are no repeating characters in the input string, the dictionary can store all unique characters in the string.\n",
-        "- Therefore, the space complexity is O(min(n, m)), where n is the length of the input string `s`, and m is the number of unique characters in the string. In the worst case, when all characters are unique, m is equal to n, so the space complexity is O(n).\n",
-        "- Additionally, there are a few integer variables used for indices and lengths, which consume constant space.\n",
-        "- Overall, the space complexity of the function is O(min(n, m)) or simply O(n) in the worst case.\n",
-        "\n",
-        "**In summary,** the time complexity of the `length_of_longest_substring` function is O(n), and the space complexity is O(n) in the worst case due to the `char_index` dictionary. This algorithm provides an efficient way to find the length of the longest substring without repeating characters in linear time."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Longest Substring with K Distinct Characters:** Modify the problem to find the length of the longest substring with exactly K distinct characters. For example, given the input string \"abcabcbb\" and K = 2, the answer would be 4 because the longest substring with two distinct characters is \"abca.\"\n",
-        "2. **Longest Substring with Unique Characters:** Write a function to find the length of the longest substring in a given string where all characters are unique. For example, given the input string \"abcabcbb,\" the answer would be 4 because \"abcd\" is the longest substring with unique characters."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb b/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb
deleted file mode 100644
index 07e96a2..0000000
--- a/_build/html/_sources/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb	
+++ /dev/null
@@ -1,195 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 424. Longest Repeating Character Replacement\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Longest Repeating Character Replacement problem on LeetCode, click here!](https://leetcode.com/problems/longest-repeating-character-replacement/)\n",
-        "\n",
-        "---\n",
-        "You are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.\n",
-        "\n",
-        "*Return the length of the longest substring containing the same letter you can get after performing the above operations.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $10^5$\n",
-        "- `s` consists of only uppercase English letters.\n",
-        "- `0 <= k <= s.length`"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 58,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def characterReplacement(s, k):\n",
-        "    max_length = 0  # Initialize the maximum length\n",
-        "    max_count = 0   # Initialize the maximum count of repeating characters\n",
-        "    start = 0       # Initialize the start of the sliding window\n",
-        "    char_count = {} # Dictionary to store the count of each character\n",
-        "\n",
-        "    for end in range(len(s)):\n",
-        "        # Update the count of the current character in the dictionary\n",
-        "        char_count[s[end]] = char_count.get(s[end], 0) + 1\n",
-        "        \n",
-        "        # Update the maximum count of repeating characters\n",
-        "        max_count = max(max_count, char_count[s[end]])\n",
-        "        \n",
-        "        # Check if the current window size is greater than k\n",
-        "        if (end - start + 1) - max_count > k:\n",
-        "            # Move the start of the window to the right\n",
-        "            char_count[s[start]] -= 1\n",
-        "            start += 1\n",
-        "        \n",
-        "        # Update the maximum length\n",
-        "        max_length = max(max_length, end - start + 1)\n",
-        "\n",
-        "    return max_length"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We initialize some variables:\n",
-        "   - `max_length` to keep track of the maximum length of the substring containing the same letter.\n",
-        "   - `max_count` to keep track of the maximum count of repeating characters within the current window.\n",
-        "   - `start` to represent the start index of the sliding window.\n",
-        "   - `char_count` is a dictionary that will store the count of each character within the current window.\n",
-        "\n",
-        "2. We use a `for` loop to iterate through the characters of the string `s`.\n",
-        "\n",
-        "3. Inside the loop, we do the following for each character at position `end`:\n",
-        "   - Update the count of the current character in the `char_count` dictionary.\n",
-        "   - Update the `max_count` to be the maximum of the current `max_count` and the count of the current character. This keeps track of the maximum count of repeating characters within the current window.\n",
-        "\n",
-        "4. We check if the current window size (the difference between `end` and `start` plus one) minus the `max_count` is greater than `k`. This condition checks whether we have exceeded the allowed number of replacements (`k`) within the current window.\n",
-        "\n",
-        "   - If we have exceeded the allowed replacements, it means we need to shrink the window from the left side. We do this by moving the `start` of the window to the right and decrementing the count of the character at `s[start]` in the `char_count` dictionary. This effectively removes characters from the left side of the window until we have a valid window again.\n",
-        "\n",
-        "   - By doing this, we ensure that the difference between the current window size and the `max_count` is always less than or equal to `k`.\n",
-        "\n",
-        "5. After each character, we update the `max_length` to be the maximum of the current `max_length` and the size of the current window (`end - start + 1`).\n",
-        "\n",
-        "6. Finally, we return `max_length`, which holds the length of the longest substring containing the same letter with at most `k` replacements."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 59,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "s1 = \"ABAB\"\n",
-        "k1 = 2\n",
-        "print(characterReplacement(s1, k1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 60,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "s2 = \"AABABBA\"\n",
-        "k2 = 1\n",
-        "print(characterReplacement(s2, k2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code uses a single `for` loop to iterate through the characters of the input string `s`. The loop runs from the beginning to the end of the string.\n",
-        "- Inside the loop, we perform constant-time operations such as updating the `char_count` dictionary and updating variables like `max_count` and `max_length`.\n",
-        "- The code's time complexity is primarily determined by the loop, which iterates through each character in the string once. Therefore, the time complexity is O(n), where n is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The code uses several variables to store information, but the space they consume is constant and does not depend on the size of the input string. These variables include `max_length`, `max_count`, `start`, and `char_count`. Therefore, the space complexity is O(1) or constant space.\n",
-        "\n",
-        "**In summary**, the time complexity of the code is O(n), where n is the length of the input string, and the space complexity is O(1), as it uses a constant amount of additional space regardless of the input size. This algorithm efficiently solves the problem with a linear time complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the code to find the longest substring containing the same letter with at most k replacements in O(n) time complexity and O(1) space complexity. Hint: You may need to update the approach to achieve this.\n",
-        "2. **Variation with Lowercase Letters:** Extend the problem to include both uppercase and lowercase English letters in the input string s. Write a function that can handle this extended input and still find the longest substring with at most k replacements efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb b/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb
deleted file mode 100644
index 33c2a44..0000000
--- a/_build/html/_sources/03. Sliding Window/76. Minimum Window Substring.ipynb	
+++ /dev/null
@@ -1,273 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 76. Minimum Window Substring\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Minimum Window Substring problem on LeetCode, click here!](https://leetcode.com/problems/minimum-window-substring/)\n",
-        "\n",
-        "---\n",
-        "Given two strings `s` and `t` of lengths `m` and `n` respectively, return *the **minimum window substring** of `s` such that every character in `t` (**including duplicates**) is included in the window. If there is no such substring, return the empty string `\"\"`.*\n",
-        "\n",
-        "The testcases will be generated such that the answer is **unique**.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `m == s.length`\n",
-        "- `n == t.length`\n",
-        "- 1 <= `m, n` <= $10^5$\n",
-        "- `s` and `t` consist of uppercase and lowercase English letters.\n",
-        "\n",
-        "**Follow up**: Could you find an algorithm that runs in $O(m + n)$ time?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 54,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def min_window_substring(s, t):\n",
-        "    if not s or not t:\n",
-        "        return \"\"\n",
-        "\n",
-        "    # Initialize dictionaries to keep track of character counts for t and the current window in s.\n",
-        "    t_dict = {}\n",
-        "    current_window_dict = {}\n",
-        "    \n",
-        "    # Populate t_dict with character counts for string t.\n",
-        "    for char in t:\n",
-        "        t_dict[char] = t_dict.get(char, 0) + 1\n",
-        "\n",
-        "    # Initialize pointers for the sliding window.\n",
-        "    left = 0\n",
-        "    min_len = float('inf')\n",
-        "    min_window = \"\"\n",
-        "    required_chars = len(t_dict)\n",
-        "\n",
-        "    # Initialize a variable to keep track of how many required characters have been found in the current window.\n",
-        "    found_chars = 0\n",
-        "\n",
-        "    # Iterate over the string s using the right pointer.\n",
-        "    for right in range(len(s)):\n",
-        "        char = s[right]\n",
-        "\n",
-        "        # Update the current_window_dict.\n",
-        "        current_window_dict[char] = current_window_dict.get(char, 0) + 1\n",
-        "\n",
-        "        # Check if the current character is a required character and if its count matches the required count.\n",
-        "        if char in t_dict and current_window_dict[char] == t_dict[char]:\n",
-        "            found_chars += 1\n",
-        "\n",
-        "        # Try to minimize the window by moving the left pointer.\n",
-        "        while left <= right and found_chars == required_chars:\n",
-        "            # Calculate the current window size.\n",
-        "            window_size = right - left + 1\n",
-        "\n",
-        "            # If the current window is smaller than the minimum found so far, update min_len and min_window.\n",
-        "            if window_size < min_len:\n",
-        "                min_len = window_size\n",
-        "                min_window = s[left:right+1]\n",
-        "\n",
-        "            # Move the left pointer to the right to shrink the window.\n",
-        "            left_char = s[left]\n",
-        "            current_window_dict[left_char] -= 1\n",
-        "\n",
-        "            # Check if the character removed from the window was a required character.\n",
-        "            if left_char in t_dict and current_window_dict[left_char] < t_dict[left_char]:\n",
-        "                found_chars -= 1\n",
-        "\n",
-        "            # Move the left pointer further to continue shrinking the window.\n",
-        "            left += 1\n",
-        "\n",
-        "    return min_window"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by checking if either of the input strings `s` or `t` is empty. If either of them is empty, we return an empty string since there can't be any valid substring in this case.\n",
-        "\n",
-        "2. We initialize two dictionaries: `t_dict` and `current_window_dict`. These dictionaries will be used to keep track of character counts in the string `t` and the current window in string `s`, respectively.\n",
-        "\n",
-        "3. We populate the `t_dict` dictionary by iterating through string `t`. For each character, we increment its count in the dictionary using `t_dict.get(char, 0) + 1`. This allows us to count the occurrences of each character in `t`.\n",
-        "\n",
-        "4. We initialize two pointers: `left` and `right`. The `left` pointer will represent the start of the current window, and the `right` pointer will represent the end of the current window. We also initialize `min_len` to store the length of the minimum window found so far, and `min_window` to store the actual minimum window substring.\n",
-        "\n",
-        "5. We determine the number of required characters in `t` (i.e., the number of distinct characters in `t`) and store it in the variable `required_chars`.\n",
-        "\n",
-        "6. We initialize a variable `found_chars` to keep track of how many required characters have been found in the current window. Initially, it is set to 0.\n",
-        "\n",
-        "7. We iterate over the string `s` using the `right` pointer. In each iteration, we do the following:\n",
-        "\n",
-        "   - Update the `current_window_dict` by incrementing the count of the current character.\n",
-        "   \n",
-        "   - Check if the current character is a required character (present in `t_dict`) and if its count in the `current_window_dict` matches the required count from `t_dict`. If so, we increment `found_chars`.\n",
-        "\n",
-        "8. After updating the window, we attempt to minimize the window size by moving the `left` pointer to the right. In this step, we:\n",
-        "\n",
-        "   - Calculate the size of the current window.\n",
-        "   \n",
-        "   - If the current window size is smaller than the minimum found so far (`min_len`), we update `min_len` and `min_window` to store the current window substring.\n",
-        "   \n",
-        "   - Move the `left` pointer to the right to shrink the window.\n",
-        "   \n",
-        "   - Check if the character being removed from the window was a required character. If it was, we decrement `found_chars`.\n",
-        "   \n",
-        "   - Continue moving the `left` pointer further to continue shrinking the window if the window still contains all required characters.\n",
-        "\n",
-        "9. Finally, we return `min_window`, which will contain the minimum window substring that contains all characters from `t`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 55,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "BANC\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(min_window_substring(\"ADOBECODEBANC\", \"ABC\"))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 56,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "a\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(min_window_substring(\"a\", \"a\")) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 57,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": []
-        },
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(min_window_substring(\"a\", \"aa\")) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `min_window_substring` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The main loop iterates through the string `s` from left to right using the `right` pointer. This loop runs in O(m) time, where 'm' is the length of string `s`.\n",
-        "\n",
-        "- Inside the loop, we have a while loop that moves the `left` pointer to the right to shrink the window as needed. In the worst case, this while loop can also run in O(m) time because in the worst case, we may have to move the `left` pointer all the way to the end of the string.\n",
-        "\n",
-        "- Within each iteration of the while loop, we perform constant time operations, such as updating dictionaries and comparing character counts.\n",
-        "\n",
-        "- The initialization of the `t_dict` dictionary takes O(n) time, where 'n' is the length of string `t`.\n",
-        "\n",
-        "Therefore, the overall time complexity of the function is O(m + n) because the dominant factor is the length of string `s`, and the length of string `t` has a smaller impact.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity is determined by the space used by the `t_dict` dictionary, the `current_window_dict` dictionary, and a few variables.\n",
-        "\n",
-        "- The `t_dict` dictionary stores character counts for string `t`. In the worst case, when all characters in `t` are distinct, this dictionary can have a maximum of 'n' key-value pairs. So, the space complexity for `t_dict` is O(n).\n",
-        "\n",
-        "- The `current_window_dict` dictionary stores character counts for the current window. In the worst case, it can have a maximum of 'm' key-value pairs. So, the space complexity for `current_window_dict` is also O(m).\n",
-        "\n",
-        "- Other variables used in the function, such as `left`, `right`, `min_len`, `min_window`, `required_chars`, and `found_chars`, are all of constant size and do not depend on the input sizes.\n",
-        "\n",
-        "**In summary**, the overall space complexity of the function is O(max(m, n)), which means it is determined by the larger of the two input strings' lengths. This is because the space used for `t_dict` and `current_window_dict` dominates the space complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Smallest Distinct Substring:** Instead of finding the minimum window substring containing all characters, find the smallest distinct (unique) substring of s that contains all characters from t. This variation adds complexity because you must find a substring with distinct characters.\n",
-        "2. **No Extra Space:** Solve the problem without using any extra space, such as dictionaries or arrays, other than a constant amount of space. This is a significant optimization challenge."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/03. Sliding Window/README.md b/_build/html/_sources/03. Sliding Window/README.md
deleted file mode 100644
index e1294ff..0000000
--- a/_build/html/_sources/03. Sliding Window/README.md	
+++ /dev/null
@@ -1,19 +0,0 @@
-# Sliding Window Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [121. Best Time to Buy And Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | Easy       |
-| [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) | Medium |
-| [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) | Medium |
-| [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)             | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb b/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb
deleted file mode 100644
index c47292f..0000000
--- a/_build/html/_sources/04. Stack/20. Valid Parentheses.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 20. Valid Parentheses\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Valid Parentheses problem on LeetCode, click here!](https://leetcode.com/problems/valid-parentheses/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given a string s containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.\n",
-        "\n",
-        "An input string is valid if:\n",
-        "\n",
-        "1. Open brackets must be closed by the same type of brackets.\n",
-        "2. Open brackets must be closed in the correct order.\n",
-        "3. Every close bracket has a corresponding open bracket of the same type.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `s.length` <= $10^4$\n",
-        "- `s` consists of parentheses only `'()[]{}'`."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 49,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def isValid(s):\n",
-        "    # Create an empty stack to store opening brackets\n",
-        "    stack = []\n",
-        "    \n",
-        "    # Define a mapping for matching brackets\n",
-        "    bracket_mapping = {')': '(', '}': '{', ']': '['}\n",
-        "    \n",
-        "    # Iterate through the characters in the input string\n",
-        "    for char in s:\n",
-        "        # If the character is an opening bracket, push it onto the stack\n",
-        "        if char in bracket_mapping.values():\n",
-        "            stack.append(char)\n",
-        "        # If the character is a closing bracket\n",
-        "        elif char in bracket_mapping.keys():\n",
-        "            # Pop the top element from the stack if it exists, or use a dummy value '#'\n",
-        "            top_element = stack.pop() if stack else '#'\n",
-        "            # If the popped element does not match the corresponding opening bracket, return False\n",
-        "            if bracket_mapping[char] != top_element:\n",
-        "                return False\n",
-        "        # If the character is not a bracket, ignore it\n",
-        "        \n",
-        "    # After processing the entire string, if there are any unmatched opening brackets left in the stack, return False\n",
-        "    return len(stack) == 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining the `isValid` function that takes a single argument `s`, which is the input string containing parentheses and brackets.\n",
-        "\n",
-        "2. Inside the function, we create an empty stack, which is a list in Python, to store opening brackets as we encounter them in the input string. The stack will help us keep track of the brackets and their order.\n",
-        "\n",
-        "3. We define a `bracket_mapping` dictionary that maps each closing bracket to its corresponding opening bracket. This mapping will be used to check if a closing bracket matches the most recent opening bracket in the stack.\n",
-        "\n",
-        "4. We iterate through each character `char` in the input string `s` using a `for` loop.\n",
-        "\n",
-        "5. If the character `char` is an opening bracket (i.e., it exists in the `values()` of `bracket_mapping`), we push it onto the stack using the `append()` method.\n",
-        "\n",
-        "6. If the character `char` is a closing bracket (i.e., it exists in the `keys()` of `bracket_mapping`), we need to check if it matches the corresponding opening bracket. To do this, we pop the top element from the stack (if it exists) and store it in `top_element`. We use a dummy value `'#'` if the stack is empty to avoid errors.\n",
-        "\n",
-        "7. We then compare `top_element` with the corresponding opening bracket using `bracket_mapping[char]`. If they do not match, it means the string is not valid, and we return `False`.\n",
-        "\n",
-        "8. If the character `char` is not a bracket, we simply ignore it and continue the loop.\n",
-        "\n",
-        "9. After processing the entire string, we check if there are any unmatched opening brackets left in the stack. If the stack is empty, it means all opening brackets have been properly matched and closed, and we return `True`. Otherwise, we return `False`.\n",
-        "\n",
-        "10. Finally, we provide some test cases at the bottom of the code to demonstrate how the function works for different input strings."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 50,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "print(isValid(\"()\"))    "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 51,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "print(isValid(\"()[]{}\")) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 52,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "print(isValid(\"(]\"))     "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `isValid` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The function iterates through each character in the input string `s` once, performing constant-time operations for each character.\n",
-        "- Pushing and popping elements from the stack (list) also takes constant time in most cases.\n",
-        "- Therefore, the overall time complexity of the function is $O(n)$, where $n$ is the length of the input string `s`.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the function is determined by the space used by the stack and the `bracket_mapping` dictionary.\n",
-        "- In the worst case, if the input string `s` consists entirely of opening brackets, the stack can potentially contain all of these opening brackets, resulting in a space complexity of $O(n)$ in terms of the stack.\n",
-        "- The `bracket_mapping` dictionary has a constant number of key-value pairs (3 pairs in this case).\n",
-        "- Therefore, the overall space complexity of the function is $O(n)$, where $n$ is the length of the input string `s`, mainly due to the stack space.\n",
-        "\n",
-        "**In summary:**\n",
-        "- The time complexity is $O(n)$ because we iterate through the string once.\n",
-        "- The space complexity is $O(n)$ due to the stack used to keep track of opening brackets."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Valid Expressions:** Modify the problem to validate not only brackets but also arithmetic expressions containing parentheses, such as \"$2 * (3 + 5) / (4 - 2)$\".\n",
-        "2. **Valid Parentheses Combinations:**\n",
-        "Write a function to generate all valid combinations of n pairs of parentheses, where $n$ is a positive integer. For example, for $n = 3$, the valid combinations are [\"((()))\", \"(()())\", \"(())()\", \"()(())\", \"()()()\"]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/04. Stack/README.md b/_build/html/_sources/04. Stack/README.md
deleted file mode 100644
index 2ed115b..0000000
--- a/_build/html/_sources/04. Stack/README.md	
+++ /dev/null
@@ -1,16 +0,0 @@
-# Stack Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)                            | Easy       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb b/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb
deleted file mode 100644
index 98f2331..0000000
--- a/_build/html/_sources/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,228 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 153. Find Minimum in Rotated Sorted Array\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Find Minimum in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:\n",
-        "\n",
-        "- `[4,5,6,7,0,1,2]` if it was rotated `4` times.\n",
-        "- `[0,1,2,4,5,6,7]` if it was rotated `7` times.\n",
-        "\n",
-        "Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.\n",
-        "\n",
-        "Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array.*\n",
-        "\n",
-        "You must write an algorithm that runs in $O(log\\ n)$ time.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `n == nums.length`\n",
-        "- `1 <= n <= 5000`\n",
-        "- `-5000 <= nums[i] <= 5000`\n",
-        "- All the integers of `nums` are **unique**.\n",
-        "- `nums` is sorted and rotated between `1` and `n` times."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 39,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def findMin(nums):\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "    \n",
-        "    while left < right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "        \n",
-        "        # If the mid element is greater than the rightmost element,\n",
-        "        # it means the minimum element is in the right half.\n",
-        "        if nums[mid] > nums[right]:\n",
-        "            left = mid + 1\n",
-        "        # If the mid element is less than or equal to the rightmost element,\n",
-        "        # it means the minimum element is in the left half or at the mid itself.\n",
-        "        else:\n",
-        "            right = mid\n",
-        "    \n",
-        "    # The loop will break when left and right converge to the minimum element.\n",
-        "    # At this point, left (or right) will be pointing to the minimum element.\n",
-        "    return nums[left]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The function `findMin(nums)` takes an input list `nums`, which represents the rotated sorted array.\n",
-        "\n",
-        "2. Initialize two pointers `left` and `right` to track the current search range. Initially, `left` is set to 0 (the beginning of the array), and `right` is set to `len(nums) - 1` (the end of the array).\n",
-        "\n",
-        "3. Enter a `while` loop with the condition `left < right`. This loop continues until `left` and `right` converge to a single element, which will be the minimum element in the rotated array.\n",
-        "\n",
-        "4. Inside the loop, calculate the middle index `mid` using integer division. This helps in finding the middle element of the current search range.\n",
-        "\n",
-        "5. Check if the element at the middle index (`nums[mid]`) is greater than the element at the rightmost index (`nums[right]`). If this condition is true, it means that the minimum element must be in the right half of the current search range. So, update `left` to `mid + 1`, effectively eliminating the left half of the search range.\n",
-        "\n",
-        "6. If the condition from step 5 is not met, it implies that the minimum element can be in the left half of the current search range or could be the element at the `mid` index itself. In this case, update `right` to `mid`, effectively eliminating the right half of the search range.\n",
-        "\n",
-        "7. Repeat steps 4-6 until `left` and `right` converge to the minimum element.\n",
-        "\n",
-        "8. When the loop exits, it means that `left` (or `right`) points to the minimum element in the array.\n",
-        "\n",
-        "9. Return `nums[left]` (or `nums[right]`) as the result, which is the minimum element in the rotated sorted array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 43,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1:\n",
-        "nums1 = [3, 4, 5, 1, 2]\n",
-        "print(findMin(nums1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 41,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "0\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "print(findMin(nums2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 42,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "11\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "nums3 = [11, 13, 15, 17]\n",
-        "print(findMin(nums3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this code is $O(log\\ n)$, where '$n$' is the length of the input array `nums`. This is because the binary search algorithm is employed, which continually divides the search range in half with each iteration.\n",
-        "\n",
-        "In each iteration of the `while` loop, the search range is halved, and this process continues until the `left` and `right` pointers converge to the minimum element. In the worst case, it will take logarithmic time to reduce the search range to a single element.\n",
-        "\n",
-        "Therefore, the binary search used in this code runs in $O(log\\ n)$ time complexity.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of this code is $O(1)$, which means it uses a constant amount of additional space that does not depend on the size of the input array.\n",
-        "\n",
-        "The algorithm uses a fixed number of variables (`left`, `right`, `mid`) to keep track of the search range and indices, but the number of these variables does not grow with the size of the input array. Hence, the space complexity is constant.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: $O(log\\ n)$\n",
-        "- Space Complexity: $O(1)$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. Solve the problem with a variation where the input array may contain duplicates.\n",
-        "\n",
-        "2. Modify the problem to return the index of the minimum element in the rotated sorted array, rather than the element itself. The algorithm should still run in $O(log\\ n)$ time.\n",
-        "\n",
-        "3. Implement a function that finds the maximum element in a rotated sorted array. How would you adapt the binary search algorithm to solve this problem efficiently in $O(log\\ n)$ time?\n",
-        "\n",
-        "4. Implement a function that can find the kth smallest element in a rotated sorted array. This is an extension of the original problem, and the algorithm should still run in $O(log\\ n)$ time."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb b/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb
deleted file mode 100644
index d8d4202..0000000
--- a/_build/html/_sources/05. Binary Search/33. Search in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 33. Search in Rotated Sorted Array\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Search in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/search-in-rotated-sorted-array/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "There is an integer array `nums` sorted in ascending order (with **distinct** values).\n",
-        "\n",
-        "Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k (1 <= k < nums.length)` such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index 3 and become `[4,5,6,7,0,1,2]`.\n",
-        "\n",
-        "Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.*\n",
-        "\n",
-        "You must write an algorithm with $O(log\\ n)$ runtime complexity.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `1 <= nums.length <= 5000`\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- All values of `nums` are **unique**.\n",
-        "- `nums` is an ascending array that is possibly rotated.\n",
-        "- $-10^4$ <= `target` <= $10^4$"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 44,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def search(nums, target):\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "\n",
-        "    # Find the pivot index using binary search\n",
-        "    while left < right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "\n",
-        "        if nums[mid] > nums[right]:\n",
-        "            left = mid + 1\n",
-        "        else:\n",
-        "            right = mid\n",
-        "\n",
-        "    pivot = left\n",
-        "    left, right = 0, len(nums) - 1\n",
-        "\n",
-        "    # Determine which part of the array to search in\n",
-        "    if target >= nums[pivot] and target <= nums[right]:\n",
-        "        left = pivot\n",
-        "    else:\n",
-        "        right = pivot\n",
-        "\n",
-        "    # Perform binary search to find the target\n",
-        "    while left <= right:\n",
-        "        mid = left + (right - left) // 2\n",
-        "\n",
-        "        if nums[mid] == target:\n",
-        "            return mid\n",
-        "        elif nums[mid] < target:\n",
-        "            left = mid + 1\n",
-        "        else:\n",
-        "            right = mid - 1\n",
-        "\n",
-        "    return -1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The `search` function takes two arguments: `nums`, which is the rotated sorted array, and `target`, which is the value we want to find in the array.\n",
-        "\n",
-        "2. We initialize two pointers, `left` and `right`, which represent the range in which we are going to perform binary search. Initially, `left` is set to 0, and `right` is set to the index of the last element in the array (`len(nums) - 1`).\n",
-        "\n",
-        "3. We start by finding the pivot index using binary search. The pivot index is the index at which the array is rotated. We use a `while` loop to continue the search until `left` is less than `right`.\n",
-        "\n",
-        "4. Inside the loop, we calculate the middle index `mid` using integer division. We then compare the value at `mid` with the value at `right`. If `nums[mid]` is greater than `nums[right]`, it means the pivot point lies to the right of `mid`, so we update `left` to `mid + 1`. Otherwise, the pivot point lies to the left of `mid`, so we update `right` to `mid`. This process continues until we find the pivot index.\n",
-        "\n",
-        "5. After finding the pivot index, we have divided the array into two parts: one part is sorted in ascending order, and the other part is also sorted in ascending order but rotated. We need to determine which part of the array contains the target value.\n",
-        "\n",
-        "6. We reset `left` and `right` pointers. If the target value is within the range `[nums[pivot], nums[right]]`, we set `left` to `pivot` (the start of the rotated part), indicating that we should search in the rotated part. Otherwise, we set `right` to `pivot` (the end of the sorted part), indicating that we should search in the sorted part.\n",
-        "\n",
-        "7. We then perform binary search again within the chosen range (`left` to `right`) to find the target element. The binary search continues until `left` is less than or equal to `right`.\n",
-        "\n",
-        "8. Inside the binary search loop, we calculate the middle index `mid` and compare `nums[mid]` with the target value. If they are equal, we return `mid` as the index of the target element. If `nums[mid]` is less than the target, we update `left` to `mid + 1` to search in the right half. If `nums[mid]` is greater than the target, we update `right` to `mid - 1` to search in the left half.\n",
-        "\n",
-        "9. If we exit the binary search loop without finding the target, we return -1 to indicate that the target is not present in the array."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 45,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n"
-          ]
-        }
-      ],
-      "source": [
-        "nums1 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "target1 = 0\n",
-        "result1 = search(nums1, target1)\n",
-        "print(result1) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 46,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "-1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-        "target2 = 3\n",
-        "result2 = search(nums2, target2)\n",
-        "print(result2) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 47,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "-1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "nums3 = [1]\n",
-        "target3 = 0\n",
-        "result3 = search(nums3, target3)\n",
-        "print(result3) "
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. Finding the pivot index using binary search takes $O(log\\ n)$ time, where '$n$' is the number of elements in the array.\n",
-        "2. After finding the pivot index, performing binary search to find the target element also takes $O(log\\ n)$ time in the worst case.\n",
-        "\n",
-        "The dominant factor in the time complexity is the binary search, and since we perform two binary searches sequentially, the overall time complexity of the code is $O(log\\ n)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is very minimal:\n",
-        "\n",
-        "1. We use a constant amount of additional space for variables such as `left`, `right`, `pivot`, and `mid`. These variables do not depend on the input size, so they contribute $O(1)$ space complexity.\n",
-        "2. The function does not use any additional data structures that scale with the input size.\n",
-        "\n",
-        "Therefore, the space complexity of the code is $O(1)$, which means it has a constant space complexity and does not depend on the size of the input array.\n",
-        "\n",
-        "**In summary:**\n",
-        "- Time Complexity: $O(log\\ n)$\n",
-        "- Space Complexity: $O(1)$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Find Minimum Element in Rotated Sorted Array**: Write an algorithm to find the minimum element in a rotated sorted array. This is a variation of the problem where you don't need to search for a target value but instead find the smallest element.\n",
-        "\n",
-        "2. **Search in a Circularly Sorted Array**: Consider an array that is sorted in a circular manner (e.g., [4, 5, 6, 7, 0, 1, 2, 3]). Adapt the search algorithm to work for circularly sorted arrays while maintaining $O(log\\ n)$ complexity."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/05. Binary Search/README.md b/_build/html/_sources/05. Binary Search/README.md
deleted file mode 100644
index 6086b9b..0000000
--- a/_build/html/_sources/05. Binary Search/README.md	
+++ /dev/null
@@ -1,17 +0,0 @@
-# Binary search Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [153. Find Minimum In Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) | Medium |
-| [33. Search In Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)   | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb b/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb
deleted file mode 100644
index 0336bc8..0000000
--- a/_build/html/_sources/06. Linked List/141. Linked List Cycle.ipynb	
+++ /dev/null
@@ -1,271 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 141. Linked List Cycle\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Linked List Cycle problem on LeetCode, click here!](https://leetcode.com/problems/linked-list-cycle/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given `head`, the head of a linked list, determine if the linked list has a cycle in it.\n",
-        "\n",
-        "There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is connected to. **Note that `pos` is not passed as a parameter.**\n",
-        "\n",
-        "Return `true` if there is a cycle in the linked list. Otherwise, return `false`.\n",
-        "\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of the nodes in the list is in the range $[0, 10^4]$.\n",
-        "- $-10^5$ <= Node.val <= $10^5$\n",
-        "- `pos` is `-1` or a **valid index** in the linked-list.\n",
-        "\n",
-        "**Follow up:** Can you solve it using $O(1)$ (i.e. constant) memory?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "    def __init__(self, val=0):\n",
-        "        self.val = val\n",
-        "        self.next = None\n",
-        "\n",
-        "class Solution:\n",
-        "    def hasCycle(self, head: ListNode) -> bool:\n",
-        "        if not head:\n",
-        "            return False\n",
-        "        \n",
-        "        slow = head\n",
-        "        fast = head\n",
-        "        \n",
-        "        while fast and fast.next:\n",
-        "            slow = slow.next\n",
-        "            fast = fast.next.next\n",
-        "            \n",
-        "            if slow == fast:\n",
-        "                return True\n",
-        "        \n",
-        "        return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The given code defines a class `Solution` with a method `hasCycle` for determining if a linked list has a cycle.\n",
-        "\n",
-        "The `ListNode` class represents nodes in the linked list, where each node has a `val` (a value associated with the node) and a `next` attribute (a reference to the next node in the list).\n",
-        "\n",
-        "The `hasCycle` method takes the `head` of a linked list as input and returns `True` if there is a cycle in the linked list, and `False` otherwise. It uses two pointers, `slow` and `fast`, to traverse the linked list. If there is a cycle, the `fast` pointer will eventually catch up to the `slow` pointer.\n",
-        "\n",
-        "The algorithm works as follows:\n",
-        "1. It checks if the input `head` is `None` (an empty list). If it is, the method returns `False` because an empty list can't have a cycle.\n",
-        "2. Two pointers, `slow` and `fast`, initially point to the `head` of the list.\n",
-        "3. The code enters a `while` loop where `fast` moves two steps at a time, and `slow` moves one step at a time.\n",
-        "4. If there is a cycle, `fast` will eventually catch up to `slow`, and the method returns `True`.\n",
-        "5. If the loop completes without finding a cycle, the method returns `False`."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list with a cycle\n",
-        "def create_linked_list_with_cycle(values, pos):\n",
-        "    dummy = ListNode()\n",
-        "    current = dummy\n",
-        "    cycle_node = None\n",
-        "    \n",
-        "    for i, val in enumerate(values):\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "        \n",
-        "        if i == pos:\n",
-        "            cycle_node = current\n",
-        "    \n",
-        "    if cycle_node:\n",
-        "        current.next = cycle_node\n",
-        "    \n",
-        "    return dummy.next"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "This helper function is designed to create a linked list with a cycle for testing purposes. It takes two arguments:\n",
-        "\n",
-        "1. `values`: A list of values representing the nodes in the linked list.\n",
-        "2. `pos`: An integer representing the index of the node where the cycle begins.\n",
-        "\n",
-        "Here's how the function works:\n",
-        "\n",
-        "- It starts by creating an empty `ListNode` called `dummy`. This `dummy` node is used to simplify the creation of the linked list.\n",
-        "- It initializes a `current` pointer to the `dummy` node. This pointer will be used to traverse the linked list.\n",
-        "- It initializes a `cycle_node` variable to `None`. This variable will hold the reference to the node where the cycle begins, if any.\n",
-        "- It then iterates through the `values` list, creating a new `ListNode` for each value and appending it to the linked list.\n",
-        "- Inside the loop, it checks if the current index `i` is equal to the specified `pos`. If they match, it sets `cycle_node` to the current node. This simulates the creation of a cycle in the linked list.\n",
-        "- After the loop, if `cycle_node` is not `None` (indicating a cycle should be created), it connects the last node in the linked list to `cycle_node`, effectively creating a cycle.\n",
-        "- Finally, it returns the reference to the first node of the linked list (not the `dummy` node).\n",
-        "\n",
-        "This helper function allows you to easily create test cases where you can specify the values for the linked list and the position where the cycle starts. It's particularly useful for testing the code's ability to detect cycles in linked lists."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "head = [3,2,0,-4]\n",
-        "pos = 1\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "head = [1,2]\n",
-        "pos = 0\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 22,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "head = [1]\n",
-        "pos = -1\n",
-        "\n",
-        "head = create_linked_list_with_cycle(head, pos)\n",
-        "solution = Solution()\n",
-        "result = solution.hasCycle(head)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's discuss the time and space complexity of the code for detecting a cycle in a linked list:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this code is $O(n)$, where \"$n$\" is the number of nodes in the linked list.\n",
-        "\n",
-        "The reason for this is that both the `slow` and `fast` pointers traverse the linked list, and they move at different speeds. In the worst case, when there is no cycle, the `fast` pointer will reach the end of the list after going through approximately $n/2$ nodes. In the case of a cycle, it may take some extra iterations for the `fast` pointer to catch up to the `slow` pointer. However, the total number of iterations is still proportional to the number of nodes in the linked list, making it $O(n)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this code is $O(1)$, which means it uses constant space.\n",
-        "\n",
-        "The reason for this is that regardless of the size of the linked list, the code only uses two additional pointers (`slow` and `fast`) to traverse the list. These pointers do not depend on the size of the input linked list, so the space complexity remains constant.\n",
-        "\n",
-        "**In summary**, this code efficiently detects cycles in a linked list with a time complexity of $O(n)$ and a space complexity of $O(1)$. It is an example of an algorithm that solves a complex problem with minimal memory usage."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/143. Reorder List.ipynb b/_build/html/_sources/06. Linked List/143. Reorder List.ipynb
deleted file mode 100644
index a83545b..0000000
--- a/_build/html/_sources/06. Linked List/143. Reorder List.ipynb	
+++ /dev/null
@@ -1,285 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 143. Reorder List\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Reorder List problem on LeetCode, click here!](https://leetcode.com/problems/reorder-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given the head of a singly linked-list. The list can be represented as:\n",
-        "\n",
-        "$L_0 → L_1 → … → L_{n - 1} → L_n$\n",
-        "\n",
-        "Reorder the list to be on the following form:\n",
-        "\n",
-        "$L_0 → L_n → L_1 → L_{n - 1} → L_2 → L_{n - 2} → …$\n",
-        "\n",
-        "You may not modify the values in the list's nodes. Only nodes themselves may be changed.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is in the range $[1, 5 * 10^4]$.\n",
-        "- `1 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 28,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def reorderList(head):\n",
-        "    if not head or not head.next:\n",
-        "        return head\n",
-        "\n",
-        "    # Step 1: Find the middle of the linked list\n",
-        "    slow, fast = head, head\n",
-        "    while fast.next and fast.next.next:\n",
-        "        slow = slow.next\n",
-        "        fast = fast.next.next\n",
-        "\n",
-        "    # Split the list into two halves\n",
-        "    first_half = head\n",
-        "    second_half = slow.next\n",
-        "    slow.next = None\n",
-        "\n",
-        "    # Step 2: Reverse the second half of the linked list\n",
-        "    prev = None\n",
-        "    current = second_half\n",
-        "    while current:\n",
-        "        next_node = current.next\n",
-        "        current.next = prev\n",
-        "        prev = current\n",
-        "        current = next_node\n",
-        "    second_half = prev\n",
-        "\n",
-        "    # Step 3: Merge the first and reversed second halves alternately\n",
-        "    p1, p2 = first_half, second_half\n",
-        "    while p2:\n",
-        "        next_p1 = p1.next\n",
-        "        next_p2 = p2.next\n",
-        "        p1.next = p2\n",
-        "        p2.next = next_p1\n",
-        "        p1 = next_p1\n",
-        "        p2 = next_p2\n",
-        "\n",
-        "    return head"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `reorderList` function is designed to reorder a singly linked list following the specific pattern described in the problem statement. Here's a detailed explanation of how the function works:\n",
-        "\n",
-        "1. **Base Cases Handling**:\n",
-        "   - The function first checks if the input linked list is empty (`not head`) or contains only one element (`not head.next`). If either of these conditions is met, the list cannot be reordered, so the function returns the original list as-is.\n",
-        "\n",
-        "2. **Finding the Middle of the Linked List**:\n",
-        "   - To reorder the list, we first need to find the middle point so that we can split the list into two halves. This is done using two pointers, `slow` and `fast`, initialized to the head of the list.\n",
-        "   - `slow` moves one step at a time while `fast` moves two steps at a time. When `fast` reaches the end of the list or the second-to-last node, `slow` will be at the middle node.\n",
-        "\n",
-        "3. **Splitting the List**:\n",
-        "   - After finding the middle node, we split the list into two halves:\n",
-        "     - The first half, `first_half`, contains nodes from the beginning up to the middle.\n",
-        "     - The second half, `second_half`, contains nodes from the middle to the end.\n",
-        "   - To split the list, we set the `next` pointer of the node before the middle node to `None`.\n",
-        "\n",
-        "4. **Reversing the Second Half**:\n",
-        "   - We reverse the second half of the linked list using the `prev`, `current`, and `next_node` pointers.\n",
-        "   - `prev` is initially `None`, and we iterate through the second half. For each node, we:\n",
-        "     - Set the `next` of the current node to `prev`, effectively reversing the next pointer direction.\n",
-        "     - Update `prev` to the current node.\n",
-        "     - Move to the next node using the `next_node`.\n",
-        "\n",
-        "5. **Merging the Two Halves Alternately**:\n",
-        "   - We now have two linked lists: `first_half` and the reversed `second_half`.\n",
-        "   - We merge these two lists alternately by adjusting the `next` pointers of the nodes.\n",
-        "   - `p1` and `p2` are pointers to the current nodes in `first_half` and `second_half`, respectively.\n",
-        "   - We iterate through both lists while reordering nodes:\n",
-        "     - Set the `next` of `p1` to `p2` to link a node from the first half to a node from the reversed second half.\n",
-        "     - Update `p1` and `p2` to their respective next nodes.\n",
-        "     - Repeat this process until we have processed all nodes in both halves.\n",
-        "\n",
-        "6. **Returning the Reordered List**:\n",
-        "   - After the merging process is complete, the linked list is reordered as specified.\n",
-        "   - The function returns the `head` of the reordered list.\n",
-        "\n",
-        "The overall result is a singly linked list that has been reordered according to the given pattern. The time complexity of this algorithm is O(N), where N is the number of nodes in the linked list, as we traverse the list once to find the middle, once to reverse the second half, and once to merge the two halves."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 29,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to print the linked list\n",
-        "def printLinkedList(head):\n",
-        "    result = []\n",
-        "    while head:\n",
-        "        result.append(head.val)\n",
-        "        head = head.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the helper functions:\n",
-        "**printLinkedList Function**:\n",
-        "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-        "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-        "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-        "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-        "\n",
-        "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 30,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 4, 2, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "head1 = ListNode(1, ListNode(2, ListNode(3, ListNode(4))))\n",
-        "reorderList(head1)\n",
-        "print(printLinkedList(head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 5, 2, 4, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "head2 = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))\n",
-        "reorderList(head2)\n",
-        "print(printLinkedList(head2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `reorderList` function:\n",
-        "\n",
-        "**Time Complexity**:\n",
-        "1. **Finding the Middle of the Linked List**: In this step, we use two pointers, one moving one step at a time (`slow`) and the other moving two steps at a time (`fast`). This process takes $O(N/2)$ time, where $N$ is the number of nodes in the linked list.\n",
-        "\n",
-        "2. **Splitting the List**: After finding the middle, we split the list into two halves by setting the `next` pointer of the node before the middle node to `None`. This step takes $O(1)$ time.\n",
-        "\n",
-        "3. **Reversing the Second Half**: We reverse the second half of the linked list using a while loop that iterates through the second half of the list once. Therefore, this step also takes $O(N/2)$ time.\n",
-        "\n",
-        "4. **Merging the Two Halves Alternately**: In this step, we merge the two halves alternately by adjusting the `next` pointers of the nodes. We iterate through both halves once, so this step takes $O(N/2)$ time.\n",
-        "\n",
-        "Overall, the time complexity of the `reorderList` function is dominated by the steps involving reversing and merging the two halves, both of which take $O(N/2)$ time. Therefore, the total time complexity is $O(N)$.\n",
-        "\n",
-        "**Space Complexity**:\n",
-        "The space complexity of the `reorderList` function is primarily determined by the variables and data structures used within the function. Let's break down the space complexity components:\n",
-        "\n",
-        "1. **Constant Space Variables**: The variables `slow`, `fast`, `prev`, `current`, `next_node`, `p1`, and `p2` are used to traverse and manipulate the linked list. These variables occupy constant space, regardless of the input size. Therefore, they contribute $O(1)$ to the space complexity.\n",
-        "\n",
-        "2. **Split Linked Lists**: In the splitting step, we create two new linked list segments (`first_half` and `second_half`) that store references to nodes from the original linked list. These segments occupy space proportional to half of the input list, $O(N/2)$. \n",
-        "\n",
-        "3. **Reversed Second Half**: During the reversal step, we reverse the second half of the linked list in-place without creating any additional data structures. Therefore, it doesn't contribute to additional space complexity.\n",
-        "\n",
-        "4. **Overall**: Combining the above components, the total space complexity is $O(N/2)$, which simplifies to $O(N)$ in terms of space complexity.\n",
-        "\n",
-        "**In summary**, the `reorderList` function has a time complexity of $O(N)$ and a space complexity of $O(N)$ due to the creation of two new linked list segments while splitting the list. The constant space variables used for traversal do not significantly impact the overall space complexity."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Reorder by K Elements:** Generalize the reorderList function to reorder the list in a different pattern, where you reorder the list by $K$ elements at a time. For example, for $K=3$, you would reorder it as ($L_0 → L_1 → L_2 → L_n → L_{n-1} → L_{n-2} → ...$).\n",
-        "You may need to handle cases where the number of nodes is not a multiple of $K$.\n",
-        "\n",
-        "2. **Reorder by Odd and Even Nodes:** Modify the reorderList function to reorder the list in a pattern where odd-indexed nodes (1-based index) come before even-indexed nodes. For example, ($L_0 → L_n → L_1 → L_{n-1} → L_2 → L_{n-2} → ...$)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb b/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb
deleted file mode 100644
index 263124e..0000000
--- a/_build/html/_sources/06. Linked List/19. Remove Nth Node From End of List.ipynb	
+++ /dev/null
@@ -1,278 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 19. Remove Nth Node From End of List\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Remove Nth Node From End of List problem on LeetCode, click here!](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the head of a linked list, remove the $n^{th}$ node from the end of the list and return its head.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is `sz`.\n",
-        "- `1 <= sz <= 30`\n",
-        "- `0 <= Node.val <= 100`\n",
-        "- `1 <= n <= sz`\n",
-        "\n",
-        "**Follow up:** Could you do this in one pass?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 23,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def removeNthFromEnd(head, n):\n",
-        "    # Create a dummy node to handle the case of removing the head node.\n",
-        "    dummy = ListNode(0)\n",
-        "    dummy.next = head\n",
-        "    first = dummy\n",
-        "    second = dummy\n",
-        "    \n",
-        "    # Advance the first pointer by n+1 nodes.\n",
-        "    for _ in range(n + 1):\n",
-        "        first = first.next\n",
-        "    \n",
-        "    # Move both pointers simultaneously until first reaches the end.\n",
-        "    while first is not None:\n",
-        "        first = first.next\n",
-        "        second = second.next\n",
-        "    \n",
-        "    # Remove the nth node by updating the next pointer of the previous node.\n",
-        "    second.next = second.next.next\n",
-        "    \n",
-        "    return dummy.next  # Return the modified head of the linked list"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a `ListNode` class to represent individual nodes in a linked list. Each node has two attributes: `val` (the value of the node) and `next` (a reference to the next node in the list).\n",
-        "\n",
-        "2. The `removeNthFromEnd` function takes the head of a linked list (`head`) and the value `n`, which represents the position from the end of the list of the node to be removed. It returns the modified head of the linked list after the removal.\n",
-        "\n",
-        "3. We create a dummy node (`dummy`) at the beginning of the list. This dummy node simplifies the code by handling cases where the head of the list needs to be removed.\n",
-        "\n",
-        "4. We initialize two pointers, `first` and `second`, both initially pointing to the dummy node.\n",
-        "\n",
-        "5. To advance the `first` pointer by `n+1` nodes, we use a `for` loop. This positions the `first` pointer `n` nodes ahead of the `second` pointer.\n",
-        "\n",
-        "6. We then move both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This ensures that the `second` pointer ends up pointing to the node that needs to be removed.\n",
-        "\n",
-        "7. To remove the nth node from the end, we update the `next` pointer of the node pointed to by the `second` pointer to skip over the node to be removed.\n",
-        "\n",
-        "8. Finally, we return `dummy.next`, which is the modified head of the linked list without the removed node.\n",
-        "\n",
-        "9. Additionally, there are two helper functions:\n",
-        "   - `createLinkedList(values)` creates a linked list from a list of values.\n",
-        "   - `convertToList(head)` converts a linked list back into a list of values."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list from a list of values.\n",
-        "def createLinkedList(values):\n",
-        "    if not values:\n",
-        "        return None\n",
-        "    head = ListNode(values[0])\n",
-        "    current = head\n",
-        "    for val in values[1:]:\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "    return head\n",
-        "\n",
-        "# Helper function to convert a linked list to a list of values.\n",
-        "def convertToList(head):\n",
-        "    result = []\n",
-        "    current = head\n",
-        "    while current:\n",
-        "        result.append(current.val)\n",
-        "        current = current.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the two helper functions:\n",
-        "\n",
-        "1. `createLinkedList(values)`: This function creates a linked list from a list of values.\n",
-        "\n",
-        "   - Input: `values` is a list containing values that you want to insert into the linked list.\n",
-        "   - Output: The function returns the head of the created linked list.\n",
-        "\n",
-        "   Explanation:\n",
-        "   - The function starts by checking if the `values` list is empty. If it is, it returns `None` to indicate an empty linked list.\n",
-        "   - If the `values` list is not empty, it creates the first node of the linked list with the value from the first element of `values`.\n",
-        "   - It then iterates through the remaining elements of `values` and creates new nodes for each value, connecting them together using the `next` attribute to form a linked list.\n",
-        "   - Finally, it returns the head of the linked list, which is the first node created.\n",
-        "   \n",
-        "\n",
-        "2. `convertToList(head)`: This function converts a linked list into a list of values.\n",
-        "\n",
-        "   - Input: `head` is the head node of the linked list that you want to convert.\n",
-        "   - Output: The function returns a list containing the values of the nodes in the linked list in order.\n",
-        "\n",
-        "   Explanation:\n",
-        "   - The function starts by initializing an empty list, `result`, to store the values of the linked list nodes.\n",
-        "   - It then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - The iteration continues until it reaches the end of the linked list (i.e., the `next` attribute of the current node becomes `None`).\n",
-        "   - After the iteration is complete, the function returns the `result` list, which now contains the values of all the nodes in the linked list in the same order as they appear in the linked list.\n",
-        "\n",
-        "These helper functions are useful for creating linked lists from lists of values and converting linked lists back into lists of values, making it easier to work with linked list examples and test cases in a more familiar list format."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 2, 3, 5]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "head1 = createLinkedList([1, 2, 3, 4, 5])\n",
-        "n1 = 2\n",
-        "new_head1 = removeNthFromEnd(head1, n1)\n",
-        "print(convertToList(new_head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 26,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "head2 = createLinkedList([1])\n",
-        "n2 = 1\n",
-        "new_head2 = removeNthFromEnd(head2, n2)\n",
-        "print(convertToList(new_head2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 27,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "head3 = createLinkedList([1, 2])\n",
-        "n3 = 1\n",
-        "new_head3 = removeNthFromEnd(head3, n3)\n",
-        "print(convertToList(new_head3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Certainly! Let's analyze the time and space complexity of the `removeNthFromEnd` function in isolation:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The `removeNthFromEnd` function performs two passes through the linked list. In the worst case, it needs to traverse the entire linked list of length $N$.\n",
-        "- The first pass positions the `first` pointer $n$ nodes ahead of the `second` pointer. This pass takes $O(N)$ time because it involves iterating through all N nodes.\n",
-        "- The second pass moves both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This also takes $O(N)$ time in the worst case.\n",
-        "- Therefore, the overall time complexity of the `removeNthFromEnd` function is $O(N)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The `removeNthFromEnd` function uses a constant amount of extra space for its variables (e.g., `first`, `second`, and `dummy`). This space usage does not depend on the size of the linked list.\n",
-        "- The space complexity is $O(1)$, indicating that the space used by the function is constant and independent of the size of the input linked list.\n",
-        "\n",
-        "**In summary**, the `removeNthFromEnd` function has a time complexity of $O(N)$ and a space complexity of $O(1)$. It efficiently removes the nth node from the end of the linked list with a single pass through the list and constant extra space usage."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb b/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb
deleted file mode 100644
index c97de2d..0000000
--- a/_build/html/_sources/06. Linked List/206. Reverse Linked List.ipynb	
+++ /dev/null
@@ -1,316 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 206. Reverse Linked List\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Reverse Linked List problem on LeetCode, click here!](https://leetcode.com/problems/reverse-linked-list/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the `head` of a singly linked list, reverse the list, and return the reversed list.\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the list is the range `[0, 5000]`.\n",
-        "- `-5000 <= Node.val <= 5000`\n",
-        "\n",
-        "**Follow up:** A linked list can be reversed either iteratively or recursively. Could you implement both?"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 33,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "# Function to reverse a linked list iteratively\n",
-        "def reverseListIterative(head):\n",
-        "    prev = None\n",
-        "    current = head\n",
-        "    \n",
-        "    while current:\n",
-        "        next_node = current.next\n",
-        "        current.next = prev\n",
-        "        prev = current\n",
-        "        current = next_node\n",
-        "    \n",
-        "    return prev\n",
-        "\n",
-        "# Function to reverse a linked list recursively\n",
-        "def reverseListRecursive(head):\n",
-        "    if not head or not head.next:\n",
-        "        return head\n",
-        "\n",
-        "    new_head = reverseListRecursive(head.next)\n",
-        "    head.next.next = head\n",
-        "    head.next = None\n",
-        "\n",
-        "    return new_head"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `reverseListIterative(head)`: This function reverses a linked list iteratively using a loop. Here's a step-by-step explanation:\n",
-        "\n",
-        "   - It takes the `head` of the input linked list as a parameter.\n",
-        "   - It initializes two pointers, `prev` and `current`, initially set to `None` and the `head` of the list, respectively.\n",
-        "   - It enters a `while` loop that continues until `current` reaches the end of the list (i.e., becomes `None`).\n",
-        "   - Inside the loop:\n",
-        "     - It stores the `next_node` which is the next node after `current`.\n",
-        "     - It updates the `next` pointer of `current` to point to the `prev` node. This effectively reverses the link direction.\n",
-        "     - It moves `prev` to `current` and `current` to `next_node`, advancing the pointers one step further in the list.\n",
-        "   - Once the loop completes, `prev` will be pointing to the new head of the reversed list (which was the last node of the original list), so it returns `prev`.\n",
-        "\n",
-        "2. `reverseListRecursive(head)`: This function reverses a linked list recursively. Here's how it works:\n",
-        "\n",
-        "   - It takes the `head` of the input linked list as a parameter.\n",
-        "   - The base case is checked: if `head` is `None` or `head.next` is `None`, meaning the list is empty or has only one node, it returns `head` as there's no need to reverse a list with zero or one element.\n",
-        "   - In the recursive case:\n",
-        "     - It calls itself with `head.next`, effectively moving down the list until it reaches the end.\n",
-        "     - Once it reaches the end, it starts reversing the links:\n",
-        "       - `head.next.next` is set to `head`, reversing the link between `head` and the next node.\n",
-        "       - `head.next` is set to `None` to avoid cycles.\n",
-        "   - Finally, it returns the new head of the reversed list, which will be the last node of the original list.\n",
-        "\n",
-        "**In summary,** `reverseListIterative` reverses the linked list by iterating through it and changing the next pointers, while `reverseListRecursive` reverses the linked list by recursively reaching the end and then reversing the links on the way back up the recursion stack. Both functions achieve the same result: reversing the linked list."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 36,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to create a linked list from a list of values\n",
-        "def createLinkedList(values):\n",
-        "    if not values:\n",
-        "        return None\n",
-        "    head = ListNode(values[0])\n",
-        "    current = head\n",
-        "    for val in values[1:]:\n",
-        "        current.next = ListNode(val)\n",
-        "        current = current.next\n",
-        "    return head\n",
-        "\n",
-        "# Helper function to convert a linked list to a list for testing\n",
-        "def linkedListToList(head):\n",
-        "    result = []\n",
-        "    current = head\n",
-        "    while current:\n",
-        "        result.append(current.val)\n",
-        "        current = current.next\n",
-        "    return result\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the two helper functions used in the provided code:\n",
-        "\n",
-        "1. `createLinkedList(values)`:\n",
-        "\n",
-        "   - Purpose: This function is used to create a linked list from a list of values. It takes a list of values as input and returns the head of the linked list.\n",
-        "\n",
-        "   - How it works:\n",
-        "     - It first checks if the input list `values` is empty. If it's empty, it returns `None` to indicate an empty linked list.\n",
-        "     - If `values` is not empty, it initializes a `head` node with the value of the first element in `values`.\n",
-        "     - It then iterates through the remaining values in `values`, creating new nodes for each value and linking them together to form a linked list.\n",
-        "     - Finally, it returns the `head` node of the newly created linked list.\n",
-        "\n",
-        "2. `linkedListToList(head)`:\n",
-        "\n",
-        "   - Purpose: This function is used to convert a linked list back into a Python list for testing and output purposes.\n",
-        "\n",
-        "   - How it works:\n",
-        "     - It takes the `head` of the linked list as input.\n",
-        "     - It initializes an empty list called `result`.\n",
-        "     - It then iterates through the linked list, starting from `head`, and appends each node's `val` (value) to the `result` list.\n",
-        "     - This process continues until the end of the linked list is reached.\n",
-        "     - Finally, it returns the `result` list, which contains the values from the linked list in the same order.\n",
-        "\n",
-        "These helper functions are used to simplify the process of creating linked lists from lists of values and converting linked lists back into lists, making it easier to work with linked lists in the provided examples and test cases."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 35,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[1, 1, 2, 3, 4, 4]"
-            ]
-          },
-          "execution_count": 35,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 1: Iterative\n",
-        "head1 = [1,2,3,4,5]\n",
-        "head1 = createLinkedList(head1)\n",
-        "reversed_head1 = reverseListIterative(head1)\n",
-        "print(linkedListToList(reversed_head1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 31,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[]"
-            ]
-          },
-          "execution_count": 37,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 2: Recursive\n",
-        "head2 = [1,2]\n",
-        "head2 = createLinkedList(head2)\n",
-        "reversed_head2 = reverseListRecursive(head2)\n",
-        "print(linkedListToList(reversed_head2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 38,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "data": {
-            "text/plain": [
-              "[0]"
-            ]
-          },
-          "execution_count": 38,
-          "metadata": {},
-          "output_type": "execute_result"
-        }
-      ],
-      "source": [
-        "# Example 3: Empty list\n",
-        "head3 = []\n",
-        "head3 = createLinkedList(head3)\n",
-        "reversed_head3 = reverseListIterative(head3)\n",
-        "print(linkedListToList(reversed_head3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the two functions:\n",
-        "\n",
-        "1. `reverseListIterative(head)`:\n",
-        "\n",
-        "   - Time Complexity: $O(n)$\n",
-        "     - The iterative function visits each node in the linked list exactly once in a single pass, where \"$n$\" is the number of nodes in the list. Therefore, the time complexity is linear in the number of nodes.\n",
-        "\n",
-        "   - Space Complexity: $O(1)$\n",
-        "     - This function uses a constant amount of extra space for the `prev`, `current`, and `next_node` pointers. Regardless of the size of the input list, the amount of additional memory used remains the same, so the space complexity is constant.\n",
-        "\n",
-        "2. `reverseListRecursive(head)`:\n",
-        "\n",
-        "   - Time Complexity: $O(n)$\n",
-        "     - The recursive function also visits each node in the linked list exactly once. It recursively traverses the list from the head to the tail, reversing the links on the way back. Therefore, like the iterative approach, the time complexity is $O(n)$, where \"$n$\" is the number of nodes.\n",
-        "\n",
-        "   - Space Complexity: $O(n)$\n",
-        "     - The recursive function uses space on the call stack for each recursive call. In the worst case, when the list has \"$n$\" nodes, it will create \"$n$\" recursive function calls on the stack, resulting in a space complexity of $O(n)$. This is because each function call stores information about its state, including the `head` pointer, until it reaches the base case and starts returning.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "- The time complexity of both functions is $O(n)$ because they both visit each node once.\n",
-        "- The space complexity of `reverseListIterative` is $O(1)$ because it uses a constant amount of extra space.\n",
-        "- The space complexity of `reverseListRecursive` is $O(n)$ due to the recursive function calls and the associated call stack space.\n",
-        "\n",
-        "Both functions are efficient in terms of time complexity, but `reverseListIterative` is more memory-efficient because it uses a constant amount of additional space, while `reverseListRecursive` uses space on the call stack proportional to the number of nodes in the list."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **K-Group Reverse:** Modify the reverseListIterative function to reverse the linked list in groups of $K$ elements. For example, if the input linked list is [1, 2, 3, 4, 5, 6], and $K$ is 3, the output should be [3, 2, 1, 6, 5, 4].\n",
-        "\n",
-        "2. **Rotate Linked List:** Write a function to rotate a linked list to the right by $K$ places. For example, if the input is [1, 2, 3, 4, 5] and $K$ is 2, the output should be [4, 5, 1, 2, 3]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb b/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb
deleted file mode 100644
index 876d633..0000000
--- a/_build/html/_sources/06. Linked List/21. Merge Two Sorted Lists.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 21. Merge Two Sorted Lists\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Merge Two Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-two-sorted-lists/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given the heads of two sorted linked lists `list1` and `list2`.\n",
-        "\n",
-        "Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.\n",
-        "\n",
-        "*Return the head of the merged linked list.*\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in both lists is in the range `[0, 50]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "- Both `list1` and `list2` are sorted in non-decreasing order."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class ListNode:\n",
-        "    def __init__(self, val=0, next=None):\n",
-        "        self.val = val\n",
-        "        self.next = next\n",
-        "\n",
-        "def mergeTwoLists(list1, list2):\n",
-        "    # Create a dummy node to simplify the code.\n",
-        "    dummy = ListNode()\n",
-        "    current = dummy\n",
-        "\n",
-        "    while list1 and list2:\n",
-        "        if list1.val < list2.val:\n",
-        "            current.next = list1\n",
-        "            list1 = list1.next\n",
-        "        else:\n",
-        "            current.next = list2\n",
-        "            list2 = list2.next\n",
-        "        current = current.next\n",
-        "\n",
-        "    # Append the remaining elements from either list.\n",
-        "    if list1:\n",
-        "        current.next = list1\n",
-        "    else:\n",
-        "        current.next = list2\n",
-        "\n",
-        "    return dummy.next"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a class `ListNode` to represent the nodes of a singly-linked list. Each node has a `val` attribute to store the value of the node and a `next` attribute to point to the next node in the list.\n",
-        "\n",
-        "2. We define the `mergeTwoLists` function that takes two linked list heads, `list1` and `list2`, as input.\n",
-        "\n",
-        "3. We create a `dummy` node at the beginning of the merged list. The `dummy` node simplifies the code by serving as a placeholder, and its `next` attribute will point to the actual merged list.\n",
-        "\n",
-        "4. We initialize a `current` pointer to point to the `dummy` node initially. This `current` pointer helps us traverse and build the merged list.\n",
-        "\n",
-        "5. We enter a `while` loop that continues until either `list1` or `list2` becomes empty. Inside the loop, we compare the values of the nodes at the current positions of `list1` and `list2`.\n",
-        "\n",
-        "6. If the `val` of the node in `list1` is smaller than the `val` of the node in `list2`, we attach the node from `list1` to the merged list by updating the `next` attribute of the `current` node to point to the node in `list1`. We then move the `list1` pointer to the next node.\n",
-        "\n",
-        "7. If the `val` of the node in `list2` is smaller than or equal to the `val` of the node in `list1`, we attach the node from `list2` to the merged list in a similar manner. We move the `list2` pointer to the next node.\n",
-        "\n",
-        "8. After attaching a node to the merged list, we move the `current` pointer to the newly added node. This step is essential for keeping track of the end of the merged list.\n",
-        "\n",
-        "9. The loop continues until either `list1` or `list2` becomes empty.\n",
-        "\n",
-        "10. Once the loop exits, we check if there are any remaining elements in `list1` or `list2`. If `list1` is not empty, we attach the remaining elements of `list1` to the merged list. If `list2` is not empty, we attach the remaining elements of `list2`.\n",
-        "\n",
-        "11. Finally, we return the `next` attribute of the `dummy` node as the head of the merged list, which represents the sorted merged list."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Helper function to print the linked list\n",
-        "def printLinkedList(head):\n",
-        "    result = []\n",
-        "    while head:\n",
-        "        result.append(head.val)\n",
-        "        head = head.next\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-      "metadata": {},
-      "source": [
-        "Let's explain the helper functions:\n",
-        "**printLinkedList Function**:\n",
-        "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-        "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-        "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-        "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-        "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-        "\n",
-        "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 1:\n",
-        "# Input: list1 = [1,2,4], list2 = [1,3,4]\n",
-        "list1 = ListNode(1, ListNode(2, ListNode(4)))\n",
-        "list2 = ListNode(1, ListNode(3, ListNode(4)))\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 2:\n",
-        "# Input: list1 = [], list2 = []\n",
-        "list1 = None\n",
-        "list2 = None\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": null,
-      "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Example 3:\n",
-        "# list1 = [], list2 = [0]\n",
-        "list1 = None\n",
-        "list2 = ListNode(0)\n",
-        "merged_list = mergeTwoLists(list1, list2)\n",
-        "printLinkedList(merged_list)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the `mergeTwoLists` function.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this function is $O(N + M$), where $N$ and $M$ are the lengths of `list1` and `list2`, respectively. Here's why:\n",
-        "\n",
-        "1. In the worst case, we need to traverse both `list1` and `list2` completely. This involves iterating through all the nodes in both lists once.\n",
-        "\n",
-        "2. The while loop runs until either `list1` or `list2` becomes empty. The number of iterations depends on the total number of nodes in both lists, which is N + M in the worst case.\n",
-        "\n",
-        "3. Inside the loop, we perform constant-time operations for each iteration, such as comparisons and updating pointers.\n",
-        "\n",
-        "Therefore, the dominant factor in the time complexity is the combined length of both input lists, resulting in a linear time complexity of $O(N + M)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this function is $O(1)$, which means it uses a constant amount of extra space regardless of the input sizes. Here's why:\n",
-        "\n",
-        "1. We create a few extra pointers (`dummy`, `current`) and temporary variables (`val`) to manage the merging process. These consume a fixed amount of memory, regardless of the input sizes. The number of these extra variables is independent of the lengths of `list1` and `list2`.\n",
-        "\n",
-        "2. We do not create a new data structure or allocate memory for the merged list. Instead, we rearrange the existing nodes from `list1` and `list2` to form the merged list. This operation does not consume additional memory proportional to the input sizes.\n",
-        "\n",
-        "**In summary**, the `mergeTwoLists` function has a time complexity of $O(N + M)$ and a space complexity of $O(1)$, making it an efficient solution for merging two sorted linked lists."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Merge K Sorted Lists:** Extend the problem to merge $K$ sorted linked lists instead of just two. You need to efficiently merge $K$ lists into one sorted list.\n",
-        "\n",
-        "2. **Merge Lists in a Zigzag Pattern:** Merge two sorted lists in a zigzag pattern. For example, given [1, 2, 4] and [1, 3, 5], the merged list should be [1, 1, 2, 3, 4, 5]."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": "sql",
-      "file_extension": "",
-      "mimetype": "",
-      "name": "sql",
-      "version": "3.32.3"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb b/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb
deleted file mode 100644
index a850734..0000000
--- a/_build/html/_sources/06. Linked List/23. Merge k Sorted Lists.ipynb	
+++ /dev/null
@@ -1,285 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 23. Merge k Sorted Lists\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Merge k Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-k-sorted-lists/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "You are given an array of $k$ linked-lists lists, each linked-list is sorted in ascending order.\n",
-        "\n",
-        "Merge all the linked-lists into one sorted linked-list and return it.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- `k == lists.length`\n",
-        "- 0 <= k <= $10^4$\n",
-        "- `0 <= lists[i].length <= 500`\n",
-        "- $-10^4$ <= lists[i][j] <= $10^4$\n",
-        "- `lists[i]` is sorted in ascending order.\n",
-        "- The sum of `lists[i].length` will not exceed $10^4$.\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for singly-linked list.\n",
-        "class ListNode:\n",
-        "     def __init__(self, val=0, next=None):\n",
-        "         self.val = val\n",
-        "         self.next = next\n",
-        "\n",
-        "class Solution:\n",
-        "    def mergeKLists(self, lists: [[ListNode]]) -> ListNode:\n",
-        "        # Check if the input list of linked lists is empty\n",
-        "        if not lists or len(lists) == 0:\n",
-        "            return None\n",
-        "\n",
-        "        # Loop until there is only one merged list left\n",
-        "        while len(lists) > 1:\n",
-        "            mergedLists = []\n",
-        "\n",
-        "            # Merge pairs of lists\n",
-        "            for i in range(0, len(lists), 2):\n",
-        "                l1 = lists[i]\n",
-        "                l2 = lists[i + 1] if (i + 1) < len(lists) else None\n",
-        "                mergedLists.append(self.mergeList(l1, l2))\n",
-        "\n",
-        "            # Update the list of lists with merged results\n",
-        "            lists = mergedLists\n",
-        "        \n",
-        "        # Return the final merged list\n",
-        "        return lists[0]\n",
-        "\n",
-        "    def mergeList(self, l1, l2):\n",
-        "        # Create a dummy node to simplify merging\n",
-        "        dummy = ListNode()\n",
-        "        tail = dummy\n",
-        "\n",
-        "        # Merge the two sorted lists\n",
-        "        while l1 and l2:\n",
-        "            if l1.val < l2.val:\n",
-        "                tail.next = l1\n",
-        "                l1 = l1.next\n",
-        "            else:\n",
-        "                tail.next = l2\n",
-        "                l2 = l2.next\n",
-        "            tail = tail.next\n",
-        "\n",
-        "        # Append any remaining elements from l1 or l2 (if any)\n",
-        "        if l1:\n",
-        "            tail.next = l1\n",
-        "        if l2:\n",
-        "            tail.next = l2\n",
-        "\n",
-        "        # Return the merged result starting from the next of the dummy node\n",
-        "        return dummy.next\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided code defines a Python class `Solution` with two methods for merging k sorted linked lists:\n",
-        "\n",
-        "1. `mergeKLists(self, lists: List[ListNode]) -> ListNode`: This method takes a list of k sorted linked lists as input and returns a single merged sorted linked list. It uses a divide-and-conquer approach to repeatedly merge pairs of lists until only one merged list remains.\n",
-        "\n",
-        "2. `mergeList(self, l1, l2)`: This method takes two sorted linked lists, `l1` and `l2`, as input and merges them into a single sorted linked list. It uses a dummy node to simplify the merging process.\n",
-        "\n",
-        "Here's a high-level overview of how the code works:\n",
-        "\n",
-        "- The `mergeKLists` method checks if the input list of linked lists is empty or contains no lists. If there are no lists, it returns `None`.\n",
-        "\n",
-        "- Inside a `while` loop, the code repeatedly merges pairs of linked lists until only one merged list remains in the `lists` array. It does this by iterating through the input lists in pairs and calling the `mergeList` method to merge each pair.\n",
-        "\n",
-        "- The `mergeList` method takes two sorted linked lists, `l1` and `l2`, and merges them into a single sorted linked list. It uses a dummy node (`dummy`) and a `tail` pointer to keep track of the merged list while comparing and merging elements from `l1` and `l2`.\n",
-        "\n",
-        "- After merging all pairs of lists and updating the `lists` array with the merged results, the loop continues until only one merged list remains in the `lists` array.\n",
-        "\n",
-        "- Finally, the `mergeKLists` method returns the merged list.\n",
-        "\n",
-        "Overall, this code efficiently merges k sorted linked lists using a divide-and-conquer strategy, resulting in a single merged sorted linked list as the output."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> "
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: lists = [[1,4,5],[1,3,4],[2,6]]\n",
-        "\n",
-        "lists1 = [\n",
-        "    ListNode(1, ListNode(4, ListNode(5))),\n",
-        "    ListNode(1, ListNode(3, ListNode(4))),\n",
-        "    ListNode(2, ListNode(6))\n",
-        "]\n",
-        "solution = Solution()\n",
-        "result1 = solution.mergeKLists(lists1)\n",
-        "\n",
-        "# Print the result\n",
-        "if result1:\n",
-        "    current = result1\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: lists = []\n",
-        "\n",
-        "lists2 = []\n",
-        "solution = Solution()\n",
-        "result2 = solution.mergeKLists(lists2)\n",
-        "\n",
-        "# Print the result\n",
-        "if result2:\n",
-        "    current = result2\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "# Input: lists = [[]]\n",
-        "\n",
-        "lists3 = [None]\n",
-        "solution = Solution()\n",
-        "result3 = solution.mergeKLists(lists3)\n",
-        "\n",
-        "# Print the result\n",
-        "if result3:\n",
-        "    current = result3\n",
-        "    while current:\n",
-        "        print(current.val, end=\" -> \")\n",
-        "        current = current.next\n",
-        "else:\n",
-        "    print(\"None\")  # Print \"None\" for input with a single None element"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. **Heap Initialization:** The code does not use a heap. Instead, it uses a divide-and-conquer approach. The initial check for empty input lists takes $O(1)$ time.\n",
-        "\n",
-        "2. **Merging:** The merging operation is performed in a divide-and-conquer fashion. In each iteration of the `while` loop, we merge pairs of linked lists, and the number of comparisons made is proportional to the total number of nodes across all the linked lists ($n$). In each merge step, we effectively process each node once. The number of iterations required to reduce k lists to 1 is $O(log\\ k)$.\n",
-        "\n",
-        "   Therefore, the overall time complexity of the code is $O(n * log\\ k)$, where n is the total number of nodes across all lists, and $k$ is the number of linked lists.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. **Heap Space:** The code doesn't use a heap data structure, so there's no additional space complexity due to a heap.\n",
-        "\n",
-        "2. **Merged Lists:** In the `mergeList` method, we create a new merged list. However, this list is not stored in memory for all lists; it's replaced with each merged pair. The space used for these merged lists is proportional to the size of the largest merged list, which is $O(n)$ in the worst case.\n",
-        "\n",
-        "3. **Additional Variables:** The code uses a few additional variables, such as `dummy` and `tail`, but these occupy a constant amount of space and don't depend on the input size.\n",
-        "\n",
-        "   Therefore, the overall space complexity of the code is $O(n)$, where n is the total number of nodes across all lists.\n",
-        "\n",
-        "In summary, the code's time complexity is $O(n * log(k))$, and its space complexity is $O(n)$. This code efficiently merges $k$ sorted linked lists using a divide-and-conquer approach with a relatively low space overhead."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/06. Linked List/README.md b/_build/html/_sources/06. Linked List/README.md
deleted file mode 100644
index fb7b3ee..0000000
--- a/_build/html/_sources/06. Linked List/README.md	
+++ /dev/null
@@ -1,21 +0,0 @@
-# Linked List Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)                       | Easy       |
-| [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)                   | Easy       |
-| [143. Reorder List](https://leetcode.com/problems/reorder-list/)                                     | Medium     |
-| [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | Medium |
-| [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)                           | Medium     |
-| [23. Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)                       | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/07. Trees/100. Same Tree.ipynb b/_build/html/_sources/07. Trees/100. Same Tree.ipynb
deleted file mode 100644
index a270fab..0000000
--- a/_build/html/_sources/07. Trees/100. Same Tree.ipynb	
+++ /dev/null
@@ -1,211 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 100: Same Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-        "\n",
-        "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-        "\n",
-        "**Constraints**\n",
-        "\n",
-        "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-        "2. $-10^4$ <= Node.val <= $10^4$\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def isSameTree(p, q):\n",
-        "    # Base case: If both p and q are None, the trees are the same.\n",
-        "    if not p and not q:\n",
-        "        return True\n",
-        "    \n",
-        "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-        "    if not p or not q:\n",
-        "        return False\n",
-        "    \n",
-        "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-        "    if p.val != q.val:\n",
-        "        return False\n",
-        "    \n",
-        "    # Recursively check the left and right subtrees of p and q.\n",
-        "    # If both subtrees are the same, the entire trees are the same.\n",
-        "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-        "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-        "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-        "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-        "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-        "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "\n",
-        "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-        "\n",
-        "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-        "print(isSameTree(p1, q1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "\n",
-        "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-        "\n",
-        "p2 = TreeNode(1, TreeNode(2), None)\n",
-        "q2 = TreeNode(1, None, TreeNode(2))\n",
-        "print(isSameTree(p2, q2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "\n",
-        "#Input: p = [1,2,1], q = [1,1,2]\n",
-        "\n",
-        "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-        "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-        "print(isSameTree(p3, q3))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time Complexity\n",
-        "\n",
-        "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-        "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-        "\n",
-        "## Space Complexity\n",
-        "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-        "\n",
-        "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-        "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-        "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-        "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-        "\n",
-        "## In summary:\n",
-        "\n",
-        "+ **Time Complexity:** $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-        "+ **Space Complexity:** $O(n)$ due to the recursive call stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb b/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb
deleted file mode 100644
index 8cb08b7..0000000
--- a/_build/html/_sources/07. Trees/102. Binary Tree Level Order Traversal.ipynb	
+++ /dev/null
@@ -1,271 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 102. Binary Tree Level Order Traversal\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Level Order Traversal problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-level-order-traversal/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, return *the level order traversal of its nodes' values*. (i.e., from left to right, level by level).\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-        "- `-1000 <= Node.val <= 1000`\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def levelOrder(root):\n",
-        "    # Check if the tree is empty, if so, return an empty list.\n",
-        "    if not root:\n",
-        "        return []\n",
-        "\n",
-        "    # Initialize an empty list to store the result.\n",
-        "    result = []\n",
-        "    \n",
-        "    # Initialize a queue with the root node to perform BFS.\n",
-        "    queue = [root]\n",
-        "\n",
-        "    while queue:\n",
-        "        # Initialize an empty list to store the values of nodes at the current level.\n",
-        "        level_values = []\n",
-        "        \n",
-        "        # Get the number of nodes at the current level.\n",
-        "        level_size = len(queue)\n",
-        "\n",
-        "        # Iterate through the nodes at the current level.\n",
-        "        for _ in range(level_size):\n",
-        "            # Dequeue the front node from the queue.\n",
-        "            node = queue.pop(0)\n",
-        "            \n",
-        "            # Append the value of the current node to the level_values list.\n",
-        "            level_values.append(node.val)\n",
-        "\n",
-        "            # Enqueue the left and right children of the current node if they exist.\n",
-        "            if node.left:\n",
-        "                queue.append(node.left)\n",
-        "            if node.right:\n",
-        "                queue.append(node.right)\n",
-        "\n",
-        "        # Append the level_values list (values at the current level) to the result list.\n",
-        "        result.append(level_values)\n",
-        "\n",
-        "    # Return the final result, which is a list of lists representing level order traversal.\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We define a `TreeNode` class that represents a node in the binary tree. Each `TreeNode` object has a value (`val`) and two child nodes: `left` and `right`.\n",
-        "\n",
-        "2. The `levelOrder` function takes the root of the binary tree as its input and returns the level order traversal of the tree as a list of lists.\n",
-        "\n",
-        "3. We start by checking if the input `root` is `None`, which indicates an empty tree. If the tree is empty, we return an empty list because there are no nodes to traverse.\n",
-        "\n",
-        "4. We initialize an empty list called `result` to store the final result, which will be a list of lists containing node values at each level.\n",
-        "\n",
-        "5. We initialize a queue called `queue` with the root node. This queue will be used for breadth-first traversal of the tree.\n",
-        "\n",
-        "6. We enter a while loop that continues until the `queue` is empty. Inside the loop, we perform the following steps:\n",
-        "\n",
-        "   - We initialize an empty list called `level_values` to store the values of nodes at the current level.\n",
-        "\n",
-        "   - We determine the number of nodes at the current level by getting the length of the `queue`. This is done to process nodes level by level.\n",
-        "\n",
-        "   - We iterate through the nodes at the current level using a for loop. For each node in the current level:\n",
-        "\n",
-        "     - We dequeue (remove) the front node from the `queue`.\n",
-        "\n",
-        "     - We append the value of the dequeued node to the `level_values` list, effectively collecting the values of nodes at the current level.\n",
-        "\n",
-        "     - If the dequeued node has a left child, we enqueue the left child to the `queue`.\n",
-        "\n",
-        "     - If the dequeued node has a right child, we enqueue the right child to the `queue`.\n",
-        "\n",
-        "   - After processing all nodes at the current level, we append the `level_values` list to the `result` list. This represents the values at the current level.\n",
-        "\n",
-        "7. The loop continues until all levels have been traversed, and the `queue` becomes empty.\n",
-        "\n",
-        "8. Finally, we return the `result` list, which contains lists of node values at each level, representing the level order traversal of the binary tree.\n",
-        "\n",
-        "The code effectively performs a breadth-first traversal of the binary tree, processing nodes level by level, and constructs the result list that represents the level order traversal."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[3], [9, 20], [15, 7]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(9)\n",
-        "root.right = TreeNode(20)\n",
-        "root.right.left = TreeNode(15)\n",
-        "root.right.right = TreeNode(7)\n",
-        "\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Using the same tree as before\n",
-        "root = TreeNode(1)\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root = None\n",
-        "result = levelOrder(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `levelOrder` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this function is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the breadth-first traversal.\n",
-        "\n",
-        "In the worst case, we have to enqueue and dequeue all nodes in the binary tree, which is proportional to the number of nodes (N). In each level, we process all nodes in that level, and since there are a total of log(N) levels in a balanced binary tree, the time complexity can also be approximated as O(N) for unbalanced trees.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this function is O(N), where N is the number of nodes in the binary tree. Here's how the space complexity breaks down:\n",
-        "\n",
-        "1. The `result` list stores the level order traversal, and in the worst case, it contains N/2 levels (for a completely unbalanced binary tree). So, the space used by `result` is O(N).\n",
-        "\n",
-        "2. The `queue` data structure is used for BFS traversal. In the worst case, it can store all nodes at the last level of the tree. In a balanced binary tree, the maximum number of nodes at any level is 2^(log(N)), which is still O(N). In the case of an unbalanced tree, it can be even worse. So, the space used by `queue` is O(N).\n",
-        "\n",
-        "Overall, the dominant factor in terms of space complexity is the `queue`, and the space complexity is O(N).\n",
-        "\n",
-        "In summary, the function's time complexity is O(N), and its space complexity is also O(N). It is an efficient and optimal solution for performing a level order traversal of a binary tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Reverse Level Order Traversal:** Modify the `levelOrder` function to return the level order traversal in reverse order (from bottom to top). For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[15, 7], [9, 20], [3]]`.\n",
-        "\n",
-        "2. **Zigzag Level Order Traversal:** Write a function that performs a level order traversal of a binary tree in a zigzag pattern. In a zigzag traversal, the nodes at even levels are traversed from left to right, and nodes at odd levels are traversed from right to left. For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[3], [20, 9], [15, 7]]`."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb b/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb
deleted file mode 100644
index 213989d..0000000
--- a/_build/html/_sources/07. Trees/104. Maximum Depth of Binary Tree.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 104. Maximum Depth of Binary Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the `root` of a binary tree, return its maximum depth.\n",
-        "\n",
-        "A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 19,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def maxDepth(root):\n",
-        "    # Base case: If the root is None, the depth is 0.\n",
-        "    if root is None:\n",
-        "        return 0\n",
-        "    \n",
-        "    # Recursively calculate the maximum depth of the left and right subtrees.\n",
-        "    left_depth = maxDepth(root.left)\n",
-        "    right_depth = maxDepth(root.right)\n",
-        "    \n",
-        "    # Return the maximum depth of the tree by adding 1 to the maximum depth of the subtrees.\n",
-        "    return max(left_depth, right_depth) + 1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's go through the code step by step to understand how it works:\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class to represent the nodes of the binary tree. Each node has three attributes:\n",
-        "   - `val`: the value stored in the node.\n",
-        "   - `left`: a reference to the left child node.\n",
-        "   - `right`: a reference to the right child node.\n",
-        "\n",
-        "2. The `maxDepth` function is the main function that calculates the maximum depth of the binary tree. It takes a single argument, `root`, which is the root node of the binary tree.\n",
-        "\n",
-        "3. In the `maxDepth` function, we have a base case:\n",
-        "   - If the `root` is `None`, it means we have reached the end of a branch of the tree (a leaf node or an empty subtree). In this case, the depth is 0 because there are no nodes to count.\n",
-        "\n",
-        "4. If the `root` is not `None`, we recursively calculate the maximum depth of the left and right subtrees:\n",
-        "   - We call `maxDepth` on the `root.left` to calculate the maximum depth of the left subtree and store it in the variable `left_depth`.\n",
-        "   - We call `maxDepth` on the `root.right` to calculate the maximum depth of the right subtree and store it in the variable `right_depth`.\n",
-        "\n",
-        "5. Finally, we return the maximum depth of the tree by taking the maximum of `left_depth` and `right_depth` and adding 1 to it. This is because we are counting the current level as well."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: root = [3,9,20,null,null,15,7]\n",
-        "\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(9)\n",
-        "root.right = TreeNode(20)\n",
-        "root.right.left = TreeNode(15)\n",
-        "root.right.right = TreeNode(7)\n",
-        "\n",
-        "print(maxDepth(root))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a230be4f-1f47-4919-87e3-62fc3f072d97",
-      "metadata": {},
-      "source": [
-        "In this example, we create a binary tree based on the given input `[3,9,20,null,null,15,7]` and calculate its maximum depth, which is `3`. The tree structure is as follows:\n",
-        "   ```\n",
-        "       3\n",
-        "      / \\\n",
-        "     9  20\n",
-        "       /  \\\n",
-        "      15   7\n",
-        "   ```\n",
-        "\n",
-        "So, the maximum depth is the length of the longest path from the root to a leaf node, which is 3 in this case."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: root = [1,null,2]\n",
-        "\n",
-        "root = TreeNode(1)\n",
-        "root.right = TreeNode(2)\n",
-        "\n",
-        "# Calculate the maximum depth of the tree and print the result\n",
-        "print(maxDepth(root))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The time complexity of the `maxDepth` function is $O(N)$, where $N$ is the number of nodes in the binary tree.\n",
-        "- This is because in the worst case, the function visits every node exactly once in a depth-first manner.\n",
-        "- The recursion explores all nodes of the tree, so the time complexity is linear with respect to the number of nodes.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity of the `maxDepth` function is $O(H)$, where H is the height of the binary tree.\n",
-        "- In the worst case, if the binary tree is completely unbalanced (skewed), the recursion stack can go as deep as the height of the tree.\n",
-        "- In the best case, if the binary tree is perfectly balanced, the height is $O(log\\ N)$, where $N$ is the number of nodes.\n",
-        "- Therefore, the space complexity can vary from $O(log\\ N)$ to $O(N)$ depending on the shape of the tree.\n",
-        "- In addition to the recursion stack, there is a small constant amount of space used for variables and function call overhead.\n",
-        "\n",
-        "**In summary:**\n",
-        "- The time complexity of the code is $O(N)$ as it visits each node once.\n",
-        "- The space complexity is $O(H)$, where H is the height of the tree, which can vary from $O(log\\ N)$ to $O(N)$ depending on the tree's shape."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb b/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb
deleted file mode 100644
index 31c9e69..0000000
--- a/_build/html/_sources/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb	
+++ /dev/null
@@ -1,313 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 105. Construct Binary Tree from Preorder and Inorder Traversal\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Construct Binary Tree from Preorder and Inorder Traversal problem on LeetCode, click here!](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)\n",
-        "\n",
-        "---\n",
-        "Given two integer arrays `preorder` and `inorder` where `preorder` is the preorder traversal of a binary tree and `inorder` is the inorder traversal of the same tree, construct and return *the binary tree.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= preorder.length <= 3000`\n",
-        "- `inorder.length == preorder.length`\n",
-        "- `-3000 <= preorder[i], inorder[i] <= 3000`\n",
-        "- `preorder` and `inorder` consist of **unique** values.\n",
-        "- Each value of `inorder` also appears in `preorder`.\n",
-        "- `preorder` is **guaranteed** to be the preorder traversal of the tree.\n",
-        "- `inorder` is **guaranteed** to be the inorder traversal of the tree.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def buildTree(preorder, inorder):\n",
-        "    if not preorder:\n",
-        "        return None\n",
-        "\n",
-        "    # The first element in the preorder traversal is the root of the current subtree\n",
-        "    root_val = preorder[0]\n",
-        "    root = TreeNode(root_val)\n",
-        "\n",
-        "    # Find the index of the root value in the inorder traversal\n",
-        "    root_idx_inorder = inorder.index(root_val)\n",
-        "\n",
-        "    # Recursively build left and right subtrees\n",
-        "    root.left = buildTree(preorder[1:1 + root_idx_inorder], inorder[:root_idx_inorder])\n",
-        "    root.right = buildTree(preorder[1 + root_idx_inorder:], inorder[root_idx_inorder + 1:])\n",
-        "\n",
-        "    return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `class TreeNode:`: This is a class definition for a binary tree node. It has three attributes:\n",
-        "   - `val`: The value of the node.\n",
-        "   - `left`: A reference to the left child node.\n",
-        "   - `right`: A reference to the right child node.\n",
-        "\n",
-        "   This class is used to create instances of binary tree nodes, which will be used to build the binary tree.\n",
-        "\n",
-        "2. `def buildTree(preorder, inorder):`: This is the main function that constructs a binary tree from its preorder and inorder traversals.\n",
-        "\n",
-        "   - `preorder`: A list representing the preorder traversal of the binary tree.\n",
-        "   - `inorder`: A list representing the inorder traversal of the binary tree.\n",
-        "\n",
-        "3. `if not preorder: return None`: This line checks if the `preorder` list is empty. If it is, it means there are no nodes to construct, so the function returns `None`.\n",
-        "\n",
-        "4. `root_val = preorder[0]`: The value of the root node is extracted from the first element of the `preorder` list.\n",
-        "\n",
-        "5. `root = TreeNode(root_val)`: A new `TreeNode` object is created with the `root_val` as its value. This represents the root of the current subtree.\n",
-        "\n",
-        "6. `root_idx_inorder = inorder.index(root_val)`: The index of the `root_val` in the `inorder` list is found. This index indicates the position of the root node in the inorder traversal.\n",
-        "\n",
-        "7. `root.left`: The left subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The left subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists up to the `root_idx_inorder`.\n",
-        "\n",
-        "8. `root.right`: The right subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The right subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists starting from `root_idx_inorder + 1`.\n",
-        "\n",
-        "9. Finally, the function returns the `root` of the subtree it just constructed.\n",
-        "\n",
-        "The recursive approach used here divides the problem of constructing the entire binary tree into smaller subproblems, starting with the root node and then recursively building the left and right subtrees until the entire tree is constructed."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "2d82857b-4fbe-474f-bf33-17b55c20147b",
-      "metadata": {},
-      "source": [
-        "## Helper Function \n",
-        "\n",
-        "Here is a helper function to represent the binary tree elements."
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "13e6f3e1-659d-479d-9521-75629b8c782f",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def treeToList(root):\n",
-        "    # Initialize an empty list to store the elements of the binary tree\n",
-        "    result = []\n",
-        "    \n",
-        "    # Initialize a queue for level-order traversal starting with the root\n",
-        "    queue = [root]\n",
-        "\n",
-        "    while queue:\n",
-        "        # Pop the front node from the queue\n",
-        "        node = queue.pop(0)\n",
-        "        \n",
-        "        # If the node is not None (i.e., a valid node), add its value to the result list\n",
-        "        if node:\n",
-        "            result.append(node.val)\n",
-        "            \n",
-        "            # Add the left and right children of the node to the queue\n",
-        "            queue.append(node.left)\n",
-        "            queue.append(node.right)\n",
-        "        else:\n",
-        "            # If the node is None, add None to the result to represent an empty node\n",
-        "            result.append(None)\n",
-        "\n",
-        "    # Remove any trailing None values from the result list\n",
-        "    while result and result[-1] is None:\n",
-        "        result.pop()\n",
-        "\n",
-        "    # Return the resulting list representing the binary tree elements\n",
-        "    return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "86cd4c9d-a120-4dd1-a1bd-530436012574",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's break down the `treeToList` helper function step by step:\n",
-        "\n",
-        "1. `def treeToList(root):`: This function takes the root of a binary tree as input and converts it into a list.\n",
-        "\n",
-        "2. `result = []`: Initialize an empty list called `result` to store the elements of the binary tree in list format.\n",
-        "\n",
-        "3. `queue = [root]`: Initialize a queue data structure for a level-order traversal, starting with the root node.\n",
-        "\n",
-        "4. `while queue:`: This starts a loop that continues until the queue is empty, indicating that all nodes have been processed.\n",
-        "\n",
-        "5. `node = queue.pop(0)`: Dequeue the first node from the queue, effectively processing it.\n",
-        "\n",
-        "6. `if node:`: Check if the `node` is not None, which means it's a valid node in the binary tree.\n",
-        "\n",
-        "    a. `result.append(node.val)`: If the node is valid, append its value (`node.val`) to the `result` list. This represents the value of the current node in the binary tree.\n",
-        "\n",
-        "    b. `queue.append(node.left)`: Enqueue the left child of the current node if it exists. This adds the left child to the queue for processing in the next iteration.\n",
-        "\n",
-        "    c. `queue.append(node.right)`: Enqueue the right child of the current node if it exists. This adds the right child to the queue for processing in the next iteration.\n",
-        "\n",
-        "7. `else:`: If the `node` is None, it represents an empty node in the binary tree.\n",
-        "\n",
-        "    a. `result.append(None)`: Append `None` to the `result` list to indicate an empty node.\n",
-        "\n",
-        "8. `while result and result[-1] is None:`: After the traversal is complete, there might be trailing `None` values in the `result` list. This loop removes any such trailing `None` values to ensure a clean representation of the tree's elements.\n",
-        "\n",
-        "9. `return result`: Return the `result` list, which now contains the elements of the binary tree in a format where `None` represents empty nodes and the order of elements reflects a level-order traversal of the tree.\n",
-        "\n",
-        "In summary, the `treeToList` function performs a level-order traversal of the binary tree using a queue, constructing a list where each element corresponds to a node's value or represents an empty node with `None`. This list represents the binary tree's elements in a structured format."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[3, 9, 20, None, None, 15, 7]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "preorder = [3, 9, 20, 15, 7]\n",
-        "inorder = [9, 3, 15, 20, 7]\n",
-        "result = buildTree(preorder, inorder)\n",
-        "\n",
-        "print(treeToList(result))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[-1]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "preorder = [-1]\n",
-        "inorder = [-1]\n",
-        "result = buildTree(preorder, inorder)\n",
-        "\n",
-        "print(treeToList(result))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `buildTree` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of the `buildTree` function can be analyzed in terms of the number of nodes in the binary tree. Let's assume there are `n` nodes in the binary tree.\n",
-        "\n",
-        "1. Finding the root value in the `preorder` traversal takes O(1) time as it's just extracting the first element from the list.\n",
-        "2. Finding the index of the root value in the `inorder` traversal using `inorder.index(root_val)` takes O(n) time in the worst case because in the worst case, it might have to search through all `n` elements in the `inorder` list to find the index.\n",
-        "3. The recursive calls to `buildTree` for the left and right subtrees are made once for each node in the tree. Therefore, the recursive calls have a combined time complexity of O(n) as they process each node once.\n",
-        "\n",
-        "The total time complexity of the `buildTree` function is O(n) due to the recursive calls and finding the root's index in the `inorder` list.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity is determined by the space used by the function's call stack during recursion and any additional data structures used. \n",
-        "\n",
-        "1. The space used by the function call stack during recursion depends on the height of the binary tree. In the worst case, where the tree is highly unbalanced (e.g., a skewed tree), the space complexity for the call stack is O(n) as it can go as deep as the number of nodes in the tree.\n",
-        "\n",
-        "2. Additionally, the function creates TreeNode objects for each node in the binary tree. Therefore, the space complexity for these objects is also O(n).\n",
-        "\n",
-        "Overall, the space complexity of the `buildTree` function is O(n) due to the space used by the call stack and the TreeNode objects.\n",
-        "\n",
-        "In summary, the time complexity of `buildTree` is O(n), and the space complexity is O(n), where 'n' is the number of nodes in the binary tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Construct Binary Tree from Postorder and Inorder Traversal:**\n",
-        "   Modify the problem to construct a binary tree from its postorder and inorder traversals. Implement a function similar to `buildTree` but for postorder and inorder traversals.\n",
-        "\n",
-        "2. **Reconstruct Binary Tree with Duplicate Values:**\n",
-        "   Extend the problem to handle binary trees with duplicate values. Ensure that your solution correctly handles scenarios where there are duplicate values in the preorder and inorder traversals."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb b/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb
deleted file mode 100644
index 59d8e7d..0000000
--- a/_build/html/_sources/07. Trees/124. Binary Tree Maximum Path Sum.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 124. Binary Tree Maximum Path Sum\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Maximum Path Sum problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-maximum-path-sum/)\n",
-        "\n",
-        "---\n",
-        "A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.\n",
-        "\n",
-        "The **path sum** of a path is the sum of the node's values in the path.\n",
-        "\n",
-        "Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[1, 3*10^4]$.\n",
-        "- `-1000 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to find the maximum path sum in a binary tree. A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. The path sum of a path is the sum of the node values in that path.\n",
-        "\n",
-        "In this problem, you are given the root of a binary tree, and you need to find the maximum path sum among all possible paths in the tree. Note that the path does not need to start at the root node or end at a leaf node; it can start and end at any nodes in the tree.\n",
-        "\n",
-        "To find the maximum path sum, you need to consider both left and right subtrees of each node while traversing the tree. At each node, you have two choices:\n",
-        "\n",
-        "1. Include the current node in the path: In this case, you add the value of the current node to the sum and continue exploring both the left and right subtrees for possible extensions of the path.\n",
-        "\n",
-        "2. Start a new path from the current node: In this case, you do not include the current node in the path sum, and you choose either the left subtree or the right subtree to start a new path.\n",
-        "\n",
-        "To solve this problem, you can use a recursive approach to traverse the binary tree. For each node, you calculate the maximum path sum that can pass through that node (option 1) and also return the maximum path sum that can be extended from that node (option 2). You keep track of the global maximum path sum encountered during the traversal.\n",
-        "\n",
-        "Ultimately, the maximum path sum among all paths in the tree will be the maximum of the global maximum path sum and the maximum path sum calculated for each node.\n",
-        "\n",
-        "The constraints for this problem include:\n",
-        "\n",
-        "- The number of nodes in the tree is in the range $[1, 3 * 10^4]$.\n",
-        "- Node values are in the range [-1000, 1000].\n",
-        "\n",
-        "By considering all possible paths through the tree, the algorithm aims to find the most significant sum of node values in any path in the binary tree.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "class Solution:\n",
-        "    def maxPathSum(self, root):\n",
-        "        def max_path_sum_helper(node):\n",
-        "            if not node:\n",
-        "                return 0\n",
-        "            \n",
-        "            # Recursively calculate the maximum path sum for left and right subtrees\n",
-        "            left_max = max(0, max_path_sum_helper(node.left))\n",
-        "            right_max = max(0, max_path_sum_helper(node.right))\n",
-        "            \n",
-        "            # Update the global maximum path sum\n",
-        "            self.max_sum = max(self.max_sum, left_max + right_max + node.val)\n",
-        "            \n",
-        "            # Return the maximum path sum starting from the current node\n",
-        "            return max(left_max, right_max) + node.val\n",
-        "        \n",
-        "        self.max_sum = float('-inf')  # Initialize the global maximum to negative infinity\n",
-        "        max_path_sum_helper(root)  # Start the recursive traversal from the root\n",
-        "        return self.max_sum"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. The code starts by defining a `TreeNode` class to represent nodes in the binary tree. Each `TreeNode` has a `val` attribute representing its value, a `left` attribute pointing to the left child, and a `right` attribute pointing to the right child.\n",
-        "\n",
-        "2. Next, a `Solution` class is defined to contain the `maxPathSum` method, which will be used to find the maximum path sum in the binary tree.\n",
-        "\n",
-        "3. Within the `maxPathSum` method, a helper function called `max_path_sum_helper` is defined. This recursive function takes a single argument, `node`, which is the current node being considered during traversal.\n",
-        "\n",
-        "4. Inside the `max_path_sum_helper` function:\n",
-        "   - If `node` is `None` (i.e., the current node is a null node), it returns 0. This is the base case of the recursion.\n",
-        "   - It calculates the maximum path sum starting from the left subtree (`left_max`) and the right subtree (`right_max`). Importantly, it uses `max(0, ...)` to ensure that negative values (which would make the path sum smaller) are not considered. If a subtree's path sum is negative, it's better to not include it in the path.\n",
-        "   - The code then updates the global maximum path sum (`self.max_sum`) by checking if the sum of `left_max`, `right_max`, and the current node's value is greater than the current maximum.\n",
-        "   - Finally, it returns the maximum path sum starting from the current node, which is the maximum of `left_max` and `right_max` plus the current node's value.\n",
-        "\n",
-        "5. Before calling the `max_path_sum_helper` function, the `max_sum` attribute of the `Solution` class is initialized to negative infinity (`float('-inf')`). This is done to ensure that any valid path sum encountered during traversal will be greater than the initial value of `max_sum`.\n",
-        "\n",
-        "6. The `max_path_sum_helper` function is called with the root of the binary tree, effectively starting the traversal from the root node.\n",
-        "\n",
-        "7. Once the traversal is complete, the method returns the value of `self.max_sum`, which contains the maximum path sum found in the binary tree.\n",
-        "\n",
-        "8. Example usage at the bottom of the code demonstrates how to create binary trees and use the `maxPathSum` method to find the maximum path sum for two different tree structures.\n",
-        "\n",
-        "In summary, the code uses a recursive approach to explore all possible paths in the binary tree, ensuring that negative path sums are not considered. It keeps track of the maximum path sum encountered and returns that maximum value as the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "6\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "root1 = TreeNode(1)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(3)\n",
-        "solution = Solution()\n",
-        "print(solution.maxPathSum(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "42\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "root2 = TreeNode(-10)\n",
-        "root2.left = TreeNode(9)\n",
-        "root2.right = TreeNode(20)\n",
-        "root2.right.left = TreeNode(15)\n",
-        "root2.right.right = TreeNode(7)\n",
-        "print(solution.maxPathSum(root2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "Time Complexity:\n",
-        "- The code uses a recursive depth-first traversal to explore each node in the binary tree exactly once.\n",
-        "- During the traversal, for each node, we perform constant time operations, such as updating the `max_sum` variable and calculating the maximum path sum starting from that node.\n",
-        "- Therefore, the time complexity of the code is O(N), where N is the number of nodes in the binary tree.\n",
-        "\n",
-        "Space Complexity:\n",
-        "- The space complexity of the code is determined by the space used in the call stack during the recursive traversal.\n",
-        "- In the worst case, the depth of the call stack can be equal to the height of the binary tree. In an unbalanced tree, this could be O(N), but in a balanced binary tree, the height is O(log N).\n",
-        "- Additionally, the code uses a constant amount of space for variables like `left_max`, `right_max`, and `max_sum`.\n",
-        "- Therefore, the overall space complexity is O(H), where H is the height of the binary tree. In the worst case, it's O(N), and in the best case (a balanced binary tree), it's O(log N).\n",
-        "\n",
-        "In summary:\n",
-        "- Time Complexity: O(N)\n",
-        "- Space Complexity: O(H), where H is the height of the binary tree, ranging from O(log N) to O(N) depending on the tree's balance."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Print the Maximum Path:**\n",
-        "   Modify the code to not only find the maximum path sum but also print the nodes involved in the maximum path. You can print the nodes in the correct order from the root to the leaf.\n",
-        "\n",
-        "2. **Path Sum Count:**\n",
-        "   Instead of finding the maximum path sum, find the count of unique paths that sum to a given target value. Each path can start and end anywhere in the tree."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb b/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb
deleted file mode 100644
index 55dc944..0000000
--- a/_build/html/_sources/07. Trees/199. Binary Tree Right Side View.ipynb	
+++ /dev/null
@@ -1,257 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 199. Binary Tree Right Side View\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Binary Tree Right Side View problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-right-side-view/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, imagine yourself standing on the **right side** of it, return *the values of the nodes you can see ordered from top to bottom*.\n",
-        "\n",
-        " \n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def rightSideView(root):\n",
-        "    if not root:\n",
-        "        return []  # If the tree is empty, return an empty list.\n",
-        "\n",
-        "    result = []  # Initialize a list to store the right side view values.\n",
-        "    queue = [root]  # Initialize a queue for level-order traversal with the root node.\n",
-        "\n",
-        "    while queue:\n",
-        "        # Get the number of nodes at the current level.\n",
-        "        level_size = len(queue)\n",
-        "\n",
-        "        # Traverse all nodes at the current level and add the rightmost node to the result.\n",
-        "        for i in range(level_size):\n",
-        "            node = queue.pop(0)  # Dequeue the first node from the queue.\n",
-        "\n",
-        "            # If it's the rightmost node at this level, add its value to the result.\n",
-        "            if i == level_size - 1:\n",
-        "                result.append(node.val)\n",
-        "\n",
-        "            # Add the children of the current node to the queue.\n",
-        "            if node.left:\n",
-        "                queue.append(node.left)\n",
-        "            if node.right:\n",
-        "                queue.append(node.right)\n",
-        "\n",
-        "    return result  # Return the list of right side view values.\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start with a binary tree represented using a TreeNode class. Each node has a value (`val`) and can have a left child (`left`) and a right child (`right`).\n",
-        "\n",
-        "2. The `rightSideView` function takes the root node of the binary tree as input and returns a list of values representing the nodes you can see when standing on the right side of the tree.\n",
-        "\n",
-        "3. We check if the `root` is `None`, which means the tree is empty. If it's empty, we return an empty list because there are no nodes to see.\n",
-        "\n",
-        "4. We initialize an empty list called `result` to store the right side view values.\n",
-        "\n",
-        "5. We initialize a queue called `queue` with the root node. This queue is used for a level-order traversal of the tree.\n",
-        "\n",
-        "6. We enter a while loop that continues until the `queue` is empty. Inside this loop, we perform the following steps for each level of the tree:\n",
-        "\n",
-        "   a. We determine the number of nodes at the current level by getting the length of the `queue`. This is important for processing nodes at the same level together.\n",
-        "\n",
-        "   b. We then iterate through the nodes at the current level using a for loop. For each node in the level, we dequeue it from the `queue` using `queue.pop(0)`.\n",
-        "\n",
-        "   c. If the node we dequeue is the rightmost node at the current level (determined by `i == level_size - 1`), we add its `val` to the `result` list. This is because, when standing on the right side of the tree, you can see the rightmost node at each level.\n",
-        "\n",
-        "   d. We enqueue the left and right children of the current node (if they exist) to the `queue`. This ensures that we process the next level in the subsequent iterations of the while loop.\n",
-        "\n",
-        "7. After processing all levels of the tree, the `result` list contains the values of the rightmost nodes in each level, ordered from top to bottom.\n",
-        "\n",
-        "8. Finally, we return the `result` list as the output of the function, which represents the right side view of the binary tree.\n",
-        "\n",
-        "The code effectively performs a level-order traversal of the binary tree while keeping track of the rightmost nodes at each level and adding them to the result list. This ensures that we obtain the correct order of nodes visible from the right side of the tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 3, 4]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root1 = TreeNode(1)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(3)\n",
-        "root1.left.right = TreeNode(5)\n",
-        "root1.right.right = TreeNode(4)\n",
-        "print(rightSideView(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[1, 3]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "root2 = TreeNode(1)\n",
-        "root2.right = TreeNode(3)\n",
-        "print(rightSideView(root2))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root = None\n",
-        "result = rightSideView(root)\n",
-        "print(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of the code is O(N), where N is the number of nodes in the binary tree. Here's why:\n",
-        "\n",
-        "1. We perform a level-order traversal of the tree using a queue. In the worst case, we visit all nodes once, which takes O(N) time since we visit each node exactly once.\n",
-        "\n",
-        "2. For each node, we perform constant-time operations like dequeuing it from the queue, checking if it's the rightmost node at the current level, and enqueuing its children (if they exist). These operations do not depend on the size of the tree, so they do not contribute to the overall time complexity.\n",
-        "\n",
-        "Therefore, the dominant factor in the time complexity is the level-order traversal, making it O(N).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is also O(N). Here's why:\n",
-        "\n",
-        "1. We use a queue to perform level-order traversal. In the worst case, the queue can contain all nodes at the maximum width of the tree, which can be up to N/2 nodes for a completely unbalanced tree. Therefore, the space required for the queue is O(N).\n",
-        "\n",
-        "2. The `result` list stores the values of the rightmost nodes. In the worst case, when the binary tree is a complete binary tree, it can have roughly N/2 rightmost nodes. Therefore, the space required for the `result` list is also O(N).\n",
-        "\n",
-        "3. Other auxiliary variables like `level_size`, `i`, and `node` require only constant space and do not depend on the size of the tree.\n",
-        "\n",
-        "Combining the space used by the queue and the `result` list, we get a space complexity of O(N).\n",
-        "\n",
-        "In summary, the code has a time complexity of O(N) and a space complexity of O(N), where N is the number of nodes in the binary tree. These complexities are efficient and scale linearly with the size of the input tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Variant with Left Side View**: Modify the code to find and return the values of nodes visible from the left side of the binary tree instead of the right side.\n",
-        "\n",
-        "2. **Zigzag Right Side View**: Extend the code to return the right side view values in a zigzag order. In a zigzag order, you alternate between starting from the rightmost node at level 0, then the leftmost at level 1, rightmost at level 2, and so on."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb b/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb
deleted file mode 100644
index 8ea4609..0000000
--- a/_build/html/_sources/07. Trees/226. Invert Binary Tree.ipynb	
+++ /dev/null
@@ -1,223 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 226: Invert Binary Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given the root of a binary tree, invert the tree, and return its root.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- The number of nodes in the tree is in the range `[0, 100]`.\n",
-        "- `-100 <= Node.val <= 100`\n",
-        "\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def invertTree(root):\n",
-        "    # Base case: If the root is None or the tree is empty, return None.\n",
-        "    if not root:\n",
-        "        return None\n",
-        "\n",
-        "    # Swap the left and right subtrees of the current node.\n",
-        "    root.left, root.right = root.right, root.left\n",
-        "\n",
-        "    # Recursively invert the left and right subtrees.\n",
-        "    invertTree(root.left)\n",
-        "    invertTree(root.right)\n",
-        "\n",
-        "    return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a TreeNode class, which represents a node in a binary tree. Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-        "\n",
-        "2. The `invertTree` function is defined to invert the binary tree. It takes the root node of the tree as an argument.\n",
-        "\n",
-        "3. In the `invertTree` function, we have a base case to handle the scenario when the root is `None` (empty tree). In such cases, we return `None` because there's nothing to invert.\n",
-        "\n",
-        "4. For non-empty trees, we swap the left and right subtrees of the current node. This effectively inverts the tree at the current node.\n",
-        "\n",
-        "5. We then recursively call `invertTree` on the left and right subtrees to invert them.\n",
-        "\n",
-        "6. Finally, we return the root of the inverted tree."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inverted Tree (Example 1): <__main__.TreeNode object at 0x270c9c0>\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 1\n",
-        "# Input: root = [4,2,7,1,3,6,9]\n",
-        "\n",
-        "root1 = TreeNode(4)\n",
-        "root1.left = TreeNode(2)\n",
-        "root1.right = TreeNode(7)\n",
-        "root1.left.left = TreeNode(1)\n",
-        "root1.left.right = TreeNode(3)\n",
-        "root1.right.left = TreeNode(6)\n",
-        "root1.right.right = TreeNode(9)\n",
-        "\n",
-        "inverted_root1 = invertTree(root1)\n",
-        "print(\"Inverted Tree (Example 1):\", inverted_root1)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inorder Traversal of Inverted Tree (Example 2): <__main__.TreeNode object at 0x2255fc0>\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 2\n",
-        "# Input: root = [2,1,3]\n",
-        "\n",
-        "root2 = TreeNode(2)\n",
-        "root2.left = TreeNode(1)\n",
-        "root2.right = TreeNode(3)\n",
-        "\n",
-        "inverted_root2 = invertTree(root2)\n",
-        "print(\"Inverted Tree (Example 2):\", inverted_root2)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Inverted Tree (Example 3): None\n"
-          ]
-        }
-      ],
-      "source": [
-        "### Example 3\n",
-        "# Input: root = []\n",
-        "\n",
-        "root3 = None  # Empty tree\n",
-        "\n",
-        "inverted_root3 = invertTree(root3)\n",
-        "print(\"Inverted Tree (Example 3):\", inverted_root3)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "Let's discuss the time and space complexity of the provided Python code to invert a binary tree.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of the `invertTree` function is $O(n)$, where $n$ is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal of the tree. In the worst case, we have to visit all nodes in the tree.\n",
-        "\n",
-        "The reason for this time complexity is the depth-first traversal of the tree, where we recursively process the left and right subtrees of each node.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of the code is determined by the function call stack during the recursion. In the worst case, the space complexity is $O(h)$, where $h$ is the height of the binary tree. \n",
-        "\n",
-        "In a completely unbalanced binary tree (essentially a linked list), the height $h$ is equal to the number of nodes $n$, resulting in a space complexity of $O(n)$. This occurs when the tree is skewed to one side.\n",
-        "\n",
-        "In a balanced binary tree, such as a full binary tree, the height $h$ is $O(log\\ n)$, and the space complexity is $O(log\\ n)$.\n",
-        "\n",
-        "The space complexity depends on how balanced the tree is. In practical scenarios, binary trees are often approximately balanced, so the space complexity is typically closer to $O(log\\ n)$.\n",
-        "\n",
-        "**In summary:**\n",
-        "\n",
-        "- Time Complexity: $O(n)$ where $n$ is the number of nodes.\n",
-        "- Space Complexity: $O(h)$, where $h$ is the height of the binary tree. In the worst case, it can be $O(n)$, and in a balanced tree, it is $O(log\\ n)$.\n",
-        "\n",
-        "Keep in mind that these complexities are based on the recursive implementation provided. Iterative solutions can achieve the same task with $O(1)$ space complexity, using auxiliary data structures like stacks or queues to mimic the recursion stack."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb b/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb
deleted file mode 100644
index 673ced2..0000000
--- a/_build/html/_sources/07. Trees/230. Kth Smallest Element in a BST.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 230. Kth Smallest Element in a BST\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Kth Smallest Element in a BST problem on LeetCode, click here!](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary search tree, and an integer `k`, return *the $k^{th}$ smallest value (**1-indexed**) of all the values of the nodes in the tree.*\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is `n`.\n",
-        "- 1 <= `k <= n` <= $10^4$\n",
-        "- 0 <= `Node.val` <= $10^4$\n",
-        "\n",
-        "**Follow up**: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def kthSmallest(root, k):\n",
-        "    stack = []  # Initialize an empty stack to simulate the traversal\n",
-        "    while root or stack:\n",
-        "        while root:\n",
-        "            stack.append(root)  # Push the current node onto the stack\n",
-        "            root = root.left  # Move to the left child\n",
-        "        root = stack.pop()  # Pop a node from the stack\n",
-        "        k -= 1  # Decrement k since we've visited a node\n",
-        "        if k == 0:\n",
-        "            return root.val  # If k becomes 0, return the current node's value as the kth smallest\n",
-        "        root = root.right  # Move to the right child to continue the traversal"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. We start by defining a `TreeNode` class, which represents a node in the binary search tree (BST). Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-        "\n",
-        "2. The `kthSmallest` function takes two parameters: `root`, which is the root of the BST, and `k`, which represents the kth smallest element we want to find.\n",
-        "\n",
-        "3. We initialize an empty stack (`stack`) to simulate the traversal of the BST. The stack will be used to keep track of nodes as we traverse the tree in an iterative manner.\n",
-        "\n",
-        "4. We enter a while loop that continues until either `root` becomes `None` (indicating we have traversed the entire tree) or the `stack` is empty.\n",
-        "\n",
-        "5. Within the loop, we start another while loop to traverse as far left as possible in the BST. We repeatedly push nodes onto the `stack` and move to their left children until we reach the leftmost leaf node. This is the smallest node in the BST.\n",
-        "\n",
-        "6. Once we have reached the leftmost leaf node (the smallest node), we pop nodes from the `stack` one by one. As we pop each node, we decrement `k` by 1 to keep track of the number of nodes we have visited.\n",
-        "\n",
-        "7. If `k` becomes 0 after decrementing, it means we have found the kth smallest element. In this case, we return the `val` of the current node as the result.\n",
-        "\n",
-        "8. If `k` is still greater than 0, it means we haven't found the kth smallest element yet. In this case, we move to the right child of the current node to continue the traversal, as the kth smallest element, if it exists, will be in the right subtree of the current node.\n",
-        "\n",
-        "9. The process continues until we find the kth smallest element or traverse the entire tree.\n",
-        "\n",
-        "10. Finally, we return the kth smallest element found.\n",
-        "\n",
-        "The key idea in this code is to perform an in-order traversal of the BST while keeping track of the kth smallest element. By visiting nodes in ascending order, we can efficiently find the kth smallest element in O(h + k) time, where h is the height of the BST and k is the desired kth element. This approach is both concise and efficient for this problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Create the tree for the first example: root = [3,1,4,null,2], k = 1\n",
-        "root1 = TreeNode(3)\n",
-        "root1.left = TreeNode(1)\n",
-        "root1.right = TreeNode(4)\n",
-        "root1.left.right = TreeNode(2)\n",
-        "k1 = 1\n",
-        "print(kthSmallest(root1, k1)) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Create the tree for the second example: root = [5,3,6,2,4,null,null,1], k = 3\n",
-        "root2 = TreeNode(5)\n",
-        "root2.left = TreeNode(3)\n",
-        "root2.right = TreeNode(6)\n",
-        "root2.left.left = TreeNode(2)\n",
-        "root2.left.right = TreeNode(4)\n",
-        "root2.left.left.left = TreeNode(1)\n",
-        "k2 = 3\n",
-        "print(kthSmallest(root2, k2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the given code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The time complexity of this code is O(h + k), where:\n",
-        "- h is the height of the binary search tree (BST).\n",
-        "- k is the desired kth smallest element we want to find.\n",
-        "\n",
-        "1. In the worst case, where the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the time complexity is O(n + k).\n",
-        "\n",
-        "2. In the best case, where the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the time complexity is O(log(n) + k).\n",
-        "\n",
-        "So, the time complexity can vary from O(log(n) + k) in the best-case scenario to O(n + k) in the worst-case scenario. Typically, for balanced BSTs, the time complexity is closer to O(log(n) + k).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "The space complexity of this code is O(h) due to the stack used for the iterative in-order traversal, where h is the height of the BST.\n",
-        "\n",
-        "1. In the worst case, when the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the space complexity is O(n) because the stack can potentially store all n nodes.\n",
-        "\n",
-        "2. In the best case, when the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the space complexity is O(log(n)) because the stack will have at most log(n) nodes.\n",
-        "\n",
-        "So, the space complexity depends on the height of the BST and can vary from O(log(n)) in the best-case scenario to O(n) in the worst-case scenario.\n",
-        "\n",
-        "In practical terms, for balanced BSTs or moderately unbalanced BSTs, the space complexity is usually close to O(log(n)), and the code is efficient in terms of space usage."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Kth Largest Element**: Modify the code to find the kth largest element in the BST instead of the kth smallest element.\n",
-        "\n",
-        "2. **Kth Smallest Element in Two BSTs**: Given two BSTs, find the kth smallest element when considering both BSTs as a single sorted list. This involves merging the two BSTs while finding the kth element efficiently."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb b/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb
deleted file mode 100644
index 666bcd9..0000000
--- a/_build/html/_sources/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 235. Lowest Common Ancestor of a Binary Search Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Lowest Common Ancestor of a Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)\n",
-        "\n",
-        "---\n",
-        "Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.\n",
-        "\n",
-        "According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[2, 10^5]$.\n",
-        "- $-10^9$ <= `Node.val` <= $10^9$\n",
-        "- All `Node.val` are **unique**.\n",
-        "- `p != q`\n",
-        "- `p` and `q` will exist in the BST.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def lowestCommonAncestor(root, p, q):\n",
-        "    # Ensure p is less than q\n",
-        "    if p.val > q.val:\n",
-        "        p, q = q, p\n",
-        "\n",
-        "    while root:\n",
-        "        # If the current node value is greater than both p and q, move left\n",
-        "        if root.val > q.val:\n",
-        "            root = root.left\n",
-        "        # If the current node value is less than both p and q, move right\n",
-        "        elif root.val < p.val:\n",
-        "            root = root.right\n",
-        "        # If the current node value is between p and q (inclusive), or it matches either p or q, it's the LCA\n",
-        "        else:\n",
-        "            return root"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. `class TreeNode:` defines a simple class for representing nodes in a binary tree. Each `TreeNode` has three attributes:\n",
-        "   - `val`: The value of the node.\n",
-        "   - `left`: A reference to the left child node.\n",
-        "   - `right`: A reference to the right child node.\n",
-        "\n",
-        "2. `def lowestCommonAncestor(root, p, q):` is a function that takes three arguments:\n",
-        "   - `root`: The root node of the BST.\n",
-        "   - `p`: A TreeNode representing one of the nodes for which we want to find the LCA.\n",
-        "   - `q`: A TreeNode representing the other node for which we want to find the LCA.\n",
-        "\n",
-        "3. `if p.val > q.val:` checks if the value of `p` is greater than the value of `q`. In a BST, it's essential to ensure that `p` represents the smaller value, and `q` represents the larger value. If `p` is greater than `q`, the code swaps their values.\n",
-        "\n",
-        "4. The main logic is inside the `while` loop, which runs until `root` becomes `None`. It performs the following steps to find the LCA:\n",
-        "   - If the current `root` node's value is greater than the value of `q`, it means both `p` and `q` are on the left subtree of the current node. So, we move to the left child of the current node by setting `root = root.left`.\n",
-        "   - If the current `root` node's value is less than the value of `p`, it means both `p` and `q` are on the right subtree of the current node. So, we move to the right child of the current node by setting `root = root.right`.\n",
-        "   - If neither of the above conditions is met, it means the current `root` node's value is between `p` and `q`, or it matches either `p` or `q`. In this case, the current node is the lowest common ancestor (LCA), so we return `root`.\n",
-        "\n",
-        "The algorithm is efficient because it takes advantage of the properties of a BST. It eliminates subtrees that cannot contain the LCA by comparing the values of `p`, `q`, and the current `root` node. Eventually, it reaches the LCA node, and that node is returned as the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "6\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Construct the tree\n",
-        "root = TreeNode(6)\n",
-        "root.left = TreeNode(2)\n",
-        "root.right = TreeNode(8)\n",
-        "root.left.left = TreeNode(0)\n",
-        "root.left.right = TreeNode(4)\n",
-        "root.right.left = TreeNode(7)\n",
-        "root.right.right = TreeNode(9)\n",
-        "root.left.right.left = TreeNode(3)\n",
-        "root.left.right.right = TreeNode(5)\n",
-        "\n",
-        "p = root.left  # Node 2\n",
-        "q = root.right  # Node 8\n",
-        "\n",
-        "result = lowestCommonAncestor(root, p, q)\n",
-        "print(result.val)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Using the same tree as before\n",
-        "p = root.left  # Node 2\n",
-        "q = root.left.right  # Node 4\n",
-        "\n",
-        "result = lowestCommonAncestor(root, p, q)\n",
-        "print(result.val) "
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3:\n",
-        "# Creating a new tree for this example\n",
-        "root2 = TreeNode(2)\n",
-        "root2.left = TreeNode(1)\n",
-        "\n",
-        "p = root2  # Node 2\n",
-        "q = root2.left  # Node 1\n",
-        "\n",
-        "result = lowestCommonAncestor(root2, p, q)\n",
-        "print(result.val)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided algorithm for finding the lowest common ancestor (LCA) in a Binary Search Tree (BST).\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this algorithm is O(h), where \"h\" is the height of the BST. In the worst case, where the tree is completely unbalanced (essentially a linked list), the height of the tree is O(n), where \"n\" is the number of nodes in the tree. However, in a well-balanced BST, the height is logarithmic, which is O(log n).\n",
-        "\n",
-        "The reason for this time complexity is that the algorithm efficiently narrows down the search space by traversing either left or right subtrees based on the values of the target nodes `p` and `q`. It eliminates entire subtrees that cannot contain the LCA, leading to a relatively quick search.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity of the algorithm is O(1). This is because it uses a constant amount of extra space, regardless of the size of the input BST. The only variables used are `root`, `p`, and `q`, and there are no data structures like stacks or queues used for additional space. The algorithm performs a simple traversal without recursion, so it does not consume extra memory as the tree depth increases.\n",
-        "\n",
-        "**In summary**, the provided algorithm for finding the LCA in a BST is both time and space-efficient. Its time complexity is O(h), where \"h\" is the height of the tree, and its space complexity is O(1), making it suitable for practical use even in large BSTs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "### Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple LCAs:** Extend the algorithm to find all the lowest common ancestors of two given nodes `p` and `q` in a BST. In some cases, there can be multiple LCAs.\n",
-        "\n",
-        "2. **LCA with k Nodes:** Given a BST and k nodes, find the lowest common ancestor of these k nodes. This is an extension of the problem where you must find the LCA of more than two nodes."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb b/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb
deleted file mode 100644
index 75c48ea..0000000
--- a/_build/html/_sources/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb	
+++ /dev/null
@@ -1,297 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 297. Serialize and Deserialize Binary Tree\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Serialize and Deserialize Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)\n",
-        "\n",
-        "---\n",
-        "Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.\n",
-        "\n",
-        "Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.\n",
-        "\n",
-        "**Clarification:** The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-        "- `-1000 <= Node.val <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "TThe problem is to design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of characters or bytes so that it can be easily stored in a file, transmitted over a network, or otherwise persisted. Deserialization is the reverse process of converting the serialized data back into the original data structure.\n",
-        "\n",
-        "In the context of this problem:\n",
-        "\n",
-        "1. **Serialization**: You are given a binary tree, and your task is to convert it into a string representation such that you can later recreate the same binary tree from this string. The format of serialization is flexible, but it should allow you to reconstruct the original binary tree accurately.\n",
-        "\n",
-        "2. **Deserialization**: Given a serialized string, you need to reconstruct the binary tree it represents, ensuring that it is identical to the original tree.\n",
-        "\n",
-        "Here are some key points to consider in solving this problem:\n",
-        "\n",
-        "- The input can include any valid binary tree, including trees with nodes having integer values within the range [-1000, 1000].\n",
-        "- You don't necessarily need to follow a specific format for serialization, but it should be designed in a way that allows unambiguous deserialization.\n",
-        "- The goal is to serialize the tree structure and its values and then deserialize it back into the same structure and values.\n",
-        "\n",
-        "A common approach for serialization is to use a traversal method like preorder traversal, where you visit nodes in the order: root, left subtree, right subtree. This way, you can serialize the tree into a string, and during deserialization, you can reconstruct the tree by parsing the string in the same order.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode(object):\n",
-        "    def __init__(self, x):\n",
-        "        self.val = x\n",
-        "        self.left = None\n",
-        "        self.right = None\n",
-        "\n",
-        "class Codec:\n",
-        "    def serialize(self, root):\n",
-        "        serialized  = []\n",
-        "\n",
-        "        def dfs(node):\n",
-        "            if not node:\n",
-        "                # If the node is None, represent it as \"Null\" in the serialized string\n",
-        "                serialized .append(\"Null\")\n",
-        "                return\n",
-        "            # Convert the node's value to a string and add it to the serialized string\n",
-        "            serialized .append(str(node.val))\n",
-        "            dfs(node.left)\n",
-        "            dfs(node.right)\n",
-        "\n",
-        "        dfs(root)\n",
-        "        # Join the serialized values with \",\"\n",
-        "        return \",\".join(serialized )\n",
-        "\n",
-        "    def deserialize(self, data):\n",
-        "        vals = data.split(\",\")\n",
-        "        self.i = 0\n",
-        "\n",
-        "        def dfs():\n",
-        "            if vals[self.i] == \"Null\":\n",
-        "                self.i += 1\n",
-        "                return None\n",
-        "            # Convert the value to an integer to create a new node\n",
-        "            node = TreeNode(int(vals[self.i]))\n",
-        "            self.i += 1\n",
-        "            node.left = dfs()\n",
-        "            node.right = dfs()\n",
-        "            return node\n",
-        "\n",
-        "        return dfs()"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Here's a step-by-step explanation of the code:\n",
-        "\n",
-        "1. **Definition of TreeNode**:\n",
-        "   - The code defines a simple `TreeNode` class that represents a node in a binary tree. Each `TreeNode` has three attributes:\n",
-        "     - `val`: The integer value stored in the node.\n",
-        "     - `left`: A reference to the left child node (or None if there is no left child).\n",
-        "     - `right`: A reference to the right child node (or None if there is no right child).\n",
-        "   - This class allows you to create binary tree nodes with integer values and left and right child references.\n",
-        "\n",
-        "2. **Serialization (serialize method)**:\n",
-        "   - The `serialize` method takes a `root` node as input, which is the root of the binary tree that needs to be serialized.\n",
-        "   - It initializes an empty list `serialized` to store the serialized elements.\n",
-        "   - The `dfs` (depth-first search) function is defined within the `serialize` method to perform a preorder traversal of the binary tree.\n",
-        "   - In the `dfs` function:\n",
-        "     - If the current `node` is `None` (i.e., a leaf node or a child of a leaf node), it appends the string \"Null\" to the `serialized` list to represent the absence of a node.\n",
-        "     - If the current `node` is not `None`, it appends the string representation of the `node.val` to the `serialized` list and then recursively calls `dfs` on the left and right children.\n",
-        "   - After the `dfs` traversal, the `serialized` list contains the serialized binary tree elements.\n",
-        "   - The method returns a string obtained by joining the elements in the `serialized` list with commas.\n",
-        "\n",
-        "3. **Deserialization (deserialize method)**:\n",
-        "   - The `deserialize` method takes a serialized string `data` as input, which represents a binary tree in the custom format.\n",
-        "   - It splits the `data` string by commas to obtain a list of elements called `vals`.\n",
-        "   - The method initializes an index `self.i` to 0. This index keeps track of the current position in the `vals` list during deserialization.\n",
-        "   - The `dfs` function is defined within the `deserialize` method to perform the deserialization process using a recursive approach.\n",
-        "   - In the `dfs` function:\n",
-        "     - If the current element in `vals` is \"Null,\" it means there is no node at this position in the binary tree, so it returns `None`.\n",
-        "     - If the current element is not \"Null,\" it converts the element to an integer to create a new `TreeNode` with that value.\n",
-        "     - It then recursively calls `dfs` to set the left and right children of the current node.\n",
-        "   - The method returns the root node of the reconstructed binary tree.\n",
-        "\n",
-        "Overall, this code demonstrates a way to serialize a binary tree into a string format and then deserialize it back into the original tree structure, allowing for the representation of both the tree structure and the values stored in the nodes."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1,2,Null,Null,3,4,Null,Null,5,Null,Null\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Serialize a binary tree\n",
-        "root = TreeNode(1)\n",
-        "root.left = TreeNode(2)\n",
-        "root.right = TreeNode(3)\n",
-        "root.right.left = TreeNode(4)\n",
-        "root.right.right = TreeNode(5)\n",
-        "\n",
-        "codec = Codec()\n",
-        "serialized_tree = codec.serialize(root)\n",
-        "print(serialized_tree)\n",
-        "\n",
-        "# Deserialize the serialized tree\n",
-        "new_root = codec.deserialize(serialized_tree)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Null\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Example usage:\n",
-        "# Serialize a binary tree\n",
-        "root = None\n",
-        "\n",
-        "codec = Codec()\n",
-        "serialized_tree = codec.serialize(root)\n",
-        "print(serialized_tree)\n",
-        "\n",
-        "# Deserialize the serialized tree\n",
-        "new_root = codec.deserialize(serialized_tree)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for serializing and deserializing a binary tree:\n",
-        "\n",
-        "1. **Serialization (serialize method)**:\n",
-        "\n",
-        "   - Time Complexity:\n",
-        "     - The `serialize` method uses a depth-first traversal (preorder) of the binary tree.\n",
-        "     - It visits each node exactly once.\n",
-        "     - At each node, it performs constant-time operations (string conversion and list append).\n",
-        "     - Therefore, the time complexity for serialization is O(N), where N is the number of nodes in the binary tree.\n",
-        "\n",
-        "   - Space Complexity:\n",
-        "     - The space complexity for serialization includes the space used by the `serialized` list and the recursive call stack.\n",
-        "     - The `serialized` list stores the serialized elements, and its space is proportional to the number of nodes.\n",
-        "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-        "     - Therefore, the space complexity for serialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-        "\n",
-        "2. **Deserialization (deserialize method)**:\n",
-        "\n",
-        "   - Time Complexity:\n",
-        "     - The `deserialize` method splits the serialized string into a list of elements, which takes O(N) time.\n",
-        "     - The `dfs` function is a recursive function that visits each element in the list exactly once.\n",
-        "     - At each element, it performs constant-time operations (string comparison, integer conversion, and recursive function calls).\n",
-        "     - Therefore, the time complexity for deserialization is O(N).\n",
-        "\n",
-        "   - Space Complexity:\n",
-        "     - The space complexity for deserialization includes the space used by the `vals` list and the recursive call stack.\n",
-        "     - The `vals` list stores the elements from the serialized string, and its space is proportional to the number of nodes.\n",
-        "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-        "     - Therefore, the space complexity for deserialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-        "\n",
-        "Overall, the provided code has a time complexity of O(N) for both serialization and deserialization and a space complexity of O(N) in the majority of cases (unless the tree is severely unbalanced, in which case the height H dominates the space complexity). It efficiently serializes and deserializes a binary tree while using a linear amount of memory."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Efficient Serialization**:\n",
-        "   Optimize the serialization process to minimize the length of the serialized string. Consider using techniques like binary encoding to represent node values more efficiently.\n",
-        "\n",
-        "2. **Custom Serialization Format**:\n",
-        "   Modify the serialization and deserialization methods to use a custom format different from the one provided. Ensure that the new format allows for accurate reconstruction of the original binary tree."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb b/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb
deleted file mode 100644
index 8a41c3b..0000000
--- a/_build/html/_sources/07. Trees/572. Subtree of Another Tree.ipynb	
+++ /dev/null
@@ -1,265 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 572: Subtree of Another Tree\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Subtree of Another Tree problem on LeetCode, click here!](https://leetcode.com/problems/subtree-of-another-tree/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "**Problem Description:**\n",
-        "\n",
-        "Given the roots of two binary trees `root` and `subRoot`, return `True` if there is a subtree of `root` with the same structure and node values as `subRoot`, and `False` otherwise.\n",
-        "\n",
-        "A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "1. The number of nodes in the `root` tree is in the range $[1, 2000]$.\n",
-        "2. The number of nodes in the `subRoot` tree is in the range $[1, 1000]$.\n",
-        "3. $-10^4$ <= root.val <= $10^4$\n",
-        "4. $-10^4$ <= subRoot.val <= $10^4$\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 17,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a binary tree node.\n",
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "class Solution:\n",
-        "    def isSubtree(self, root, subRoot):\n",
-        "        if not root:\n",
-        "            return False\n",
-        "        \n",
-        "        # Check if the current subtree is equal to the subRoot\n",
-        "        if self.isSameTree(root, subRoot):\n",
-        "            return True\n",
-        "        \n",
-        "        # Recursively check the left and right subtrees\n",
-        "        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)\n",
-        "    \n",
-        "    def isSameTree(self, tree1, tree2):\n",
-        "        if not tree1 and not tree2:\n",
-        "            return True\n",
-        "        if not tree1 or not tree2:\n",
-        "            return False\n",
-        "        \n",
-        "        return (\n",
-        "            tree1.val == tree2.val and\n",
-        "            self.isSameTree(tree1.left, tree2.left) and\n",
-        "            self.isSameTree(tree1.right, tree2.right)\n",
-        "        )"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "This code defines a TreeNode class for the binary tree nodes and a Solution class with two methods:\n",
-        "\n",
-        "1. `isSubtree`: This method checks if there is a subtree of the `root` tree with the same structure and node values as the `subRoot` tree. It uses a helper function `isSameTree` to compare two trees for equality.\n",
-        "\n",
-        "2. `isSameTree`: This helper method recursively compares two trees to check if they are the same in structure and node values.\n",
-        "\n",
-        "### Here's a detailed explanation of the code:\n",
-        "\n",
-        "1. **TreeNode Class:**\n",
-        "\n",
-        "The `TreeNode` class is defined to represent nodes in a binary tree. Each node has a `val` (the node's value) and may have a `left` and `right` child.\n",
-        "\n",
-        "2. **Solution Class:**\n",
-        "\n",
-        "    + The `Solution` class contains the solution for the problem and defines two important methods:\n",
-        "\n",
-        "    + `isSubtree(self, root, subRoot)`:\n",
-        "\n",
-        "        - This method checks whether `subRoot` is a subtree of `root`. It takes two tree nodes, `root` and `subRoot`, as input arguments.\n",
-        "        - If `root` is `None`, it returns `False` because there is no subtree to search.\n",
-        "        - It then checks if the current subtree with `root` as its root is equal to `subRoot` using the `isSameTree` method. If they are the same, it returns `True`.\n",
-        "        - If the current subtree is not the same as `subRoot`, it recursively checks the left and right subtrees of `root` to see if `subRoot` is a subtree of any of them.\n",
-        "        - It returns `True` if `subRoot` is found in either the left or right subtree; otherwise, it returns `False`.\n",
-        "\n",
-        "    + `isSameTree(self, tree1, tree2)`:\n",
-        "\n",
-        "        - This method checks whether two trees, `tree1` and `tree2`, are the same.\n",
-        "        - If both `tree1` and `tree2` are `None`, they are considered the same tree, so it returns `True`.\n",
-        "        - If only one of them is `None` (but not both), they are different trees, so it returns `False`.\n",
-        "        - If both `tree1` and `tree2` have values, it checks if their values are equal and recursively checks if their left and right subtrees are the same.\n",
-        "        - It returns `True` if the trees are the same; otherwise, it returns `False`.\n",
-        "\n",
-        "The code effectively uses recursion to traverse the binary trees and check for subtree equality. The `isSubtree` method starts the recursive search, and the `isSameTree` method is used to compare individual subtrees. The approach is efficient and avoids unnecessary checks when possible.\n",
-        "\n",
-        "This solution allows you to determine if there exists a subtree within the `root` tree that matches the structure and node values of `subRoot`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Is subRoot a subtree of root? True\n"
-          ]
-        }
-      ],
-      "source": [
-        "#root = [3,4,5,1,2]\n",
-        "#subRoot = [4,1,2]\n",
-        "\n",
-        "# Example input\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(4)\n",
-        "root.right = TreeNode(5)\n",
-        "root.left.left = TreeNode(1)\n",
-        "root.left.right = TreeNode(2)\n",
-        "\n",
-        "subRoot = TreeNode(4)\n",
-        "subRoot.left = TreeNode(1)\n",
-        "subRoot.right = TreeNode(2)\n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test the isSubtree method\n",
-        "result = solution.isSubtree(root, subRoot)\n",
-        "print(\"Is subRoot a subtree of root?\", result)"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Is subRoot a subtree of root? False\n"
-          ]
-        }
-      ],
-      "source": [
-        "#root = [3,4,5,1,2,null,null,null,null,0] \n",
-        "#subRoot = [4,1,2]\n",
-        "\n",
-        "\n",
-        "# Example input\n",
-        "root = TreeNode(3)\n",
-        "root.left = TreeNode(4)\n",
-        "root.right = TreeNode(5)\n",
-        "root.left.left = TreeNode(1)\n",
-        "root.left.right = TreeNode(2)\n",
-        "root.left.right.left = TreeNode(0)\n",
-        "\n",
-        "subRoot = TreeNode(4)\n",
-        "subRoot.left = TreeNode(1)\n",
-        "subRoot.right = TreeNode(2)\n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test the isSubtree method\n",
-        "result = solution.isSubtree(root, subRoot)\n",
-        "print(\"Is subRoot a subtree of root?\", result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "86a34f0d-0789-498c-818c-2d30f84754d0",
-      "metadata": {},
-      "source": [
-        "Let's analyze the time and space complexity of the provided code for the \"Subtree of Another Tree\" problem:\n",
-        "\n",
-        "**Time Complexity**\n",
-        "\n",
-        "The time complexity of the code primarily depends on the recursive traversal of the binary tree. In both the `isSubtree` and `isSameTree` functions, we visit each node in the binary trees once. Let's break down the time complexity:\n",
-        "\n",
-        "1. The `isSubtree` function:\n",
-        "   - In the worst case, we need to visit every node in the `root` tree.\n",
-        "   - For each node in the `root` tree, we call the `isSameTree` function, which has its own traversal.\n",
-        "   - So, the total time complexity is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree.\n",
-        "\n",
-        "2. The `isSameTree` function:\n",
-        "   - In the worst case, we visit every node in both `tree1` and `tree2`.\n",
-        "   - The number of recursive calls made is proportional to the number of nodes in the trees.\n",
-        "   - So, the time complexity of this function is $O(max(n, m))$, where $n$ and $m$ are the numbers of nodes in `tree1` and `tree2`, respectively.\n",
-        "\n",
-        "Overall, the time complexity of the entire code is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree. In practice, it may be less than $O(n * m)$ if a subtree mismatch is detected early during the traversal.\n",
-        "\n",
-        "**Space Complexity**\n",
-        "\n",
-        "The space complexity of the code is determined by the function call stack during recursion and the space used by the recursive functions. Let's analyze the space complexity:\n",
-        "\n",
-        "1. The `isSubtree` function:\n",
-        "   - It uses the call stack for recursion.\n",
-        "   - The maximum depth of the recursion is equal to the height of the `root` tree, which can be $O(n)$ in the worst case (unbalanced tree).\n",
-        "   - Additionally, the function doesn't use any significant extra space other than the recursion stack.\n",
-        "\n",
-        "2. The `isSameTree` function:\n",
-        "   - It also uses the call stack for recursion.\n",
-        "   - The maximum depth of the recursion is equal to the height of the `tree1` or `tree2`, whichever is greater.\n",
-        "   - So, the maximum space used for the call stack is $O(max(n, m))$.\n",
-        "\n",
-        "In summary, the space complexity of the code is $O(max(n, m))$ due to the function call stack. It scales with the maximum height of the trees being compared.\n",
-        "\n",
-        "Overall, the code is efficient and works well for trees with moderate sizes. However, it's important to keep in mind that the worst-case time complexity is $O(n * m)$, so for very large trees, the performance may degrade."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb b/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb
deleted file mode 100644
index ad86882..0000000
--- a/_build/html/_sources/07. Trees/98. Validate Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 98. Validate Binary Search Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Validate Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/validate-binary-search-tree/)\n",
-        "\n",
-        "---\n",
-        "Given the `root` of a binary tree, *determine if it is a valid binary search tree (BST)*.\n",
-        "\n",
-        "A **valid BST** is defined as follows:\n",
-        "\n",
-        "- The left subtree of a node contains only nodes with keys **less than** the node's key.\n",
-        "- The right subtree of a node contains only nodes with keys **greater than** the node's key.\n",
-        "- Both the left and right subtrees must also be binary search trees. \n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the tree is in the range $[1, 10^4]$.\n",
-        "- $-2^{31}$ <= `Node.val` <= $2^{31} - 1$\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to solve this problem:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TreeNode:\n",
-        "    def __init__(self, val=0, left=None, right=None):\n",
-        "        self.val = val\n",
-        "        self.left = left\n",
-        "        self.right = right\n",
-        "\n",
-        "def is_valid_BST(root):\n",
-        "    def is_valid(node, min_val, max_val):\n",
-        "        # Base case: If the node is None, it's a valid BST.\n",
-        "        if node is None:\n",
-        "            return True\n",
-        "        \n",
-        "        # Check if the current node's value is within the valid range.\n",
-        "        if not (min_val < node.val < max_val):\n",
-        "            return False\n",
-        "        \n",
-        "        # Recursively check the left and right subtrees with updated ranges.\n",
-        "        return (is_valid(node.left, min_val, node.val) and\n",
-        "                is_valid(node.right, node.val, max_val))\n",
-        "    \n",
-        "    # Call the helper function starting with a wide range for root node.\n",
-        "    return is_valid(root, float('-inf'), float('inf'))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- It defines a `TreeNode` class to represent nodes in the binary tree.\n",
-        "\n",
-        "- The `is_valid_BST` function takes the root node of the binary tree as input and returns `True` if the tree is a valid BST, and `False` otherwise.\n",
-        "\n",
-        "- Inside the `is_valid_BST` function, there is a helper function called `is_valid` that performs the actual validation using a recursive approach.\n",
-        "\n",
-        "- The `is_valid` function checks each node in the tree to ensure that it satisfies the properties of a BST:\n",
-        "  - The value of the current node must be within a valid range defined by `min_val` and `max_val`.\n",
-        "  - The left subtree of the current node should contain values less than the current node's value.\n",
-        "  - The right subtree of the current node should contain values greater than the current node's value.\n",
-        "\n",
-        "- If any of these conditions are violated, the function returns `False`, indicating that the tree is not a valid BST.\n",
-        "\n",
-        "- If all nodes satisfy these conditions, the function returns `True`, indicating that the tree is a valid BST.\n",
-        "\n",
-        "- The code calls the `is_valid` function with the root node and initial range values of negative infinity to positive infinity to start the validation.\n",
-        "\n",
-        "- The time complexity of the code is O(N), where N is the number of nodes in the tree, as it traverses each node once.\n",
-        "\n",
-        "- The space complexity depends on the height of the tree. In the average case for a balanced BST, the space complexity is O(log N), but in the worst case (skewed tree), it can be O(N) due to the recursive call stack.\n",
-        "\n",
-        "In summary, this code checks whether a given binary tree is a valid BST by recursively validating each node's value and its left and right subtrees while maintaining valid value ranges. If all nodes satisfy the BST properties, the tree is considered a valid BST."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "# Create the tree for the first example: [2, 1, 3]\n",
-        "root1 = TreeNode(2)\n",
-        "root1.left = TreeNode(1)\n",
-        "root1.right = TreeNode(3)\n",
-        "print(is_valid_BST(root1))"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "\n",
-        "# Create the tree for the second example: [5, 1, 4, None, None, 3, 6]\n",
-        "root2 = TreeNode(5)\n",
-        "root2.left = TreeNode(1)\n",
-        "root2.right = TreeNode(4)\n",
-        "root2.right.left = TreeNode(3)\n",
-        "root2.right.right = TreeNode(6)\n",
-        "print(is_valid_BST(root2))"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `is_valid_BST` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The time complexity of the function primarily depends on the number of nodes in the binary tree.\n",
-        "- In the worst case, we may have to visit every node in the tree once to validate whether it's part of a valid BST.\n",
-        "- Since we are performing a depth-first search (DFS) traversal of the tree, the time complexity is O(N), where N is the number of nodes in the tree.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The space complexity is determined by the space used by the recursive call stack during the DFS traversal.\n",
-        "- In the worst case, if the tree is completely unbalanced (a skewed tree), the maximum depth of the call stack would be equal to the height of the tree.\n",
-        "- In a balanced BST, the height is approximately log2(N), where N is the number of nodes.\n",
-        "- Therefore, the space complexity of the call stack is O(log N) for a balanced BST.\n",
-        "- In the worst case (skewed tree), the space complexity can be O(N) as the height of the tree can be equal to N.\n",
-        "\n",
-        "**Overall:**\n",
-        "- Time Complexity: O(N)\n",
-        "- Space Complexity: O(log N) on average for a balanced BST, and O(N) in the worst case for a skewed tree.\n",
-        "\n",
-        "The space complexity is typically dominated by the recursive call stack, and it varies depending on the shape of the binary tree. In practice, for balanced binary trees, the space complexity is often close to O(log N), which is quite efficient."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Complex Constraints**: Modify the problem to have more complex constraints on the tree. For example, consider a binary tree with constraints like \"The left subtree of a node contains only nodes with keys less than or equal to the node's key,\" and \"The right subtree of a node contains only nodes with keys greater than or equal to the node's key.\" How would you adapt the code to handle these constraints?\n",
-        "\n",
-        "2. **Handling Duplicates**: Modify the code to handle binary search trees that allow duplicate values. In a standard BST, each value is unique, but in this case, multiple nodes can have the same value. The tree should still be considered valid if it follows the BST property."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/07. Trees/README.md b/_build/html/_sources/07. Trees/README.md
deleted file mode 100644
index 1eba5d4..0000000
--- a/_build/html/_sources/07. Trees/README.md	
+++ /dev/null
@@ -1,26 +0,0 @@
-# Trees Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)                          | Easy       |
-| [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)     | Easy       |
-| [100. Same Tree](https://leetcode.com/problems/same-tree/)                                           | Easy       |
-| [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)               | Easy       |
-| [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/) | Easy |
-| [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/) | Medium |
-| [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)       | Medium     |
-| [230. Kth Smallest Element In a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)   | Medium     |
-| [105. Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) | Medium |
-| [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)     | Hard       |
-| [297. Serialize And Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/) | Hard |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb b/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb
deleted file mode 100644
index 9da2a17..0000000
--- a/_build/html/_sources/08. Tries/208. Implement Trie (Prefix Tree).ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 208. Implement Trie (Prefix Tree)\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Implement Trie problem on LeetCode, click here!](https://leetcode.com/problems/implement-trie-prefix-tree/)\n",
-        "\n",
-        "---\n",
-        "A **trie** (pronounced as \"try\") or **prefix tree** is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.\n",
-        "\n",
-        "Implement the Trie class:\n",
-        "\n",
-        "- `Trie()` Initializes the trie object.\n",
-        "- `void insert(String word)` Inserts the string `word` into the trie.\n",
-        "- `boolean search(String word)` Returns `true` if the string `word` is in the trie (i.e., was inserted before), and `false` otherwise.\n",
-        "- `boolean startsWith(String prefix)` Returns `true` if there is a previously inserted string `word` that has the prefix `prefix`, and `false` otherwise.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= word.length, prefix.length <= 2000`\n",
-        "- `word` and `prefix` consist only of lowercase English letters.\n",
-        "- At most $3 * 10^4$ calls in total will be made to `insert`, `search`, and `startsWith`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to implement a data structure known as a Trie (pronounced \"try\") or Prefix Tree. A Trie is a tree-like data structure that is used to efficiently store and retrieve a set of strings or words. It's particularly useful for tasks involving string manipulation, such as autocomplete and spell-checking. The problem statement defines the following operations for implementing the Trie class:\n",
-        "\n",
-        "1. `Trie()`: This is the constructor that initializes the Trie object.\n",
-        "\n",
-        "2. `void insert(String word)`: This method inserts the given string `word` into the Trie. It effectively adds the characters of the word to the Trie's structure, creating nodes for each character if they don't already exist. At the end of the word, a special flag is set to indicate that this node represents the end of a valid word.\n",
-        "\n",
-        "3. `boolean search(String word)`: This method checks if the given string `word` is present in the Trie. It starts from the root of the Trie and traverses the Trie by following the characters in the word. If it successfully traverses the Trie and reaches the end of the word, it returns `true`, indicating that the word exists in the Trie; otherwise, it returns `false`.\n",
-        "\n",
-        "4. `boolean startsWith(String prefix)`: This method checks if there is any previously inserted word in the Trie that has the given `prefix`. It's similar to the `search` method but does not require that the prefix be a complete word. If there is any word in the Trie that starts with the given prefix, it returns `true`; otherwise, it returns `false`.\n",
-        "\n",
-        "The problem also provides an example scenario where these operations are called, demonstrating the expected output for each operation.\n",
-        "\n",
-        "In summary, the goal is to create a data structure (Trie) that efficiently stores a set of strings and provides methods to insert new strings, search for complete words, and check if a given prefix exists in the stored set of strings.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        self.children = {}           # A dictionary to store child nodes\n",
-        "        self.is_end_of_word = False  # Indicates if a word ends at this node\n",
-        "\n",
-        "class Trie:\n",
-        "    def __init__(self):\n",
-        "        self.root = TrieNode()       # Initialize the Trie with a root node\n",
-        "\n",
-        "    def insert(self, word):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                node.children[char] = TrieNode()  # Create a new node if the character is not present\n",
-        "            node = node.children[char]           # Move to the child node\n",
-        "        node.is_end_of_word = True                # Mark the end of the inserted word\n",
-        "\n",
-        "    def search(self, word):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                return False         # If the character is not found, the word is not in the Trie\n",
-        "            node = node.children[char]  # Move to the child node\n",
-        "        return node.is_end_of_word    # Check if the node represents the end of a valid word\n",
-        "\n",
-        "    def startsWith(self, prefix):\n",
-        "        node = self.root             # Start from the root\n",
-        "        for char in prefix:\n",
-        "            if char not in node.children:\n",
-        "                return False         # If the character is not found, no word starts with the prefix\n",
-        "            node = node.children[char]  # Move to the child node\n",
-        "        return True                  # Prefix found in the Trie"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "Let's explain how the code works step by step:\n",
-        "\n",
-        "1. `TrieNode` Class:\n",
-        "   - `TrieNode` is a class that represents nodes in the Trie. Each node has two attributes:\n",
-        "     - `children`: This is a dictionary that stores child nodes. Each key in the dictionary represents a character, and the corresponding value is the child node for that character.\n",
-        "     - `is_end_of_word`: This boolean flag indicates whether the node represents the end of a complete word. Initially, it's set to `False` for all nodes.\n",
-        "\n",
-        "2. `Trie` Class:\n",
-        "   - `Trie` is the main class that implements the Trie data structure. It has the following methods:\n",
-        "\n",
-        "   - `__init__(self)`: The constructor initializes the Trie object by creating a root node, which serves as the starting point for all Trie operations.\n",
-        "\n",
-        "   - `insert(self, word)`: This method inserts a string `word` into the Trie. It starts from the root node and iterates through each character in the word.\n",
-        "     - If a character is not present as a child node, it creates a new node for that character.\n",
-        "     - It then moves to the child node and continues the process until the entire word is inserted.\n",
-        "     - Finally, it sets the `is_end_of_word` flag to `True` for the last node to mark the end of the inserted word.\n",
-        "\n",
-        "   - `search(self, word)`: This method checks if a complete word (string) exists in the Trie. It starts from the root node and iterates through each character in the word.\n",
-        "     - If a character is not found as a child node, it immediately returns `False` because the word is not in the Trie.\n",
-        "     - It continues to move to the child node for each character.\n",
-        "     - After reaching the end of the word, it checks if the `is_end_of_word` flag is `True` for the last node to confirm the presence of the word.\n",
-        "\n",
-        "   - `startsWith(self, prefix)`: This method checks if there is any previously inserted word in the Trie that starts with a given `prefix`. It follows the same logic as the `search` method but does not require the entire word to be present.\n",
-        "     - If the prefix is found in the Trie, it returns `True`; otherwise, it returns `False`.\n",
-        "\n",
-        "3. Example Usage:\n",
-        "   - The code demonstrates how to create a Trie object, insert words (\"apple\" and \"app\"), and perform operations on the Trie:\n",
-        "     - It inserts the word \"apple\" into the Trie.\n",
-        "     - It checks if \"apple\" is in the Trie, which returns `True`.\n",
-        "     - It checks if \"app\" is in the Trie, which returns `False`.\n",
-        "     - It checks if there is any word in the Trie that starts with \"app,\" which returns `True`.\n",
-        "     - It inserts the word \"app\" into the Trie.\n",
-        "     - It checks if \"app\" is in the Trie again, which now returns `True`.\n",
-        "\n",
-        "In summary, the code efficiently implements a Trie data structure with the ability to insert words, search for complete words, and check for the existence of words with a given prefix. The Trie is organized as a tree of nodes, with each node representing a character in the words being stored."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n",
-            "False\n",
-            "True\n",
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "\n",
-        "trie = Trie()\n",
-        "trie.insert(\"apple\")\n",
-        "print(trie.search(\"apple\"))   # Output: True\n",
-        "print(trie.search(\"app\"))     # Output: False\n",
-        "print(trie.startsWith(\"app\")) # Output: True\n",
-        "trie.insert(\"app\")\n",
-        "print(trie.search(\"app\"))     # Output: True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the Trie implementation:\n",
-        "\n",
-        "1. Time Complexity:\n",
-        "   - Insertion (`insert` method): The time complexity of inserting a word into the Trie is O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to insert it into the Trie.\n",
-        "   - Search (`search` method): The time complexity of searching for a word in the Trie is also O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to determine if it exists in the Trie.\n",
-        "   - Starts with (`startsWith` method): The time complexity of checking if a prefix exists in the Trie is O(p), where p is the length of the prefix. In the worst case, you may need to traverse the entire prefix to determine its presence.\n",
-        "\n",
-        "   Therefore, for each operation (insertion, search, or starts with), the time complexity is O(m) or O(p), where m is the length of the word being operated on, and p is the length of the prefix.\n",
-        "\n",
-        "2. Space Complexity:\n",
-        "   - The space complexity of the Trie is determined by the number of nodes and characters stored in it. In the worst case, where none of the inserted words share common prefixes, the space complexity is O(N), where N is the total number of characters across all inserted words.\n",
-        "   - Each node in the Trie represents a character, and the number of nodes is directly related to the total number of characters.\n",
-        "\n",
-        "   In practice, the space complexity can be less than O(N) because common prefixes are shared among words in the Trie, leading to space optimization.\n",
-        "\n",
-        "In summary:\n",
-        "- Time complexity for each operation (insertion, search, starts with) is O(m) or O(p), where m is the length of the word and p is the length of the prefix.\n",
-        "- Space complexity is O(N) in the worst case, where N is the total number of characters across all inserted words. However, space optimization occurs due to common prefix sharing in practice, potentially reducing the actual space used."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Count Prefixes**:\n",
-        "   Implement a method to count the number of words in the Trie that have a specific prefix. This exercise requires you to maintain additional data in the Trie nodes to keep track of the count.\n",
-        "\n",
-        "2. **Auto-Complete Suggestions**:\n",
-        "   Implement an auto-complete feature using the Trie. Given a prefix, the program should return a list of suggested words based on the words previously inserted into the Trie."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb b/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb
deleted file mode 100644
index 2de472c..0000000
--- a/_build/html/_sources/08. Tries/211. Design Add and Search Words Data Structure.ipynb	
+++ /dev/null
@@ -1,275 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 211. Design Add and Search Words Data Structure\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Design Add and Search Words Data Structure problem on LeetCode, click here!](https://leetcode.com/problems/design-add-and-search-words-data-structure/)\n",
-        "\n",
-        "---\n",
-        "Design a data structure that supports adding new words and finding if a string matches any previously added string.\n",
-        "\n",
-        "Implement the `WordDictionary` class:\n",
-        "\n",
-        "- `WordDictionary()` Initializes the object.\n",
-        "- `void addWord(word)` Adds `word` to the data structure, it can be matched later.\n",
-        "- `bool search(word)` Returns `true` if there is any string in the data structure that matches `word` or `false` otherwise. `word` may contain dots `'.'` where dots can be matched with any letter.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= word.length <= 25`\n",
-        "- `word` in `addWord` consists of lowercase English letters.\n",
-        "- `word` in `search` consist of `'.'` or lowercase English letters.\n",
-        "- There will be at most `2` dots in `word` for `search` queries.\n",
-        "- At most $10^4$ calls will be made to `addWord` and `search`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to design a data structure called `WordDictionary` that allows you to add words and search for words efficiently. The key feature of this data structure is that it should support wildcard search, where a dot ('.') can match any letter.\n",
-        "\n",
-        "Here are the specific requirements and explanations for the problem:\n",
-        "\n",
-        "1. Initialize the `WordDictionary` object using the constructor `WordDictionary()`.\n",
-        "\n",
-        "2. The `addWord(word)` method allows you to add a word to the data structure. Once added, you should be able to search for this word later. The words consist of lowercase English letters, and each word has a length between 1 and 25 characters.\n",
-        "\n",
-        "3. The `search(word)` method allows you to search for a word in the data structure. This method returns `True` if there is any string in the data structure that matches the given word or `False` otherwise. The word you are searching for may contain dots ('.'), where each dot can match any letter.\n",
-        "\n",
-        "4. The wildcard character ('.') in the `search` method allows for partial or fuzzy searching. For example, if the word dictionary contains the words \"bad,\" \"dad,\" and \"mad,\" then:\n",
-        "   - Searching for \"pad\" should return `False` because there is no exact match.\n",
-        "   - Searching for \"bad\" should return `True` because \"bad\" is in the dictionary.\n",
-        "   - Searching for \".ad\" should return `True` because it can match \"bad,\" \"dad,\" or \"mad.\"\n",
-        "   - Searching for \"b..\" should return `True` because it can match \"bad,\" \"bed,\" \"bet,\" etc.\n",
-        "\n",
-        "5. The problem specifies that there will be at most two dots ('.') in search queries, and there will be at most 10^4 calls to both `addWord` and `search`.\n",
-        "\n",
-        "In summary, you need to implement the `WordDictionary` class that efficiently supports adding words and searching for words, where the search can include wildcard characters ('.') that match any letter.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        # Initialize a TrieNode with children and a flag to indicate the end of a word.\n",
-        "        self.children = {}\n",
-        "        self.is_end_of_word = False\n",
-        "\n",
-        "class WordDictionary:\n",
-        "\n",
-        "    def __init__(self):\n",
-        "        # Initialize the WordDictionary with a root node.\n",
-        "        self.root = TrieNode()\n",
-        "\n",
-        "    def addWord(self, word: str) -> None:\n",
-        "        node = self.root\n",
-        "        # Iterate through the characters in the word.\n",
-        "        for char in word:\n",
-        "            if char not in node.children:\n",
-        "                # Create a new node for the character if it doesn't exist.\n",
-        "                node.children[char] = TrieNode()\n",
-        "            # Move to the next node.\n",
-        "            node = node.children[char]\n",
-        "        # Mark the end of the word by setting the flag to True.\n",
-        "        node.is_end_of_word = True\n",
-        "\n",
-        "    def search_helper(self, node: TrieNode, word: str) -> bool:\n",
-        "        if not word:\n",
-        "            # If there are no characters left in the word, check if we reached the end of a word in the trie.\n",
-        "            return node.is_end_of_word\n",
-        "        \n",
-        "        char = word[0]\n",
-        "        if char == '.':\n",
-        "            # If the character is a dot, explore all possible child nodes.\n",
-        "            for child in node.children.values():\n",
-        "                if self.search_helper(child, word[1:]):\n",
-        "                    return True\n",
-        "        elif char in node.children:\n",
-        "            # If the character is in the children of the current node, move to the next node.\n",
-        "            return self.search_helper(node.children[char], word[1:])\n",
-        "        \n",
-        "        # If the character is not found and is not a dot, the word is not in the trie.\n",
-        "        return False\n",
-        "\n",
-        "    def search(self, word: str) -> bool:\n",
-        "        # Start the search from the root of the trie.\n",
-        "        return self.search_helper(self.root, word)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "This code defines a `WordDictionary` class that uses a Trie data structure to efficiently store and search for words. Here's a step-by-step explanation:\n",
-        "\n",
-        "1. `TrieNode` class: This class represents nodes in the Trie data structure. Each node has two attributes: `children` (a dictionary to store child nodes) and `is_end_of_word` (a flag to indicate whether the node marks the end of a word).\n",
-        "\n",
-        "2. `WordDictionary` class: This class initializes with an empty Trie by creating a root node.\n",
-        "\n",
-        "3. `addWord` method: This method adds a word to the Trie. It iterates through each character in the word and creates Trie nodes as necessary. It sets the `is_end_of_word` flag to `True` for the final character of the word to mark the end of the word.\n",
-        "\n",
-        "4. `search_helper` method: This is a recursive helper function for searching words in the Trie. It takes a Trie node (`node`) and a word to search (`word`) as input. If there are no characters left in the word (`not word`), it checks if the current node marks the end of a word. If it does, it returns `True`.\n",
-        "\n",
-        "5. If the first character of the word is a dot ('.'), the function explores all possible child nodes by iterating through `node.children.values()` and recursively calls `search_helper` for each child with the remaining part of the word (`word[1:]`).\n",
-        "\n",
-        "6. If the first character of the word is not a dot and is found in the children of the current node, the function recursively moves to the next node by calling `search_helper` on the child node and the remaining part of the word.\n",
-        "\n",
-        "7. If the character is not found in the children and is not a dot, the word is not in the Trie, so the function returns `False`.\n",
-        "\n",
-        "8. `search` method: This is the public method for searching words. It initiates the search process by calling `search_helper` starting from the root node of the Trie.\n",
-        "\n",
-        "Overall, this code efficiently implements a Trie-based data structure that allows adding and searching for words, including support for wildcard characters represented by dots ('.')."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n",
-            "True\n",
-            "True\n",
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "\n",
-        "wordDictionary = WordDictionary()\n",
-        "wordDictionary.addWord(\"bad\")\n",
-        "wordDictionary.addWord(\"dad\")\n",
-        "wordDictionary.addWord(\"mad\")\n",
-        "print(wordDictionary.search(\"pad\"))  # Output: False\n",
-        "print(wordDictionary.search(\"bad\"))  # Output: True\n",
-        "print(wordDictionary.search(\".ad\"))  # Output: True\n",
-        "print(wordDictionary.search(\"b..\"))  # Output: True"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `WordDictionary` class methods:\n",
-        "\n",
-        "1. `addWord` Method:\n",
-        "\n",
-        "   - Time Complexity: O(L)\n",
-        "     - L is the length of the word being added.\n",
-        "     - In the worst case, we need to iterate through each character in the word and create nodes in the Trie. This takes O(L) time.\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for storing the word in the Trie is also O(L) because we create nodes for each character in the word.\n",
-        "\n",
-        "2. `search_helper` Method (called by `search`):\n",
-        "\n",
-        "   - Time Complexity: O(2^M * L)\n",
-        "     - M is the maximum number of dots ('.') in the search word.\n",
-        "     - In the worst case, when there are M dots in the search word, we may explore all possible child nodes at each dot. This results in a branching factor of 26 (for lowercase letters) at each dot.\n",
-        "     - So, the time complexity is exponential in the number of dots and linear in the length of the word being searched (L).\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for the recursive call stack is O(L) in the worst case because we may have to recurse to a depth equal to the length of the search word.\n",
-        "\n",
-        "3. `search` Method:\n",
-        "\n",
-        "   - Time Complexity: O(2^M * L)\n",
-        "     - The `search` method calls the `search_helper` method, which has the same time complexity explained above.\n",
-        "\n",
-        "   - Space Complexity: O(L)\n",
-        "     - The space complexity for the recursive call stack is O(L) in the worst case, as explained earlier.\n",
-        "\n",
-        "Overall, the time complexity of the `search` method is mainly affected by the number of dots ('.') in the search word. In the worst case, when there are multiple dots and many possibilities to explore at each dot, the time complexity can be exponential. However, for most practical cases, it performs reasonably well.\n",
-        "\n",
-        "The space complexity primarily depends on the space required to store the words in the Trie. It is proportional to the total number of characters in all added words and is linear with respect to the length of the words added.\n",
-        "\n",
-        "In summary:\n",
-        "\n",
-        "- Time Complexity for `addWord`: O(L)\n",
-        "- Time Complexity for `search`: O(2^M * L)\n",
-        "- Space Complexity: O(L) for storing words in the Trie, O(L) for the recursive call stack during search.\n",
-        "\n",
-        "Note: The space complexity analysis assumes that the dictionary contains a reasonable number of words and does not account for the overhead of Python objects or system-level memory allocation."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Dictionary Auto-Completion**: Extend the `WordDictionary` class to provide auto-completion suggestions based on the prefix of a word. Implement a method that returns a list of words that match the given prefix.\n",
-        "\n",
-        "2. **Support Word Deletion**: Extend the `WordDictionary` class to support word deletion. Add a method `deleteWord(word)` that removes a word from the data structure. Ensure that the search operation still works correctly after deletion."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/08. Tries/212. Word Search II.ipynb b/_build/html/_sources/08. Tries/212. Word Search II.ipynb
deleted file mode 100644
index 71597b8..0000000
--- a/_build/html/_sources/08. Tries/212. Word Search II.ipynb	
+++ /dev/null
@@ -1,316 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 212. Word Search II\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Word Search II problem on LeetCode, click here!](https://leetcode.com/problems/word-search-ii/)\n",
-        "\n",
-        "---\n",
-        "Given an `m x n` `board` of characters and a list of strings `words`, the task is to return *all words on the board*.\n",
-        "\n",
-        "Each word must be constructed from letters of sequentially **adjacent cells**, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == board.length`\n",
-        "- `n == board[i].length`\n",
-        "- `1 <= m, n <= 12`\n",
-        "- `board[i][j]` is a lowercase English letter.\n",
-        "- 1 <= `words.length` <= $3 * 10^4$\n",
-        "- `1 <= words[i].length <= 10`\n",
-        "- `words[i]` consists of lowercase English letters.\n",
-        "- All the strings of `words` are unique."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "To solve the \"Word Search II\" problem, you can use a Trie data structure and perform a depth-first search (DFS) on the board. Here's a step-by-step approach to solve the problem:\n",
-        "\n",
-        "1. Build a Trie from the given list of words:\n",
-        "   - Create a Trie data structure to store all the words from the input list.\n",
-        "   - For each word in the input list, insert it into the Trie, character by character. Make sure to mark the end of each word in the Trie.\n",
-        "\n",
-        "2. Perform DFS on the board:\n",
-        "   - Iterate through each cell (i, j) on the board.\n",
-        "   - For each cell, start a DFS traversal from that cell to search for words.\n",
-        "   - During the DFS traversal, maintain a current path string.\n",
-        "   - At each cell, check if the current path string, concatenated with the character in the cell, matches any word prefix in the Trie. If it does, continue the DFS.\n",
-        "   - If the current path string matches a word in the Trie, add it to the result.\n",
-        "\n",
-        "3. Implement DFS with backtracking:\n",
-        "   - During the DFS traversal, you will explore neighboring cells (horizontally and vertically) if they are within bounds and haven't been visited before.\n",
-        "   - To prevent revisiting the same cell within the same word, mark the cell as visited (e.g., change its character to a special character like '#') before exploring it and restore its original character after the DFS traversal.\n",
-        "\n",
-        "4. Return the result:\n",
-        "   - After completing the DFS traversal for all cells on the board, you will have collected all the valid words in the result set.\n",
-        "   - Convert the result set to a list and return it as the final output.\n",
-        "\n",
-        "This approach efficiently finds all words on the board by utilizing the Trie data structure to prune unnecessary DFS paths and avoiding duplicate visits to the same cell within the same word. It ensures that each word is constructed from letters of sequentially adjacent cells without using the same letter cell more than once in a word.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Define a TrieNode class to represent nodes in the Trie.\n",
-        "class TrieNode:\n",
-        "    def __init__(self):\n",
-        "        self.children = {}  # Dictionary to store child nodes.\n",
-        "        self.is_end_of_word = False  # Flag to mark the end of a word.\n",
-        "\n",
-        "# Main function to find words on the board.\n",
-        "def findWords(board, words):\n",
-        "    # DFS function to search for words starting from a given cell.\n",
-        "    def dfs(node, i, j, path):\n",
-        "        char = board[i][j]  # Get the character at the current cell.\n",
-        "        curr_node = node.children.get(char)  # Find the corresponding Trie node.\n",
-        "\n",
-        "        if not curr_node:\n",
-        "            return  # If no matching Trie node, stop the search.\n",
-        "\n",
-        "        path += char  # Add the character to the current path.\n",
-        "\n",
-        "        if curr_node.is_end_of_word:\n",
-        "            result.add(path)  # If the path matches a word, add it to the result.\n",
-        "\n",
-        "        temp, board[i][j] = board[i][j], \"#\"  # Mark the cell as visited.\n",
-        "\n",
-        "        # Explore neighboring cells (up, down, left, right).\n",
-        "        if i > 0:\n",
-        "            dfs(curr_node, i - 1, j, path)\n",
-        "        if i < m - 1:\n",
-        "            dfs(curr_node, i + 1, j, path)\n",
-        "        if j > 0:\n",
-        "            dfs(curr_node, i, j - 1, path)\n",
-        "        if j < n - 1:\n",
-        "            dfs(curr_node, i, j + 1, path)\n",
-        "\n",
-        "        board[i][j] = temp  # Restore the cell's original character.\n",
-        "\n",
-        "    # Function to build a Trie from the list of words.\n",
-        "    def buildTrie():\n",
-        "        root = TrieNode()  # Create a root node.\n",
-        "        for word in words:\n",
-        "            node = root\n",
-        "            for char in word:\n",
-        "                if char not in node.children:\n",
-        "                    node.children[char] = TrieNode()  # Create a new node if needed.\n",
-        "                node = node.children[char]\n",
-        "            node.is_end_of_word = True  # Mark the end of the word.\n",
-        "        return root\n",
-        "\n",
-        "    m, n = len(board), len(board[0])  # Get the dimensions of the board.\n",
-        "    root = buildTrie()  # Build the Trie from the list of words.\n",
-        "    result = set()  # Use a set to store unique words found on the board.\n",
-        "\n",
-        "    # Iterate through each cell on the board and start DFS from there.\n",
-        "    for i in range(m):\n",
-        "        for j in range(n):\n",
-        "            dfs(root, i, j, \"\")  # Start DFS from each cell on the board.\n",
-        "\n",
-        "    return list(result)  # Convert the set to a list and return the result."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code solves the \"Word Search II\" problem, where you have a grid of characters and a list of words. The goal is to find all words from the list that can be constructed by traversing adjacent cells (horizontally or vertically) in the grid, without using the same cell more than once for a single word.\n",
-        "\n",
-        "The code does the following:\n",
-        "\n",
-        "1. Defines a `TrieNode` class to represent nodes in a Trie data structure. Each node has children (other characters in the word) and a flag to mark the end of a word.\n",
-        "\n",
-        "2. Defines the `findWords` function, which takes the grid (`board`) and a list of words (`words`) as input.\n",
-        "\n",
-        "3. Inside the `findWords` function, there is a `dfs` (depth-first search) function. This function is used to explore the grid starting from a specific cell. It checks if the current path matches any word in the Trie and adds it to the result if found.\n",
-        "\n",
-        "4. The code also includes a `buildTrie` function that constructs a Trie data structure from the list of words. It iterates through each word, creating nodes for each character in the Trie and marking the end of words.\n",
-        "\n",
-        "5. The dimensions of the grid are obtained (m x n).\n",
-        "\n",
-        "6. A Trie is built using the `buildTrie` function.\n",
-        "\n",
-        "7. A set called `result` is used to store unique words found in the grid.\n",
-        "\n",
-        "8. The code iterates through each cell on the grid (using nested loops) and starts a DFS search from each cell.\n",
-        "\n",
-        "9. During the DFS, it explores neighboring cells (up, down, left, right) if they haven't been visited before and if the current path matches a prefix in the Trie.\n",
-        "\n",
-        "10. When a word is found during the DFS traversal, it is added to the `result` set.\n",
-        "\n",
-        "11. The grid cell is marked as visited during the DFS by changing its character to \"#\", and it is restored to its original character after the DFS traversal.\n",
-        "\n",
-        "12. Finally, the set of unique words found is converted to a list and returned as the output.\n",
-        "\n",
-        "The code effectively uses a Trie data structure to optimize the search and ensures that each word is constructed from adjacent cells without reusing the same cell for the same word."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "['eat', 'oath']\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "board1 = [[\"o\",\"a\",\"a\",\"n\"],[\"e\",\"t\",\"a\",\"e\"],[\"i\",\"h\",\"k\",\"r\"],[\"i\",\"f\",\"l\",\"v\"]]\n",
-        "words1 = [\"oath\",\"pea\",\"eat\",\"rain\"]\n",
-        "print(findWords(board1, words1))  # Output: [\"eat\", \"oath\"]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "c748b879-a2d8-4298-a42c-c7f61f9c7fc6",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "\n",
-        "board2 = [[\"a\",\"b\"],[\"c\",\"d\"]]\n",
-        "words2 = [\"abcb\"]\n",
-        "print(findWords(board2, words2))  # Output: []"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. **Building the Trie (`buildTrie` function):** \n",
-        "   - Suppose there are `N` words in the input list, and the average length of a word is `L`.\n",
-        "   - Building the Trie takes O(N * L) time, as we iterate through each word and insert each character into the Trie.\n",
-        "\n",
-        "2. **DFS Traversal (`dfs` function):** \n",
-        "   - In the worst case, the DFS function explores all possible paths in the grid.\n",
-        "   - The maximum number of recursive calls for each cell is 4 (up, down, left, right).\n",
-        "   - The maximum depth of the DFS is limited by the length of the longest word in the Trie, which is at most `L`.\n",
-        "   - Therefore, the DFS traversal for each cell takes $O(4^L)$ time.\n",
-        "\n",
-        "3. **Iterating Through the Grid (`findWords` function):**\n",
-        "   - In the worst case, we iterate through all `m` rows and `n` columns in the grid.\n",
-        "   - For each cell, we start a DFS traversal.\n",
-        "   - Therefore, iterating through the entire grid takes $O(m * n * 4^L)$ time.\n",
-        "\n",
-        "Overall, the time complexity of the code is $O(N * L + m * n * 4^L)$.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. **Trie Data Structure:**\n",
-        "   - The space required to store the Trie depends on the number of unique characters in all the words.\n",
-        "   - In the worst case, where all words are unique and have no common prefixes, the space complexity of the Trie is O(N * L).\n",
-        "   - In practice, it may be less if there are common prefixes among words.\n",
-        "\n",
-        "2. **DFS Stack (Recursive Calls):**\n",
-        "   - The depth of the recursive call stack during DFS is at most `L`, where `L` is the length of the longest word in the Trie.\n",
-        "   - Therefore, the space complexity for the recursive call stack is O(L).\n",
-        "\n",
-        "3. **Result Set (`result` set):**\n",
-        "   - The space used to store the result set depends on the number of valid words found in the grid.\n",
-        "   - In the worst case, when all words are found, the space complexity of the result set is O(N * L).\n",
-        "\n",
-        "Overall, the space complexity of the code is O(N * L) for the Trie and O(L) for the recursive call stack.\n",
-        "\n",
-        "In summary, the time complexity is dominated by the DFS traversal and is influenced by the number of words, their lengths, and the size of the grid. The space complexity is mainly determined by the Trie structure and the recursive call stack depth during DFS."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Word Search Paths:**\n",
-        "   Extend the basic word search exercise to find all possible paths (sequences of cells) that spell out a given word on the board. Return a list of paths if the word can be formed, otherwise an empty list.\n",
-        "\n",
-        "\n",
-        "2. **Reverse Word Search:**\n",
-        "   Modify the code to search for words in reverse order on the board. Given a 2D grid of characters and a list of words, find all words from the list that can be formed by traversing the board in reverse (right to left, bottom to top, etc.)."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/08. Tries/README.md b/_build/html/_sources/08. Tries/README.md
deleted file mode 100644
index 22c3bcd..0000000
--- a/_build/html/_sources/08. Tries/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Trie Problems - Blind 75 LeetCode
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [208. Implement Trie Prefix Tree](https://leetcode.com/problems/implement-trie-prefix-tree/)         | Medium     |
-| [211. Design Add And Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/) | Medium |
-| [212. Word Search II](https://leetcode.com/problems/word-search-ii/)                                 | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb b/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb
deleted file mode 100644
index cb66b6a..0000000
--- a/_build/html/_sources/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb	
+++ /dev/null
@@ -1,254 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 1046. Last Stone Weight\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Last Stone Weight problem on LeetCode, click here!](https://leetcode.com/problems/last-stone-weight/)\n",
-        "\n",
-        "---\n",
-        "You are given an array of integers `stones` where `stones[i]` is the weight of the `i`th stone.\n",
-        "\n",
-        "We are playing a game with the stones. On each turn, you choose the **heaviest two stones** and smash them together. Suppose the heaviest two stones have weights `x` and `y` with `x <= y`. The result of this smash is:\n",
-        "\n",
-        "1. If `x == y`, both stones are destroyed.\n",
-        "2. If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - x`.\n",
-        "\n",
-        "At the end of the game, there is **at most one** stone left.\n",
-        "\n",
-        "Return *the weight of the last remaining stone*. If there are no stones left, return `0`.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= stones.length <= 30`\n",
-        "- `1 <= stones[i] <= 1000`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The \"Last Stone Weight\" problem involves a game played with a collection of stones, each stone having a certain weight. The goal is to find the weight of the last remaining stone after applying a specific set of rules.\n",
-        "\n",
-        "Here's how the game works:\n",
-        "\n",
-        "1. You start with an array of stones, where each stone is represented by its weight. The weights of the stones are given in the input as an array.\n",
-        "\n",
-        "2. In each turn of the game, you select the two heaviest stones from the remaining stones. If there is only one stone left, you've found the answer.\n",
-        "\n",
-        "3. If there are two stones with weights `x` and `y`, where `x` is less than or equal to `y`, the following happens:\n",
-        "   - If `x` is equal to `y`, both stones are completely destroyed, and they are removed from the array of stones.\n",
-        "   - If `x` is not equal to `y`, the stone with weight `x` is destroyed, and the stone with weight `y` now has a new weight of `y - x`. The stone with weight `x` is removed from the array, and the modified stone with weight `y - x` remains in the array.\n",
-        "\n",
-        "4. The game continues with the modified array of stones until there is at most one stone left.\n",
-        "\n",
-        "5. The goal is to find the weight of the last remaining stone, or if there are no stones left, return `0`.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "\n",
-        "def lastStoneWeight(stones):\n",
-        "    # Convert the input list to a max heap (negate the values to simulate a max heap)\n",
-        "    max_heap = [-stone for stone in stones]\n",
-        "    heapq.heapify(max_heap)\n",
-        "\n",
-        "    # Continue as long as there are more than 1 stone in the heap\n",
-        "    while len(max_heap) > 1:\n",
-        "        # Get the two heaviest stones\n",
-        "        y = -heapq.heappop(max_heap)  # Get the heaviest stone and negate it back to positive\n",
-        "        x = -heapq.heappop(max_heap)  # Get the second heaviest stone and negate it back to positive\n",
-        "\n",
-        "        # Calculate the new stone's weight after smashing\n",
-        "        if x != y:\n",
-        "            new_stone = y - x\n",
-        "            heapq.heappush(max_heap, -new_stone)  # Negate the value and push it back to the heap\n",
-        "\n",
-        "    # If there is one stone left, return its weight (negate it back to positive)\n",
-        "    if max_heap:\n",
-        "        return -max_heap[0]\n",
-        "    else:\n",
-        "        return 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a Python function called `lastStoneWeight` that takes a list of stone weights as input. It aims to find the weight of the last remaining stone after playing the stone smashing game described in the problem statement.\n",
-        "\n",
-        "Here's a step-by-step explanation of the code:\n",
-        "\n",
-        "1. We import the `heapq` library, which allows us to create and manipulate a heap (priority queue).\n",
-        "\n",
-        "2. Inside the `lastStoneWeight` function:\n",
-        "   - We create a max heap (negating stone weights to simulate a max heap) using the `heapq.heapify` function. This heap will keep track of the heaviest stones.\n",
-        "   \n",
-        "   - We enter a loop that continues as long as there are more than 1 stone in the heap.\n",
-        "   \n",
-        "   - Inside the loop, we:\n",
-        "     - Retrieve and negate the heaviest stone (`y`) from the heap.\n",
-        "     - Retrieve and negate the second heaviest stone (`x`) from the heap.\n",
-        "   \n",
-        "     - Calculate the new stone's weight after smashing (`y - x`), and negate it back to simulate a max heap.\n",
-        "     - Add the new stone back to the heap using `heapq.heappush`.\n",
-        "   \n",
-        "3. After the loop, if there is one stone left in the heap, we retrieve and negate it to get its actual weight, and return it as the result.\n",
-        "   \n",
-        "4. If there are no stones left (the heap is empty), we return `0`, indicating that no stones remain.\n",
-        "\n",
-        "5. Finally, we provide two test cases to demonstrate the function's usage and correctness.\n",
-        "\n",
-        "In summary, the code efficiently implements the stone smashing game by maintaining a max heap, selecting the heaviest stones in each turn, smashing them, and updating the heap until there is at most one stone left. It then returns the weight of the last remaining stone or 0 if there are no stones left."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "stones1 = [2, 7, 4, 1, 8, 1]\n",
-        "print(lastStoneWeight(stones1))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 16,
-      "id": "b7e79514-4821-48d0-bb9d-0f2de8150945",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2: \n",
-        "\n",
-        "stones2 = [1]\n",
-        "print(lastStoneWeight(stones2))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided Python code for the \"Last Stone Weight\" problem:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. The `heapq.heapify` operation to create the max heap from the input stone weights has a time complexity of O(n), where n is the number of stones.\n",
-        "\n",
-        "2. The loop that continues until there are more than 1 stone in the heap performs the following operations in each iteration:\n",
-        "   - Retrieving and negating the heaviest stone (`y`) from the heap, which takes O(log n) time.\n",
-        "   - Retrieving and negating the second heaviest stone (`x`) from the heap, also taking O(log n) time.\n",
-        "   - Calculating the new stone's weight and pushing it back into the heap using `heapq.heappush`, which takes O(log n) time.\n",
-        "\n",
-        "3. In the worst case, the loop runs until there is only one stone left, which requires approximately (n-1) iterations.\n",
-        "\n",
-        "Overall, the dominant time complexity is O(n) for the initial heap creation, and within the loop, each iteration takes O(log n) time. Therefore, the overall time complexity of the algorithm is O(n log n).\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The space complexity is primarily determined by the space required for the max heap. In the worst case, this heap can contain all the stones, so the space complexity for the heap is O(n).\n",
-        "\n",
-        "2. The rest of the variables used in the function (e.g., `x`, `y`, `new_stone`) require only constant space, so they do not significantly contribute to the space complexity.\n",
-        "\n",
-        "Therefore, the overall space complexity of the algorithm is O(n) due to the space used for the max heap.\n",
-        "\n",
-        "In summary:\n",
-        "- Time Complexity: $O(n\\ log\\ n)$\n",
-        "- Space Complexity: $O(n)$\n",
-        "\n",
-        "The algorithm is efficient enough to handle the problem's constraints, as the worst-case time and space complexities are both linearithmic in terms of the number of stones."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Kth Last Stone Weight**: Modify the problem to find the weight of the Kth last remaining stone, where K is an integer input. Your function should work efficiently for large values of K.\n",
-        "\n",
-        "\n",
-        "2. **Multiple Stones Smash**: Modify the problem so that instead of smashing two stones at a time, you can smash up to K stones at each turn, where K is an integer input. Determine the weight of the last remaining stone in this variation."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb b/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb
deleted file mode 100644
index 847dd60..0000000
--- a/_build/html/_sources/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 295. Find Median from Data Stream\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Find Median from Data Stream problem on LeetCode, click here!](https://leetcode.com/problems/find-median-from-data-stream/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.\n",
-        "\n",
-        "Implement the MedianFinder class:\n",
-        "\n",
-        "- `MedianFinder()` initializes the `MedianFinder` object.\n",
-        "- `void addNum(int num)` adds the integer `num` from the data stream to the data structure.\n",
-        "- `double findMedian()` returns the median of all elements so far. Answers within $10^{-5}$ of the actual answer will be accepted.\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- $-10^{5}$ <= `num` <= $10^{5}$\n",
-        "- There will be at least one element in the data structure before calling `findMedian`.\n",
-        "- At most $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-        "\n",
-        "**Follow-up:**\n",
-        "\n",
-        "1. If all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?\n",
-        "2. If `99%` of all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem, \"Find Median from Data Stream,\" deals with efficiently computing the median of a sequence of numbers as they are incrementally provided. The median is the middle value in an ordered list of numbers. When the list has an even number of elements, the median is the average of the two middle values.\n",
-        "\n",
-        "Here's a breakdown of the problem:\n",
-        "\n",
-        "1. **Initialization**: You are asked to implement a `MedianFinder` class that supports the following operations:\n",
-        "\n",
-        "    - `MedianFinder()`: Initializes the MedianFinder object.\n",
-        "    - `addNum(int num)`: Adds an integer `num` from the data stream to the data structure.\n",
-        "    - `findMedian()`: Returns the median of all elements added to the data structure.\n",
-        "\n",
-        "2. **Median Calculation**: The `findMedian` operation should efficiently compute the median, and the result should have a precision of at least $10^{-5}$ (i.e., answers within this range will be considered correct).\n",
-        "\n",
-        "3. **Examples**:\n",
-        "   \n",
-        "   - If you add numbers [1, 2], the median should be 1.5 because (1 + 2) / 2 = 1.5.\n",
-        "   - If you add numbers [1, 2, 3], the median should be 2 because it's the middle value.\n",
-        "   \n",
-        "4. **Constraints**:\n",
-        "\n",
-        "   - The integers provided in the data stream are in the range from $-10^5$ to $10^5$.\n",
-        "   - There will be at least one element in the data structure before calling `findMedian`.\n",
-        "   - At most, $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-        "\n",
-        "The problem can be solved by using two priority queues (heaps): one max-heap to store the smaller half of the numbers and one min-heap to store the larger half of the numbers. The max-heap ensures that the largest number in the smaller half is at the top, and the min-heap ensures that the smallest number in the larger half is at the top. These heaps are balanced to efficiently find the median when requested.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "\n",
-        "class MedianFinder:\n",
-        "\n",
-        "    def __init__(self):\n",
-        "        # Initialize two heaps: a max-heap (small_half) for the smaller half of numbers,\n",
-        "        # and a min-heap (large_half) for the larger half of numbers.\n",
-        "        self.small_half = []  # Max-heap for the smaller half of the numbers\n",
-        "        self.large_half = []  # Min-heap for the larger half of the numbers\n",
-        "\n",
-        "    def addNum(self, num: int) -> None:\n",
-        "        # Add the number to the appropriate heap based on its value\n",
-        "        if not self.small_half or num <= -self.small_half[0]:\n",
-        "            # If the number is less than or equal to the current maximum in the smaller half,\n",
-        "            # add it to the smaller half (max-heap)\n",
-        "            heapq.heappush(self.small_half, -num)\n",
-        "        else:\n",
-        "            # Otherwise, add it to the larger half (min-heap)\n",
-        "            heapq.heappush(self.large_half, num)\n",
-        "\n",
-        "        # Balance the heaps if necessary to ensure the size difference is at most 1\n",
-        "        if len(self.small_half) > len(self.large_half) + 1:\n",
-        "            # If the size of the smaller half is more than one greater than the size of the larger half,\n",
-        "            # move the maximum from the smaller half to the larger half to balance them.\n",
-        "            heapq.heappush(self.large_half, -heapq.heappop(self.small_half))\n",
-        "        elif len(self.large_half) > len(self.small_half):\n",
-        "            # If the size of the larger half is greater than the size of the smaller half,\n",
-        "            # move the minimum from the larger half to the smaller half to balance them.\n",
-        "            heapq.heappush(self.small_half, -heapq.heappop(self.large_half))\n",
-        "\n",
-        "    def findMedian(self) -> float:\n",
-        "        if len(self.small_half) == len(self.large_half):\n",
-        "            # If both halves have the same size, there's an even number of elements,\n",
-        "            # so the median is the average of the top elements in both heaps.\n",
-        "            return (-self.small_half[0] + self.large_half[0]) / 2.0\n",
-        "        else:\n",
-        "            # If the smaller half has more elements, it contains the median (odd number of elements).\n",
-        "            return -self.small_half[0]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a `MedianFinder` class for efficiently finding the median of a stream of numbers. It uses two heaps: a max-heap (`small_half`) to store the smaller half of the numbers and a min-heap (`large_half`) to store the larger half. Here's how the code works:\n",
-        "\n",
-        "1. The `__init__` method initializes the two heaps.\n",
-        "\n",
-        "2. The `addNum` method adds a number to the appropriate heap, ensuring that the heaps remain balanced. It does this by comparing the number to the current maximum in the smaller half and adding it to the appropriate heap. Then, it checks the sizes of the two heaps and rebalances them if necessary.\n",
-        "\n",
-        "3. The `findMedian` method calculates and returns the median. If both halves have the same size (even number of elements), it computes the average of the tops of both heaps. If the smaller half has more elements (odd number of elements), it returns the top of the max-heap, which is the median.\n",
-        "\n",
-        "In summary, this code efficiently maintains two heaps to divide the elements into smaller and larger halves, allowing for quick median retrieval. It offers a time complexity of O(log N) for adding elements and O(1) for finding the median, where N is the total number of elements processed. The space complexity is O(1) as the space used by the two heaps is independent of the input size."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1.5\n",
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "medianFinder = MedianFinder()\n",
-        "medianFinder.addNum(1)\n",
-        "medianFinder.addNum(2)\n",
-        "print(medianFinder.findMedian())  # Output: 1.5\n",
-        "medianFinder.addNum(3)\n",
-        "print(medianFinder.findMedian())  # Output: 2.0\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `MedianFinder` class:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. `MedianFinder()` Initialization: This operation has a time complexity of O(1). It simply initializes the two heaps.\n",
-        "\n",
-        "2. `addNum(int num)`: The time complexity for adding a number is O(log N), where N is the total number of elements processed so far. This is because, in the worst case, you might need to perform logarithmic operations on the heaps to maintain their balance.\n",
-        "\n",
-        "3. `findMedian()`: Finding the median has a time complexity of O(1). It simply involves accessing the tops of the two heaps, which takes constant time.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. `MedianFinder()` Initialization: The space complexity for initialization is O(1). It involves creating two empty heaps, which do not depend on the size of the data stream.\n",
-        "\n",
-        "2. `addNum(int num)`: The space complexity for adding a number is also O(1). It involves adding the number to one of the two heaps, which do not significantly affect the overall space complexity.\n",
-        "\n",
-        "3. `findMedian()`: The space complexity for finding the median is O(1). It doesn't require any additional data structures that grow with the input size.\n",
-        "\n",
-        "In summary, the time complexity of the `addNum` operation is O(log N), and the time complexity of the `findMedian` operation is O(1). The space complexity is O(1) as well, as the space used by the two heaps is independent of the size of the data stream. This solution efficiently maintains and retrieves the median, making it suitable for the given problem constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Generalized kth Element**: Extend the `MedianFinder` class to support finding the kth element in the data stream efficiently. This means finding the kth smallest or kth largest element in the stream.\n",
-        "\n",
-        "2. **Sliding Window Median**: Given an array and a sliding window size, find the median within the window as it slides over the array. This is an extension of the problem to a moving window scenario."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
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-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb b/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb
deleted file mode 100644
index 2e1522d..0000000
--- a/_build/html/_sources/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb	
+++ /dev/null
@@ -1,239 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 703. Kth Largest Element in a Stream\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Kth Largest Element in a Stream problem on LeetCode, click here!](https://leetcode.com/problems/kth-largest-element-in-a-stream/)\n",
-        "\n",
-        "---\n",
-        "Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.\n",
-        "\n",
-        "Implement the `KthLargest` class with the following methods:\n",
-        "\n",
-        "- `KthLargest(int k, int[] nums)`: Initializes the object with the integer `k` and the stream of integers `nums`.\n",
-        "- `int add(int val)`: Appends the integer `val` to the stream and returns the element representing the `kth` largest element in the stream.\n",
-        "\n",
-        "\n",
-        "**Constraints:**\n",
-        "\n",
-        "- 1 <= `k` <= $10^4$\n",
-        "- 0 <= `nums.length` <= $10^4$\n",
-        "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-        "- $-10^4$ <= `val` <= $10^4$\n",
-        "- At most $10^4$ calls will be made to `add`.\n",
-        "- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem you're trying to solve is to design a class called `KthLargest` that can efficiently find the kth largest element in a stream of integers. You need to support two main operations:\n",
-        "\n",
-        "To solve this problem efficiently, you can use a min-heap data structure. Here's how the approach works:\n",
-        "\n",
-        "1. Initialize a min-heap to store the k largest elements. Initially, it's empty.\n",
-        "\n",
-        "2. In the `__init__` method, populate the min-heap with the first `k` elements from `nums`. This ensures that you have the k largest elements seen so far.\n",
-        "\n",
-        "3. Whenever you add a new element to the stream using the `add` method, follow these steps:\n",
-        "   - Add the new element to the min-heap.\n",
-        "   - If the size of the min-heap exceeds `k`, remove the smallest element from the min-heap. This ensures that you always have the k largest elements in the min-heap.\n",
-        "   - The smallest element in the min-heap (the root) will always represent the kth largest element in the stream.\n",
-        "\n",
-        "Here's why this approach works efficiently:\n",
-        "\n",
-        "- By using a min-heap, you can quickly maintain the k largest elements, and finding the smallest element in the heap (the root) takes constant time.\n",
-        "\n",
-        "- When you add a new element, the min-heap's size is kept at most `k`, which ensures that you only track the k largest elements and discard the smaller ones.\n",
-        "\n",
-        "- The time complexity for adding an element is O(log k), which is very efficient compared to sorting the entire stream, which would be O(n log n).\n",
-        "\n",
-        "- This approach meets the constraints of the problem, including handling large streams and having a low time complexity for both initialization and adding elements.\n",
-        "\n",
-        "In summary, the min-heap approach efficiently tracks the kth largest element in the stream by maintaining a heap of the k largest elements seen so far, updating it as new elements are added. This approach provides a fast and scalable solution to the problem.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "import heapq\n",
-        "from typing import List  # Import the List type from the typing module\n",
-        "\n",
-        "class KthLargest:\n",
-        "\n",
-        "    def __init__(self, k: int, nums: List[int]):\n",
-        "        # Initialize the KthLargest object with k and nums.\n",
-        "        self.k = k\n",
-        "        # Create a min-heap to store the k largest elements.\n",
-        "        self.min_heap = []\n",
-        "        # Populate the min-heap with the first k elements from nums.\n",
-        "        for num in nums:\n",
-        "            self.add(num)\n",
-        "\n",
-        "    def add(self, val: int) -> int:\n",
-        "        # Add val to the min-heap.\n",
-        "        heapq.heappush(self.min_heap, val)\n",
-        "        # If the size of the min-heap exceeds k, remove the smallest element.\n",
-        "        if len(self.min_heap) > self.k:\n",
-        "            heapq.heappop(self.min_heap)\n",
-        "        # The root of the min-heap is the kth largest element.\n",
-        "        return self.min_heap[0]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "- We start by importing the necessary modules:\n",
-        "  - `heapq`: This module provides functions to create and manipulate heaps.\n",
-        "  - `List` from the `typing` module: This is used to specify the type of the `nums` parameter.\n",
-        "\n",
-        "- We define the `KthLargest` class:\n",
-        "  - The `__init__` method initializes the object with the integer `k` and the list of integers `nums`.\n",
-        "  - It also creates an empty min-heap called `self.min_heap` to store the k largest elements.\n",
-        "  - It populates the min-heap with the first `k` elements from `nums` by calling the `add` method.\n",
-        "\n",
-        "- The `add` method:\n",
-        "  - Adds the new integer `val` to the min-heap using `heapq.heappush`. This maintains the min-heap property.\n",
-        "  - Checks if the size of the min-heap exceeds `k`. If it does, it removes the smallest element (the k+1th largest) using `heapq.heappop`.\n",
-        "  - Finally, it returns the smallest element in the min-heap, which is always the kth largest element.\n",
-        "\n",
-        "Overall, this code implements a class that efficiently finds the kth largest element in a stream of integers by maintaining a min-heap of the k largest elements seen so far.\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "4\n",
-            "5\n",
-            "5\n",
-            "8\n",
-            "8\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example usage:\n",
-        "kthLargest = KthLargest(3, [4, 5, 8, 2])\n",
-        "print(kthLargest.add(3))  # Output: 4\n",
-        "print(kthLargest.add(5))  # Output: 5\n",
-        "print(kthLargest.add(10)) # Output: 5\n",
-        "print(kthLargest.add(9))  # Output: 8\n",
-        "print(kthLargest.add(4))  # Output: 8"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `KthLargest` class using the min-heap approach:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. `__init__` method:\n",
-        "   - In the `__init__` method, you iterate over the first `k` elements in `nums` and add them to the min-heap. Each `heapq.heappush` operation takes O(log k) time.\n",
-        "   - Therefore, the time complexity of the `__init__` method is O(k * log k).\n",
-        "\n",
-        "2. `add` method:\n",
-        "   - In the `add` method, you perform the following operations:\n",
-        "     - `heapq.heappush`: O(log k) to add the new element to the min-heap.\n",
-        "     - If the size of the min-heap exceeds `k`, you perform `heapq.heappop`, which is also O(log k).\n",
-        "     - Finally, you return the smallest element from the min-heap, which is O(1) because it's always the root.\n",
-        "   - Overall, the time complexity of the `add` method is O(log k).\n",
-        "\n",
-        "3. Overall, if you make `n` calls to the `add` method, the total time complexity is O(n * log k), where `n` is the total number of elements added to the stream.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The space complexity is determined by the space used to store the min-heap and the instance variables.\n",
-        "2. The min-heap stores at most `k` elements at any given time, so its space complexity is O(k).\n",
-        "3. The instance variables (such as `self.k` and `self.min_heap`) have constant space requirements.\n",
-        "4. Therefore, the overall space complexity of the `KthLargest` class is O(k).\n",
-        "\n",
-        "In summary, the time complexity of the `KthLargest` class is O(n * log k) for `n` add operations, and the space complexity is O(k), where `k` is the parameter passed during initialization. This implementation efficiently maintains the kth largest element in the stream while meeting the problem's constraints."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Dynamic k:** Modify the `KthLargest` class to support dynamic changes in the value of `k`. Implement a method `update_k` that allows changing the value of `k` during the lifetime of the object. Ensure that the object can still correctly find the kth largest element based on the updated value of `k`.\n",
-        "\n",
-        "\n",
-        "2. **Implement kth Smallest:** Create a new class called `KthSmallest` that finds the kth smallest element in a stream instead of the kth largest. You may need to modify the data structure used in the implementation."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/09. Heap - Priority Queue/README.md b/_build/html/_sources/09. Heap - Priority Queue/README.md
deleted file mode 100644
index 2eac70c..0000000
--- a/_build/html/_sources/09. Heap - Priority Queue/README.md	
+++ /dev/null
@@ -1,18 +0,0 @@
-# Heap / Priority Queue Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [703. Kth Largest Element in a Stream](https://leetcode.com/problems/kth-largest-element-in-a-stream/)     | Easy       |
-| [1046. Last Stone Weight](https://leetcode.com/problems/last-stone-weight/)     | Easy       |
-| [295. Find Median From Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)     | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
diff --git a/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb b/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb
deleted file mode 100644
index 228a62e..0000000
--- a/_build/html/_sources/10. Backtracking/39. Combination Sum.ipynb	
+++ /dev/null
@@ -1,282 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 39. Combination Sum\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Combination Sum problem on LeetCode, click here!](https://leetcode.com/problems/combination-sum/)\n",
-        "\n",
-        "---\n",
-        "Given an array of **distinct** integers `candidates` and a target integer `target`, return *a list of all **unique combinations** of `candidates` where the chosen numbers sum to the `target`*. You may return the combinations in **any order**. \n",
-        "\n",
-        "The **same** number may be chosen from `candidates` an **unlimited number** of times. Two combinations are unique if the **frequency** of at least one of the chosen numbers is different.\n",
-        "\n",
-        "The test cases are generated such that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= candidates.length <= 30`\n",
-        "- `2 <= candidates[i] <= 40`\n",
-        "- All elements of `candidates` are **distinct**.\n",
-        "- `1 <= target <= 40`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to find all unique combinations of numbers from a given array (candidates) such that their sum equals a target number. Here are the details of the problem:\n",
-        "\n",
-        "- You are given an array of distinct integers called \"candidates.\"\n",
-        "- You are also given a target integer called \"target.\"\n",
-        "\n",
-        "The goal is to find all unique combinations of numbers from the candidates array where the sum of the selected numbers is equal to the target. You can use the same number from the candidates array an unlimited number of times. A combination is considered unique if it has a different frequency (i.e., a different number of occurrences) of at least one chosen number compared to other combinations.\n",
-        "\n",
-        "For example:\n",
-        "\n",
-        "**Example 1:**\n",
-        "```python\n",
-        "Input: candidates = [2, 3, 6, 7], target = 7\n",
-        "Output: [[2, 2, 3], [7]]\n",
-        "```\n",
-        "In this example, there are two unique combinations that sum up to the target:\n",
-        "- 2 + 2 + 3 = 7\n",
-        "- 7 = 7\n",
-        "\n",
-        "**Example 2:**\n",
-        "```python\n",
-        "Input: candidates = [2, 3, 5], target = 8\n",
-        "Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n",
-        "```\n",
-        "Here, there are three unique combinations that sum up to the target.\n",
-        "\n",
-        "**Example 3:**\n",
-        "```python\n",
-        "Input: candidates = [2], target = 1\n",
-        "Output: []\n",
-        "```\n",
-        "In this case, there are no combinations that can be formed from the candidates to reach the target of 1, so the output is an empty list.\n",
-        "\n",
-        "The problem asks you to find and return these combinations in any order.\n",
-        "\n",
-        "The constraints for this problem include the length of the candidates array, the values of candidates, and the target value.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def combinationSum(candidates, target):\n",
-        "    # Define a recursive DFS function to find combinations\n",
-        "    def dfs(remaining, start, path):\n",
-        "        # If the remaining target is 0, we found a valid combination\n",
-        "        if remaining == 0:\n",
-        "            result.append(path)\n",
-        "            return\n",
-        "        # If the remaining target is negative, this path is invalid\n",
-        "        if remaining < 0:\n",
-        "            return\n",
-        "        # Iterate through candidates starting from 'start' to avoid duplicates\n",
-        "        for i in range(start, len(candidates)):\n",
-        "            # Explore the current candidate by subtracting it from the remaining target\n",
-        "            # Add the current candidate to the path\n",
-        "            dfs(remaining - candidates[i], i, path + [candidates[i]])\n",
-        "\n",
-        "    result = []  # Initialize an empty list to store the result\n",
-        "    candidates.sort()  # Sort the candidates for deduplication and early stopping\n",
-        "    dfs(target, 0, [])  # Start the DFS from the target value with an empty path\n",
-        "    return result  # Return the list of unique combinations"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `combinationSum` function takes an array of distinct integers `candidates` and a target integer `target`. It returns a list of all unique combinations of `candidates` where the chosen numbers sum to the `target`. Each number in the `candidates` array can be used an unlimited number of times.\n",
-        "\n",
-        "Here's an overview of how the code works:\n",
-        "\n",
-        "1. The `combinationSum` function is defined, and it takes two arguments: `candidates` and `target`.\n",
-        "\n",
-        "2. Inside the function, there is a nested helper function called `backtrack`. This function is responsible for the actual combination generation using a recursive approach.\n",
-        "\n",
-        "3. The `backtrack` function is called with three arguments: `start`, `target`, and `path`. The `start` variable helps keep track of the current index in the `candidates` array, `target` keeps track of the remaining sum to be achieved, and `path` is a list that stores the current combination.\n",
-        "\n",
-        "4. Within the `backtrack` function, there are three main conditional statements:\n",
-        "   - If `target` becomes zero, it means we have found a combination that sums up to the target, so we add `path` to the `result` list.\n",
-        "   - If `target` becomes negative, it means the current combination doesn't work, so we return without doing anything.\n",
-        "   - Otherwise, we enter a loop that iterates over the `candidates` array, starting from the current index `start`.\n",
-        "\n",
-        "5. In the loop, the `backtrack` function is called recursively with the updated `target` and `path` after including the current candidate. This process explores different combinations by considering the current candidate or moving to the next candidate.\n",
-        "\n",
-        "6. After the loop completes, the `result` list contains all unique combinations that sum to the target.\n",
-        "\n",
-        "7. The `candidates` array is sorted to optimize the search. Sorting helps in avoiding unnecessary recursive branches and reduces the number of explored combinations.\n",
-        "\n",
-        "8. Finally, the `backtrack` function is initially called from the `combinationSum` function with `start` set to 0, `target` set to the original target value, and an empty `path`. The result is returned as a list of unique combinations.\n",
-        "\n",
-        "The code includes a few example cases to demonstrate how the function works with different inputs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[2, 2, 3], [7]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Example 1\n",
-        "candidates1 = [2, 3, 6, 7]\n",
-        "target1 = 7\n",
-        "print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "candidates2 = [2, 3, 5]\n",
-        "target2 = 8\n",
-        "print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "candidates3 = [2]\n",
-        "target3 = 1\n",
-        "print(combinationSum(candidates3, target3))  # Output: []"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `combinationSum` function.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "The time complexity of this function is influenced by the number of recursive calls made by the `backtrack` function and the amount of work done within each call. In the worst case, the function will explore all possible combinations.\n",
-        "\n",
-        "The worst-case scenario occurs when we have many combinations to reach the target. The time complexity can be expressed as $O(2^n)$, where 'n' is the maximum number of recursive calls. In this context, 'n' corresponds to the number of elements in the `candidates` list, and it may be up to 30 (as per the constraints). However, the actual number of combinations explored is generally much smaller because we eliminate branches when the target becomes negative, and we skip over candidates that cannot lead to a solution. The sorting step also helps to optimize the search.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "The space complexity is determined by the auxiliary space used for storing the results and the stack space for the recursive calls.\n",
-        "\n",
-        "1. The space required for the `result` list can be up to $O(2^n)$ in the worst case because each combination is stored.\n",
-        "\n",
-        "2. The stack space for the recursive calls can also be up to O(n) in the worst case, where 'n' is the number of elements in the `candidates` list.\n",
-        "\n",
-        "Therefore, the overall space complexity of the function is $O(2^n + n)$. However, in practice, the space used for the results list is often the dominant factor, and the actual space used may be much smaller than 2^n because not all combinations are explored.\n",
-        "\n",
-        "In summary, the time complexity is exponential but typically smaller in practice due to optimization, and the space complexity is influenced by the number of results and the depth of the recursion."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Find the Number of Unique Combinations:** Instead of listing all unique combinations, modify the function to return the count of unique combinations that sum to the target.\n",
-        "\n",
-        "2. **Combinations with a Maximum Value:** Add a maximum value constraint for each combination, so they cannot exceed a certain value. Modify the code to find combinations respecting this constraint."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/10. Backtracking/79. Word Search.ipynb b/_build/html/_sources/10. Backtracking/79. Word Search.ipynb
deleted file mode 100644
index cee509b..0000000
--- a/_build/html/_sources/10. Backtracking/79. Word Search.ipynb	
+++ /dev/null
@@ -1,298 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 79. Word Search\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Word Search problem on LeetCode, click here!](https://leetcode.com/problems/word-search/)\n",
-        "\n",
-        "---\n",
-        "Given an `m x n` grid of characters `board` and a string `word`, return *`true` if `word` exists in the grid.*\n",
-        "\n",
-        "The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == board.length`\n",
-        "- `n = board[i].length`\n",
-        "- `1 <= m, n <= 6`\n",
-        "- `1 <= word.length <= 15`\n",
-        "- `board` and `word` consists of only lowercase and uppercase English letters.\n",
-        "\n",
-        "**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether a given word can be constructed from characters in an `m x n` grid of characters. The word can be formed by moving sequentially through horizontally or vertically neighboring cells in the grid, and you cannot use the same cell more than once.\n",
-        "\n",
-        "Here are the specifics of the problem:\n",
-        "\n",
-        "- You are given an `m x n` grid, which is a 2D board of characters.\n",
-        "- You are also given a string `word` consisting of lowercase and uppercase English letters.\n",
-        "- Your task is to determine if the word can be found in the grid by moving from one cell to an adjacent cell (horizontally or vertically) and collecting the characters sequentially to form the word.\n",
-        "- You can't use the same cell more than once when forming the word.\n",
-        "- If it's possible to form the word in the grid, return `True`; otherwise, return `False`.\n",
-        "\n",
-        "For example, consider the following grid and word:\n",
-        "\n",
-        "```python\n",
-        "board = [[\"A\", \"B\", \"C\", \"E\"],\n",
-        "         [\"S\", \"F\", \"C\", \"S\"],\n",
-        "         [\"A\", \"D\", \"E\", \"E\"]]         \n",
-        "word = \"ABCCED\"\n",
-        "```\n",
-        "\n",
-        "In this case, the word \"ABCCED\" can be formed by moving through the cells (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) in the grid. Therefore, the function should return `True`.\n",
-        "\n",
-        "Conversely, if the word is impossible to construct, such as in the case of the word \"ABCB\" for the same grid, where there's no valid path that forms the word, the function should return `False`.\n",
-        "\n",
-        "The problem involves searching for a path through the grid that matches the characters in the word while respecting the constraints mentioned. It can be solved using depth-first search (DFS) or backtracking.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 3,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def exist(self, board: List[List[str]], word: str) -> bool:\n",
-        "        def dfs(i, j, k):\n",
-        "            # Check if the current position is out of bounds or doesn't match the character in the word.\n",
-        "            if not (0 <= i < len(board)) or not (0 <= j < len(board[0])) or board[i][j] != word[k]:\n",
-        "                return False\n",
-        "            \n",
-        "            # If we have matched all characters in the word, return True.\n",
-        "            if k == len(word) - 1:\n",
-        "                return True\n",
-        "\n",
-        "            # Temporarily mark the current cell to prevent revisiting.\n",
-        "            tmp, board[i][j] = board[i][j], \"/\"\n",
-        "\n",
-        "            # Explore adjacent cells by making recursive calls in all four directions.\n",
-        "            found = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)\n",
-        "\n",
-        "            # Restore the original character in the cell.\n",
-        "            board[i][j] = tmp\n",
-        "\n",
-        "            return found\n",
-        "\n",
-        "        # Iterate through all cells in the board.\n",
-        "        for i in range(len(board)):\n",
-        "            for j in range(len(board[0])):\n",
-        "                # Check if the 'dfs' function starting from this cell returns True.\n",
-        "                if dfs(i, j, 0):\n",
-        "                    return True\n",
-        "\n",
-        "        # If no match is found after exploring all cells, return False.\n",
-        "        return False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "1. Import the necessary module:\n",
-        "   - `from typing import List`: This line imports the `List` type from the `typing` module, which is used to specify the type of the `board` parameter in the method signature.\n",
-        "\n",
-        "2. Define the `Solution` class:\n",
-        "   - `class Solution:`: This line defines a class called `Solution` to encapsulate the solution for the word search problem.\n",
-        "\n",
-        "3. Define the `exist` method:\n",
-        "   - `def exist(self, board: List[List[str]], word: str) -> bool:`: This method is a member of the `Solution` class and takes two parameters: `board`, which is a 2D list of characters, and `word`, which is a string. It returns a boolean value, indicating whether the word exists in the board.\n",
-        "\n",
-        "4. Implement the inner `dfs` function:\n",
-        "   - `def dfs(i, j, k):`: This is a helper function used for depth-first search (DFS). It is called recursively to explore the board and search for the word. It takes three parameters: `i` and `j` representing the current position on the board, and `k` representing the current index in the `word` string.\n",
-        "\n",
-        "   - Inside `dfs`, it checks if the current position is out of bounds or if the character at the current position on the board does not match the character at the current index in the `word`. If either of these conditions is met, it returns `False`.\n",
-        "\n",
-        "   - If `k` is equal to the length of the `word` minus one, it means we have successfully matched all characters in the `word`, so it returns `True`.\n",
-        "\n",
-        "   - The function then temporarily replaces the character at the current position with a marker, \"/\", to prevent revisiting the same cell. It then explores adjacent cells by making recursive calls in all four directions: up, down, left, and right. If any of these recursive calls return `True`, it propagates the `True` result up the call stack. After the exploration is complete, it restores the original character in the cell.\n",
-        "\n",
-        "5. Iterate through the board:\n",
-        "   - The outer loop iterates through all cells in the `board` using two nested loops. For each cell, it checks if the `dfs` function starting from that cell (with `k` initially set to 0) returns `True`. If it does, it means the word exists, so the method returns `True`.\n",
-        "\n",
-        "6. Return `False`:\n",
-        "   - If the loop completes without finding the word in the board, the method returns `False`.\n",
-        "\n",
-        "To use this code, you can create an instance of the `Solution` class and call the `exist` method on that instance, passing the `board` and the `word` as arguments to check if the word exists in the board."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 4,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1: \n",
-        "\n",
-        "# Create an instance of the Solution class\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Test case 1: Word \"ABCCED\" can be found in the grid\n",
-        "board1 = [[\"A\",\"B\",\"C\",\"E\"],\n",
-        "          [\"S\",\"F\",\"C\",\"S\"],\n",
-        "          [\"A\",\"D\",\"E\",\"E\"]]\n",
-        "word1 = \"ABCCED\"\n",
-        "print(solution.exist(board1, word1))  # Expected output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "\n",
-        "# Test case 2: Word \"SEE\" can be found in the grid\n",
-        "word2 = \"SEE\"\n",
-        "print(solution.exist(board1, word2))  # Expected output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 3\n",
-        "\n",
-        "# Test case 3: Word \"ABCB\" cannot be found in the grid\n",
-        "word3 = \"ABCB\"\n",
-        "print(solution.exist(board1, word3))  # Expected output: False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `exist` method in the given code.\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "The primary work in this code is done by the `dfs` function, which performs a depth-first search through the grid. The time complexity of the `dfs` function is $O(4^(n))$, where \"n\" is the length of the `word`. This is because, in the worst case, the function explores all possible directions (up, down, left, right) for each character in the `word`.\n",
-        "\n",
-        "The outer loops iterate through all cells in the `board`, which has dimensions `m x n`. So, the total number of iterations of `dfs` is O(m * n). However, since the `dfs` function has a higher time complexity, the overall time complexity is still dominated by the `dfs` function.\n",
-        "\n",
-        "Therefore, the overall time complexity of the `exist` method is $O(m * n * 4^n)$, where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. The primary space usage comes from the recursive calls in the `dfs` function. The depth of the recursion can be at most equal to the length of the `word`, which is O(n).\n",
-        "\n",
-        "2. Additionally, the `tmp` variable and the recursive calls on the call stack consume a small amount of additional memory.\n",
-        "\n",
-        "3. The marking of visited cells with the \"/\" character temporarily modifies the `board` in-place but doesn't significantly contribute to the space complexity.\n",
-        "\n",
-        "Therefore, the overall space complexity is O(n) due to the recursive call stack.\n",
-        "\n",
-        "In summary, the time complexity is $O(m * n * 4^n)$, and the space complexity is O(n), where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word. The time complexity is exponential in the worst case due to the depth-first search through all possible paths in the grid."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the code to find all unique words that can be constructed from the grid. Instead of returning `True` or `False`, return a list of words found. Ensure that the same cell is not used more than once in constructing each word.\n",
-        "\n",
-        "2. **Board Generation:** Create a function to generate random grids of characters and then use the `exist` method to solve them. Test the efficiency and correctness of your code by generating larger boards and words."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/10. Backtracking/README.md b/_build/html/_sources/10. Backtracking/README.md
deleted file mode 100644
index 0d1ea13..0000000
--- a/_build/html/_sources/10. Backtracking/README.md	
+++ /dev/null
@@ -1,17 +0,0 @@
-# Backtracking Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [39. Combination Sum](https://leetcode.com/problems/combination-sum/)        | Medium     |
-| [79. Word Search](https://leetcode.com/problems/word-search/)                | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb b/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb
deleted file mode 100644
index 07351b6..0000000
--- a/_build/html/_sources/11. Graphs/133. Clone Graph.ipynb	
+++ /dev/null
@@ -1,317 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 133. Clone Graph\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Clone Graph problem on LeetCode, click here!](https://leetcode.com/problems/clone-graph/)\n",
-        "\n",
-        "---\n",
-        "Given a reference of a node in a **connected** undirected graph.\n",
-        "\n",
-        "Return a **deep copy** (clone) of the graph.\n",
-        "\n",
-        "Each node in the graph contains a value (`int`) and a list (`List[Node]`) of its neighbors.\n",
-        "\n",
-        "```\n",
-        "class Node {\n",
-        "    public int val;\n",
-        "    public List<Node> neighbors;\n",
-        "}\n",
-        "```\n",
-        "\n",
-        "**Test case format:**\n",
-        "\n",
-        "For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with `val == 1`, the second node with `val == 2`, and so on. The graph is represented in the test case using an adjacency list.\n",
-        "\n",
-        "**An adjacency list** is a collection of unordered **lists** used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.\n",
-        "\n",
-        "The given node will always be the first node with `val = 1`. You must return the **copy of the given node** as a reference to the cloned graph.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- The number of nodes in the graph is in the range `[0, 100]`.\n",
-        "- `1 <= Node.val <= 100`\n",
-        "- `Node.val` is unique for each node.\n",
-        "- There are no repeated edges and no self-loops in the graph.\n",
-        "- The Graph is connected and all nodes can be visited starting from the given node."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to clone a connected undirected graph, which means we need to create a deep copy of the original graph while preserving its structure and relationships between nodes. The graph is represented using adjacency lists, where each node has a value (an integer) and a list of its neighbors.\n",
-        "\n",
-        "Here are the key aspects of the problem:\n",
-        "\n",
-        "1. **Input**: You are given a reference to one of the nodes in the original graph. This reference node serves as the entry point to the graph. The entire graph can be explored starting from this node.\n",
-        "\n",
-        "2. **Output**: The goal is to create a deep copy of the graph and return a reference to the cloned graph. The cloned graph should have the same structure and relationships as the original one but should be a separate instance in memory.\n",
-        "\n",
-        "3. **Graph Structure**: The graph is composed of nodes (vertices) and edges. Nodes have values (integers) and are connected to other nodes through edges (undirected connections). Each node has a list of neighbors, which are other nodes it is connected to.\n",
-        "\n",
-        "4. **Connected Graph**: The problem assumes that the graph is connected, meaning that you can start from the reference node provided and traverse the entire graph by following the edges. This ensures that there are no isolated components in the graph.\n",
-        "\n",
-        "To solve this problem, you typically need to perform a graph traversal (e.g., depth-first search or breadth-first search) to visit all the nodes in the original graph, create corresponding nodes in the clone, and establish the same connections between nodes in the cloned graph as in the original one. To avoid duplication of nodes, you also need to keep track of visited nodes to ensure that the cloning process is efficient and that nodes are not duplicated in the clone.\n",
-        "\n",
-        "The problem can be challenging because you need to create a deep copy of the graph while handling cyclic dependencies between nodes and ensuring that the structure and relationships of the original graph are preserved in the clone. The solution requires recursion or a data structure to maintain the mapping between original nodes and their clones to achieve an efficient and accurate cloning process.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "# Definition for a Node.\n",
-        "class Node:\n",
-        "    def __init__(self, val=0, neighbors=None):\n",
-        "        self.val = val\n",
-        "        self.neighbors = neighbors if neighbors is not None else []\n",
-        "\n",
-        "class Solution:\n",
-        "    def cloneGraph(self, node: 'Node') -> 'Node':\n",
-        "        if not node:\n",
-        "            return None\n",
-        "\n",
-        "        visited = {}  # Dictionary to keep track of visited nodes\n",
-        "\n",
-        "        def dfs(original_node):\n",
-        "            if original_node in visited:\n",
-        "                return visited[original_node]\n",
-        "\n",
-        "            # Create a copy of the original node\n",
-        "            copy_node = Node(original_node.val)\n",
-        "            visited[original_node] = copy_node\n",
-        "\n",
-        "            # Recursively clone the neighbors of the original node\n",
-        "            for neighbor in original_node.neighbors:\n",
-        "                copy_node.neighbors.append(dfs(neighbor))\n",
-        "\n",
-        "            return copy_node\n",
-        "\n",
-        "        return dfs(node)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The given code defines a Python class, `Node`, and a solution class, `Solution`, to clone a connected undirected graph represented by nodes and their neighbors. Here's an explanation of the code:\n",
-        "\n",
-        "1. The `Node` class defines a node in the graph. Each node has a value (`val`) which is an integer and a list of neighbors (`neighbors`) that are other nodes in the graph.\n",
-        "\n",
-        "2. The `Solution` class contains the method `cloneGraph`, which takes a reference to the first node of the graph (`node`) as input and returns a deep copy of the entire graph, preserving the structure and relationships between nodes.\n",
-        "\n",
-        "3. Inside the `cloneGraph` method:\n",
-        "   - It first checks if the input `node` is `None`. If it's `None`, it means there is no graph to clone, and the method returns `None`.\n",
-        "\n",
-        "   - It initializes a `visited` dictionary to keep track of visited nodes. This dictionary is used to ensure that each node is cloned only once to avoid duplication.\n",
-        "\n",
-        "   - The `dfs` (depth-first search) function is defined within the `cloneGraph` method. It takes an original node as input and returns its corresponding clone. This function performs the actual cloning of the graph.\n",
-        "\n",
-        "   - Inside the `dfs` function:\n",
-        "     - It checks if the original node has been visited before by looking it up in the `visited` dictionary. If it has been visited, it returns the corresponding clone from the `visited` dictionary.\n",
-        "\n",
-        "     - If the original node is not visited, it creates a new node, `copy_node`, with the same value as the original node. This is the first step in creating the clone of the original node.\n",
-        "\n",
-        "     - It then recursively clones the neighbors of the original node by iterating through the `neighbors` list of the original node and appending the corresponding cloned neighbors to the `neighbors` list of the `copy_node`. This step ensures that the relationships between nodes are preserved in the clone.\n",
-        "\n",
-        "     - The `copy_node` is added to the `visited` dictionary with the original node as the key and the clone as the value.\n",
-        "\n",
-        "     - Finally, the `copy_node` is returned as the result of the DFS for the original node.\n",
-        "\n",
-        "4. The `cloneGraph` method returns the result of the DFS on the input `node`, which is the clone of the entire graph with its structure and relationships intact.\n",
-        "\n",
-        "In summary, the code uses depth-first search (DFS) to traverse the original graph, creating a clone for each node and its neighbors. It ensures that each node is cloned only once to avoid duplication and returns a reference to the cloned graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1, 2, 4]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "def get_adjacency_list(node):\n",
-        "    visited = {}  # Dictionary to keep track of visited nodes\n",
-        "\n",
-        "    def dfs(node):\n",
-        "        if not node:\n",
-        "            return []\n",
-        "\n",
-        "        if node in visited:\n",
-        "            return visited[node]\n",
-        "\n",
-        "        neighbors = [n.val for n in node.neighbors]\n",
-        "        visited[node] = neighbors\n",
-        "\n",
-        "        for neighbor in node.neighbors:\n",
-        "            dfs(neighbor)\n",
-        "\n",
-        "    dfs(node)\n",
-        "    result = []\n",
-        "\n",
-        "    def get_node_and_neighbors(node):\n",
-        "        result.append([node.val] + visited[node])\n",
-        "\n",
-        "    get_node_and_neighbors(node)\n",
-        "    return result\n",
-        "\n",
-        "\n",
-        "# Testing the code with the provided examples\n",
-        "adjList1 = [[2, 4], [1, 3], [2, 4], [1, 3]]\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Create the graph from adjList\n",
-        "nodes = [Node(i) for i in range(1, len(adjList1) + 1)]\n",
-        "for i, adj in enumerate(adjList1):\n",
-        "    nodes[i].neighbors = [nodes[j - 1] for j in adj]\n",
-        "\n",
-        "# Clone the graph and get its adjacency list\n",
-        "cloned_node1 = solution.cloneGraph(nodes[0])\n",
-        "cloned_adjList1 = get_adjacency_list(cloned_node1)\n",
-        "\n",
-        "# Print the cloned graph as an adjacency list\n",
-        "print(cloned_adjList1)\n",
-        "\n",
-        "# Output should match adjList1\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[1]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "# Input: adjList = [[]]\n",
-        "\n",
-        "adjList2 = [[]]\n",
-        "solution = Solution()\n",
-        "\n",
-        "# Create the graph from adjList\n",
-        "node1 = Node(1)\n",
-        "\n",
-        "# Clone the graph and get its adjacency list\n",
-        "cloned_node1 = solution.cloneGraph(node1)\n",
-        "cloned_adjList2 = get_adjacency_list(cloned_node1)\n",
-        "\n",
-        "# Print the cloned graph as an adjacency list\n",
-        "print(cloned_adjList2)\n",
-        "\n",
-        "# Output should be [[]], which represents an empty graph"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for cloning a graph using depth-first search (DFS).\n",
-        "\n",
-        "Time Complexity:\n",
-        "1. Creating the graph: The time complexity to create the original graph from `adjList` is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. We iterate through each node and its neighbors in `adjList`.\n",
-        "\n",
-        "2. Cloning the graph (DFS): The depth-first search (DFS) is used to traverse the original graph and create a clone. The time complexity of DFS is O(V + E), as we visit each node and each edge once.\n",
-        "\n",
-        "Overall, the time complexity is O(V + E) for both creating the original graph and cloning it.\n",
-        "\n",
-        "Space Complexity:\n",
-        "1. Storage for the original graph: We need to store the original graph, which includes all nodes and their adjacency lists. In the worst case, this requires O(V + E) space, where V is the number of nodes, and E is the number of edges.\n",
-        "\n",
-        "2. Storage for the visited dictionary: The `visited` dictionary stores mappings from original nodes to their corresponding clones. In the worst case, this requires O(V) space because each node is visited once.\n",
-        "\n",
-        "3. Recursive stack for DFS: The space used for the recursive call stack during DFS can be up to O(V) in the worst case, where V is the number of nodes.\n",
-        "\n",
-        "Overall, the space complexity is O(V + E) for storing the original graph, and O(V) for auxiliary data structures, including the `visited` dictionary and the DFS call stack. Therefore, the overall space complexity is O(V + E).\n",
-        "\n",
-        "In summary, the time and space complexities of the provided code for cloning a graph are both O(V + E), where V is the number of nodes, and E is the number of edges in the graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Cyclic Graph**: Design a graph with cycles (nodes connected in a cycle). Test if the code can correctly clone graphs with cycles and doesn't fall into an infinite loop during traversal.\n",
-        "\n",
-        "\n",
-        "2. **Partial Graph Cloning**: Modify the code to clone only a subset of the graph starting from a given reference node, rather than the entire graph. Implement this as an extension to the problem."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb b/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb
deleted file mode 100644
index e0235d6..0000000
--- a/_build/html/_sources/11. Graphs/178. Graph Valid Tree.ipynb	
+++ /dev/null
@@ -1,248 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 178. Graph Valid Tree\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Graph Valid Tree problem on LintCode, click here!](https://www.lintcode.com/problem/178/)\n",
-        "\n",
-        "---\n",
-        "Given `n` nodes labeled from `0` to `n - 1` and a list of `undirected` edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- You can assume that no duplicate edges will appear in edges. Since all edges are `undirected`, `[0, 1]` is the same as `[1, 0]` and thus will not appear together in edges."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether a given set of undirected edges can form a valid tree. In this context, a \"tree\" is a specific type of graph with the following properties:\n",
-        "\n",
-        "1. It is a connected graph, meaning there is a path between any pair of nodes in the graph.\n",
-        "2. It has no cycles, which means there are no closed loops or circuits in the graph.\n",
-        "3. It has exactly `n - 1` edges, where `n` is the number of nodes in the tree.\n",
-        "\n",
-        "The problem is typically given with two pieces of information:\n",
-        "\n",
-        "1. The number of nodes (`n`), which are usually labeled from 0 to `n-1`.\n",
-        "2. A list of undirected edges, where each edge is represented as a pair of nodes.\n",
-        "\n",
-        "The task is to write a function or algorithm to determine whether the provided set of edges can be arranged to form a valid tree based on the properties mentioned above.\n",
-        "\n",
-        "To solve this problem, you need to check the following conditions:\n",
-        "\n",
-        "1. There should be exactly `n - 1` edges to ensure that the graph is connected and tree-like.\n",
-        "2. There should be no cycles in the graph. If there are any cycles, the set of edges cannot form a tree.\n",
-        "3. The graph should be connected, which means you can reach any node from any other node through a series of edges.\n",
-        "\n",
-        "The solution typically involves using graph traversal algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the graph and check these conditions. If all conditions are met, the given edges form a valid tree; otherwise, they do not.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 10,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def validTree(n, edges):\n",
-        "    if len(edges) != n - 1:\n",
-        "        return False  # A tree with n nodes must have n-1 edges.\n",
-        "\n",
-        "    # Create an adjacency list to represent the graph.\n",
-        "    adj_list = {i: [] for i in range(n)}\n",
-        "    for u, v in edges:\n",
-        "        adj_list[u].append(v)\n",
-        "        adj_list[v].append(u)\n",
-        "\n",
-        "    visited = set()\n",
-        "\n",
-        "    # Define a recursive DFS function.\n",
-        "    def dfs(node, parent):\n",
-        "        if node in visited:\n",
-        "            return False  # If we encounter a visited node, there's a cycle.\n",
-        "        visited.add(node)\n",
-        "\n",
-        "        for neighbor in adj_list[node]:\n",
-        "            if neighbor != parent:\n",
-        "                if not dfs(neighbor, node):\n",
-        "                    return False\n",
-        "\n",
-        "        return True\n",
-        "\n",
-        "    # Start DFS from any node (e.g., node 0).\n",
-        "    if not dfs(0, -1):\n",
-        "        return False  # If there's a cycle, it's not a valid tree.\n",
-        "\n",
-        "    return len(visited) == n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The `validTree` function checks if a given set of undirected edges forms a valid tree. A valid tree must satisfy these conditions:\n",
-        "\n",
-        "1. It has exactly `n - 1` edges, where `n` is the number of nodes.\n",
-        "2. There are no cycles in the graph.\n",
-        "3. The graph is connected, meaning all nodes are reachable from any starting node.\n",
-        "\n",
-        "The code accomplishes this as follows:\n",
-        "\n",
-        "- It first checks if the number of edges is equal to `n - 1`. If not, it immediately returns `False` since a tree with `n` nodes must have `n - 1` edges to be connected.\n",
-        "\n",
-        "- It then constructs an adjacency list (`adj_list`) to represent the graph efficiently. This data structure stores the neighbors of each node.\n",
-        "\n",
-        "- The code uses a `visited` set to keep track of nodes visited during depth-first search (DFS).\n",
-        "\n",
-        "- The `dfs` function is a recursive function that checks for cycles in the graph. It does this by visiting nodes and marking them as visited. If it encounters a node that has already been visited (indicating a cycle), it returns `False`.\n",
-        "\n",
-        "- The DFS function explores the neighbors of each node, avoiding revisiting the parent node to prevent cycles.\n",
-        "\n",
-        "- After the DFS is complete, the code checks if the number of visited nodes is equal to `n`, ensuring the graph is connected.\n",
-        "\n",
-        "- Finally, if all conditions are met, the code returns `True`, indicating that the given edges form a valid tree. Otherwise, it returns `False`.\n",
-        "\n",
-        "The code provides examples to demonstrate how to use the `validTree` function to check whether a set of edges forms a valid tree or not."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 11,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "n1 = 5\n",
-        "edges1 = [[0, 1], [0, 2], [0, 3], [1, 4]]\n",
-        "print(validTree(n1, edges1))  # Output: true"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 12,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "n2 = 5\n",
-        "edges2 = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]\n",
-        "print(validTree(n2, edges2))  # Output: false"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexities of the `validTree` function:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "1. Constructing the adjacency list (`adj_list`) takes O(E) time, where E is the number of edges in the input graph (|edges|).\n",
-        "\n",
-        "2. The DFS function, when considering all nodes, has a time complexity of O(V + E), where V is the number of nodes in the graph. This is because, in the worst case, the DFS function visits each node and each edge once.\n",
-        "\n",
-        "3. Checking if the number of visited nodes is equal to `n` takes O(1) time.\n",
-        "\n",
-        "Overall, the time complexity of the `validTree` function is O(E + V) in the worst case.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "1. The adjacency list (`adj_list`) consumes additional space to store the graph structure, taking O(V + E) space. This is because it stores information about each node and its adjacent nodes.\n",
-        "\n",
-        "2. The `visited` set keeps track of visited nodes, and its space complexity is O(V) in the worst case, as there can be at most V unique nodes in the set.\n",
-        "\n",
-        "3. The depth of the function call stack during DFS can be at most V, which contributes O(V) space to the space complexity.\n",
-        "\n",
-        "Overall, the space complexity of the `validTree` function is O(V + E) due to the adjacency list and O(V) due to the `visited` set and DFS call stack.\n",
-        "\n",
-        "In summary, the time complexity is O(E + V), and the space complexity is O(V + E). These complexities are based on the worst-case scenario, where all nodes and edges are considered. In practical cases, the actual complexities may be smaller, depending on the specific structure of the graph."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Minimum Spanning Tree (MST)**: Implement an algorithm to find the minimum spanning tree of a connected graph using either Kruskal's or Prim's algorithm. Compare the MST to the original graph to determine if the original graph is a tree. This exercise reinforces the concept of tree properties.\n",
-        "\n",
-        "2. **Detecting Cycles**: Modify the code to not only check if the input forms a tree but also to identify and print out any cycles present in the graph if it's not a tree. This exercise enhances your ability to detect and visualize cycles in a graph."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb b/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb
deleted file mode 100644
index a8a7727..0000000
--- a/_build/html/_sources/11. Graphs/200. Number of Islands.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 200. Number of Islands\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Number of Islands problem on LeetCode, click here!](https://leetcode.com/problems/number-of-islands/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "Given an `m x n` 2D binary grid `grid` which represents a map of `'1'`s (land) and `'0'`s (water), return *the number of islands*. \n",
-        "\n",
-        "An **island** is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == grid.length`\n",
-        "- `n == grid[i].length`\n",
-        "- `1 <= m, n <= 300`\n",
-        "- `grid[i][j]` is `'0'` or `'1'`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is about counting the number of islands in a 2D binary grid. In this grid, '1' represents land, and '0' represents water. An island is defined as a group of '1' cells that are adjacent to each other either horizontally or vertically. It's important to note that diagonal connections do not count. Additionally, it's assumed that all four edges of the grid are surrounded by water, meaning the grid is finite and doesn't \"wrap around.\"\n",
-        "\n",
-        "The objective is to determine how many distinct islands exist in the grid. To clarify, an island consists of connected '1' cells, and these connections can only occur horizontally or vertically. If there are no '1' cells in the grid, the number of islands would be zero.\n",
-        "\n",
-        "The problem often involves using graph traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to identify and count these islands. You start at one '1' cell and explore its adjacent cells to determine the extent of the island. Once you've visited all the '1' cells in an island, you move on to the next unvisited '1' cell and repeat the process until you've counted all the islands in the grid.\n",
-        "\n",
-        "In essence, the problem asks you to analyze the grid to find groups of connected '1' cells and count them as individual islands. The goal is to return the total count of islands in the given grid.\n",
-        "\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def numIslands(self, grid: List[List[str]]) -> int:\n",
-        "        if not grid:\n",
-        "            return 0\n",
-        "\n",
-        "        m, n = len(grid), len(grid[0])  # Get the dimensions of the grid\n",
-        "        num_islands = 0  # Initialize the count of islands\n",
-        "\n",
-        "        def dfs(row, col):\n",
-        "            if row < 0 or row >= m or col < 0 or col >= n or grid[row][col] == '0':\n",
-        "                return\n",
-        "\n",
-        "            grid[row][col] = '0'  # Mark the current cell as visited by changing it to '0'\n",
-        "            directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Define four possible directions to move\n",
-        "\n",
-        "            for dr, dc in directions:\n",
-        "                # Explore adjacent cells\n",
-        "                dfs(row + dr, col + dc)\n",
-        "\n",
-        "        for i in range(m):\n",
-        "            for j in range(n):\n",
-        "                if grid[i][j] == '1':\n",
-        "                    num_islands += 1  # Increment the island count\n",
-        "                    dfs(i, j)  # Start a DFS from the current '1' cell to explore the entire island\n",
-        "\n",
-        "        return num_islands\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided code defines a Python class called `Solution` with a method called `numIslands`. This class is designed to count the number of islands in a 2D binary grid.\n",
-        "\n",
-        "The `numIslands` method takes a 2D grid as input, where '1' represents land, and '0' represents water. It uses a depth-first search (DFS) approach to identify and count islands. The key steps in the code include:\n",
-        "\n",
-        "1. Checking if the input grid is empty, and if so, returning 0 (indicating there are no islands).\n",
-        "\n",
-        "2. Determining the dimensions of the grid (number of rows and columns).\n",
-        "\n",
-        "3. Initializing a variable `num_islands` to keep track of the count of islands.\n",
-        "\n",
-        "4. Defining a nested function `dfs` to perform depth-first search. This function recursively explores adjacent land cells connected to the current cell and marks them as visited by changing '1' to '0'.\n",
-        "\n",
-        "5. In the `dfs` function, it checks for boundary conditions and whether the current cell is '0'. If these conditions are met, it returns without further exploration.\n",
-        "\n",
-        "6. The `dfs` function explores neighboring cells in all four directions (up, down, left, right) using a loop.\n",
-        "\n",
-        "7. The main loop iterates through each cell of the grid. When a '1' cell is encountered, it increments the `num_islands` count and starts a DFS from that cell to explore the entire island.\n",
-        "\n",
-        "8. Finally, the method returns the total count of islands found in the grid.\n",
-        "\n",
-        "This code encapsulates the solution in a class, allowing you to create an instance of the `Solution` class and call the `numIslands` method on it to count the number of islands in a given grid."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 1 Output: 1\n"
-          ]
-        }
-      ],
-      "source": [
-        "grid1 = [\n",
-        "  [\"1\", \"1\", \"1\", \"1\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"1\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"0\", \"0\", \"0\"]\n",
-        "]\n",
-        "\n",
-        "solution = Solution()\n",
-        "result1 = solution.numIslands(grid1)\n",
-        "print(\"Example 1 Output:\", result1)  # Expected output: 1\n"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 6,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 2 Output: 3\n"
-          ]
-        }
-      ],
-      "source": [
-        "grid2 = [\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"1\", \"0\", \"0\"],\n",
-        "  [\"0\", \"0\", \"0\", \"1\", \"1\"]\n",
-        "]\n",
-        "\n",
-        "result2 = solution.numIslands(grid2)\n",
-        "print(\"Example 2 Output:\", result2)  # Expected output: 3\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code:\n",
-        "\n",
-        "Time Complexity:\n",
-        "- The code uses a depth-first search (DFS) to explore the grid and count the number of islands.\n",
-        "- In the worst case, it visits every cell in the grid exactly once.\n",
-        "- The DFS function explores neighboring cells in four directions (up, down, left, right), so it has a time complexity of $O(4^{max(m, n)})$, where m is the number of rows and n is the number of columns.\n",
-        "- Therefore, the overall time complexity of the code is O(m * n), where m is the number of rows, and n is the number of columns.\n",
-        "\n",
-        "Space Complexity:\n",
-        "- The space complexity is determined by the recursive calls and the depth of the DFS stack.\n",
-        "- In the worst case, when the entire grid consists of '1's, the depth of the DFS recursion can be as deep as max(m, n).\n",
-        "- Therefore, the space complexity of the code is O(max(m, n)), representing the maximum depth of the DFS stack.\n",
-        "\n",
-        "In practical terms, for reasonably sized grids, the time and space complexity is linear, and the code should perform well. However, it's important to keep in mind that the worst-case time complexity is O(m * n), and the space complexity can be O(max(m, n))."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Identify Largest Island:** In addition to counting the number of islands, modify the code to identify the largest island in the grid and return its size.\n",
-        "\n",
-        "2. **Optimize for Space:** Modify the solution to reduce its space complexity to O(1) without modifying the input grid. This is a more memory-efficient version of the problem."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb b/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb
deleted file mode 100644
index 37624f6..0000000
--- a/_build/html/_sources/11. Graphs/207. Course Schedule.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 207. Course Schedule\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Course Schedule problem on LeetCode, click here!](https://leetcode.com/problems/course-schedule/)\n",
-        "\n",
-        "---\n",
-        "There are a total of `numCourses` courses to take, labeled from 0 to `numCourses - 1`. You are given an array `prerequisites` where `prerequisites[i] = [ai, bi]` indicates that you **must** take course `bi` first if you want to take course `ai`. \n",
-        "\n",
-        "- For example, the pair `[0, 1]`, indicates that to take course `0` you have to first take course `1`.\n",
-        "\n",
-        "Return `true` if you can finish all courses. Otherwise, return `false`.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- 1 <= `numCourses` <= 2000\n",
-        "- 0 <= `prerequisites.length` <= 5000\n",
-        "- `prerequisites[i].length` == 2\n",
-        "- 0 <= `ai`, `bi` < `numCourses`\n",
-        "- All the pairs `prerequisites[i]` are **unique**."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem is to determine whether it's possible to complete a set of courses given a list of prerequisites. You are given the following:\n",
-        "\n",
-        "- `numCourses`: The total number of courses to be taken, labeled from 0 to `numCourses - 1`.\n",
-        "- `prerequisites`: A list of prerequisite courses, where each element `prerequisites[i]` is a pair `[ai, bi]`, indicating that you must take course `bi` before you can take course `ai`.\n",
-        "\n",
-        "The goal is to check whether it's possible to complete all the courses while respecting the prerequisite requirements. In other words, you need to determine if there are any circular dependencies in the course prerequisites that would prevent you from taking all the courses.\n",
-        "\n",
-        "For example:\n",
-        "\n",
-        "- If `numCourses` is 2, and `prerequisites` is `[[1, 0]]`, it means you have two courses, and you must finish course 0 before you can take course 1. In this case, it's possible to complete all courses, so the function should return `True`.\n",
-        "\n",
-        "- If `numCourses` is 2, and `prerequisites` is `[[1, 0], [0, 1]]`, it means you have two courses, but there's a circular dependency where course 1 requires course 0 and course 0 requires course 1. In this case, it's impossible to complete all courses, so the function should return `False`.\n",
-        "\n",
-        "The problem is essentially about checking for the presence of cycles in a directed graph, where the courses are nodes, and the prerequisites represent directed edges between them. If there are no cycles, it's possible to complete all courses; otherwise, it's not possible.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 18,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict, deque\n",
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:\n",
-        "        # Create a graph using an adjacency list to represent the courses and their prerequisites.\n",
-        "        graph = defaultdict(list)  # Initialize an empty adjacency list.\n",
-        "        in_degree = [0] * numCourses  # Initialize an array to store in-degrees of courses.\n",
-        "\n",
-        "        # Populate the graph and calculate in-degrees.\n",
-        "        for course, prerequisite in prerequisites:\n",
-        "            graph[prerequisite].append(course)  # Add the course as a neighbor to its prerequisite.\n",
-        "            in_degree[course] += 1  # Increment the in-degree of the course.\n",
-        "\n",
-        "        # Initialize a queue with courses that have no prerequisites.\n",
-        "        queue = deque()\n",
-        "        for course in range(numCourses):\n",
-        "            if in_degree[course] == 0:\n",
-        "                queue.append(course)\n",
-        "\n",
-        "        # Perform topological sorting.\n",
-        "        while queue:\n",
-        "            course = queue.popleft()  # Take a course with no prerequisites.\n",
-        "            numCourses -= 1  # Decrement the count of remaining courses.\n",
-        "\n",
-        "            for neighbor in graph[course]:\n",
-        "                in_degree[neighbor] -= 1  # Remove the prerequisite relationship.\n",
-        "                if in_degree[neighbor] == 0:\n",
-        "                    queue.append(neighbor)  # If no more prerequisites, add to the queue.\n",
-        "\n",
-        "        # If all courses were successfully taken (numCourses becomes 0), return True.\n",
-        "        return numCourses == 0"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "This Python code is designed to determine whether it is possible to complete all the required courses given a set of prerequisites. It uses a topological sorting algorithm to accomplish this. Here's a brief explanation of the code:\n",
-        "\n",
-        "1. The code defines a `Solution` class with a `canFinish` method that takes two parameters: `numCourses` (the total number of courses) and `prerequisites` (a list of pairs indicating which course must be taken before another).\n",
-        "\n",
-        "2. It initializes an empty graph as an adjacency list using a `defaultdict` to represent the courses and their prerequisites. It also initializes a list called `in_degree` to keep track of the in-degrees of courses, which is used in the topological sorting process.\n",
-        "\n",
-        "3. The code then populates the graph by iterating through the `prerequisites` list. For each pair `(ai, bi)`, it adds course `ai` as a neighbor to its prerequisite course `bi` in the graph and increments the in-degree of course `ai`.\n",
-        "\n",
-        "4. Next, it initializes a queue with courses that have no prerequisites. It does this by iterating through all the courses and adding those with an in-degree of 0 to the queue.\n",
-        "\n",
-        "5. The code performs a topological sorting of the courses using a while loop. In each iteration of the loop, it dequeues a course from the front of the queue (a course with no prerequisites).\n",
-        "\n",
-        "6. For each neighbor (course that depends on the dequeued course), it decrements the in-degree of the neighbor. If the neighbor's in-degree becomes 0, it means all of its prerequisites have been taken, so the neighbor is added to the queue.\n",
-        "\n",
-        "7. The loop continues until there are no more courses with no prerequisites to dequeue, and the in-degrees are updated accordingly.\n",
-        "\n",
-        "8. After the loop, if all courses were successfully taken (i.e., `numCourses` becomes 0), it means that there were no circular dependencies, and it's possible to finish all courses. In this case, the method returns `True`. Otherwise, if there are remaining courses (indicating a circular dependency), it returns `False`."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 20,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "True\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "numCourses1 = 2\n",
-        "prerequisites1 = [[1, 0]]\n",
-        "sol = Solution()\n",
-        "print(sol.canFinish(numCourses1, prerequisites1))  # Output: True"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 21,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "False\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "numCourses2 = 2\n",
-        "prerequisites2 = [[1, 0], [0, 1]]\n",
-        "print(sol.canFinish(numCourses2, prerequisites2))  # Output: False"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the `canFinish` method:\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "\n",
-        "1. Constructing the graph: In the first loop where we iterate through the `prerequisites` list, we populate the graph with prerequisites. This loop has a time complexity of O(E), where E is the number of prerequisites (edges).\n",
-        "\n",
-        "2. Initializing in-degrees and finding courses with no prerequisites: In the worst case, we iterate through all `numCourses` courses to initialize in-degrees and find the courses with no prerequisites. This has a time complexity of O(V), where V is the number of courses (vertices).\n",
-        "\n",
-        "3. Topological Sorting: In the while loop, we perform topological sorting, processing each course once and its outgoing edges. In the worst case, each course is processed once. This has a time complexity of O(V + E).\n",
-        "\n",
-        "The overall time complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "\n",
-        "1. Graph Representation: The space complexity for storing the graph as an adjacency list is O(E), where E is the number of prerequisites.\n",
-        "\n",
-        "2. In-Degree Array: The space complexity for storing the in-degrees of courses is O(V), where V is the number of courses.\n",
-        "\n",
-        "3. Queue: The space complexity for the queue is O(V), as it may contain all the courses.\n",
-        "\n",
-        "The overall space complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-        "\n",
-        "In most practical cases, the number of prerequisites (E) is much smaller than the number of courses (V), so the time and space complexity can often be considered as O(V)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimizing Course Scheduling**: Given a list of courses with their durations and prerequisites, find the most efficient way to schedule these courses to minimize the time it takes to complete all of them.\n",
-        "\n",
-        "2. **Detecting Cycles in Course Dependencies**: Modify the original problem to not just determine if you can finish all courses but also to identify the specific courses that form a cycle in the prerequisite dependencies."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb b/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb
deleted file mode 100644
index fb186f2..0000000
--- a/_build/html/_sources/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb	
+++ /dev/null
@@ -1,236 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 3651. Number of Connected Components in an Undirected Graph\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Number of Connected Components in an Undirected Graph problem on LintCode, click here!](https://www.lintcode.com/problem/3651/)\n",
-        "\n",
-        "---\n",
-        "In this problem, there is an undirected graph with `n` nodes. There is also an `edges` array. Where `edges[i] = [a, b]` means that there is an edge between *node a* and *node b* in the graph.\n",
-        "\n",
-        "You need to return *the number of connected components in that graph*."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand is to determine the number of connected components in an undirected graph. Here's a more detailed explanation of the problem:\n",
-        "\n",
-        "- You are given an undirected graph, which consists of nodes and edges. The graph may have multiple distinct connected components.\n",
-        "\n",
-        "- A connected component is a group of nodes within the graph where each node can be reached from every other node in the same component. In other words, there is a path between any two nodes in the same connected component, but there are no paths connecting nodes from different components.\n",
-        "\n",
-        "- Your task is to write a function that takes as input the number of nodes `n` and a list of edges `edges`, where each edge is represented as a pair of nodes `[a, b]` indicating an edge between node `a` and node `b`.\n",
-        "\n",
-        "- The goal is to determine the number of distinct connected components in the graph.\n",
-        "\n",
-        "- You need to implement an algorithm that can identify these connected components and count how many there are in the given graph.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 13,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "def countConnectedComponents(n, edges):\n",
-        "    # Define a helper function for depth-first search (DFS).\n",
-        "    def dfs(node):\n",
-        "        if not visited[node]:  # If the node has not been visited yet:\n",
-        "            visited[node] = True  # Mark it as visited.\n",
-        "            for neighbor in graph[node]:  # Explore all neighbors of the current node.\n",
-        "                dfs(neighbor)  # Recursively call DFS on the neighbor.\n",
-        "\n",
-        "    # Create an empty adjacency list to represent the graph.\n",
-        "    graph = [[] for _ in range(n)]\n",
-        "\n",
-        "    # Initialize a boolean list to keep track of visited nodes.\n",
-        "    visited = [False] * n\n",
-        "\n",
-        "    # Populate the adjacency list with edges from the input.\n",
-        "    for a, b in edges:\n",
-        "        graph[a].append(b)\n",
-        "        graph[b].append(a)  # Since the graph is undirected, we add edges in both directions.\n",
-        "\n",
-        "    # Initialize a counter to keep track of the number of connected components.\n",
-        "    count = 0\n",
-        "\n",
-        "    # Iterate through all nodes in the graph.\n",
-        "    for i in range(n):\n",
-        "        if not visited[i]:  # If the node has not been visited:\n",
-        "            count += 1  # Increment the count since we've found a new connected component.\n",
-        "            dfs(i)  # Start a depth-first search from the unvisited node to explore the component.\n",
-        "\n",
-        "    return count"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The provided Python code calculates the number of connected components in an undirected graph. Here's a high-level explanation of how the code works:\n",
-        "\n",
-        "1. The `countConnectedComponents` function takes two arguments: `n` (the number of nodes in the graph) and `edges` (a list of edge pairs representing connections between nodes).\n",
-        "\n",
-        "2. Inside the function, there is a helper function called `dfs` (depth-first search), which is used to explore and mark nodes as visited.\n",
-        "\n",
-        "3. An empty adjacency list called `graph` is created to represent the graph. The `graph` is a list of lists where each element represents a node, and the list associated with each node contains its neighboring nodes.\n",
-        "\n",
-        "4. Another list called `visited` is created to keep track of visited nodes. Initially, all nodes are marked as not visited (`False`).\n",
-        "\n",
-        "5. The code populates the `graph` by iterating through the `edges` list and adding each edge to the adjacency list. Since the graph is undirected, both directions of each edge are added.\n",
-        "\n",
-        "6. A counter variable called `count` is initialized to keep track of the number of connected components.\n",
-        "\n",
-        "7. The code iterates through all nodes from `0` to `n-1`. For each node, it checks whether it has been visited yet. If it hasn't been visited, it means a new connected component is found. The counter `count` is incremented, and a depth-first search (DFS) is initiated from this unvisited node to explore the connected component.\n",
-        "\n",
-        "8. The `dfs` function recursively visits all nodes in the connected component, marking them as visited. It does this by checking each neighbor of the current node and recursively calling `dfs` on unvisited neighbors.\n",
-        "\n",
-        "9. After the loop through all nodes, the `count` variable contains the total number of connected components in the graph.\n",
-        "\n",
-        "10. The function returns the value of `count`, which represents the number of connected components.\n",
-        "\n",
-        "11. Two examples are provided at the end to demonstrate how to use the function with different inputs."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 14,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "1\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "n1 = 3\n",
-        "edges1 = [[0, 1], [0, 2]]\n",
-        "print(countConnectedComponents(n1, edges1))  # Output: 1"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 15,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "n2 = 6\n",
-        "edges2 = [[0, 1], [1, 2], [2, 3], [4, 5]]\n",
-        "print(countConnectedComponents(n2, edges2))  # Output: 2"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the code for counting the number of connected components in an undirected graph using depth-first search (DFS).\n",
-        "\n",
-        "**Time Complexity:**\n",
-        "- The code iterates through all nodes in the graph (from 0 to `n-1`).\n",
-        "- For each unvisited node, it initiates a depth-first search (DFS) from that node to explore the connected component.\n",
-        "- In the worst case, every node may belong to a separate connected component, and you perform a DFS for each node.\n",
-        "- The time complexity of the DFS for a single connected component is O(V + E), where V is the number of nodes in the component, and E is the number of edges.\n",
-        "- In the worst case, you perform O(n) DFS calls, each with its own O(V + E) complexity.\n",
-        "\n",
-        "So, the overall time complexity of the code is O(n * (V + E)). In the worst case, if the graph is connected (a single connected component), this simplifies to O(n * (n + m)), where m is the number of edges.\n",
-        "\n",
-        "**Space Complexity:**\n",
-        "- The main data structures used are the `graph` (adjacency list), `visited` (boolean array), and the recursive call stack for DFS.\n",
-        "- The `graph` has a space complexity of O(n + m), where n is the number of nodes, and m is the number of edges.\n",
-        "- The `visited` boolean array also has a space complexity of O(n) to keep track of whether nodes have been visited.\n",
-        "- The depth of the recursive call stack for DFS can go up to O(n) in the worst case, when all nodes are part of a single connected component.\n",
-        "\n",
-        "So, the overall space complexity of the code is O(n + m) for data structures and O(n) for the call stack. In the worst case, this simplifies to O(n + m).\n",
-        "\n",
-        "In summary, the time complexity is O(n * (V + E)), and the space complexity is O(n + m)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Disconnected Components with Multiple Queries:** You are given an undirected graph with `n` nodes and `m` edges. Initially, the graph is disconnected, and you need to process `q` queries. Each query consists of adding an edge to connect two nodes. After each query, you need to output the number of connected components in the updated graph.\n",
-        "\n",
-        "2. **Maximum Size of Connected Components:** Given an undirected graph with `n` nodes and `m` edges, find the size of the largest connected component in the graph. Additionally, identify the nodes that belong to this largest component. You can assume there is at least one connected component in the graph.\n"
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb b/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb
deleted file mode 100644
index eb34497..0000000
--- a/_build/html/_sources/11. Graphs/417. Pacific Atlantic Water Flow.ipynb	
+++ /dev/null
@@ -1,261 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 417. Pacific Atlantic Water Flow\n",
-        "\n",
-        "\n",
-        "**Difficulty:** Medium\n",
-        "\n",
-        "**Link to Problem:** [To see the Pacific Atlantic Water Flow problem on LeetCode, click here!](https://leetcode.com/problems/pacific-atlantic-water-flow/)\n",
-        "\n",
-        "---\n",
-        "There is an `m x n` rectangular island that borders both the **Pacific Ocean** and **Atlantic Ocean**. The **Pacific Ocea**n touches the island's left and top edges, and the **Atlantic Ocean** touches the island's right and bottom edges.\n",
-        "\n",
-        "The island is partitioned into a grid of square cells. You are given an `m x n` integer matrix `heights` where `heights[r][c]` represents the **height above sea level** of the cell at coordinate `(r, c)`.\n",
-        "\n",
-        "The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is **less than or equal to** the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.\n",
-        "\n",
-        "Return *a **2D list** of grid coordinates `result` where $result[i] = [r_i, c_i]$ denotes that rain water can flow from cell $(r_i, c_i)$ to **both** the Pacific and Atlantic oceans*.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `m == heights.length`\n",
-        "- `n == heights[r].length`\n",
-        "- `1 <= m, n <= 200`\n",
-        "- 0 <= `heights[r][c]` <= $10^5$"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand involves a scenario with a rectangular island surrounded by both the Pacific Ocean and the Atlantic Ocean. The Pacific Ocean is in contact with the island's left and top edges, while the Atlantic Ocean is in contact with the island's right and bottom edges. The goal is to determine which cells on the island can allow rainwater to flow to both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "Here are the key details and constraints of the problem:\n",
-        "\n",
-        "1. The island is divided into a grid of square cells.\n",
-        "\n",
-        "2. Each cell is represented by an integer value in a matrix `heights`. The value at `heights[r][c]` represents the height above sea level of the cell at coordinates `(r, c)` on the island.\n",
-        "\n",
-        "3. Rainwater can flow from a cell to neighboring cells directly north, south, east, or west only if the neighboring cell's height is less than or equal to the current cell's height.\n",
-        "\n",
-        "4. Water can flow from any cell adjacent to the ocean into the ocean. This means water can flow from cells on the edge of the island to the Pacific Ocean and the Atlantic Ocean.\n",
-        "\n",
-        "The task is to find and return a 2D list of grid coordinates where rainwater can flow from a cell to both the Pacific and Atlantic Oceans. In other words, you need to identify the cells that can send water to both the left and top edges (Pacific) and the right and bottom edges (Atlantic) of the island.\n",
-        "\n",
-        "The problem is solved by performing a depth-first search (DFS) starting from the ocean edges and marking the cells that can flow to each ocean. Then, the algorithm identifies cells that are reachable from both oceans and returns their coordinates as the result.\n",
-        "\n",
-        "The problem involves both traversal of the island and backtracking to find the solution, which makes it a classic graph traversal problem.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 24,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from typing import List\n",
-        "\n",
-        "class Solution:\n",
-        "    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:\n",
-        "        # Check if the input matrix is empty\n",
-        "        if not heights:\n",
-        "            return []\n",
-        "        \n",
-        "        m, n = len(heights), len(heights[0])\n",
-        "        \n",
-        "        # Initialize boolean matrices to keep track of cells reachable from each ocean\n",
-        "        pacific_reachable = [[False] * n for _ in range(m)]\n",
-        "        atlantic_reachable = [[False] * n for _ in range(m)]\n",
-        "        \n",
-        "        # Depth-First Search (DFS) function to mark cells that can be reached from an ocean\n",
-        "        def dfs(x, y, reachable):\n",
-        "            if reachable[x][y]:\n",
-        "                return\n",
-        "            reachable[x][y] = True\n",
-        "            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:\n",
-        "                new_x, new_y = x + dx, y + dy\n",
-        "                if 0 <= new_x < m and 0 <= new_y < n and heights[new_x][new_y] >= heights[x][y]:\n",
-        "                    dfs(new_x, new_y, reachable)\n",
-        "        \n",
-        "        # Start DFS from the Pacific Ocean (left and top edges)\n",
-        "        for i in range(m):\n",
-        "            dfs(i, 0, pacific_reachable)\n",
-        "            dfs(i, n - 1, atlantic_reachable)\n",
-        "        \n",
-        "        for j in range(n):\n",
-        "            dfs(0, j, pacific_reachable)\n",
-        "            dfs(m - 1, j, atlantic_reachable)\n",
-        "        \n",
-        "        result = []\n",
-        "        \n",
-        "        # Find cells that can flow to both the Pacific and Atlantic oceans\n",
-        "        for i in range(m):\n",
-        "            for j in range(n):\n",
-        "                if pacific_reachable[i][j] and atlantic_reachable[i][j]:\n",
-        "                    result.append([i, j])\n",
-        "        \n",
-        "        return result"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a class called `Solution` with a method named `pacificAtlantic`. This method takes a 2D matrix `heights`, where each element represents the height above sea level of a cell on a rectangular island. The goal of the method is to find and return a list of grid coordinates (cells) from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "The code uses a Depth-First Search (DFS) algorithm to traverse the island, starting from the ocean edges, and marks cells that are reachable from either the Pacific or Atlantic Ocean. It does this by creating two boolean matrices, `pacific_reachable` and `atlantic_reachable`, to keep track of cells that can be reached from each ocean.\n",
-        "\n",
-        "The code iterates through the entire island, initiating DFS searches from the edges of the island. If a cell is reachable from an ocean (either Pacific or Atlantic), it is marked as reachable in the corresponding boolean matrix. The DFS function ensures that the traversal follows the rule that water can flow to a neighboring cell if the neighboring cell's height is less than or equal to the current cell's height.\n",
-        "\n",
-        "After marking all reachable cells from both oceans, the code then looks for cells that are reachable from both the Pacific and Atlantic Oceans. If a cell satisfies this condition, it is added to the `result` list.\n",
-        "\n",
-        "Finally, the code returns the `result` list, which contains the grid coordinates of cells from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-        "\n",
-        "The code provides a solution to the problem of finding cells on the island where water can reach both oceans, demonstrating a traversal algorithm that explores the connectivity of cells on the island."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 25,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "heights1 = [\n",
-        "    [1,2,2,3,5],\n",
-        "    [3,2,3,4,4],\n",
-        "    [2,4,5,3,1],\n",
-        "    [6,7,1,4,5],\n",
-        "    [5,1,1,2,4]\n",
-        "]\n",
-        "\n",
-        "solution = Solution()\n",
-        "print(solution.pacificAtlantic(heights1)) # Expected Output: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 26,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "[[0, 0]]\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "heights2 = [[1]]\n",
-        "print(solution.pacificAtlantic(heights2)) # Expected Output: [[0, 0]]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Time Complexity:\n",
-        "The time complexity of this solution is O(m * n), where 'm' is the number of rows in the input matrix and 'n' is the number of columns. Here's a breakdown of the time complexity:\n",
-        "\n",
-        "1. Creating the `pacific_reachable` and `atlantic_reachable` matrices with dimensions m x n takes O(m * n) time.\n",
-        "\n",
-        "2. Initiating Depth-First Search (DFS) from the edges of the island takes O(m * n) time because it processes each cell once.\n",
-        "\n",
-        "3. The DFS function visits each cell at most once, and its time complexity is O(1) for each cell. The number of cells visited by the DFS in the worst case is m * n.\n",
-        "\n",
-        "4. The final step of identifying and appending cells to the result list takes O(m * n) time in the worst case.\n",
-        "\n",
-        "Overall, the time complexity is dominated by the DFS and is O(m * n).\n",
-        "\n",
-        "Space Complexity:\n",
-        "The space complexity of this solution is also O(m * n). Here's how the space is used:\n",
-        "\n",
-        "1. Two boolean matrices, `pacific_reachable` and `atlantic_reachable`, are created with dimensions m x n. Each matrix takes O(m * n) space, resulting in O(2 * m * n) space usage.\n",
-        "\n",
-        "2. The depth-first search (DFS) stack space during the recursive calls can go as deep as the diagonal of the grid, which is at most min(m, n). This additional space for the call stack is relatively small compared to the boolean matrices and can be considered O(min(m, n)).\n",
-        "\n",
-        "3. The `result` list stores the grid coordinates, and its size is determined by the number of cells that can flow to both oceans. In the worst case, this list can contain all m * n cells, so it takes O(m * n) space.\n",
-        "\n",
-        "The overall space complexity is the sum of these components, which is O(m * n) + O(min(m, n)) + O(m * n). In big O notation, we can simplify this to O(m * n)."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Optimization Challenge:** Modify the algorithm to find the maximum flow of water from any cell to both the Pacific and Atlantic Oceans. In other words, find the cell(s) that can send the most water to both oceans.\n",
-        "\n",
-        "2. **Efficiency Challenge:** Optimize the time and space complexity of the algorithm while maintaining correctness. Aim to reduce both time and space complexity as much as possible."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/11. Graphs/README.md b/_build/html/_sources/11. Graphs/README.md
deleted file mode 100644
index aad262c..0000000
--- a/_build/html/_sources/11. Graphs/README.md	
+++ /dev/null
@@ -1,23 +0,0 @@
-# Graphs Problems - Blind 75 LeetCode
-
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)                           | Medium     |
-| [133. Clone Graph](https://leetcode.com/problems/clone-graph/)                                       | Medium     |
-| [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)       | Medium     |
-| [207. Course Schedule](https://leetcode.com/problems/course-schedule/)                                 | Medium     |
-| [323. Number of Connected Components In An Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/) | Medium |
-| [261. Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)                               | Medium     |
-
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb b/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb
deleted file mode 100644
index 81f0e5b..0000000
--- a/_build/html/_sources/12. Advanced Graphs/892. Alien Dictionary.ipynb	
+++ /dev/null
@@ -1,256 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {
-        "jp-MarkdownHeadingCollapsed": true
-      },
-      "source": [
-        "# 892. Alien Dictionary\n",
-        "\n",
-        "**Difficulty:** Hard\n",
-        "\n",
-        "**Link to Problem:** [To see the Alien Dictionary problem on LintCode, click here!](https://www.lintcode.com/problem/892/)\n",
-        "\n",
-        "---\n",
-        "\n",
-        "There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of **non-empty** words from the dictionary, where words are **sorted lexicographically by the rules of this new language**. Derive the order of letters in this language.\n",
-        "\n",
-        "**Constraints:**\n",
-        "- You may assume all letters are in lowercase.\n",
-        "- The dictionary is invalid, if string a is prefix of string b and b is appear before a.\n",
-        "- If the order is invalid, return an empty string.\n",
-        "- There may be multiple valid order of letters, return the smallest in normal lexicographical order.\n",
-        "- The letters in **one** string are of the same rank by default and are sorted in Human dictionary order."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The problem at hand involves determining the order of letters in a new alien language that uses the Latin alphabet. However, the order of letters in this alien language is unknown. To solve this problem, we are given a list of non-empty words from the dictionary, where the words are sorted lexicographically based on the rules of the alien language. The goal is to derive the correct order of letters in this new language.\n",
-        "\n",
-        "Here's a more detailed explanation of the problem:\n",
-        "\n",
-        "1. The alphabet: The new alien language uses the Latin alphabet, which consists of lowercase letters.\n",
-        "\n",
-        "2. Dictionary: We are provided with a list of non-empty words. These words are sorted in lexicographical order based on the rules of the alien language. The order of letters in the words reflects the correct order in the new language.\n",
-        "\n",
-        "3. Invalid order: The problem specifies that the dictionary is considered invalid if there exists a situation where one string 'a' is a prefix of another string 'b,' and 'b' appears before 'a' in the dictionary. This condition ensures that there are no inconsistencies in the order of letters.\n",
-        "\n",
-        "4. Multiple valid orders: It's possible that there are multiple valid orders of letters that satisfy the provided dictionary. However, we are instructed to return the smallest valid order in normal lexicographical order.\n",
-        "\n",
-        "To solve this problem, you need to analyze the given list of words and determine the correct order of letters in the alien language while adhering to the specified constraints. You should return the order as a string in lexicographical order if it's valid, and an empty string if the order is invalid.\n",
-        "\n",
-        "The key to solving this problem is to construct a directed graph of letter relationships based on the order information provided by the words in the dictionary. Then, perform a topological sort on this graph to derive the order of letters, ensuring that the order is consistent and valid according to the given constraints.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 7,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "from collections import defaultdict, deque\n",
-        "\n",
-        "def alienOrder(words):\n",
-        "    # Create an adjacency list to represent the graph\n",
-        "    graph = defaultdict(list)\n",
-        "    \n",
-        "    # Create a dictionary to store the in-degrees of each letter\n",
-        "    in_degree = {}\n",
-        "    \n",
-        "    # Initialize in-degrees to 0 for all letters\n",
-        "    for word in words:\n",
-        "        for char in word:\n",
-        "            if char not in in_degree:\n",
-        "                in_degree[char] = 0\n",
-        "    \n",
-        "    # Build the graph and update in-degrees\n",
-        "    for i in range(1, len(words)):\n",
-        "        word1, word2 = words[i-1], words[i]\n",
-        "        min_length = min(len(word1), len(word2))\n",
-        "        \n",
-        "        # Compare characters in the two words\n",
-        "        j = 0\n",
-        "        while j < min_length and word1[j] == word2[j]:\n",
-        "            j += 1\n",
-        "        \n",
-        "        if j < min_length:\n",
-        "            # If word1[j] is lexicographically before word2[j], add an edge\n",
-        "            graph[word1[j]].append(word2[j])\n",
-        "            in_degree[word2[j]] += 1\n",
-        "            \n",
-        "            # Break the loop to avoid further comparisons\n",
-        "            break\n",
-        "    \n",
-        "    # Perform topological sorting using Kahn's algorithm\n",
-        "    result = []\n",
-        "    queue = deque([char for char in in_degree if in_degree[char] == 0])\n",
-        "    \n",
-        "    while queue:\n",
-        "        char = queue.popleft()\n",
-        "        result.append(char)\n",
-        "        \n",
-        "        for neighbor in graph[char]:\n",
-        "            in_degree[neighbor] -= 1\n",
-        "            if in_degree[neighbor] == 0:\n",
-        "                queue.append(neighbor)\n",
-        "    \n",
-        "    # Check if there is a valid order\n",
-        "    if len(result) < len(in_degree):\n",
-        "        return \"\"\n",
-        "    \n",
-        "    return \"\".join(result)"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code is designed to determine the order of letters in an unknown alien language that uses the Latin alphabet. It takes a list of non-empty words from a dictionary as input, where the words are sorted lexicographically based on the rules of this alien language.\n",
-        "\n",
-        "The code follows these main steps:\n",
-        "\n",
-        "1. Create a data structure to represent the graph of letters and their relationships in the alien language. It uses a defaultdict to store a list of letters that come after each letter. Additionally, it uses a dictionary called `in_degree` to keep track of the in-degrees (the number of letters that precede a letter) for each letter.\n",
-        "\n",
-        "2. Initialize the in-degrees of all letters to 0. This is done by iterating through all the words and characters in the words, adding each character to the `in_degree` dictionary with an initial in-degree of 0.\n",
-        "\n",
-        "3. Build the graph by comparing adjacent words in the dictionary. For each pair of adjacent words, the code finds the first differing character between the two words. If such a character exists, it means there's an order relationship between the letters represented by these characters. The code updates the graph and in-degrees accordingly.\n",
-        "\n",
-        "4. After constructing the graph, the code performs a topological sorting of the letters using Kahn's algorithm. It starts by initializing an empty result list and a queue containing letters with in-degrees of 0. The algorithm repeatedly removes a letter with an in-degree of 0 from the queue, adds it to the result list, and updates the in-degrees of its neighbors. This process continues until the queue is empty.\n",
-        "\n",
-        "5. Finally, the code checks if the topological sorting was successful by comparing the length of the result list with the number of unique letters in the input. If the lengths differ, it means there's a cycle in the graph, and the order is invalid. In such a case, an empty string is returned. Otherwise, the result list is joined together to form the smallest order of letters in normal lexicographical order, and it is returned as the output.\n",
-        "\n",
-        "The code provides a way to determine the order of letters in the alien language while handling various edge cases, including checking for invalid order conditions."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 8,
-      "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "wrtef\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1\n",
-        "words1 = [\"wrt\", \"wrf\", \"er\", \"ett\", \"rftt\"]\n",
-        "print(alienOrder(words1))  # Output: \"wertf\""
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 9,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "zx\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2\n",
-        "words2 = [\"z\", \"x\"]\n",
-        "print(alienOrder(words2))  # Output: \"zx\""
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Let's analyze the time and space complexity of the provided code for determining the order of letters in an alien language.\n",
-        "\n",
-        "Time Complexity:\n",
-        "1. Building the graph: The code iterates through the list of words and compares adjacent words. In the worst case, where all words have the same prefix, it takes O(N) time to build the graph, where N is the total number of characters in all the words.\n",
-        "2. Topological sorting: Performing a topological sort on the graph takes O(V + E) time, where V is the number of unique letters (vertices) and E is the number of relationships between letters (edges) in the graph.\n",
-        "\n",
-        "Overall, the time complexity is O(N + V + E), where N is the total number of characters in the words, V is the number of unique letters, and E is the number of order relationships between letters.\n",
-        "\n",
-        "Space Complexity:\n",
-        "1. Graph and In-degrees: The code uses data structures to represent the graph and in-degrees. The space complexity for these data structures is O(V + E), where V is the number of unique letters, and E is the number of order relationships between letters in the words.\n",
-        "2. Result List: The result list stores the order of letters, which can have a maximum size of V, where V is the number of unique letters.\n",
-        "3. Queue: The space used by the queue during topological sorting is also O(V).\n",
-        "\n",
-        "Overall, the space complexity is O(V + E) due to the data structures used for graph representation and the space needed for storing the result and the queue.\n",
-        "\n",
-        "In the worst case, where there are many unique letters and many order relationships, the space complexity is dominated by the number of unique letters and their relationships.\n",
-        "\n",
-        "In summary, the time complexity of the code is O(N + V + E), and the space complexity is O(V + E). The actual performance will depend on the specific input data, such as the number of unique letters and the relationships between them in the alien language."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Multiple Valid Orders**: Extend the problem to handle cases where there are multiple valid orders of letters in the alien language. Modify the code to return all valid orders instead of just one. Be mindful of performance.\n",
-        "\n",
-        "2. **Detect Invalid Dictionary**: Given a dictionary of words, write a function to detect whether the dictionary is valid or not based on the constraints mentioned in the problem statement."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/12. Advanced Graphs/README.md b/_build/html/_sources/12. Advanced Graphs/README.md
deleted file mode 100644
index f362de7..0000000
--- a/_build/html/_sources/12. Advanced Graphs/README.md	
+++ /dev/null
@@ -1,16 +0,0 @@
-# Advanced Graphs Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [269. Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)                               | Hard       |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb b/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb
deleted file mode 100644
index 8d32e21..0000000
--- a/_build/html/_sources/13. One-D Dynamic Programming/198. House Robber.ipynb	
+++ /dev/null
@@ -1,103 +0,0 @@
-{
-  "metadata": {
-    "kernelspec": {
-      "name": "python",
-      "display_name": "Python (Pyodide)",
-      "language": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat_minor": 5,
-  "nbformat": 4,
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "source": "# House Robber\n**Difficulty:** Medium\n\n---\nYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and **it will automatically contact the police if two adjacent houses were broken into on the same night.**\n\nGiven an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight **without alerting the police**.*\n\n**Constraints:**\n- `1 <= nums.length <= 100`\n- `0 <= nums[i] <= 400`",
-      "metadata": {},
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Probelm Explanation:\n\nThe problem is a classic dynamic programming challenge known as the \"House Robber\" problem. The scenario is that you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, but the constraint is that adjacent houses have security systems connected. If two adjacent houses are broken into on the same night, the security systems will automatically contact the police.\n\nThe goal is to determine the maximum amount of money you can rob tonight without alerting the police. You need to find a strategy to maximize your earnings while avoiding robbing two adjacent houses.\n\nThe input to the problem is an array `nums`, where `nums[i]` represents the amount of money in the `i`-th house. The task is to return the maximum amount of money that can be robbed without triggering the security system.\n\nFor example:\n```\nInput: nums = [1,2,3,1]\nOutput: 4\nExplanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).\nTotal amount you can rob = 1 + 3 = 4.\n\nInput: nums = [2,7,9,3,1]\nOutput: 12\nExplanation: Rob house 1 (money = 2), rob house 3 (money = 9), and rob house 5 (money = 1).\nTotal amount you can rob = 2 + 9 + 1 = 12.\n```\n\nThe challenge involves designing an algorithm to efficiently solve this problem and find the optimal strategy for robbing houses to maximize the stolen money while adhering to the constraint of not robbing adjacent houses. The provided solution utilizes dynamic programming to achieve this.\n\n## Solution:\nHere's a Python function to implement this algorithm:",
-      "metadata": {},
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76"
-    },
-    {
-      "cell_type": "code",
-      "source": "from typing import List\n\nclass Solution:\n    def rob(self, nums: List[int]) -> int:\n        # Initialize variables to keep track of the maximum amount of money robbed at the previous and current houses\n        prev, curr = 0, 0\n\n        # Iterate through the houses\n        for n in nums:\n            # Calculate the maximum amount of money that can be robbed at the current house\n            # It is the maximum of either robbing the current house and the money robbed at the house before the previous one (prev),\n            # or skipping the current house and taking the maximum amount from the previous house (curr)\n            temp = max(n + prev, curr)\n\n            # Update the variables for the next iteration\n            prev, curr = curr, temp\n\n        # The final result is the maximum amount of money that can be robbed at the last house\n        return curr",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 4,
-      "outputs": [],
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Explanation:\n\nThe provided code is a Python implementation of the \"House Robber\" problem using a dynamic programming approach. It's part of a class `Solution` and includes a method `rob` that takes a list of integers `nums` as input and returns the maximum amount of money that can be robbed without alerting the police.\n\nThe key idea of the algorithm is to maintain two variables, `prev` and `curr`, which represent the maximum amount of money that can be robbed up to the previous house and the current house, respectively. The algorithm iterates through the houses, updating these variables based on the maximum amount that can be robbed at the current house, considering the constraint of not robbing adjacent houses.\n\nThe loop calculates a temporary variable `temp` that represents the maximum amount of money that can be robbed at the current house. It considers two scenarios: either robbing the current house and adding the money robbed at the house before the previous one (`n + prev`), or skipping the current house and taking the maximum amount from the previous house (`curr`). The maximum of these two scenarios is assigned to `temp`.\n\nThe `prev` and `curr` variables are then updated for the next iteration. `prev` is set to the previous value of `curr`, and `curr` is set to the calculated `temp`.\n\nThe final result is the value stored in the `curr` variable, representing the maximum amount of money that can be robbed at the last house. This value is returned as the result of the `rob` method.\n\nThe provided test cases at the end demonstrate the usage of the `rob` method with different input arrays.",
-      "metadata": {},
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Test cases:\n\nHere's how you can use this solution:",
-      "metadata": {},
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 1:\nsol = Solution()\n\nnums1 = [1, 2, 3, 1]\nprint(sol.rob(nums1))  # Output: 4",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 5,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "4\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums2 = [2, 7, 9, 3, 1]\nprint(sol.rob(nums2))  # Output: 12",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 6,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "12\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Time and Space Complexity Analysis\n\nLet's analyze the time and space complexity of the provided code.\n\n### Time Complexity:\n\nThe time complexity of the code is O(n), where n is the number of houses. The algorithm iterates through the array once, and in each iteration, it performs constant time operations. The loop runs for each house, and the number of operations inside the loop is independent of the input size. Therefore, the overall time complexity is linear, O(n).\n\n### Space Complexity:\n\nThe space complexity is O(1), constant space. The algorithm uses only a constant amount of extra space regardless of the input size. The variables `prev`, `curr`, and `temp` are the only additional variables used, and they do not depend on the size of the input array. Thus, the space complexity is constant, O(1).\n\nIn summary:\n- Time Complexity: O(n)\n- Space Complexity: O(1)",
-      "metadata": {},
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Challenging Exercises:\n\n1. **Variation with Maximum Number of Houses to Rob:** Modify the problem to include an additional constraint: you are allowed to rob at most 'k' houses without triggering the alarm. Find the maximum amount you can rob under this new constraint.\n\n2. **Robbing in a Circular Street:** Extend the problem to a circular street where the first and last houses are adjacent. You cannot rob both the first and last houses simultaneously. Find the maximum amount you can rob in this circular setting.\n\n\n**Link to Problem:** [To see the House Robber problem on LeetCode, click here!](https://leetcode.com/problems/house-robber/description/)\n",
-      "metadata": {},
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6"
-    }
-  ]
-}
\ No newline at end of file
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb b/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb
deleted file mode 100644
index b1080e8..0000000
--- a/_build/html/_sources/13. One-D Dynamic Programming/213. House Robber II.ipynb	
+++ /dev/null
@@ -1,119 +0,0 @@
-{
-  "metadata": {
-    "kernelspec": {
-      "name": "python",
-      "display_name": "Python (Pyodide)",
-      "language": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat_minor": 5,
-  "nbformat": 4,
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "source": "# House Robber II\n**Difficulty:** Medium\n\n---\nYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle**. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.\n\nGiven an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight without alerting the police*.\n\n**Constraints:**\n- `1 <= nums.length <= 100`\n- `0 <= nums[i] <= 1000`",
-      "metadata": {},
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Probelm Explanation:\n\nCertainly! The problem statement describes a scenario where you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. However, there is a security system in place: adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.\n\nThe twist in this problem is that the houses are arranged in a circle. That means the first house is the neighbor of the last one. So, robbing the first house has implications for the last house, and vice versa.\n\nThe goal is to find the maximum amount of money you can rob tonight without alerting the police. You need to implement a function that takes an integer array `nums` representing the amount of money in each house and returns the maximum amount of money that can be robbed without triggering the security system.\n\nHere are a few examples to illustrate the problem:\n\n1. **Example 1:**\n   ```python\n   Input: nums = [2, 3, 2]\n   Output: 3\n   Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2) because they are adjacent houses. The optimal strategy is to rob house 2 (money = 3).\n   ```\n\n2. **Example 2:**\n   ```python\n   Input: nums = [1, 2, 3, 1]\n   Output: 4\n   Explanation: Robbing house 1 (money = 1) and then robbing house 3 (money = 3) results in a total amount of 4. Robbing house 2 is skipped to avoid alerting the police.\n   ```\n\n3. **Example 3:**\n   ```python\n   Input: nums = [1, 2, 3]\n   Output: 3\n   Explanation: Since the houses are in a circle, you can choose either house 1, 2, or 3. The optimal strategy is to rob house 3 for a total amount of 3.\n   ```\n\nThe constraints for the problem are that the length of the `nums` array is between 1 and 100, and the amount of money in each house (`nums[i]`) is between 0 and 1000.\n\n## Solution:\nHere's a Python function to implement this algorithm:",
-      "metadata": {},
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76"
-    },
-    {
-      "cell_type": "code",
-      "source": "from typing import List\n\nclass Solution:\n    def rob(self, nums: List[int]) -> int:\n        def simple_rob(nums):\n            # Helper function for basic house robbery problem\n            prev, curr = 0, 0\n            for num in nums:\n                # Calculate the maximum amount of money that can be robbed\n                # without alerting the police for each house\n                prev, curr = curr, max(prev + num, curr)\n            return curr\n\n        # Check the base cases where the length of nums is 1 or 2\n        if len(nums) == 1:\n            return nums[0]\n        elif len(nums) == 2:\n            return max(nums[0], nums[1])\n\n        # Case 1: First house is robbed, last house is not\n        case1 = simple_rob(nums[:-1])\n\n        # Case 2: First house is not robbed, last house is\n        case2 = simple_rob(nums[1:])\n\n        # Maximum of the two cases\n        return max(case1, case2)",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 7,
-      "outputs": [],
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Explanation:\n\nThe provided code defines a class `Solution` with a method `rob`, which aims to determine the maximum amount of money a professional robber can steal from houses arranged in a circle without alerting the police. Each house contains a certain amount of money, and adjacent houses have a security system that triggers if both are robbed on the same night.\n\nThe code uses dynamic programming to solve the problem efficiently. It contains a helper function `simple_rob` that computes the maximum amount of money that can be robbed from a linear arrangement of houses. The main method `rob` handles the circular arrangement by considering two cases: robbing the first house and not robbing the last, and not robbing the first house and robbing the last. The result is the maximum of these two cases.\n\nThe class is instantiated as `solution`, and the `rob` method is applied to three different scenarios (test cases) to demonstrate its functionality. The final output represents the maximum amount of money the robber can steal without triggering the security system.",
-      "metadata": {},
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Test cases:\n\nHere's how you can use this solution:",
-      "metadata": {},
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 1:\n\n# Create an instance of the Solution class\nsolution = Solution()\n\nnums1 = [2, 3, 2]\nprint(solution.rob(nums1))  # Output: 3",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 8,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "3\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums2 = [1, 2, 3, 1]\nprint(solution.rob(nums2))  # Output: 4",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 9,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "4\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3"
-    },
-    {
-      "cell_type": "code",
-      "source": "# Example 2:\nnums3 = [1, 2, 3]\nprint(solution.rob(nums3))  # Output: 3",
-      "metadata": {
-        "trusted": true
-      },
-      "execution_count": 10,
-      "outputs": [
-        {
-          "name": "stdout",
-          "text": "3\n",
-          "output_type": "stream"
-        }
-      ],
-      "id": "e6e1554f-ae77-410c-8af2-e59e5428166c"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Time and Space Complexity Analysis\n\nLet's analyze the time and space complexity of the provided code:\n\n### Time Complexity:\nThe time complexity is primarily determined by the `simple_rob` helper function, which has a time complexity of O(N), where N is the number of houses in the input array. The main `rob` function calls `simple_rob` twice, once for the case of robbing the first house and not robbing the last, and once for the case of not robbing the first house and robbing the last. Therefore, the overall time complexity is O(N).\n\n### Space Complexity:\nThe space complexity is O(1) because the space used by the algorithm is constant, regardless of the size of the input array. The only variables used are `prev` and `curr` in the `simple_rob` function, which represent the previous and current maximum amounts of money that can be robbed without alerting the police. These variables are updated iteratively, and no additional data structures are used that scale with the input size. Therefore, the space complexity is constant, or O(1).\n\nIn summary:\n- Time Complexity: O(N)\n- Space Complexity: O(1)",
-      "metadata": {},
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2"
-    },
-    {
-      "cell_type": "markdown",
-      "source": "## Challenging Exercises:\n\n1. **Optimization Challenge:** Optimize the space complexity of the solution. Can you solve the problem with O(1) space complexity without sacrificing the time complexity?\n\n2. **Circular Robbery Variations:** Consider variations of the circular house robbery problem, such as allowing the robber to skip a certain number of houses or restricting the number of consecutive houses that can be robbed. Modify the solution to accommodate these variations.\n\n**Link to Problem:** [To see the House Robber II problem on LeetCode, click here!](https://leetcode.com/problems/house-robber-ii/)",
-      "metadata": {},
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6"
-    }
-  ]
-}
\ No newline at end of file
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb b/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb
deleted file mode 100644
index b727489..0000000
--- a/_build/html/_sources/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb	
+++ /dev/null
@@ -1,210 +0,0 @@
-{
-  "cells": [
-    {
-      "cell_type": "markdown",
-      "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-      "metadata": {},
-      "source": [
-        "# 70. Climbing Stairs\n",
-        "\n",
-        "**Difficulty:** Easy\n",
-        "\n",
-        "**Link to Problem:** [To see the Climbing Stairs problem on LeetCode, click here!](https://leetcode.com/problems/climbing-stairs/)\n",
-        "\n",
-        "---\n",
-        "You are climbing a staircase. It takes `n` steps to reach the top.\n",
-        "\n",
-        "Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?\n",
-        "\n",
-        "**Constraints:**\n",
-        "- `1 <= n <= 45`"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-      "metadata": {},
-      "source": [
-        "## Probelm Explanation:\n",
-        "\n",
-        "The \"Climbing Stairs\" problem is a classic algorithmic problem that asks you to find the number of distinct ways to climb a staircase with a given number of steps. Here's the problem statement:\n",
-        "\n",
-        "You are given a staircase with 'n' steps. Each time you can either climb 1 step or 2 steps. You need to find out how many distinct ways there are to reach the top of the staircase.\n",
-        "\n",
-        "For example, if 'n' is 2, there are two ways to climb to the top:\n",
-        "\n",
-        "1. Climbing two steps at once.\n",
-        "2. Climbing one step, and then another step.\n",
-        "\n",
-        "If 'n' is 3, there are three ways:\n",
-        "\n",
-        "1. Climbing one step, then one step, and finally one step.\n",
-        "2. Climbing one step, then two steps.\n",
-        "3. Climbing two steps, then one step.\n",
-        "\n",
-        "The problem essentially asks you to find a solution that counts the possible combinations of 1-step and 2-step climbs that can be taken to reach the top of the staircase. It's a classic example of a dynamic programming problem where you can break it down into smaller subproblems and build a solution incrementally, as shown in the code I provided earlier.\n",
-        "\n",
-        "## Solution:\n",
-        "Here's a Python function to implement this algorithm:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 1,
-      "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [],
-      "source": [
-        "class Solution:\n",
-        "    def climbStairs(self, n: int) -> int:\n",
-        "        # If there are 1 or 2 steps, there are exactly n ways to reach the top.\n",
-        "        if n <= 2:\n",
-        "            return n\n",
-        "\n",
-        "        # Create an array to store the number of ways to reach each step.\n",
-        "        dp = [0] * (n + 1)\n",
-        "\n",
-        "        # There is one way to reach the first step (base case).\n",
-        "        dp[1] = 1\n",
-        "\n",
-        "        # There are two ways to reach the second step (base case).\n",
-        "        dp[2] = 2\n",
-        "\n",
-        "        # Calculate the number of ways to reach each step from the previous steps.\n",
-        "        for i in range(3, n + 1):\n",
-        "            # The number of ways to reach step 'i' is the sum of the ways to reach\n",
-        "            # the previous two steps, as you can take 1 or 2 steps at a time.\n",
-        "            dp[i] = dp[i - 1] + dp[i - 2]\n",
-        "\n",
-        "        # The result is the number of ways to reach the top, which is stored in dp[n].\n",
-        "        return dp[n]"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-      "metadata": {},
-      "source": [
-        "## Explanation:\n",
-        "\n",
-        "The code defines a class `Solution` with a method `climbStairs` that takes an integer `n` as input and returns the number of distinct ways to climb a staircase with `n` steps.\n",
-        "\n",
-        "1. If `n` is 1 or 2, the function returns `n` directly because there are only 1 or 2 distinct ways to reach the top in those cases.\n",
-        "\n",
-        "2. For `n` greater than 2, the code uses dynamic programming to calculate the number of ways to reach each step. It creates an array `dp` of length `n + 1` to store these values.\n",
-        "\n",
-        "3. The `dp[1]` and `dp[2]` are initialized to 1 and 2, respectively. This is because there is only one way to reach the first step (by taking 1 step), and there are two ways to reach the second step (by taking 2 steps or two 1-step climbs).\n",
-        "\n",
-        "4. The code then uses a loop to fill in the `dp` array for steps 3 and above. For each step `i` from 3 to `n`, it calculates the number of ways to reach that step by adding the number of ways to reach the previous two steps, as you can either take 1 step or 2 steps at a time.\n",
-        "\n",
-        "5. Finally, the function returns the value stored in `dp[n]`, which represents the total number of distinct ways to climb the staircase with `n` steps."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-      "metadata": {},
-      "source": [
-        "## Test cases:\n",
-        "\n",
-        "Here's how you can use this solution:"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 2,
-      "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 1 - Input: n = 2, Output: 2\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 1:\n",
-        "n = 2\n",
-        "solution = Solution()\n",
-        "print(f\"Example 1 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 2"
-      ]
-    },
-    {
-      "cell_type": "code",
-      "execution_count": 5,
-      "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-      "metadata": {
-        "trusted": true
-      },
-      "outputs": [
-        {
-          "name": "stdout",
-          "output_type": "stream",
-          "text": [
-            "Example 2 - Input: n = 3, Output: 3\n"
-          ]
-        }
-      ],
-      "source": [
-        "# Example 2:\n",
-        "n = 3\n",
-        "print(f\"Example 2 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 3\n"
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-      "metadata": {},
-      "source": [
-        "## Time and Space Complexity Analysis\n",
-        "\n",
-        "Time Complexity:\n",
-        "The time complexity of the solution is O(n) because we use a single for loop that iterates from 3 to n to fill the `dp` array. In each iteration, we perform constant time operations (addition and assignment). Therefore, the time complexity is linear with respect to the input value `n`.\n",
-        "\n",
-        "Space Complexity:\n",
-        "The space complexity of the solution is O(n) as well. We use an array `dp` of length `n + 1` to store the number of ways to reach each step. The size of this array is directly proportional to the input value `n`, which makes the space complexity linear in terms of `n`. The rest of the variables used in the function (e.g., `i`) occupy constant space and do not contribute significantly to the overall space complexity.\n",
-        "\n",
-        "In summary, the time complexity is O(n), and the space complexity is O(n) for this dynamic programming solution to the \"Climbing Stairs\" problem."
-      ]
-    },
-    {
-      "cell_type": "markdown",
-      "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-      "metadata": {},
-      "source": [
-        "## Challenging Exercises:\n",
-        "\n",
-        "1. **Generalized Step Sizes**: Instead of being limited to 1 step or 2 steps, consider a scenario where you can take steps of sizes `[1, 2, 3]`. Implement a function that calculates the number of ways to climb the staircase with these different step sizes.\n",
-        "\n",
-        "2. **Minimize Steps**: Given an integer `n`, find the minimum number of steps required to reach the top of the staircase. Return both the minimum number of steps and the specific sequence of steps taken."
-      ]
-    }
-  ],
-  "metadata": {
-    "kernelspec": {
-      "display_name": "Python (Pyodide)",
-      "language": "python",
-      "name": "python"
-    },
-    "language_info": {
-      "codemirror_mode": {
-        "name": "python",
-        "version": 3
-      },
-      "file_extension": ".py",
-      "mimetype": "text/x-python",
-      "name": "python",
-      "nbconvert_exporter": "python",
-      "pygments_lexer": "ipython3",
-      "version": "3.8"
-    }
-  },
-  "nbformat": 4,
-  "nbformat_minor": 5
-}
diff --git a/_build/html/_sources/13. One-D Dynamic Programming/README.md b/_build/html/_sources/13. One-D Dynamic Programming/README.md
deleted file mode 100644
index 73f4de3..0000000
--- a/_build/html/_sources/13. One-D Dynamic Programming/README.md	
+++ /dev/null
@@ -1,25 +0,0 @@
-# 1-D Dynamic Programming Problems - Blind 75 LeetCode 
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)                                   | Easy       |
-| [198. House Robber](https://leetcode.com/problems/house-robber/)                                       | Medium     |
-| [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)                                 | Medium     |
-| [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)       | Medium     |
-| [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)                   | Medium     |
-| [91. Decode Ways](https://leetcode.com/problems/decode-ways/)                                         | Medium     |
-| [322. Coin Change](https://leetcode.com/problems/coin-change/)                                         | Medium     |
-| [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)             | Medium     |
-| [139. Word Break](https://leetcode.com/problems/word-break/)                                           | Medium     |
-| [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | Medium     |
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/14. Two-D Dynamic Programming/README.md b/_build/html/_sources/14. Two-D Dynamic Programming/README.md
deleted file mode 100644
index 8bab876..0000000
--- a/_build/html/_sources/14. Two-D Dynamic Programming/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### 2-D Dynamic Programming
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [62. Unique Paths](https://leetcode.com/problems/unique-paths/)                                         | Medium     |
-| [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)       | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/15. Greedy/README.md b/_build/html/_sources/15. Greedy/README.md
deleted file mode 100644
index 82b6543..0000000
--- a/_build/html/_sources/15. Greedy/README.md	
+++ /dev/null
@@ -1,47 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Greedy
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)                                   | Easy       |
-| [55. Jump Game](https://leetcode.com/problems/jump-game/)                                                 | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/16. Intervals/README.md b/_build/html/_sources/16. Intervals/README.md
deleted file mode 100644
index a57713e..0000000
--- a/_build/html/_sources/16. Intervals/README.md	
+++ /dev/null
@@ -1,50 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Intervals
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [57. Insert Interval](https://leetcode.com/problems/insert-interval/)                                     | Hard       |
-| [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)                                   | Medium     |
-| [435. Non Overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)             | Medium     |
-| [252. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)                                       | Easy       |
-| [253. Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)                                 | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/17. Math Geometry/README.md b/_build/html/_sources/17. Math Geometry/README.md
deleted file mode 100644
index 8b2f180..0000000
--- a/_build/html/_sources/17. Math Geometry/README.md	
+++ /dev/null
@@ -1,48 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Math & Geometry
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [48. Rotate Image](https://leetcode.com/problems/rotate-image/)                                           | Medium     |
-| [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)                                         | Medium     |
-| [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)                               | Medium     |
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/18. Bit Manipulation/README.md b/_build/html/_sources/18. Bit Manipulation/README.md
deleted file mode 100644
index 05b2fef..0000000
--- a/_build/html/_sources/18. Bit Manipulation/README.md	
+++ /dev/null
@@ -1,51 +0,0 @@
-# Blind 75 LeetCode Problems
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Bit Manupulation
-
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)                                   | Easy       |
-| [338. Counting Bits](https://leetcode.com/problems/counting-bits/)                                         | Medium     |
-| [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)                                           | Easy       |
-| [268. Missing Number](https://leetcode.com/problems/missing-number/)                                       | Easy       |
-| [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)                             | Medium     |
-
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-Happy coding!
\ No newline at end of file
diff --git a/_build/html/_sources/CODE_OF_CONDUCT.md b/_build/html/_sources/CODE_OF_CONDUCT.md
deleted file mode 100644
index 5fe5041..0000000
--- a/_build/html/_sources/CODE_OF_CONDUCT.md
+++ /dev/null
@@ -1,128 +0,0 @@
-# Contributor Covenant Code of Conduct
-
-## Our Pledge
-
-We as members, contributors, and leaders pledge to make participation in our
-community a harassment-free experience for everyone, regardless of age, body
-size, visible or invisible disability, ethnicity, sex characteristics, gender
-identity and expression, level of experience, education, socio-economic status,
-nationality, personal appearance, race, religion, or sexual identity
-and orientation.
-
-We pledge to act and interact in ways that contribute to an open, welcoming,
-diverse, inclusive, and healthy community.
-
-## Our Standards
-
-Examples of behavior that contributes to a positive environment for our
-community include:
-
-* Demonstrating empathy and kindness toward other people
-* Being respectful of differing opinions, viewpoints, and experiences
-* Giving and gracefully accepting constructive feedback
-* Accepting responsibility and apologizing to those affected by our mistakes,
-  and learning from the experience
-* Focusing on what is best not just for us as individuals, but for the
-  overall community
-
-Examples of unacceptable behavior include:
-
-* The use of sexualized language or imagery, and sexual attention or
-  advances of any kind
-* Trolling, insulting or derogatory comments, and personal or political attacks
-* Public or private harassment
-* Publishing others' private information, such as a physical or email
-  address, without their explicit permission
-* Other conduct which could reasonably be considered inappropriate in a
-  professional setting
-
-## Enforcement Responsibilities
-
-Community leaders are responsible for clarifying and enforcing our standards of
-acceptable behavior and will take appropriate and fair corrective action in
-response to any behavior that they deem inappropriate, threatening, offensive,
-or harmful.
-
-Community leaders have the right and responsibility to remove, edit, or reject
-comments, commits, code, wiki edits, issues, and other contributions that are
-not aligned to this Code of Conduct, and will communicate reasons for moderation
-decisions when appropriate.
-
-## Scope
-
-This Code of Conduct applies within all community spaces, and also applies when
-an individual is officially representing the community in public spaces.
-Examples of representing our community include using an official e-mail address,
-posting via an official social media account, or acting as an appointed
-representative at an online or offline event.
-
-## Enforcement
-
-Instances of abusive, harassing, or otherwise unacceptable behavior may be
-reported to the community leaders responsible for enforcement at
-Email.
-All complaints will be reviewed and investigated promptly and fairly.
-
-All community leaders are obligated to respect the privacy and security of the
-reporter of any incident.
-
-## Enforcement Guidelines
-
-Community leaders will follow these Community Impact Guidelines in determining
-the consequences for any action they deem in violation of this Code of Conduct:
-
-### 1. Correction
-
-**Community Impact**: Use of inappropriate language or other behavior deemed
-unprofessional or unwelcome in the community.
-
-**Consequence**: A private, written warning from community leaders, providing
-clarity around the nature of the violation and an explanation of why the
-behavior was inappropriate. A public apology may be requested.
-
-### 2. Warning
-
-**Community Impact**: A violation through a single incident or series
-of actions.
-
-**Consequence**: A warning with consequences for continued behavior. No
-interaction with the people involved, including unsolicited interaction with
-those enforcing the Code of Conduct, for a specified period of time. This
-includes avoiding interactions in community spaces as well as external channels
-like social media. Violating these terms may lead to a temporary or
-permanent ban.
-
-### 3. Temporary Ban
-
-**Community Impact**: A serious violation of community standards, including
-sustained inappropriate behavior.
-
-**Consequence**: A temporary ban from any sort of interaction or public
-communication with the community for a specified period of time. No public or
-private interaction with the people involved, including unsolicited interaction
-with those enforcing the Code of Conduct, is allowed during this period.
-Violating these terms may lead to a permanent ban.
-
-### 4. Permanent Ban
-
-**Community Impact**: Demonstrating a pattern of violation of community
-standards, including sustained inappropriate behavior,  harassment of an
-individual, or aggression toward or disparagement of classes of individuals.
-
-**Consequence**: A permanent ban from any sort of public interaction within
-the community.
-
-## Attribution
-
-This Code of Conduct is adapted from the [Contributor Covenant][homepage],
-version 2.0, available at
-https://www.contributor-covenant.org/version/2/0/code_of_conduct.html.
-
-Community Impact Guidelines were inspired by [Mozilla's code of conduct
-enforcement ladder](https://github.com/mozilla/diversity).
-
-[homepage]: https://www.contributor-covenant.org
-
-For answers to common questions about this code of conduct, see the FAQ at
-https://www.contributor-covenant.org/faq. Translations are available at
-https://www.contributor-covenant.org/translations.
diff --git a/_build/html/_sources/CONTRIBUTING.md b/_build/html/_sources/CONTRIBUTING.md
deleted file mode 100644
index ca794ef..0000000
--- a/_build/html/_sources/CONTRIBUTING.md
+++ /dev/null
@@ -1,79 +0,0 @@
-# Contributing Guidelines
-
-Thank you for considering contributing to the Blind 75 LeetCode repository! We welcome contributions from the community to help improve this resource and make it even more valuable.
-
-Please take a moment to review this document to understand how you can contribute.
-
-## Table of Contents
-- [Ways to Contribute](#ways-to-contribute)
-- [Getting Started](#getting-started)
-- [Contributing Process](#contributing-process)
-- [Code of Conduct](#code-of-conduct)
-
-## Ways to Contribute
-
-There are several ways you can contribute to this repository:
-
-1. **Suggest Improvements**: If you have ideas for making the Blind 75 LeetCode repository better, please open an issue to discuss your suggestions.
-
-2. **Report Issues**: If you encounter any issues with the existing content or code, please open an issue and provide as much detail as possible.
-
-3. **Contribute Solutions**: You can contribute solutions to the problems listed in the repository. Please follow the guidelines mentioned below when submitting your solutions.
-
-4. **Enhance Documentation**: Improve documentation, including explanations and time/space complexity analysis, for existing solutions.
-
-5. **Fix Bugs**: If you find and fix any bugs in the existing code or documentation, you are welcome to submit a pull request.
-
-## Getting Started
-
-Before you start contributing, please ensure you have the following prerequisites:
-
-- A GitHub account.
-- Familiarity with Git and GitHub.
-- Code editor or IDE for making changes.
-
-## Contributing Process
-
-Here's the general process for contributing:
-
-1. **Fork the Repository**: Click the "Fork" button at the top of this repository to create your copy.
-
-2. **Clone Your Fork**: Clone your fork to your local machine.
-
-    ```
-    git clone https://github.com/your-username/Blind-75-LeetCode.git
-    ```
-
-3. **Create a Branch**: Create a new branch for your contribution.
-
-    ```
-    git checkout -b feature/your-feature-name
-    ```
-
-4. **Make Changes**: Implement your changes, following the guidelines provided.
-
-5. **Test Your Changes**: Ensure that your changes do not introduce new issues and are consistent with the repository's style and structure.
-
-6. **Commit Changes**: Commit your changes with a meaningful commit message.
-
-    ```
-    git commit -m "Add solution to problem X"
-    ```
-
-7. **Push Changes**: Push your changes to your fork on GitHub.
-
-    ```
-    git push origin feature/your-feature-name
-    ```
-
-8. **Create a Pull Request**: Go to the original repository and create a pull request. Please provide a detailed description of your changes.
-
-9. **Review and Collaboration**: Your pull request will be reviewed, and you may need to make further changes. Be responsive to comments and feedback.
-
-10. **Merge**: Once your pull request is approved, it will be merged into the main branch.
-
-## Code of Conduct
-
-Please note that we have a [Code of Conduct](CODE_OF_CONDUCT.md) that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.
-
-We appreciate your contributions to the Blind 75 LeetCode repository and look forward to working together to make it an even better resource for the community!
diff --git a/_build/html/_sources/README.md b/_build/html/_sources/README.md
deleted file mode 100644
index a9b8b2e..0000000
--- a/_build/html/_sources/README.md
+++ /dev/null
@@ -1,216 +0,0 @@
-# Blind 75 LeetCode Solutions
-
-![Blind 75 LeetCode](https://img.shields.io/badge/Blind_75_LeetCode-Solutions-blue?labelColor=red)
-
-This repository contains a curated list of 75 LeetCode problems that are commonly known as the "Blind 75." These problems cover a wide range of data structures and algorithms and are frequently asked in technical interviews. Solving these problems can help you build a strong foundation in algorithmic problem-solving.
-
-## Problem List
-
-### Array & Hashing
-![Array & Hashing](https://img.shields.io/badge/Array_&_Hashing-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [217. Contains Duplicate](https://leetcode.com/problems/contains-duplicate/)                        | Easy       |
-| [242. Valid Anagram](https://leetcode.com/problems/valid-anagram/)                                  | Easy       |
-| [1. Two Sum](https://leetcode.com/problems/two-sum/)                                              | Easy       |
-| [49. Group Anagrams](https://leetcode.com/problems/group-anagrams/)                                | Medium     |
-| [347. Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/)              | Medium     |
-| [238. Product of Array Except Self](https://leetcode.com/problems/product-of-array-except-self/)    | Medium     |
-| [659. Encode and Decode Strings](https://leetcode.com/problems/encode-and-decode-strings/)          | Medium     |
-| [128. Longest Consecutive Sequence](https://leetcode.com/problems/longest-consecutive-sequence/)    | Hard       |
-
-### Two Pointers
-![Array & Hashing](https://img.shields.io/badge/Two_Pointers-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/)                            | Easy       |
-| [15. 3Sum](https://leetcode.com/problems/3sum/)                                                       | Medium     |
-| [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)           | Medium     |
-
-### Sliding Window
-![Array & Hashing](https://img.shields.io/badge/Sliding_Window-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [121. Best Time to Buy And Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/) | Easy       |
-| [3. Longest Substring Without Repeating Characters](https://leetcode.com/problems/longest-substring-without-repeating-characters/) | Medium |
-| [424. Longest Repeating Character Replacement](https://leetcode.com/problems/longest-repeating-character-replacement/) | Medium |
-| [76. Minimum Window Substring](https://leetcode.com/problems/minimum-window-substring/)             | Hard       |
-
-### Stack
-![Array & Hashing](https://img.shields.io/badge/Stack-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/)                            | Easy       |
-
-### Binary Search
-![Array & Hashing](https://img.shields.io/badge/Binary_Search-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [153. Find Minimum In Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/) | Medium |
-| [33. Search In Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/)   | Medium     |
-
-### Linked List
-![Array & Hashing](https://img.shields.io/badge/Linked_List-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)                       | Easy       |
-| [21. Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/)                   | Easy       |
-| [143. Reorder List](https://leetcode.com/problems/reorder-list/)                                     | Medium     |
-| [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | Medium |
-| [141. Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/)                           | Medium     |
-| [23. Merge K Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)                       | Hard       |
-
-### Trees
-![Array & Hashing](https://img.shields.io/badge/Trees-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [226. Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/)                          | Easy       |
-| [104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)     | Easy       |
-| [100. Same Tree](https://leetcode.com/problems/same-tree/)                                           | Easy       |
-| [572. Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/)               | Easy       |
-| [235. Lowest Common Ancestor of a Binary Search Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/) | Easy |
-| [102. Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/) | Medium |
-| [98. Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)       | Medium     |
-| [230. Kth Smallest Element In a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)   | Medium     |
-| [105. Construct Binary Tree From Preorder And Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/) | Medium |
-| [124. Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/)     | Hard       |
-| [297. Serialize And Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/) | Hard |
-
-### Tries
-![Array & Hashing](https://img.shields.io/badge/Tries-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [208. Implement Trie Prefix Tree](https://leetcode.com/problems/implement-trie-prefix-tree/)         | Medium     |
-| [211. Design Add And Search Words Data Structure](https://leetcode.com/problems/design-add-and-search-words-data-structure/) | Medium |
-| [212. Word Search II](https://leetcode.com/problems/word-search-ii/)                                 | Hard       |
-
-### Heap / Priority Queue
-![Array & Hashing](https://img.shields.io/badge/Heap_/_Priority_Queue-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [295. Find Median From Data Stream](https://leetcode.com/problems/find-median-from-data-stream/)     | Hard       |
-
-### Backtracking
-![Array & Hashing](https://img.shields.io/badge/Backtracking-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [39. Combination Sum](https://leetcode.com/problems/combination-sum/)                                 | Medium     |
-| [79. Word Search](https://leetcode.com/problems/word-search/)                                         | Medium     |
-
-### Graphs
-![Array & Hashing](https://img.shields.io/badge/Graphs-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [200. Number of Islands](https://leetcode.com/problems/number-of-islands/)                           | Medium     |
-| [133. Clone Graph](https://leetcode.com/problems/clone-graph/)                                       | Medium     |
-| [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)       | Medium     |
-| [207. Course Schedule](https://leetcode.com/problems/course-schedule/)                                 | Medium     |
-| [323. Number of Connected Components In An Undirected Graph](https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/) | Medium |
-| [261. Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)                               | Medium     |
-
-### Advanced Graphs
-![Array & Hashing](https://img.shields.io/badge/Advanced_Graphs-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [269. Alien Dictionary](https://leetcode.com/problems/alien-dictionary/)                               | Hard       |
-
-### 1-D Dynamic Programming
-![Array & Hashing](https://img.shields.io/badge/1_D_Dynamic_Programming-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [70. Climbing Stairs](https://leetcode.com/problems/climbing-stairs/)                                   | Easy       |
-| [198. House Robber](https://leetcode.com/problems/house-robber/)                                       | Easy       |
-| [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)                                 | Medium     |
-| [5. Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/)       | Medium     |
-| [647. Palindromic Substrings](https://leetcode.com/problems/palindromic-substrings/)                   | Medium     |
-| [91. Decode Ways](https://leetcode.com/problems/decode-ways/)                                         | Medium     |
-| [322. Coin Change](https://leetcode.com/problems/coin-change/)                                         | Medium     |
-| [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)             | Medium     |
-| [139. Word Break](https://leetcode.com/problems/word-break/)                                           | Medium     |
-| [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | Medium     |
-
-### 2-D Dynamic Programming
-![Array & Hashing](https://img.shields.io/badge/2_D_Dynamic_Programming-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [62. Unique Paths](https://leetcode.com/problems/unique-paths/)                                         | Medium     |
-| [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/)       | Medium     |
-
-### Greedy
-![Array & Hashing](https://img.shields.io/badge/Greedy-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [53. Maximum Subarray](https://leetcode.com/problems/maximum-subarray/)                                   | Easy       |
-| [55. Jump Game](https://leetcode.com/problems/jump-game/)                                                 | Medium     |
-
-### Intervals
-![Array & Hashing](https://img.shields.io/badge/Intervals-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [57. Insert Interval](https://leetcode.com/problems/insert-interval/)                                     | Hard       |
-| [56. Merge Intervals](https://leetcode.com/problems/merge-intervals/)                                   | Medium     |
-| [435. Non Overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)             | Medium     |
-| [252. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)                                       | Easy       |
-| [253. Meeting Rooms II](https://leetcode.com/problems/meeting-rooms-ii/)                                 | Medium     |
-
-### Math & Geometry
-![Array & Hashing](https://img.shields.io/badge/Math_&_Geometry-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [48. Rotate Image](https://leetcode.com/problems/rotate-image/)                                           | Medium     |
-| [54. Spiral Matrix](https://leetcode.com/problems/spiral-matrix/)                                         | Medium     |
-| [73. Set Matrix Zeroes](https://leetcode.com/problems/set-matrix-zeroes/)                               | Medium     |
-
-### Bit Manipulation
-![Array & Hashing](https://img.shields.io/badge/Bit_Manipulation-Solutions-blue?labelColor=red)
-| Problem Name                                    | Difficulty |
-| ----------------------------------------------- | ---------- |
-| [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/)                                   | Easy       |
-| [338. Counting Bits](https://leetcode.com/problems/counting-bits/)                                         | Medium     |
-| [190. Reverse Bits](https://leetcode.com/problems/reverse-bits/)                                           | Easy       |
-| [268. Missing Number](https://leetcode.com/problems/missing-number/)                                       | Easy       |
-| [371. Sum of Two Integers](https://leetcode.com/problems/sum-of-two-integers/)                             | Medium     |
-
-
-## How to Use
-
-1. Clone or download this repository to your local machine.
-
-```bash
-git clone https://github.com/mohsentabibian/LeetCode-Solutions/Blind-75-LeetCode.git
-```
-
-2. Navigate to the directory of the specific problem you want to solve.
-
-```bash
-cd Blind-75-LeetCode/Problem-Directory
-```
-
-3. Open the problem's directory to find its description, code template, and solution(s).
-
-4. Solve the problem using your preferred programming language.
-
-5. Check the solution(s) provided in the directory for reference and to compare your solution.
-
-## Contribute
-
-If you have a better solution or want to add more problems to the collection, feel free to contribute. You can create a pull request or open an issue to discuss additions or improvements.
-
-## Resources
-
-- [LeetCode](https://leetcode.com/): The official LeetCode website.
-- [LeetCode Discuss](https://leetcode.com/discuss/): LeetCode's discussion forum for each problem. You can find helpful hints and discussions here.
-
-## Disclaimer
-
-This repository is meant for educational purposes and to help individuals prepare for technical interviews. Please respect LeetCode's terms of use and guidelines when using this repository.
-
-## Code of Conduct
-
-We have a [Code of Conduct](CODE_OF_CONDUCT.md) that we expect all contributors to adhere to. By participating in this project, you agree to follow this code.
-
-## Security Policy
-
-We take security seriously. Please review our [Security Policy](SECURITY.md) for information on reporting security vulnerabilities.
-
-Happy coding!
diff --git a/_build/html/_sources/SECURITY.md b/_build/html/_sources/SECURITY.md
deleted file mode 100644
index 7c19993..0000000
--- a/_build/html/_sources/SECURITY.md
+++ /dev/null
@@ -1,42 +0,0 @@
-# Security Policy
-
-## Reporting a Vulnerability
-
-We take the security of the Blind 75 LeetCode repository seriously. If you believe you have found a security vulnerability, please help us by following these steps:
-
-1. **Do Not Disclose Publicly**: Please do not disclose the issue publicly until it has been addressed by our team.
-
-2. **Report Privately**: Create a GitHub issue with the title "Security Vulnerability Report" and provide a detailed description of the vulnerability. You can use a sample report template as follows:
-
-    ```
-    **Description**
-    A clear and concise description of the security vulnerability.
-
-    **Steps to Reproduce**
-    Provide detailed steps on how to reproduce the vulnerability.
-
-    **Expected Behavior**
-    Describe what you expected to happen.
-
-    **Actual Behavior**
-    Describe what actually happened.
-
-    **Additional Information**
-    Any additional information that can help us understand and reproduce the issue, such as environment details, configuration, or screenshots.
-    ```
-
-3. **Secure Communication**: If the vulnerability involves sensitive data, please encrypt your communication with our PGP key. Contact us for our PGP key information.
-
-4. **Response**: We will acknowledge your report within 5 business days and provide an estimated timeline for the resolution.
-
-5. **Fix and Disclosure**: We will work to fix the issue promptly and privately. Once resolved, we will provide a security advisory and give credit to the reporter if desired.
-
-6. **Responsible Disclosure**: We encourage responsible disclosure. Please do not take advantage of the vulnerability or attempt to exploit it for malicious purposes.
-
-## Security Updates
-
-We are committed to promptly addressing and resolving any security issues. We recommend that you stay up-to-date with the latest releases of this repository, as they may include security updates and fixes.
-
-## Contact
-
-If you have any questions or need further assistance regarding security, you can contact us at mohsentabibian@gmail.com.
diff --git a/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb b/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb
deleted file mode 100644
index 5d55ca7..0000000
--- a/_build/jupyter_execute/01. Array Hashing/1. Two Sum.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 100: Same Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-    "\n",
-    "---\n",
-    "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-    "\n",
-    "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-    "\n",
-    "**Constraints**\n",
-    "\n",
-    "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-    "2. $-10^4$ <= `Node.val` <= $10^4$"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def isSameTree(p, q):\n",
-    "    # Base case: If both p and q are None, the trees are the same.\n",
-    "    if not p and not q:\n",
-    "        return True\n",
-    "    \n",
-    "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-    "    if not p or not q:\n",
-    "        return False\n",
-    "    \n",
-    "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-    "    if p.val != q.val:\n",
-    "        return False\n",
-    "    \n",
-    "    # Recursively check the left and right subtrees of p and q.\n",
-    "    # If both subtrees are the same, the entire trees are the same.\n",
-    "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-    "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-    "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-    "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-    "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-    "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "\n",
-    "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-    "\n",
-    "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "print(isSameTree(p1, q1)) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "\n",
-    "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-    "\n",
-    "p2 = TreeNode(1, TreeNode(2), None)\n",
-    "q2 = TreeNode(1, None, TreeNode(2))\n",
-    "print(isSameTree(p2, q2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "\n",
-    "#Input: p = [1,2,1], q = [1,1,2]\n",
-    "\n",
-    "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-    "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-    "print(isSameTree(p3, q3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "**Time Complexity**\n",
-    "\n",
-    "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-    "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-    "\n",
-    "**Space Complexity**\n",
-    "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-    "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-    "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-    "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-    "\n",
-    "**In summary:**\n",
-    "\n",
-    "+ Time Complexity: $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-    "+ Space Complexity: $O(n)$ due to the recursive call stack."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb b/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb
deleted file mode 100644
index 9256d0e..0000000
--- a/_build/jupyter_execute/01. Array Hashing/128. Longest Consecutive Sequence.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 128. Longest Consecutive Sequence\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Longest Consecutive Sequence problem on LeetCode, click here!](https://leetcode.com/problems/longest-consecutive-sequence/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given an unsorted array of integers `nums`, return *the length of the longest consecutive elements sequence.*\n",
-    "\n",
-    "You must write an algorithm that runs in `O(n)` time.\n",
-    "\n",
-    "*Constraints:**\n",
-    "- 0 <= `nums.length` <= $10^5$\n",
-    "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def longestConsecutive(nums):\n",
-    "    if not nums:\n",
-    "        return 0\n",
-    "\n",
-    "    num_set = set(nums)\n",
-    "    max_length = 0\n",
-    "\n",
-    "    for num in num_set:\n",
-    "        if num - 1 not in num_set:\n",
-    "            current_num = num\n",
-    "            current_length = 1\n",
-    "\n",
-    "            while current_num + 1 in num_set:\n",
-    "                current_num += 1\n",
-    "                current_length += 1\n",
-    "\n",
-    "            max_length = max(max_length, current_length)\n",
-    "\n",
-    "    return max_length"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining the `longestConsecutive` function that takes the `nums` array as input.\n",
-    "\n",
-    "2. We check if the `nums` array is empty. If it's empty, there are no consecutive elements, so we return 0.\n",
-    "\n",
-    "3. We create a Python set called `num_set` and insert all elements from the `nums` array into it. Using a set allows us to efficiently check for the existence of elements in O(1) time.\n",
-    "\n",
-    "4. We initialize a variable `max_length` to 0. This variable will keep track of the maximum length of consecutive elements sequence found.\n",
-    "\n",
-    "5. We iterate through the elements of the `num_set`. For each element `num`, we check if `num - 1` exists in the `num_set`. If it doesn't exist, it means `num` is the starting element of a potential consecutive sequence.\n",
-    "\n",
-    "6. Inside the loop, we initialize two variables: `current_num` to the current element `num` and `current_length` to 1. We start with a length of 1 because `num` itself is part of the sequence.\n",
-    "\n",
-    "7. We then enter a while loop that continues as long as `current_num + 1` exists in the `num_set`. This means we are incrementing the consecutive sequence.\n",
-    "\n",
-    "8. Inside the while loop, we increment `current_num` by 1 and also increment `current_length` by 1 to account for the next consecutive element.\n",
-    "\n",
-    "9. We compare `current_length` with the `max_length` and update `max_length` if the current sequence is longer.\n",
-    "\n",
-    "10. After the while loop, we move to the next element in the outer loop and repeat the process.\n",
-    "\n",
-    "11. Finally, we return the `max_length` as the result, which represents the length of the longest consecutive elements sequence in the `nums` array.\n",
-    "\n",
-    "The key idea here is to use a set to efficiently check for the existence of elements and to iterate through the elements, considering each element as the potential start of a consecutive sequence. By doing this, we can find the longest consecutive sequence in O(n) time complexity, where n is the number of elements in the array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "nums1 = [100, 4, 200, 1, 3, 2]\n",
-    "print(longestConsecutive(nums1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "9\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "nums2 = [0, 3, 7, 2, 5, 8, 4, 6, 0, 1]\n",
-    "print(longestConsecutive(nums2)) "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The code is designed to run in O(n) time complexity, where n is the number of elements in the `nums` array. Here's the breakdown:\n",
-    "\n",
-    "1. Constructing the `num_set` by inserting all elements from `nums` into it takes O(n) time because we perform an insertion operation for each element in `nums`.\n",
-    "\n",
-    "2. The main loop iterates through the elements in `num_set`. In the worst case, each element is visited only once, so the loop itself takes O(n) time.\n",
-    "\n",
-    "3. Within the loop, we have a while loop that may also take O(n) time in the worst case. However, this while loop is nested inside the main loop, so its overall time complexity remains O(n).\n",
-    "\n",
-    "Therefore, the overall time complexity of the code is O(n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity of the code is determined by the space used by the `num_set` and a few additional variables. Here's the breakdown:\n",
-    "\n",
-    "1. `num_set` is a set that stores the unique elements from `nums`. In the worst case, it can store all n elements from `nums`, so the space complexity is O(n).\n",
-    "\n",
-    "2. The additional variables used, such as `max_length`, `current_num`, and `current_length`, have constant space requirements and do not depend on the size of the input array.\n",
-    "\n",
-    "Therefore, the overall space complexity of the code is O(n) due to the space used by the `num_set`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. Write an algorithm that not only returns the length of the longest consecutive elements sequence but also returns the actual consecutive sequence itself.\n",
-    "\n",
-    "2. Extend the problem to allow elements to be considered consecutive if they are within a certain absolute difference (e.g., less than or equal to k) instead of exactly 1. Write an algorithm that finds the longest such sequence and runs in O(n) time."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb b/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb
deleted file mode 100644
index f7dbbe9..0000000
--- a/_build/jupyter_execute/01. Array Hashing/217. Contains Duplicate.ipynb	
+++ /dev/null
@@ -1,210 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 217. Contains Duplicate\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Contains Duplicate problem on LeetCode, click here!](https://leetcode.com/problems/contains-duplicate/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `nums.length` <= $10^5$\n",
-    "- $-10^9$ <= `nums[i]` <= $10^9$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def containsDuplicate(nums):\n",
-    "    # Create an empty set to store unique elements\n",
-    "    unique_elements = set()\n",
-    "    \n",
-    "    # Iterate through the array\n",
-    "    for num in nums:\n",
-    "        # If the element is already in the set, return True\n",
-    "        if num in unique_elements:\n",
-    "            return True\n",
-    "        # Otherwise, add it to the set\n",
-    "        else:\n",
-    "            unique_elements.add(num)\n",
-    "    \n",
-    "    # If the loop completes without finding duplicates, return False\n",
-    "    return False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The `containsDuplicate` function takes a single argument, `nums`, which is the input integer array.\n",
-    "\n",
-    "2. Inside the function, an empty set called `unique_elements` is created. This set will be used to keep track of unique elements in the input array.\n",
-    "\n",
-    "3. The function then iterates through the input array `nums` using a `for` loop.\n",
-    "\n",
-    "4. For each element `num` in the array, it checks whether `num` is already in the `unique_elements` set using the `if num in unique_elements:` condition.\n",
-    "\n",
-    "5. If `num` is already in the set, it means there is a duplicate element in the array, and the function immediately returns `True`.\n",
-    "\n",
-    "6. If `num` is not in the set, it is added to the `unique_elements` set using `unique_elements.add(num)`.\n",
-    "\n",
-    "7. The loop continues to the next element, and the process repeats.\n",
-    "\n",
-    "8. If the loop completes without finding any duplicates, it means that all elements in the array are distinct, and the function returns `False`.\n",
-    "\n",
-    "The code efficiently utilizes a set data structure to keep track of unique elements while iterating through the array, allowing it to quickly detect duplicate elements. This code meets the problem's requirements and constraints, as explained earlier."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "nums1 = [1, 2, 3, 1]\n",
-    "print(containsDuplicate(nums1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "nums2 = [1, 2, 3, 4]\n",
-    "print(containsDuplicate(nums2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "nums3 = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]\n",
-    "print(containsDuplicate(nums3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for the \"Contains Duplicate\" problem:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The primary operation that affects the time complexity is the `for` loop that iterates through the input array `nums`. In the worst case, the loop will iterate through all `n` elements of the array, where `n` is the length of the input array.\n",
-    "\n",
-    "- Iterating through the array: O(n)\n",
-    "\n",
-    "Inside the loop, we perform two operations:\n",
-    "\n",
-    "1. Checking whether an element exists in the `unique_elements` set (`if num in unique_elements`). This operation has an average time complexity of O(1) for a set.\n",
-    "2. Adding an element to the `unique_elements` set (`unique_elements.add(num)`), which also has an average time complexity of O(1) for a set.\n",
-    "\n",
-    "Since these operations are performed for each element in the array, the overall time complexity remains O(n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity is determined by the additional data structures used in the code, which are the `unique_elements` set.\n",
-    "\n",
-    "- `unique_elements` set: This set stores unique elements from the input array. In the worst case, if all elements in the input array are distinct, the set will store all `n` elements.\n",
-    "\n",
-    "Therefore, the space complexity is O(n) because, in the worst case, the set's size will grow linearly with the input array's size.\n",
-    "\n",
-    "**In summary**, the time complexity of the code is O(n), where `n` is the length of the input array, and the space complexity is also O(n) in the worst case. This code is efficient and meets the constraints of the problem."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **K Duplicates:** Modify the problem to return `true` if there are exactly K duplicate elements in the array. Write a function that takes an additional integer parameter K and returns `true` if there are exactly K duplicates.\n",
-    "2. **Frequency Count:** Write a function that returns a list of all the elements that appear more than once in the array, along with their frequencies."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb b/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb
deleted file mode 100644
index c5a69c0..0000000
--- a/_build/jupyter_execute/01. Array Hashing/238. Product of Array Except Self.ipynb	
+++ /dev/null
@@ -1,191 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 238. Product of Array Except Self\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Product of Array Except Self problem on LeetCode, click here!](https://leetcode.com/problems/product-of-array-except-self/)\n",
-    "\n",
-    "---\n",
-    "Given an integer array `nums`, return *an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.*\n",
-    "\n",
-    "The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-    "\n",
-    "You must write an algorithm that runs in O(n) time and without using the division operation.\n",
-    "\n",
-    "Constraints:\n",
-    "- 2 <= `nums.length` <= $10^5$\n",
-    "- `-30 <= nums[i] <= 30`\n",
-    "- The product of any prefix or suffix of `nums` is **guaranteed** to fit in a **32-bit** integer.\n",
-    "\n",
-    "**Follow-up:** Can you solve the problem in O(1) extra space complexity? (The output array **does not** count as extra space for space complexity analysis.)\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def productExceptSelf(nums):\n",
-    "    n = len(nums)\n",
-    "    \n",
-    "    # Initialize two arrays to store products to the left and right of each element\n",
-    "    left_products = [1] * n\n",
-    "    right_products = [1] * n\n",
-    "    \n",
-    "    # Calculate products to the left of each element\n",
-    "    left_product = 1\n",
-    "    for i in range(n):\n",
-    "        left_products[i] = left_product\n",
-    "        left_product *= nums[i]\n",
-    "    \n",
-    "    # Calculate products to the right of each element\n",
-    "    right_product = 1\n",
-    "    for i in range(n - 1, -1, -1):\n",
-    "        right_products[i] = right_product\n",
-    "        right_product *= nums[i]\n",
-    "    \n",
-    "    # Calculate the final answer using left and right products\n",
-    "    answer = [left_products[i] * right_products[i] for i in range(n)]\n",
-    "    \n",
-    "    return answer"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `n` is initialized as the length of the `nums` array.\n",
-    "2. Two arrays, `left_products` and `right_products`, are created, each with a length of `n` and filled with ones initially.\n",
-    "3. `left_product` is initialized to 1.\n",
-    "4. A loop iterates through the `nums` array, from left to right. For each element at index `i`, it stores the product of all elements to its left (including itself) in the `left_products` array and updates `left_product` accordingly.\n",
-    "5. `right_product` is initialized to 1.\n",
-    "6. Another loop iterates through the `nums` array in reverse order, from right to left. For each element at index `i`, it stores the product of all elements to its right (including itself) in the `right_products` array and updates `right_product`.\n",
-    "7. The final `answer` list is constructed by multiplying corresponding elements from the `left_products` and `right_products` arrays for each index `i`.\n",
-    "8. The function returns the `answer` list, which contains the desired results.\n",
-    "\n",
-    "Overall, this algorithm efficiently computes the product of all elements except the current element, as required. It runs in O(n) time complexity and uses O(n) extra space, which is within the problem's constraints."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[24, 12, 8, 6]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "nums = [1,2,3,4]\n",
-    "result1 = productExceptSelf(nums)\n",
-    "print(result1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[0, 0, 9, 0, 0]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums = [-1,1,0,-3,3]\n",
-    "result2 = productExceptSelf(nums)\n",
-    "print(result2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. The first loop iterates through the `nums` array to calculate the products to the left of each element. This loop runs in O(n) time, where n is the length of the `nums` array.\n",
-    "2. The second loop also iterates through the `nums` array in reverse order to calculate the products to the right of each element. This loop also runs in O(n) time.\n",
-    "3. The final loop constructs the `answer` list by multiplying elements from the `left_products` and `right_products` arrays. This loop runs in O(n) time.\n",
-    "\n",
-    "Since all the loops are independent and sequential, the overall time complexity is O(n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. Two additional arrays, `left_products` and `right_products`, are created with a length of `n`, where `n` is the length of the `nums` array. Therefore, these arrays consume O(n) extra space.\n",
-    "2. The `left_product` and `right_product` variables are used to keep track of the product to the left and right of each element, respectively. These variables occupy O(1) extra space.\n",
-    "3. The `answer` list, which contains the final results, also consumes O(n) space, but it's not counted toward the extra space complexity analysis as per the problem's constraints.\n",
-    "\n",
-    "Overall, the space complexity is O(n) for this solution."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Handle Zeros in the Input**: Modify the solution to handle cases where the input array contains zeros. For example, if `nums = [0, 1, 2, 3, 0]`, the output should be `[0, 0, 0, 0, 0]` because any element multiplied by zero results in zero.\n",
-    "\n",
-    "2. **In-Place Modification**: Attempt to solve the problem with in-place modification of the `nums` array, where the output is stored directly in the input array without using any additional space."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb b/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb
deleted file mode 100644
index 5649123..0000000
--- a/_build/jupyter_execute/01. Array Hashing/242. Valid Anagram.ipynb	
+++ /dev/null
@@ -1,192 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 242. Valid Anagram\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Valid Anagram problem on LeetCode, click here!](https://leetcode.com/problems/valid-anagram/)\n",
-    "\n",
-    "---\n",
-    "Given two strings `s` and `t`, return `true` if `t` is an anagram of `s`, and `false` otherwise.\n",
-    "\n",
-    "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `s.length`, `t.length` <= $5 * 10^4$\n",
-    "- `s` and `t` consist of lowercase English letters.\n",
-    "\n",
-    "**Follow-up:** What if the inputs contain Unicode characters? How would you adapt your solution to such a case?\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def isAnagram(s: str, t: str) -> bool:\n",
-    "    # Check if the lengths of s and t are not equal\n",
-    "    if len(s) != len(t):\n",
-    "        return False\n",
-    "\n",
-    "    # Create dictionaries to store character frequencies for s and t\n",
-    "    countS, countT = {}, {}\n",
-    "\n",
-    "    # Count character frequencies in s\n",
-    "    for i in range(len(s)):\n",
-    "        countS[s[i]] = 1 + countS.get(s[i], 0)\n",
-    "\n",
-    "    # Count character frequencies in t\n",
-    "    for i in range(len(t)):\n",
-    "        countT[t[i]] = 1 + countT.get(t[i], 0)\n",
-    "\n",
-    "    # Check if the character frequencies in s and t are the same\n",
-    "    return countS == countT"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The code defines a class called `Solution` with a method `isAnagram` that takes two input strings, `s` and `t`, and returns a boolean (`True` or `False`) based on whether `t` is an anagram of `s`.\n",
-    "\n",
-    "2. The first check performed in the method is whether the lengths of `s` and `t` are equal. If they are not equal, it immediately returns `False` because strings of different lengths cannot be anagrams of each other.\n",
-    "\n",
-    "3. Two dictionaries, `countS` and `countT`, are created to store the frequency of characters in `s` and `t`, respectively. These dictionaries will be used to count the occurrences of each character in the strings.\n",
-    "\n",
-    "4. The code then enters a loop to iterate through each character in string `s` using a `for` loop. Inside the loop, it updates the `countS` dictionary. The line `countS[s[i]] = 1 + countS.get(s[i], 0)` increments the count of character `s[i]` in `countS` by 1. If the character is not already in the dictionary, it initializes the count to 1.\n",
-    "\n",
-    "5. Similarly, the code iterates through each character in string `t` and updates the `countT` dictionary in the same way.\n",
-    "\n",
-    "6. After counting the character frequencies in both strings, the code compares the two dictionaries using `countS == countT`. If the dictionaries are equal, it means that both strings have the same character frequencies, and therefore, `t` is an anagram of `s`. In this case, the method returns `True`.\n",
-    "\n",
-    "7. If the dictionaries are not equal, the method returns `False`, indicating that `t` is not an anagram of `s`.\n",
-    "\n",
-    "In summary, this code efficiently determines whether two input strings are anagrams by counting the character frequencies in each string using dictionaries and then comparing these counts. If the character frequencies match, the strings are considered anagrams, and the method returns `True`; otherwise, it returns `False`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "s = \"anagram\"\n",
-    "t = \"nagaram\"\n",
-    "print(isAnagram(s, t))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "s = \"rat\"\n",
-    "t = \"car\"\n",
-    "print(isAnagram(s, t))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. The code first checks whether the lengths of the input strings `s` and `t` are equal, which takes constant time. This check has a time complexity of O(1).\n",
-    "\n",
-    "2. The code then iterates through both strings, `s` and `t`, once to count the character frequencies. Since both strings have a maximum length of 5 * 10^4, the worst-case scenario is that the code iterates through 5 * 10^4 characters in each string. This results in a linear time complexity of O(n), where n is the length of the longer of the two input strings (`s` or `t`).\n",
-    "\n",
-    "3. Finally, the code compares the two dictionaries `countS` and `countT` to check if the character frequencies match. This comparison takes O(n) time in the worst case, where n is the length of the longer string.\n",
-    "\n",
-    "Overall, the time complexity of the code is O(n), where n is the length of the longer input string.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. The code uses two dictionaries, `countS` and `countT`, to store the character frequencies of the input strings `s` and `t`. In the worst case, both dictionaries can contain all unique characters from the input strings. Since the input strings can have a maximum length of 5 * 10^4, the space complexity for these dictionaries is O(5 * 10^4) or simply O(n), where n is the length of the longer input string.\n",
-    "\n",
-    "2. The code uses a few additional variables for bookkeeping, but these variables have constant space requirements and do not depend on the input size. Therefore, they do not significantly impact the overall space complexity.\n",
-    "\n",
-    "**In summary**, the space complexity of the code is O(n), where n is the length of the longer input string."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Anagram Chains**: Given a list of words, find the longest chain of anagrams, where each word in the chain is an anagram of the previous word. For example, given [\"bat\", \"tab\", \"cat\", \"tac\", \"dog\"], the longest anagram chain is [\"bat\", \"tab\", \"cat\", \"tac\"].\n",
-    "\n",
-    "2. **Minimum Deletions to Make Anagrams**: Given two strings `s` and `t`, find the minimum number of character deletions required in both strings to make them anagrams of each other. For example, if `s = \"hello\"` and `t = \"billion\"`, you can remove the characters \"heo\" from `s` and \"billi\" from `t` to make them anagrams (\"lo\" and \"on\" are left in the two strings)."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb b/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb
deleted file mode 100644
index 1b11faf..0000000
--- a/_build/jupyter_execute/01. Array Hashing/347. Top K Frequent Elements.ipynb	
+++ /dev/null
@@ -1,209 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 347. Top K Frequent Elements\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Top K Frequent Elements problem on LeetCode, click here!](https://leetcode.com/problems/top-k-frequent-elements/)\n",
-    "\n",
-    "---\n",
-    "Given an integer array `nums` and an integer `k`, return the `k` most frequent elements. You may return the answer in **any order**.\n",
-    "\n",
-    "Constraints:\n",
-    "- 1 <= `nums.length` <= $10^5$\n",
-    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-    "- `k` is in the range `[1, the number of unique elements in the array]`.\n",
-    "- It is **guaranteed** that the answer is **unique**.\n",
-    "\n",
-    "**Follow-up:** Your algorithm's time complexity must be better than $O(n\\ log\\ n)$, where n is the array's size.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def topKFrequent(nums, k):\n",
-    "    # Create a dictionary to store the frequency of each element\n",
-    "    count = {}\n",
-    "    # Create a list of lists to store elements with the same frequency\n",
-    "    frequency = [[] for i in range(len(nums) + 1)]\n",
-    "\n",
-    "    # Count the frequency of each element in nums\n",
-    "    for n in nums:\n",
-    "        count[n] = 1 + count.get(n, 0)\n",
-    "\n",
-    "    # Place elements in the freq list according to their frequency\n",
-    "    for n, c in count.items():\n",
-    "        frequency[c].append(n)\n",
-    "\n",
-    "    res = []\n",
-    "    # Traverse freq list from the end (higher frequencies)\n",
-    "    for i in range(len(frequency) - 1, 0, -1):\n",
-    "        for n in frequency[i]:\n",
-    "            res.append(n)\n",
-    "            if len(res) == k:\n",
-    "                return res"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `def topKFrequent(nums, k):`: This is a function that takes two arguments: `nums`, which is the input array of integers, and `k`, which is the number of most frequent elements to return.\n",
-    "\n",
-    "2. `count = {}`: This dictionary `count` will be used to store the frequency of each unique element in the `nums` array. The keys are elements from the input array, and the values are their corresponding frequencies.\n",
-    "\n",
-    "3. `frequency = [[] for i in range(len(nums) + 1)]`: This creates a list of empty lists called `frequency`. It's used to store elements based on their frequencies, similar to the `freq` list in the previous code. The size of this list is set to be one greater than the length of the input `nums`.\n",
-    "\n",
-    "4. `for n in nums:`: This loop iterates through each element `n` in the input `nums` array.\n",
-    "\n",
-    "5. `count[n] = 1 + count.get(n, 0)`: This line counts the frequency of each element `n` in the `nums` array. It uses the `count.get(n, 0)` method to retrieve the current count of `n` from the `count` dictionary. If `n` is not in the dictionary, it defaults to 0. It then increments the count by 1.\n",
-    "\n",
-    "6. After the above loop, the `count` dictionary will contain counts of each unique element in the `nums` array.\n",
-    "\n",
-    "7. `for n, c in count.items():`: This loop iterates through the items (key-value pairs) of the `count` dictionary. `n` represents the element, and `c` represents its frequency.\n",
-    "\n",
-    "8. `frequency[c].append(n)`: This line places the element `n` into the bucket corresponding to its frequency `c`. Buckets are represented by the `frequency` list. For example, if an element `n` has a frequency of 3, it will be added to `frequency[3]`.\n",
-    "\n",
-    "9. After this loop, the `frequency` list will contain buckets of elements grouped by their frequencies.\n",
-    "\n",
-    "10. `res = []`: This list `res` will be used to store the k most frequent elements.\n",
-    "\n",
-    "11. `for i in range(len(freq) - 1, 0, -1):`: This loop iterates in reverse order through the `frequency` list, starting from the highest frequency and going down to 1. Note that there is a typo here; it should be `frequency` instead of `freq`.\n",
-    "\n",
-    "12. `for n in frequency[i]:`: For each element `n` in the current bucket (i.e., elements with frequency `i`), it appends `n` to the result list `res`.\n",
-    "\n",
-    "13. `if len(res) == k:`: Once `res` contains k elements, the code exits and returns the result.\n",
-    "\n",
-    "The code efficiently finds the k most frequent elements in the input array without using sorting, similar to the previous explanation, with the only difference being the variable names (e.g., `frequency` instead of `freq`)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 2]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "nums = [1,1,1,2,2,3]\n",
-    "k = 2\n",
-    "result1 = topKFrequent(nums, k)\n",
-    "print(result1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums = [1]\n",
-    "k = 1\n",
-    "result2 = topKFrequent(nums, k)\n",
-    "print(result2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. The first loop that counts the frequency of elements by iterating through `nums` has a time complexity of O(n), where n is the number of elements in the input array `nums`.\n",
-    "\n",
-    "2. The second loop iterates through the keys in the `count` dictionary, which has at most `k` unique elements (as per the constraints). Therefore, the second loop also has a time complexity of O(k).\n",
-    "\n",
-    "3. The third loop iterates through the `frequency` list, which has a length equal to the maximum frequency of elements in `nums`. In the worst case, this loop can have a time complexity of O(n), where n is the number of elements in `nums`.\n",
-    "\n",
-    "Overall, the dominant factor in terms of time complexity is the loop that iterates through the `count` dictionary. So, the total time complexity of the code is O(n + k).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The `count` dictionary stores the frequency of each unique element in `nums`. In the worst case, it can have at most `k` unique elements, so the space complexity for `count` is O(k).\n",
-    "\n",
-    "2. The `frequency` list is used to store elements grouped by their frequencies. It has a length of `len(nums) + 1`, which can be at most 105 based on the constraints. Therefore, the space complexity for `frequency` is O(105).\n",
-    "\n",
-    "3. The `res` list stores the k most frequent elements, which can be at most `k` elements, so the space complexity for `res` is O(k).\n",
-    "\n",
-    "**In summary**, the space complexity is dominated by the `count` dictionary and the `frequency` list, both of which have a space complexity of O(k + 105)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimized k Most Frequent Elements:** Modify the code to find the k most frequent elements in an array while ensuring that the time complexity is O(n + klogk). You can use a priority queue (heap) to achieve this.\n",
-    "\n",
-    "2. **Handling Duplicate Frequencies:** Extend the code to handle cases where multiple elements have the same frequency and are among the k most frequent elements. Ensure that the output contains exactly k elements."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb b/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb
deleted file mode 100644
index 256e8f7..0000000
--- a/_build/jupyter_execute/01. Array Hashing/49. Group Anagrams.ipynb	
+++ /dev/null
@@ -1,221 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 49. Group Anagrams\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Group Anagrams problem on LeetCode, click here!](https://leetcode.com/problems/group-anagrams/)\n",
-    "\n",
-    "---\n",
-    "Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.\n",
-    "\n",
-    "An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `strs.length <= $10^4$\n",
-    "- `0 <= strs[i].length <= 100`\n",
-    "- `strs[i]` consists of lowercase English letters.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from collections import defaultdict\n",
-    "\n",
-    "\n",
-    "def groupAnagrams(strs):\n",
-    "    ans = defaultdict(list)\n",
-    "\n",
-    "    # Iterate through the list of input strings\n",
-    "    for s in strs:\n",
-    "        # Initialize a list to represent character counts for each character (a-z)\n",
-    "        count = [0] * 26\n",
-    "\n",
-    "        # Count the occurrences of each character in the current word\n",
-    "        for c in s:\n",
-    "            count[ord(c) - ord(\"a\")] += 1\n",
-    "\n",
-    "        # Use a tuple of character counts as the key and append the word to the anagram group\n",
-    "        ans[tuple(count)].append(s)\n",
-    "\n",
-    "    # Convert the values (lists of anagrams) to a list of lists\n",
-    "    return list(ans.values())"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. Inside the method, a `defaultdict` called `ans` is created to store anagram groups. This dictionary will have a list as its default value, meaning that each key in the dictionary will be associated with an empty list by default.\n",
-    "\n",
-    "2. The code then iterates through the list of input strings, `strs`, using a `for` loop. For each word `s` in `strs`, it performs the following steps:\n",
-    "\n",
-    "   - It initializes a list `count` of length 26, where each element represents the count of a specific character (a-z) in the word `s`.\n",
-    "\n",
-    "   - The code then iterates through the characters of `s`. For each character `c`, it increments the corresponding count in the `count` list based on its ASCII value.\n",
-    "\n",
-    "   - After counting the characters in `s`, it converts the `count` list into a tuple `tuple(count)` to use as a key for the `ans` dictionary.\n",
-    "\n",
-    "   - It appends the word `s` to the list associated with the key (tuple) in the `ans` dictionary. This groups all anagrams of the same word together under the same key.\n",
-    "\n",
-    "3. After processing all words in `strs`, the code converts the values of the `ans` dictionary (which are lists of anagrams) to a list of lists using the `list()` constructor.\n",
-    "\n",
-    "4. Finally, the code returns the list of anagram groups, which is the result of grouping the anagrams in the input list.\n",
-    "\n",
-    "In summary, the code efficiently groups anagrams together by counting the characters in each word and using a dictionary to store them under the same key. The result is a list of lists, where each inner list represents a group of anagrams."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: Group anagrams from a list of words\n",
-    "strs1 = [\"eat\", \"tea\", \"tan\", \"ate\", \"nat\", \"bat\"]\n",
-    "result1 = groupAnagrams(strs1)\n",
-    "print(result1)\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[['']]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2: Group anagrams from a list with an empty string\n",
-    "strs2 = [\"\"]\n",
-    "result2 = groupAnagrams(strs2)\n",
-    "print(result2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "c8367e2a-e7d5-4e18-8235-2870e341a04b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[['a']]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3: Group anagrams from a list with a single word\n",
-    "strs3 = [\"a\"]\n",
-    "result3 = groupAnagrams(strs3)\n",
-    "print(result3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `groupAnagrams` method:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. First, we create an instance of the `Solution` class, which is an O(1) operation.\n",
-    "\n",
-    "2. Inside the `groupAnagrams` method, we iterate through the list of input strings `strs` once in a loop:\n",
-    "   - For each word in `strs`, we perform character counting, which is done in O(K) time, where K is the maximum length of a word in the list.\n",
-    "\n",
-    "3. The overall time complexity of the method is O(N * K), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. We use a `defaultdict` called `ans` to store the anagram groups. In the worst case, each word belongs to a distinct anagram group, so the space complexity for `ans` is O(N).\n",
-    "\n",
-    "2. Within the loop, we create a `count` list of length 26 to store character counts for each word. This is a fixed-size list and does not depend on the input size, so it has a constant space complexity of O(26), which is effectively O(1).\n",
-    "\n",
-    "3. The space used for other variables is also constant and does not depend on the input size.\n",
-    "\n",
-    "4. The final result, `result`, is a list of lists that contains the grouped anagrams. In the worst case, each word is a distinct anagram group, so the space complexity for `result` is O(N).\n",
-    "\n",
-    "5. The overall space complexity of the method is O(N) due to the space used by `ans` and `result`. The space used by the `count` list and other variables is constant and does not significantly affect the overall space complexity.\n",
-    "\n",
-    "**In summary,** the time complexity of the `groupAnagrams` method is O(N * K), and the space complexity is O(N), where N is the number of words in the input list `strs`, and K is the maximum length of a word in `strs`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Anagramic Palindromes**: Given a list of strings, find all the groups of anagrams where each group contains words that can be rearranged into a palindrome. For example, in the list [\"bat\", \"tab\", \"act\", \"tac\", \"dog\"], there are two groups: [\"bat\", \"tab\", \"act\", \"tac\"] and [\"dog\"].\n",
-    "\n",
-    "2. **Largest Anagram Group**: Given a list of strings, find the largest group of anagrams. Return the list of anagrams in that group. If there are multiple largest groups, return all of them."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb b/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb
deleted file mode 100644
index 2fae2a5..0000000
--- a/_build/jupyter_execute/01. Array Hashing/659. Encode and Decode Strings.ipynb	
+++ /dev/null
@@ -1,198 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 659· Encode and Decode Strings\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Encode and Decode Strings problem on LintCode, click here!](https://www.lintcode.com/problem/659/)\n",
-    "\n",
-    "---\n",
-    "Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.\n",
-    "\n",
-    "Please implement `encode` and `decode`.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class Codec:\n",
-    "\n",
-    "    def encode(self, strs):\n",
-    "        \"\"\"Encodes a list of strings to a single string.\n",
-    "        \n",
-    "        :type strs: List[str]\n",
-    "        :rtype: str\n",
-    "        \"\"\"\n",
-    "        encoded = \"\"\n",
-    "        for s in strs:\n",
-    "            encoded += s.replace(':', '::') + ':;'\n",
-    "        return encoded\n",
-    "\n",
-    "    def decode(self, s):\n",
-    "        \"\"\"Decodes a single string to a list of strings.\n",
-    "        \n",
-    "        :type s: str\n",
-    "        :rtype: List[str]\n",
-    "        \"\"\"\n",
-    "        decoded = []\n",
-    "        i = 0\n",
-    "        while i < len(s):\n",
-    "            end = s.find(':;', i)\n",
-    "            if end == -1:\n",
-    "                end = len(s)\n",
-    "            decoded.append(s[i:end].replace('::', ':'))\n",
-    "            i = end + 2\n",
-    "        return decoded"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We define a class called `Codec` to encapsulate the encoding and decoding operations for a list of strings.\n",
-    "\n",
-    "2. In the `encode` method, we take a list of strings (`strs`) as input and return a single encoded string. The encoding process involves concatenating the strings together with a delimiter `:;`. We iterate through each string in the input list, replace any occurrence of `:` with `::` (to avoid conflicts with the delimiter), and then append `:;` to indicate the end of that string.\n",
-    "\n",
-    "3. In the `decode` method, we take an encoded string (`s`) as input and return a list of strings. The decoding process involves splitting the encoded string using the `:;` delimiter. We start from the beginning of the encoded string (`i = 0`) and repeatedly find the next occurrence of `:;`. We extract the substring between the current position (`i`) and the next delimiter (`end`). Before adding it to the result list, we replace any `::` with `:` to revert the encoding. We then update the current position `i` to be after the delimiter, so we can find the next substring in the next iteration.\n",
-    "\n",
-    "4. Finally, we create an instance of the `Codec` class and test it with two examples:\n",
-    "\n",
-    "By using the `:;` delimiter and handling the `::` escaping, this code can encode and decode lists of strings, preserving the original content even if it contains colons or the delimiter itself."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "['lint', 'code', 'love', 'you']\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "codec = Codec()\n",
-    "\n",
-    "input1 = [\"lint\", \"code\", \"love\", \"you\"]\n",
-    "encoded1 = codec.encode(input1)\n",
-    "decoded1 = codec.decode(encoded1)\n",
-    "print(decoded1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "['we', 'say', ':', 'yes']\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "codec = Codec() \n",
-    "\n",
-    "input2 = [\"we\", \"say\", \":\", \"yes\"]\n",
-    "encoded2 = codec.encode(input2)\n",
-    "decoded2 = codec.decode(encoded2)\n",
-    "print(decoded2)  "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `encode` and `decode` methods in the provided code.\n",
-    "\n",
-    "**Encode Method (`encode`):**\n",
-    "\n",
-    "- Time Complexity: O(n * m)\n",
-    "  - Here, `n` is the number of strings in the input list `strs`, and `m` is the average length of these strings.\n",
-    "  - In the worst case, for each string in the list, we iterate over its characters once to replace any colons (`:`) with double colons (`::`) and then append `:;`. This is done for each string in the list.\n",
-    "\n",
-    "- Space Complexity: O(n * m)\n",
-    "  - The space complexity is also O(n * m) because we build the encoded string by concatenating the modified strings. In the worst case, the encoded string can be of the same length as the original strings.\n",
-    "\n",
-    "**Decode Method (`decode`):**\n",
-    "\n",
-    "- Time Complexity: O(n * m)\n",
-    "  - Similar to the encode method, we iterate through the encoded string in the decode method. In the worst case, we may have to scan each character in the encoded string to find the `:;` delimiter.\n",
-    "\n",
-    "- Space Complexity: O(n * m)\n",
-    "  - The space complexity of the decode method is also O(n * m) because we build the list of decoded strings, which can be of the same length as the encoded string.\n",
-    "\n",
-    "Overall, both the `encode` and `decode` methods have time and space complexities of O(n * m), where `n` is the number of strings in the input list, and `m` is the average length of these strings. The space complexity arises from storing the encoded or decoded strings, and the time complexity arises from iterating through the characters in these strings."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple Delimiters**: Modify the code to allow for multiple delimiters, not just one. Each string could have its own delimiter, and the code should be able to handle this complexity.\n",
-    "\n",
-    "2. **Optimize Encoding**: Modify the `encode` method to achieve a more efficient encoding algorithm. Try to minimize the length of the encoded string while ensuring that it can be correctly decoded."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb b/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb
deleted file mode 100644
index e330eb0..0000000
--- a/_build/jupyter_execute/02. Two Pointers/11. Container With Most Water.ipynb	
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 11. Container With Most Water\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Container With Most Water problem on LeetCode, click here!](https://leetcode.com/problems/container-with-most-water/)\n",
-    "\n",
-    "---\n",
-    "You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the $i^{th}$ line are `(i, 0)` and `(i, height[i])`.\n",
-    "\n",
-    "Find two lines that together with the x-axis form a container, such that the container contains the most water.\n",
-    "\n",
-    "Return *the maximum amount of water a container can store.*\n",
-    "\n",
-    "**Notice** that you may not slant the container.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `n == height.length`\n",
-    "- 2 <= `n` <= $10^5$\n",
-    "- 0 <= `height[i]` <= $10^4$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def maxArea(height):\n",
-    "    # Initialize the maximum area to 0.\n",
-    "    max_area = 0\n",
-    "    # Initialize two pointers, one at the beginning (left) and one at the end (right) of the array.\n",
-    "    left = 0\n",
-    "    right = len(height) - 1\n",
-    "\n",
-    "    # Iterate until the left pointer is less than the right pointer.\n",
-    "    while left < right:\n",
-    "        # Calculate the height of the container, which is the minimum height of the two lines.\n",
-    "        h = min(height[left], height[right])\n",
-    "        # Calculate the width of the container, which is the distance between the two pointers.\n",
-    "        width = right - left\n",
-    "        # Calculate the area of the container using height and width and update max_area if it's greater.\n",
-    "        max_area = max(max_area, h * width)\n",
-    "\n",
-    "        # Move the pointer with the shorter line inward.\n",
-    "        if height[left] < height[right]:\n",
-    "            left += 1\n",
-    "        else:\n",
-    "            right -= 1\n",
-    "\n",
-    "    # Continue this process until left < right, and then return the maximum area.\n",
-    "    return max_area"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start with initializing `max_area` to 0, which will store the maximum area of water that can be contained by the lines.\n",
-    "\n",
-    "2. Two pointers, `left` and `right`, are initialized to the beginning and end of the input `height` array.\n",
-    "\n",
-    "3. The main loop runs while `left` is less than `right`. This ensures that we are checking all possible pairs of lines in a systematic way.\n",
-    "\n",
-    "4. Inside the loop:\n",
-    "   - We calculate the height of the container `h` as the minimum of the heights at the `left` and `right` pointers. This represents the height of the water the container can hold.\n",
-    "   - We calculate the width of the container as the difference between `right` and `left`. This represents the distance between the two lines.\n",
-    "   - We calculate the area of the container using `h * width` and update `max_area` if the calculated area is greater than the current `max_area`.\n",
-    "\n",
-    "5. After calculating the area and updating `max_area`, we move one of the pointers. We move the pointer pointing to the shorter line inward because moving the pointer pointing to the taller line won't increase the height and will only decrease the width, which will result in a smaller area.\n",
-    "\n",
-    "6. We continue this process until `left` is no longer less than `right`, indicating that we have checked all possible pairs of lines.\n",
-    "\n",
-    "7. Finally, we return the `max_area` which contains the maximum area of water that can be contained by two lines from the input array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output: 49\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "height1 = [1, 8, 6, 2, 5, 4, 8, 3, 7]\n",
-    "result1 = maxArea(height1)\n",
-    "print(\"Output:\", result1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output: 1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "height2 = [1, 1]\n",
-    "result2 = maxArea(height2)\n",
-    "print(\"Output:\", result2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Output: 45\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Additional Example\n",
-    "height3 = [3, 9, 3, 4, 7, 2, 12, 6]\n",
-    "result3 = maxArea(height3)\n",
-    "print(\"Output:\", result3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `maxArea` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The function uses a two-pointer approach to find the maximum area of water. In the worst case, both the left and right pointers traverse the entire input array once. Therefore, the time complexity is O(n), where n is the number of elements in the `height` array.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The function uses a constant amount of extra space for variables like `max_area`, `left`, `right`, `h`, and `width`. It does not use any data structures whose space consumption depends on the input size. As a result, the space complexity is O(1), which means it has constant space complexity.\n",
-    "\n",
-    "**In summary,** the `maxArea` function has a time complexity of O(n) and a space complexity of O(1), making it an efficient algorithm for finding the maximum area of water that can be contained by two lines in the input array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the maxArea function to not only return the maximum area but also the indices of the two lines that form the container with the maximum area.\n",
-    "2. **Variant Problem:** Instead of finding the maximum area of water, find the minimum amount of water required to fill all the gaps between the lines. You'll need to return the total water volume needed."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb b/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb
deleted file mode 100644
index 12c7daa..0000000
--- a/_build/jupyter_execute/02. Two Pointers/125. Valid Palindrome.ipynb	
+++ /dev/null
@@ -1,210 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 125. Valid Palindrome\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Valid Palindrome problem on LeetCode, click here!](https://leetcode.com/problems/valid-palindrome/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "A phrase is a **palindrome** if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.\n",
-    "\n",
-    "Given a string `s`, *return `true` if it is a **palindrome**, or `false` otherwise.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `s.length` <= $2*10^5$\n",
-    "- `s` consists only of printable ASCII characters.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def isPalindrome(s):\n",
-    "    # Step 1: Convert the input string to lowercase\n",
-    "    s = s.lower()\n",
-    "    \n",
-    "    # Step 2: Remove all non-alphanumeric characters from the string\n",
-    "    cleaned_s = ''.join(c for c in s if c.isalnum())\n",
-    "    \n",
-    "    # Step 3: Initialize two pointers, one at the beginning and one at the end of the cleaned string\n",
-    "    left, right = 0, len(cleaned_s) - 1\n",
-    "    \n",
-    "    # Step 4: Compare characters using the two pointers while moving them towards each other\n",
-    "    while left < right:\n",
-    "        if cleaned_s[left] != cleaned_s[right]:\n",
-    "            return False  # If characters don't match, it's not a palindrome\n",
-    "        left += 1\n",
-    "        right -= 1\n",
-    "    \n",
-    "    # Step 5: If the loop completes without returning False, it's a palindrome\n",
-    "    return True"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The `isPalindrome` function takes an input string `s`.\n",
-    "\n",
-    "2. It starts by converting the input string to lowercase using `s.lower()`. This step ensures that the function is case-insensitive when checking for palindromes.\n",
-    "\n",
-    "3. Next, it removes all non-alphanumeric characters from the string and stores the result in the `cleaned_s` variable. It does this by iterating through each character in the input string and only keeping characters that are alphanumeric (letters and digits). This step removes spaces, punctuation, and other non-alphanumeric characters.\n",
-    "\n",
-    "4. The function then initializes two pointers, `left` and `right`, which will be used to compare characters from the beginning and end of the cleaned string.\n",
-    "\n",
-    "5. In a while loop, it compares characters using the two pointers while moving them towards each other. The loop continues until the `left` pointer is less than the `right` pointer. This is done to check all characters in the cleaned string.\n",
-    "\n",
-    "6. Inside the loop, it checks if the characters at the `left` and `right` pointers don't match. If they don't match, it means the input is not a palindrome, so the function returns `False`.\n",
-    "\n",
-    "7. If the characters at the `left` and `right` pointers match, the pointers are moved closer to each other by incrementing `left` and decrementing `right`.\n",
-    "\n",
-    "8. The loop continues to compare characters until it either finds a mismatch (returning `False`) or until the `left` pointer is greater than or equal to the `right` pointer. If the loop completes without finding a mismatch, it means the input is a palindrome, so the function returns `True`.\n",
-    "\n",
-    "In summary, the code converts the input string to lowercase, removes non-alphanumeric characters, and then uses two pointers to compare characters from the cleaned string's beginning and end. If all characters match during the comparison, the function returns `True`, indicating that the input is a palindrome. If any characters do not match, it returns `False`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "input1 = \"A man, a plan, a canal: Panama\"\n",
-    "output1 = isPalindrome(input1)\n",
-    "print(output1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "input2 = \"race a car\"\n",
-    "output2 = isPalindrome(input2)\n",
-    "print(output2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "input3 = \"\"\n",
-    "output3 = isPalindrome(input3)\n",
-    "print(output3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `isPalindrome` function that uses two pointers:\n",
-    "\n",
-    "1. **Time Complexity:**\n",
-    "   - Converting the input string to lowercase using `s.lower()` takes O(n) time, where 'n' is the length of the input string.\n",
-    "   - Removing non-alphanumeric characters using the list comprehension `''.join(c for c in s if c.isalnum())` also takes O(n) time because it iterates through each character in the string once.\n",
-    "   - The comparison of characters using the two pointers (`left` and `right`) is done in a loop that runs until `left` is less than `right`. In the worst case, the loop iterates through half of the string, so it takes O(n/2) time, which is still considered O(n).\n",
-    "\n",
-    "   Therefore, the overall time complexity of the `isPalindrome` function is O(n), where 'n' is the length of the input string.\n",
-    "\n",
-    "2. **Space Complexity:**\n",
-    "   - The space complexity is primarily determined by the additional string `cleaned_s` created to store the cleaned version of the input string. In the worst case, if all characters in the input string are alphanumeric, this cleaned string can be as large as the original input string.\n",
-    "   - Additionally, the function uses two pointers, `left` and `right`, which consume constant space and do not depend on the size of the input.\n",
-    "\n",
-    "**In summary,** the overall space complexity of the `isPalindrome` function is O(n) in the worst case, where 'n' is the length of the input string. The space complexity is dominated by the space required for `cleaned_s`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Longest Palindromic Substring:** Given a string `s`, find and return the longest palindromic substring within `s`. For example, if `s = \"babad\"`, the function should return `\"bab\"` or `\"aba\"`.\n",
-    "2. **Palindrome with Limited Characters:** Given a string `s` and a list of characters `allowed_chars`, check if `s` is a palindrome considering only the characters from `allowed_chars`. Ignore all other characters. For example, if `s = \"A man, a plan, a canal: Panama\"` and `allowed_chars = ['a', 'm', 'n', 'p']`, the function should return `True`."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb b/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb
deleted file mode 100644
index 15d189a..0000000
--- a/_build/jupyter_execute/02. Two Pointers/15. 3Sum.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 15. 3Sum\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the 3Sum problem on LeetCode, click here!](https://leetcode.com/problems/3sum/)\n",
-    "\n",
-    "---\n",
-    "Given an integer array nums, *return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.*\n",
-    "\n",
-    "**Notice** that the solution set must not contain duplicate triplets.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `3 <= nums.length <= 3000`\n",
-    "- $-10^5$ <= `nums[i]` <= $10^5$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def threeSum(nums):\n",
-    "    # Sort the input array in ascending order\n",
-    "    nums.sort()\n",
-    "    triplets = []\n",
-    "    \n",
-    "    for i in range(len(nums) - 2):\n",
-    "        # Skip duplicates to avoid duplicate triplets\n",
-    "        if i > 0 and nums[i] == nums[i - 1]:\n",
-    "            continue\n",
-    "        \n",
-    "        left, right = i + 1, len(nums) - 1\n",
-    "        \n",
-    "        while left < right:\n",
-    "            # Calculate the total sum of the current triplet\n",
-    "            total = nums[i] + nums[left] + nums[right]\n",
-    "            \n",
-    "            if total == 0:\n",
-    "                # If the total is zero, we found a valid triplet\n",
-    "                triplets.append([nums[i], nums[left], nums[right]])\n",
-    "                \n",
-    "                # Skip duplicates of left and right pointers\n",
-    "                while left < right and nums[left] == nums[left + 1]:\n",
-    "                    left += 1\n",
-    "                while left < right and nums[right] == nums[right - 1]:\n",
-    "                    right -= 1\n",
-    "                \n",
-    "                # Move the pointers to the next unique elements\n",
-    "                left += 1\n",
-    "                right -= 1\n",
-    "            elif total < 0:\n",
-    "                # If the total is negative, we need to increase the sum by moving the left pointer to the right\n",
-    "                left += 1\n",
-    "            else:\n",
-    "                # If the total is positive, we need to decrease the sum by moving the right pointer to the left\n",
-    "                right -= 1\n",
-    "    \n",
-    "    return triplets"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. It starts by sorting the input array `nums` in ascending order. Sorting helps in efficiently finding triplets that sum up to zero.\n",
-    "\n",
-    "2. It initializes an empty list `triplets` to store the unique triplets that meet the given conditions.\n",
-    "\n",
-    "3. The code then iterates through the `nums` array using a loop with index `i`. This loop will consider each element of the array as a potential starting point for a triplet.\n",
-    "\n",
-    "4. Inside the loop, it checks for duplicates and skips them. This is done to ensure that the solution set does not contain duplicate triplets. If `nums[i]` is the same as the previous element `nums[i-1]`, it continues to the next iteration of the loop.\n",
-    "\n",
-    "5. Two pointers, `left` and `right`, are initialized. `left` starts just after the current element `nums[i]`, and `right` starts at the end of the array.\n",
-    "\n",
-    "6. The code enters another loop with `left` and `right` pointers, trying to find a triplet that sums up to zero.\n",
-    "\n",
-    "7. It calculates the `total` sum of the current triplet as `nums[i] + nums[left] + nums[right]`.\n",
-    "\n",
-    "8. If `total` is zero, it means a valid triplet is found, so it appends `[nums[i], nums[left], nums[right]]` to the `triplets` list. Then, it moves both the `left` and `right` pointers to their next unique elements while skipping duplicates.\n",
-    "\n",
-    "9. If `total` is less than zero, it means the sum is negative, so it increments the `left` pointer to move towards a larger sum.\n",
-    "\n",
-    "10. If `total` is greater than zero, it means the sum is positive, so it decrements the `right` pointer to move towards a smaller sum.\n",
-    "\n",
-    "11. This process continues until all possible triplets have been considered.\n",
-    "\n",
-    "12. Finally, the function returns the `triplets` list, which contains all the unique triplets that sum up to zero in the sorted `nums` array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[-1, -1, 2], [-1, 0, 1]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "nums = [-1, 0, 1, 2, -1, -4]\n",
-    "result = threeSum(nums)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "nums = [0, 1, 1]\n",
-    "result = threeSum(nums)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e191cf7e-95f9-4e5c-a99b-146c2a779c0b",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[0, 0, 0]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "nums = [0, 0, 0]\n",
-    "result = threeSum(nums)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `threeSum` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. Sorting the input array `nums` takes O(n log n) time, where n is the length of the array.\n",
-    "\n",
-    "2. The outer loop runs for each element of the array, so it iterates O(n) times.\n",
-    "\n",
-    "3. Inside the outer loop, we have a while loop with two pointers (`left` and `right`). In the worst case, the while loop can iterate O(n) times (when all elements are unique).\n",
-    "\n",
-    "4. Inside the while loop, we have constant time operations (additions, comparisons, and list appends).\n",
-    "\n",
-    "Overall, the time complexity of the `threeSum` function is dominated by the sorting step, so it is O(n log n) due to the sorting. The other operations inside the loops contribute linearly but are dominated by the sorting step.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The space used by the `triplets` list to store the output triplets is O(k), where k is the number of unique triplets that sum up to zero. In the worst case, this can be O(n^2/3) because there can be roughly O(n^2) possible triplets, and many of them may sum up to zero.\n",
-    "\n",
-    "2. Other than the `triplets` list, the function uses only a constant amount of additional space for variables like `i`, `left`, `right`, and `total`.\n",
-    "\n",
-    "Overall, the space complexity of the `threeSum` function is O(k), where k is the number of unique triplets that meet the criteria. It's important to note that the space complexity does not depend on the size of the input array but rather on the number of valid triplets found.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: O(n log n) due to sorting (dominant factor) and O(n^2) in the worst case for the nested loops.\n",
-    "- Space Complexity: O(k) for storing the output triplets, where k is the number of unique triplets that meet the criteria."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **K-Sum Problem:** Generalize the problem to the K-Sum problem, where instead of finding triplets that sum to zero, you need to find K elements that sum to a target value. Implement a function for this generalization.\n",
-    "2. **Count the Number of Unique Triplets:** Modify the algorithm to count the number of unique triplets that satisfy the conditions instead of returning the triplets themselves. This will require some modifications to handle duplicates efficiently."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb b/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb
deleted file mode 100644
index 451a21d..0000000
--- a/_build/jupyter_execute/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb	
+++ /dev/null
@@ -1,185 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 121. Best Time to Buy and Sell Stock\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Best Time to Buy and Sell Stock problem on LeetCode, click here!](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given an array `prices` where `prices[i]` is the price of a given stock on the $i^{th}$ day.\n",
-    "\n",
-    "You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.\n",
-    "\n",
-    "*Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `prices.length` <= $10^5$\n",
-    "- 0 <= `prices[i]` <= $10^4$\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def maxProfit(prices):\n",
-    "    if not prices:\n",
-    "        return 0\n",
-    "    \n",
-    "    min_price = prices[0]\n",
-    "    max_profit = 0\n",
-    "    \n",
-    "    for price in prices:\n",
-    "        # Update the minimum price if a lower price is encountered\n",
-    "        min_price = min(min_price, price)\n",
-    "        # Calculate the profit if selling at the current price\n",
-    "        max_profit = max(max_profit, price - min_price)\n",
-    "    \n",
-    "    return max_profit"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "- The `maxProfit` function takes a single argument, `prices`, which is a list of stock prices.\n",
-    "- The first line of the function checks if the `prices` list is empty. If it's empty, there are no prices to analyze, so the function returns 0 (no profit can be made).\n",
-    "\n",
-    "- Then, we initialize two variables:\n",
-    "  - `min_price`: This variable will keep track of the minimum price seen so far. We initialize it with the price of the first day in the list (`prices[0]`).\n",
-    "  - `max_profit`: This variable will keep track of the maximum profit we can achieve. We initialize it to 0 because initially, we haven't made any profit.\n",
-    "\n",
-    "- We start iterating through the `prices` list, where `price` represents the stock price for the current day.\n",
-    "\n",
-    "- In each iteration, we check if the current `price` is lower than the `min_price` we've seen so far. If it is, we update `min_price` to the current `price`. This is because we want to buy the stock at the lowest possible price.\n",
-    "\n",
-    "- Next, we calculate the profit we would make if we sold the stock at the current `price` (assuming we bought it at `min_price`). We do this by subtracting `min_price` from the current `price`. This represents the profit for the current day.\n",
-    "\n",
-    "- We also use the `max` function to keep track of the maximum profit seen so far. If the profit calculated for the current day is greater than the `max_profit` we've seen previously, we update `max_profit` with this higher value.\n",
-    "\n",
-    "- Finally, after iterating through all the days in the `prices` list, we return `max_profit`, which represents the maximum profit that can be achieved by buying and selling a single stock."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "5\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "prices1 = [7, 1, 5, 3, 6, 4]\n",
-    "print(maxProfit(prices1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "0\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "prices2 = [7, 6, 4, 3, 1]\n",
-    "print(maxProfit(prices2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The code iterates through the `prices` array once, examining each price exactly once.\n",
-    "- Within the loop, there are constant-time operations such as comparisons, additions, and subtractions.\n",
-    "- Therefore, the time complexity of this code is O(n), where n is the length of the `prices` array. It performs a linear number of operations relative to the size of the input.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity of the code is constant, O(1).\n",
-    "- Regardless of the size of the input `prices` array, the code only uses a fixed amount of additional memory to store two variables (`min_price` and `max_profit`) and a few loop variables.\n",
-    "- The memory usage does not depend on the size of the input, so it is considered constant space complexity.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: O(n) where n is the length of the `prices` array.\n",
-    "- Space Complexity: O(1) (constant space usage)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple Transactions Allowed:** Modify the problem to allow multiple transactions (buy and sell multiple times), but you must sell before buying again. Find the maximum profit in this scenario.\n",
-    "2. **K Transactions Allowed:** Extend the problem to allow at most k transactions. Find the maximum profit considering this constraint.\n",
-    "3. **With Transaction Fee:** Introduce a transaction fee for each buy/sell operation. Modify the code to maximize profit while considering the transaction fee."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb b/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb
deleted file mode 100644
index ded03f3..0000000
--- a/_build/jupyter_execute/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb	
+++ /dev/null
@@ -1,206 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 3. Longest Substring Without Repeating Characters\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Longest Substring Without Repeating Characters problem on LeetCode, click here!](https://leetcode.com/problems/longest-substring-without-repeating-characters/)\n",
-    "\n",
-    "---\n",
-    "Given a string `s`, find the length of the **longest substring** without repeating characters.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 0 <= `s.length` <= $5 * 10^4$\n",
-    "- `s` consists of English letters, digits, symbols and spaces.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def length_of_longest_substring(s):\n",
-    "    # Create a dictionary to store the index of each character's last occurrence.\n",
-    "    char_index = {}\n",
-    "    \n",
-    "    # Initialize variables to keep track of the start and end of the current substring.\n",
-    "    start = 0\n",
-    "    max_length = 0\n",
-    "    \n",
-    "    for end in range(len(s)):\n",
-    "        # If the character is in the dictionary and its last occurrence is after the start of the current substring,\n",
-    "        # update the start of the substring to the next character after its last occurrence.\n",
-    "        if s[end] in char_index and char_index[s[end]] >= start:\n",
-    "            start = char_index[s[end]] + 1\n",
-    "        \n",
-    "        # Update the last occurrence index of the current character.\n",
-    "        char_index[s[end]] = end\n",
-    "        \n",
-    "        # Calculate the length of the current substring and update the maximum length if necessary.\n",
-    "        current_length = end - start + 1\n",
-    "        max_length = max(max_length, current_length)\n",
-    "    \n",
-    "    return max_length"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a function called `length_of_longest_substring` that takes a single argument `s`, which is the input string for which we want to find the length of the longest substring without repeating characters.\n",
-    "\n",
-    "2. We create a dictionary called `char_index` to store the index of the last occurrence of each character in the input string. This dictionary will help us keep track of where each character was last seen.\n",
-    "\n",
-    "3. We initialize two variables, `start` and `max_length`. \n",
-    "   - `start` represents the start index of the current substring without repeating characters. It starts at 0.\n",
-    "   - `max_length` is used to keep track of the maximum length found so far and is initially set to 0.\n",
-    "\n",
-    "4. We then iterate through the input string `s` using a for loop, where the loop variable `end` represents the current end index of the substring we are considering.\n",
-    "\n",
-    "5. Inside the loop:\n",
-    "   - We check if the character `s[end]` is already in the `char_index` dictionary and if its last occurrence is within or after the current substring. If so, it means that we've encountered a repeating character, and we need to update the `start` of the substring to the next character after the last occurrence of `s[end]`. This ensures that we have a new valid substring without repeating characters.\n",
-    "   - We then update the `char_index` dictionary by storing the current index `end` as the last occurrence index of the character `s[end]`.\n",
-    "\n",
-    "6. Next, we calculate the length of the current substring without repeating characters, which is `current_length = end - start + 1`. We add 1 to account for the fact that the indices are zero-based.\n",
-    "\n",
-    "7. We update the `max_length` with the maximum of its current value and the `current_length`. This step ensures that we keep track of the longest valid substring we have encountered so far.\n",
-    "\n",
-    "8. The loop continues to iterate through the input string, and at the end, we return the `max_length`, which represents the length of the longest substring without repeating characters."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "print(length_of_longest_substring(\"abcabcbb\"))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "print(length_of_longest_substring(\"bbbbb\"))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "cf04eba0-0ead-4e8b-ba33-4316ea64a9d3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "print(length_of_longest_substring(\"pwwkew\"))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "Let's analyze the time and space complexity of the `length_of_longest_substring` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The function iterates through the input string `s` using a single for loop. The loop runs from the beginning of the string to the end once.\n",
-    "- Inside the loop, we perform constant-time operations such as dictionary lookups and updates, comparisons, and arithmetic operations.\n",
-    "- Therefore, the time complexity of the function is O(n), where n is the length of the input string `s`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The primary data structure that consumes space in this function is the `char_index` dictionary.\n",
-    "- In the worst case, if there are no repeating characters in the input string, the dictionary can store all unique characters in the string.\n",
-    "- Therefore, the space complexity is O(min(n, m)), where n is the length of the input string `s`, and m is the number of unique characters in the string. In the worst case, when all characters are unique, m is equal to n, so the space complexity is O(n).\n",
-    "- Additionally, there are a few integer variables used for indices and lengths, which consume constant space.\n",
-    "- Overall, the space complexity of the function is O(min(n, m)) or simply O(n) in the worst case.\n",
-    "\n",
-    "**In summary,** the time complexity of the `length_of_longest_substring` function is O(n), and the space complexity is O(n) in the worst case due to the `char_index` dictionary. This algorithm provides an efficient way to find the length of the longest substring without repeating characters in linear time."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Longest Substring with K Distinct Characters:** Modify the problem to find the length of the longest substring with exactly K distinct characters. For example, given the input string \"abcabcbb\" and K = 2, the answer would be 4 because the longest substring with two distinct characters is \"abca.\"\n",
-    "2. **Longest Substring with Unique Characters:** Write a function to find the length of the longest substring in a given string where all characters are unique. For example, given the input string \"abcabcbb,\" the answer would be 4 because \"abcd\" is the longest substring with unique characters."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb b/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb
deleted file mode 100644
index df4b0d3..0000000
--- a/_build/jupyter_execute/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb	
+++ /dev/null
@@ -1,189 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 424. Longest Repeating Character Replacement\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Longest Repeating Character Replacement problem on LeetCode, click here!](https://leetcode.com/problems/longest-repeating-character-replacement/)\n",
-    "\n",
-    "---\n",
-    "You are given a string `s` and an integer `k`. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most `k` times.\n",
-    "\n",
-    "*Return the length of the longest substring containing the same letter you can get after performing the above operations.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `s.length` <= $10^5$\n",
-    "- `s` consists of only uppercase English letters.\n",
-    "- `0 <= k <= s.length`"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def characterReplacement(s, k):\n",
-    "    max_length = 0  # Initialize the maximum length\n",
-    "    max_count = 0   # Initialize the maximum count of repeating characters\n",
-    "    start = 0       # Initialize the start of the sliding window\n",
-    "    char_count = {} # Dictionary to store the count of each character\n",
-    "\n",
-    "    for end in range(len(s)):\n",
-    "        # Update the count of the current character in the dictionary\n",
-    "        char_count[s[end]] = char_count.get(s[end], 0) + 1\n",
-    "        \n",
-    "        # Update the maximum count of repeating characters\n",
-    "        max_count = max(max_count, char_count[s[end]])\n",
-    "        \n",
-    "        # Check if the current window size is greater than k\n",
-    "        if (end - start + 1) - max_count > k:\n",
-    "            # Move the start of the window to the right\n",
-    "            char_count[s[start]] -= 1\n",
-    "            start += 1\n",
-    "        \n",
-    "        # Update the maximum length\n",
-    "        max_length = max(max_length, end - start + 1)\n",
-    "\n",
-    "    return max_length"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We initialize some variables:\n",
-    "   - `max_length` to keep track of the maximum length of the substring containing the same letter.\n",
-    "   - `max_count` to keep track of the maximum count of repeating characters within the current window.\n",
-    "   - `start` to represent the start index of the sliding window.\n",
-    "   - `char_count` is a dictionary that will store the count of each character within the current window.\n",
-    "\n",
-    "2. We use a `for` loop to iterate through the characters of the string `s`.\n",
-    "\n",
-    "3. Inside the loop, we do the following for each character at position `end`:\n",
-    "   - Update the count of the current character in the `char_count` dictionary.\n",
-    "   - Update the `max_count` to be the maximum of the current `max_count` and the count of the current character. This keeps track of the maximum count of repeating characters within the current window.\n",
-    "\n",
-    "4. We check if the current window size (the difference between `end` and `start` plus one) minus the `max_count` is greater than `k`. This condition checks whether we have exceeded the allowed number of replacements (`k`) within the current window.\n",
-    "\n",
-    "   - If we have exceeded the allowed replacements, it means we need to shrink the window from the left side. We do this by moving the `start` of the window to the right and decrementing the count of the character at `s[start]` in the `char_count` dictionary. This effectively removes characters from the left side of the window until we have a valid window again.\n",
-    "\n",
-    "   - By doing this, we ensure that the difference between the current window size and the `max_count` is always less than or equal to `k`.\n",
-    "\n",
-    "5. After each character, we update the `max_length` to be the maximum of the current `max_length` and the size of the current window (`end - start + 1`).\n",
-    "\n",
-    "6. Finally, we return `max_length`, which holds the length of the longest substring containing the same letter with at most `k` replacements."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "s1 = \"ABAB\"\n",
-    "k1 = 2\n",
-    "print(characterReplacement(s1, k1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "s2 = \"AABABBA\"\n",
-    "k2 = 1\n",
-    "print(characterReplacement(s2, k2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The code uses a single `for` loop to iterate through the characters of the input string `s`. The loop runs from the beginning to the end of the string.\n",
-    "- Inside the loop, we perform constant-time operations such as updating the `char_count` dictionary and updating variables like `max_count` and `max_length`.\n",
-    "- The code's time complexity is primarily determined by the loop, which iterates through each character in the string once. Therefore, the time complexity is O(n), where n is the length of the input string `s`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The code uses several variables to store information, but the space they consume is constant and does not depend on the size of the input string. These variables include `max_length`, `max_count`, `start`, and `char_count`. Therefore, the space complexity is O(1) or constant space.\n",
-    "\n",
-    "**In summary**, the time complexity of the code is O(n), where n is the length of the input string, and the space complexity is O(1), as it uses a constant amount of additional space regardless of the input size. This algorithm efficiently solves the problem with a linear time complexity."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the code to find the longest substring containing the same letter with at most k replacements in O(n) time complexity and O(1) space complexity. Hint: You may need to update the approach to achieve this.\n",
-    "2. **Variation with Lowercase Letters:** Extend the problem to include both uppercase and lowercase English letters in the input string s. Write a function that can handle this extended input and still find the longest substring with at most k replacements efficiently."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb b/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb
deleted file mode 100644
index cf22c86..0000000
--- a/_build/jupyter_execute/03. Sliding Window/76. Minimum Window Substring.ipynb	
+++ /dev/null
@@ -1,260 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 76. Minimum Window Substring\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Minimum Window Substring problem on LeetCode, click here!](https://leetcode.com/problems/minimum-window-substring/)\n",
-    "\n",
-    "---\n",
-    "Given two strings `s` and `t` of lengths `m` and `n` respectively, return *the **minimum window substring** of `s` such that every character in `t` (**including duplicates**) is included in the window. If there is no such substring, return the empty string `\"\"`.*\n",
-    "\n",
-    "The testcases will be generated such that the answer is **unique**.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `m == s.length`\n",
-    "- `n == t.length`\n",
-    "- 1 <= `m, n` <= $10^5$\n",
-    "- `s` and `t` consist of uppercase and lowercase English letters.\n",
-    "\n",
-    "**Follow up**: Could you find an algorithm that runs in $O(m + n)$ time?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def min_window_substring(s, t):\n",
-    "    if not s or not t:\n",
-    "        return \"\"\n",
-    "\n",
-    "    # Initialize dictionaries to keep track of character counts for t and the current window in s.\n",
-    "    t_dict = {}\n",
-    "    current_window_dict = {}\n",
-    "    \n",
-    "    # Populate t_dict with character counts for string t.\n",
-    "    for char in t:\n",
-    "        t_dict[char] = t_dict.get(char, 0) + 1\n",
-    "\n",
-    "    # Initialize pointers for the sliding window.\n",
-    "    left = 0\n",
-    "    min_len = float('inf')\n",
-    "    min_window = \"\"\n",
-    "    required_chars = len(t_dict)\n",
-    "\n",
-    "    # Initialize a variable to keep track of how many required characters have been found in the current window.\n",
-    "    found_chars = 0\n",
-    "\n",
-    "    # Iterate over the string s using the right pointer.\n",
-    "    for right in range(len(s)):\n",
-    "        char = s[right]\n",
-    "\n",
-    "        # Update the current_window_dict.\n",
-    "        current_window_dict[char] = current_window_dict.get(char, 0) + 1\n",
-    "\n",
-    "        # Check if the current character is a required character and if its count matches the required count.\n",
-    "        if char in t_dict and current_window_dict[char] == t_dict[char]:\n",
-    "            found_chars += 1\n",
-    "\n",
-    "        # Try to minimize the window by moving the left pointer.\n",
-    "        while left <= right and found_chars == required_chars:\n",
-    "            # Calculate the current window size.\n",
-    "            window_size = right - left + 1\n",
-    "\n",
-    "            # If the current window is smaller than the minimum found so far, update min_len and min_window.\n",
-    "            if window_size < min_len:\n",
-    "                min_len = window_size\n",
-    "                min_window = s[left:right+1]\n",
-    "\n",
-    "            # Move the left pointer to the right to shrink the window.\n",
-    "            left_char = s[left]\n",
-    "            current_window_dict[left_char] -= 1\n",
-    "\n",
-    "            # Check if the character removed from the window was a required character.\n",
-    "            if left_char in t_dict and current_window_dict[left_char] < t_dict[left_char]:\n",
-    "                found_chars -= 1\n",
-    "\n",
-    "            # Move the left pointer further to continue shrinking the window.\n",
-    "            left += 1\n",
-    "\n",
-    "    return min_window"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by checking if either of the input strings `s` or `t` is empty. If either of them is empty, we return an empty string since there can't be any valid substring in this case.\n",
-    "\n",
-    "2. We initialize two dictionaries: `t_dict` and `current_window_dict`. These dictionaries will be used to keep track of character counts in the string `t` and the current window in string `s`, respectively.\n",
-    "\n",
-    "3. We populate the `t_dict` dictionary by iterating through string `t`. For each character, we increment its count in the dictionary using `t_dict.get(char, 0) + 1`. This allows us to count the occurrences of each character in `t`.\n",
-    "\n",
-    "4. We initialize two pointers: `left` and `right`. The `left` pointer will represent the start of the current window, and the `right` pointer will represent the end of the current window. We also initialize `min_len` to store the length of the minimum window found so far, and `min_window` to store the actual minimum window substring.\n",
-    "\n",
-    "5. We determine the number of required characters in `t` (i.e., the number of distinct characters in `t`) and store it in the variable `required_chars`.\n",
-    "\n",
-    "6. We initialize a variable `found_chars` to keep track of how many required characters have been found in the current window. Initially, it is set to 0.\n",
-    "\n",
-    "7. We iterate over the string `s` using the `right` pointer. In each iteration, we do the following:\n",
-    "\n",
-    "   - Update the `current_window_dict` by incrementing the count of the current character.\n",
-    "   \n",
-    "   - Check if the current character is a required character (present in `t_dict`) and if its count in the `current_window_dict` matches the required count from `t_dict`. If so, we increment `found_chars`.\n",
-    "\n",
-    "8. After updating the window, we attempt to minimize the window size by moving the `left` pointer to the right. In this step, we:\n",
-    "\n",
-    "   - Calculate the size of the current window.\n",
-    "   \n",
-    "   - If the current window size is smaller than the minimum found so far (`min_len`), we update `min_len` and `min_window` to store the current window substring.\n",
-    "   \n",
-    "   - Move the `left` pointer to the right to shrink the window.\n",
-    "   \n",
-    "   - Check if the character being removed from the window was a required character. If it was, we decrement `found_chars`.\n",
-    "   \n",
-    "   - Continue moving the `left` pointer further to continue shrinking the window if the window still contains all required characters.\n",
-    "\n",
-    "9. Finally, we return `min_window`, which will contain the minimum window substring that contains all characters from `t`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "BANC\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "print(min_window_substring(\"ADOBECODEBANC\", \"ABC\"))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "a\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "print(min_window_substring(\"a\", \"a\")) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "print(min_window_substring(\"a\", \"aa\")) "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `min_window_substring` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The main loop iterates through the string `s` from left to right using the `right` pointer. This loop runs in O(m) time, where 'm' is the length of string `s`.\n",
-    "\n",
-    "- Inside the loop, we have a while loop that moves the `left` pointer to the right to shrink the window as needed. In the worst case, this while loop can also run in O(m) time because in the worst case, we may have to move the `left` pointer all the way to the end of the string.\n",
-    "\n",
-    "- Within each iteration of the while loop, we perform constant time operations, such as updating dictionaries and comparing character counts.\n",
-    "\n",
-    "- The initialization of the `t_dict` dictionary takes O(n) time, where 'n' is the length of string `t`.\n",
-    "\n",
-    "Therefore, the overall time complexity of the function is O(m + n) because the dominant factor is the length of string `s`, and the length of string `t` has a smaller impact.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity is determined by the space used by the `t_dict` dictionary, the `current_window_dict` dictionary, and a few variables.\n",
-    "\n",
-    "- The `t_dict` dictionary stores character counts for string `t`. In the worst case, when all characters in `t` are distinct, this dictionary can have a maximum of 'n' key-value pairs. So, the space complexity for `t_dict` is O(n).\n",
-    "\n",
-    "- The `current_window_dict` dictionary stores character counts for the current window. In the worst case, it can have a maximum of 'm' key-value pairs. So, the space complexity for `current_window_dict` is also O(m).\n",
-    "\n",
-    "- Other variables used in the function, such as `left`, `right`, `min_len`, `min_window`, `required_chars`, and `found_chars`, are all of constant size and do not depend on the input sizes.\n",
-    "\n",
-    "**In summary**, the overall space complexity of the function is O(max(m, n)), which means it is determined by the larger of the two input strings' lengths. This is because the space used for `t_dict` and `current_window_dict` dominates the space complexity."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Smallest Distinct Substring:** Instead of finding the minimum window substring containing all characters, find the smallest distinct (unique) substring of s that contains all characters from t. This variation adds complexity because you must find a substring with distinct characters.\n",
-    "2. **No Extra Space:** Solve the problem without using any extra space, such as dictionaries or arrays, other than a constant amount of space. This is a significant optimization challenge."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb b/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb
deleted file mode 100644
index 04c0280..0000000
--- a/_build/jupyter_execute/04. Stack/20. Valid Parentheses.ipynb	
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 20. Valid Parentheses\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Valid Parentheses problem on LeetCode, click here!](https://leetcode.com/problems/valid-parentheses/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given a string s containing just the characters `'('`, `')'`, `'{'`, `'}'`, `'['` and `']'`, determine if the input string is valid.\n",
-    "\n",
-    "An input string is valid if:\n",
-    "\n",
-    "1. Open brackets must be closed by the same type of brackets.\n",
-    "2. Open brackets must be closed in the correct order.\n",
-    "3. Every close bracket has a corresponding open bracket of the same type.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `s.length` <= $10^4$\n",
-    "- `s` consists of parentheses only `'()[]{}'`."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def isValid(s):\n",
-    "    # Create an empty stack to store opening brackets\n",
-    "    stack = []\n",
-    "    \n",
-    "    # Define a mapping for matching brackets\n",
-    "    bracket_mapping = {')': '(', '}': '{', ']': '['}\n",
-    "    \n",
-    "    # Iterate through the characters in the input string\n",
-    "    for char in s:\n",
-    "        # If the character is an opening bracket, push it onto the stack\n",
-    "        if char in bracket_mapping.values():\n",
-    "            stack.append(char)\n",
-    "        # If the character is a closing bracket\n",
-    "        elif char in bracket_mapping.keys():\n",
-    "            # Pop the top element from the stack if it exists, or use a dummy value '#'\n",
-    "            top_element = stack.pop() if stack else '#'\n",
-    "            # If the popped element does not match the corresponding opening bracket, return False\n",
-    "            if bracket_mapping[char] != top_element:\n",
-    "                return False\n",
-    "        # If the character is not a bracket, ignore it\n",
-    "        \n",
-    "    # After processing the entire string, if there are any unmatched opening brackets left in the stack, return False\n",
-    "    return len(stack) == 0"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining the `isValid` function that takes a single argument `s`, which is the input string containing parentheses and brackets.\n",
-    "\n",
-    "2. Inside the function, we create an empty stack, which is a list in Python, to store opening brackets as we encounter them in the input string. The stack will help us keep track of the brackets and their order.\n",
-    "\n",
-    "3. We define a `bracket_mapping` dictionary that maps each closing bracket to its corresponding opening bracket. This mapping will be used to check if a closing bracket matches the most recent opening bracket in the stack.\n",
-    "\n",
-    "4. We iterate through each character `char` in the input string `s` using a `for` loop.\n",
-    "\n",
-    "5. If the character `char` is an opening bracket (i.e., it exists in the `values()` of `bracket_mapping`), we push it onto the stack using the `append()` method.\n",
-    "\n",
-    "6. If the character `char` is a closing bracket (i.e., it exists in the `keys()` of `bracket_mapping`), we need to check if it matches the corresponding opening bracket. To do this, we pop the top element from the stack (if it exists) and store it in `top_element`. We use a dummy value `'#'` if the stack is empty to avoid errors.\n",
-    "\n",
-    "7. We then compare `top_element` with the corresponding opening bracket using `bracket_mapping[char]`. If they do not match, it means the string is not valid, and we return `False`.\n",
-    "\n",
-    "8. If the character `char` is not a bracket, we simply ignore it and continue the loop.\n",
-    "\n",
-    "9. After processing the entire string, we check if there are any unmatched opening brackets left in the stack. If the stack is empty, it means all opening brackets have been properly matched and closed, and we return `True`. Otherwise, we return `False`.\n",
-    "\n",
-    "10. Finally, we provide some test cases at the bottom of the code to demonstrate how the function works for different input strings."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "print(isValid(\"()\"))    "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "print(isValid(\"()[]{}\")) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "print(isValid(\"(]\"))     "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `isValid` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The function iterates through each character in the input string `s` once, performing constant-time operations for each character.\n",
-    "- Pushing and popping elements from the stack (list) also takes constant time in most cases.\n",
-    "- Therefore, the overall time complexity of the function is $O(n)$, where $n$ is the length of the input string `s`.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity of the function is determined by the space used by the stack and the `bracket_mapping` dictionary.\n",
-    "- In the worst case, if the input string `s` consists entirely of opening brackets, the stack can potentially contain all of these opening brackets, resulting in a space complexity of $O(n)$ in terms of the stack.\n",
-    "- The `bracket_mapping` dictionary has a constant number of key-value pairs (3 pairs in this case).\n",
-    "- Therefore, the overall space complexity of the function is $O(n)$, where $n$ is the length of the input string `s`, mainly due to the stack space.\n",
-    "\n",
-    "**In summary:**\n",
-    "- The time complexity is $O(n)$ because we iterate through the string once.\n",
-    "- The space complexity is $O(n)$ due to the stack used to keep track of opening brackets."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Valid Expressions:** Modify the problem to validate not only brackets but also arithmetic expressions containing parentheses, such as \"$2 * (3 + 5) / (4 - 2)$\".\n",
-    "2. **Valid Parentheses Combinations:**\n",
-    "Write a function to generate all valid combinations of n pairs of parentheses, where $n$ is a positive integer. For example, for $n = 3$, the valid combinations are [\"((()))\", \"(()())\", \"(())()\", \"()(())\", \"()()()\"]."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb b/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb
deleted file mode 100644
index eb4858b..0000000
--- a/_build/jupyter_execute/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,220 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 153. Find Minimum in Rotated Sorted Array\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Find Minimum in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:\n",
-    "\n",
-    "- `[4,5,6,7,0,1,2]` if it was rotated `4` times.\n",
-    "- `[0,1,2,4,5,6,7]` if it was rotated `7` times.\n",
-    "\n",
-    "Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.\n",
-    "\n",
-    "Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array.*\n",
-    "\n",
-    "You must write an algorithm that runs in $O(log\\ n)$ time.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `n == nums.length`\n",
-    "- `1 <= n <= 5000`\n",
-    "- `-5000 <= nums[i] <= 5000`\n",
-    "- All the integers of `nums` are **unique**.\n",
-    "- `nums` is sorted and rotated between `1` and `n` times."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def findMin(nums):\n",
-    "    left, right = 0, len(nums) - 1\n",
-    "    \n",
-    "    while left < right:\n",
-    "        mid = left + (right - left) // 2\n",
-    "        \n",
-    "        # If the mid element is greater than the rightmost element,\n",
-    "        # it means the minimum element is in the right half.\n",
-    "        if nums[mid] > nums[right]:\n",
-    "            left = mid + 1\n",
-    "        # If the mid element is less than or equal to the rightmost element,\n",
-    "        # it means the minimum element is in the left half or at the mid itself.\n",
-    "        else:\n",
-    "            right = mid\n",
-    "    \n",
-    "    # The loop will break when left and right converge to the minimum element.\n",
-    "    # At this point, left (or right) will be pointing to the minimum element.\n",
-    "    return nums[left]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The function `findMin(nums)` takes an input list `nums`, which represents the rotated sorted array.\n",
-    "\n",
-    "2. Initialize two pointers `left` and `right` to track the current search range. Initially, `left` is set to 0 (the beginning of the array), and `right` is set to `len(nums) - 1` (the end of the array).\n",
-    "\n",
-    "3. Enter a `while` loop with the condition `left < right`. This loop continues until `left` and `right` converge to a single element, which will be the minimum element in the rotated array.\n",
-    "\n",
-    "4. Inside the loop, calculate the middle index `mid` using integer division. This helps in finding the middle element of the current search range.\n",
-    "\n",
-    "5. Check if the element at the middle index (`nums[mid]`) is greater than the element at the rightmost index (`nums[right]`). If this condition is true, it means that the minimum element must be in the right half of the current search range. So, update `left` to `mid + 1`, effectively eliminating the left half of the search range.\n",
-    "\n",
-    "6. If the condition from step 5 is not met, it implies that the minimum element can be in the left half of the current search range or could be the element at the `mid` index itself. In this case, update `right` to `mid`, effectively eliminating the right half of the search range.\n",
-    "\n",
-    "7. Repeat steps 4-6 until `left` and `right` converge to the minimum element.\n",
-    "\n",
-    "8. When the loop exits, it means that `left` (or `right`) points to the minimum element in the array.\n",
-    "\n",
-    "9. Return `nums[left]` (or `nums[right]`) as the result, which is the minimum element in the rotated sorted array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1:\n",
-    "nums1 = [3, 4, 5, 1, 2]\n",
-    "print(findMin(nums1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "0\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-    "print(findMin(nums2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "11\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "nums3 = [11, 13, 15, 17]\n",
-    "print(findMin(nums3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of this code is $O(log\\ n)$, where '$n$' is the length of the input array `nums`. This is because the binary search algorithm is employed, which continually divides the search range in half with each iteration.\n",
-    "\n",
-    "In each iteration of the `while` loop, the search range is halved, and this process continues until the `left` and `right` pointers converge to the minimum element. In the worst case, it will take logarithmic time to reduce the search range to a single element.\n",
-    "\n",
-    "Therefore, the binary search used in this code runs in $O(log\\ n)$ time complexity.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity of this code is $O(1)$, which means it uses a constant amount of additional space that does not depend on the size of the input array.\n",
-    "\n",
-    "The algorithm uses a fixed number of variables (`left`, `right`, `mid`) to keep track of the search range and indices, but the number of these variables does not grow with the size of the input array. Hence, the space complexity is constant.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: $O(log\\ n)$\n",
-    "- Space Complexity: $O(1)$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. Solve the problem with a variation where the input array may contain duplicates.\n",
-    "\n",
-    "2. Modify the problem to return the index of the minimum element in the rotated sorted array, rather than the element itself. The algorithm should still run in $O(log\\ n)$ time.\n",
-    "\n",
-    "3. Implement a function that finds the maximum element in a rotated sorted array. How would you adapt the binary search algorithm to solve this problem efficiently in $O(log\\ n)$ time?\n",
-    "\n",
-    "4. Implement a function that can find the kth smallest element in a rotated sorted array. This is an extension of the original problem, and the algorithm should still run in $O(log\\ n)$ time."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb b/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb
deleted file mode 100644
index 7b3139d..0000000
--- a/_build/jupyter_execute/05. Binary Search/33. Search in Rotated Sorted Array.ipynb	
+++ /dev/null
@@ -1,237 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 33. Search in Rotated Sorted Array\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Search in Rotated Sorted Array problem on LeetCode, click here!](https://leetcode.com/problems/search-in-rotated-sorted-array/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "There is an integer array `nums` sorted in ascending order (with **distinct** values).\n",
-    "\n",
-    "Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k (1 <= k < nums.length)` such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index 3 and become `[4,5,6,7,0,1,2]`.\n",
-    "\n",
-    "Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.*\n",
-    "\n",
-    "You must write an algorithm with $O(log\\ n)$ runtime complexity.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `1 <= nums.length <= 5000`\n",
-    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-    "- All values of `nums` are **unique**.\n",
-    "- `nums` is an ascending array that is possibly rotated.\n",
-    "- $-10^4$ <= `target` <= $10^4$"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def search(nums, target):\n",
-    "    left, right = 0, len(nums) - 1\n",
-    "\n",
-    "    # Find the pivot index using binary search\n",
-    "    while left < right:\n",
-    "        mid = left + (right - left) // 2\n",
-    "\n",
-    "        if nums[mid] > nums[right]:\n",
-    "            left = mid + 1\n",
-    "        else:\n",
-    "            right = mid\n",
-    "\n",
-    "    pivot = left\n",
-    "    left, right = 0, len(nums) - 1\n",
-    "\n",
-    "    # Determine which part of the array to search in\n",
-    "    if target >= nums[pivot] and target <= nums[right]:\n",
-    "        left = pivot\n",
-    "    else:\n",
-    "        right = pivot\n",
-    "\n",
-    "    # Perform binary search to find the target\n",
-    "    while left <= right:\n",
-    "        mid = left + (right - left) // 2\n",
-    "\n",
-    "        if nums[mid] == target:\n",
-    "            return mid\n",
-    "        elif nums[mid] < target:\n",
-    "            left = mid + 1\n",
-    "        else:\n",
-    "            right = mid - 1\n",
-    "\n",
-    "    return -1"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The `search` function takes two arguments: `nums`, which is the rotated sorted array, and `target`, which is the value we want to find in the array.\n",
-    "\n",
-    "2. We initialize two pointers, `left` and `right`, which represent the range in which we are going to perform binary search. Initially, `left` is set to 0, and `right` is set to the index of the last element in the array (`len(nums) - 1`).\n",
-    "\n",
-    "3. We start by finding the pivot index using binary search. The pivot index is the index at which the array is rotated. We use a `while` loop to continue the search until `left` is less than `right`.\n",
-    "\n",
-    "4. Inside the loop, we calculate the middle index `mid` using integer division. We then compare the value at `mid` with the value at `right`. If `nums[mid]` is greater than `nums[right]`, it means the pivot point lies to the right of `mid`, so we update `left` to `mid + 1`. Otherwise, the pivot point lies to the left of `mid`, so we update `right` to `mid`. This process continues until we find the pivot index.\n",
-    "\n",
-    "5. After finding the pivot index, we have divided the array into two parts: one part is sorted in ascending order, and the other part is also sorted in ascending order but rotated. We need to determine which part of the array contains the target value.\n",
-    "\n",
-    "6. We reset `left` and `right` pointers. If the target value is within the range `[nums[pivot], nums[right]]`, we set `left` to `pivot` (the start of the rotated part), indicating that we should search in the rotated part. Otherwise, we set `right` to `pivot` (the end of the sorted part), indicating that we should search in the sorted part.\n",
-    "\n",
-    "7. We then perform binary search again within the chosen range (`left` to `right`) to find the target element. The binary search continues until `left` is less than or equal to `right`.\n",
-    "\n",
-    "8. Inside the binary search loop, we calculate the middle index `mid` and compare `nums[mid]` with the target value. If they are equal, we return `mid` as the index of the target element. If `nums[mid]` is less than the target, we update `left` to `mid + 1` to search in the right half. If `nums[mid]` is greater than the target, we update `right` to `mid - 1` to search in the left half.\n",
-    "\n",
-    "9. If we exit the binary search loop without finding the target, we return -1 to indicate that the target is not present in the array."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "nums1 = [4, 5, 6, 7, 0, 1, 2]\n",
-    "target1 = 0\n",
-    "result1 = search(nums1, target1)\n",
-    "print(result1) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [4, 5, 6, 7, 0, 1, 2]\n",
-    "target2 = 3\n",
-    "result2 = search(nums2, target2)\n",
-    "print(result2) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "-1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "nums3 = [1]\n",
-    "target3 = 0\n",
-    "result3 = search(nums3, target3)\n",
-    "print(result3) "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. Finding the pivot index using binary search takes $O(log\\ n)$ time, where '$n$' is the number of elements in the array.\n",
-    "2. After finding the pivot index, performing binary search to find the target element also takes $O(log\\ n)$ time in the worst case.\n",
-    "\n",
-    "The dominant factor in the time complexity is the binary search, and since we perform two binary searches sequentially, the overall time complexity of the code is $O(log\\ n)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of the code is very minimal:\n",
-    "\n",
-    "1. We use a constant amount of additional space for variables such as `left`, `right`, `pivot`, and `mid`. These variables do not depend on the input size, so they contribute $O(1)$ space complexity.\n",
-    "2. The function does not use any additional data structures that scale with the input size.\n",
-    "\n",
-    "Therefore, the space complexity of the code is $O(1)$, which means it has a constant space complexity and does not depend on the size of the input array.\n",
-    "\n",
-    "**In summary:**\n",
-    "- Time Complexity: $O(log\\ n)$\n",
-    "- Space Complexity: $O(1)$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Find Minimum Element in Rotated Sorted Array**: Write an algorithm to find the minimum element in a rotated sorted array. This is a variation of the problem where you don't need to search for a target value but instead find the smallest element.\n",
-    "\n",
-    "2. **Search in a Circularly Sorted Array**: Consider an array that is sorted in a circular manner (e.g., [4, 5, 6, 7, 0, 1, 2, 3]). Adapt the search algorithm to work for circularly sorted arrays while maintaining $O(log\\ n)$ complexity."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb b/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb
deleted file mode 100644
index c428c9f..0000000
--- a/_build/jupyter_execute/06. Linked List/141. Linked List Cycle.ipynb	
+++ /dev/null
@@ -1,261 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 141. Linked List Cycle\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Linked List Cycle problem on LeetCode, click here!](https://leetcode.com/problems/linked-list-cycle/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given `head`, the head of a linked list, determine if the linked list has a cycle in it.\n",
-    "\n",
-    "There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the `next` pointer. Internally, `pos` is used to denote the index of the node that tail's `next` pointer is connected to. **Note that `pos` is not passed as a parameter.**\n",
-    "\n",
-    "Return `true` if there is a cycle in the linked list. Otherwise, return `false`.\n",
-    "\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of the nodes in the list is in the range $[0, 10^4]$.\n",
-    "- $-10^5$ <= Node.val <= $10^5$\n",
-    "- `pos` is `-1` or a **valid index** in the linked-list.\n",
-    "\n",
-    "**Follow up:** Can you solve it using $O(1)$ (i.e. constant) memory?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for singly-linked list.\n",
-    "class ListNode:\n",
-    "    def __init__(self, val=0):\n",
-    "        self.val = val\n",
-    "        self.next = None\n",
-    "\n",
-    "class Solution:\n",
-    "    def hasCycle(self, head: ListNode) -> bool:\n",
-    "        if not head:\n",
-    "            return False\n",
-    "        \n",
-    "        slow = head\n",
-    "        fast = head\n",
-    "        \n",
-    "        while fast and fast.next:\n",
-    "            slow = slow.next\n",
-    "            fast = fast.next.next\n",
-    "            \n",
-    "            if slow == fast:\n",
-    "                return True\n",
-    "        \n",
-    "        return False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The given code defines a class `Solution` with a method `hasCycle` for determining if a linked list has a cycle.\n",
-    "\n",
-    "The `ListNode` class represents nodes in the linked list, where each node has a `val` (a value associated with the node) and a `next` attribute (a reference to the next node in the list).\n",
-    "\n",
-    "The `hasCycle` method takes the `head` of a linked list as input and returns `True` if there is a cycle in the linked list, and `False` otherwise. It uses two pointers, `slow` and `fast`, to traverse the linked list. If there is a cycle, the `fast` pointer will eventually catch up to the `slow` pointer.\n",
-    "\n",
-    "The algorithm works as follows:\n",
-    "1. It checks if the input `head` is `None` (an empty list). If it is, the method returns `False` because an empty list can't have a cycle.\n",
-    "2. Two pointers, `slow` and `fast`, initially point to the `head` of the list.\n",
-    "3. The code enters a `while` loop where `fast` moves two steps at a time, and `slow` moves one step at a time.\n",
-    "4. If there is a cycle, `fast` will eventually catch up to `slow`, and the method returns `True`.\n",
-    "5. If the loop completes without finding a cycle, the method returns `False`."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to create a linked list with a cycle\n",
-    "def create_linked_list_with_cycle(values, pos):\n",
-    "    dummy = ListNode()\n",
-    "    current = dummy\n",
-    "    cycle_node = None\n",
-    "    \n",
-    "    for i, val in enumerate(values):\n",
-    "        current.next = ListNode(val)\n",
-    "        current = current.next\n",
-    "        \n",
-    "        if i == pos:\n",
-    "            cycle_node = current\n",
-    "    \n",
-    "    if cycle_node:\n",
-    "        current.next = cycle_node\n",
-    "    \n",
-    "    return dummy.next"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "This helper function is designed to create a linked list with a cycle for testing purposes. It takes two arguments:\n",
-    "\n",
-    "1. `values`: A list of values representing the nodes in the linked list.\n",
-    "2. `pos`: An integer representing the index of the node where the cycle begins.\n",
-    "\n",
-    "Here's how the function works:\n",
-    "\n",
-    "- It starts by creating an empty `ListNode` called `dummy`. This `dummy` node is used to simplify the creation of the linked list.\n",
-    "- It initializes a `current` pointer to the `dummy` node. This pointer will be used to traverse the linked list.\n",
-    "- It initializes a `cycle_node` variable to `None`. This variable will hold the reference to the node where the cycle begins, if any.\n",
-    "- It then iterates through the `values` list, creating a new `ListNode` for each value and appending it to the linked list.\n",
-    "- Inside the loop, it checks if the current index `i` is equal to the specified `pos`. If they match, it sets `cycle_node` to the current node. This simulates the creation of a cycle in the linked list.\n",
-    "- After the loop, if `cycle_node` is not `None` (indicating a cycle should be created), it connects the last node in the linked list to `cycle_node`, effectively creating a cycle.\n",
-    "- Finally, it returns the reference to the first node of the linked list (not the `dummy` node).\n",
-    "\n",
-    "This helper function allows you to easily create test cases where you can specify the values for the linked list and the position where the cycle starts. It's particularly useful for testing the code's ability to detect cycles in linked lists."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "head = [3,2,0,-4]\n",
-    "pos = 1\n",
-    "\n",
-    "head = create_linked_list_with_cycle(head, pos)\n",
-    "solution = Solution()\n",
-    "result = solution.hasCycle(head)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "head = [1,2]\n",
-    "pos = 0\n",
-    "\n",
-    "head = create_linked_list_with_cycle(head, pos)\n",
-    "solution = Solution()\n",
-    "result = solution.hasCycle(head)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "head = [1]\n",
-    "pos = -1\n",
-    "\n",
-    "head = create_linked_list_with_cycle(head, pos)\n",
-    "solution = Solution()\n",
-    "result = solution.hasCycle(head)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's discuss the time and space complexity of the code for detecting a cycle in a linked list:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this code is $O(n)$, where \"$n$\" is the number of nodes in the linked list.\n",
-    "\n",
-    "The reason for this is that both the `slow` and `fast` pointers traverse the linked list, and they move at different speeds. In the worst case, when there is no cycle, the `fast` pointer will reach the end of the list after going through approximately $n/2$ nodes. In the case of a cycle, it may take some extra iterations for the `fast` pointer to catch up to the `slow` pointer. However, the total number of iterations is still proportional to the number of nodes in the linked list, making it $O(n)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this code is $O(1)$, which means it uses constant space.\n",
-    "\n",
-    "The reason for this is that regardless of the size of the linked list, the code only uses two additional pointers (`slow` and `fast`) to traverse the list. These pointers do not depend on the size of the input linked list, so the space complexity remains constant.\n",
-    "\n",
-    "**In summary**, this code efficiently detects cycles in a linked list with a time complexity of $O(n)$ and a space complexity of $O(1)$. It is an example of an algorithm that solves a complex problem with minimal memory usage."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb b/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb
deleted file mode 100644
index 4ec3f3b..0000000
--- a/_build/jupyter_execute/06. Linked List/143. Reorder List.ipynb	
+++ /dev/null
@@ -1,277 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 143. Reorder List\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Reorder List problem on LeetCode, click here!](https://leetcode.com/problems/reorder-list/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given the head of a singly linked-list. The list can be represented as:\n",
-    "\n",
-    "$L_0 → L_1 → … → L_{n - 1} → L_n$\n",
-    "\n",
-    "Reorder the list to be on the following form:\n",
-    "\n",
-    "$L_0 → L_n → L_1 → L_{n - 1} → L_2 → L_{n - 2} → …$\n",
-    "\n",
-    "You may not modify the values in the list's nodes. Only nodes themselves may be changed.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the list is in the range $[1, 5 * 10^4]$.\n",
-    "- `1 <= Node.val <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "def reorderList(head):\n",
-    "    if not head or not head.next:\n",
-    "        return head\n",
-    "\n",
-    "    # Step 1: Find the middle of the linked list\n",
-    "    slow, fast = head, head\n",
-    "    while fast.next and fast.next.next:\n",
-    "        slow = slow.next\n",
-    "        fast = fast.next.next\n",
-    "\n",
-    "    # Split the list into two halves\n",
-    "    first_half = head\n",
-    "    second_half = slow.next\n",
-    "    slow.next = None\n",
-    "\n",
-    "    # Step 2: Reverse the second half of the linked list\n",
-    "    prev = None\n",
-    "    current = second_half\n",
-    "    while current:\n",
-    "        next_node = current.next\n",
-    "        current.next = prev\n",
-    "        prev = current\n",
-    "        current = next_node\n",
-    "    second_half = prev\n",
-    "\n",
-    "    # Step 3: Merge the first and reversed second halves alternately\n",
-    "    p1, p2 = first_half, second_half\n",
-    "    while p2:\n",
-    "        next_p1 = p1.next\n",
-    "        next_p2 = p2.next\n",
-    "        p1.next = p2\n",
-    "        p2.next = next_p1\n",
-    "        p1 = next_p1\n",
-    "        p2 = next_p2\n",
-    "\n",
-    "    return head"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The `reorderList` function is designed to reorder a singly linked list following the specific pattern described in the problem statement. Here's a detailed explanation of how the function works:\n",
-    "\n",
-    "1. **Base Cases Handling**:\n",
-    "   - The function first checks if the input linked list is empty (`not head`) or contains only one element (`not head.next`). If either of these conditions is met, the list cannot be reordered, so the function returns the original list as-is.\n",
-    "\n",
-    "2. **Finding the Middle of the Linked List**:\n",
-    "   - To reorder the list, we first need to find the middle point so that we can split the list into two halves. This is done using two pointers, `slow` and `fast`, initialized to the head of the list.\n",
-    "   - `slow` moves one step at a time while `fast` moves two steps at a time. When `fast` reaches the end of the list or the second-to-last node, `slow` will be at the middle node.\n",
-    "\n",
-    "3. **Splitting the List**:\n",
-    "   - After finding the middle node, we split the list into two halves:\n",
-    "     - The first half, `first_half`, contains nodes from the beginning up to the middle.\n",
-    "     - The second half, `second_half`, contains nodes from the middle to the end.\n",
-    "   - To split the list, we set the `next` pointer of the node before the middle node to `None`.\n",
-    "\n",
-    "4. **Reversing the Second Half**:\n",
-    "   - We reverse the second half of the linked list using the `prev`, `current`, and `next_node` pointers.\n",
-    "   - `prev` is initially `None`, and we iterate through the second half. For each node, we:\n",
-    "     - Set the `next` of the current node to `prev`, effectively reversing the next pointer direction.\n",
-    "     - Update `prev` to the current node.\n",
-    "     - Move to the next node using the `next_node`.\n",
-    "\n",
-    "5. **Merging the Two Halves Alternately**:\n",
-    "   - We now have two linked lists: `first_half` and the reversed `second_half`.\n",
-    "   - We merge these two lists alternately by adjusting the `next` pointers of the nodes.\n",
-    "   - `p1` and `p2` are pointers to the current nodes in `first_half` and `second_half`, respectively.\n",
-    "   - We iterate through both lists while reordering nodes:\n",
-    "     - Set the `next` of `p1` to `p2` to link a node from the first half to a node from the reversed second half.\n",
-    "     - Update `p1` and `p2` to their respective next nodes.\n",
-    "     - Repeat this process until we have processed all nodes in both halves.\n",
-    "\n",
-    "6. **Returning the Reordered List**:\n",
-    "   - After the merging process is complete, the linked list is reordered as specified.\n",
-    "   - The function returns the `head` of the reordered list.\n",
-    "\n",
-    "The overall result is a singly linked list that has been reordered according to the given pattern. The time complexity of this algorithm is O(N), where N is the number of nodes in the linked list, as we traverse the list once to find the middle, once to reverse the second half, and once to merge the two halves."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to print the linked list\n",
-    "def printLinkedList(head):\n",
-    "    result = []\n",
-    "    while head:\n",
-    "        result.append(head.val)\n",
-    "        head = head.next\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the helper functions:\n",
-    "**printLinkedList Function**:\n",
-    "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-    "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-    "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-    "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-    "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-    "\n",
-    "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 4, 2, 3]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "head1 = ListNode(1, ListNode(2, ListNode(3, ListNode(4))))\n",
-    "reorderList(head1)\n",
-    "print(printLinkedList(head1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 5, 2, 4, 3]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "head2 = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))\n",
-    "reorderList(head2)\n",
-    "print(printLinkedList(head2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `reorderList` function:\n",
-    "\n",
-    "**Time Complexity**:\n",
-    "1. **Finding the Middle of the Linked List**: In this step, we use two pointers, one moving one step at a time (`slow`) and the other moving two steps at a time (`fast`). This process takes $O(N/2)$ time, where $N$ is the number of nodes in the linked list.\n",
-    "\n",
-    "2. **Splitting the List**: After finding the middle, we split the list into two halves by setting the `next` pointer of the node before the middle node to `None`. This step takes $O(1)$ time.\n",
-    "\n",
-    "3. **Reversing the Second Half**: We reverse the second half of the linked list using a while loop that iterates through the second half of the list once. Therefore, this step also takes $O(N/2)$ time.\n",
-    "\n",
-    "4. **Merging the Two Halves Alternately**: In this step, we merge the two halves alternately by adjusting the `next` pointers of the nodes. We iterate through both halves once, so this step takes $O(N/2)$ time.\n",
-    "\n",
-    "Overall, the time complexity of the `reorderList` function is dominated by the steps involving reversing and merging the two halves, both of which take $O(N/2)$ time. Therefore, the total time complexity is $O(N)$.\n",
-    "\n",
-    "**Space Complexity**:\n",
-    "The space complexity of the `reorderList` function is primarily determined by the variables and data structures used within the function. Let's break down the space complexity components:\n",
-    "\n",
-    "1. **Constant Space Variables**: The variables `slow`, `fast`, `prev`, `current`, `next_node`, `p1`, and `p2` are used to traverse and manipulate the linked list. These variables occupy constant space, regardless of the input size. Therefore, they contribute $O(1)$ to the space complexity.\n",
-    "\n",
-    "2. **Split Linked Lists**: In the splitting step, we create two new linked list segments (`first_half` and `second_half`) that store references to nodes from the original linked list. These segments occupy space proportional to half of the input list, $O(N/2)$. \n",
-    "\n",
-    "3. **Reversed Second Half**: During the reversal step, we reverse the second half of the linked list in-place without creating any additional data structures. Therefore, it doesn't contribute to additional space complexity.\n",
-    "\n",
-    "4. **Overall**: Combining the above components, the total space complexity is $O(N/2)$, which simplifies to $O(N)$ in terms of space complexity.\n",
-    "\n",
-    "**In summary**, the `reorderList` function has a time complexity of $O(N)$ and a space complexity of $O(N)$ due to the creation of two new linked list segments while splitting the list. The constant space variables used for traversal do not significantly impact the overall space complexity."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Reorder by K Elements:** Generalize the reorderList function to reorder the list in a different pattern, where you reorder the list by $K$ elements at a time. For example, for $K=3$, you would reorder it as ($L_0 → L_1 → L_2 → L_n → L_{n-1} → L_{n-2} → ...$).\n",
-    "You may need to handle cases where the number of nodes is not a multiple of $K$.\n",
-    "\n",
-    "2. **Reorder by Odd and Even Nodes:** Modify the reorderList function to reorder the list in a pattern where odd-indexed nodes (1-based index) come before even-indexed nodes. For example, ($L_0 → L_n → L_1 → L_{n-1} → L_2 → L_{n-2} → ...$)."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb b/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb
deleted file mode 100644
index 73dbf66..0000000
--- a/_build/jupyter_execute/06. Linked List/19. Remove Nth Node From End of List.ipynb	
+++ /dev/null
@@ -1,268 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 19. Remove Nth Node From End of List\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Remove Nth Node From End of List problem on LeetCode, click here!](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the head of a linked list, remove the $n^{th}$ node from the end of the list and return its head.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the list is `sz`.\n",
-    "- `1 <= sz <= 30`\n",
-    "- `0 <= Node.val <= 100`\n",
-    "- `1 <= n <= sz`\n",
-    "\n",
-    "**Follow up:** Could you do this in one pass?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "def removeNthFromEnd(head, n):\n",
-    "    # Create a dummy node to handle the case of removing the head node.\n",
-    "    dummy = ListNode(0)\n",
-    "    dummy.next = head\n",
-    "    first = dummy\n",
-    "    second = dummy\n",
-    "    \n",
-    "    # Advance the first pointer by n+1 nodes.\n",
-    "    for _ in range(n + 1):\n",
-    "        first = first.next\n",
-    "    \n",
-    "    # Move both pointers simultaneously until first reaches the end.\n",
-    "    while first is not None:\n",
-    "        first = first.next\n",
-    "        second = second.next\n",
-    "    \n",
-    "    # Remove the nth node by updating the next pointer of the previous node.\n",
-    "    second.next = second.next.next\n",
-    "    \n",
-    "    return dummy.next  # Return the modified head of the linked list"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a `ListNode` class to represent individual nodes in a linked list. Each node has two attributes: `val` (the value of the node) and `next` (a reference to the next node in the list).\n",
-    "\n",
-    "2. The `removeNthFromEnd` function takes the head of a linked list (`head`) and the value `n`, which represents the position from the end of the list of the node to be removed. It returns the modified head of the linked list after the removal.\n",
-    "\n",
-    "3. We create a dummy node (`dummy`) at the beginning of the list. This dummy node simplifies the code by handling cases where the head of the list needs to be removed.\n",
-    "\n",
-    "4. We initialize two pointers, `first` and `second`, both initially pointing to the dummy node.\n",
-    "\n",
-    "5. To advance the `first` pointer by `n+1` nodes, we use a `for` loop. This positions the `first` pointer `n` nodes ahead of the `second` pointer.\n",
-    "\n",
-    "6. We then move both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This ensures that the `second` pointer ends up pointing to the node that needs to be removed.\n",
-    "\n",
-    "7. To remove the nth node from the end, we update the `next` pointer of the node pointed to by the `second` pointer to skip over the node to be removed.\n",
-    "\n",
-    "8. Finally, we return `dummy.next`, which is the modified head of the linked list without the removed node.\n",
-    "\n",
-    "9. Additionally, there are two helper functions:\n",
-    "   - `createLinkedList(values)` creates a linked list from a list of values.\n",
-    "   - `convertToList(head)` converts a linked list back into a list of values."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to create a linked list from a list of values.\n",
-    "def createLinkedList(values):\n",
-    "    if not values:\n",
-    "        return None\n",
-    "    head = ListNode(values[0])\n",
-    "    current = head\n",
-    "    for val in values[1:]:\n",
-    "        current.next = ListNode(val)\n",
-    "        current = current.next\n",
-    "    return head\n",
-    "\n",
-    "# Helper function to convert a linked list to a list of values.\n",
-    "def convertToList(head):\n",
-    "    result = []\n",
-    "    current = head\n",
-    "    while current:\n",
-    "        result.append(current.val)\n",
-    "        current = current.next\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the two helper functions:\n",
-    "\n",
-    "1. `createLinkedList(values)`: This function creates a linked list from a list of values.\n",
-    "\n",
-    "   - Input: `values` is a list containing values that you want to insert into the linked list.\n",
-    "   - Output: The function returns the head of the created linked list.\n",
-    "\n",
-    "   Explanation:\n",
-    "   - The function starts by checking if the `values` list is empty. If it is, it returns `None` to indicate an empty linked list.\n",
-    "   - If the `values` list is not empty, it creates the first node of the linked list with the value from the first element of `values`.\n",
-    "   - It then iterates through the remaining elements of `values` and creates new nodes for each value, connecting them together using the `next` attribute to form a linked list.\n",
-    "   - Finally, it returns the head of the linked list, which is the first node created.\n",
-    "   \n",
-    "\n",
-    "2. `convertToList(head)`: This function converts a linked list into a list of values.\n",
-    "\n",
-    "   - Input: `head` is the head node of the linked list that you want to convert.\n",
-    "   - Output: The function returns a list containing the values of the nodes in the linked list in order.\n",
-    "\n",
-    "   Explanation:\n",
-    "   - The function starts by initializing an empty list, `result`, to store the values of the linked list nodes.\n",
-    "   - It then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-    "   - The iteration continues until it reaches the end of the linked list (i.e., the `next` attribute of the current node becomes `None`).\n",
-    "   - After the iteration is complete, the function returns the `result` list, which now contains the values of all the nodes in the linked list in the same order as they appear in the linked list.\n",
-    "\n",
-    "These helper functions are useful for creating linked lists from lists of values and converting linked lists back into lists of values, making it easier to work with linked list examples and test cases in a more familiar list format."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 2, 3, 5]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "head1 = createLinkedList([1, 2, 3, 4, 5])\n",
-    "n1 = 2\n",
-    "new_head1 = removeNthFromEnd(head1, n1)\n",
-    "print(convertToList(new_head1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "head2 = createLinkedList([1])\n",
-    "n2 = 1\n",
-    "new_head2 = removeNthFromEnd(head2, n2)\n",
-    "print(convertToList(new_head2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "head3 = createLinkedList([1, 2])\n",
-    "n3 = 1\n",
-    "new_head3 = removeNthFromEnd(head3, n3)\n",
-    "print(convertToList(new_head3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Certainly! Let's analyze the time and space complexity of the `removeNthFromEnd` function in isolation:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The `removeNthFromEnd` function performs two passes through the linked list. In the worst case, it needs to traverse the entire linked list of length $N$.\n",
-    "- The first pass positions the `first` pointer $n$ nodes ahead of the `second` pointer. This pass takes $O(N)$ time because it involves iterating through all N nodes.\n",
-    "- The second pass moves both `first` and `second` pointers simultaneously until `first` reaches the end of the list. This also takes $O(N)$ time in the worst case.\n",
-    "- Therefore, the overall time complexity of the `removeNthFromEnd` function is $O(N)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The `removeNthFromEnd` function uses a constant amount of extra space for its variables (e.g., `first`, `second`, and `dummy`). This space usage does not depend on the size of the linked list.\n",
-    "- The space complexity is $O(1)$, indicating that the space used by the function is constant and independent of the size of the input linked list.\n",
-    "\n",
-    "**In summary**, the `removeNthFromEnd` function has a time complexity of $O(N)$ and a space complexity of $O(1)$. It efficiently removes the nth node from the end of the linked list with a single pass through the list and constant extra space usage."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb b/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb
deleted file mode 100644
index 09de0a3..0000000
--- a/_build/jupyter_execute/06. Linked List/206. Reverse Linked List.ipynb	
+++ /dev/null
@@ -1,297 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 206. Reverse Linked List\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Reverse Linked List problem on LeetCode, click here!](https://leetcode.com/problems/reverse-linked-list/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the `head` of a singly linked list, reverse the list, and return the reversed list.\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the list is the range `[0, 5000]`.\n",
-    "- `-5000 <= Node.val <= 5000`\n",
-    "\n",
-    "**Follow up:** A linked list can be reversed either iteratively or recursively. Could you implement both?"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for singly-linked list.\n",
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "# Function to reverse a linked list iteratively\n",
-    "def reverseListIterative(head):\n",
-    "    prev = None\n",
-    "    current = head\n",
-    "    \n",
-    "    while current:\n",
-    "        next_node = current.next\n",
-    "        current.next = prev\n",
-    "        prev = current\n",
-    "        current = next_node\n",
-    "    \n",
-    "    return prev\n",
-    "\n",
-    "# Function to reverse a linked list recursively\n",
-    "def reverseListRecursive(head):\n",
-    "    if not head or not head.next:\n",
-    "        return head\n",
-    "\n",
-    "    new_head = reverseListRecursive(head.next)\n",
-    "    head.next.next = head\n",
-    "    head.next = None\n",
-    "\n",
-    "    return new_head"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `reverseListIterative(head)`: This function reverses a linked list iteratively using a loop. Here's a step-by-step explanation:\n",
-    "\n",
-    "   - It takes the `head` of the input linked list as a parameter.\n",
-    "   - It initializes two pointers, `prev` and `current`, initially set to `None` and the `head` of the list, respectively.\n",
-    "   - It enters a `while` loop that continues until `current` reaches the end of the list (i.e., becomes `None`).\n",
-    "   - Inside the loop:\n",
-    "     - It stores the `next_node` which is the next node after `current`.\n",
-    "     - It updates the `next` pointer of `current` to point to the `prev` node. This effectively reverses the link direction.\n",
-    "     - It moves `prev` to `current` and `current` to `next_node`, advancing the pointers one step further in the list.\n",
-    "   - Once the loop completes, `prev` will be pointing to the new head of the reversed list (which was the last node of the original list), so it returns `prev`.\n",
-    "\n",
-    "2. `reverseListRecursive(head)`: This function reverses a linked list recursively. Here's how it works:\n",
-    "\n",
-    "   - It takes the `head` of the input linked list as a parameter.\n",
-    "   - The base case is checked: if `head` is `None` or `head.next` is `None`, meaning the list is empty or has only one node, it returns `head` as there's no need to reverse a list with zero or one element.\n",
-    "   - In the recursive case:\n",
-    "     - It calls itself with `head.next`, effectively moving down the list until it reaches the end.\n",
-    "     - Once it reaches the end, it starts reversing the links:\n",
-    "       - `head.next.next` is set to `head`, reversing the link between `head` and the next node.\n",
-    "       - `head.next` is set to `None` to avoid cycles.\n",
-    "   - Finally, it returns the new head of the reversed list, which will be the last node of the original list.\n",
-    "\n",
-    "**In summary,** `reverseListIterative` reverses the linked list by iterating through it and changing the next pointers, while `reverseListRecursive` reverses the linked list by recursively reaching the end and then reversing the links on the way back up the recursion stack. Both functions achieve the same result: reversing the linked list."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to create a linked list from a list of values\n",
-    "def createLinkedList(values):\n",
-    "    if not values:\n",
-    "        return None\n",
-    "    head = ListNode(values[0])\n",
-    "    current = head\n",
-    "    for val in values[1:]:\n",
-    "        current.next = ListNode(val)\n",
-    "        current = current.next\n",
-    "    return head\n",
-    "\n",
-    "# Helper function to convert a linked list to a list for testing\n",
-    "def linkedListToList(head):\n",
-    "    result = []\n",
-    "    current = head\n",
-    "    while current:\n",
-    "        result.append(current.val)\n",
-    "        current = current.next\n",
-    "    return result\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the two helper functions used in the provided code:\n",
-    "\n",
-    "1. `createLinkedList(values)`:\n",
-    "\n",
-    "   - Purpose: This function is used to create a linked list from a list of values. It takes a list of values as input and returns the head of the linked list.\n",
-    "\n",
-    "   - How it works:\n",
-    "     - It first checks if the input list `values` is empty. If it's empty, it returns `None` to indicate an empty linked list.\n",
-    "     - If `values` is not empty, it initializes a `head` node with the value of the first element in `values`.\n",
-    "     - It then iterates through the remaining values in `values`, creating new nodes for each value and linking them together to form a linked list.\n",
-    "     - Finally, it returns the `head` node of the newly created linked list.\n",
-    "\n",
-    "2. `linkedListToList(head)`:\n",
-    "\n",
-    "   - Purpose: This function is used to convert a linked list back into a Python list for testing and output purposes.\n",
-    "\n",
-    "   - How it works:\n",
-    "     - It takes the `head` of the linked list as input.\n",
-    "     - It initializes an empty list called `result`.\n",
-    "     - It then iterates through the linked list, starting from `head`, and appends each node's `val` (value) to the `result` list.\n",
-    "     - This process continues until the end of the linked list is reached.\n",
-    "     - Finally, it returns the `result` list, which contains the values from the linked list in the same order.\n",
-    "\n",
-    "These helper functions are used to simplify the process of creating linked lists from lists of values and converting linked lists back into lists, making it easier to work with linked lists in the provided examples and test cases."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[5, 4, 3, 2, 1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: Iterative\n",
-    "head1 = [1,2,3,4,5]\n",
-    "head1 = createLinkedList(head1)\n",
-    "reversed_head1 = reverseListIterative(head1)\n",
-    "print(linkedListToList(reversed_head1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[2, 1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2: Recursive\n",
-    "head2 = [1,2]\n",
-    "head2 = createLinkedList(head2)\n",
-    "reversed_head2 = reverseListRecursive(head2)\n",
-    "print(linkedListToList(reversed_head2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3: Empty list\n",
-    "head3 = []\n",
-    "head3 = createLinkedList(head3)\n",
-    "reversed_head3 = reverseListIterative(head3)\n",
-    "print(linkedListToList(reversed_head3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the two functions:\n",
-    "\n",
-    "1. `reverseListIterative(head)`:\n",
-    "\n",
-    "   - Time Complexity: $O(n)$\n",
-    "     - The iterative function visits each node in the linked list exactly once in a single pass, where \"$n$\" is the number of nodes in the list. Therefore, the time complexity is linear in the number of nodes.\n",
-    "\n",
-    "   - Space Complexity: $O(1)$\n",
-    "     - This function uses a constant amount of extra space for the `prev`, `current`, and `next_node` pointers. Regardless of the size of the input list, the amount of additional memory used remains the same, so the space complexity is constant.\n",
-    "\n",
-    "2. `reverseListRecursive(head)`:\n",
-    "\n",
-    "   - Time Complexity: $O(n)$\n",
-    "     - The recursive function also visits each node in the linked list exactly once. It recursively traverses the list from the head to the tail, reversing the links on the way back. Therefore, like the iterative approach, the time complexity is $O(n)$, where \"$n$\" is the number of nodes.\n",
-    "\n",
-    "   - Space Complexity: $O(n)$\n",
-    "     - The recursive function uses space on the call stack for each recursive call. In the worst case, when the list has \"$n$\" nodes, it will create \"$n$\" recursive function calls on the stack, resulting in a space complexity of $O(n)$. This is because each function call stores information about its state, including the `head` pointer, until it reaches the base case and starts returning.\n",
-    "\n",
-    "**In summary:**\n",
-    "\n",
-    "- The time complexity of both functions is $O(n)$ because they both visit each node once.\n",
-    "- The space complexity of `reverseListIterative` is $O(1)$ because it uses a constant amount of extra space.\n",
-    "- The space complexity of `reverseListRecursive` is $O(n)$ due to the recursive function calls and the associated call stack space.\n",
-    "\n",
-    "Both functions are efficient in terms of time complexity, but `reverseListIterative` is more memory-efficient because it uses a constant amount of additional space, while `reverseListRecursive` uses space on the call stack proportional to the number of nodes in the list."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **K-Group Reverse:** Modify the reverseListIterative function to reverse the linked list in groups of $K$ elements. For example, if the input linked list is [1, 2, 3, 4, 5, 6], and $K$ is 3, the output should be [3, 2, 1, 6, 5, 4].\n",
-    "\n",
-    "2. **Rotate Linked List:** Write a function to rotate a linked list to the right by $K$ places. For example, if the input is [1, 2, 3, 4, 5] and $K$ is 2, the output should be [4, 5, 1, 2, 3]."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb b/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb
deleted file mode 100644
index b433cf3..0000000
--- a/_build/jupyter_execute/06. Linked List/21. Merge Two Sorted Lists.ipynb	
+++ /dev/null
@@ -1,277 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 21. Merge Two Sorted Lists\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Merge Two Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-two-sorted-lists/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given the heads of two sorted linked lists `list1` and `list2`.\n",
-    "\n",
-    "Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.\n",
-    "\n",
-    "*Return the head of the merged linked list.*\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in both lists is in the range `[0, 50]`.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "- Both `list1` and `list2` are sorted in non-decreasing order."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class ListNode:\n",
-    "    def __init__(self, val=0, next=None):\n",
-    "        self.val = val\n",
-    "        self.next = next\n",
-    "\n",
-    "def mergeTwoLists(list1, list2):\n",
-    "    # Create a dummy node to simplify the code.\n",
-    "    dummy = ListNode()\n",
-    "    current = dummy\n",
-    "\n",
-    "    while list1 and list2:\n",
-    "        if list1.val < list2.val:\n",
-    "            current.next = list1\n",
-    "            list1 = list1.next\n",
-    "        else:\n",
-    "            current.next = list2\n",
-    "            list2 = list2.next\n",
-    "        current = current.next\n",
-    "\n",
-    "    # Append the remaining elements from either list.\n",
-    "    if list1:\n",
-    "        current.next = list1\n",
-    "    else:\n",
-    "        current.next = list2\n",
-    "\n",
-    "    return dummy.next"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We define a class `ListNode` to represent the nodes of a singly-linked list. Each node has a `val` attribute to store the value of the node and a `next` attribute to point to the next node in the list.\n",
-    "\n",
-    "2. We define the `mergeTwoLists` function that takes two linked list heads, `list1` and `list2`, as input.\n",
-    "\n",
-    "3. We create a `dummy` node at the beginning of the merged list. The `dummy` node simplifies the code by serving as a placeholder, and its `next` attribute will point to the actual merged list.\n",
-    "\n",
-    "4. We initialize a `current` pointer to point to the `dummy` node initially. This `current` pointer helps us traverse and build the merged list.\n",
-    "\n",
-    "5. We enter a `while` loop that continues until either `list1` or `list2` becomes empty. Inside the loop, we compare the values of the nodes at the current positions of `list1` and `list2`.\n",
-    "\n",
-    "6. If the `val` of the node in `list1` is smaller than the `val` of the node in `list2`, we attach the node from `list1` to the merged list by updating the `next` attribute of the `current` node to point to the node in `list1`. We then move the `list1` pointer to the next node.\n",
-    "\n",
-    "7. If the `val` of the node in `list2` is smaller than or equal to the `val` of the node in `list1`, we attach the node from `list2` to the merged list in a similar manner. We move the `list2` pointer to the next node.\n",
-    "\n",
-    "8. After attaching a node to the merged list, we move the `current` pointer to the newly added node. This step is essential for keeping track of the end of the merged list.\n",
-    "\n",
-    "9. The loop continues until either `list1` or `list2` becomes empty.\n",
-    "\n",
-    "10. Once the loop exits, we check if there are any remaining elements in `list1` or `list2`. If `list1` is not empty, we attach the remaining elements of `list1` to the merged list. If `list2` is not empty, we attach the remaining elements of `list2`.\n",
-    "\n",
-    "11. Finally, we return the `next` attribute of the `dummy` node as the head of the merged list, which represents the sorted merged list."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "01849ced-1bbf-4956-8e54-451b358cb4fe",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Helper function to print the linked list\n",
-    "def printLinkedList(head):\n",
-    "    result = []\n",
-    "    while head:\n",
-    "        result.append(head.val)\n",
-    "        head = head.next\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "f7490d01-d64e-4f20-b178-e14490d6dd6f",
-   "metadata": {},
-   "source": [
-    "Let's explain the helper functions:\n",
-    "**printLinkedList Function**:\n",
-    "   - The `printLinkedList` function takes the `head` of a linked list as input and returns a list of values representing the nodes in the linked list.\n",
-    "   - Inside the function, a `result` list is initialized to store the values of the linked list nodes.\n",
-    "   - The function then iterates through the linked list starting from the `head` node and appends the `val` attribute of each node to the `result` list.\n",
-    "   - As it iterates through the list, it moves to the next node using the `next` attribute of each node.\n",
-    "   - Finally, the function returns the `result` list containing the values of the linked list nodes in the order they appear in the linked list.\n",
-    "\n",
-    "   This function is useful for debugging and displaying the contents of a linked list. It allows you to easily convert a linked list into a regular Python list for visualization and testing purposes."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "[1, 1, 2, 3, 4, 4]"
-      ]
-     },
-     "execution_count": 3,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "# Input: list1 = [1,2,4], list2 = [1,3,4]\n",
-    "list1 = ListNode(1, ListNode(2, ListNode(4)))\n",
-    "list2 = ListNode(1, ListNode(3, ListNode(4)))\n",
-    "merged_list = mergeTwoLists(list1, list2)\n",
-    "printLinkedList(merged_list)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "[]"
-      ]
-     },
-     "execution_count": 4,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "# Input: list1 = [], list2 = []\n",
-    "list1 = None\n",
-    "list2 = None\n",
-    "merged_list = mergeTwoLists(list1, list2)\n",
-    "printLinkedList(merged_list)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "id": "6c69a707-02af-4212-b2e7-1881d62a62d2",
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "[0]"
-      ]
-     },
-     "execution_count": 5,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# list1 = [], list2 = [0]\n",
-    "list1 = None\n",
-    "list2 = ListNode(0)\n",
-    "merged_list = mergeTwoLists(list1, list2)\n",
-    "printLinkedList(merged_list)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the `mergeTwoLists` function.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this function is $O(N + M$), where $N$ and $M$ are the lengths of `list1` and `list2`, respectively. Here's why:\n",
-    "\n",
-    "1. In the worst case, we need to traverse both `list1` and `list2` completely. This involves iterating through all the nodes in both lists once.\n",
-    "\n",
-    "2. The while loop runs until either `list1` or `list2` becomes empty. The number of iterations depends on the total number of nodes in both lists, which is N + M in the worst case.\n",
-    "\n",
-    "3. Inside the loop, we perform constant-time operations for each iteration, such as comparisons and updating pointers.\n",
-    "\n",
-    "Therefore, the dominant factor in the time complexity is the combined length of both input lists, resulting in a linear time complexity of $O(N + M)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this function is $O(1)$, which means it uses a constant amount of extra space regardless of the input sizes. Here's why:\n",
-    "\n",
-    "1. We create a few extra pointers (`dummy`, `current`) and temporary variables (`val`) to manage the merging process. These consume a fixed amount of memory, regardless of the input sizes. The number of these extra variables is independent of the lengths of `list1` and `list2`.\n",
-    "\n",
-    "2. We do not create a new data structure or allocate memory for the merged list. Instead, we rearrange the existing nodes from `list1` and `list2` to form the merged list. This operation does not consume additional memory proportional to the input sizes.\n",
-    "\n",
-    "**In summary**, the `mergeTwoLists` function has a time complexity of $O(N + M)$ and a space complexity of $O(1)$, making it an efficient solution for merging two sorted linked lists."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Merge K Sorted Lists:** Extend the problem to merge $K$ sorted linked lists instead of just two. You need to efficiently merge $K$ lists into one sorted list.\n",
-    "\n",
-    "2. **Merge Lists in a Zigzag Pattern:** Merge two sorted lists in a zigzag pattern. For example, given [1, 2, 4] and [1, 3, 5], the merged list should be [1, 1, 2, 3, 4, 5]."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb b/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb
deleted file mode 100644
index 0568867..0000000
--- a/_build/jupyter_execute/06. Linked List/23. Merge k Sorted Lists.ipynb	
+++ /dev/null
@@ -1,277 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 23. Merge k Sorted Lists\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Merge k Sorted Lists problem on LeetCode, click here!](https://leetcode.com/problems/merge-k-sorted-lists/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "You are given an array of $k$ linked-lists lists, each linked-list is sorted in ascending order.\n",
-    "\n",
-    "Merge all the linked-lists into one sorted linked-list and return it.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- `k == lists.length`\n",
-    "- 0 <= k <= $10^4$\n",
-    "- `0 <= lists[i].length <= 500`\n",
-    "- $-10^4$ <= lists[i][j] <= $10^4$\n",
-    "- `lists[i]` is sorted in ascending order.\n",
-    "- The sum of `lists[i].length` will not exceed $10^4$.\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for singly-linked list.\n",
-    "class ListNode:\n",
-    "     def __init__(self, val=0, next=None):\n",
-    "         self.val = val\n",
-    "         self.next = next\n",
-    "\n",
-    "class Solution:\n",
-    "    def mergeKLists(self, lists: [[ListNode]]) -> ListNode:\n",
-    "        # Check if the input list of linked lists is empty\n",
-    "        if not lists or len(lists) == 0:\n",
-    "            return None\n",
-    "\n",
-    "        # Loop until there is only one merged list left\n",
-    "        while len(lists) > 1:\n",
-    "            mergedLists = []\n",
-    "\n",
-    "            # Merge pairs of lists\n",
-    "            for i in range(0, len(lists), 2):\n",
-    "                l1 = lists[i]\n",
-    "                l2 = lists[i + 1] if (i + 1) < len(lists) else None\n",
-    "                mergedLists.append(self.mergeList(l1, l2))\n",
-    "\n",
-    "            # Update the list of lists with merged results\n",
-    "            lists = mergedLists\n",
-    "        \n",
-    "        # Return the final merged list\n",
-    "        return lists[0]\n",
-    "\n",
-    "    def mergeList(self, l1, l2):\n",
-    "        # Create a dummy node to simplify merging\n",
-    "        dummy = ListNode()\n",
-    "        tail = dummy\n",
-    "\n",
-    "        # Merge the two sorted lists\n",
-    "        while l1 and l2:\n",
-    "            if l1.val < l2.val:\n",
-    "                tail.next = l1\n",
-    "                l1 = l1.next\n",
-    "            else:\n",
-    "                tail.next = l2\n",
-    "                l2 = l2.next\n",
-    "            tail = tail.next\n",
-    "\n",
-    "        # Append any remaining elements from l1 or l2 (if any)\n",
-    "        if l1:\n",
-    "            tail.next = l1\n",
-    "        if l2:\n",
-    "            tail.next = l2\n",
-    "\n",
-    "        # Return the merged result starting from the next of the dummy node\n",
-    "        return dummy.next\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code defines a Python class `Solution` with two methods for merging k sorted linked lists:\n",
-    "\n",
-    "1. `mergeKLists(self, lists: List[ListNode]) -> ListNode`: This method takes a list of k sorted linked lists as input and returns a single merged sorted linked list. It uses a divide-and-conquer approach to repeatedly merge pairs of lists until only one merged list remains.\n",
-    "\n",
-    "2. `mergeList(self, l1, l2)`: This method takes two sorted linked lists, `l1` and `l2`, as input and merges them into a single sorted linked list. It uses a dummy node to simplify the merging process.\n",
-    "\n",
-    "Here's a high-level overview of how the code works:\n",
-    "\n",
-    "- The `mergeKLists` method checks if the input list of linked lists is empty or contains no lists. If there are no lists, it returns `None`.\n",
-    "\n",
-    "- Inside a `while` loop, the code repeatedly merges pairs of linked lists until only one merged list remains in the `lists` array. It does this by iterating through the input lists in pairs and calling the `mergeList` method to merge each pair.\n",
-    "\n",
-    "- The `mergeList` method takes two sorted linked lists, `l1` and `l2`, and merges them into a single sorted linked list. It uses a dummy node (`dummy`) and a `tail` pointer to keep track of the merged list while comparing and merging elements from `l1` and `l2`.\n",
-    "\n",
-    "- After merging all pairs of lists and updating the `lists` array with the merged results, the loop continues until only one merged list remains in the `lists` array.\n",
-    "\n",
-    "- Finally, the `mergeKLists` method returns the merged list.\n",
-    "\n",
-    "Overall, this code efficiently merges k sorted linked lists using a divide-and-conquer strategy, resulting in a single merged sorted linked list as the output."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6 -> "
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "# Input: lists = [[1,4,5],[1,3,4],[2,6]]\n",
-    "\n",
-    "lists1 = [\n",
-    "    ListNode(1, ListNode(4, ListNode(5))),\n",
-    "    ListNode(1, ListNode(3, ListNode(4))),\n",
-    "    ListNode(2, ListNode(6))\n",
-    "]\n",
-    "solution = Solution()\n",
-    "result1 = solution.mergeKLists(lists1)\n",
-    "\n",
-    "# Print the result\n",
-    "if result1:\n",
-    "    current = result1\n",
-    "    while current:\n",
-    "        print(current.val, end=\" -> \")\n",
-    "        current = current.next\n",
-    "else:\n",
-    "    print(\"None\")  # Print \"None\" for input with a single None element"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "None\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "# Input: lists = []\n",
-    "\n",
-    "lists2 = []\n",
-    "solution = Solution()\n",
-    "result2 = solution.mergeKLists(lists2)\n",
-    "\n",
-    "# Print the result\n",
-    "if result2:\n",
-    "    current = result2\n",
-    "    while current:\n",
-    "        print(current.val, end=\" -> \")\n",
-    "        current = current.next\n",
-    "else:\n",
-    "    print(\"None\")  # Print \"None\" for input with a single None element"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e140b956-7b3d-40d5-b6b0-9615860af67e",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "None\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "# Input: lists = [[]]\n",
-    "\n",
-    "lists3 = [None]\n",
-    "solution = Solution()\n",
-    "result3 = solution.mergeKLists(lists3)\n",
-    "\n",
-    "# Print the result\n",
-    "if result3:\n",
-    "    current = result3\n",
-    "    while current:\n",
-    "        print(current.val, end=\" -> \")\n",
-    "        current = current.next\n",
-    "else:\n",
-    "    print(\"None\")  # Print \"None\" for input with a single None element"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. **Heap Initialization:** The code does not use a heap. Instead, it uses a divide-and-conquer approach. The initial check for empty input lists takes $O(1)$ time.\n",
-    "\n",
-    "2. **Merging:** The merging operation is performed in a divide-and-conquer fashion. In each iteration of the `while` loop, we merge pairs of linked lists, and the number of comparisons made is proportional to the total number of nodes across all the linked lists ($n$). In each merge step, we effectively process each node once. The number of iterations required to reduce k lists to 1 is $O(log\\ k)$.\n",
-    "\n",
-    "   Therefore, the overall time complexity of the code is $O(n * log\\ k)$, where n is the total number of nodes across all lists, and $k$ is the number of linked lists.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. **Heap Space:** The code doesn't use a heap data structure, so there's no additional space complexity due to a heap.\n",
-    "\n",
-    "2. **Merged Lists:** In the `mergeList` method, we create a new merged list. However, this list is not stored in memory for all lists; it's replaced with each merged pair. The space used for these merged lists is proportional to the size of the largest merged list, which is $O(n)$ in the worst case.\n",
-    "\n",
-    "3. **Additional Variables:** The code uses a few additional variables, such as `dummy` and `tail`, but these occupy a constant amount of space and don't depend on the input size.\n",
-    "\n",
-    "   Therefore, the overall space complexity of the code is $O(n)$, where n is the total number of nodes across all lists.\n",
-    "\n",
-    "In summary, the code's time complexity is $O(n * log(k))$, and its space complexity is $O(n)$. This code efficiently merges $k$ sorted linked lists using a divide-and-conquer approach with a relatively low space overhead."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb b/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb
deleted file mode 100644
index 2238b48..0000000
--- a/_build/jupyter_execute/07. Trees/100. Same Tree.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 100: Same Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Same Tree problem on LeetCode, click here!](https://leetcode.com/problems/same-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the roots of two binary trees `p` and `q`, write a function to check if they are the same or not.\n",
-    "\n",
-    "Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.\n",
-    "\n",
-    "**Constraints**\n",
-    "\n",
-    "1. The number of nodes in both trees is in the range `[0, 100]`.\n",
-    "2. $-10^4$ <= Node.val <= $10^4$\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        # Initialize a TreeNode with a value (val), left child, and right child.\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def isSameTree(p, q):\n",
-    "    # Base case: If both p and q are None, the trees are the same.\n",
-    "    if not p and not q:\n",
-    "        return True\n",
-    "    \n",
-    "    # Base case: If either p or q is None (but not both), the trees are different.\n",
-    "    if not p or not q:\n",
-    "        return False\n",
-    "    \n",
-    "    # Check if the values of the current nodes (p.val and q.val) are equal.\n",
-    "    if p.val != q.val:\n",
-    "        return False\n",
-    "    \n",
-    "    # Recursively check the left and right subtrees of p and q.\n",
-    "    # If both subtrees are the same, the entire trees are the same.\n",
-    "    return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "In this code, we define a `TreeNode` class to represent binary tree nodes and a `isSameTree` function to check if two binary trees are the same. The function uses recursive traversal to compare the trees' structures and values.\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class, which represents a node in a binary tree. Each node has a `val` (the node's value), a `left` child, and a right child. This class will help us create and work with binary trees.\n",
-    "2. Next, we define the `isSameTree` function, which checks if two binary trees (`p` and `q`) are the same.\n",
-    "      + The base case for the recursion is when both `p` and `q` are `None`. In this case, they are considered the same, so we return `True`.\n",
-    "      + If either `p` or `q` is `None` (but not both), they cannot be the same, so we return `False`.\n",
-    "      + If the values of the current nodes `p.val` and `q.val` are not equal, we return `False` because the trees cannot be the same.\n",
-    "      + Finally, we recursively check the left and right subtrees of `p` and `q` to see if they are the same."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "\n",
-    "#Input: `p = [1,2,3]`, `q = [1,2,3]`\n",
-    "\n",
-    "p1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "q1 = TreeNode(1, TreeNode(2), TreeNode(3))\n",
-    "print(isSameTree(p1, q1)) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "\n",
-    "#Input: `p = [1,2]`, `q = [1,null,2]`\n",
-    "\n",
-    "p2 = TreeNode(1, TreeNode(2), None)\n",
-    "q2 = TreeNode(1, None, TreeNode(2))\n",
-    "print(isSameTree(p2, q2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "\n",
-    "#Input: p = [1,2,1], q = [1,1,2]\n",
-    "\n",
-    "p3 = TreeNode(1, TreeNode(2), TreeNode(1))\n",
-    "q3 = TreeNode(1, TreeNode(1), TreeNode(2))\n",
-    "print(isSameTree(p3, q3))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time Complexity\n",
-    "\n",
-    "The time complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "In the worst case, the function needs to visit every node in both trees once to determine if they are the same.\n",
-    "Since each node is visited exactly once, the time complexity is $O(n)$, where $n$ is the total number of nodes in the input trees.\n",
-    "\n",
-    "## Space Complexity\n",
-    "The space complexity of the `isSameTree` function can be analyzed as follows:\n",
-    "\n",
-    "The space used by the function's call stack during recursion is proportional to the maximum depth of the binary trees.\n",
-    "In the worst case, when the trees are completely unbalanced (all nodes form a single branch), the maximum depth will be $n$, where $n$ is the total number of nodes in the input trees.\n",
-    "Therefore, the space complexity is $O(n)$ due to the recursive call stack.\n",
-    "In addition to the call stack, there is a small constant amount of space used for variables and comparisons within each recursive call, but this space is not significant in terms of the overall space complexity.\n",
-    "\n",
-    "## In summary:\n",
-    "\n",
-    "+ **Time Complexity:** $O(n)$ where $n$ is the total number of nodes in the input trees.\n",
-    "+ **Space Complexity:** $O(n)$ due to the recursive call stack."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb b/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb
deleted file mode 100644
index 9d2cc7f..0000000
--- a/_build/jupyter_execute/07. Trees/102. Binary Tree Level Order Traversal.ipynb	
+++ /dev/null
@@ -1,263 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 102. Binary Tree Level Order Traversal\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Binary Tree Level Order Traversal problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-level-order-traversal/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary tree, return *the level order traversal of its nodes' values*. (i.e., from left to right, level by level).\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-    "- `-1000 <= Node.val <= 1000`\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def levelOrder(root):\n",
-    "    # Check if the tree is empty, if so, return an empty list.\n",
-    "    if not root:\n",
-    "        return []\n",
-    "\n",
-    "    # Initialize an empty list to store the result.\n",
-    "    result = []\n",
-    "    \n",
-    "    # Initialize a queue with the root node to perform BFS.\n",
-    "    queue = [root]\n",
-    "\n",
-    "    while queue:\n",
-    "        # Initialize an empty list to store the values of nodes at the current level.\n",
-    "        level_values = []\n",
-    "        \n",
-    "        # Get the number of nodes at the current level.\n",
-    "        level_size = len(queue)\n",
-    "\n",
-    "        # Iterate through the nodes at the current level.\n",
-    "        for _ in range(level_size):\n",
-    "            # Dequeue the front node from the queue.\n",
-    "            node = queue.pop(0)\n",
-    "            \n",
-    "            # Append the value of the current node to the level_values list.\n",
-    "            level_values.append(node.val)\n",
-    "\n",
-    "            # Enqueue the left and right children of the current node if they exist.\n",
-    "            if node.left:\n",
-    "                queue.append(node.left)\n",
-    "            if node.right:\n",
-    "                queue.append(node.right)\n",
-    "\n",
-    "        # Append the level_values list (values at the current level) to the result list.\n",
-    "        result.append(level_values)\n",
-    "\n",
-    "    # Return the final result, which is a list of lists representing level order traversal.\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We define a `TreeNode` class that represents a node in the binary tree. Each `TreeNode` object has a value (`val`) and two child nodes: `left` and `right`.\n",
-    "\n",
-    "2. The `levelOrder` function takes the root of the binary tree as its input and returns the level order traversal of the tree as a list of lists.\n",
-    "\n",
-    "3. We start by checking if the input `root` is `None`, which indicates an empty tree. If the tree is empty, we return an empty list because there are no nodes to traverse.\n",
-    "\n",
-    "4. We initialize an empty list called `result` to store the final result, which will be a list of lists containing node values at each level.\n",
-    "\n",
-    "5. We initialize a queue called `queue` with the root node. This queue will be used for breadth-first traversal of the tree.\n",
-    "\n",
-    "6. We enter a while loop that continues until the `queue` is empty. Inside the loop, we perform the following steps:\n",
-    "\n",
-    "   - We initialize an empty list called `level_values` to store the values of nodes at the current level.\n",
-    "\n",
-    "   - We determine the number of nodes at the current level by getting the length of the `queue`. This is done to process nodes level by level.\n",
-    "\n",
-    "   - We iterate through the nodes at the current level using a for loop. For each node in the current level:\n",
-    "\n",
-    "     - We dequeue (remove) the front node from the `queue`.\n",
-    "\n",
-    "     - We append the value of the dequeued node to the `level_values` list, effectively collecting the values of nodes at the current level.\n",
-    "\n",
-    "     - If the dequeued node has a left child, we enqueue the left child to the `queue`.\n",
-    "\n",
-    "     - If the dequeued node has a right child, we enqueue the right child to the `queue`.\n",
-    "\n",
-    "   - After processing all nodes at the current level, we append the `level_values` list to the `result` list. This represents the values at the current level.\n",
-    "\n",
-    "7. The loop continues until all levels have been traversed, and the `queue` becomes empty.\n",
-    "\n",
-    "8. Finally, we return the `result` list, which contains lists of node values at each level, representing the level order traversal of the binary tree.\n",
-    "\n",
-    "The code effectively performs a breadth-first traversal of the binary tree, processing nodes level by level, and constructs the result list that represents the level order traversal."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[3], [9, 20], [15, 7]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Construct the tree\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(9)\n",
-    "root.right = TreeNode(20)\n",
-    "root.right.left = TreeNode(15)\n",
-    "root.right.right = TreeNode(7)\n",
-    "\n",
-    "result = levelOrder(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[1]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Using the same tree as before\n",
-    "root = TreeNode(1)\n",
-    "result = levelOrder(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# Creating a new tree for this example\n",
-    "root = None\n",
-    "result = levelOrder(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `levelOrder` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this function is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the breadth-first traversal.\n",
-    "\n",
-    "In the worst case, we have to enqueue and dequeue all nodes in the binary tree, which is proportional to the number of nodes (N). In each level, we process all nodes in that level, and since there are a total of log(N) levels in a balanced binary tree, the time complexity can also be approximated as O(N) for unbalanced trees.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this function is O(N), where N is the number of nodes in the binary tree. Here's how the space complexity breaks down:\n",
-    "\n",
-    "1. The `result` list stores the level order traversal, and in the worst case, it contains N/2 levels (for a completely unbalanced binary tree). So, the space used by `result` is O(N).\n",
-    "\n",
-    "2. The `queue` data structure is used for BFS traversal. In the worst case, it can store all nodes at the last level of the tree. In a balanced binary tree, the maximum number of nodes at any level is 2^(log(N)), which is still O(N). In the case of an unbalanced tree, it can be even worse. So, the space used by `queue` is O(N).\n",
-    "\n",
-    "Overall, the dominant factor in terms of space complexity is the `queue`, and the space complexity is O(N).\n",
-    "\n",
-    "In summary, the function's time complexity is O(N), and its space complexity is also O(N). It is an efficient and optimal solution for performing a level order traversal of a binary tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Reverse Level Order Traversal:** Modify the `levelOrder` function to return the level order traversal in reverse order (from bottom to top). For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[15, 7], [9, 20], [3]]`.\n",
-    "\n",
-    "2. **Zigzag Level Order Traversal:** Write a function that performs a level order traversal of a binary tree in a zigzag pattern. In a zigzag traversal, the nodes at even levels are traversed from left to right, and nodes at odd levels are traversed from right to left. For example, if the input tree is `[3, 9, 20, null, null, 15, 7]`, the output should be `[[3], [20, 9], [15, 7]]`."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb b/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb
deleted file mode 100644
index 87c5eaa..0000000
--- a/_build/jupyter_execute/07. Trees/104. Maximum Depth of Binary Tree.ipynb	
+++ /dev/null
@@ -1,203 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 104. Maximum Depth of Binary Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the `root` of a binary tree, return its maximum depth.\n",
-    "\n",
-    "A binary tree's **maximum depth** is the number of nodes along the longest path from the root node down to the farthest leaf node.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def maxDepth(root):\n",
-    "    # Base case: If the root is None, the depth is 0.\n",
-    "    if root is None:\n",
-    "        return 0\n",
-    "    \n",
-    "    # Recursively calculate the maximum depth of the left and right subtrees.\n",
-    "    left_depth = maxDepth(root.left)\n",
-    "    right_depth = maxDepth(root.right)\n",
-    "    \n",
-    "    # Return the maximum depth of the tree by adding 1 to the maximum depth of the subtrees.\n",
-    "    return max(left_depth, right_depth) + 1"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Let's go through the code step by step to understand how it works:\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class to represent the nodes of the binary tree. Each node has three attributes:\n",
-    "   - `val`: the value stored in the node.\n",
-    "   - `left`: a reference to the left child node.\n",
-    "   - `right`: a reference to the right child node.\n",
-    "\n",
-    "2. The `maxDepth` function is the main function that calculates the maximum depth of the binary tree. It takes a single argument, `root`, which is the root node of the binary tree.\n",
-    "\n",
-    "3. In the `maxDepth` function, we have a base case:\n",
-    "   - If the `root` is `None`, it means we have reached the end of a branch of the tree (a leaf node or an empty subtree). In this case, the depth is 0 because there are no nodes to count.\n",
-    "\n",
-    "4. If the `root` is not `None`, we recursively calculate the maximum depth of the left and right subtrees:\n",
-    "   - We call `maxDepth` on the `root.left` to calculate the maximum depth of the left subtree and store it in the variable `left_depth`.\n",
-    "   - We call `maxDepth` on the `root.right` to calculate the maximum depth of the right subtree and store it in the variable `right_depth`.\n",
-    "\n",
-    "5. Finally, we return the maximum depth of the tree by taking the maximum of `left_depth` and `right_depth` and adding 1 to it. This is because we are counting the current level as well."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "# Input: root = [3,9,20,null,null,15,7]\n",
-    "\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(9)\n",
-    "root.right = TreeNode(20)\n",
-    "root.right.left = TreeNode(15)\n",
-    "root.right.right = TreeNode(7)\n",
-    "\n",
-    "print(maxDepth(root))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a230be4f-1f47-4919-87e3-62fc3f072d97",
-   "metadata": {},
-   "source": [
-    "In this example, we create a binary tree based on the given input `[3,9,20,null,null,15,7]` and calculate its maximum depth, which is `3`. The tree structure is as follows:\n",
-    "   ```\n",
-    "       3\n",
-    "      / \\\n",
-    "     9  20\n",
-    "       /  \\\n",
-    "      15   7\n",
-    "   ```\n",
-    "\n",
-    "So, the maximum depth is the length of the longest path from the root to a leaf node, which is 3 in this case."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "# Input: root = [1,null,2]\n",
-    "\n",
-    "root = TreeNode(1)\n",
-    "root.right = TreeNode(2)\n",
-    "\n",
-    "# Calculate the maximum depth of the tree and print the result\n",
-    "print(maxDepth(root))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The time complexity of the `maxDepth` function is $O(N)$, where $N$ is the number of nodes in the binary tree.\n",
-    "- This is because in the worst case, the function visits every node exactly once in a depth-first manner.\n",
-    "- The recursion explores all nodes of the tree, so the time complexity is linear with respect to the number of nodes.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity of the `maxDepth` function is $O(H)$, where H is the height of the binary tree.\n",
-    "- In the worst case, if the binary tree is completely unbalanced (skewed), the recursion stack can go as deep as the height of the tree.\n",
-    "- In the best case, if the binary tree is perfectly balanced, the height is $O(log\\ N)$, where $N$ is the number of nodes.\n",
-    "- Therefore, the space complexity can vary from $O(log\\ N)$ to $O(N)$ depending on the shape of the tree.\n",
-    "- In addition to the recursion stack, there is a small constant amount of space used for variables and function call overhead.\n",
-    "\n",
-    "**In summary:**\n",
-    "- The time complexity of the code is $O(N)$ as it visits each node once.\n",
-    "- The space complexity is $O(H)$, where H is the height of the tree, which can vary from $O(log\\ N)$ to $O(N)$ depending on the tree's shape."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb b/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb
deleted file mode 100644
index 0d5bbd6..0000000
--- a/_build/jupyter_execute/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb	
+++ /dev/null
@@ -1,305 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 105. Construct Binary Tree from Preorder and Inorder Traversal\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Construct Binary Tree from Preorder and Inorder Traversal problem on LeetCode, click here!](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)\n",
-    "\n",
-    "---\n",
-    "Given two integer arrays `preorder` and `inorder` where `preorder` is the preorder traversal of a binary tree and `inorder` is the inorder traversal of the same tree, construct and return *the binary tree.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= preorder.length <= 3000`\n",
-    "- `inorder.length == preorder.length`\n",
-    "- `-3000 <= preorder[i], inorder[i] <= 3000`\n",
-    "- `preorder` and `inorder` consist of **unique** values.\n",
-    "- Each value of `inorder` also appears in `preorder`.\n",
-    "- `preorder` is **guaranteed** to be the preorder traversal of the tree.\n",
-    "- `inorder` is **guaranteed** to be the inorder traversal of the tree.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def buildTree(preorder, inorder):\n",
-    "    if not preorder:\n",
-    "        return None\n",
-    "\n",
-    "    # The first element in the preorder traversal is the root of the current subtree\n",
-    "    root_val = preorder[0]\n",
-    "    root = TreeNode(root_val)\n",
-    "\n",
-    "    # Find the index of the root value in the inorder traversal\n",
-    "    root_idx_inorder = inorder.index(root_val)\n",
-    "\n",
-    "    # Recursively build left and right subtrees\n",
-    "    root.left = buildTree(preorder[1:1 + root_idx_inorder], inorder[:root_idx_inorder])\n",
-    "    root.right = buildTree(preorder[1 + root_idx_inorder:], inorder[root_idx_inorder + 1:])\n",
-    "\n",
-    "    return root"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `class TreeNode:`: This is a class definition for a binary tree node. It has three attributes:\n",
-    "   - `val`: The value of the node.\n",
-    "   - `left`: A reference to the left child node.\n",
-    "   - `right`: A reference to the right child node.\n",
-    "\n",
-    "   This class is used to create instances of binary tree nodes, which will be used to build the binary tree.\n",
-    "\n",
-    "2. `def buildTree(preorder, inorder):`: This is the main function that constructs a binary tree from its preorder and inorder traversals.\n",
-    "\n",
-    "   - `preorder`: A list representing the preorder traversal of the binary tree.\n",
-    "   - `inorder`: A list representing the inorder traversal of the binary tree.\n",
-    "\n",
-    "3. `if not preorder: return None`: This line checks if the `preorder` list is empty. If it is, it means there are no nodes to construct, so the function returns `None`.\n",
-    "\n",
-    "4. `root_val = preorder[0]`: The value of the root node is extracted from the first element of the `preorder` list.\n",
-    "\n",
-    "5. `root = TreeNode(root_val)`: A new `TreeNode` object is created with the `root_val` as its value. This represents the root of the current subtree.\n",
-    "\n",
-    "6. `root_idx_inorder = inorder.index(root_val)`: The index of the `root_val` in the `inorder` list is found. This index indicates the position of the root node in the inorder traversal.\n",
-    "\n",
-    "7. `root.left`: The left subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The left subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists up to the `root_idx_inorder`.\n",
-    "\n",
-    "8. `root.right`: The right subtree is constructed recursively by calling the `buildTree` function with the appropriate sublists of `preorder` and `inorder`. The right subtree's preorder and inorder traversals are the slices of `preorder` and `inorder` lists starting from `root_idx_inorder + 1`.\n",
-    "\n",
-    "9. Finally, the function returns the `root` of the subtree it just constructed.\n",
-    "\n",
-    "The recursive approach used here divides the problem of constructing the entire binary tree into smaller subproblems, starting with the root node and then recursively building the left and right subtrees until the entire tree is constructed."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "2d82857b-4fbe-474f-bf33-17b55c20147b",
-   "metadata": {},
-   "source": [
-    "## Helper Function \n",
-    "\n",
-    "Here is a helper function to represent the binary tree elements."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "13e6f3e1-659d-479d-9521-75629b8c782f",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def treeToList(root):\n",
-    "    # Initialize an empty list to store the elements of the binary tree\n",
-    "    result = []\n",
-    "    \n",
-    "    # Initialize a queue for level-order traversal starting with the root\n",
-    "    queue = [root]\n",
-    "\n",
-    "    while queue:\n",
-    "        # Pop the front node from the queue\n",
-    "        node = queue.pop(0)\n",
-    "        \n",
-    "        # If the node is not None (i.e., a valid node), add its value to the result list\n",
-    "        if node:\n",
-    "            result.append(node.val)\n",
-    "            \n",
-    "            # Add the left and right children of the node to the queue\n",
-    "            queue.append(node.left)\n",
-    "            queue.append(node.right)\n",
-    "        else:\n",
-    "            # If the node is None, add None to the result to represent an empty node\n",
-    "            result.append(None)\n",
-    "\n",
-    "    # Remove any trailing None values from the result list\n",
-    "    while result and result[-1] is None:\n",
-    "        result.pop()\n",
-    "\n",
-    "    # Return the resulting list representing the binary tree elements\n",
-    "    return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "86cd4c9d-a120-4dd1-a1bd-530436012574",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Let's break down the `treeToList` helper function step by step:\n",
-    "\n",
-    "1. `def treeToList(root):`: This function takes the root of a binary tree as input and converts it into a list.\n",
-    "\n",
-    "2. `result = []`: Initialize an empty list called `result` to store the elements of the binary tree in list format.\n",
-    "\n",
-    "3. `queue = [root]`: Initialize a queue data structure for a level-order traversal, starting with the root node.\n",
-    "\n",
-    "4. `while queue:`: This starts a loop that continues until the queue is empty, indicating that all nodes have been processed.\n",
-    "\n",
-    "5. `node = queue.pop(0)`: Dequeue the first node from the queue, effectively processing it.\n",
-    "\n",
-    "6. `if node:`: Check if the `node` is not None, which means it's a valid node in the binary tree.\n",
-    "\n",
-    "    a. `result.append(node.val)`: If the node is valid, append its value (`node.val`) to the `result` list. This represents the value of the current node in the binary tree.\n",
-    "\n",
-    "    b. `queue.append(node.left)`: Enqueue the left child of the current node if it exists. This adds the left child to the queue for processing in the next iteration.\n",
-    "\n",
-    "    c. `queue.append(node.right)`: Enqueue the right child of the current node if it exists. This adds the right child to the queue for processing in the next iteration.\n",
-    "\n",
-    "7. `else:`: If the `node` is None, it represents an empty node in the binary tree.\n",
-    "\n",
-    "    a. `result.append(None)`: Append `None` to the `result` list to indicate an empty node.\n",
-    "\n",
-    "8. `while result and result[-1] is None:`: After the traversal is complete, there might be trailing `None` values in the `result` list. This loop removes any such trailing `None` values to ensure a clean representation of the tree's elements.\n",
-    "\n",
-    "9. `return result`: Return the `result` list, which now contains the elements of the binary tree in a format where `None` represents empty nodes and the order of elements reflects a level-order traversal of the tree.\n",
-    "\n",
-    "In summary, the `treeToList` function performs a level-order traversal of the binary tree using a queue, constructing a list where each element corresponds to a node's value or represents an empty node with `None`. This list represents the binary tree's elements in a structured format."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[3, 9, 20, None, None, 15, 7]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "preorder = [3, 9, 20, 15, 7]\n",
-    "inorder = [9, 3, 15, 20, 7]\n",
-    "result = buildTree(preorder, inorder)\n",
-    "\n",
-    "print(treeToList(result))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[-1]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "preorder = [-1]\n",
-    "inorder = [-1]\n",
-    "result = buildTree(preorder, inorder)\n",
-    "\n",
-    "print(treeToList(result))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `buildTree` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of the `buildTree` function can be analyzed in terms of the number of nodes in the binary tree. Let's assume there are `n` nodes in the binary tree.\n",
-    "\n",
-    "1. Finding the root value in the `preorder` traversal takes O(1) time as it's just extracting the first element from the list.\n",
-    "2. Finding the index of the root value in the `inorder` traversal using `inorder.index(root_val)` takes O(n) time in the worst case because in the worst case, it might have to search through all `n` elements in the `inorder` list to find the index.\n",
-    "3. The recursive calls to `buildTree` for the left and right subtrees are made once for each node in the tree. Therefore, the recursive calls have a combined time complexity of O(n) as they process each node once.\n",
-    "\n",
-    "The total time complexity of the `buildTree` function is O(n) due to the recursive calls and finding the root's index in the `inorder` list.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity is determined by the space used by the function's call stack during recursion and any additional data structures used. \n",
-    "\n",
-    "1. The space used by the function call stack during recursion depends on the height of the binary tree. In the worst case, where the tree is highly unbalanced (e.g., a skewed tree), the space complexity for the call stack is O(n) as it can go as deep as the number of nodes in the tree.\n",
-    "\n",
-    "2. Additionally, the function creates TreeNode objects for each node in the binary tree. Therefore, the space complexity for these objects is also O(n).\n",
-    "\n",
-    "Overall, the space complexity of the `buildTree` function is O(n) due to the space used by the call stack and the TreeNode objects.\n",
-    "\n",
-    "In summary, the time complexity of `buildTree` is O(n), and the space complexity is O(n), where 'n' is the number of nodes in the binary tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Construct Binary Tree from Postorder and Inorder Traversal:**\n",
-    "   Modify the problem to construct a binary tree from its postorder and inorder traversals. Implement a function similar to `buildTree` but for postorder and inorder traversals.\n",
-    "\n",
-    "2. **Reconstruct Binary Tree with Duplicate Values:**\n",
-    "   Extend the problem to handle binary trees with duplicate values. Ensure that your solution correctly handles scenarios where there are duplicate values in the preorder and inorder traversals."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb b/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb
deleted file mode 100644
index b661fa1..0000000
--- a/_build/jupyter_execute/07. Trees/124. Binary Tree Maximum Path Sum.ipynb	
+++ /dev/null
@@ -1,243 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 124. Binary Tree Maximum Path Sum\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Binary Tree Maximum Path Sum problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-maximum-path-sum/)\n",
-    "\n",
-    "---\n",
-    "A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.\n",
-    "\n",
-    "The **path sum** of a path is the sum of the node's values in the path.\n",
-    "\n",
-    "Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[1, 3*10^4]$.\n",
-    "- `-1000 <= Node.val <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to find the maximum path sum in a binary tree. A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. The path sum of a path is the sum of the node values in that path.\n",
-    "\n",
-    "In this problem, you are given the root of a binary tree, and you need to find the maximum path sum among all possible paths in the tree. Note that the path does not need to start at the root node or end at a leaf node; it can start and end at any nodes in the tree.\n",
-    "\n",
-    "To find the maximum path sum, you need to consider both left and right subtrees of each node while traversing the tree. At each node, you have two choices:\n",
-    "\n",
-    "1. Include the current node in the path: In this case, you add the value of the current node to the sum and continue exploring both the left and right subtrees for possible extensions of the path.\n",
-    "\n",
-    "2. Start a new path from the current node: In this case, you do not include the current node in the path sum, and you choose either the left subtree or the right subtree to start a new path.\n",
-    "\n",
-    "To solve this problem, you can use a recursive approach to traverse the binary tree. For each node, you calculate the maximum path sum that can pass through that node (option 1) and also return the maximum path sum that can be extended from that node (option 2). You keep track of the global maximum path sum encountered during the traversal.\n",
-    "\n",
-    "Ultimately, the maximum path sum among all paths in the tree will be the maximum of the global maximum path sum and the maximum path sum calculated for each node.\n",
-    "\n",
-    "The constraints for this problem include:\n",
-    "\n",
-    "- The number of nodes in the tree is in the range $[1, 3 * 10^4]$.\n",
-    "- Node values are in the range [-1000, 1000].\n",
-    "\n",
-    "By considering all possible paths through the tree, the algorithm aims to find the most significant sum of node values in any path in the binary tree.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "class Solution:\n",
-    "    def maxPathSum(self, root):\n",
-    "        def max_path_sum_helper(node):\n",
-    "            if not node:\n",
-    "                return 0\n",
-    "            \n",
-    "            # Recursively calculate the maximum path sum for left and right subtrees\n",
-    "            left_max = max(0, max_path_sum_helper(node.left))\n",
-    "            right_max = max(0, max_path_sum_helper(node.right))\n",
-    "            \n",
-    "            # Update the global maximum path sum\n",
-    "            self.max_sum = max(self.max_sum, left_max + right_max + node.val)\n",
-    "            \n",
-    "            # Return the maximum path sum starting from the current node\n",
-    "            return max(left_max, right_max) + node.val\n",
-    "        \n",
-    "        self.max_sum = float('-inf')  # Initialize the global maximum to negative infinity\n",
-    "        max_path_sum_helper(root)  # Start the recursive traversal from the root\n",
-    "        return self.max_sum"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. The code starts by defining a `TreeNode` class to represent nodes in the binary tree. Each `TreeNode` has a `val` attribute representing its value, a `left` attribute pointing to the left child, and a `right` attribute pointing to the right child.\n",
-    "\n",
-    "2. Next, a `Solution` class is defined to contain the `maxPathSum` method, which will be used to find the maximum path sum in the binary tree.\n",
-    "\n",
-    "3. Within the `maxPathSum` method, a helper function called `max_path_sum_helper` is defined. This recursive function takes a single argument, `node`, which is the current node being considered during traversal.\n",
-    "\n",
-    "4. Inside the `max_path_sum_helper` function:\n",
-    "   - If `node` is `None` (i.e., the current node is a null node), it returns 0. This is the base case of the recursion.\n",
-    "   - It calculates the maximum path sum starting from the left subtree (`left_max`) and the right subtree (`right_max`). Importantly, it uses `max(0, ...)` to ensure that negative values (which would make the path sum smaller) are not considered. If a subtree's path sum is negative, it's better to not include it in the path.\n",
-    "   - The code then updates the global maximum path sum (`self.max_sum`) by checking if the sum of `left_max`, `right_max`, and the current node's value is greater than the current maximum.\n",
-    "   - Finally, it returns the maximum path sum starting from the current node, which is the maximum of `left_max` and `right_max` plus the current node's value.\n",
-    "\n",
-    "5. Before calling the `max_path_sum_helper` function, the `max_sum` attribute of the `Solution` class is initialized to negative infinity (`float('-inf')`). This is done to ensure that any valid path sum encountered during traversal will be greater than the initial value of `max_sum`.\n",
-    "\n",
-    "6. The `max_path_sum_helper` function is called with the root of the binary tree, effectively starting the traversal from the root node.\n",
-    "\n",
-    "7. Once the traversal is complete, the method returns the value of `self.max_sum`, which contains the maximum path sum found in the binary tree.\n",
-    "\n",
-    "8. Example usage at the bottom of the code demonstrates how to create binary trees and use the `maxPathSum` method to find the maximum path sum for two different tree structures.\n",
-    "\n",
-    "In summary, the code uses a recursive approach to explore all possible paths in the binary tree, ensuring that negative path sums are not considered. It keeps track of the maximum path sum encountered and returns that maximum value as the result."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "6\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "root1 = TreeNode(1)\n",
-    "root1.left = TreeNode(2)\n",
-    "root1.right = TreeNode(3)\n",
-    "solution = Solution()\n",
-    "print(solution.maxPathSum(root1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "42\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "root2 = TreeNode(-10)\n",
-    "root2.left = TreeNode(9)\n",
-    "root2.right = TreeNode(20)\n",
-    "root2.right.left = TreeNode(15)\n",
-    "root2.right.right = TreeNode(7)\n",
-    "print(solution.maxPathSum(root2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the given code:\n",
-    "\n",
-    "Time Complexity:\n",
-    "- The code uses a recursive depth-first traversal to explore each node in the binary tree exactly once.\n",
-    "- During the traversal, for each node, we perform constant time operations, such as updating the `max_sum` variable and calculating the maximum path sum starting from that node.\n",
-    "- Therefore, the time complexity of the code is O(N), where N is the number of nodes in the binary tree.\n",
-    "\n",
-    "Space Complexity:\n",
-    "- The space complexity of the code is determined by the space used in the call stack during the recursive traversal.\n",
-    "- In the worst case, the depth of the call stack can be equal to the height of the binary tree. In an unbalanced tree, this could be O(N), but in a balanced binary tree, the height is O(log N).\n",
-    "- Additionally, the code uses a constant amount of space for variables like `left_max`, `right_max`, and `max_sum`.\n",
-    "- Therefore, the overall space complexity is O(H), where H is the height of the binary tree. In the worst case, it's O(N), and in the best case (a balanced binary tree), it's O(log N).\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: O(N)\n",
-    "- Space Complexity: O(H), where H is the height of the binary tree, ranging from O(log N) to O(N) depending on the tree's balance."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Print the Maximum Path:**\n",
-    "   Modify the code to not only find the maximum path sum but also print the nodes involved in the maximum path. You can print the nodes in the correct order from the root to the leaf.\n",
-    "\n",
-    "2. **Path Sum Count:**\n",
-    "   Instead of finding the maximum path sum, find the count of unique paths that sum to a given target value. Each path can start and end anywhere in the tree."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb b/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb
deleted file mode 100644
index 5bbdf42..0000000
--- a/_build/jupyter_execute/07. Trees/199. Binary Tree Right Side View.ipynb	
+++ /dev/null
@@ -1,249 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 199. Binary Tree Right Side View\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Binary Tree Right Side View problem on LeetCode, click here!](https://leetcode.com/problems/binary-tree-right-side-view/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary tree, imagine yourself standing on the **right side** of it, return *the values of the nodes you can see ordered from top to bottom*.\n",
-    "\n",
-    " \n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range `[0, 2000]`.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def rightSideView(root):\n",
-    "    if not root:\n",
-    "        return []  # If the tree is empty, return an empty list.\n",
-    "\n",
-    "    result = []  # Initialize a list to store the right side view values.\n",
-    "    queue = [root]  # Initialize a queue for level-order traversal with the root node.\n",
-    "\n",
-    "    while queue:\n",
-    "        # Get the number of nodes at the current level.\n",
-    "        level_size = len(queue)\n",
-    "\n",
-    "        # Traverse all nodes at the current level and add the rightmost node to the result.\n",
-    "        for i in range(level_size):\n",
-    "            node = queue.pop(0)  # Dequeue the first node from the queue.\n",
-    "\n",
-    "            # If it's the rightmost node at this level, add its value to the result.\n",
-    "            if i == level_size - 1:\n",
-    "                result.append(node.val)\n",
-    "\n",
-    "            # Add the children of the current node to the queue.\n",
-    "            if node.left:\n",
-    "                queue.append(node.left)\n",
-    "            if node.right:\n",
-    "                queue.append(node.right)\n",
-    "\n",
-    "    return result  # Return the list of right side view values.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start with a binary tree represented using a TreeNode class. Each node has a value (`val`) and can have a left child (`left`) and a right child (`right`).\n",
-    "\n",
-    "2. The `rightSideView` function takes the root node of the binary tree as input and returns a list of values representing the nodes you can see when standing on the right side of the tree.\n",
-    "\n",
-    "3. We check if the `root` is `None`, which means the tree is empty. If it's empty, we return an empty list because there are no nodes to see.\n",
-    "\n",
-    "4. We initialize an empty list called `result` to store the right side view values.\n",
-    "\n",
-    "5. We initialize a queue called `queue` with the root node. This queue is used for a level-order traversal of the tree.\n",
-    "\n",
-    "6. We enter a while loop that continues until the `queue` is empty. Inside this loop, we perform the following steps for each level of the tree:\n",
-    "\n",
-    "   a. We determine the number of nodes at the current level by getting the length of the `queue`. This is important for processing nodes at the same level together.\n",
-    "\n",
-    "   b. We then iterate through the nodes at the current level using a for loop. For each node in the level, we dequeue it from the `queue` using `queue.pop(0)`.\n",
-    "\n",
-    "   c. If the node we dequeue is the rightmost node at the current level (determined by `i == level_size - 1`), we add its `val` to the `result` list. This is because, when standing on the right side of the tree, you can see the rightmost node at each level.\n",
-    "\n",
-    "   d. We enqueue the left and right children of the current node (if they exist) to the `queue`. This ensures that we process the next level in the subsequent iterations of the while loop.\n",
-    "\n",
-    "7. After processing all levels of the tree, the `result` list contains the values of the rightmost nodes in each level, ordered from top to bottom.\n",
-    "\n",
-    "8. Finally, we return the `result` list as the output of the function, which represents the right side view of the binary tree.\n",
-    "\n",
-    "The code effectively performs a level-order traversal of the binary tree while keeping track of the rightmost nodes at each level and adding them to the result list. This ensures that we obtain the correct order of nodes visible from the right side of the tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 3, 4]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Construct the tree\n",
-    "root1 = TreeNode(1)\n",
-    "root1.left = TreeNode(2)\n",
-    "root1.right = TreeNode(3)\n",
-    "root1.left.right = TreeNode(5)\n",
-    "root1.right.right = TreeNode(4)\n",
-    "print(rightSideView(root1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[1, 3]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "root2 = TreeNode(1)\n",
-    "root2.right = TreeNode(3)\n",
-    "print(rightSideView(root2))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# Creating a new tree for this example\n",
-    "root = None\n",
-    "result = rightSideView(root)\n",
-    "print(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the given code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of the code is O(N), where N is the number of nodes in the binary tree. Here's why:\n",
-    "\n",
-    "1. We perform a level-order traversal of the tree using a queue. In the worst case, we visit all nodes once, which takes O(N) time since we visit each node exactly once.\n",
-    "\n",
-    "2. For each node, we perform constant-time operations like dequeuing it from the queue, checking if it's the rightmost node at the current level, and enqueuing its children (if they exist). These operations do not depend on the size of the tree, so they do not contribute to the overall time complexity.\n",
-    "\n",
-    "Therefore, the dominant factor in the time complexity is the level-order traversal, making it O(N).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of the code is also O(N). Here's why:\n",
-    "\n",
-    "1. We use a queue to perform level-order traversal. In the worst case, the queue can contain all nodes at the maximum width of the tree, which can be up to N/2 nodes for a completely unbalanced tree. Therefore, the space required for the queue is O(N).\n",
-    "\n",
-    "2. The `result` list stores the values of the rightmost nodes. In the worst case, when the binary tree is a complete binary tree, it can have roughly N/2 rightmost nodes. Therefore, the space required for the `result` list is also O(N).\n",
-    "\n",
-    "3. Other auxiliary variables like `level_size`, `i`, and `node` require only constant space and do not depend on the size of the tree.\n",
-    "\n",
-    "Combining the space used by the queue and the `result` list, we get a space complexity of O(N).\n",
-    "\n",
-    "In summary, the code has a time complexity of O(N) and a space complexity of O(N), where N is the number of nodes in the binary tree. These complexities are efficient and scale linearly with the size of the input tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Variant with Left Side View**: Modify the code to find and return the values of nodes visible from the left side of the binary tree instead of the right side.\n",
-    "\n",
-    "2. **Zigzag Right Side View**: Extend the code to return the right side view values in a zigzag order. In a zigzag order, you alternate between starting from the rightmost node at level 0, then the leftmost at level 1, rightmost at level 2, and so on."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb b/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb
deleted file mode 100644
index 3e542c9..0000000
--- a/_build/jupyter_execute/07. Trees/226. Invert Binary Tree.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 226: Invert Binary Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Invert Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/invert-binary-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given the root of a binary tree, invert the tree, and return its root.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- The number of nodes in the tree is in the range `[0, 100]`.\n",
-    "- `-100 <= Node.val <= 100`\n",
-    "\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def invertTree(root):\n",
-    "    # Base case: If the root is None or the tree is empty, return None.\n",
-    "    if not root:\n",
-    "        return None\n",
-    "\n",
-    "    # Swap the left and right subtrees of the current node.\n",
-    "    root.left, root.right = root.right, root.left\n",
-    "\n",
-    "    # Recursively invert the left and right subtrees.\n",
-    "    invertTree(root.left)\n",
-    "    invertTree(root.right)\n",
-    "\n",
-    "    return root"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a TreeNode class, which represents a node in a binary tree. Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-    "\n",
-    "2. The `invertTree` function is defined to invert the binary tree. It takes the root node of the tree as an argument.\n",
-    "\n",
-    "3. In the `invertTree` function, we have a base case to handle the scenario when the root is `None` (empty tree). In such cases, we return `None` because there's nothing to invert.\n",
-    "\n",
-    "4. For non-empty trees, we swap the left and right subtrees of the current node. This effectively inverts the tree at the current node.\n",
-    "\n",
-    "5. We then recursively call `invertTree` on the left and right subtrees to invert them.\n",
-    "\n",
-    "6. Finally, we return the root of the inverted tree."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Inverted Tree (Example 1): <__main__.TreeNode object at 0x7faa5c445ab0>\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 1\n",
-    "# Input: root = [4,2,7,1,3,6,9]\n",
-    "\n",
-    "root1 = TreeNode(4)\n",
-    "root1.left = TreeNode(2)\n",
-    "root1.right = TreeNode(7)\n",
-    "root1.left.left = TreeNode(1)\n",
-    "root1.left.right = TreeNode(3)\n",
-    "root1.right.left = TreeNode(6)\n",
-    "root1.right.right = TreeNode(9)\n",
-    "\n",
-    "inverted_root1 = invertTree(root1)\n",
-    "print(\"Inverted Tree (Example 1):\", inverted_root1)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Inverted Tree (Example 2): <__main__.TreeNode object at 0x7faa5c444310>\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 2\n",
-    "# Input: root = [2,1,3]\n",
-    "\n",
-    "root2 = TreeNode(2)\n",
-    "root2.left = TreeNode(1)\n",
-    "root2.right = TreeNode(3)\n",
-    "\n",
-    "inverted_root2 = invertTree(root2)\n",
-    "print(\"Inverted Tree (Example 2):\", inverted_root2)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "5897c29d-26f8-486a-878c-43c09ff25ce4",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Inverted Tree (Example 3): None\n"
-     ]
-    }
-   ],
-   "source": [
-    "### Example 3\n",
-    "# Input: root = []\n",
-    "\n",
-    "root3 = None  # Empty tree\n",
-    "\n",
-    "inverted_root3 = invertTree(root3)\n",
-    "print(\"Inverted Tree (Example 3):\", inverted_root3)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "Let's discuss the time and space complexity of the provided Python code to invert a binary tree.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of the `invertTree` function is $O(n)$, where $n$ is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal of the tree. In the worst case, we have to visit all nodes in the tree.\n",
-    "\n",
-    "The reason for this time complexity is the depth-first traversal of the tree, where we recursively process the left and right subtrees of each node.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of the code is determined by the function call stack during the recursion. In the worst case, the space complexity is $O(h)$, where $h$ is the height of the binary tree. \n",
-    "\n",
-    "In a completely unbalanced binary tree (essentially a linked list), the height $h$ is equal to the number of nodes $n$, resulting in a space complexity of $O(n)$. This occurs when the tree is skewed to one side.\n",
-    "\n",
-    "In a balanced binary tree, such as a full binary tree, the height $h$ is $O(log\\ n)$, and the space complexity is $O(log\\ n)$.\n",
-    "\n",
-    "The space complexity depends on how balanced the tree is. In practical scenarios, binary trees are often approximately balanced, so the space complexity is typically closer to $O(log\\ n)$.\n",
-    "\n",
-    "**In summary:**\n",
-    "\n",
-    "- Time Complexity: $O(n)$ where $n$ is the number of nodes.\n",
-    "- Space Complexity: $O(h)$, where $h$ is the height of the binary tree. In the worst case, it can be $O(n)$, and in a balanced tree, it is $O(log\\ n)$.\n",
-    "\n",
-    "Keep in mind that these complexities are based on the recursive implementation provided. Iterative solutions can achieve the same task with $O(1)$ space complexity, using auxiliary data structures like stacks or queues to mimic the recursion stack."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb b/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb
deleted file mode 100644
index fdad9ba..0000000
--- a/_build/jupyter_execute/07. Trees/230. Kth Smallest Element in a BST.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 230. Kth Smallest Element in a BST\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Kth Smallest Element in a BST problem on LeetCode, click here!](https://leetcode.com/problems/kth-smallest-element-in-a-bst/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary search tree, and an integer `k`, return *the $k^{th}$ smallest value (**1-indexed**) of all the values of the nodes in the tree.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is `n`.\n",
-    "- 1 <= `k <= n` <= $10^4$\n",
-    "- 0 <= `Node.val` <= $10^4$\n",
-    "\n",
-    "**Follow up**: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def kthSmallest(root, k):\n",
-    "    stack = []  # Initialize an empty stack to simulate the traversal\n",
-    "    while root or stack:\n",
-    "        while root:\n",
-    "            stack.append(root)  # Push the current node onto the stack\n",
-    "            root = root.left  # Move to the left child\n",
-    "        root = stack.pop()  # Pop a node from the stack\n",
-    "        k -= 1  # Decrement k since we've visited a node\n",
-    "        if k == 0:\n",
-    "            return root.val  # If k becomes 0, return the current node's value as the kth smallest\n",
-    "        root = root.right  # Move to the right child to continue the traversal"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. We start by defining a `TreeNode` class, which represents a node in the binary search tree (BST). Each node has a value (`val`), a left child (`left`), and a right child (`right`).\n",
-    "\n",
-    "2. The `kthSmallest` function takes two parameters: `root`, which is the root of the BST, and `k`, which represents the kth smallest element we want to find.\n",
-    "\n",
-    "3. We initialize an empty stack (`stack`) to simulate the traversal of the BST. The stack will be used to keep track of nodes as we traverse the tree in an iterative manner.\n",
-    "\n",
-    "4. We enter a while loop that continues until either `root` becomes `None` (indicating we have traversed the entire tree) or the `stack` is empty.\n",
-    "\n",
-    "5. Within the loop, we start another while loop to traverse as far left as possible in the BST. We repeatedly push nodes onto the `stack` and move to their left children until we reach the leftmost leaf node. This is the smallest node in the BST.\n",
-    "\n",
-    "6. Once we have reached the leftmost leaf node (the smallest node), we pop nodes from the `stack` one by one. As we pop each node, we decrement `k` by 1 to keep track of the number of nodes we have visited.\n",
-    "\n",
-    "7. If `k` becomes 0 after decrementing, it means we have found the kth smallest element. In this case, we return the `val` of the current node as the result.\n",
-    "\n",
-    "8. If `k` is still greater than 0, it means we haven't found the kth smallest element yet. In this case, we move to the right child of the current node to continue the traversal, as the kth smallest element, if it exists, will be in the right subtree of the current node.\n",
-    "\n",
-    "9. The process continues until we find the kth smallest element or traverse the entire tree.\n",
-    "\n",
-    "10. Finally, we return the kth smallest element found.\n",
-    "\n",
-    "The key idea in this code is to perform an in-order traversal of the BST while keeping track of the kth smallest element. By visiting nodes in ascending order, we can efficiently find the kth smallest element in O(h + k) time, where h is the height of the BST and k is the desired kth element. This approach is both concise and efficient for this problem."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "# Create the tree for the first example: root = [3,1,4,null,2], k = 1\n",
-    "root1 = TreeNode(3)\n",
-    "root1.left = TreeNode(1)\n",
-    "root1.right = TreeNode(4)\n",
-    "root1.left.right = TreeNode(2)\n",
-    "k1 = 1\n",
-    "print(kthSmallest(root1, k1)) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Create the tree for the second example: root = [5,3,6,2,4,null,null,1], k = 3\n",
-    "root2 = TreeNode(5)\n",
-    "root2.left = TreeNode(3)\n",
-    "root2.right = TreeNode(6)\n",
-    "root2.left.left = TreeNode(2)\n",
-    "root2.left.right = TreeNode(4)\n",
-    "root2.left.left.left = TreeNode(1)\n",
-    "k2 = 3\n",
-    "print(kthSmallest(root2, k2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the given code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The time complexity of this code is O(h + k), where:\n",
-    "- h is the height of the binary search tree (BST).\n",
-    "- k is the desired kth smallest element we want to find.\n",
-    "\n",
-    "1. In the worst case, where the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the time complexity is O(n + k).\n",
-    "\n",
-    "2. In the best case, where the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the time complexity is O(log(n) + k).\n",
-    "\n",
-    "So, the time complexity can vary from O(log(n) + k) in the best-case scenario to O(n + k) in the worst-case scenario. Typically, for balanced BSTs, the time complexity is closer to O(log(n) + k).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "The space complexity of this code is O(h) due to the stack used for the iterative in-order traversal, where h is the height of the BST.\n",
-    "\n",
-    "1. In the worst case, when the BST is highly unbalanced and resembles a linked list, the height (h) of the tree can be equal to the number of nodes (n) in the tree. In this case, the space complexity is O(n) because the stack can potentially store all n nodes.\n",
-    "\n",
-    "2. In the best case, when the BST is perfectly balanced, the height (h) is log(n), where n is the number of nodes in the tree. In this case, the space complexity is O(log(n)) because the stack will have at most log(n) nodes.\n",
-    "\n",
-    "So, the space complexity depends on the height of the BST and can vary from O(log(n)) in the best-case scenario to O(n) in the worst-case scenario.\n",
-    "\n",
-    "In practical terms, for balanced BSTs or moderately unbalanced BSTs, the space complexity is usually close to O(log(n)), and the code is efficient in terms of space usage."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Kth Largest Element**: Modify the code to find the kth largest element in the BST instead of the kth smallest element.\n",
-    "\n",
-    "2. **Kth Smallest Element in Two BSTs**: Given two BSTs, find the kth smallest element when considering both BSTs as a single sorted list. This involves merging the two BSTs while finding the kth element efficiently."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb b/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb
deleted file mode 100644
index 0179ca4..0000000
--- a/_build/jupyter_execute/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,237 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 235. Lowest Common Ancestor of a Binary Search Tree\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Lowest Common Ancestor of a Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)\n",
-    "\n",
-    "---\n",
-    "Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.\n",
-    "\n",
-    "According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[2, 10^5]$.\n",
-    "- $-10^9$ <= `Node.val` <= $10^9$\n",
-    "- All `Node.val` are **unique**.\n",
-    "- `p != q`\n",
-    "- `p` and `q` will exist in the BST.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def lowestCommonAncestor(root, p, q):\n",
-    "    # Ensure p is less than q\n",
-    "    if p.val > q.val:\n",
-    "        p, q = q, p\n",
-    "\n",
-    "    while root:\n",
-    "        # If the current node value is greater than both p and q, move left\n",
-    "        if root.val > q.val:\n",
-    "            root = root.left\n",
-    "        # If the current node value is less than both p and q, move right\n",
-    "        elif root.val < p.val:\n",
-    "            root = root.right\n",
-    "        # If the current node value is between p and q (inclusive), or it matches either p or q, it's the LCA\n",
-    "        else:\n",
-    "            return root"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. `class TreeNode:` defines a simple class for representing nodes in a binary tree. Each `TreeNode` has three attributes:\n",
-    "   - `val`: The value of the node.\n",
-    "   - `left`: A reference to the left child node.\n",
-    "   - `right`: A reference to the right child node.\n",
-    "\n",
-    "2. `def lowestCommonAncestor(root, p, q):` is a function that takes three arguments:\n",
-    "   - `root`: The root node of the BST.\n",
-    "   - `p`: A TreeNode representing one of the nodes for which we want to find the LCA.\n",
-    "   - `q`: A TreeNode representing the other node for which we want to find the LCA.\n",
-    "\n",
-    "3. `if p.val > q.val:` checks if the value of `p` is greater than the value of `q`. In a BST, it's essential to ensure that `p` represents the smaller value, and `q` represents the larger value. If `p` is greater than `q`, the code swaps their values.\n",
-    "\n",
-    "4. The main logic is inside the `while` loop, which runs until `root` becomes `None`. It performs the following steps to find the LCA:\n",
-    "   - If the current `root` node's value is greater than the value of `q`, it means both `p` and `q` are on the left subtree of the current node. So, we move to the left child of the current node by setting `root = root.left`.\n",
-    "   - If the current `root` node's value is less than the value of `p`, it means both `p` and `q` are on the right subtree of the current node. So, we move to the right child of the current node by setting `root = root.right`.\n",
-    "   - If neither of the above conditions is met, it means the current `root` node's value is between `p` and `q`, or it matches either `p` or `q`. In this case, the current node is the lowest common ancestor (LCA), so we return `root`.\n",
-    "\n",
-    "The algorithm is efficient because it takes advantage of the properties of a BST. It eliminates subtrees that cannot contain the LCA by comparing the values of `p`, `q`, and the current `root` node. Eventually, it reaches the LCA node, and that node is returned as the result."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "6\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Construct the tree\n",
-    "root = TreeNode(6)\n",
-    "root.left = TreeNode(2)\n",
-    "root.right = TreeNode(8)\n",
-    "root.left.left = TreeNode(0)\n",
-    "root.left.right = TreeNode(4)\n",
-    "root.right.left = TreeNode(7)\n",
-    "root.right.right = TreeNode(9)\n",
-    "root.left.right.left = TreeNode(3)\n",
-    "root.left.right.right = TreeNode(5)\n",
-    "\n",
-    "p = root.left  # Node 2\n",
-    "q = root.right  # Node 8\n",
-    "\n",
-    "result = lowestCommonAncestor(root, p, q)\n",
-    "print(result.val)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Using the same tree as before\n",
-    "p = root.left  # Node 2\n",
-    "q = root.left.right  # Node 4\n",
-    "\n",
-    "result = lowestCommonAncestor(root, p, q)\n",
-    "print(result.val) "
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "7fc71180-e586-4c37-9fcf-2f574cb2b9d6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3:\n",
-    "# Creating a new tree for this example\n",
-    "root2 = TreeNode(2)\n",
-    "root2.left = TreeNode(1)\n",
-    "\n",
-    "p = root2  # Node 2\n",
-    "q = root2.left  # Node 1\n",
-    "\n",
-    "result = lowestCommonAncestor(root2, p, q)\n",
-    "print(result.val)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided algorithm for finding the lowest common ancestor (LCA) in a Binary Search Tree (BST).\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of this algorithm is O(h), where \"h\" is the height of the BST. In the worst case, where the tree is completely unbalanced (essentially a linked list), the height of the tree is O(n), where \"n\" is the number of nodes in the tree. However, in a well-balanced BST, the height is logarithmic, which is O(log n).\n",
-    "\n",
-    "The reason for this time complexity is that the algorithm efficiently narrows down the search space by traversing either left or right subtrees based on the values of the target nodes `p` and `q`. It eliminates entire subtrees that cannot contain the LCA, leading to a relatively quick search.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity of the algorithm is O(1). This is because it uses a constant amount of extra space, regardless of the size of the input BST. The only variables used are `root`, `p`, and `q`, and there are no data structures like stacks or queues used for additional space. The algorithm performs a simple traversal without recursion, so it does not consume extra memory as the tree depth increases.\n",
-    "\n",
-    "**In summary**, the provided algorithm for finding the LCA in a BST is both time and space-efficient. Its time complexity is O(h), where \"h\" is the height of the tree, and its space complexity is O(1), making it suitable for practical use even in large BSTs."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "### Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple LCAs:** Extend the algorithm to find all the lowest common ancestors of two given nodes `p` and `q` in a BST. In some cases, there can be multiple LCAs.\n",
-    "\n",
-    "2. **LCA with k Nodes:** Given a BST and k nodes, find the lowest common ancestor of these k nodes. This is an extension of the problem where you must find the LCA of more than two nodes."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb b/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb
deleted file mode 100644
index 78a78b5..0000000
--- a/_build/jupyter_execute/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb	
+++ /dev/null
@@ -1,291 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 297. Serialize and Deserialize Binary Tree\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Serialize and Deserialize Binary Tree problem on LeetCode, click here!](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)\n",
-    "\n",
-    "---\n",
-    "Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.\n",
-    "\n",
-    "Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.\n",
-    "\n",
-    "**Clarification:** The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[0, 10^4]$.\n",
-    "- `-1000 <= Node.val <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "TThe problem is to design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of characters or bytes so that it can be easily stored in a file, transmitted over a network, or otherwise persisted. Deserialization is the reverse process of converting the serialized data back into the original data structure.\n",
-    "\n",
-    "In the context of this problem:\n",
-    "\n",
-    "1. **Serialization**: You are given a binary tree, and your task is to convert it into a string representation such that you can later recreate the same binary tree from this string. The format of serialization is flexible, but it should allow you to reconstruct the original binary tree accurately.\n",
-    "\n",
-    "2. **Deserialization**: Given a serialized string, you need to reconstruct the binary tree it represents, ensuring that it is identical to the original tree.\n",
-    "\n",
-    "Here are some key points to consider in solving this problem:\n",
-    "\n",
-    "- The input can include any valid binary tree, including trees with nodes having integer values within the range [-1000, 1000].\n",
-    "- You don't necessarily need to follow a specific format for serialization, but it should be designed in a way that allows unambiguous deserialization.\n",
-    "- The goal is to serialize the tree structure and its values and then deserialize it back into the same structure and values.\n",
-    "\n",
-    "A common approach for serialization is to use a traversal method like preorder traversal, where you visit nodes in the order: root, left subtree, right subtree. This way, you can serialize the tree into a string, and during deserialization, you can reconstruct the tree by parsing the string in the same order.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode(object):\n",
-    "    def __init__(self, x):\n",
-    "        self.val = x\n",
-    "        self.left = None\n",
-    "        self.right = None\n",
-    "\n",
-    "class Codec:\n",
-    "    def serialize(self, root):\n",
-    "        serialized  = []\n",
-    "\n",
-    "        def dfs(node):\n",
-    "            if not node:\n",
-    "                # If the node is None, represent it as \"Null\" in the serialized string\n",
-    "                serialized .append(\"Null\")\n",
-    "                return\n",
-    "            # Convert the node's value to a string and add it to the serialized string\n",
-    "            serialized .append(str(node.val))\n",
-    "            dfs(node.left)\n",
-    "            dfs(node.right)\n",
-    "\n",
-    "        dfs(root)\n",
-    "        # Join the serialized values with \",\"\n",
-    "        return \",\".join(serialized )\n",
-    "\n",
-    "    def deserialize(self, data):\n",
-    "        vals = data.split(\",\")\n",
-    "        self.i = 0\n",
-    "\n",
-    "        def dfs():\n",
-    "            if vals[self.i] == \"Null\":\n",
-    "                self.i += 1\n",
-    "                return None\n",
-    "            # Convert the value to an integer to create a new node\n",
-    "            node = TreeNode(int(vals[self.i]))\n",
-    "            self.i += 1\n",
-    "            node.left = dfs()\n",
-    "            node.right = dfs()\n",
-    "            return node\n",
-    "\n",
-    "        return dfs()"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Here's a step-by-step explanation of the code:\n",
-    "\n",
-    "1. **Definition of TreeNode**:\n",
-    "   - The code defines a simple `TreeNode` class that represents a node in a binary tree. Each `TreeNode` has three attributes:\n",
-    "     - `val`: The integer value stored in the node.\n",
-    "     - `left`: A reference to the left child node (or None if there is no left child).\n",
-    "     - `right`: A reference to the right child node (or None if there is no right child).\n",
-    "   - This class allows you to create binary tree nodes with integer values and left and right child references.\n",
-    "\n",
-    "2. **Serialization (serialize method)**:\n",
-    "   - The `serialize` method takes a `root` node as input, which is the root of the binary tree that needs to be serialized.\n",
-    "   - It initializes an empty list `serialized` to store the serialized elements.\n",
-    "   - The `dfs` (depth-first search) function is defined within the `serialize` method to perform a preorder traversal of the binary tree.\n",
-    "   - In the `dfs` function:\n",
-    "     - If the current `node` is `None` (i.e., a leaf node or a child of a leaf node), it appends the string \"Null\" to the `serialized` list to represent the absence of a node.\n",
-    "     - If the current `node` is not `None`, it appends the string representation of the `node.val` to the `serialized` list and then recursively calls `dfs` on the left and right children.\n",
-    "   - After the `dfs` traversal, the `serialized` list contains the serialized binary tree elements.\n",
-    "   - The method returns a string obtained by joining the elements in the `serialized` list with commas.\n",
-    "\n",
-    "3. **Deserialization (deserialize method)**:\n",
-    "   - The `deserialize` method takes a serialized string `data` as input, which represents a binary tree in the custom format.\n",
-    "   - It splits the `data` string by commas to obtain a list of elements called `vals`.\n",
-    "   - The method initializes an index `self.i` to 0. This index keeps track of the current position in the `vals` list during deserialization.\n",
-    "   - The `dfs` function is defined within the `deserialize` method to perform the deserialization process using a recursive approach.\n",
-    "   - In the `dfs` function:\n",
-    "     - If the current element in `vals` is \"Null,\" it means there is no node at this position in the binary tree, so it returns `None`.\n",
-    "     - If the current element is not \"Null,\" it converts the element to an integer to create a new `TreeNode` with that value.\n",
-    "     - It then recursively calls `dfs` to set the left and right children of the current node.\n",
-    "   - The method returns the root node of the reconstructed binary tree.\n",
-    "\n",
-    "Overall, this code demonstrates a way to serialize a binary tree into a string format and then deserialize it back into the original tree structure, allowing for the representation of both the tree structure and the values stored in the nodes."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1,2,Null,Null,3,4,Null,Null,5,Null,Null\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "# Serialize a binary tree\n",
-    "root = TreeNode(1)\n",
-    "root.left = TreeNode(2)\n",
-    "root.right = TreeNode(3)\n",
-    "root.right.left = TreeNode(4)\n",
-    "root.right.right = TreeNode(5)\n",
-    "\n",
-    "codec = Codec()\n",
-    "serialized_tree = codec.serialize(root)\n",
-    "print(serialized_tree)\n",
-    "\n",
-    "# Deserialize the serialized tree\n",
-    "new_root = codec.deserialize(serialized_tree)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Null\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Example usage:\n",
-    "# Serialize a binary tree\n",
-    "root = None\n",
-    "\n",
-    "codec = Codec()\n",
-    "serialized_tree = codec.serialize(root)\n",
-    "print(serialized_tree)\n",
-    "\n",
-    "# Deserialize the serialized tree\n",
-    "new_root = codec.deserialize(serialized_tree)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for serializing and deserializing a binary tree:\n",
-    "\n",
-    "1. **Serialization (serialize method)**:\n",
-    "\n",
-    "   - Time Complexity:\n",
-    "     - The `serialize` method uses a depth-first traversal (preorder) of the binary tree.\n",
-    "     - It visits each node exactly once.\n",
-    "     - At each node, it performs constant-time operations (string conversion and list append).\n",
-    "     - Therefore, the time complexity for serialization is O(N), where N is the number of nodes in the binary tree.\n",
-    "\n",
-    "   - Space Complexity:\n",
-    "     - The space complexity for serialization includes the space used by the `serialized` list and the recursive call stack.\n",
-    "     - The `serialized` list stores the serialized elements, and its space is proportional to the number of nodes.\n",
-    "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-    "     - Therefore, the space complexity for serialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-    "\n",
-    "2. **Deserialization (deserialize method)**:\n",
-    "\n",
-    "   - Time Complexity:\n",
-    "     - The `deserialize` method splits the serialized string into a list of elements, which takes O(N) time.\n",
-    "     - The `dfs` function is a recursive function that visits each element in the list exactly once.\n",
-    "     - At each element, it performs constant-time operations (string comparison, integer conversion, and recursive function calls).\n",
-    "     - Therefore, the time complexity for deserialization is O(N).\n",
-    "\n",
-    "   - Space Complexity:\n",
-    "     - The space complexity for deserialization includes the space used by the `vals` list and the recursive call stack.\n",
-    "     - The `vals` list stores the elements from the serialized string, and its space is proportional to the number of nodes.\n",
-    "     - The recursive call stack depth is determined by the height of the binary tree, and in the worst case (completely unbalanced tree), it can be O(N).\n",
-    "     - Therefore, the space complexity for deserialization is O(N) due to the list and O(H) due to the recursive call stack, where H is the height of the tree. In most cases, the dominant factor is O(N).\n",
-    "\n",
-    "Overall, the provided code has a time complexity of O(N) for both serialization and deserialization and a space complexity of O(N) in the majority of cases (unless the tree is severely unbalanced, in which case the height H dominates the space complexity). It efficiently serializes and deserializes a binary tree while using a linear amount of memory."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Efficient Serialization**:\n",
-    "   Optimize the serialization process to minimize the length of the serialized string. Consider using techniques like binary encoding to represent node values more efficiently.\n",
-    "\n",
-    "2. **Custom Serialization Format**:\n",
-    "   Modify the serialization and deserialization methods to use a custom format different from the one provided. Ensure that the new format allows for accurate reconstruction of the original binary tree."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/572. Subtree of Another Tree.ipynb b/_build/jupyter_execute/07. Trees/572. Subtree of Another Tree.ipynb
deleted file mode 100644
index 425f22d..0000000
--- a/_build/jupyter_execute/07. Trees/572. Subtree of Another Tree.ipynb	
+++ /dev/null
@@ -1,259 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 572: Subtree of Another Tree\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Subtree of Another Tree problem on LeetCode, click here!](https://leetcode.com/problems/subtree-of-another-tree/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "**Problem Description:**\n",
-    "\n",
-    "Given the roots of two binary trees `root` and `subRoot`, return `True` if there is a subtree of `root` with the same structure and node values as `subRoot`, and `False` otherwise.\n",
-    "\n",
-    "A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "1. The number of nodes in the `root` tree is in the range $[1, 2000]$.\n",
-    "2. The number of nodes in the `subRoot` tree is in the range $[1, 1000]$.\n",
-    "3. $-10^4$ <= root.val <= $10^4$\n",
-    "4. $-10^4$ <= subRoot.val <= $10^4$\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a binary tree node.\n",
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "class Solution:\n",
-    "    def isSubtree(self, root, subRoot):\n",
-    "        if not root:\n",
-    "            return False\n",
-    "        \n",
-    "        # Check if the current subtree is equal to the subRoot\n",
-    "        if self.isSameTree(root, subRoot):\n",
-    "            return True\n",
-    "        \n",
-    "        # Recursively check the left and right subtrees\n",
-    "        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)\n",
-    "    \n",
-    "    def isSameTree(self, tree1, tree2):\n",
-    "        if not tree1 and not tree2:\n",
-    "            return True\n",
-    "        if not tree1 or not tree2:\n",
-    "            return False\n",
-    "        \n",
-    "        return (\n",
-    "            tree1.val == tree2.val and\n",
-    "            self.isSameTree(tree1.left, tree2.left) and\n",
-    "            self.isSameTree(tree1.right, tree2.right)\n",
-    "        )"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "This code defines a TreeNode class for the binary tree nodes and a Solution class with two methods:\n",
-    "\n",
-    "1. `isSubtree`: This method checks if there is a subtree of the `root` tree with the same structure and node values as the `subRoot` tree. It uses a helper function `isSameTree` to compare two trees for equality.\n",
-    "\n",
-    "2. `isSameTree`: This helper method recursively compares two trees to check if they are the same in structure and node values.\n",
-    "\n",
-    "### Here's a detailed explanation of the code:\n",
-    "\n",
-    "1. **TreeNode Class:**\n",
-    "\n",
-    "The `TreeNode` class is defined to represent nodes in a binary tree. Each node has a `val` (the node's value) and may have a `left` and `right` child.\n",
-    "\n",
-    "2. **Solution Class:**\n",
-    "\n",
-    "    + The `Solution` class contains the solution for the problem and defines two important methods:\n",
-    "\n",
-    "    + `isSubtree(self, root, subRoot)`:\n",
-    "\n",
-    "        - This method checks whether `subRoot` is a subtree of `root`. It takes two tree nodes, `root` and `subRoot`, as input arguments.\n",
-    "        - If `root` is `None`, it returns `False` because there is no subtree to search.\n",
-    "        - It then checks if the current subtree with `root` as its root is equal to `subRoot` using the `isSameTree` method. If they are the same, it returns `True`.\n",
-    "        - If the current subtree is not the same as `subRoot`, it recursively checks the left and right subtrees of `root` to see if `subRoot` is a subtree of any of them.\n",
-    "        - It returns `True` if `subRoot` is found in either the left or right subtree; otherwise, it returns `False`.\n",
-    "\n",
-    "    + `isSameTree(self, tree1, tree2)`:\n",
-    "\n",
-    "        - This method checks whether two trees, `tree1` and `tree2`, are the same.\n",
-    "        - If both `tree1` and `tree2` are `None`, they are considered the same tree, so it returns `True`.\n",
-    "        - If only one of them is `None` (but not both), they are different trees, so it returns `False`.\n",
-    "        - If both `tree1` and `tree2` have values, it checks if their values are equal and recursively checks if their left and right subtrees are the same.\n",
-    "        - It returns `True` if the trees are the same; otherwise, it returns `False`.\n",
-    "\n",
-    "The code effectively uses recursion to traverse the binary trees and check for subtree equality. The `isSubtree` method starts the recursive search, and the `isSameTree` method is used to compare individual subtrees. The approach is efficient and avoids unnecessary checks when possible.\n",
-    "\n",
-    "This solution allows you to determine if there exists a subtree within the `root` tree that matches the structure and node values of `subRoot`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Is subRoot a subtree of root? True\n"
-     ]
-    }
-   ],
-   "source": [
-    "#root = [3,4,5,1,2]\n",
-    "#subRoot = [4,1,2]\n",
-    "\n",
-    "# Example input\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(4)\n",
-    "root.right = TreeNode(5)\n",
-    "root.left.left = TreeNode(1)\n",
-    "root.left.right = TreeNode(2)\n",
-    "\n",
-    "subRoot = TreeNode(4)\n",
-    "subRoot.left = TreeNode(1)\n",
-    "subRoot.right = TreeNode(2)\n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Test the isSubtree method\n",
-    "result = solution.isSubtree(root, subRoot)\n",
-    "print(\"Is subRoot a subtree of root?\", result)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Is subRoot a subtree of root? False\n"
-     ]
-    }
-   ],
-   "source": [
-    "#root = [3,4,5,1,2,null,null,null,null,0] \n",
-    "#subRoot = [4,1,2]\n",
-    "\n",
-    "\n",
-    "# Example input\n",
-    "root = TreeNode(3)\n",
-    "root.left = TreeNode(4)\n",
-    "root.right = TreeNode(5)\n",
-    "root.left.left = TreeNode(1)\n",
-    "root.left.right = TreeNode(2)\n",
-    "root.left.right.left = TreeNode(0)\n",
-    "\n",
-    "subRoot = TreeNode(4)\n",
-    "subRoot.left = TreeNode(1)\n",
-    "subRoot.right = TreeNode(2)\n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Test the isSubtree method\n",
-    "result = solution.isSubtree(root, subRoot)\n",
-    "print(\"Is subRoot a subtree of root?\", result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "86a34f0d-0789-498c-818c-2d30f84754d0",
-   "metadata": {},
-   "source": [
-    "Let's analyze the time and space complexity of the provided code for the \"Subtree of Another Tree\" problem:\n",
-    "\n",
-    "**Time Complexity**\n",
-    "\n",
-    "The time complexity of the code primarily depends on the recursive traversal of the binary tree. In both the `isSubtree` and `isSameTree` functions, we visit each node in the binary trees once. Let's break down the time complexity:\n",
-    "\n",
-    "1. The `isSubtree` function:\n",
-    "   - In the worst case, we need to visit every node in the `root` tree.\n",
-    "   - For each node in the `root` tree, we call the `isSameTree` function, which has its own traversal.\n",
-    "   - So, the total time complexity is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree.\n",
-    "\n",
-    "2. The `isSameTree` function:\n",
-    "   - In the worst case, we visit every node in both `tree1` and `tree2`.\n",
-    "   - The number of recursive calls made is proportional to the number of nodes in the trees.\n",
-    "   - So, the time complexity of this function is $O(max(n, m))$, where $n$ and $m$ are the numbers of nodes in `tree1` and `tree2`, respectively.\n",
-    "\n",
-    "Overall, the time complexity of the entire code is $O(n * m)$, where $n$ is the number of nodes in the `root` tree, and $m$ is the number of nodes in the `subRoot` tree. In practice, it may be less than $O(n * m)$ if a subtree mismatch is detected early during the traversal.\n",
-    "\n",
-    "**Space Complexity**\n",
-    "\n",
-    "The space complexity of the code is determined by the function call stack during recursion and the space used by the recursive functions. Let's analyze the space complexity:\n",
-    "\n",
-    "1. The `isSubtree` function:\n",
-    "   - It uses the call stack for recursion.\n",
-    "   - The maximum depth of the recursion is equal to the height of the `root` tree, which can be $O(n)$ in the worst case (unbalanced tree).\n",
-    "   - Additionally, the function doesn't use any significant extra space other than the recursion stack.\n",
-    "\n",
-    "2. The `isSameTree` function:\n",
-    "   - It also uses the call stack for recursion.\n",
-    "   - The maximum depth of the recursion is equal to the height of the `tree1` or `tree2`, whichever is greater.\n",
-    "   - So, the maximum space used for the call stack is $O(max(n, m))$.\n",
-    "\n",
-    "In summary, the space complexity of the code is $O(max(n, m))$ due to the function call stack. It scales with the maximum height of the trees being compared.\n",
-    "\n",
-    "Overall, the code is efficient and works well for trees with moderate sizes. However, it's important to keep in mind that the worst-case time complexity is $O(n * m)$, so for very large trees, the performance may degrade."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb b/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb
deleted file mode 100644
index 0cbe2fc..0000000
--- a/_build/jupyter_execute/07. Trees/98. Validate Binary Search Tree.ipynb	
+++ /dev/null
@@ -1,215 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 98. Validate Binary Search Tree\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Validate Binary Search Tree problem on LeetCode, click here!](https://leetcode.com/problems/validate-binary-search-tree/)\n",
-    "\n",
-    "---\n",
-    "Given the `root` of a binary tree, *determine if it is a valid binary search tree (BST)*.\n",
-    "\n",
-    "A **valid BST** is defined as follows:\n",
-    "\n",
-    "- The left subtree of a node contains only nodes with keys **less than** the node's key.\n",
-    "- The right subtree of a node contains only nodes with keys **greater than** the node's key.\n",
-    "- Both the left and right subtrees must also be binary search trees. \n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the tree is in the range $[1, 10^4]$.\n",
-    "- $-2^{31}$ <= `Node.val` <= $2^{31} - 1$\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to solve this problem:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TreeNode:\n",
-    "    def __init__(self, val=0, left=None, right=None):\n",
-    "        self.val = val\n",
-    "        self.left = left\n",
-    "        self.right = right\n",
-    "\n",
-    "def is_valid_BST(root):\n",
-    "    def is_valid(node, min_val, max_val):\n",
-    "        # Base case: If the node is None, it's a valid BST.\n",
-    "        if node is None:\n",
-    "            return True\n",
-    "        \n",
-    "        # Check if the current node's value is within the valid range.\n",
-    "        if not (min_val < node.val < max_val):\n",
-    "            return False\n",
-    "        \n",
-    "        # Recursively check the left and right subtrees with updated ranges.\n",
-    "        return (is_valid(node.left, min_val, node.val) and\n",
-    "                is_valid(node.right, node.val, max_val))\n",
-    "    \n",
-    "    # Call the helper function starting with a wide range for root node.\n",
-    "    return is_valid(root, float('-inf'), float('inf'))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "- It defines a `TreeNode` class to represent nodes in the binary tree.\n",
-    "\n",
-    "- The `is_valid_BST` function takes the root node of the binary tree as input and returns `True` if the tree is a valid BST, and `False` otherwise.\n",
-    "\n",
-    "- Inside the `is_valid_BST` function, there is a helper function called `is_valid` that performs the actual validation using a recursive approach.\n",
-    "\n",
-    "- The `is_valid` function checks each node in the tree to ensure that it satisfies the properties of a BST:\n",
-    "  - The value of the current node must be within a valid range defined by `min_val` and `max_val`.\n",
-    "  - The left subtree of the current node should contain values less than the current node's value.\n",
-    "  - The right subtree of the current node should contain values greater than the current node's value.\n",
-    "\n",
-    "- If any of these conditions are violated, the function returns `False`, indicating that the tree is not a valid BST.\n",
-    "\n",
-    "- If all nodes satisfy these conditions, the function returns `True`, indicating that the tree is a valid BST.\n",
-    "\n",
-    "- The code calls the `is_valid` function with the root node and initial range values of negative infinity to positive infinity to start the validation.\n",
-    "\n",
-    "- The time complexity of the code is O(N), where N is the number of nodes in the tree, as it traverses each node once.\n",
-    "\n",
-    "- The space complexity depends on the height of the tree. In the average case for a balanced BST, the space complexity is O(log N), but in the worst case (skewed tree), it can be O(N) due to the recursive call stack.\n",
-    "\n",
-    "In summary, this code checks whether a given binary tree is a valid BST by recursively validating each node's value and its left and right subtrees while maintaining valid value ranges. If all nodes satisfy the BST properties, the tree is considered a valid BST."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "# Create the tree for the first example: [2, 1, 3]\n",
-    "root1 = TreeNode(2)\n",
-    "root1.left = TreeNode(1)\n",
-    "root1.right = TreeNode(3)\n",
-    "print(is_valid_BST(root1))"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "192912fb-9d12-4911-9a4f-bfd37043e11d",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "\n",
-    "# Create the tree for the second example: [5, 1, 4, None, None, 3, 6]\n",
-    "root2 = TreeNode(5)\n",
-    "root2.left = TreeNode(1)\n",
-    "root2.right = TreeNode(4)\n",
-    "root2.right.left = TreeNode(3)\n",
-    "root2.right.right = TreeNode(6)\n",
-    "print(is_valid_BST(root2))"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `is_valid_BST` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The time complexity of the function primarily depends on the number of nodes in the binary tree.\n",
-    "- In the worst case, we may have to visit every node in the tree once to validate whether it's part of a valid BST.\n",
-    "- Since we are performing a depth-first search (DFS) traversal of the tree, the time complexity is O(N), where N is the number of nodes in the tree.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The space complexity is determined by the space used by the recursive call stack during the DFS traversal.\n",
-    "- In the worst case, if the tree is completely unbalanced (a skewed tree), the maximum depth of the call stack would be equal to the height of the tree.\n",
-    "- In a balanced BST, the height is approximately log2(N), where N is the number of nodes.\n",
-    "- Therefore, the space complexity of the call stack is O(log N) for a balanced BST.\n",
-    "- In the worst case (skewed tree), the space complexity can be O(N) as the height of the tree can be equal to N.\n",
-    "\n",
-    "**Overall:**\n",
-    "- Time Complexity: O(N)\n",
-    "- Space Complexity: O(log N) on average for a balanced BST, and O(N) in the worst case for a skewed tree.\n",
-    "\n",
-    "The space complexity is typically dominated by the recursive call stack, and it varies depending on the shape of the binary tree. In practice, for balanced binary trees, the space complexity is often close to O(log N), which is quite efficient."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Complex Constraints**: Modify the problem to have more complex constraints on the tree. For example, consider a binary tree with constraints like \"The left subtree of a node contains only nodes with keys less than or equal to the node's key,\" and \"The right subtree of a node contains only nodes with keys greater than or equal to the node's key.\" How would you adapt the code to handle these constraints?\n",
-    "\n",
-    "2. **Handling Duplicates**: Modify the code to handle binary search trees that allow duplicate values. In a standard BST, each value is unique, but in this case, multiple nodes can have the same value. The tree should still be considered valid if it follows the BST property."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb b/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb
deleted file mode 100644
index 5c54c25..0000000
--- a/_build/jupyter_execute/08. Tries/208. Implement Trie (Prefix Tree).ipynb	
+++ /dev/null
@@ -1,245 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 208. Implement Trie (Prefix Tree)\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Implement Trie problem on LeetCode, click here!](https://leetcode.com/problems/implement-trie-prefix-tree/)\n",
-    "\n",
-    "---\n",
-    "A **trie** (pronounced as \"try\") or **prefix tree** is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.\n",
-    "\n",
-    "Implement the Trie class:\n",
-    "\n",
-    "- `Trie()` Initializes the trie object.\n",
-    "- `void insert(String word)` Inserts the string `word` into the trie.\n",
-    "- `boolean search(String word)` Returns `true` if the string `word` is in the trie (i.e., was inserted before), and `false` otherwise.\n",
-    "- `boolean startsWith(String prefix)` Returns `true` if there is a previously inserted string `word` that has the prefix `prefix`, and `false` otherwise.\n",
-    "\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= word.length, prefix.length <= 2000`\n",
-    "- `word` and `prefix` consist only of lowercase English letters.\n",
-    "- At most $3 * 10^4$ calls in total will be made to `insert`, `search`, and `startsWith`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand is to implement a data structure known as a Trie (pronounced \"try\") or Prefix Tree. A Trie is a tree-like data structure that is used to efficiently store and retrieve a set of strings or words. It's particularly useful for tasks involving string manipulation, such as autocomplete and spell-checking. The problem statement defines the following operations for implementing the Trie class:\n",
-    "\n",
-    "1. `Trie()`: This is the constructor that initializes the Trie object.\n",
-    "\n",
-    "2. `void insert(String word)`: This method inserts the given string `word` into the Trie. It effectively adds the characters of the word to the Trie's structure, creating nodes for each character if they don't already exist. At the end of the word, a special flag is set to indicate that this node represents the end of a valid word.\n",
-    "\n",
-    "3. `boolean search(String word)`: This method checks if the given string `word` is present in the Trie. It starts from the root of the Trie and traverses the Trie by following the characters in the word. If it successfully traverses the Trie and reaches the end of the word, it returns `true`, indicating that the word exists in the Trie; otherwise, it returns `false`.\n",
-    "\n",
-    "4. `boolean startsWith(String prefix)`: This method checks if there is any previously inserted word in the Trie that has the given `prefix`. It's similar to the `search` method but does not require that the prefix be a complete word. If there is any word in the Trie that starts with the given prefix, it returns `true`; otherwise, it returns `false`.\n",
-    "\n",
-    "The problem also provides an example scenario where these operations are called, demonstrating the expected output for each operation.\n",
-    "\n",
-    "In summary, the goal is to create a data structure (Trie) that efficiently stores a set of strings and provides methods to insert new strings, search for complete words, and check if a given prefix exists in the stored set of strings.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TrieNode:\n",
-    "    def __init__(self):\n",
-    "        self.children = {}           # A dictionary to store child nodes\n",
-    "        self.is_end_of_word = False  # Indicates if a word ends at this node\n",
-    "\n",
-    "class Trie:\n",
-    "    def __init__(self):\n",
-    "        self.root = TrieNode()       # Initialize the Trie with a root node\n",
-    "\n",
-    "    def insert(self, word):\n",
-    "        node = self.root             # Start from the root\n",
-    "        for char in word:\n",
-    "            if char not in node.children:\n",
-    "                node.children[char] = TrieNode()  # Create a new node if the character is not present\n",
-    "            node = node.children[char]           # Move to the child node\n",
-    "        node.is_end_of_word = True                # Mark the end of the inserted word\n",
-    "\n",
-    "    def search(self, word):\n",
-    "        node = self.root             # Start from the root\n",
-    "        for char in word:\n",
-    "            if char not in node.children:\n",
-    "                return False         # If the character is not found, the word is not in the Trie\n",
-    "            node = node.children[char]  # Move to the child node\n",
-    "        return node.is_end_of_word    # Check if the node represents the end of a valid word\n",
-    "\n",
-    "    def startsWith(self, prefix):\n",
-    "        node = self.root             # Start from the root\n",
-    "        for char in prefix:\n",
-    "            if char not in node.children:\n",
-    "                return False         # If the character is not found, no word starts with the prefix\n",
-    "            node = node.children[char]  # Move to the child node\n",
-    "        return True                  # Prefix found in the Trie"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "Let's explain how the code works step by step:\n",
-    "\n",
-    "1. `TrieNode` Class:\n",
-    "   - `TrieNode` is a class that represents nodes in the Trie. Each node has two attributes:\n",
-    "     - `children`: This is a dictionary that stores child nodes. Each key in the dictionary represents a character, and the corresponding value is the child node for that character.\n",
-    "     - `is_end_of_word`: This boolean flag indicates whether the node represents the end of a complete word. Initially, it's set to `False` for all nodes.\n",
-    "\n",
-    "2. `Trie` Class:\n",
-    "   - `Trie` is the main class that implements the Trie data structure. It has the following methods:\n",
-    "\n",
-    "   - `__init__(self)`: The constructor initializes the Trie object by creating a root node, which serves as the starting point for all Trie operations.\n",
-    "\n",
-    "   - `insert(self, word)`: This method inserts a string `word` into the Trie. It starts from the root node and iterates through each character in the word.\n",
-    "     - If a character is not present as a child node, it creates a new node for that character.\n",
-    "     - It then moves to the child node and continues the process until the entire word is inserted.\n",
-    "     - Finally, it sets the `is_end_of_word` flag to `True` for the last node to mark the end of the inserted word.\n",
-    "\n",
-    "   - `search(self, word)`: This method checks if a complete word (string) exists in the Trie. It starts from the root node and iterates through each character in the word.\n",
-    "     - If a character is not found as a child node, it immediately returns `False` because the word is not in the Trie.\n",
-    "     - It continues to move to the child node for each character.\n",
-    "     - After reaching the end of the word, it checks if the `is_end_of_word` flag is `True` for the last node to confirm the presence of the word.\n",
-    "\n",
-    "   - `startsWith(self, prefix)`: This method checks if there is any previously inserted word in the Trie that starts with a given `prefix`. It follows the same logic as the `search` method but does not require the entire word to be present.\n",
-    "     - If the prefix is found in the Trie, it returns `True`; otherwise, it returns `False`.\n",
-    "\n",
-    "3. Example Usage:\n",
-    "   - The code demonstrates how to create a Trie object, insert words (\"apple\" and \"app\"), and perform operations on the Trie:\n",
-    "     - It inserts the word \"apple\" into the Trie.\n",
-    "     - It checks if \"apple\" is in the Trie, which returns `True`.\n",
-    "     - It checks if \"app\" is in the Trie, which returns `False`.\n",
-    "     - It checks if there is any word in the Trie that starts with \"app,\" which returns `True`.\n",
-    "     - It inserts the word \"app\" into the Trie.\n",
-    "     - It checks if \"app\" is in the Trie again, which now returns `True`.\n",
-    "\n",
-    "In summary, the code efficiently implements a Trie data structure with the ability to insert words, search for complete words, and check for the existence of words with a given prefix. The Trie is organized as a tree of nodes, with each node representing a character in the words being stored."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n",
-      "False\n",
-      "True\n",
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "\n",
-    "trie = Trie()\n",
-    "trie.insert(\"apple\")\n",
-    "print(trie.search(\"apple\"))   # Output: True\n",
-    "print(trie.search(\"app\"))     # Output: False\n",
-    "print(trie.startsWith(\"app\")) # Output: True\n",
-    "trie.insert(\"app\")\n",
-    "print(trie.search(\"app\"))     # Output: True"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the Trie implementation:\n",
-    "\n",
-    "1. Time Complexity:\n",
-    "   - Insertion (`insert` method): The time complexity of inserting a word into the Trie is O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to insert it into the Trie.\n",
-    "   - Search (`search` method): The time complexity of searching for a word in the Trie is also O(m), where m is the length of the word. In the worst case, you may need to traverse the entire word to determine if it exists in the Trie.\n",
-    "   - Starts with (`startsWith` method): The time complexity of checking if a prefix exists in the Trie is O(p), where p is the length of the prefix. In the worst case, you may need to traverse the entire prefix to determine its presence.\n",
-    "\n",
-    "   Therefore, for each operation (insertion, search, or starts with), the time complexity is O(m) or O(p), where m is the length of the word being operated on, and p is the length of the prefix.\n",
-    "\n",
-    "2. Space Complexity:\n",
-    "   - The space complexity of the Trie is determined by the number of nodes and characters stored in it. In the worst case, where none of the inserted words share common prefixes, the space complexity is O(N), where N is the total number of characters across all inserted words.\n",
-    "   - Each node in the Trie represents a character, and the number of nodes is directly related to the total number of characters.\n",
-    "\n",
-    "   In practice, the space complexity can be less than O(N) because common prefixes are shared among words in the Trie, leading to space optimization.\n",
-    "\n",
-    "In summary:\n",
-    "- Time complexity for each operation (insertion, search, starts with) is O(m) or O(p), where m is the length of the word and p is the length of the prefix.\n",
-    "- Space complexity is O(N) in the worst case, where N is the total number of characters across all inserted words. However, space optimization occurs due to common prefix sharing in practice, potentially reducing the actual space used."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Count Prefixes**:\n",
-    "   Implement a method to count the number of words in the Trie that have a specific prefix. This exercise requires you to maintain additional data in the Trie nodes to keep track of the count.\n",
-    "\n",
-    "2. **Auto-Complete Suggestions**:\n",
-    "   Implement an auto-complete feature using the Trie. Given a prefix, the program should return a list of suggested words based on the words previously inserted into the Trie."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
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-   "file_extension": ".py",
-   "mimetype": "text/x-python",
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-   "nbconvert_exporter": "python",
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\ No newline at end of file
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@@ -1,271 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 211. Design Add and Search Words Data Structure\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Design Add and Search Words Data Structure problem on LeetCode, click here!](https://leetcode.com/problems/design-add-and-search-words-data-structure/)\n",
-    "\n",
-    "---\n",
-    "Design a data structure that supports adding new words and finding if a string matches any previously added string.\n",
-    "\n",
-    "Implement the `WordDictionary` class:\n",
-    "\n",
-    "- `WordDictionary()` Initializes the object.\n",
-    "- `void addWord(word)` Adds `word` to the data structure, it can be matched later.\n",
-    "- `bool search(word)` Returns `true` if there is any string in the data structure that matches `word` or `false` otherwise. `word` may contain dots `'.'` where dots can be matched with any letter.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= word.length <= 25`\n",
-    "- `word` in `addWord` consists of lowercase English letters.\n",
-    "- `word` in `search` consist of `'.'` or lowercase English letters.\n",
-    "- There will be at most `2` dots in `word` for `search` queries.\n",
-    "- At most $10^4$ calls will be made to `addWord` and `search`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to design a data structure called `WordDictionary` that allows you to add words and search for words efficiently. The key feature of this data structure is that it should support wildcard search, where a dot ('.') can match any letter.\n",
-    "\n",
-    "Here are the specific requirements and explanations for the problem:\n",
-    "\n",
-    "1. Initialize the `WordDictionary` object using the constructor `WordDictionary()`.\n",
-    "\n",
-    "2. The `addWord(word)` method allows you to add a word to the data structure. Once added, you should be able to search for this word later. The words consist of lowercase English letters, and each word has a length between 1 and 25 characters.\n",
-    "\n",
-    "3. The `search(word)` method allows you to search for a word in the data structure. This method returns `True` if there is any string in the data structure that matches the given word or `False` otherwise. The word you are searching for may contain dots ('.'), where each dot can match any letter.\n",
-    "\n",
-    "4. The wildcard character ('.') in the `search` method allows for partial or fuzzy searching. For example, if the word dictionary contains the words \"bad,\" \"dad,\" and \"mad,\" then:\n",
-    "   - Searching for \"pad\" should return `False` because there is no exact match.\n",
-    "   - Searching for \"bad\" should return `True` because \"bad\" is in the dictionary.\n",
-    "   - Searching for \".ad\" should return `True` because it can match \"bad,\" \"dad,\" or \"mad.\"\n",
-    "   - Searching for \"b..\" should return `True` because it can match \"bad,\" \"bed,\" \"bet,\" etc.\n",
-    "\n",
-    "5. The problem specifies that there will be at most two dots ('.') in search queries, and there will be at most 10^4 calls to both `addWord` and `search`.\n",
-    "\n",
-    "In summary, you need to implement the `WordDictionary` class that efficiently supports adding words and searching for words, where the search can include wildcard characters ('.') that match any letter.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class TrieNode:\n",
-    "    def __init__(self):\n",
-    "        # Initialize a TrieNode with children and a flag to indicate the end of a word.\n",
-    "        self.children = {}\n",
-    "        self.is_end_of_word = False\n",
-    "\n",
-    "class WordDictionary:\n",
-    "\n",
-    "    def __init__(self):\n",
-    "        # Initialize the WordDictionary with a root node.\n",
-    "        self.root = TrieNode()\n",
-    "\n",
-    "    def addWord(self, word: str) -> None:\n",
-    "        node = self.root\n",
-    "        # Iterate through the characters in the word.\n",
-    "        for char in word:\n",
-    "            if char not in node.children:\n",
-    "                # Create a new node for the character if it doesn't exist.\n",
-    "                node.children[char] = TrieNode()\n",
-    "            # Move to the next node.\n",
-    "            node = node.children[char]\n",
-    "        # Mark the end of the word by setting the flag to True.\n",
-    "        node.is_end_of_word = True\n",
-    "\n",
-    "    def search_helper(self, node: TrieNode, word: str) -> bool:\n",
-    "        if not word:\n",
-    "            # If there are no characters left in the word, check if we reached the end of a word in the trie.\n",
-    "            return node.is_end_of_word\n",
-    "        \n",
-    "        char = word[0]\n",
-    "        if char == '.':\n",
-    "            # If the character is a dot, explore all possible child nodes.\n",
-    "            for child in node.children.values():\n",
-    "                if self.search_helper(child, word[1:]):\n",
-    "                    return True\n",
-    "        elif char in node.children:\n",
-    "            # If the character is in the children of the current node, move to the next node.\n",
-    "            return self.search_helper(node.children[char], word[1:])\n",
-    "        \n",
-    "        # If the character is not found and is not a dot, the word is not in the trie.\n",
-    "        return False\n",
-    "\n",
-    "    def search(self, word: str) -> bool:\n",
-    "        # Start the search from the root of the trie.\n",
-    "        return self.search_helper(self.root, word)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "This code defines a `WordDictionary` class that uses a Trie data structure to efficiently store and search for words. Here's a step-by-step explanation:\n",
-    "\n",
-    "1. `TrieNode` class: This class represents nodes in the Trie data structure. Each node has two attributes: `children` (a dictionary to store child nodes) and `is_end_of_word` (a flag to indicate whether the node marks the end of a word).\n",
-    "\n",
-    "2. `WordDictionary` class: This class initializes with an empty Trie by creating a root node.\n",
-    "\n",
-    "3. `addWord` method: This method adds a word to the Trie. It iterates through each character in the word and creates Trie nodes as necessary. It sets the `is_end_of_word` flag to `True` for the final character of the word to mark the end of the word.\n",
-    "\n",
-    "4. `search_helper` method: This is a recursive helper function for searching words in the Trie. It takes a Trie node (`node`) and a word to search (`word`) as input. If there are no characters left in the word (`not word`), it checks if the current node marks the end of a word. If it does, it returns `True`.\n",
-    "\n",
-    "5. If the first character of the word is a dot ('.'), the function explores all possible child nodes by iterating through `node.children.values()` and recursively calls `search_helper` for each child with the remaining part of the word (`word[1:]`).\n",
-    "\n",
-    "6. If the first character of the word is not a dot and is found in the children of the current node, the function recursively moves to the next node by calling `search_helper` on the child node and the remaining part of the word.\n",
-    "\n",
-    "7. If the character is not found in the children and is not a dot, the word is not in the Trie, so the function returns `False`.\n",
-    "\n",
-    "8. `search` method: This is the public method for searching words. It initiates the search process by calling `search_helper` starting from the root node of the Trie.\n",
-    "\n",
-    "Overall, this code efficiently implements a Trie-based data structure that allows adding and searching for words, including support for wildcard characters represented by dots ('.')."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n",
-      "True\n",
-      "True\n",
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "\n",
-    "wordDictionary = WordDictionary()\n",
-    "wordDictionary.addWord(\"bad\")\n",
-    "wordDictionary.addWord(\"dad\")\n",
-    "wordDictionary.addWord(\"mad\")\n",
-    "print(wordDictionary.search(\"pad\"))  # Output: False\n",
-    "print(wordDictionary.search(\"bad\"))  # Output: True\n",
-    "print(wordDictionary.search(\".ad\"))  # Output: True\n",
-    "print(wordDictionary.search(\"b..\"))  # Output: True"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `WordDictionary` class methods:\n",
-    "\n",
-    "1. `addWord` Method:\n",
-    "\n",
-    "   - Time Complexity: O(L)\n",
-    "     - L is the length of the word being added.\n",
-    "     - In the worst case, we need to iterate through each character in the word and create nodes in the Trie. This takes O(L) time.\n",
-    "\n",
-    "   - Space Complexity: O(L)\n",
-    "     - The space complexity for storing the word in the Trie is also O(L) because we create nodes for each character in the word.\n",
-    "\n",
-    "2. `search_helper` Method (called by `search`):\n",
-    "\n",
-    "   - Time Complexity: O(2^M * L)\n",
-    "     - M is the maximum number of dots ('.') in the search word.\n",
-    "     - In the worst case, when there are M dots in the search word, we may explore all possible child nodes at each dot. This results in a branching factor of 26 (for lowercase letters) at each dot.\n",
-    "     - So, the time complexity is exponential in the number of dots and linear in the length of the word being searched (L).\n",
-    "\n",
-    "   - Space Complexity: O(L)\n",
-    "     - The space complexity for the recursive call stack is O(L) in the worst case because we may have to recurse to a depth equal to the length of the search word.\n",
-    "\n",
-    "3. `search` Method:\n",
-    "\n",
-    "   - Time Complexity: O(2^M * L)\n",
-    "     - The `search` method calls the `search_helper` method, which has the same time complexity explained above.\n",
-    "\n",
-    "   - Space Complexity: O(L)\n",
-    "     - The space complexity for the recursive call stack is O(L) in the worst case, as explained earlier.\n",
-    "\n",
-    "Overall, the time complexity of the `search` method is mainly affected by the number of dots ('.') in the search word. In the worst case, when there are multiple dots and many possibilities to explore at each dot, the time complexity can be exponential. However, for most practical cases, it performs reasonably well.\n",
-    "\n",
-    "The space complexity primarily depends on the space required to store the words in the Trie. It is proportional to the total number of characters in all added words and is linear with respect to the length of the words added.\n",
-    "\n",
-    "In summary:\n",
-    "\n",
-    "- Time Complexity for `addWord`: O(L)\n",
-    "- Time Complexity for `search`: O(2^M * L)\n",
-    "- Space Complexity: O(L) for storing words in the Trie, O(L) for the recursive call stack during search.\n",
-    "\n",
-    "Note: The space complexity analysis assumes that the dictionary contains a reasonable number of words and does not account for the overhead of Python objects or system-level memory allocation."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Dictionary Auto-Completion**: Extend the `WordDictionary` class to provide auto-completion suggestions based on the prefix of a word. Implement a method that returns a list of words that match the given prefix.\n",
-    "\n",
-    "2. **Support Word Deletion**: Extend the `WordDictionary` class to support word deletion. Add a method `deleteWord(word)` that removes a word from the data structure. Ensure that the search operation still works correctly after deletion."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb b/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb
deleted file mode 100644
index e582b9d..0000000
--- a/_build/jupyter_execute/08. Tries/212. Word Search II.ipynb	
+++ /dev/null
@@ -1,310 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 212. Word Search II\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Word Search II problem on LeetCode, click here!](https://leetcode.com/problems/word-search-ii/)\n",
-    "\n",
-    "---\n",
-    "Given an `m x n` `board` of characters and a list of strings `words`, the task is to return *all words on the board*.\n",
-    "\n",
-    "Each word must be constructed from letters of sequentially **adjacent cells**, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.\n",
-    "\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == board.length`\n",
-    "- `n == board[i].length`\n",
-    "- `1 <= m, n <= 12`\n",
-    "- `board[i][j]` is a lowercase English letter.\n",
-    "- 1 <= `words.length` <= $3 * 10^4$\n",
-    "- `1 <= words[i].length <= 10`\n",
-    "- `words[i]` consists of lowercase English letters.\n",
-    "- All the strings of `words` are unique."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "To solve the \"Word Search II\" problem, you can use a Trie data structure and perform a depth-first search (DFS) on the board. Here's a step-by-step approach to solve the problem:\n",
-    "\n",
-    "1. Build a Trie from the given list of words:\n",
-    "   - Create a Trie data structure to store all the words from the input list.\n",
-    "   - For each word in the input list, insert it into the Trie, character by character. Make sure to mark the end of each word in the Trie.\n",
-    "\n",
-    "2. Perform DFS on the board:\n",
-    "   - Iterate through each cell (i, j) on the board.\n",
-    "   - For each cell, start a DFS traversal from that cell to search for words.\n",
-    "   - During the DFS traversal, maintain a current path string.\n",
-    "   - At each cell, check if the current path string, concatenated with the character in the cell, matches any word prefix in the Trie. If it does, continue the DFS.\n",
-    "   - If the current path string matches a word in the Trie, add it to the result.\n",
-    "\n",
-    "3. Implement DFS with backtracking:\n",
-    "   - During the DFS traversal, you will explore neighboring cells (horizontally and vertically) if they are within bounds and haven't been visited before.\n",
-    "   - To prevent revisiting the same cell within the same word, mark the cell as visited (e.g., change its character to a special character like '#') before exploring it and restore its original character after the DFS traversal.\n",
-    "\n",
-    "4. Return the result:\n",
-    "   - After completing the DFS traversal for all cells on the board, you will have collected all the valid words in the result set.\n",
-    "   - Convert the result set to a list and return it as the final output.\n",
-    "\n",
-    "This approach efficiently finds all words on the board by utilizing the Trie data structure to prune unnecessary DFS paths and avoiding duplicate visits to the same cell within the same word. It ensures that each word is constructed from letters of sequentially adjacent cells without using the same letter cell more than once in a word.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Define a TrieNode class to represent nodes in the Trie.\n",
-    "class TrieNode:\n",
-    "    def __init__(self):\n",
-    "        self.children = {}  # Dictionary to store child nodes.\n",
-    "        self.is_end_of_word = False  # Flag to mark the end of a word.\n",
-    "\n",
-    "# Main function to find words on the board.\n",
-    "def findWords(board, words):\n",
-    "    # DFS function to search for words starting from a given cell.\n",
-    "    def dfs(node, i, j, path):\n",
-    "        char = board[i][j]  # Get the character at the current cell.\n",
-    "        curr_node = node.children.get(char)  # Find the corresponding Trie node.\n",
-    "\n",
-    "        if not curr_node:\n",
-    "            return  # If no matching Trie node, stop the search.\n",
-    "\n",
-    "        path += char  # Add the character to the current path.\n",
-    "\n",
-    "        if curr_node.is_end_of_word:\n",
-    "            result.add(path)  # If the path matches a word, add it to the result.\n",
-    "\n",
-    "        temp, board[i][j] = board[i][j], \"#\"  # Mark the cell as visited.\n",
-    "\n",
-    "        # Explore neighboring cells (up, down, left, right).\n",
-    "        if i > 0:\n",
-    "            dfs(curr_node, i - 1, j, path)\n",
-    "        if i < m - 1:\n",
-    "            dfs(curr_node, i + 1, j, path)\n",
-    "        if j > 0:\n",
-    "            dfs(curr_node, i, j - 1, path)\n",
-    "        if j < n - 1:\n",
-    "            dfs(curr_node, i, j + 1, path)\n",
-    "\n",
-    "        board[i][j] = temp  # Restore the cell's original character.\n",
-    "\n",
-    "    # Function to build a Trie from the list of words.\n",
-    "    def buildTrie():\n",
-    "        root = TrieNode()  # Create a root node.\n",
-    "        for word in words:\n",
-    "            node = root\n",
-    "            for char in word:\n",
-    "                if char not in node.children:\n",
-    "                    node.children[char] = TrieNode()  # Create a new node if needed.\n",
-    "                node = node.children[char]\n",
-    "            node.is_end_of_word = True  # Mark the end of the word.\n",
-    "        return root\n",
-    "\n",
-    "    m, n = len(board), len(board[0])  # Get the dimensions of the board.\n",
-    "    root = buildTrie()  # Build the Trie from the list of words.\n",
-    "    result = set()  # Use a set to store unique words found on the board.\n",
-    "\n",
-    "    # Iterate through each cell on the board and start DFS from there.\n",
-    "    for i in range(m):\n",
-    "        for j in range(n):\n",
-    "            dfs(root, i, j, \"\")  # Start DFS from each cell on the board.\n",
-    "\n",
-    "    return list(result)  # Convert the set to a list and return the result."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code solves the \"Word Search II\" problem, where you have a grid of characters and a list of words. The goal is to find all words from the list that can be constructed by traversing adjacent cells (horizontally or vertically) in the grid, without using the same cell more than once for a single word.\n",
-    "\n",
-    "The code does the following:\n",
-    "\n",
-    "1. Defines a `TrieNode` class to represent nodes in a Trie data structure. Each node has children (other characters in the word) and a flag to mark the end of a word.\n",
-    "\n",
-    "2. Defines the `findWords` function, which takes the grid (`board`) and a list of words (`words`) as input.\n",
-    "\n",
-    "3. Inside the `findWords` function, there is a `dfs` (depth-first search) function. This function is used to explore the grid starting from a specific cell. It checks if the current path matches any word in the Trie and adds it to the result if found.\n",
-    "\n",
-    "4. The code also includes a `buildTrie` function that constructs a Trie data structure from the list of words. It iterates through each word, creating nodes for each character in the Trie and marking the end of words.\n",
-    "\n",
-    "5. The dimensions of the grid are obtained (m x n).\n",
-    "\n",
-    "6. A Trie is built using the `buildTrie` function.\n",
-    "\n",
-    "7. A set called `result` is used to store unique words found in the grid.\n",
-    "\n",
-    "8. The code iterates through each cell on the grid (using nested loops) and starts a DFS search from each cell.\n",
-    "\n",
-    "9. During the DFS, it explores neighboring cells (up, down, left, right) if they haven't been visited before and if the current path matches a prefix in the Trie.\n",
-    "\n",
-    "10. When a word is found during the DFS traversal, it is added to the `result` set.\n",
-    "\n",
-    "11. The grid cell is marked as visited during the DFS by changing its character to \"#\", and it is restored to its original character after the DFS traversal.\n",
-    "\n",
-    "12. Finally, the set of unique words found is converted to a list and returned as the output.\n",
-    "\n",
-    "The code effectively uses a Trie data structure to optimize the search and ensures that each word is constructed from adjacent cells without reusing the same cell for the same word."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "['eat', 'oath']\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "board1 = [[\"o\",\"a\",\"a\",\"n\"],[\"e\",\"t\",\"a\",\"e\"],[\"i\",\"h\",\"k\",\"r\"],[\"i\",\"f\",\"l\",\"v\"]]\n",
-    "words1 = [\"oath\",\"pea\",\"eat\",\"rain\"]\n",
-    "print(findWords(board1, words1))  # Output: [\"eat\", \"oath\"]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "c748b879-a2d8-4298-a42c-c7f61f9c7fc6",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "\n",
-    "board2 = [[\"a\",\"b\"],[\"c\",\"d\"]]\n",
-    "words2 = [\"abcb\"]\n",
-    "print(findWords(board2, words2))  # Output: []"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. **Building the Trie (`buildTrie` function):** \n",
-    "   - Suppose there are `N` words in the input list, and the average length of a word is `L`.\n",
-    "   - Building the Trie takes O(N * L) time, as we iterate through each word and insert each character into the Trie.\n",
-    "\n",
-    "2. **DFS Traversal (`dfs` function):** \n",
-    "   - In the worst case, the DFS function explores all possible paths in the grid.\n",
-    "   - The maximum number of recursive calls for each cell is 4 (up, down, left, right).\n",
-    "   - The maximum depth of the DFS is limited by the length of the longest word in the Trie, which is at most `L`.\n",
-    "   - Therefore, the DFS traversal for each cell takes $O(4^L)$ time.\n",
-    "\n",
-    "3. **Iterating Through the Grid (`findWords` function):**\n",
-    "   - In the worst case, we iterate through all `m` rows and `n` columns in the grid.\n",
-    "   - For each cell, we start a DFS traversal.\n",
-    "   - Therefore, iterating through the entire grid takes $O(m * n * 4^L)$ time.\n",
-    "\n",
-    "Overall, the time complexity of the code is $O(N * L + m * n * 4^L)$.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. **Trie Data Structure:**\n",
-    "   - The space required to store the Trie depends on the number of unique characters in all the words.\n",
-    "   - In the worst case, where all words are unique and have no common prefixes, the space complexity of the Trie is O(N * L).\n",
-    "   - In practice, it may be less if there are common prefixes among words.\n",
-    "\n",
-    "2. **DFS Stack (Recursive Calls):**\n",
-    "   - The depth of the recursive call stack during DFS is at most `L`, where `L` is the length of the longest word in the Trie.\n",
-    "   - Therefore, the space complexity for the recursive call stack is O(L).\n",
-    "\n",
-    "3. **Result Set (`result` set):**\n",
-    "   - The space used to store the result set depends on the number of valid words found in the grid.\n",
-    "   - In the worst case, when all words are found, the space complexity of the result set is O(N * L).\n",
-    "\n",
-    "Overall, the space complexity of the code is O(N * L) for the Trie and O(L) for the recursive call stack.\n",
-    "\n",
-    "In summary, the time complexity is dominated by the DFS traversal and is influenced by the number of words, their lengths, and the size of the grid. The space complexity is mainly determined by the Trie structure and the recursive call stack depth during DFS."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Word Search Paths:**\n",
-    "   Extend the basic word search exercise to find all possible paths (sequences of cells) that spell out a given word on the board. Return a list of paths if the word can be formed, otherwise an empty list.\n",
-    "\n",
-    "\n",
-    "2. **Reverse Word Search:**\n",
-    "   Modify the code to search for words in reverse order on the board. Given a 2D grid of characters and a list of words, find all words from the list that can be formed by traversing the board in reverse (right to left, bottom to top, etc.)."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
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-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb b/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb
deleted file mode 100644
index f82ad98..0000000
--- a/_build/jupyter_execute/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb	
+++ /dev/null
@@ -1,248 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 1046. Last Stone Weight\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Last Stone Weight problem on LeetCode, click here!](https://leetcode.com/problems/last-stone-weight/)\n",
-    "\n",
-    "---\n",
-    "You are given an array of integers `stones` where `stones[i]` is the weight of the `i`th stone.\n",
-    "\n",
-    "We are playing a game with the stones. On each turn, you choose the **heaviest two stones** and smash them together. Suppose the heaviest two stones have weights `x` and `y` with `x <= y`. The result of this smash is:\n",
-    "\n",
-    "1. If `x == y`, both stones are destroyed.\n",
-    "2. If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - x`.\n",
-    "\n",
-    "At the end of the game, there is **at most one** stone left.\n",
-    "\n",
-    "Return *the weight of the last remaining stone*. If there are no stones left, return `0`.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= stones.length <= 30`\n",
-    "- `1 <= stones[i] <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The \"Last Stone Weight\" problem involves a game played with a collection of stones, each stone having a certain weight. The goal is to find the weight of the last remaining stone after applying a specific set of rules.\n",
-    "\n",
-    "Here's how the game works:\n",
-    "\n",
-    "1. You start with an array of stones, where each stone is represented by its weight. The weights of the stones are given in the input as an array.\n",
-    "\n",
-    "2. In each turn of the game, you select the two heaviest stones from the remaining stones. If there is only one stone left, you've found the answer.\n",
-    "\n",
-    "3. If there are two stones with weights `x` and `y`, where `x` is less than or equal to `y`, the following happens:\n",
-    "   - If `x` is equal to `y`, both stones are completely destroyed, and they are removed from the array of stones.\n",
-    "   - If `x` is not equal to `y`, the stone with weight `x` is destroyed, and the stone with weight `y` now has a new weight of `y - x`. The stone with weight `x` is removed from the array, and the modified stone with weight `y - x` remains in the array.\n",
-    "\n",
-    "4. The game continues with the modified array of stones until there is at most one stone left.\n",
-    "\n",
-    "5. The goal is to find the weight of the last remaining stone, or if there are no stones left, return `0`.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import heapq\n",
-    "\n",
-    "def lastStoneWeight(stones):\n",
-    "    # Convert the input list to a max heap (negate the values to simulate a max heap)\n",
-    "    max_heap = [-stone for stone in stones]\n",
-    "    heapq.heapify(max_heap)\n",
-    "\n",
-    "    # Continue as long as there are more than 1 stone in the heap\n",
-    "    while len(max_heap) > 1:\n",
-    "        # Get the two heaviest stones\n",
-    "        y = -heapq.heappop(max_heap)  # Get the heaviest stone and negate it back to positive\n",
-    "        x = -heapq.heappop(max_heap)  # Get the second heaviest stone and negate it back to positive\n",
-    "\n",
-    "        # Calculate the new stone's weight after smashing\n",
-    "        if x != y:\n",
-    "            new_stone = y - x\n",
-    "            heapq.heappush(max_heap, -new_stone)  # Negate the value and push it back to the heap\n",
-    "\n",
-    "    # If there is one stone left, return its weight (negate it back to positive)\n",
-    "    if max_heap:\n",
-    "        return -max_heap[0]\n",
-    "    else:\n",
-    "        return 0"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a Python function called `lastStoneWeight` that takes a list of stone weights as input. It aims to find the weight of the last remaining stone after playing the stone smashing game described in the problem statement.\n",
-    "\n",
-    "Here's a step-by-step explanation of the code:\n",
-    "\n",
-    "1. We import the `heapq` library, which allows us to create and manipulate a heap (priority queue).\n",
-    "\n",
-    "2. Inside the `lastStoneWeight` function:\n",
-    "   - We create a max heap (negating stone weights to simulate a max heap) using the `heapq.heapify` function. This heap will keep track of the heaviest stones.\n",
-    "   \n",
-    "   - We enter a loop that continues as long as there are more than 1 stone in the heap.\n",
-    "   \n",
-    "   - Inside the loop, we:\n",
-    "     - Retrieve and negate the heaviest stone (`y`) from the heap.\n",
-    "     - Retrieve and negate the second heaviest stone (`x`) from the heap.\n",
-    "   \n",
-    "     - Calculate the new stone's weight after smashing (`y - x`), and negate it back to simulate a max heap.\n",
-    "     - Add the new stone back to the heap using `heapq.heappush`.\n",
-    "   \n",
-    "3. After the loop, if there is one stone left in the heap, we retrieve and negate it to get its actual weight, and return it as the result.\n",
-    "   \n",
-    "4. If there are no stones left (the heap is empty), we return `0`, indicating that no stones remain.\n",
-    "\n",
-    "5. Finally, we provide two test cases to demonstrate the function's usage and correctness.\n",
-    "\n",
-    "In summary, the code efficiently implements the stone smashing game by maintaining a max heap, selecting the heaviest stones in each turn, smashing them, and updating the heap until there is at most one stone left. It then returns the weight of the last remaining stone or 0 if there are no stones left."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "stones1 = [2, 7, 4, 1, 8, 1]\n",
-    "print(lastStoneWeight(stones1))  # Output: 1"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "b7e79514-4821-48d0-bb9d-0f2de8150945",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2: \n",
-    "\n",
-    "stones2 = [1]\n",
-    "print(lastStoneWeight(stones2))  # Output: 1"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided Python code for the \"Last Stone Weight\" problem:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. The `heapq.heapify` operation to create the max heap from the input stone weights has a time complexity of O(n), where n is the number of stones.\n",
-    "\n",
-    "2. The loop that continues until there are more than 1 stone in the heap performs the following operations in each iteration:\n",
-    "   - Retrieving and negating the heaviest stone (`y`) from the heap, which takes O(log n) time.\n",
-    "   - Retrieving and negating the second heaviest stone (`x`) from the heap, also taking O(log n) time.\n",
-    "   - Calculating the new stone's weight and pushing it back into the heap using `heapq.heappush`, which takes O(log n) time.\n",
-    "\n",
-    "3. In the worst case, the loop runs until there is only one stone left, which requires approximately (n-1) iterations.\n",
-    "\n",
-    "Overall, the dominant time complexity is O(n) for the initial heap creation, and within the loop, each iteration takes O(log n) time. Therefore, the overall time complexity of the algorithm is O(n log n).\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The space complexity is primarily determined by the space required for the max heap. In the worst case, this heap can contain all the stones, so the space complexity for the heap is O(n).\n",
-    "\n",
-    "2. The rest of the variables used in the function (e.g., `x`, `y`, `new_stone`) require only constant space, so they do not significantly contribute to the space complexity.\n",
-    "\n",
-    "Therefore, the overall space complexity of the algorithm is O(n) due to the space used for the max heap.\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: $O(n\\ log\\ n)$\n",
-    "- Space Complexity: $O(n)$\n",
-    "\n",
-    "The algorithm is efficient enough to handle the problem's constraints, as the worst-case time and space complexities are both linearithmic in terms of the number of stones."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Kth Last Stone Weight**: Modify the problem to find the weight of the Kth last remaining stone, where K is an integer input. Your function should work efficiently for large values of K.\n",
-    "\n",
-    "\n",
-    "2. **Multiple Stones Smash**: Modify the problem so that instead of smashing two stones at a time, you can smash up to K stones at each turn, where K is an integer input. Determine the weight of the last remaining stone in this variation."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb b/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb
deleted file mode 100644
index 6c43a8d..0000000
--- a/_build/jupyter_execute/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb	
+++ /dev/null
@@ -1,238 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 295. Find Median from Data Stream\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Find Median from Data Stream problem on LeetCode, click here!](https://leetcode.com/problems/find-median-from-data-stream/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.\n",
-    "\n",
-    "Implement the MedianFinder class:\n",
-    "\n",
-    "- `MedianFinder()` initializes the `MedianFinder` object.\n",
-    "- `void addNum(int num)` adds the integer `num` from the data stream to the data structure.\n",
-    "- `double findMedian()` returns the median of all elements so far. Answers within $10^{-5}$ of the actual answer will be accepted.\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- $-10^{5}$ <= `num` <= $10^{5}$\n",
-    "- There will be at least one element in the data structure before calling `findMedian`.\n",
-    "- At most $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-    "\n",
-    "**Follow-up:**\n",
-    "\n",
-    "1. If all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?\n",
-    "2. If `99%` of all integer numbers from the stream are in the range `[0, 100]`, how would you optimize your solution?"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem, \"Find Median from Data Stream,\" deals with efficiently computing the median of a sequence of numbers as they are incrementally provided. The median is the middle value in an ordered list of numbers. When the list has an even number of elements, the median is the average of the two middle values.\n",
-    "\n",
-    "Here's a breakdown of the problem:\n",
-    "\n",
-    "1. **Initialization**: You are asked to implement a `MedianFinder` class that supports the following operations:\n",
-    "\n",
-    "    - `MedianFinder()`: Initializes the MedianFinder object.\n",
-    "    - `addNum(int num)`: Adds an integer `num` from the data stream to the data structure.\n",
-    "    - `findMedian()`: Returns the median of all elements added to the data structure.\n",
-    "\n",
-    "2. **Median Calculation**: The `findMedian` operation should efficiently compute the median, and the result should have a precision of at least $10^{-5}$ (i.e., answers within this range will be considered correct).\n",
-    "\n",
-    "3. **Examples**:\n",
-    "   \n",
-    "   - If you add numbers [1, 2], the median should be 1.5 because (1 + 2) / 2 = 1.5.\n",
-    "   - If you add numbers [1, 2, 3], the median should be 2 because it's the middle value.\n",
-    "   \n",
-    "4. **Constraints**:\n",
-    "\n",
-    "   - The integers provided in the data stream are in the range from $-10^5$ to $10^5$.\n",
-    "   - There will be at least one element in the data structure before calling `findMedian`.\n",
-    "   - At most, $5 * 10^4$ calls will be made to `addNum` and `findMedian`.\n",
-    "\n",
-    "The problem can be solved by using two priority queues (heaps): one max-heap to store the smaller half of the numbers and one min-heap to store the larger half of the numbers. The max-heap ensures that the largest number in the smaller half is at the top, and the min-heap ensures that the smallest number in the larger half is at the top. These heaps are balanced to efficiently find the median when requested.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import heapq\n",
-    "\n",
-    "class MedianFinder:\n",
-    "\n",
-    "    def __init__(self):\n",
-    "        # Initialize two heaps: a max-heap (small_half) for the smaller half of numbers,\n",
-    "        # and a min-heap (large_half) for the larger half of numbers.\n",
-    "        self.small_half = []  # Max-heap for the smaller half of the numbers\n",
-    "        self.large_half = []  # Min-heap for the larger half of the numbers\n",
-    "\n",
-    "    def addNum(self, num: int) -> None:\n",
-    "        # Add the number to the appropriate heap based on its value\n",
-    "        if not self.small_half or num <= -self.small_half[0]:\n",
-    "            # If the number is less than or equal to the current maximum in the smaller half,\n",
-    "            # add it to the smaller half (max-heap)\n",
-    "            heapq.heappush(self.small_half, -num)\n",
-    "        else:\n",
-    "            # Otherwise, add it to the larger half (min-heap)\n",
-    "            heapq.heappush(self.large_half, num)\n",
-    "\n",
-    "        # Balance the heaps if necessary to ensure the size difference is at most 1\n",
-    "        if len(self.small_half) > len(self.large_half) + 1:\n",
-    "            # If the size of the smaller half is more than one greater than the size of the larger half,\n",
-    "            # move the maximum from the smaller half to the larger half to balance them.\n",
-    "            heapq.heappush(self.large_half, -heapq.heappop(self.small_half))\n",
-    "        elif len(self.large_half) > len(self.small_half):\n",
-    "            # If the size of the larger half is greater than the size of the smaller half,\n",
-    "            # move the minimum from the larger half to the smaller half to balance them.\n",
-    "            heapq.heappush(self.small_half, -heapq.heappop(self.large_half))\n",
-    "\n",
-    "    def findMedian(self) -> float:\n",
-    "        if len(self.small_half) == len(self.large_half):\n",
-    "            # If both halves have the same size, there's an even number of elements,\n",
-    "            # so the median is the average of the top elements in both heaps.\n",
-    "            return (-self.small_half[0] + self.large_half[0]) / 2.0\n",
-    "        else:\n",
-    "            # If the smaller half has more elements, it contains the median (odd number of elements).\n",
-    "            return -self.small_half[0]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a `MedianFinder` class for efficiently finding the median of a stream of numbers. It uses two heaps: a max-heap (`small_half`) to store the smaller half of the numbers and a min-heap (`large_half`) to store the larger half. Here's how the code works:\n",
-    "\n",
-    "1. The `__init__` method initializes the two heaps.\n",
-    "\n",
-    "2. The `addNum` method adds a number to the appropriate heap, ensuring that the heaps remain balanced. It does this by comparing the number to the current maximum in the smaller half and adding it to the appropriate heap. Then, it checks the sizes of the two heaps and rebalances them if necessary.\n",
-    "\n",
-    "3. The `findMedian` method calculates and returns the median. If both halves have the same size (even number of elements), it computes the average of the tops of both heaps. If the smaller half has more elements (odd number of elements), it returns the top of the max-heap, which is the median.\n",
-    "\n",
-    "In summary, this code efficiently maintains two heaps to divide the elements into smaller and larger halves, allowing for quick median retrieval. It offers a time complexity of O(log N) for adding elements and O(1) for finding the median, where N is the total number of elements processed. The space complexity is O(1) as the space used by the two heaps is independent of the input size."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1.5\n",
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "medianFinder = MedianFinder()\n",
-    "medianFinder.addNum(1)\n",
-    "medianFinder.addNum(2)\n",
-    "print(medianFinder.findMedian())  # Output: 1.5\n",
-    "medianFinder.addNum(3)\n",
-    "print(medianFinder.findMedian())  # Output: 2.0\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `MedianFinder` class:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. `MedianFinder()` Initialization: This operation has a time complexity of O(1). It simply initializes the two heaps.\n",
-    "\n",
-    "2. `addNum(int num)`: The time complexity for adding a number is O(log N), where N is the total number of elements processed so far. This is because, in the worst case, you might need to perform logarithmic operations on the heaps to maintain their balance.\n",
-    "\n",
-    "3. `findMedian()`: Finding the median has a time complexity of O(1). It simply involves accessing the tops of the two heaps, which takes constant time.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. `MedianFinder()` Initialization: The space complexity for initialization is O(1). It involves creating two empty heaps, which do not depend on the size of the data stream.\n",
-    "\n",
-    "2. `addNum(int num)`: The space complexity for adding a number is also O(1). It involves adding the number to one of the two heaps, which do not significantly affect the overall space complexity.\n",
-    "\n",
-    "3. `findMedian()`: The space complexity for finding the median is O(1). It doesn't require any additional data structures that grow with the input size.\n",
-    "\n",
-    "In summary, the time complexity of the `addNum` operation is O(log N), and the time complexity of the `findMedian` operation is O(1). The space complexity is O(1) as well, as the space used by the two heaps is independent of the size of the data stream. This solution efficiently maintains and retrieves the median, making it suitable for the given problem constraints."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Generalized kth Element**: Extend the `MedianFinder` class to support finding the kth element in the data stream efficiently. This means finding the kth smallest or kth largest element in the stream.\n",
-    "\n",
-    "2. **Sliding Window Median**: Given an array and a sliding window size, find the median within the window as it slides over the array. This is an extension of the problem to a moving window scenario."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb b/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb
deleted file mode 100644
index e6ab9e3..0000000
--- a/_build/jupyter_execute/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb	
+++ /dev/null
@@ -1,235 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 703. Kth Largest Element in a Stream\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Kth Largest Element in a Stream problem on LeetCode, click here!](https://leetcode.com/problems/kth-largest-element-in-a-stream/)\n",
-    "\n",
-    "---\n",
-    "Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.\n",
-    "\n",
-    "Implement the `KthLargest` class with the following methods:\n",
-    "\n",
-    "- `KthLargest(int k, int[] nums)`: Initializes the object with the integer `k` and the stream of integers `nums`.\n",
-    "- `int add(int val)`: Appends the integer `val` to the stream and returns the element representing the `kth` largest element in the stream.\n",
-    "\n",
-    "\n",
-    "**Constraints:**\n",
-    "\n",
-    "- 1 <= `k` <= $10^4$\n",
-    "- 0 <= `nums.length` <= $10^4$\n",
-    "- $-10^4$ <= `nums[i]` <= $10^4$\n",
-    "- $-10^4$ <= `val` <= $10^4$\n",
-    "- At most $10^4$ calls will be made to `add`.\n",
-    "- It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem you're trying to solve is to design a class called `KthLargest` that can efficiently find the kth largest element in a stream of integers. You need to support two main operations:\n",
-    "\n",
-    "To solve this problem efficiently, you can use a min-heap data structure. Here's how the approach works:\n",
-    "\n",
-    "1. Initialize a min-heap to store the k largest elements. Initially, it's empty.\n",
-    "\n",
-    "2. In the `__init__` method, populate the min-heap with the first `k` elements from `nums`. This ensures that you have the k largest elements seen so far.\n",
-    "\n",
-    "3. Whenever you add a new element to the stream using the `add` method, follow these steps:\n",
-    "   - Add the new element to the min-heap.\n",
-    "   - If the size of the min-heap exceeds `k`, remove the smallest element from the min-heap. This ensures that you always have the k largest elements in the min-heap.\n",
-    "   - The smallest element in the min-heap (the root) will always represent the kth largest element in the stream.\n",
-    "\n",
-    "Here's why this approach works efficiently:\n",
-    "\n",
-    "- By using a min-heap, you can quickly maintain the k largest elements, and finding the smallest element in the heap (the root) takes constant time.\n",
-    "\n",
-    "- When you add a new element, the min-heap's size is kept at most `k`, which ensures that you only track the k largest elements and discard the smaller ones.\n",
-    "\n",
-    "- The time complexity for adding an element is O(log k), which is very efficient compared to sorting the entire stream, which would be O(n log n).\n",
-    "\n",
-    "- This approach meets the constraints of the problem, including handling large streams and having a low time complexity for both initialization and adding elements.\n",
-    "\n",
-    "In summary, the min-heap approach efficiently tracks the kth largest element in the stream by maintaining a heap of the k largest elements seen so far, updating it as new elements are added. This approach provides a fast and scalable solution to the problem.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import heapq\n",
-    "from typing import List  # Import the List type from the typing module\n",
-    "\n",
-    "class KthLargest:\n",
-    "\n",
-    "    def __init__(self, k: int, nums: List[int]):\n",
-    "        # Initialize the KthLargest object with k and nums.\n",
-    "        self.k = k\n",
-    "        # Create a min-heap to store the k largest elements.\n",
-    "        self.min_heap = []\n",
-    "        # Populate the min-heap with the first k elements from nums.\n",
-    "        for num in nums:\n",
-    "            self.add(num)\n",
-    "\n",
-    "    def add(self, val: int) -> int:\n",
-    "        # Add val to the min-heap.\n",
-    "        heapq.heappush(self.min_heap, val)\n",
-    "        # If the size of the min-heap exceeds k, remove the smallest element.\n",
-    "        if len(self.min_heap) > self.k:\n",
-    "            heapq.heappop(self.min_heap)\n",
-    "        # The root of the min-heap is the kth largest element.\n",
-    "        return self.min_heap[0]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "- We start by importing the necessary modules:\n",
-    "  - `heapq`: This module provides functions to create and manipulate heaps.\n",
-    "  - `List` from the `typing` module: This is used to specify the type of the `nums` parameter.\n",
-    "\n",
-    "- We define the `KthLargest` class:\n",
-    "  - The `__init__` method initializes the object with the integer `k` and the list of integers `nums`.\n",
-    "  - It also creates an empty min-heap called `self.min_heap` to store the k largest elements.\n",
-    "  - It populates the min-heap with the first `k` elements from `nums` by calling the `add` method.\n",
-    "\n",
-    "- The `add` method:\n",
-    "  - Adds the new integer `val` to the min-heap using `heapq.heappush`. This maintains the min-heap property.\n",
-    "  - Checks if the size of the min-heap exceeds `k`. If it does, it removes the smallest element (the k+1th largest) using `heapq.heappop`.\n",
-    "  - Finally, it returns the smallest element in the min-heap, which is always the kth largest element.\n",
-    "\n",
-    "Overall, this code implements a class that efficiently finds the kth largest element in a stream of integers by maintaining a min-heap of the k largest elements seen so far.\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n",
-      "5\n",
-      "5\n",
-      "8\n",
-      "8\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example usage:\n",
-    "kthLargest = KthLargest(3, [4, 5, 8, 2])\n",
-    "print(kthLargest.add(3))  # Output: 4\n",
-    "print(kthLargest.add(5))  # Output: 5\n",
-    "print(kthLargest.add(10)) # Output: 5\n",
-    "print(kthLargest.add(9))  # Output: 8\n",
-    "print(kthLargest.add(4))  # Output: 8"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `KthLargest` class using the min-heap approach:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. `__init__` method:\n",
-    "   - In the `__init__` method, you iterate over the first `k` elements in `nums` and add them to the min-heap. Each `heapq.heappush` operation takes O(log k) time.\n",
-    "   - Therefore, the time complexity of the `__init__` method is O(k * log k).\n",
-    "\n",
-    "2. `add` method:\n",
-    "   - In the `add` method, you perform the following operations:\n",
-    "     - `heapq.heappush`: O(log k) to add the new element to the min-heap.\n",
-    "     - If the size of the min-heap exceeds `k`, you perform `heapq.heappop`, which is also O(log k).\n",
-    "     - Finally, you return the smallest element from the min-heap, which is O(1) because it's always the root.\n",
-    "   - Overall, the time complexity of the `add` method is O(log k).\n",
-    "\n",
-    "3. Overall, if you make `n` calls to the `add` method, the total time complexity is O(n * log k), where `n` is the total number of elements added to the stream.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. The space complexity is determined by the space used to store the min-heap and the instance variables.\n",
-    "2. The min-heap stores at most `k` elements at any given time, so its space complexity is O(k).\n",
-    "3. The instance variables (such as `self.k` and `self.min_heap`) have constant space requirements.\n",
-    "4. Therefore, the overall space complexity of the `KthLargest` class is O(k).\n",
-    "\n",
-    "In summary, the time complexity of the `KthLargest` class is O(n * log k) for `n` add operations, and the space complexity is O(k), where `k` is the parameter passed during initialization. This implementation efficiently maintains the kth largest element in the stream while meeting the problem's constraints."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Dynamic k:** Modify the `KthLargest` class to support dynamic changes in the value of `k`. Implement a method `update_k` that allows changing the value of `k` during the lifetime of the object. Ensure that the object can still correctly find the kth largest element based on the updated value of `k`.\n",
-    "\n",
-    "\n",
-    "2. **Implement kth Smallest:** Create a new class called `KthSmallest` that finds the kth smallest element in a stream instead of the kth largest. You may need to modify the data structure used in the implementation."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb b/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb
deleted file mode 100644
index 85d59ac..0000000
--- a/_build/jupyter_execute/10. Backtracking/39. Combination Sum.ipynb	
+++ /dev/null
@@ -1,274 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 39. Combination Sum\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Combination Sum problem on LeetCode, click here!](https://leetcode.com/problems/combination-sum/)\n",
-    "\n",
-    "---\n",
-    "Given an array of **distinct** integers `candidates` and a target integer `target`, return *a list of all **unique combinations** of `candidates` where the chosen numbers sum to the `target`*. You may return the combinations in **any order**. \n",
-    "\n",
-    "The **same** number may be chosen from `candidates` an **unlimited number** of times. Two combinations are unique if the **frequency** of at least one of the chosen numbers is different.\n",
-    "\n",
-    "The test cases are generated such that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= candidates.length <= 30`\n",
-    "- `2 <= candidates[i] <= 40`\n",
-    "- All elements of `candidates` are **distinct**.\n",
-    "- `1 <= target <= 40`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to find all unique combinations of numbers from a given array (candidates) such that their sum equals a target number. Here are the details of the problem:\n",
-    "\n",
-    "- You are given an array of distinct integers called \"candidates.\"\n",
-    "- You are also given a target integer called \"target.\"\n",
-    "\n",
-    "The goal is to find all unique combinations of numbers from the candidates array where the sum of the selected numbers is equal to the target. You can use the same number from the candidates array an unlimited number of times. A combination is considered unique if it has a different frequency (i.e., a different number of occurrences) of at least one chosen number compared to other combinations.\n",
-    "\n",
-    "For example:\n",
-    "\n",
-    "**Example 1:**\n",
-    "```python\n",
-    "Input: candidates = [2, 3, 6, 7], target = 7\n",
-    "Output: [[2, 2, 3], [7]]\n",
-    "```\n",
-    "In this example, there are two unique combinations that sum up to the target:\n",
-    "- 2 + 2 + 3 = 7\n",
-    "- 7 = 7\n",
-    "\n",
-    "**Example 2:**\n",
-    "```python\n",
-    "Input: candidates = [2, 3, 5], target = 8\n",
-    "Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n",
-    "```\n",
-    "Here, there are three unique combinations that sum up to the target.\n",
-    "\n",
-    "**Example 3:**\n",
-    "```python\n",
-    "Input: candidates = [2], target = 1\n",
-    "Output: []\n",
-    "```\n",
-    "In this case, there are no combinations that can be formed from the candidates to reach the target of 1, so the output is an empty list.\n",
-    "\n",
-    "The problem asks you to find and return these combinations in any order.\n",
-    "\n",
-    "The constraints for this problem include the length of the candidates array, the values of candidates, and the target value.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def combinationSum(candidates, target):\n",
-    "    # Define a recursive DFS function to find combinations\n",
-    "    def dfs(remaining, start, path):\n",
-    "        # If the remaining target is 0, we found a valid combination\n",
-    "        if remaining == 0:\n",
-    "            result.append(path)\n",
-    "            return\n",
-    "        # If the remaining target is negative, this path is invalid\n",
-    "        if remaining < 0:\n",
-    "            return\n",
-    "        # Iterate through candidates starting from 'start' to avoid duplicates\n",
-    "        for i in range(start, len(candidates)):\n",
-    "            # Explore the current candidate by subtracting it from the remaining target\n",
-    "            # Add the current candidate to the path\n",
-    "            dfs(remaining - candidates[i], i, path + [candidates[i]])\n",
-    "\n",
-    "    result = []  # Initialize an empty list to store the result\n",
-    "    candidates.sort()  # Sort the candidates for deduplication and early stopping\n",
-    "    dfs(target, 0, [])  # Start the DFS from the target value with an empty path\n",
-    "    return result  # Return the list of unique combinations"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The `combinationSum` function takes an array of distinct integers `candidates` and a target integer `target`. It returns a list of all unique combinations of `candidates` where the chosen numbers sum to the `target`. Each number in the `candidates` array can be used an unlimited number of times.\n",
-    "\n",
-    "Here's an overview of how the code works:\n",
-    "\n",
-    "1. The `combinationSum` function is defined, and it takes two arguments: `candidates` and `target`.\n",
-    "\n",
-    "2. Inside the function, there is a nested helper function called `backtrack`. This function is responsible for the actual combination generation using a recursive approach.\n",
-    "\n",
-    "3. The `backtrack` function is called with three arguments: `start`, `target`, and `path`. The `start` variable helps keep track of the current index in the `candidates` array, `target` keeps track of the remaining sum to be achieved, and `path` is a list that stores the current combination.\n",
-    "\n",
-    "4. Within the `backtrack` function, there are three main conditional statements:\n",
-    "   - If `target` becomes zero, it means we have found a combination that sums up to the target, so we add `path` to the `result` list.\n",
-    "   - If `target` becomes negative, it means the current combination doesn't work, so we return without doing anything.\n",
-    "   - Otherwise, we enter a loop that iterates over the `candidates` array, starting from the current index `start`.\n",
-    "\n",
-    "5. In the loop, the `backtrack` function is called recursively with the updated `target` and `path` after including the current candidate. This process explores different combinations by considering the current candidate or moving to the next candidate.\n",
-    "\n",
-    "6. After the loop completes, the `result` list contains all unique combinations that sum to the target.\n",
-    "\n",
-    "7. The `candidates` array is sorted to optimize the search. Sorting helps in avoiding unnecessary recursive branches and reduces the number of explored combinations.\n",
-    "\n",
-    "8. Finally, the `backtrack` function is initially called from the `combinationSum` function with `start` set to 0, `target` set to the original target value, and an empty `path`. The result is returned as a list of unique combinations.\n",
-    "\n",
-    "The code includes a few example cases to demonstrate how the function works with different inputs."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[2, 2, 3], [7]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Example 1\n",
-    "candidates1 = [2, 3, 6, 7]\n",
-    "target1 = 7\n",
-    "print(combinationSum(candidates1, target1))  # Output: [[2, 2, 3], [7]]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[2, 2, 2, 2], [2, 3, 3], [3, 5]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "candidates2 = [2, 3, 5]\n",
-    "target2 = 8\n",
-    "print(combinationSum(candidates2, target2))  # Output: [[2, 2, 2, 2], [2, 3, 3], [3, 5]]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "candidates3 = [2]\n",
-    "target3 = 1\n",
-    "print(combinationSum(candidates3, target3))  # Output: []"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `combinationSum` function.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "The time complexity of this function is influenced by the number of recursive calls made by the `backtrack` function and the amount of work done within each call. In the worst case, the function will explore all possible combinations.\n",
-    "\n",
-    "The worst-case scenario occurs when we have many combinations to reach the target. The time complexity can be expressed as $O(2^n)$, where 'n' is the maximum number of recursive calls. In this context, 'n' corresponds to the number of elements in the `candidates` list, and it may be up to 30 (as per the constraints). However, the actual number of combinations explored is generally much smaller because we eliminate branches when the target becomes negative, and we skip over candidates that cannot lead to a solution. The sorting step also helps to optimize the search.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "The space complexity is determined by the auxiliary space used for storing the results and the stack space for the recursive calls.\n",
-    "\n",
-    "1. The space required for the `result` list can be up to $O(2^n)$ in the worst case because each combination is stored.\n",
-    "\n",
-    "2. The stack space for the recursive calls can also be up to O(n) in the worst case, where 'n' is the number of elements in the `candidates` list.\n",
-    "\n",
-    "Therefore, the overall space complexity of the function is $O(2^n + n)$. However, in practice, the space used for the results list is often the dominant factor, and the actual space used may be much smaller than 2^n because not all combinations are explored.\n",
-    "\n",
-    "In summary, the time complexity is exponential but typically smaller in practice due to optimization, and the space complexity is influenced by the number of results and the depth of the recursion."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Find the Number of Unique Combinations:** Instead of listing all unique combinations, modify the function to return the count of unique combinations that sum to the target.\n",
-    "\n",
-    "2. **Combinations with a Maximum Value:** Add a maximum value constraint for each combination, so they cannot exceed a certain value. Modify the code to find combinations respecting this constraint."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb b/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb
deleted file mode 100644
index 8d97d20..0000000
--- a/_build/jupyter_execute/10. Backtracking/79. Word Search.ipynb	
+++ /dev/null
@@ -1,290 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 79. Word Search\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Word Search problem on LeetCode, click here!](https://leetcode.com/problems/word-search/)\n",
-    "\n",
-    "---\n",
-    "Given an `m x n` grid of characters `board` and a string `word`, return *`true` if `word` exists in the grid.*\n",
-    "\n",
-    "The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == board.length`\n",
-    "- `n = board[i].length`\n",
-    "- `1 <= m, n <= 6`\n",
-    "- `1 <= word.length <= 15`\n",
-    "- `board` and `word` consists of only lowercase and uppercase English letters.\n",
-    "\n",
-    "**Follow up:** Could you use search pruning to make your solution faster with a larger `board`?"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to determine whether a given word can be constructed from characters in an `m x n` grid of characters. The word can be formed by moving sequentially through horizontally or vertically neighboring cells in the grid, and you cannot use the same cell more than once.\n",
-    "\n",
-    "Here are the specifics of the problem:\n",
-    "\n",
-    "- You are given an `m x n` grid, which is a 2D board of characters.\n",
-    "- You are also given a string `word` consisting of lowercase and uppercase English letters.\n",
-    "- Your task is to determine if the word can be found in the grid by moving from one cell to an adjacent cell (horizontally or vertically) and collecting the characters sequentially to form the word.\n",
-    "- You can't use the same cell more than once when forming the word.\n",
-    "- If it's possible to form the word in the grid, return `True`; otherwise, return `False`.\n",
-    "\n",
-    "For example, consider the following grid and word:\n",
-    "\n",
-    "```python\n",
-    "board = [[\"A\", \"B\", \"C\", \"E\"],\n",
-    "         [\"S\", \"F\", \"C\", \"S\"],\n",
-    "         [\"A\", \"D\", \"E\", \"E\"]]         \n",
-    "word = \"ABCCED\"\n",
-    "```\n",
-    "\n",
-    "In this case, the word \"ABCCED\" can be formed by moving through the cells (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) in the grid. Therefore, the function should return `True`.\n",
-    "\n",
-    "Conversely, if the word is impossible to construct, such as in the case of the word \"ABCB\" for the same grid, where there's no valid path that forms the word, the function should return `False`.\n",
-    "\n",
-    "The problem involves searching for a path through the grid that matches the characters in the word while respecting the constraints mentioned. It can be solved using depth-first search (DFS) or backtracking.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def exist(self, board: List[List[str]], word: str) -> bool:\n",
-    "        def dfs(i, j, k):\n",
-    "            # Check if the current position is out of bounds or doesn't match the character in the word.\n",
-    "            if not (0 <= i < len(board)) or not (0 <= j < len(board[0])) or board[i][j] != word[k]:\n",
-    "                return False\n",
-    "            \n",
-    "            # If we have matched all characters in the word, return True.\n",
-    "            if k == len(word) - 1:\n",
-    "                return True\n",
-    "\n",
-    "            # Temporarily mark the current cell to prevent revisiting.\n",
-    "            tmp, board[i][j] = board[i][j], \"/\"\n",
-    "\n",
-    "            # Explore adjacent cells by making recursive calls in all four directions.\n",
-    "            found = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1)\n",
-    "\n",
-    "            # Restore the original character in the cell.\n",
-    "            board[i][j] = tmp\n",
-    "\n",
-    "            return found\n",
-    "\n",
-    "        # Iterate through all cells in the board.\n",
-    "        for i in range(len(board)):\n",
-    "            for j in range(len(board[0])):\n",
-    "                # Check if the 'dfs' function starting from this cell returns True.\n",
-    "                if dfs(i, j, 0):\n",
-    "                    return True\n",
-    "\n",
-    "        # If no match is found after exploring all cells, return False.\n",
-    "        return False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "1. Import the necessary module:\n",
-    "   - `from typing import List`: This line imports the `List` type from the `typing` module, which is used to specify the type of the `board` parameter in the method signature.\n",
-    "\n",
-    "2. Define the `Solution` class:\n",
-    "   - `class Solution:`: This line defines a class called `Solution` to encapsulate the solution for the word search problem.\n",
-    "\n",
-    "3. Define the `exist` method:\n",
-    "   - `def exist(self, board: List[List[str]], word: str) -> bool:`: This method is a member of the `Solution` class and takes two parameters: `board`, which is a 2D list of characters, and `word`, which is a string. It returns a boolean value, indicating whether the word exists in the board.\n",
-    "\n",
-    "4. Implement the inner `dfs` function:\n",
-    "   - `def dfs(i, j, k):`: This is a helper function used for depth-first search (DFS). It is called recursively to explore the board and search for the word. It takes three parameters: `i` and `j` representing the current position on the board, and `k` representing the current index in the `word` string.\n",
-    "\n",
-    "   - Inside `dfs`, it checks if the current position is out of bounds or if the character at the current position on the board does not match the character at the current index in the `word`. If either of these conditions is met, it returns `False`.\n",
-    "\n",
-    "   - If `k` is equal to the length of the `word` minus one, it means we have successfully matched all characters in the `word`, so it returns `True`.\n",
-    "\n",
-    "   - The function then temporarily replaces the character at the current position with a marker, \"/\", to prevent revisiting the same cell. It then explores adjacent cells by making recursive calls in all four directions: up, down, left, and right. If any of these recursive calls return `True`, it propagates the `True` result up the call stack. After the exploration is complete, it restores the original character in the cell.\n",
-    "\n",
-    "5. Iterate through the board:\n",
-    "   - The outer loop iterates through all cells in the `board` using two nested loops. For each cell, it checks if the `dfs` function starting from that cell (with `k` initially set to 0) returns `True`. If it does, it means the word exists, so the method returns `True`.\n",
-    "\n",
-    "6. Return `False`:\n",
-    "   - If the loop completes without finding the word in the board, the method returns `False`.\n",
-    "\n",
-    "To use this code, you can create an instance of the `Solution` class and call the `exist` method on that instance, passing the `board` and the `word` as arguments to check if the word exists in the board."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1: \n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Test case 1: Word \"ABCCED\" can be found in the grid\n",
-    "board1 = [[\"A\",\"B\",\"C\",\"E\"],\n",
-    "          [\"S\",\"F\",\"C\",\"S\"],\n",
-    "          [\"A\",\"D\",\"E\",\"E\"]]\n",
-    "word1 = \"ABCCED\"\n",
-    "print(solution.exist(board1, word1))  # Expected output: True"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "\n",
-    "# Test case 2: Word \"SEE\" can be found in the grid\n",
-    "word2 = \"SEE\"\n",
-    "print(solution.exist(board1, word2))  # Expected output: True"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "b5fca8ab-8c98-45b1-b44b-0503309121c9",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 3\n",
-    "\n",
-    "# Test case 3: Word \"ABCB\" cannot be found in the grid\n",
-    "word3 = \"ABCB\"\n",
-    "print(solution.exist(board1, word3))  # Expected output: False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `exist` method in the given code.\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "The primary work in this code is done by the `dfs` function, which performs a depth-first search through the grid. The time complexity of the `dfs` function is $O(4^(n))$, where \"n\" is the length of the `word`. This is because, in the worst case, the function explores all possible directions (up, down, left, right) for each character in the `word`.\n",
-    "\n",
-    "The outer loops iterate through all cells in the `board`, which has dimensions `m x n`. So, the total number of iterations of `dfs` is O(m * n). However, since the `dfs` function has a higher time complexity, the overall time complexity is still dominated by the `dfs` function.\n",
-    "\n",
-    "Therefore, the overall time complexity of the `exist` method is $O(m * n * 4^n)$, where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. The primary space usage comes from the recursive calls in the `dfs` function. The depth of the recursion can be at most equal to the length of the `word`, which is O(n).\n",
-    "\n",
-    "2. Additionally, the `tmp` variable and the recursive calls on the call stack consume a small amount of additional memory.\n",
-    "\n",
-    "3. The marking of visited cells with the \"/\" character temporarily modifies the `board` in-place but doesn't significantly contribute to the space complexity.\n",
-    "\n",
-    "Therefore, the overall space complexity is O(n) due to the recursive call stack.\n",
-    "\n",
-    "In summary, the time complexity is $O(m * n * 4^n)$, and the space complexity is O(n), where \"m\" and \"n\" are the dimensions of the board, and \"n\" is the length of the word. The time complexity is exponential in the worst case due to the depth-first search through all possible paths in the grid."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the code to find all unique words that can be constructed from the grid. Instead of returning `True` or `False`, return a list of words found. Ensure that the same cell is not used more than once in constructing each word.\n",
-    "\n",
-    "2. **Board Generation:** Create a function to generate random grids of characters and then use the `exist` method to solve them. Test the efficiency and correctness of your code by generating larger boards and words."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb b/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb
deleted file mode 100644
index 56b7647..0000000
--- a/_build/jupyter_execute/11. Graphs/133. Clone Graph.ipynb	
+++ /dev/null
@@ -1,311 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 133. Clone Graph\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Clone Graph problem on LeetCode, click here!](https://leetcode.com/problems/clone-graph/)\n",
-    "\n",
-    "---\n",
-    "Given a reference of a node in a **connected** undirected graph.\n",
-    "\n",
-    "Return a **deep copy** (clone) of the graph.\n",
-    "\n",
-    "Each node in the graph contains a value (`int`) and a list (`List[Node]`) of its neighbors.\n",
-    "\n",
-    "```\n",
-    "class Node {\n",
-    "    public int val;\n",
-    "    public List<Node> neighbors;\n",
-    "}\n",
-    "```\n",
-    "\n",
-    "**Test case format:**\n",
-    "\n",
-    "For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with `val == 1`, the second node with `val == 2`, and so on. The graph is represented in the test case using an adjacency list.\n",
-    "\n",
-    "**An adjacency list** is a collection of unordered **lists** used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.\n",
-    "\n",
-    "The given node will always be the first node with `val = 1`. You must return the **copy of the given node** as a reference to the cloned graph.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- The number of nodes in the graph is in the range `[0, 100]`.\n",
-    "- `1 <= Node.val <= 100`\n",
-    "- `Node.val` is unique for each node.\n",
-    "- There are no repeated edges and no self-loops in the graph.\n",
-    "- The Graph is connected and all nodes can be visited starting from the given node."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand is to clone a connected undirected graph, which means we need to create a deep copy of the original graph while preserving its structure and relationships between nodes. The graph is represented using adjacency lists, where each node has a value (an integer) and a list of its neighbors.\n",
-    "\n",
-    "Here are the key aspects of the problem:\n",
-    "\n",
-    "1. **Input**: You are given a reference to one of the nodes in the original graph. This reference node serves as the entry point to the graph. The entire graph can be explored starting from this node.\n",
-    "\n",
-    "2. **Output**: The goal is to create a deep copy of the graph and return a reference to the cloned graph. The cloned graph should have the same structure and relationships as the original one but should be a separate instance in memory.\n",
-    "\n",
-    "3. **Graph Structure**: The graph is composed of nodes (vertices) and edges. Nodes have values (integers) and are connected to other nodes through edges (undirected connections). Each node has a list of neighbors, which are other nodes it is connected to.\n",
-    "\n",
-    "4. **Connected Graph**: The problem assumes that the graph is connected, meaning that you can start from the reference node provided and traverse the entire graph by following the edges. This ensures that there are no isolated components in the graph.\n",
-    "\n",
-    "To solve this problem, you typically need to perform a graph traversal (e.g., depth-first search or breadth-first search) to visit all the nodes in the original graph, create corresponding nodes in the clone, and establish the same connections between nodes in the cloned graph as in the original one. To avoid duplication of nodes, you also need to keep track of visited nodes to ensure that the cloning process is efficient and that nodes are not duplicated in the clone.\n",
-    "\n",
-    "The problem can be challenging because you need to create a deep copy of the graph while handling cyclic dependencies between nodes and ensuring that the structure and relationships of the original graph are preserved in the clone. The solution requires recursion or a data structure to maintain the mapping between original nodes and their clones to achieve an efficient and accurate cloning process.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Definition for a Node.\n",
-    "class Node:\n",
-    "    def __init__(self, val=0, neighbors=None):\n",
-    "        self.val = val\n",
-    "        self.neighbors = neighbors if neighbors is not None else []\n",
-    "\n",
-    "class Solution:\n",
-    "    def cloneGraph(self, node: 'Node') -> 'Node':\n",
-    "        if not node:\n",
-    "            return None\n",
-    "\n",
-    "        visited = {}  # Dictionary to keep track of visited nodes\n",
-    "\n",
-    "        def dfs(original_node):\n",
-    "            if original_node in visited:\n",
-    "                return visited[original_node]\n",
-    "\n",
-    "            # Create a copy of the original node\n",
-    "            copy_node = Node(original_node.val)\n",
-    "            visited[original_node] = copy_node\n",
-    "\n",
-    "            # Recursively clone the neighbors of the original node\n",
-    "            for neighbor in original_node.neighbors:\n",
-    "                copy_node.neighbors.append(dfs(neighbor))\n",
-    "\n",
-    "            return copy_node\n",
-    "\n",
-    "        return dfs(node)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The given code defines a Python class, `Node`, and a solution class, `Solution`, to clone a connected undirected graph represented by nodes and their neighbors. Here's an explanation of the code:\n",
-    "\n",
-    "1. The `Node` class defines a node in the graph. Each node has a value (`val`) which is an integer and a list of neighbors (`neighbors`) that are other nodes in the graph.\n",
-    "\n",
-    "2. The `Solution` class contains the method `cloneGraph`, which takes a reference to the first node of the graph (`node`) as input and returns a deep copy of the entire graph, preserving the structure and relationships between nodes.\n",
-    "\n",
-    "3. Inside the `cloneGraph` method:\n",
-    "   - It first checks if the input `node` is `None`. If it's `None`, it means there is no graph to clone, and the method returns `None`.\n",
-    "\n",
-    "   - It initializes a `visited` dictionary to keep track of visited nodes. This dictionary is used to ensure that each node is cloned only once to avoid duplication.\n",
-    "\n",
-    "   - The `dfs` (depth-first search) function is defined within the `cloneGraph` method. It takes an original node as input and returns its corresponding clone. This function performs the actual cloning of the graph.\n",
-    "\n",
-    "   - Inside the `dfs` function:\n",
-    "     - It checks if the original node has been visited before by looking it up in the `visited` dictionary. If it has been visited, it returns the corresponding clone from the `visited` dictionary.\n",
-    "\n",
-    "     - If the original node is not visited, it creates a new node, `copy_node`, with the same value as the original node. This is the first step in creating the clone of the original node.\n",
-    "\n",
-    "     - It then recursively clones the neighbors of the original node by iterating through the `neighbors` list of the original node and appending the corresponding cloned neighbors to the `neighbors` list of the `copy_node`. This step ensures that the relationships between nodes are preserved in the clone.\n",
-    "\n",
-    "     - The `copy_node` is added to the `visited` dictionary with the original node as the key and the clone as the value.\n",
-    "\n",
-    "     - Finally, the `copy_node` is returned as the result of the DFS for the original node.\n",
-    "\n",
-    "4. The `cloneGraph` method returns the result of the DFS on the input `node`, which is the clone of the entire graph with its structure and relationships intact.\n",
-    "\n",
-    "In summary, the code uses depth-first search (DFS) to traverse the original graph, creating a clone for each node and its neighbors. It ensures that each node is cloned only once to avoid duplication and returns a reference to the cloned graph."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[1, 2, 4]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "def get_adjacency_list(node):\n",
-    "    visited = {}  # Dictionary to keep track of visited nodes\n",
-    "\n",
-    "    def dfs(node):\n",
-    "        if not node:\n",
-    "            return []\n",
-    "\n",
-    "        if node in visited:\n",
-    "            return visited[node]\n",
-    "\n",
-    "        neighbors = [n.val for n in node.neighbors]\n",
-    "        visited[node] = neighbors\n",
-    "\n",
-    "        for neighbor in node.neighbors:\n",
-    "            dfs(neighbor)\n",
-    "\n",
-    "    dfs(node)\n",
-    "    result = []\n",
-    "\n",
-    "    def get_node_and_neighbors(node):\n",
-    "        result.append([node.val] + visited[node])\n",
-    "\n",
-    "    get_node_and_neighbors(node)\n",
-    "    return result\n",
-    "\n",
-    "\n",
-    "# Testing the code with the provided examples\n",
-    "adjList1 = [[2, 4], [1, 3], [2, 4], [1, 3]]\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Create the graph from adjList\n",
-    "nodes = [Node(i) for i in range(1, len(adjList1) + 1)]\n",
-    "for i, adj in enumerate(adjList1):\n",
-    "    nodes[i].neighbors = [nodes[j - 1] for j in adj]\n",
-    "\n",
-    "# Clone the graph and get its adjacency list\n",
-    "cloned_node1 = solution.cloneGraph(nodes[0])\n",
-    "cloned_adjList1 = get_adjacency_list(cloned_node1)\n",
-    "\n",
-    "# Print the cloned graph as an adjacency list\n",
-    "print(cloned_adjList1)\n",
-    "\n",
-    "# Output should match adjList1\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[1]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "# Input: adjList = [[]]\n",
-    "\n",
-    "adjList2 = [[]]\n",
-    "solution = Solution()\n",
-    "\n",
-    "# Create the graph from adjList\n",
-    "node1 = Node(1)\n",
-    "\n",
-    "# Clone the graph and get its adjacency list\n",
-    "cloned_node1 = solution.cloneGraph(node1)\n",
-    "cloned_adjList2 = get_adjacency_list(cloned_node1)\n",
-    "\n",
-    "# Print the cloned graph as an adjacency list\n",
-    "print(cloned_adjList2)\n",
-    "\n",
-    "# Output should be [[]], which represents an empty graph"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for cloning a graph using depth-first search (DFS).\n",
-    "\n",
-    "Time Complexity:\n",
-    "1. Creating the graph: The time complexity to create the original graph from `adjList` is O(V + E), where V is the number of vertices (nodes) and E is the number of edges. We iterate through each node and its neighbors in `adjList`.\n",
-    "\n",
-    "2. Cloning the graph (DFS): The depth-first search (DFS) is used to traverse the original graph and create a clone. The time complexity of DFS is O(V + E), as we visit each node and each edge once.\n",
-    "\n",
-    "Overall, the time complexity is O(V + E) for both creating the original graph and cloning it.\n",
-    "\n",
-    "Space Complexity:\n",
-    "1. Storage for the original graph: We need to store the original graph, which includes all nodes and their adjacency lists. In the worst case, this requires O(V + E) space, where V is the number of nodes, and E is the number of edges.\n",
-    "\n",
-    "2. Storage for the visited dictionary: The `visited` dictionary stores mappings from original nodes to their corresponding clones. In the worst case, this requires O(V) space because each node is visited once.\n",
-    "\n",
-    "3. Recursive stack for DFS: The space used for the recursive call stack during DFS can be up to O(V) in the worst case, where V is the number of nodes.\n",
-    "\n",
-    "Overall, the space complexity is O(V + E) for storing the original graph, and O(V) for auxiliary data structures, including the `visited` dictionary and the DFS call stack. Therefore, the overall space complexity is O(V + E).\n",
-    "\n",
-    "In summary, the time and space complexities of the provided code for cloning a graph are both O(V + E), where V is the number of nodes, and E is the number of edges in the graph."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Cyclic Graph**: Design a graph with cycles (nodes connected in a cycle). Test if the code can correctly clone graphs with cycles and doesn't fall into an infinite loop during traversal.\n",
-    "\n",
-    "\n",
-    "2. **Partial Graph Cloning**: Modify the code to clone only a subset of the graph starting from a given reference node, rather than the entire graph. Implement this as an extension to the problem."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
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-   "nbconvert_exporter": "python",
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-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb b/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb
deleted file mode 100644
index 0dd53e9..0000000
--- a/_build/jupyter_execute/11. Graphs/178. Graph Valid Tree.ipynb	
+++ /dev/null
@@ -1,242 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 178. Graph Valid Tree\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Graph Valid Tree problem on LintCode, click here!](https://www.lintcode.com/problem/178/)\n",
-    "\n",
-    "---\n",
-    "Given `n` nodes labeled from `0` to `n - 1` and a list of `undirected` edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- You can assume that no duplicate edges will appear in edges. Since all edges are `undirected`, `[0, 1]` is the same as `[1, 0]` and thus will not appear together in edges."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to determine whether a given set of undirected edges can form a valid tree. In this context, a \"tree\" is a specific type of graph with the following properties:\n",
-    "\n",
-    "1. It is a connected graph, meaning there is a path between any pair of nodes in the graph.\n",
-    "2. It has no cycles, which means there are no closed loops or circuits in the graph.\n",
-    "3. It has exactly `n - 1` edges, where `n` is the number of nodes in the tree.\n",
-    "\n",
-    "The problem is typically given with two pieces of information:\n",
-    "\n",
-    "1. The number of nodes (`n`), which are usually labeled from 0 to `n-1`.\n",
-    "2. A list of undirected edges, where each edge is represented as a pair of nodes.\n",
-    "\n",
-    "The task is to write a function or algorithm to determine whether the provided set of edges can be arranged to form a valid tree based on the properties mentioned above.\n",
-    "\n",
-    "To solve this problem, you need to check the following conditions:\n",
-    "\n",
-    "1. There should be exactly `n - 1` edges to ensure that the graph is connected and tree-like.\n",
-    "2. There should be no cycles in the graph. If there are any cycles, the set of edges cannot form a tree.\n",
-    "3. The graph should be connected, which means you can reach any node from any other node through a series of edges.\n",
-    "\n",
-    "The solution typically involves using graph traversal algorithms like Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the graph and check these conditions. If all conditions are met, the given edges form a valid tree; otherwise, they do not.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def validTree(n, edges):\n",
-    "    if len(edges) != n - 1:\n",
-    "        return False  # A tree with n nodes must have n-1 edges.\n",
-    "\n",
-    "    # Create an adjacency list to represent the graph.\n",
-    "    adj_list = {i: [] for i in range(n)}\n",
-    "    for u, v in edges:\n",
-    "        adj_list[u].append(v)\n",
-    "        adj_list[v].append(u)\n",
-    "\n",
-    "    visited = set()\n",
-    "\n",
-    "    # Define a recursive DFS function.\n",
-    "    def dfs(node, parent):\n",
-    "        if node in visited:\n",
-    "            return False  # If we encounter a visited node, there's a cycle.\n",
-    "        visited.add(node)\n",
-    "\n",
-    "        for neighbor in adj_list[node]:\n",
-    "            if neighbor != parent:\n",
-    "                if not dfs(neighbor, node):\n",
-    "                    return False\n",
-    "\n",
-    "        return True\n",
-    "\n",
-    "    # Start DFS from any node (e.g., node 0).\n",
-    "    if not dfs(0, -1):\n",
-    "        return False  # If there's a cycle, it's not a valid tree.\n",
-    "\n",
-    "    return len(visited) == n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The `validTree` function checks if a given set of undirected edges forms a valid tree. A valid tree must satisfy these conditions:\n",
-    "\n",
-    "1. It has exactly `n - 1` edges, where `n` is the number of nodes.\n",
-    "2. There are no cycles in the graph.\n",
-    "3. The graph is connected, meaning all nodes are reachable from any starting node.\n",
-    "\n",
-    "The code accomplishes this as follows:\n",
-    "\n",
-    "- It first checks if the number of edges is equal to `n - 1`. If not, it immediately returns `False` since a tree with `n` nodes must have `n - 1` edges to be connected.\n",
-    "\n",
-    "- It then constructs an adjacency list (`adj_list`) to represent the graph efficiently. This data structure stores the neighbors of each node.\n",
-    "\n",
-    "- The code uses a `visited` set to keep track of nodes visited during depth-first search (DFS).\n",
-    "\n",
-    "- The `dfs` function is a recursive function that checks for cycles in the graph. It does this by visiting nodes and marking them as visited. If it encounters a node that has already been visited (indicating a cycle), it returns `False`.\n",
-    "\n",
-    "- The DFS function explores the neighbors of each node, avoiding revisiting the parent node to prevent cycles.\n",
-    "\n",
-    "- After the DFS is complete, the code checks if the number of visited nodes is equal to `n`, ensuring the graph is connected.\n",
-    "\n",
-    "- Finally, if all conditions are met, the code returns `True`, indicating that the given edges form a valid tree. Otherwise, it returns `False`.\n",
-    "\n",
-    "The code provides examples to demonstrate how to use the `validTree` function to check whether a set of edges forms a valid tree or not."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "n1 = 5\n",
-    "edges1 = [[0, 1], [0, 2], [0, 3], [1, 4]]\n",
-    "print(validTree(n1, edges1))  # Output: true"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "n2 = 5\n",
-    "edges2 = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]\n",
-    "print(validTree(n2, edges2))  # Output: false"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexities of the `validTree` function:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "1. Constructing the adjacency list (`adj_list`) takes O(E) time, where E is the number of edges in the input graph (|edges|).\n",
-    "\n",
-    "2. The DFS function, when considering all nodes, has a time complexity of O(V + E), where V is the number of nodes in the graph. This is because, in the worst case, the DFS function visits each node and each edge once.\n",
-    "\n",
-    "3. Checking if the number of visited nodes is equal to `n` takes O(1) time.\n",
-    "\n",
-    "Overall, the time complexity of the `validTree` function is O(E + V) in the worst case.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "1. The adjacency list (`adj_list`) consumes additional space to store the graph structure, taking O(V + E) space. This is because it stores information about each node and its adjacent nodes.\n",
-    "\n",
-    "2. The `visited` set keeps track of visited nodes, and its space complexity is O(V) in the worst case, as there can be at most V unique nodes in the set.\n",
-    "\n",
-    "3. The depth of the function call stack during DFS can be at most V, which contributes O(V) space to the space complexity.\n",
-    "\n",
-    "Overall, the space complexity of the `validTree` function is O(V + E) due to the adjacency list and O(V) due to the `visited` set and DFS call stack.\n",
-    "\n",
-    "In summary, the time complexity is O(E + V), and the space complexity is O(V + E). These complexities are based on the worst-case scenario, where all nodes and edges are considered. In practical cases, the actual complexities may be smaller, depending on the specific structure of the graph."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Minimum Spanning Tree (MST)**: Implement an algorithm to find the minimum spanning tree of a connected graph using either Kruskal's or Prim's algorithm. Compare the MST to the original graph to determine if the original graph is a tree. This exercise reinforces the concept of tree properties.\n",
-    "\n",
-    "2. **Detecting Cycles**: Modify the code to not only check if the input forms a tree but also to identify and print out any cycles present in the graph if it's not a tree. This exercise enhances your ability to detect and visualize cycles in a graph."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb b/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb
deleted file mode 100644
index 5a878a4..0000000
--- a/_build/jupyter_execute/11. Graphs/200. Number of Islands.ipynb	
+++ /dev/null
@@ -1,236 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 200. Number of Islands\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Number of Islands problem on LeetCode, click here!](https://leetcode.com/problems/number-of-islands/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "Given an `m x n` 2D binary grid `grid` which represents a map of `'1'`s (land) and `'0'`s (water), return *the number of islands*. \n",
-    "\n",
-    "An **island** is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == grid.length`\n",
-    "- `n == grid[i].length`\n",
-    "- `1 <= m, n <= 300`\n",
-    "- `grid[i][j]` is `'0'` or `'1'`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is about counting the number of islands in a 2D binary grid. In this grid, '1' represents land, and '0' represents water. An island is defined as a group of '1' cells that are adjacent to each other either horizontally or vertically. It's important to note that diagonal connections do not count. Additionally, it's assumed that all four edges of the grid are surrounded by water, meaning the grid is finite and doesn't \"wrap around.\"\n",
-    "\n",
-    "The objective is to determine how many distinct islands exist in the grid. To clarify, an island consists of connected '1' cells, and these connections can only occur horizontally or vertically. If there are no '1' cells in the grid, the number of islands would be zero.\n",
-    "\n",
-    "The problem often involves using graph traversal techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to identify and count these islands. You start at one '1' cell and explore its adjacent cells to determine the extent of the island. Once you've visited all the '1' cells in an island, you move on to the next unvisited '1' cell and repeat the process until you've counted all the islands in the grid.\n",
-    "\n",
-    "In essence, the problem asks you to analyze the grid to find groups of connected '1' cells and count them as individual islands. The goal is to return the total count of islands in the given grid.\n",
-    "\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def numIslands(self, grid: List[List[str]]) -> int:\n",
-    "        if not grid:\n",
-    "            return 0\n",
-    "\n",
-    "        m, n = len(grid), len(grid[0])  # Get the dimensions of the grid\n",
-    "        num_islands = 0  # Initialize the count of islands\n",
-    "\n",
-    "        def dfs(row, col):\n",
-    "            if row < 0 or row >= m or col < 0 or col >= n or grid[row][col] == '0':\n",
-    "                return\n",
-    "\n",
-    "            grid[row][col] = '0'  # Mark the current cell as visited by changing it to '0'\n",
-    "            directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]  # Define four possible directions to move\n",
-    "\n",
-    "            for dr, dc in directions:\n",
-    "                # Explore adjacent cells\n",
-    "                dfs(row + dr, col + dc)\n",
-    "\n",
-    "        for i in range(m):\n",
-    "            for j in range(n):\n",
-    "                if grid[i][j] == '1':\n",
-    "                    num_islands += 1  # Increment the island count\n",
-    "                    dfs(i, j)  # Start a DFS from the current '1' cell to explore the entire island\n",
-    "\n",
-    "        return num_islands\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code defines a Python class called `Solution` with a method called `numIslands`. This class is designed to count the number of islands in a 2D binary grid.\n",
-    "\n",
-    "The `numIslands` method takes a 2D grid as input, where '1' represents land, and '0' represents water. It uses a depth-first search (DFS) approach to identify and count islands. The key steps in the code include:\n",
-    "\n",
-    "1. Checking if the input grid is empty, and if so, returning 0 (indicating there are no islands).\n",
-    "\n",
-    "2. Determining the dimensions of the grid (number of rows and columns).\n",
-    "\n",
-    "3. Initializing a variable `num_islands` to keep track of the count of islands.\n",
-    "\n",
-    "4. Defining a nested function `dfs` to perform depth-first search. This function recursively explores adjacent land cells connected to the current cell and marks them as visited by changing '1' to '0'.\n",
-    "\n",
-    "5. In the `dfs` function, it checks for boundary conditions and whether the current cell is '0'. If these conditions are met, it returns without further exploration.\n",
-    "\n",
-    "6. The `dfs` function explores neighboring cells in all four directions (up, down, left, right) using a loop.\n",
-    "\n",
-    "7. The main loop iterates through each cell of the grid. When a '1' cell is encountered, it increments the `num_islands` count and starts a DFS from that cell to explore the entire island.\n",
-    "\n",
-    "8. Finally, the method returns the total count of islands found in the grid.\n",
-    "\n",
-    "This code encapsulates the solution in a class, allowing you to create an instance of the `Solution` class and call the `numIslands` method on it to count the number of islands in a given grid."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 1 Output: 1\n"
-     ]
-    }
-   ],
-   "source": [
-    "grid1 = [\n",
-    "  [\"1\", \"1\", \"1\", \"1\", \"0\"],\n",
-    "  [\"1\", \"1\", \"0\", \"1\", \"0\"],\n",
-    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-    "  [\"0\", \"0\", \"0\", \"0\", \"0\"]\n",
-    "]\n",
-    "\n",
-    "solution = Solution()\n",
-    "result1 = solution.numIslands(grid1)\n",
-    "print(\"Example 1 Output:\", result1)  # Expected output: 1\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 2 Output: 3\n"
-     ]
-    }
-   ],
-   "source": [
-    "grid2 = [\n",
-    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-    "  [\"1\", \"1\", \"0\", \"0\", \"0\"],\n",
-    "  [\"0\", \"0\", \"1\", \"0\", \"0\"],\n",
-    "  [\"0\", \"0\", \"0\", \"1\", \"1\"]\n",
-    "]\n",
-    "\n",
-    "result2 = solution.numIslands(grid2)\n",
-    "print(\"Example 2 Output:\", result2)  # Expected output: 3\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "Time Complexity:\n",
-    "- The code uses a depth-first search (DFS) to explore the grid and count the number of islands.\n",
-    "- In the worst case, it visits every cell in the grid exactly once.\n",
-    "- The DFS function explores neighboring cells in four directions (up, down, left, right), so it has a time complexity of $O(4^{max(m, n)})$, where m is the number of rows and n is the number of columns.\n",
-    "- Therefore, the overall time complexity of the code is O(m * n), where m is the number of rows, and n is the number of columns.\n",
-    "\n",
-    "Space Complexity:\n",
-    "- The space complexity is determined by the recursive calls and the depth of the DFS stack.\n",
-    "- In the worst case, when the entire grid consists of '1's, the depth of the DFS recursion can be as deep as max(m, n).\n",
-    "- Therefore, the space complexity of the code is O(max(m, n)), representing the maximum depth of the DFS stack.\n",
-    "\n",
-    "In practical terms, for reasonably sized grids, the time and space complexity is linear, and the code should perform well. However, it's important to keep in mind that the worst-case time complexity is O(m * n), and the space complexity can be O(max(m, n))."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Identify Largest Island:** In addition to counting the number of islands, modify the code to identify the largest island in the grid and return its size.\n",
-    "\n",
-    "2. **Optimize for Space:** Modify the solution to reduce its space complexity to O(1) without modifying the input grid. This is a more memory-efficient version of the problem."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb b/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb
deleted file mode 100644
index 4176cab..0000000
--- a/_build/jupyter_execute/11. Graphs/207. Course Schedule.ipynb	
+++ /dev/null
@@ -1,243 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 207. Course Schedule\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Course Schedule problem on LeetCode, click here!](https://leetcode.com/problems/course-schedule/)\n",
-    "\n",
-    "---\n",
-    "There are a total of `numCourses` courses to take, labeled from 0 to `numCourses - 1`. You are given an array `prerequisites` where `prerequisites[i] = [ai, bi]` indicates that you **must** take course `bi` first if you want to take course `ai`. \n",
-    "\n",
-    "- For example, the pair `[0, 1]`, indicates that to take course `0` you have to first take course `1`.\n",
-    "\n",
-    "Return `true` if you can finish all courses. Otherwise, return `false`.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- 1 <= `numCourses` <= 2000\n",
-    "- 0 <= `prerequisites.length` <= 5000\n",
-    "- `prerequisites[i].length` == 2\n",
-    "- 0 <= `ai`, `bi` < `numCourses`\n",
-    "- All the pairs `prerequisites[i]` are **unique**."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is to determine whether it's possible to complete a set of courses given a list of prerequisites. You are given the following:\n",
-    "\n",
-    "- `numCourses`: The total number of courses to be taken, labeled from 0 to `numCourses - 1`.\n",
-    "- `prerequisites`: A list of prerequisite courses, where each element `prerequisites[i]` is a pair `[ai, bi]`, indicating that you must take course `bi` before you can take course `ai`.\n",
-    "\n",
-    "The goal is to check whether it's possible to complete all the courses while respecting the prerequisite requirements. In other words, you need to determine if there are any circular dependencies in the course prerequisites that would prevent you from taking all the courses.\n",
-    "\n",
-    "For example:\n",
-    "\n",
-    "- If `numCourses` is 2, and `prerequisites` is `[[1, 0]]`, it means you have two courses, and you must finish course 0 before you can take course 1. In this case, it's possible to complete all courses, so the function should return `True`.\n",
-    "\n",
-    "- If `numCourses` is 2, and `prerequisites` is `[[1, 0], [0, 1]]`, it means you have two courses, but there's a circular dependency where course 1 requires course 0 and course 0 requires course 1. In this case, it's impossible to complete all courses, so the function should return `False`.\n",
-    "\n",
-    "The problem is essentially about checking for the presence of cycles in a directed graph, where the courses are nodes, and the prerequisites represent directed edges between them. If there are no cycles, it's possible to complete all courses; otherwise, it's not possible.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from collections import defaultdict, deque\n",
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:\n",
-    "        # Create a graph using an adjacency list to represent the courses and their prerequisites.\n",
-    "        graph = defaultdict(list)  # Initialize an empty adjacency list.\n",
-    "        in_degree = [0] * numCourses  # Initialize an array to store in-degrees of courses.\n",
-    "\n",
-    "        # Populate the graph and calculate in-degrees.\n",
-    "        for course, prerequisite in prerequisites:\n",
-    "            graph[prerequisite].append(course)  # Add the course as a neighbor to its prerequisite.\n",
-    "            in_degree[course] += 1  # Increment the in-degree of the course.\n",
-    "\n",
-    "        # Initialize a queue with courses that have no prerequisites.\n",
-    "        queue = deque()\n",
-    "        for course in range(numCourses):\n",
-    "            if in_degree[course] == 0:\n",
-    "                queue.append(course)\n",
-    "\n",
-    "        # Perform topological sorting.\n",
-    "        while queue:\n",
-    "            course = queue.popleft()  # Take a course with no prerequisites.\n",
-    "            numCourses -= 1  # Decrement the count of remaining courses.\n",
-    "\n",
-    "            for neighbor in graph[course]:\n",
-    "                in_degree[neighbor] -= 1  # Remove the prerequisite relationship.\n",
-    "                if in_degree[neighbor] == 0:\n",
-    "                    queue.append(neighbor)  # If no more prerequisites, add to the queue.\n",
-    "\n",
-    "        # If all courses were successfully taken (numCourses becomes 0), return True.\n",
-    "        return numCourses == 0"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "This Python code is designed to determine whether it is possible to complete all the required courses given a set of prerequisites. It uses a topological sorting algorithm to accomplish this. Here's a brief explanation of the code:\n",
-    "\n",
-    "1. The code defines a `Solution` class with a `canFinish` method that takes two parameters: `numCourses` (the total number of courses) and `prerequisites` (a list of pairs indicating which course must be taken before another).\n",
-    "\n",
-    "2. It initializes an empty graph as an adjacency list using a `defaultdict` to represent the courses and their prerequisites. It also initializes a list called `in_degree` to keep track of the in-degrees of courses, which is used in the topological sorting process.\n",
-    "\n",
-    "3. The code then populates the graph by iterating through the `prerequisites` list. For each pair `(ai, bi)`, it adds course `ai` as a neighbor to its prerequisite course `bi` in the graph and increments the in-degree of course `ai`.\n",
-    "\n",
-    "4. Next, it initializes a queue with courses that have no prerequisites. It does this by iterating through all the courses and adding those with an in-degree of 0 to the queue.\n",
-    "\n",
-    "5. The code performs a topological sorting of the courses using a while loop. In each iteration of the loop, it dequeues a course from the front of the queue (a course with no prerequisites).\n",
-    "\n",
-    "6. For each neighbor (course that depends on the dequeued course), it decrements the in-degree of the neighbor. If the neighbor's in-degree becomes 0, it means all of its prerequisites have been taken, so the neighbor is added to the queue.\n",
-    "\n",
-    "7. The loop continues until there are no more courses with no prerequisites to dequeue, and the in-degrees are updated accordingly.\n",
-    "\n",
-    "8. After the loop, if all courses were successfully taken (i.e., `numCourses` becomes 0), it means that there were no circular dependencies, and it's possible to finish all courses. In this case, the method returns `True`. Otherwise, if there are remaining courses (indicating a circular dependency), it returns `False`."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "True\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "numCourses1 = 2\n",
-    "prerequisites1 = [[1, 0]]\n",
-    "sol = Solution()\n",
-    "print(sol.canFinish(numCourses1, prerequisites1))  # Output: True"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "False\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "numCourses2 = 2\n",
-    "prerequisites2 = [[1, 0], [0, 1]]\n",
-    "print(sol.canFinish(numCourses2, prerequisites2))  # Output: False"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the `canFinish` method:\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "\n",
-    "1. Constructing the graph: In the first loop where we iterate through the `prerequisites` list, we populate the graph with prerequisites. This loop has a time complexity of O(E), where E is the number of prerequisites (edges).\n",
-    "\n",
-    "2. Initializing in-degrees and finding courses with no prerequisites: In the worst case, we iterate through all `numCourses` courses to initialize in-degrees and find the courses with no prerequisites. This has a time complexity of O(V), where V is the number of courses (vertices).\n",
-    "\n",
-    "3. Topological Sorting: In the while loop, we perform topological sorting, processing each course once and its outgoing edges. In the worst case, each course is processed once. This has a time complexity of O(V + E).\n",
-    "\n",
-    "The overall time complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "\n",
-    "1. Graph Representation: The space complexity for storing the graph as an adjacency list is O(E), where E is the number of prerequisites.\n",
-    "\n",
-    "2. In-Degree Array: The space complexity for storing the in-degrees of courses is O(V), where V is the number of courses.\n",
-    "\n",
-    "3. Queue: The space complexity for the queue is O(V), as it may contain all the courses.\n",
-    "\n",
-    "The overall space complexity is O(V + E), where V is the number of courses, and E is the number of prerequisites.\n",
-    "\n",
-    "In most practical cases, the number of prerequisites (E) is much smaller than the number of courses (V), so the time and space complexity can often be considered as O(V)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimizing Course Scheduling**: Given a list of courses with their durations and prerequisites, find the most efficient way to schedule these courses to minimize the time it takes to complete all of them.\n",
-    "\n",
-    "2. **Detecting Cycles in Course Dependencies**: Modify the original problem to not just determine if you can finish all courses but also to identify the specific courses that form a cycle in the prerequisite dependencies."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb b/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb
deleted file mode 100644
index a2789c4..0000000
--- a/_build/jupyter_execute/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb	
+++ /dev/null
@@ -1,230 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 3651. Number of Connected Components in an Undirected Graph\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Number of Connected Components in an Undirected Graph problem on LintCode, click here!](https://www.lintcode.com/problem/3651/)\n",
-    "\n",
-    "---\n",
-    "In this problem, there is an undirected graph with `n` nodes. There is also an `edges` array. Where `edges[i] = [a, b]` means that there is an edge between *node a* and *node b* in the graph.\n",
-    "\n",
-    "You need to return *the number of connected components in that graph*."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand is to determine the number of connected components in an undirected graph. Here's a more detailed explanation of the problem:\n",
-    "\n",
-    "- You are given an undirected graph, which consists of nodes and edges. The graph may have multiple distinct connected components.\n",
-    "\n",
-    "- A connected component is a group of nodes within the graph where each node can be reached from every other node in the same component. In other words, there is a path between any two nodes in the same connected component, but there are no paths connecting nodes from different components.\n",
-    "\n",
-    "- Your task is to write a function that takes as input the number of nodes `n` and a list of edges `edges`, where each edge is represented as a pair of nodes `[a, b]` indicating an edge between node `a` and node `b`.\n",
-    "\n",
-    "- The goal is to determine the number of distinct connected components in the graph.\n",
-    "\n",
-    "- You need to implement an algorithm that can identify these connected components and count how many there are in the given graph.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "def countConnectedComponents(n, edges):\n",
-    "    # Define a helper function for depth-first search (DFS).\n",
-    "    def dfs(node):\n",
-    "        if not visited[node]:  # If the node has not been visited yet:\n",
-    "            visited[node] = True  # Mark it as visited.\n",
-    "            for neighbor in graph[node]:  # Explore all neighbors of the current node.\n",
-    "                dfs(neighbor)  # Recursively call DFS on the neighbor.\n",
-    "\n",
-    "    # Create an empty adjacency list to represent the graph.\n",
-    "    graph = [[] for _ in range(n)]\n",
-    "\n",
-    "    # Initialize a boolean list to keep track of visited nodes.\n",
-    "    visited = [False] * n\n",
-    "\n",
-    "    # Populate the adjacency list with edges from the input.\n",
-    "    for a, b in edges:\n",
-    "        graph[a].append(b)\n",
-    "        graph[b].append(a)  # Since the graph is undirected, we add edges in both directions.\n",
-    "\n",
-    "    # Initialize a counter to keep track of the number of connected components.\n",
-    "    count = 0\n",
-    "\n",
-    "    # Iterate through all nodes in the graph.\n",
-    "    for i in range(n):\n",
-    "        if not visited[i]:  # If the node has not been visited:\n",
-    "            count += 1  # Increment the count since we've found a new connected component.\n",
-    "            dfs(i)  # Start a depth-first search from the unvisited node to explore the component.\n",
-    "\n",
-    "    return count"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided Python code calculates the number of connected components in an undirected graph. Here's a high-level explanation of how the code works:\n",
-    "\n",
-    "1. The `countConnectedComponents` function takes two arguments: `n` (the number of nodes in the graph) and `edges` (a list of edge pairs representing connections between nodes).\n",
-    "\n",
-    "2. Inside the function, there is a helper function called `dfs` (depth-first search), which is used to explore and mark nodes as visited.\n",
-    "\n",
-    "3. An empty adjacency list called `graph` is created to represent the graph. The `graph` is a list of lists where each element represents a node, and the list associated with each node contains its neighboring nodes.\n",
-    "\n",
-    "4. Another list called `visited` is created to keep track of visited nodes. Initially, all nodes are marked as not visited (`False`).\n",
-    "\n",
-    "5. The code populates the `graph` by iterating through the `edges` list and adding each edge to the adjacency list. Since the graph is undirected, both directions of each edge are added.\n",
-    "\n",
-    "6. A counter variable called `count` is initialized to keep track of the number of connected components.\n",
-    "\n",
-    "7. The code iterates through all nodes from `0` to `n-1`. For each node, it checks whether it has been visited yet. If it hasn't been visited, it means a new connected component is found. The counter `count` is incremented, and a depth-first search (DFS) is initiated from this unvisited node to explore the connected component.\n",
-    "\n",
-    "8. The `dfs` function recursively visits all nodes in the connected component, marking them as visited. It does this by checking each neighbor of the current node and recursively calling `dfs` on unvisited neighbors.\n",
-    "\n",
-    "9. After the loop through all nodes, the `count` variable contains the total number of connected components in the graph.\n",
-    "\n",
-    "10. The function returns the value of `count`, which represents the number of connected components.\n",
-    "\n",
-    "11. Two examples are provided at the end to demonstrate how to use the function with different inputs."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "1\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "n1 = 3\n",
-    "edges1 = [[0, 1], [0, 2]]\n",
-    "print(countConnectedComponents(n1, edges1))  # Output: 1"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "n2 = 6\n",
-    "edges2 = [[0, 1], [1, 2], [2, 3], [4, 5]]\n",
-    "print(countConnectedComponents(n2, edges2))  # Output: 2"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the code for counting the number of connected components in an undirected graph using depth-first search (DFS).\n",
-    "\n",
-    "**Time Complexity:**\n",
-    "- The code iterates through all nodes in the graph (from 0 to `n-1`).\n",
-    "- For each unvisited node, it initiates a depth-first search (DFS) from that node to explore the connected component.\n",
-    "- In the worst case, every node may belong to a separate connected component, and you perform a DFS for each node.\n",
-    "- The time complexity of the DFS for a single connected component is O(V + E), where V is the number of nodes in the component, and E is the number of edges.\n",
-    "- In the worst case, you perform O(n) DFS calls, each with its own O(V + E) complexity.\n",
-    "\n",
-    "So, the overall time complexity of the code is O(n * (V + E)). In the worst case, if the graph is connected (a single connected component), this simplifies to O(n * (n + m)), where m is the number of edges.\n",
-    "\n",
-    "**Space Complexity:**\n",
-    "- The main data structures used are the `graph` (adjacency list), `visited` (boolean array), and the recursive call stack for DFS.\n",
-    "- The `graph` has a space complexity of O(n + m), where n is the number of nodes, and m is the number of edges.\n",
-    "- The `visited` boolean array also has a space complexity of O(n) to keep track of whether nodes have been visited.\n",
-    "- The depth of the recursive call stack for DFS can go up to O(n) in the worst case, when all nodes are part of a single connected component.\n",
-    "\n",
-    "So, the overall space complexity of the code is O(n + m) for data structures and O(n) for the call stack. In the worst case, this simplifies to O(n + m).\n",
-    "\n",
-    "In summary, the time complexity is O(n * (V + E)), and the space complexity is O(n + m)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Disconnected Components with Multiple Queries:** You are given an undirected graph with `n` nodes and `m` edges. Initially, the graph is disconnected, and you need to process `q` queries. Each query consists of adding an edge to connect two nodes. After each query, you need to output the number of connected components in the updated graph.\n",
-    "\n",
-    "2. **Maximum Size of Connected Components:** Given an undirected graph with `n` nodes and `m` edges, find the size of the largest connected component in the graph. Additionally, identify the nodes that belong to this largest component. You can assume there is at least one connected component in the graph.\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb b/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb
deleted file mode 100644
index c685b23..0000000
--- a/_build/jupyter_execute/11. Graphs/417. Pacific Atlantic Water Flow.ipynb	
+++ /dev/null
@@ -1,255 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 417. Pacific Atlantic Water Flow\n",
-    "\n",
-    "\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "**Link to Problem:** [To see the Pacific Atlantic Water Flow problem on LeetCode, click here!](https://leetcode.com/problems/pacific-atlantic-water-flow/)\n",
-    "\n",
-    "---\n",
-    "There is an `m x n` rectangular island that borders both the **Pacific Ocean** and **Atlantic Ocean**. The **Pacific Ocea**n touches the island's left and top edges, and the **Atlantic Ocean** touches the island's right and bottom edges.\n",
-    "\n",
-    "The island is partitioned into a grid of square cells. You are given an `m x n` integer matrix `heights` where `heights[r][c]` represents the **height above sea level** of the cell at coordinate `(r, c)`.\n",
-    "\n",
-    "The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is **less than or equal to** the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.\n",
-    "\n",
-    "Return *a **2D list** of grid coordinates `result` where $result[i] = [r_i, c_i]$ denotes that rain water can flow from cell $(r_i, c_i)$ to **both** the Pacific and Atlantic oceans*.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `m == heights.length`\n",
-    "- `n == heights[r].length`\n",
-    "- `1 <= m, n <= 200`\n",
-    "- 0 <= `heights[r][c]` <= $10^5$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand involves a scenario with a rectangular island surrounded by both the Pacific Ocean and the Atlantic Ocean. The Pacific Ocean is in contact with the island's left and top edges, while the Atlantic Ocean is in contact with the island's right and bottom edges. The goal is to determine which cells on the island can allow rainwater to flow to both the Pacific and Atlantic Oceans.\n",
-    "\n",
-    "Here are the key details and constraints of the problem:\n",
-    "\n",
-    "1. The island is divided into a grid of square cells.\n",
-    "\n",
-    "2. Each cell is represented by an integer value in a matrix `heights`. The value at `heights[r][c]` represents the height above sea level of the cell at coordinates `(r, c)` on the island.\n",
-    "\n",
-    "3. Rainwater can flow from a cell to neighboring cells directly north, south, east, or west only if the neighboring cell's height is less than or equal to the current cell's height.\n",
-    "\n",
-    "4. Water can flow from any cell adjacent to the ocean into the ocean. This means water can flow from cells on the edge of the island to the Pacific Ocean and the Atlantic Ocean.\n",
-    "\n",
-    "The task is to find and return a 2D list of grid coordinates where rainwater can flow from a cell to both the Pacific and Atlantic Oceans. In other words, you need to identify the cells that can send water to both the left and top edges (Pacific) and the right and bottom edges (Atlantic) of the island.\n",
-    "\n",
-    "The problem is solved by performing a depth-first search (DFS) starting from the ocean edges and marking the cells that can flow to each ocean. Then, the algorithm identifies cells that are reachable from both oceans and returns their coordinates as the result.\n",
-    "\n",
-    "The problem involves both traversal of the island and backtracking to find the solution, which makes it a classic graph traversal problem.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:\n",
-    "        # Check if the input matrix is empty\n",
-    "        if not heights:\n",
-    "            return []\n",
-    "        \n",
-    "        m, n = len(heights), len(heights[0])\n",
-    "        \n",
-    "        # Initialize boolean matrices to keep track of cells reachable from each ocean\n",
-    "        pacific_reachable = [[False] * n for _ in range(m)]\n",
-    "        atlantic_reachable = [[False] * n for _ in range(m)]\n",
-    "        \n",
-    "        # Depth-First Search (DFS) function to mark cells that can be reached from an ocean\n",
-    "        def dfs(x, y, reachable):\n",
-    "            if reachable[x][y]:\n",
-    "                return\n",
-    "            reachable[x][y] = True\n",
-    "            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:\n",
-    "                new_x, new_y = x + dx, y + dy\n",
-    "                if 0 <= new_x < m and 0 <= new_y < n and heights[new_x][new_y] >= heights[x][y]:\n",
-    "                    dfs(new_x, new_y, reachable)\n",
-    "        \n",
-    "        # Start DFS from the Pacific Ocean (left and top edges)\n",
-    "        for i in range(m):\n",
-    "            dfs(i, 0, pacific_reachable)\n",
-    "            dfs(i, n - 1, atlantic_reachable)\n",
-    "        \n",
-    "        for j in range(n):\n",
-    "            dfs(0, j, pacific_reachable)\n",
-    "            dfs(m - 1, j, atlantic_reachable)\n",
-    "        \n",
-    "        result = []\n",
-    "        \n",
-    "        # Find cells that can flow to both the Pacific and Atlantic oceans\n",
-    "        for i in range(m):\n",
-    "            for j in range(n):\n",
-    "                if pacific_reachable[i][j] and atlantic_reachable[i][j]:\n",
-    "                    result.append([i, j])\n",
-    "        \n",
-    "        return result"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a class called `Solution` with a method named `pacificAtlantic`. This method takes a 2D matrix `heights`, where each element represents the height above sea level of a cell on a rectangular island. The goal of the method is to find and return a list of grid coordinates (cells) from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-    "\n",
-    "The code uses a Depth-First Search (DFS) algorithm to traverse the island, starting from the ocean edges, and marks cells that are reachable from either the Pacific or Atlantic Ocean. It does this by creating two boolean matrices, `pacific_reachable` and `atlantic_reachable`, to keep track of cells that can be reached from each ocean.\n",
-    "\n",
-    "The code iterates through the entire island, initiating DFS searches from the edges of the island. If a cell is reachable from an ocean (either Pacific or Atlantic), it is marked as reachable in the corresponding boolean matrix. The DFS function ensures that the traversal follows the rule that water can flow to a neighboring cell if the neighboring cell's height is less than or equal to the current cell's height.\n",
-    "\n",
-    "After marking all reachable cells from both oceans, the code then looks for cells that are reachable from both the Pacific and Atlantic Oceans. If a cell satisfies this condition, it is added to the `result` list.\n",
-    "\n",
-    "Finally, the code returns the `result` list, which contains the grid coordinates of cells from which rainwater can flow into both the Pacific and Atlantic Oceans.\n",
-    "\n",
-    "The code provides a solution to the problem of finding cells on the island where water can reach both oceans, demonstrating a traversal algorithm that explores the connectivity of cells on the island."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "heights1 = [\n",
-    "    [1,2,2,3,5],\n",
-    "    [3,2,3,4,4],\n",
-    "    [2,4,5,3,1],\n",
-    "    [6,7,1,4,5],\n",
-    "    [5,1,1,2,4]\n",
-    "]\n",
-    "\n",
-    "solution = Solution()\n",
-    "print(solution.pacificAtlantic(heights1)) # Expected Output: [[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[[0, 0]]\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "heights2 = [[1]]\n",
-    "print(solution.pacificAtlantic(heights2)) # Expected Output: [[0, 0]]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Time Complexity:\n",
-    "The time complexity of this solution is O(m * n), where 'm' is the number of rows in the input matrix and 'n' is the number of columns. Here's a breakdown of the time complexity:\n",
-    "\n",
-    "1. Creating the `pacific_reachable` and `atlantic_reachable` matrices with dimensions m x n takes O(m * n) time.\n",
-    "\n",
-    "2. Initiating Depth-First Search (DFS) from the edges of the island takes O(m * n) time because it processes each cell once.\n",
-    "\n",
-    "3. The DFS function visits each cell at most once, and its time complexity is O(1) for each cell. The number of cells visited by the DFS in the worst case is m * n.\n",
-    "\n",
-    "4. The final step of identifying and appending cells to the result list takes O(m * n) time in the worst case.\n",
-    "\n",
-    "Overall, the time complexity is dominated by the DFS and is O(m * n).\n",
-    "\n",
-    "Space Complexity:\n",
-    "The space complexity of this solution is also O(m * n). Here's how the space is used:\n",
-    "\n",
-    "1. Two boolean matrices, `pacific_reachable` and `atlantic_reachable`, are created with dimensions m x n. Each matrix takes O(m * n) space, resulting in O(2 * m * n) space usage.\n",
-    "\n",
-    "2. The depth-first search (DFS) stack space during the recursive calls can go as deep as the diagonal of the grid, which is at most min(m, n). This additional space for the call stack is relatively small compared to the boolean matrices and can be considered O(min(m, n)).\n",
-    "\n",
-    "3. The `result` list stores the grid coordinates, and its size is determined by the number of cells that can flow to both oceans. In the worst case, this list can contain all m * n cells, so it takes O(m * n) space.\n",
-    "\n",
-    "The overall space complexity is the sum of these components, which is O(m * n) + O(min(m, n)) + O(m * n). In big O notation, we can simplify this to O(m * n)."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Modify the algorithm to find the maximum flow of water from any cell to both the Pacific and Atlantic Oceans. In other words, find the cell(s) that can send the most water to both oceans.\n",
-    "\n",
-    "2. **Efficiency Challenge:** Optimize the time and space complexity of the algorithm while maintaining correctness. Aim to reduce both time and space complexity as much as possible."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb b/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb
deleted file mode 100644
index 19db846..0000000
--- a/_build/jupyter_execute/12. Advanced Graphs/892. Alien Dictionary.ipynb	
+++ /dev/null
@@ -1,250 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {
-    "jp-MarkdownHeadingCollapsed": true
-   },
-   "source": [
-    "# 892. Alien Dictionary\n",
-    "\n",
-    "**Difficulty:** Hard\n",
-    "\n",
-    "**Link to Problem:** [To see the Alien Dictionary problem on LintCode, click here!](https://www.lintcode.com/problem/892/)\n",
-    "\n",
-    "---\n",
-    "\n",
-    "There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of **non-empty** words from the dictionary, where words are **sorted lexicographically by the rules of this new language**. Derive the order of letters in this language.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- You may assume all letters are in lowercase.\n",
-    "- The dictionary is invalid, if string a is prefix of string b and b is appear before a.\n",
-    "- If the order is invalid, return an empty string.\n",
-    "- There may be multiple valid order of letters, return the smallest in normal lexicographical order.\n",
-    "- The letters in **one** string are of the same rank by default and are sorted in Human dictionary order."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem at hand involves determining the order of letters in a new alien language that uses the Latin alphabet. However, the order of letters in this alien language is unknown. To solve this problem, we are given a list of non-empty words from the dictionary, where the words are sorted lexicographically based on the rules of the alien language. The goal is to derive the correct order of letters in this new language.\n",
-    "\n",
-    "Here's a more detailed explanation of the problem:\n",
-    "\n",
-    "1. The alphabet: The new alien language uses the Latin alphabet, which consists of lowercase letters.\n",
-    "\n",
-    "2. Dictionary: We are provided with a list of non-empty words. These words are sorted in lexicographical order based on the rules of the alien language. The order of letters in the words reflects the correct order in the new language.\n",
-    "\n",
-    "3. Invalid order: The problem specifies that the dictionary is considered invalid if there exists a situation where one string 'a' is a prefix of another string 'b,' and 'b' appears before 'a' in the dictionary. This condition ensures that there are no inconsistencies in the order of letters.\n",
-    "\n",
-    "4. Multiple valid orders: It's possible that there are multiple valid orders of letters that satisfy the provided dictionary. However, we are instructed to return the smallest valid order in normal lexicographical order.\n",
-    "\n",
-    "To solve this problem, you need to analyze the given list of words and determine the correct order of letters in the alien language while adhering to the specified constraints. You should return the order as a string in lexicographical order if it's valid, and an empty string if the order is invalid.\n",
-    "\n",
-    "The key to solving this problem is to construct a directed graph of letter relationships based on the order information provided by the words in the dictionary. Then, perform a topological sort on this graph to derive the order of letters, ensuring that the order is consistent and valid according to the given constraints.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from collections import defaultdict, deque\n",
-    "\n",
-    "def alienOrder(words):\n",
-    "    # Create an adjacency list to represent the graph\n",
-    "    graph = defaultdict(list)\n",
-    "    \n",
-    "    # Create a dictionary to store the in-degrees of each letter\n",
-    "    in_degree = {}\n",
-    "    \n",
-    "    # Initialize in-degrees to 0 for all letters\n",
-    "    for word in words:\n",
-    "        for char in word:\n",
-    "            if char not in in_degree:\n",
-    "                in_degree[char] = 0\n",
-    "    \n",
-    "    # Build the graph and update in-degrees\n",
-    "    for i in range(1, len(words)):\n",
-    "        word1, word2 = words[i-1], words[i]\n",
-    "        min_length = min(len(word1), len(word2))\n",
-    "        \n",
-    "        # Compare characters in the two words\n",
-    "        j = 0\n",
-    "        while j < min_length and word1[j] == word2[j]:\n",
-    "            j += 1\n",
-    "        \n",
-    "        if j < min_length:\n",
-    "            # If word1[j] is lexicographically before word2[j], add an edge\n",
-    "            graph[word1[j]].append(word2[j])\n",
-    "            in_degree[word2[j]] += 1\n",
-    "            \n",
-    "            # Break the loop to avoid further comparisons\n",
-    "            break\n",
-    "    \n",
-    "    # Perform topological sorting using Kahn's algorithm\n",
-    "    result = []\n",
-    "    queue = deque([char for char in in_degree if in_degree[char] == 0])\n",
-    "    \n",
-    "    while queue:\n",
-    "        char = queue.popleft()\n",
-    "        result.append(char)\n",
-    "        \n",
-    "        for neighbor in graph[char]:\n",
-    "            in_degree[neighbor] -= 1\n",
-    "            if in_degree[neighbor] == 0:\n",
-    "                queue.append(neighbor)\n",
-    "    \n",
-    "    # Check if there is a valid order\n",
-    "    if len(result) < len(in_degree):\n",
-    "        return \"\"\n",
-    "    \n",
-    "    return \"\".join(result)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code is designed to determine the order of letters in an unknown alien language that uses the Latin alphabet. It takes a list of non-empty words from a dictionary as input, where the words are sorted lexicographically based on the rules of this alien language.\n",
-    "\n",
-    "The code follows these main steps:\n",
-    "\n",
-    "1. Create a data structure to represent the graph of letters and their relationships in the alien language. It uses a defaultdict to store a list of letters that come after each letter. Additionally, it uses a dictionary called `in_degree` to keep track of the in-degrees (the number of letters that precede a letter) for each letter.\n",
-    "\n",
-    "2. Initialize the in-degrees of all letters to 0. This is done by iterating through all the words and characters in the words, adding each character to the `in_degree` dictionary with an initial in-degree of 0.\n",
-    "\n",
-    "3. Build the graph by comparing adjacent words in the dictionary. For each pair of adjacent words, the code finds the first differing character between the two words. If such a character exists, it means there's an order relationship between the letters represented by these characters. The code updates the graph and in-degrees accordingly.\n",
-    "\n",
-    "4. After constructing the graph, the code performs a topological sorting of the letters using Kahn's algorithm. It starts by initializing an empty result list and a queue containing letters with in-degrees of 0. The algorithm repeatedly removes a letter with an in-degree of 0 from the queue, adds it to the result list, and updates the in-degrees of its neighbors. This process continues until the queue is empty.\n",
-    "\n",
-    "5. Finally, the code checks if the topological sorting was successful by comparing the length of the result list with the number of unique letters in the input. If the lengths differ, it means there's a cycle in the graph, and the order is invalid. In such a case, an empty string is returned. Otherwise, the result list is joined together to form the smallest order of letters in normal lexicographical order, and it is returned as the output.\n",
-    "\n",
-    "The code provides a way to determine the order of letters in the alien language while handling various edge cases, including checking for invalid order conditions."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "355f5bfd-94e1-4c5b-9914-cc20775f4160",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "wrtef\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1\n",
-    "words1 = [\"wrt\", \"wrf\", \"er\", \"ett\", \"rftt\"]\n",
-    "print(alienOrder(words1))  # Output: \"wertf\""
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "zx\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2\n",
-    "words2 = [\"z\", \"x\"]\n",
-    "print(alienOrder(words2))  # Output: \"zx\""
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code for determining the order of letters in an alien language.\n",
-    "\n",
-    "Time Complexity:\n",
-    "1. Building the graph: The code iterates through the list of words and compares adjacent words. In the worst case, where all words have the same prefix, it takes O(N) time to build the graph, where N is the total number of characters in all the words.\n",
-    "2. Topological sorting: Performing a topological sort on the graph takes O(V + E) time, where V is the number of unique letters (vertices) and E is the number of relationships between letters (edges) in the graph.\n",
-    "\n",
-    "Overall, the time complexity is O(N + V + E), where N is the total number of characters in the words, V is the number of unique letters, and E is the number of order relationships between letters.\n",
-    "\n",
-    "Space Complexity:\n",
-    "1. Graph and In-degrees: The code uses data structures to represent the graph and in-degrees. The space complexity for these data structures is O(V + E), where V is the number of unique letters, and E is the number of order relationships between letters in the words.\n",
-    "2. Result List: The result list stores the order of letters, which can have a maximum size of V, where V is the number of unique letters.\n",
-    "3. Queue: The space used by the queue during topological sorting is also O(V).\n",
-    "\n",
-    "Overall, the space complexity is O(V + E) due to the data structures used for graph representation and the space needed for storing the result and the queue.\n",
-    "\n",
-    "In the worst case, where there are many unique letters and many order relationships, the space complexity is dominated by the number of unique letters and their relationships.\n",
-    "\n",
-    "In summary, the time complexity of the code is O(N + V + E), and the space complexity is O(V + E). The actual performance will depend on the specific input data, such as the number of unique letters and the relationships between them in the alien language."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Multiple Valid Orders**: Extend the problem to handle cases where there are multiple valid orders of letters in the alien language. Modify the code to return all valid orders instead of just one. Be mindful of performance.\n",
-    "\n",
-    "2. **Detect Invalid Dictionary**: Given a dictionary of words, write a function to detect whether the dictionary is valid or not based on the constraints mentioned in the problem statement."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb b/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb
deleted file mode 100644
index 9539363..0000000
--- a/_build/jupyter_execute/13. One-D Dynamic Programming/198. House Robber.ipynb	
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# House Robber\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "---\n",
-    "You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and **it will automatically contact the police if two adjacent houses were broken into on the same night.**\n",
-    "\n",
-    "Given an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight **without alerting the police**.*\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= nums.length <= 100`\n",
-    "- `0 <= nums[i] <= 400`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The problem is a classic dynamic programming challenge known as the \"House Robber\" problem. The scenario is that you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, but the constraint is that adjacent houses have security systems connected. If two adjacent houses are broken into on the same night, the security systems will automatically contact the police.\n",
-    "\n",
-    "The goal is to determine the maximum amount of money you can rob tonight without alerting the police. You need to find a strategy to maximize your earnings while avoiding robbing two adjacent houses.\n",
-    "\n",
-    "The input to the problem is an array `nums`, where `nums[i]` represents the amount of money in the `i`-th house. The task is to return the maximum amount of money that can be robbed without triggering the security system.\n",
-    "\n",
-    "For example:\n",
-    "```\n",
-    "Input: nums = [1,2,3,1]\n",
-    "Output: 4\n",
-    "Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).\n",
-    "Total amount you can rob = 1 + 3 = 4.\n",
-    "\n",
-    "Input: nums = [2,7,9,3,1]\n",
-    "Output: 12\n",
-    "Explanation: Rob house 1 (money = 2), rob house 3 (money = 9), and rob house 5 (money = 1).\n",
-    "Total amount you can rob = 2 + 9 + 1 = 12.\n",
-    "```\n",
-    "\n",
-    "The challenge involves designing an algorithm to efficiently solve this problem and find the optimal strategy for robbing houses to maximize the stolen money while adhering to the constraint of not robbing adjacent houses. The provided solution utilizes dynamic programming to achieve this.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def rob(self, nums: List[int]) -> int:\n",
-    "        # Initialize variables to keep track of the maximum amount of money robbed at the previous and current houses\n",
-    "        prev, curr = 0, 0\n",
-    "\n",
-    "        # Iterate through the houses\n",
-    "        for n in nums:\n",
-    "            # Calculate the maximum amount of money that can be robbed at the current house\n",
-    "            # It is the maximum of either robbing the current house and the money robbed at the house before the previous one (prev),\n",
-    "            # or skipping the current house and taking the maximum amount from the previous house (curr)\n",
-    "            temp = max(n + prev, curr)\n",
-    "\n",
-    "            # Update the variables for the next iteration\n",
-    "            prev, curr = curr, temp\n",
-    "\n",
-    "        # The final result is the maximum amount of money that can be robbed at the last house\n",
-    "        return curr"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code is a Python implementation of the \"House Robber\" problem using a dynamic programming approach. It's part of a class `Solution` and includes a method `rob` that takes a list of integers `nums` as input and returns the maximum amount of money that can be robbed without alerting the police.\n",
-    "\n",
-    "The key idea of the algorithm is to maintain two variables, `prev` and `curr`, which represent the maximum amount of money that can be robbed up to the previous house and the current house, respectively. The algorithm iterates through the houses, updating these variables based on the maximum amount that can be robbed at the current house, considering the constraint of not robbing adjacent houses.\n",
-    "\n",
-    "The loop calculates a temporary variable `temp` that represents the maximum amount of money that can be robbed at the current house. It considers two scenarios: either robbing the current house and adding the money robbed at the house before the previous one (`n + prev`), or skipping the current house and taking the maximum amount from the previous house (`curr`). The maximum of these two scenarios is assigned to `temp`.\n",
-    "\n",
-    "The `prev` and `curr` variables are then updated for the next iteration. `prev` is set to the previous value of `curr`, and `curr` is set to the calculated `temp`.\n",
-    "\n",
-    "The final result is the value stored in the `curr` variable, representing the maximum amount of money that can be robbed at the last house. This value is returned as the result of the `rob` method.\n",
-    "\n",
-    "The provided test cases at the end demonstrate the usage of the `rob` method with different input arrays."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "sol = Solution()\n",
-    "\n",
-    "nums1 = [1, 2, 3, 1]\n",
-    "print(sol.rob(nums1))  # Output: 4"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "12\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [2, 7, 9, 3, 1]\n",
-    "print(sol.rob(nums2))  # Output: 12"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code.\n",
-    "\n",
-    "### Time Complexity:\n",
-    "\n",
-    "The time complexity of the code is O(n), where n is the number of houses. The algorithm iterates through the array once, and in each iteration, it performs constant time operations. The loop runs for each house, and the number of operations inside the loop is independent of the input size. Therefore, the overall time complexity is linear, O(n).\n",
-    "\n",
-    "### Space Complexity:\n",
-    "\n",
-    "The space complexity is O(1), constant space. The algorithm uses only a constant amount of extra space regardless of the input size. The variables `prev`, `curr`, and `temp` are the only additional variables used, and they do not depend on the size of the input array. Thus, the space complexity is constant, O(1).\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: O(n)\n",
-    "- Space Complexity: O(1)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Variation with Maximum Number of Houses to Rob:** Modify the problem to include an additional constraint: you are allowed to rob at most 'k' houses without triggering the alarm. Find the maximum amount you can rob under this new constraint.\n",
-    "\n",
-    "2. **Robbing in a Circular Street:** Extend the problem to a circular street where the first and last houses are adjacent. You cannot rob both the first and last houses simultaneously. Find the maximum amount you can rob in this circular setting.\n",
-    "\n",
-    "\n",
-    "**Link to Problem:** [To see the House Robber problem on LeetCode, click here!](https://leetcode.com/problems/house-robber/description/)\n"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb b/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb
deleted file mode 100644
index d210017..0000000
--- a/_build/jupyter_execute/13. One-D Dynamic Programming/213. House Robber II.ipynb	
+++ /dev/null
@@ -1,243 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# House Robber II\n",
-    "**Difficulty:** Medium\n",
-    "\n",
-    "---\n",
-    "You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle**. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.\n",
-    "\n",
-    "Given an integer array `nums` representing the amount of money of each house, return *the maximum amount of money you can rob tonight without alerting the police*.\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= nums.length <= 100`\n",
-    "- `0 <= nums[i] <= 1000`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "Certainly! The problem statement describes a scenario where you are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. However, there is a security system in place: adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.\n",
-    "\n",
-    "The twist in this problem is that the houses are arranged in a circle. That means the first house is the neighbor of the last one. So, robbing the first house has implications for the last house, and vice versa.\n",
-    "\n",
-    "The goal is to find the maximum amount of money you can rob tonight without alerting the police. You need to implement a function that takes an integer array `nums` representing the amount of money in each house and returns the maximum amount of money that can be robbed without triggering the security system.\n",
-    "\n",
-    "Here are a few examples to illustrate the problem:\n",
-    "\n",
-    "1. **Example 1:**\n",
-    "   ```python\n",
-    "   Input: nums = [2, 3, 2]\n",
-    "   Output: 3\n",
-    "   Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2) because they are adjacent houses. The optimal strategy is to rob house 2 (money = 3).\n",
-    "   ```\n",
-    "\n",
-    "2. **Example 2:**\n",
-    "   ```python\n",
-    "   Input: nums = [1, 2, 3, 1]\n",
-    "   Output: 4\n",
-    "   Explanation: Robbing house 1 (money = 1) and then robbing house 3 (money = 3) results in a total amount of 4. Robbing house 2 is skipped to avoid alerting the police.\n",
-    "   ```\n",
-    "\n",
-    "3. **Example 3:**\n",
-    "   ```python\n",
-    "   Input: nums = [1, 2, 3]\n",
-    "   Output: 3\n",
-    "   Explanation: Since the houses are in a circle, you can choose either house 1, 2, or 3. The optimal strategy is to rob house 3 for a total amount of 3.\n",
-    "   ```\n",
-    "\n",
-    "The constraints for the problem are that the length of the `nums` array is between 1 and 100, and the amount of money in each house (`nums[i]`) is between 0 and 1000.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from typing import List\n",
-    "\n",
-    "class Solution:\n",
-    "    def rob(self, nums: List[int]) -> int:\n",
-    "        def simple_rob(nums):\n",
-    "            # Helper function for basic house robbery problem\n",
-    "            prev, curr = 0, 0\n",
-    "            for num in nums:\n",
-    "                # Calculate the maximum amount of money that can be robbed\n",
-    "                # without alerting the police for each house\n",
-    "                prev, curr = curr, max(prev + num, curr)\n",
-    "            return curr\n",
-    "\n",
-    "        # Check the base cases where the length of nums is 1 or 2\n",
-    "        if len(nums) == 1:\n",
-    "            return nums[0]\n",
-    "        elif len(nums) == 2:\n",
-    "            return max(nums[0], nums[1])\n",
-    "\n",
-    "        # Case 1: First house is robbed, last house is not\n",
-    "        case1 = simple_rob(nums[:-1])\n",
-    "\n",
-    "        # Case 2: First house is not robbed, last house is\n",
-    "        case2 = simple_rob(nums[1:])\n",
-    "\n",
-    "        # Maximum of the two cases\n",
-    "        return max(case1, case2)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The provided code defines a class `Solution` with a method `rob`, which aims to determine the maximum amount of money a professional robber can steal from houses arranged in a circle without alerting the police. Each house contains a certain amount of money, and adjacent houses have a security system that triggers if both are robbed on the same night.\n",
-    "\n",
-    "The code uses dynamic programming to solve the problem efficiently. It contains a helper function `simple_rob` that computes the maximum amount of money that can be robbed from a linear arrangement of houses. The main method `rob` handles the circular arrangement by considering two cases: robbing the first house and not robbing the last, and not robbing the first house and robbing the last. The result is the maximum of these two cases.\n",
-    "\n",
-    "The class is instantiated as `solution`, and the `rob` method is applied to three different scenarios (test cases) to demonstrate its functionality. The final output represents the maximum amount of money the robber can steal without triggering the security system."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "\n",
-    "# Create an instance of the Solution class\n",
-    "solution = Solution()\n",
-    "\n",
-    "nums1 = [2, 3, 2]\n",
-    "print(solution.rob(nums1))  # Output: 3"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "4\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums2 = [1, 2, 3, 1]\n",
-    "print(solution.rob(nums2))  # Output: 4"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "id": "e6e1554f-ae77-410c-8af2-e59e5428166c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "nums3 = [1, 2, 3]\n",
-    "print(solution.rob(nums3))  # Output: 3"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Let's analyze the time and space complexity of the provided code:\n",
-    "\n",
-    "### Time Complexity:\n",
-    "The time complexity is primarily determined by the `simple_rob` helper function, which has a time complexity of O(N), where N is the number of houses in the input array. The main `rob` function calls `simple_rob` twice, once for the case of robbing the first house and not robbing the last, and once for the case of not robbing the first house and robbing the last. Therefore, the overall time complexity is O(N).\n",
-    "\n",
-    "### Space Complexity:\n",
-    "The space complexity is O(1) because the space used by the algorithm is constant, regardless of the size of the input array. The only variables used are `prev` and `curr` in the `simple_rob` function, which represent the previous and current maximum amounts of money that can be robbed without alerting the police. These variables are updated iteratively, and no additional data structures are used that scale with the input size. Therefore, the space complexity is constant, or O(1).\n",
-    "\n",
-    "In summary:\n",
-    "- Time Complexity: O(N)\n",
-    "- Space Complexity: O(1)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Optimization Challenge:** Optimize the space complexity of the solution. Can you solve the problem with O(1) space complexity without sacrificing the time complexity?\n",
-    "\n",
-    "2. **Circular Robbery Variations:** Consider variations of the circular house robbery problem, such as allowing the robber to skip a certain number of houses or restricting the number of consecutive houses that can be robbed. Modify the solution to accommodate these variations.\n",
-    "\n",
-    "**Link to Problem:** [To see the House Robber II problem on LeetCode, click here!](https://leetcode.com/problems/house-robber-ii/)"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
diff --git a/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb b/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb
deleted file mode 100644
index 32c5da3..0000000
--- a/_build/jupyter_execute/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb	
+++ /dev/null
@@ -1,204 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "id": "d6161252-e815-48d5-a2a3-5444c12e3b49",
-   "metadata": {},
-   "source": [
-    "# 70. Climbing Stairs\n",
-    "\n",
-    "**Difficulty:** Easy\n",
-    "\n",
-    "**Link to Problem:** [To see the Climbing Stairs problem on LeetCode, click here!](https://leetcode.com/problems/climbing-stairs/)\n",
-    "\n",
-    "---\n",
-    "You are climbing a staircase. It takes `n` steps to reach the top.\n",
-    "\n",
-    "Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?\n",
-    "\n",
-    "**Constraints:**\n",
-    "- `1 <= n <= 45`"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "fe9e71ae-2099-4eb9-b7a1-ec795ce92c76",
-   "metadata": {},
-   "source": [
-    "## Probelm Explanation:\n",
-    "\n",
-    "The \"Climbing Stairs\" problem is a classic algorithmic problem that asks you to find the number of distinct ways to climb a staircase with a given number of steps. Here's the problem statement:\n",
-    "\n",
-    "You are given a staircase with 'n' steps. Each time you can either climb 1 step or 2 steps. You need to find out how many distinct ways there are to reach the top of the staircase.\n",
-    "\n",
-    "For example, if 'n' is 2, there are two ways to climb to the top:\n",
-    "\n",
-    "1. Climbing two steps at once.\n",
-    "2. Climbing one step, and then another step.\n",
-    "\n",
-    "If 'n' is 3, there are three ways:\n",
-    "\n",
-    "1. Climbing one step, then one step, and finally one step.\n",
-    "2. Climbing one step, then two steps.\n",
-    "3. Climbing two steps, then one step.\n",
-    "\n",
-    "The problem essentially asks you to find a solution that counts the possible combinations of 1-step and 2-step climbs that can be taken to reach the top of the staircase. It's a classic example of a dynamic programming problem where you can break it down into smaller subproblems and build a solution incrementally, as shown in the code I provided earlier.\n",
-    "\n",
-    "## Solution:\n",
-    "Here's a Python function to implement this algorithm:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "id": "aabf7095-07dc-4ab7-876c-ec7213ed0de5",
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "class Solution:\n",
-    "    def climbStairs(self, n: int) -> int:\n",
-    "        # If there are 1 or 2 steps, there are exactly n ways to reach the top.\n",
-    "        if n <= 2:\n",
-    "            return n\n",
-    "\n",
-    "        # Create an array to store the number of ways to reach each step.\n",
-    "        dp = [0] * (n + 1)\n",
-    "\n",
-    "        # There is one way to reach the first step (base case).\n",
-    "        dp[1] = 1\n",
-    "\n",
-    "        # There are two ways to reach the second step (base case).\n",
-    "        dp[2] = 2\n",
-    "\n",
-    "        # Calculate the number of ways to reach each step from the previous steps.\n",
-    "        for i in range(3, n + 1):\n",
-    "            # The number of ways to reach step 'i' is the sum of the ways to reach\n",
-    "            # the previous two steps, as you can take 1 or 2 steps at a time.\n",
-    "            dp[i] = dp[i - 1] + dp[i - 2]\n",
-    "\n",
-    "        # The result is the number of ways to reach the top, which is stored in dp[n].\n",
-    "        return dp[n]"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "ece39324-cd9e-47d7-8677-a001878494d9",
-   "metadata": {},
-   "source": [
-    "## Explanation:\n",
-    "\n",
-    "The code defines a class `Solution` with a method `climbStairs` that takes an integer `n` as input and returns the number of distinct ways to climb a staircase with `n` steps.\n",
-    "\n",
-    "1. If `n` is 1 or 2, the function returns `n` directly because there are only 1 or 2 distinct ways to reach the top in those cases.\n",
-    "\n",
-    "2. For `n` greater than 2, the code uses dynamic programming to calculate the number of ways to reach each step. It creates an array `dp` of length `n + 1` to store these values.\n",
-    "\n",
-    "3. The `dp[1]` and `dp[2]` are initialized to 1 and 2, respectively. This is because there is only one way to reach the first step (by taking 1 step), and there are two ways to reach the second step (by taking 2 steps or two 1-step climbs).\n",
-    "\n",
-    "4. The code then uses a loop to fill in the `dp` array for steps 3 and above. For each step `i` from 3 to `n`, it calculates the number of ways to reach that step by adding the number of ways to reach the previous two steps, as you can either take 1 step or 2 steps at a time.\n",
-    "\n",
-    "5. Finally, the function returns the value stored in `dp[n]`, which represents the total number of distinct ways to climb the staircase with `n` steps."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "a54ea5b7-c53f-46c0-a44f-ef769ac68244",
-   "metadata": {},
-   "source": [
-    "## Test cases:\n",
-    "\n",
-    "Here's how you can use this solution:"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "id": "f12c5a63-b811-4be7-83c2-34ee396b546c",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 1 - Input: n = 2, Output: 2\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 1:\n",
-    "n = 2\n",
-    "solution = Solution()\n",
-    "print(f\"Example 1 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 2"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "id": "364aa061-a452-45eb-96db-36f7dbdbdcb3",
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example 2 - Input: n = 3, Output: 3\n"
-     ]
-    }
-   ],
-   "source": [
-    "# Example 2:\n",
-    "n = 3\n",
-    "print(f\"Example 2 - Input: n = {n}, Output: {solution.climbStairs(n)}\")  # Output should be 3\n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "7d9dd448-c5f5-4192-8aa5-7c70ec7f42e2",
-   "metadata": {},
-   "source": [
-    "## Time and Space Complexity Analysis\n",
-    "\n",
-    "Time Complexity:\n",
-    "The time complexity of the solution is O(n) because we use a single for loop that iterates from 3 to n to fill the `dp` array. In each iteration, we perform constant time operations (addition and assignment). Therefore, the time complexity is linear with respect to the input value `n`.\n",
-    "\n",
-    "Space Complexity:\n",
-    "The space complexity of the solution is O(n) as well. We use an array `dp` of length `n + 1` to store the number of ways to reach each step. The size of this array is directly proportional to the input value `n`, which makes the space complexity linear in terms of `n`. The rest of the variables used in the function (e.g., `i`) occupy constant space and do not contribute significantly to the overall space complexity.\n",
-    "\n",
-    "In summary, the time complexity is O(n), and the space complexity is O(n) for this dynamic programming solution to the \"Climbing Stairs\" problem."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "037b210a-555b-4da3-8130-b74d1e4f33f6",
-   "metadata": {},
-   "source": [
-    "## Challenging Exercises:\n",
-    "\n",
-    "1. **Generalized Step Sizes**: Instead of being limited to 1 step or 2 steps, consider a scenario where you can take steps of sizes `[1, 2, 3]`. Implement a function that calculates the number of ways to climb the staircase with these different step sizes.\n",
-    "\n",
-    "2. **Minimize Steps**: Given an integer `n`, find the minimum number of steps required to reach the top of the staircase. Return both the minimum number of steps and the specific sequence of steps taken."
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python (Pyodide)",
-   "language": "python",
-   "name": "python"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 5
-}
\ No newline at end of file
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@@ -1,914 +0,0 @@
-{
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-      "Requirement already satisfied: zipp>=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from importlib_metadata->myst-nb~=0.17.1->jupyter-book) (3.17.0)\r\n"
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-     ]
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-      "Requirement already satisfied: backcall in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.2.0)\r\n",
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-     ]
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-     ]
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-      "Requirement already satisfied: platformdirs>=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-core!=5.0.*,>=4.12->nbclient->myst-nb~=0.17.1->jupyter-book) (3.11.0)\r\n"
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-      "Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py<3.0.0,>=1.0.0->myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.1.2)\r\n",
-      "Requirement already satisfied: ptyprocess>=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect>4.3->ipython->myst-nb~=0.17.1->jupyter-book) (0.7.0)\r\n",
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-     ]
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-      "Requirement already satisfied: greenlet!=0.4.17 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sqlalchemy<3,>=1.3.12->jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (3.0.1)\r\n"
-     ]
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-     "text": [
-      "Requirement already satisfied: soupsieve>1.2 in /home/codespace/.local/lib/python3.10/site-packages (from beautifulsoup4->pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (2.5)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Requirement already satisfied: executing>=1.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.0.0)\r\n",
-      "Requirement already satisfied: asttokens>=2.1.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.4.0)\r\n",
-      "Requirement already satisfied: pure-eval in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.2)\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Note: you may need to restart the kernel to use updated packages.\n"
-     ]
-    }
-   ],
-   "source": [
-    "!python -m pip install --upgrade pip\n",
-    "%pip install jupyter-book"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions\n"
-     ]
-    }
-   ],
-   "source": [
-    "%cd /workspaces/LeetCode-Solutions"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Traceback (most recent call last):\r\n",
-      "  File \"/home/codespace/.python/current/bin/jupyter-book\", line 8, in <module>\r\n",
-      "    sys.exit(main())\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1157, in __call__\r\n",
-      "    return self.main(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1078, in main\r\n",
-      "    rv = self.invoke(ctx)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
-      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
-      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1434, in invoke\r\n",
-      "    return ctx.invoke(self.callback, **ctx.params)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 783, in invoke\r\n",
-      "    return __callback(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx_external_toc/cli.py\", line 139, in migrate_toc\r\n",
-      "    toc = migrate_jupyter_book(Path(toc_file))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx_external_toc/tools.py\", line 327, in migrate_jupyter_book\r\n",
-      "    raise MalformedError(\"no top-level 'file' key found\")\r\n",
-      "sphinx_external_toc.parsing.MalformedError: no top-level 'file' key found\r\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book toc migrate /workspaces/LeetCode-Solutions/_toc.yml -o /workspaces/LeetCode-Solutions/_toc.yml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions\r\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\r\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\r\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "[etoc] Changing master_doc to 'README'\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\r\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\r\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\r\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 77 source files that are out of date\r\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 77 added, 0 changed, 0 removed\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[  1%] \u001b[35m01. Array Hashing/1. Two Sum\u001b[39;49;00m                          \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.07 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  2%] \u001b[35m01. Array Hashing/128. Longest Consecutive Sequence\u001b[39;49;00m   \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.17 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[  3%] \u001b[35m01. Array Hashing/217. Contains Duplicate\u001b[39;49;00m             \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.19 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[  5%] \u001b[35m01. Array Hashing/238. Product of Array Except Self\u001b[39;49;00m   \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 1.26 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  6%] \u001b[35m01. Array Hashing/242. Valid Anagram\u001b[39;49;00m                  \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 1.18 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  7%] \u001b[35m01. Array Hashing/347. Top K Frequent Elements\u001b[39;49;00m        \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.13 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[  9%] \u001b[35m01. Array Hashing/49. Group Anagrams\u001b[39;49;00m                  \r",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 4.02 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 10%] \u001b[35m01. Array Hashing/659. Encode and Decode Strings\u001b[39;49;00m      \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 11%] \u001b[35m01. Array Hashing/README\u001b[39;49;00m                              \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 12%] \u001b[35m02. Two Pointers/11. Container With Most Water\u001b[39;49;00m        \r",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.26 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 14%] \u001b[35m02. Two Pointers/125. Valid Palindrome\u001b[39;49;00m                \r",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 1.07 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 15%] \u001b[35m02. Two Pointers/15. 3Sum\u001b[39;49;00m                             \r",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.00 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 16%] \u001b[35m02. Two Pointers/README\u001b[39;49;00m                               \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 18%] \u001b[35m03. Sliding Window/121. Best Time to Buy and Sell Stock\u001b[39;49;00m\r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.89 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 19%] \u001b[35m03. Sliding Window/3. Longest Substring Without Repeating Characters\u001b[39;49;00m\r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 1.16 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 20%] \u001b[35m03. Sliding Window/424. Longest Repeating Character Replacement\u001b[39;49;00m\r",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 22%] \u001b[35m03. Sliding Window/76. Minimum Window Substring\u001b[39;49;00m       \r",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 0.87 seconds [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 23%] \u001b[35m03. Sliding Window/README\u001b[39;49;00m                             \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 24%] \u001b[35m04. Stack/20. Valid Parentheses\u001b[39;49;00m                       \r",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 4.02 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 25%] \u001b[35m04. Stack/README\u001b[39;49;00m                                      \r",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 27%] \u001b[35m05. Binary Search/153. Find Minimum in Rotated Sorted Array\u001b[39;49;00m\r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 0.98 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 28%] \u001b[35m05. Binary Search/33. Search in Rotated Sorted Array\u001b[39;49;00m  \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.05 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 29%] \u001b[35m05. Binary Search/README\u001b[39;49;00m                              \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 31%] \u001b[35m06. Linked List/141. Linked List Cycle\u001b[39;49;00m                \r",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 0.96 seconds [mystnb]\r\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[ 32%] \u001b[35m06. Linked List/143. Reorder List\u001b[39;49;00m                     \r"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]\r\n"
-     ]
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "^C\r\n",
-      "\r\n",
-      "\u001b[91mException occurred:\u001b[39;49;00m\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 512, in _async_cleanup_kernel\r\n",
-      "    await ensure_async(self.km.cleanup_resources())\r\n",
-      "AttributeError: 'NoneType' object has no attribute 'cleanup_resources'\r\n",
-      "\u001b[91mThe full traceback has been saved in /tmp/sphinx-err-tl19fjln.log, if you want to report the issue to the developers.\u001b[39;49;00m\r\n",
-      "Please also report this if it was a user error, so that a better error message can be provided next time.\r\n",
-      "A bug report can be filed in the tracker at <https://github.com/sphinx-doc/sphinx/issues>. Thanks!\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py\", line 89, in wrapper\r\n",
-      "    self._ready.set_result(None)\r\n",
-      "asyncio.exceptions.InvalidStateError: invalid state\r\n",
-      "\r\n",
-      "During handling of the above exception, another exception occurred:\r\n",
-      "\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 505, in _async_cleanup_kernel\r\n",
-      "    await ensure_async(self.km.shutdown_kernel(now=now))\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py\", line 189, in ensure_async\r\n",
-      "    result = await obj\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py\", line 92, in wrapper\r\n",
-      "    self._ready.set_exception(e)\r\n",
-      "asyncio.exceptions.InvalidStateError: invalid state\r\n",
-      "\r\n",
-      "During handling of the above exception, another exception occurred:\r\n",
-      "\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/sphinx.py\", line 171, in build_sphinx\r\n",
-      "    app.build(force_all, filenames)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/application.py\", line 329, in build\r\n",
-      "    self.builder.build_update()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 288, in build_update\r\n",
-      "    self.build(to_build,\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 302, in build\r\n",
-      "    updated_docnames = set(self.read())\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 409, in read\r\n",
-      "    self._read_serial(docnames)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 430, in _read_serial\r\n",
-      "    self.read_doc(docname)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py\", line 483, in read_doc\r\n",
-      "    publisher.publish()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/core.py\", line 217, in publish\r\n",
-      "    self.document = self.reader.read(self.source, self.parser,\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/io.py\", line 103, in read\r\n",
-      "    self.parse()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/readers/__init__.py\", line 78, in parse\r\n",
-      "    self.parser.parse(self.input, document)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/sphinx_.py\", line 150, in parse\r\n",
-      "    with create_client(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/base.py\", line 83, in __enter__\r\n",
-      "    self.start_client()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/direct.py\", line 40, in start_client\r\n",
-      "    result = single_nb_execution(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_cache/executors/utils.py\", line 58, in single_nb_execution\r\n",
-      "    executenb(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 1305, in execute\r\n",
-      "    return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute()\r\n",
-      "  File \"/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py\", line 173, in wrapped\r\n",
-      "    return loop.run_until_complete(inner)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/asyncio/base_events.py\", line 649, in run_until_complete\r\n",
-      "    return future.result()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 689, in async_execute\r\n",
-      "    async with self.async_setup_kernel(**kwargs):\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/contextlib.py\", line 206, in __aexit__\r\n",
-      "    await anext(self.gen)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 656, in async_setup_kernel\r\n",
-      "    await self._async_cleanup_kernel()\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py\", line 512, in _async_cleanup_kernel\r\n",
-      "    await ensure_async(self.km.cleanup_resources())\r\n",
-      "AttributeError: 'NoneType' object has no attribute 'cleanup_resources'\r\n",
-      "\r\n",
-      "The above exception was the direct cause of the following exception:\r\n",
-      "\r\n",
-      "Traceback (most recent call last):\r\n",
-      "  File \"/home/codespace/.python/current/bin/jupyter-book\", line 8, in <module>\r\n",
-      "    sys.exit(main())\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1157, in __call__\r\n",
-      "    return self.main(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1078, in main\r\n",
-      "    rv = self.invoke(ctx)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1688, in invoke\r\n",
-      "    return _process_result(sub_ctx.command.invoke(sub_ctx))\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 1434, in invoke\r\n",
-      "    return ctx.invoke(self.callback, **ctx.params)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py\", line 783, in invoke\r\n",
-      "    return __callback(*args, **kwargs)\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py\", line 317, in build\r\n",
-      "    builder_specific_actions(\r\n",
-      "  File \"/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py\", line 525, in builder_specific_actions\r\n",
-      "    raise RuntimeError(_message_box(msg, color=\"red\", doprint=False)) from result\r\n",
-      "RuntimeError: \r\n",
-      "\u001b[91m===============================================================================\u001b[0m\r\n",
-      "\r\n",
-      "There was an error in building your book. Look above for the cause.\r\n",
-      "\r\n",
-      "\u001b[91m===============================================================================\u001b[0m\r\n",
-      "\r\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build ."
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book build /workspaces/LeetCode-Solutions/ --builder pdfhtml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/html": [
-       "<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>"
-      ],
-      "text/plain": [
-       "<IPython.core.display.HTML object>"
-      ]
-     },
-     "execution_count": 5,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "from IPython.display import HTML\n",
-    "HTML('<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>')\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Your book template can be found at\n",
-      "\n",
-      "    mybook/\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book create mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions/mybook\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_build/html\n",
-      "[sphinxcontrib-bibtex] Beware that docutils versions 0.18 and 0.19 (you are running 0.18.1) are known to generate invalid html for citations. If this issue affects you, please use docutils<0.18 (or >=0.20 once released) instead. For more details, see https://sourceforge.net/p/docutils/patches/195/\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "[etoc] Changing master_doc to 'intro'\n",
-      "checking bibtex cache... out of date\n",
-      "parsing bibtex file /workspaces/LeetCode-Solutions/mybook/references.bib... parsed 5 entries\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='force', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/mybook/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 14 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 14 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.35 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.97 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 0.83 seconds [mystnb]\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mintro\u001b[39;49;00m                                                 \n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mintro\u001b[39;49;00m                                                  \n",
-      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
-      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
-      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
-      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
-      "[etoc] missing index.html written as redirect to 'intro.html'\n",
-      "\u001b[01mbuild succeeded, 2 warnings.\u001b[39;49;00m\n",
-      "\n",
-      "The HTML pages are in mybook/_build/html.\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Finished generating HTML for book.\n",
-      "Your book's HTML pages are here:\n",
-      "    mybook/_build/html/\n",
-      "You can look at your book by opening this file in a browser:\n",
-      "    mybook/_build/html/index.html\n",
-      "Or paste this line directly into your browser bar:\n",
-      "    file:///workspaces/LeetCode-Solutions/mybook/_build/html/index.html            \n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book build mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 31,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "format: jb-book\n",
-      "root: intro\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book toc from-project /workspaces/LeetCode-Solutions/mybook/ -f jb-book"
-   ]
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "Python 3",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.8"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 2
-}
\ No newline at end of file
diff --git a/_config.yml b/_config.yml
deleted file mode 100644
index 3def3c7..0000000
--- a/_config.yml
+++ /dev/null
@@ -1,31 +0,0 @@
-# Jupyter Book configuration
-
-# Project information
-title: "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges" # Python Problem-Solving Pro: Conquering the Blind 75 LeetCode Challenges
-author: Mohsen Tabibian and Saeed Samadidana
-email: mohsentabibian@gmail.com
-description: Explore 75 essential problems frequently asked in tech interviews. Each solution includes detailed explanations and time/space complexity analysis. Boost your coding skills and interview preparedness!
-
-repository: 
-   url: https://github.com/mohsentabibian/LeetCode-Solutions
-   path: docs #LeetCode-Solutions/
-   branch: main
-
-# Execute notebooks when building (set to false if you don't want to run the code)
-execute:
-  execute_notebooks: auto
-
-# HTML settings
-html:
-  theme: jekyll-theme-Architect #jupyter
-  logo: logo.png
-  toc:
-    display: true
-  use_issues_button: true
-  use_repository_button: true
-
-#Build settings
-#repository_url: https://github.com/mohsentabibian/LeetCode-Solutions  # URL to the GitHub repository
-
-# Add GitHub buttons to the book
-# See https://jupyterbook.org/customize/config.html #add-a-link-to-your-repository
\ No newline at end of file
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diff --git a/_toc.yml b/_toc.yml
deleted file mode 100644
index 0e4e1db..0000000
--- a/_toc.yml
+++ /dev/null
@@ -1,88 +0,0 @@
-format: jb-book
-root: README
-chapters:
-  - file: 01. Array Hashing/README
-    sections:
-      - file: 01. Array Hashing/1. Two Sum
-      - file: 01. Array Hashing/128. Longest Consecutive Sequence
-      - file: 01. Array Hashing/217. Contains Duplicate
-      - file: 01. Array Hashing/238. Product of Array Except Self
-      - file: 01. Array Hashing/242. Valid Anagram
-      - file: 01. Array Hashing/347. Top K Frequent Elements
-      - file: 01. Array Hashing/49. Group Anagrams
-      - file: 01. Array Hashing/659. Encode and Decode Strings
-  - file: 02. Two Pointers/README
-    sections:
-      - file: 02. Two Pointers/11. Container With Most Water
-      - file: 02. Two Pointers/125. Valid Palindrome
-      - file: 02. Two Pointers/15. 3Sum
-  - file: 03. Sliding Window/README
-    sections:
-      - file: 03. Sliding Window/121. Best Time to Buy and Sell Stock
-      - file: 03. Sliding Window/3. Longest Substring Without Repeating Characters
-      - file: 03. Sliding Window/424. Longest Repeating Character Replacement
-      - file: 03. Sliding Window/76. Minimum Window Substring
-  - file: 04. Stack/README
-    sections:
-      - file: 04. Stack/20. Valid Parentheses
-  - file: 05. Binary Search/README
-    sections:
-      - file: 05. Binary Search/153. Find Minimum in Rotated Sorted Array
-      - file: 05. Binary Search/33. Search in Rotated Sorted Array
-  - file: 06. Linked List/README
-    sections:
-      - file: 06. Linked List/141. Linked List Cycle
-      - file: 06. Linked List/143. Reorder List
-      - file: 06. Linked List/19. Remove Nth Node From End of List
-      - file: 06. Linked List/206. Reverse Linked List
-      - file: 06. Linked List/21. Merge Two Sorted Lists
-      - file: 06. Linked List/23. Merge k Sorted Lists
-  - file: 07. Trees/README
-    sections:
-      - file: 07. Trees/100. Same Tree
-      - file: 07. Trees/102. Binary Tree Level Order Traversal
-      - file: 07. Trees/104. Maximum Depth of Binary Tree
-      - file: 07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal
-      - file: 07. Trees/124. Binary Tree Maximum Path Sum
-      - file: 07. Trees/199. Binary Tree Right Side View
-      - file: 07. Trees/226. Invert Binary Tree
-      - file: 07. Trees/230. Kth Smallest Element in a BST
-      - file: 07. Trees/235. Lowest Common Ancestor of a Binary Search Tree
-      - file: 07. Trees/297. Serialize and Deserialize Binary Tree
-      - file: 07. Trees/572. Subtree of Another Tree
-      - file: 07. Trees/98. Validate Binary Search Tree
-  - file: 08. Tries/README
-    sections:
-      - file: 08. Tries/208. Implement Trie (Prefix Tree)
-      - file: 08. Tries/211. Design Add and Search Words Data Structure
-      - file: 08. Tries/212. Word Search II
-  - file: 09. Heap - Priority Queue/README
-    sections:
-      - file: 09. Heap - Priority Queue/1046. Last Stone Weight
-      - file: 09. Heap - Priority Queue/295. Find Median from Data Stream
-      - file: 09. Heap - Priority Queue/703. Kth Largest Element in a Stream
-  - file: 10. Backtracking/README
-    sections:
-      - file: 10. Backtracking/39. Combination Sum
-      - file: 10. Backtracking/79. Word Search
-  - file: 11. Graphs/README
-    sections:
-      - file: 11. Graphs/133. Clone Graph
-      - file: 11. Graphs/178. Graph Valid Tree
-      - file: 11. Graphs/200. Number of Islands
-      - file: 11. Graphs/207. Course Schedule
-      - file: 11. Graphs/3651. Number of Connected Components in an Undirected Graph
-      - file: 11. Graphs/417. Pacific Atlantic Water Flow
-  - file: 12. Advanced Graphs/README
-    sections:
-      - file: 12. Advanced Graphs/892. Alien Dictionary
-  - file: 13. One-D Dynamic Programming/README
-    sections:
-      - file: 13. One-D Dynamic Programming/70. Climbing Stairs
-      - file: 13. One-D Dynamic Programming/198. House Robber
-      - file: 13. One-D Dynamic Programming/213. House Robber II
-  - file: 14. Two-D Dynamic Programming/README
-  - file: 15. Greedy/README
-  - file: 16. Intervals/README
-  - file: 17. Math Geometry/README
-  - file: 18. Bit Manipulation/README
\ No newline at end of file
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diff --git a/logo.png b/logo.png
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diff --git a/_build/html/objects.inv b/objects.inv
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rename from _build/html/objects.inv
rename to objects.inv
diff --git a/_build/html/reports/Book.err.log b/reports/Book.err.log
similarity index 100%
rename from _build/html/reports/Book.err.log
rename to reports/Book.err.log
diff --git a/_build/html/search.html b/search.html
similarity index 100%
rename from _build/html/search.html
rename to search.html
diff --git a/_build/html/searchindex.js b/searchindex.js
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rename from _build/html/searchindex.js
rename to searchindex.js

From f40f6c98ed0772397057137963bca45f71ea5329 Mon Sep 17 00:00:00 2001
From: Mohsen Tabibian <36014419+mohsentabibian@users.noreply.github.com>
Date: Tue, 14 Nov 2023 17:55:29 +0000
Subject: [PATCH 5/7] Update documentation

---
 01. Array Hashing/1. Two Sum.html             | 117 +---
 .../128. Longest Consecutive Sequence.html    | 117 +---
 .../217. Contains Duplicate.html              | 117 +---
 .../238. Product of Array Except Self.html    | 117 +---
 01. Array Hashing/242. Valid Anagram.html     | 117 +---
 .../347. Top K Frequent Elements.html         | 117 +---
 01. Array Hashing/49. Group Anagrams.html     | 117 +---
 .../659. Encode and Decode Strings.html       | 117 +---
 01. Array Hashing/README.html                 | 119 +---
 .../11. Container With Most Water.html        | 117 +---
 02. Two Pointers/125. Valid Palindrome.html   | 117 +---
 02. Two Pointers/15. 3Sum.html                | 117 +---
 02. Two Pointers/README.html                  | 119 +---
 .../121. Best Time to Buy and Sell Stock.html | 117 +---
 ...ubstring Without Repeating Characters.html | 117 +---
 ...ngest Repeating Character Replacement.html | 117 +---
 .../76. Minimum Window Substring.html         | 117 +---
 03. Sliding Window/README.html                | 119 +---
 04. Stack/20. Valid Parentheses.html          | 117 +---
 04. Stack/README.html                         | 119 +---
 ... Find Minimum in Rotated Sorted Array.html | 117 +---
 .../33. Search in Rotated Sorted Array.html   | 117 +---
 05. Binary Search/README.html                 | 119 +---
 06. Linked List/141. Linked List Cycle.html   | 117 +---
 06. Linked List/143. Reorder List.html        | 117 +---
 .../19. Remove Nth Node From End of List.html | 117 +---
 06. Linked List/206. Reverse Linked List.html | 117 +---
 .../21. Merge Two Sorted Lists.html           | 117 +---
 06. Linked List/23. Merge k Sorted Lists.html | 117 +---
 06. Linked List/README.html                   | 119 +---
 07. Trees/100. Same Tree.html                 | 117 +---
 ...02. Binary Tree Level Order Traversal.html | 117 +---
 .../104. Maximum Depth of Binary Tree.html    | 117 +---
 ...e from Preorder and Inorder Traversal.html | 117 +---
 .../124. Binary Tree Maximum Path Sum.html    | 117 +---
 .../199. Binary Tree Right Side View.html     | 117 +---
 07. Trees/226. Invert Binary Tree.html        | 121 +---
 .../230. Kth Smallest Element in a BST.html   | 117 +---
 ...mmon Ancestor of a Binary Search Tree.html | 117 +---
 ...Serialize and Deserialize Binary Tree.html | 117 +---
 07. Trees/572. Subtree of Another Tree.html   | 117 +---
 .../98. Validate Binary Search Tree.html      | 117 +---
 07. Trees/README.html                         | 119 +---
 .../208. Implement Trie (Prefix Tree).html    | 117 +---
 ...n Add and Search Words Data Structure.html | 117 +---
 08. Tries/212. Word Search II.html            | 117 +---
 08. Tries/README.html                         | 119 +---
 .../1046. Last Stone Weight.html              | 117 +---
 .../295. Find Median from Data Stream.html    | 117 +---
 .../703. Kth Largest Element in a Stream.html | 117 +---
 09. Heap - Priority Queue/README.html         | 119 +---
 10. Backtracking/39. Combination Sum.html     | 117 +---
 10. Backtracking/79. Word Search.html         | 117 +---
 10. Backtracking/README.html                  | 119 +---
 11. Graphs/133. Clone Graph.html              | 117 +---
 11. Graphs/178. Graph Valid Tree.html         | 117 +---
 11. Graphs/200. Number of Islands.html        | 117 +---
 11. Graphs/207. Course Schedule.html          | 117 +---
 ...ted Components in an Undirected Graph.html | 117 +---
 .../417. Pacific Atlantic Water Flow.html     | 117 +---
 11. Graphs/README.html                        | 119 +---
 .../892. Alien Dictionary.html                | 117 +---
 12. Advanced Graphs/README.html               | 119 +---
 .../198. House Robber.html                    |  95 +---
 .../213. House Robber II.html                 |  95 +---
 .../70. Climbing Stairs.html                  |  95 +---
 13. One-D Dynamic Programming/README.html     | 101 +---
 14. Two-D Dynamic Programming/README.html     |  95 +---
 15. Greedy/README.html                        |  95 +---
 16. Intervals/README.html                     |  95 +---
 17. Math Geometry/README.html                 |  95 +---
 18. Bit Manipulation/README.html              |  95 +---
 Book.html                                     | 466 ++++------------
 CODE_OF_CONDUCT.html                          |  95 +---
 CONTRIBUTING.html                             |  95 +---
 README.html                                   | 109 +---
 SECURITY.html                                 |  95 +---
 _sources/Book.ipynb                           | 397 ++++++--------
 _sources/intro.md                             |  23 +
 genindex.html                                 |  95 +---
 intro.html                                    | 517 ++++++++++++++++++
 objects.inv                                   | Bin 2686 -> 2700 bytes
 search.html                                   |  95 +---
 searchindex.js                                |   2 +-
 84 files changed, 1197 insertions(+), 9052 deletions(-)
 create mode 100644 _sources/intro.md
 create mode 100644 intro.html

diff --git a/01. Array Hashing/1. Two Sum.html b/01. Array Hashing/1. Two Sum.html
index e4849aa..d7504ca 100644
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-        <ul class="current nav bd-sidenav">
-<li class="toctree-l1 current active has-children"><a class="reference internal" href="README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -620,24 +531,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
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-</ul>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -627,24 +538,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/README.html">Linked List Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -515,8 +426,6 @@ <h2>Resources<a class="headerlink" href="#resources" title="Permalink to this he
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diff --git a/02. Two Pointers/11. Container With Most Water.html b/02. Two Pointers/11. Container With Most Water.html
index d158b88..6a15259 100644
--- a/02. Two Pointers/11. Container With Most Water.html	
+++ b/02. Two Pointers/11. Container With Most Water.html	
@@ -67,8 +67,6 @@
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@@ -143,7 +141,7 @@
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+<a class="navbar-brand logo" href="../intro.html">
   
   
   
@@ -158,100 +156,13 @@
         
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-                <a class="reference internal" href="../README.html">
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+                <a class="reference internal" href="../intro.html">
+                    Introduction
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-        <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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diff --git a/02. Two Pointers/15. 3Sum.html b/02. Two Pointers/15. 3Sum.html
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
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+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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@@ -498,8 +409,6 @@ <h2>Resources<a class="headerlink" href="#resources" title="Permalink to this he
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--- a/03. Sliding Window/121. Best Time to Buy and Sell Stock.html	
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+                    Introduction
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
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diff --git a/03. Sliding Window/424. Longest Repeating Character Replacement.html b/03. Sliding Window/424. Longest Repeating Character Replacement.html
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-<li class="toctree-l2 current active"><a class="current reference internal" href="#">424. Longest Repeating Character Replacement</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -623,24 +534,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -669,24 +580,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/03. Sliding Window/README.html b/03. Sliding Window/README.html
index 7dcd69b..397bb7f 100644
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+++ b/03. Sliding Window/README.html	
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+<a class="navbar-brand logo" href="../intro.html">
   
   
   
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             <li class="toctree-l1">
-                <a class="reference internal" href="../README.html">
-                    Blind 75 LeetCode Solutions
+                <a class="reference internal" href="../intro.html">
+                    Introduction
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-        <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
-</li>
-<li class="toctree-l1 current active has-children"><a class="current reference internal" href="#">Sliding Window Problems - Blind 75 LeetCode</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/README.html">Stack Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
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@@ -501,8 +412,6 @@ <h2>Resources<a class="headerlink" href="#resources" title="Permalink to this he
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/README.html">Stack Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/README.html">Binary search Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
-</li>
-<li class="toctree-l1 current active has-children"><a class="reference internal" href="README.html">Linked List Problems - Blind 75 LeetCode</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
-<li class="toctree-l2 current active"><a class="current reference internal" href="#">141. Linked List Cycle</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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-        <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
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-<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
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-</ul>
-</li>
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-<li class="toctree-l2"><a class="reference internal" href="141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2 current active"><a class="current reference internal" href="#">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -706,24 +617,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2 current active"><a class="current reference internal" href="#">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -663,24 +574,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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+                    Introduction
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
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-<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2 current active"><a class="current reference internal" href="#">23. Merge k Sorted Lists</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
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+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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@@ -664,24 +575,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l2 current active"><a class="current reference internal" href="#">199. Binary Tree Right Side View</a></li>
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-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -655,24 +566,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
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-<li class="toctree-l2 current active"><a class="current reference internal" href="#">226: Invert Binary Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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-</ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -537,7 +448,7 @@ <h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this
 </div>
 </div>
 <div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Inverted Tree (Example 1): &lt;__main__.TreeNode object at 0x7faa5c445ab0&gt;
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Inverted Tree (Example 1): &lt;__main__.TreeNode object at 0x7f9db83bdb10&gt;
 </pre></div>
 </div>
 </div>
@@ -557,7 +468,7 @@ <h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this
 </div>
 </div>
 <div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Inverted Tree (Example 2): &lt;__main__.TreeNode object at 0x7faa5c444310&gt;
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Inverted Tree (Example 2): &lt;__main__.TreeNode object at 0x7f9db83bc2e0&gt;
 </pre></div>
 </div>
 </div>
@@ -628,24 +539,6 @@ <h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this
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diff --git a/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.html b/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.html
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
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-</ul>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
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-<li class="toctree-l2 current active"><a class="current reference internal" href="#">297. Serialize and Deserialize Binary Tree</a></li>
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diff --git a/07. Trees/98. Validate Binary Search Tree.html b/07. Trees/98. Validate Binary Search Tree.html
index 0d90f35..3a458e0 100644
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+<a class="navbar-brand logo" href="../intro.html">
   
   
   
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -646,24 +557,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/07. Trees/README.html b/07. Trees/README.html
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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diff --git a/08. Tries/README.html b/08. Tries/README.html
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+                    Introduction
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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@@ -498,8 +409,6 @@ <h2>Resources<a class="headerlink" href="#resources" title="Permalink to this he
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-                <a class="reference internal" href="../README.html">
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+                    Introduction
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-        <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
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-</ul>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
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-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -675,24 +586,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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+                    Introduction
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
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+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 current active has-children"><a class="reference internal" href="README.html">Graphs Problems - Blind 75 LeetCode</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
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-<li class="toctree-l2"><a class="reference internal" href="207.%20Course%20Schedule.html">207. Course Schedule</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
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+        <ul class="nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -716,24 +627,6 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/README.html">Binary search Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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diff --git a/12. Advanced Graphs/892. Alien Dictionary.html b/12. Advanced Graphs/892. Alien Dictionary.html
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
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diff --git a/13. One-D Dynamic Programming/213. House Robber II.html b/13. One-D Dynamic Programming/213. House Robber II.html
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
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-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l1 current active has-children"><a class="reference internal" href="README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
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 <li class="toctree-l2"><a class="reference internal" href="198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2 current active"><a class="current reference internal" href="#">House Robber II</a></li>
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+<a class="navbar-brand logo" href="../intro.html">
   
   
   
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             <li class="toctree-l1">
-                <a class="reference internal" href="../README.html">
-                    Blind 75 LeetCode Solutions
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+                    Introduction
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         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
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-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
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diff --git a/13. One-D Dynamic Programming/README.html b/13. One-D Dynamic Programming/README.html
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/README.html">Stack Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/README.html">Binary search Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/README.html">Linked List Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/README.html">Trie Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/15. Greedy/README.html b/15. Greedy/README.html
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         <div class="sidebar-primary-item">
   
 
-<a class="navbar-brand logo" href="../README.html">
+<a class="navbar-brand logo" href="../intro.html">
   
   
   
@@ -156,100 +156,13 @@
         
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
-                <a class="reference internal" href="../README.html">
-                    Blind 75 LeetCode Solutions
+                <a class="reference internal" href="../intro.html">
+                    Introduction
                 </a>
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/README.html">Stack Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/README.html">Binary search Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/README.html">Linked List Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/README.html">Trie Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/16. Intervals/README.html b/16. Intervals/README.html
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-<a class="navbar-brand logo" href="../README.html">
+<a class="navbar-brand logo" href="../intro.html">
   
   
   
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
-                <a class="reference internal" href="../README.html">
-                    Blind 75 LeetCode Solutions
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+                    Introduction
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/17. Math Geometry/README.html b/17. Math Geometry/README.html
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@@ -156,100 +156,13 @@
         
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-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
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-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/README.html">Binary search Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/README.html">Linked List Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
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-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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         <div class="sidebar-primary-item">
   
 
-<a class="navbar-brand logo" href="../README.html">
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
-                <a class="reference internal" href="../README.html">
-                    Blind 75 LeetCode Solutions
+                <a class="reference internal" href="../intro.html">
+                    Introduction
                 </a>
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/README.html">Stack Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/README.html">Binary search Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
-</li>
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-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/README.html">Trie Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/Book.html b/Book.html
index b11e1fe..858c3b4 100644
--- a/Book.html
+++ b/Book.html
@@ -139,7 +139,7 @@
         <div class="sidebar-primary-item">
   
 
-<a class="navbar-brand logo" href="README.html">
+<a class="navbar-brand logo" href="intro.html">
   
   
   
@@ -154,100 +154,13 @@
         
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
-                <a class="reference internal" href="README.html">
-                    Blind 75 LeetCode Solutions
+                <a class="reference internal" href="intro.html">
+                    Introduction
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/README.html">Stack Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/README.html">Binary search Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
-<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/README.html">Linked List Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/README.html">Trie Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -495,16 +408,14 @@ <h2> Contents </h2>
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: uc-micro-py in /usr/local/python/3.10.8/lib/python3.10/site-packages (from linkify-it-py~=2.0.0-&gt;jupyter-book) (1.0.2)
+Requirement already satisfied: importlib_metadata in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (6.8.0)
 Requirement already satisfied: ipython in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (8.16.1)
 Requirement already satisfied: jupyter-cache&lt;0.7,&gt;=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (0.6.1)
 Requirement already satisfied: nbclient in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (0.7.4)
@@ -512,7 +423,9 @@ <h2> Contents </h2>
 Requirement already satisfied: nbformat~=5.0 in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (5.9.2)
 Requirement already satisfied: typing-extensions in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (4.8.0)
 Requirement already satisfied: ipykernel in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (6.25.2)
-Requirement already satisfied: sphinxcontrib-applehelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.0.7)
+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: sphinxcontrib-applehelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.0.7)
 Requirement already satisfied: sphinxcontrib-devhelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.0.5)
 Requirement already satisfied: sphinxcontrib-jsmath in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.0.1)
 Requirement already satisfied: sphinxcontrib-htmlhelp&gt;=2.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (2.0.4)
@@ -526,9 +439,7 @@ <h2> Contents </h2>
 Requirement already satisfied: requests&gt;=2.5.0 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (2.31.0)
 Requirement already satisfied: packaging in /home/codespace/.local/lib/python3.10/site-packages (from sphinx&lt;6,&gt;=4-&gt;jupyter-book) (23.2)
 Requirement already satisfied: MarkupSafe&gt;=2.0 in /home/codespace/.local/lib/python3.10/site-packages (from Jinja2-&gt;jupyter-book) (2.1.3)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: pydata-sphinx-theme&gt;=0.13.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx-book-theme~=1.0.0-&gt;jupyter-book) (0.14.3)
+Requirement already satisfied: pydata-sphinx-theme&gt;=0.13.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx-book-theme~=1.0.0-&gt;jupyter-book) (0.14.3)
 </pre></div>
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 <div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: pybtex&gt;=0.24 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex&lt;=2.5.0,&gt;=2.2.0-&gt;jupyter-book) (0.24.0)
@@ -545,28 +456,30 @@ <h2> Contents </h2>
 </div>
 <div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: markdown-it-py&lt;3.0.0,&gt;=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.2.0)
 Requirement already satisfied: mdit-py-plugins~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.3.5)
-Requirement already satisfied: jupyter-client&gt;=6.1.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (8.4.0)
+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: jupyter-client&gt;=6.1.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (8.4.0)
 Requirement already satisfied: jupyter-core!=5.0.*,&gt;=4.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (5.4.0)
 Requirement already satisfied: traitlets&gt;=5.3 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (5.11.2)
-Requirement already satisfied: fastjsonschema in /home/codespace/.local/lib/python3.10/site-packages (from nbformat~=5.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.18.1)
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: latexcodec&gt;=1.0.4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pybtex&gt;=0.24-&gt;sphinxcontrib-bibtex&lt;=2.5.0,&gt;=2.2.0-&gt;jupyter-book) (2.0.1)
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: fastjsonschema in /home/codespace/.local/lib/python3.10/site-packages (from nbformat~=5.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.18.1)
+Requirement already satisfied: latexcodec&gt;=1.0.4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pybtex&gt;=0.24-&gt;sphinxcontrib-bibtex&lt;=2.5.0,&gt;=2.2.0-&gt;jupyter-book) (2.0.1)
 Requirement already satisfied: six in /home/codespace/.local/lib/python3.10/site-packages (from pybtex&gt;=0.24-&gt;sphinxcontrib-bibtex&lt;=2.5.0,&gt;=2.2.0-&gt;jupyter-book) (1.16.0)
-Requirement already satisfied: beautifulsoup4 in /home/codespace/.local/lib/python3.10/site-packages (from pydata-sphinx-theme&gt;=0.13.3-&gt;sphinx-book-theme~=1.0.0-&gt;jupyter-book) (4.12.2)
-Requirement already satisfied: accessible-pygments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pydata-sphinx-theme&gt;=0.13.3-&gt;sphinx-book-theme~=1.0.0-&gt;jupyter-book) (0.0.4)
 </pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: charset-normalizer&lt;4,&gt;=2 in /home/codespace/.local/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (3.3.0)
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: beautifulsoup4 in /home/codespace/.local/lib/python3.10/site-packages (from pydata-sphinx-theme&gt;=0.13.3-&gt;sphinx-book-theme~=1.0.0-&gt;jupyter-book) (4.12.2)
+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: accessible-pygments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pydata-sphinx-theme&gt;=0.13.3-&gt;sphinx-book-theme~=1.0.0-&gt;jupyter-book) (0.0.4)
+Requirement already satisfied: charset-normalizer&lt;4,&gt;=2 in /home/codespace/.local/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (3.3.0)
 Requirement already satisfied: idna&lt;4,&gt;=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (3.4)
 Requirement already satisfied: urllib3&lt;3,&gt;=1.21.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.26.18)
 Requirement already satisfied: certifi&gt;=2017.4.17 in /home/codespace/.local/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (2023.7.22)
 </pre></div>
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 <div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: zipp&gt;=0.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from importlib_metadata-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (3.17.0)
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: comm&gt;=0.1.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.1.4)
+Requirement already satisfied: comm&gt;=0.1.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.1.4)
 Requirement already satisfied: debugpy&gt;=1.6.5 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (1.8.0)
 Requirement already satisfied: matplotlib-inline&gt;=0.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.1.6)
 Requirement already satisfied: nest-asyncio in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (1.5.8)
@@ -586,14 +499,14 @@ <h2> Contents </h2>
 </pre></div>
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 <div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: parso&lt;0.9.0,&gt;=0.8.3 in /home/codespace/.local/lib/python3.10/site-packages (from jedi&gt;=0.16-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.8.3)
+Requirement already satisfied: python-dateutil&gt;=2.8.2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-client&gt;=6.1.12-&gt;nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.8.2)
 </pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: python-dateutil&gt;=2.8.2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-client&gt;=6.1.12-&gt;nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.8.2)
-Requirement already satisfied: platformdirs&gt;=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-core!=5.0.*,&gt;=4.12-&gt;nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (3.11.0)
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: platformdirs&gt;=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-core!=5.0.*,&gt;=4.12-&gt;nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (3.11.0)
+Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py&lt;3.0.0,&gt;=1.0.0-&gt;myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.1.2)
 </pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py&lt;3.0.0,&gt;=1.0.0-&gt;myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.1.2)
-Requirement already satisfied: ptyprocess&gt;=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect&gt;4.3-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.7.0)
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: ptyprocess&gt;=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect&gt;4.3-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.7.0)
 Requirement already satisfied: wcwidth in /home/codespace/.local/lib/python3.10/site-packages (from prompt-toolkit!=3.0.37,&lt;3.1.0,&gt;=3.0.30-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.2.8)
 </pre></div>
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@@ -672,118 +585,118 @@ <h2> Contents </h2>
 <span class=" -Color -Color-Bold">Running Sphinx v5.0.2</span>
 </pre></div>
 </div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[etoc] Changing master_doc to &#39;README&#39;
-</pre></div>
-</div>
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">myst v0.18.1:</span> MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=[&#39;colon_fence&#39;, &#39;dollarmath&#39;, &#39;linkify&#39;, &#39;substitution&#39;, &#39;tasklist&#39;], disable_syntax=[], all_links_external=False, url_schemes=[&#39;mailto&#39;, &#39;http&#39;, &#39;https&#39;], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=(&#39;{&#39;, &#39;}&#39;), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes=&#39;tex2jax_process|mathjax_process|math|output_area&#39;)
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[etoc] Changing master_doc to &#39;intro&#39;
+<span class=" -Color -Color-Bold">loading pickled environment... </span>done
+<span class=" -Color -Color-Bold">myst v0.18.1:</span> MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=[&#39;colon_fence&#39;, &#39;dollarmath&#39;, &#39;linkify&#39;, &#39;substitution&#39;, &#39;tasklist&#39;], disable_syntax=[], all_links_external=False, url_schemes=[&#39;mailto&#39;, &#39;http&#39;, &#39;https&#39;], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=(&#39;{&#39;, &#39;}&#39;), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes=&#39;tex2jax_process|mathjax_process|math|output_area&#39;)
 <span class=" -Color -Color-Bold">myst-nb v0.17.2:</span> NbParserConfig(custom_formats={}, metadata_key=&#39;mystnb&#39;, cell_metadata_key=&#39;mystnb&#39;, kernel_rgx_aliases={}, execution_mode=&#39;auto&#39;, execution_cache_path=&#39;&#39;, execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin=&#39;default&#39;, remove_code_source=False, remove_code_outputs=False, code_prompt_show=&#39;Show code cell {type}&#39;, code_prompt_hide=&#39;Hide code cell {type}&#39;, number_source_lines=False, output_stderr=&#39;show&#39;, render_text_lexer=&#39;myst-ansi&#39;, render_error_lexer=&#39;ipythontb&#39;, render_image_options={}, render_figure_options={}, render_markdown_format=&#39;commonmark&#39;, output_folder=&#39;build&#39;, append_css=True, metadata_to_fm=False)
 Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache
 </pre></div>
 </div>
 <div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">building [mo]: </span>targets for 0 po files that are out of date
-<span class=" -Color -Color-Bold">building [html]: </span>targets for 77 source files that are out of date
-<span class=" -Color -Color-Bold">updating environment: </span>[new config] 77 added, 0 changed, 0 removed
+<span class=" -Color -Color-Bold">building [html]: </span>targets for 1 source files that are out of date
+<span class=" -Color -Color-Bold">updating environment: </span>
+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[config changed (&#39;master_doc&#39;)] 78 added, 1 changed, 0 removed
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.07 seconds [mystnb]
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.20 seconds [mystnb]
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.17 seconds [mystnb]
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 0.88 seconds [mystnb]
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.19 seconds [mystnb]
-<span class=" -Color -Color-Bold">reading sources... </span>[  5%] <span class=" -Color -Color-Magenta">01. Array Hashing/238. Product of Array Except Self</span>   
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.05 seconds [mystnb]
+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[  5%] <span class=" -Color -Color-Magenta">01. Array Hashing/238. Product of Array Except Self</span>   
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 0.98 seconds [mystnb]
+<span class=" -Color -Color-Bold">reading sources... </span>[  6%] <span class=" -Color -Color-Magenta">01. Array Hashing/242. Valid Anagram</span>                  
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-/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>^C
-
-Exception occurred:
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 512, in _async_cleanup_kernel
-    await ensure_async(self.km.cleanup_resources())
-AttributeError: &#39;NoneType&#39; object has no attribute &#39;cleanup_resources&#39;
-The full traceback has been saved in /tmp/sphinx-err-tl19fjln.log, if you want to report the issue to the developers.
-Please also report this if it was a user error, so that a better error message can be provided next time.
-A bug report can be filed in the tracker at &lt;https://github.com/sphinx-doc/sphinx/issues&gt;. Thanks!
-Traceback (most recent call last):
-  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py&quot;, line 89, in wrapper
-    self._ready.set_result(None)
-asyncio.exceptions.InvalidStateError: invalid state
-
-During handling of the above exception, another exception occurred:
-
-Traceback (most recent call last):
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 505, in _async_cleanup_kernel
-    await ensure_async(self.km.shutdown_kernel(now=now))
-  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py&quot;, line 189, in ensure_async
-    result = await obj
-  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/manager.py&quot;, line 92, in wrapper
-    self._ready.set_exception(e)
-asyncio.exceptions.InvalidStateError: invalid state
-
-During handling of the above exception, another exception occurred:
-
-Traceback (most recent call last):
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/sphinx.py&quot;, line 171, in build_sphinx
-    app.build(force_all, filenames)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/application.py&quot;, line 329, in build
-    self.builder.build_update()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 288, in build_update
-    self.build(to_build,
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 302, in build
-    updated_docnames = set(self.read())
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 409, in read
-    self._read_serial(docnames)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 430, in _read_serial
-    self.read_doc(docname)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 483, in read_doc
-    publisher.publish()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/core.py&quot;, line 217, in publish
-    self.document = self.reader.read(self.source, self.parser,
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/io.py&quot;, line 103, in read
-    self.parse()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/readers/__init__.py&quot;, line 78, in parse
-    self.parser.parse(self.input, document)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/sphinx_.py&quot;, line 150, in parse
-    with create_client(
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/base.py&quot;, line 83, in __enter__
-    self.start_client()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/direct.py&quot;, line 40, in start_client
-    result = single_nb_execution(
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_cache/executors/utils.py&quot;, line 58, in single_nb_execution
-    executenb(
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 1305, in execute
-    return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute()
-  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py&quot;, line 173, in wrapped
-    return loop.run_until_complete(inner)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/asyncio/base_events.py&quot;, line 649, in run_until_complete
-    return future.result()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 689, in async_execute
-    async with self.async_setup_kernel(**kwargs):
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/contextlib.py&quot;, line 206, in __aexit__
-    await anext(self.gen)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 656, in async_setup_kernel
-    await self._async_cleanup_kernel()
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 512, in _async_cleanup_kernel
-    await ensure_async(self.km.cleanup_resources())
-AttributeError: &#39;NoneType&#39; object has no attribute &#39;cleanup_resources&#39;
-
-The above exception was the direct cause of the following exception:
-
-Traceback (most recent call last):
-  File &quot;/home/codespace/.python/current/bin/jupyter-book&quot;, line 8, in &lt;module&gt;
-    sys.exit(main())
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1157, in __call__
-    return self.main(*args, **kwargs)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1078, in main
-    rv = self.invoke(ctx)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1688, in invoke
-    return _process_result(sub_ctx.command.invoke(sub_ctx))
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1434, in invoke
-    return ctx.invoke(self.callback, **ctx.params)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 783, in invoke
-    return __callback(*args, **kwargs)
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py&quot;, line 317, in build
-    builder_specific_actions(
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py&quot;, line 525, in builder_specific_actions
-    raise RuntimeError(_message_box(msg, color=&quot;red&quot;, doprint=False)) from result
-RuntimeError: 
-===============================================================================
-
-There was an error in building your book. Look above for the cause.
-
-===============================================================================
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-<div class="cell docutils container">
-<div class="cell_input docutils container">
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-/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.10 seconds [mystnb]
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-/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.12 seconds [mystnb]
-/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]
-/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]
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-<span class=" -Color -Color-Bold">reading sources... </span>[100%] <span class=" -Color -Color-Magenta">intro</span>                                                 
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-<span class=" -Color -Color-Bold">dumping object inventory... </span>done
-[etoc] missing index.html written as redirect to &#39;intro.html&#39;
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-
-The HTML pages are in mybook/_build/html.
-
-===============================================================================
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-Finished generating HTML for book.
-Your book&#39;s HTML pages are here:
-    mybook/_build/html/
-You can look at your book by opening this file in a browser:
-    mybook/_build/html/index.html
-Or paste this line directly into your browser bar:
-    file:///workspaces/LeetCode-Solutions/mybook/_build/html/index.html            
-
-===============================================================================
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>^C
+Exception ignored in: &lt;function WeakSet.__init__.&lt;locals&gt;._remove at 0x7f861ec2d2d0&gt;
+Traceback (most recent call last):
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/_weakrefset.py&quot;, line 39, in _remove
+    def _remove(item, selfref=ref(self)):
+KeyboardInterrupt: 
+<span class=" -Color -Color-Bold">reading sources... </span>[ 37%] <span class=" -Color -Color-Magenta">06. Linked List/23. Merge k Sorted Lists</span>              
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+<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="o">%</span><span class="k">pip</span> install ghp-import
 </pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>format: jb-book
-root: intro
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: ghp-import in /usr/local/python/3.10.8/lib/python3.10/site-packages (2.1.0)
+Requirement already satisfied: python-dateutil&gt;=2.8.1 in /home/codespace/.local/lib/python3.10/site-packages (from ghp-import) (2.8.2)
+Requirement already satisfied: six&gt;=1.5 in /home/codespace/.local/lib/python3.10/site-packages (from python-dateutil&gt;=2.8.1-&gt;ghp-import) (1.16.0)
+Note: you may need to restart the kernel to use updated packages.
 </pre></div>
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diff --git a/CODE_OF_CONDUCT.html b/CODE_OF_CONDUCT.html
index e1f3fd1..9cad015 100644
--- a/CODE_OF_CONDUCT.html
+++ b/CODE_OF_CONDUCT.html
@@ -139,7 +139,7 @@
         <div class="sidebar-primary-item">
   
 
-<a class="navbar-brand logo" href="README.html">
+<a class="navbar-brand logo" href="intro.html">
   
   
   
@@ -154,100 +154,13 @@
         
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
-                <a class="reference internal" href="README.html">
-                    Blind 75 LeetCode Solutions
+                <a class="reference internal" href="intro.html">
+                    Introduction
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/README.html">Stack Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/README.html">Binary search Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
-<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/README.html">Linked List Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/README.html">Trie Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/CONTRIBUTING.html b/CONTRIBUTING.html
index 9254de9..1a3aa7a 100644
--- a/CONTRIBUTING.html
+++ b/CONTRIBUTING.html
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         <div class="sidebar-primary-item">
   
 
-<a class="navbar-brand logo" href="README.html">
+<a class="navbar-brand logo" href="intro.html">
   
   
   
@@ -154,100 +154,13 @@
         
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
-                <a class="reference internal" href="README.html">
-                    Blind 75 LeetCode Solutions
+                <a class="reference internal" href="intro.html">
+                    Introduction
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         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/README.html">Stack Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/README.html">Binary search Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
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-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/README.html">Linked List Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/README.html">Trie Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
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-</ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/README.html">Trie Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/_sources/Book.ipynb b/_sources/Book.ipynb
index d400192..59960cb 100644
--- a/_sources/Book.ipynb
+++ b/_sources/Book.ipynb
@@ -2,12 +2,116 @@
  "cells": [
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 2,
    "metadata": {},
-   "outputs": [],
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Requirement already satisfied: pip in /usr/local/python/3.10.8/lib/python3.10/site-packages (23.3.1)\n",
+      "Requirement already satisfied: jupyter-book in /usr/local/python/3.10.8/lib/python3.10/site-packages (0.15.1)\n",
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+      "Requirement already satisfied: docutils<0.19,>=0.15 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.18.1)\n",
+      "Requirement already satisfied: Jinja2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (3.1.2)\n",
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+      "Requirement already satisfied: linkify-it-py~=2.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (2.0.2)\n",
+      "Requirement already satisfied: myst-nb~=0.17.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.17.2)\n",
+      "Requirement already satisfied: pyyaml in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-book) (6.0.1)\n",
+      "Requirement already satisfied: sphinx<6,>=4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (5.0.2)\n",
+      "Requirement already satisfied: sphinx-comments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.0.3)\n",
+      "Requirement already satisfied: sphinx-copybutton in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.5.2)\n",
+      "Requirement already satisfied: sphinx-external-toc~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.1)\n",
+      "Requirement already satisfied: sphinx-jupyterbook-latex~=0.5.2 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.5.2)\n",
+      "Requirement already satisfied: sphinx-design~=0.3.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.0)\n",
+      "Requirement already satisfied: sphinx-thebe~=0.2.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.2.1)\n",
+      "Requirement already satisfied: sphinx-book-theme~=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (1.0.1)\n",
+      "Requirement already satisfied: sphinx_togglebutton in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.3.2)\n",
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+      "Requirement already satisfied: sphinx-multitoc-numbering~=0.1.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.1.3)\n",
+      "Requirement already satisfied: attrs>=22.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (23.1.0)\n",
+      "Requirement already satisfied: jsonschema-specifications>=2023.03.6 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (2023.7.1)\n",
+      "Requirement already satisfied: referencing>=0.28.4 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (0.30.2)\n",
+      "Requirement already satisfied: rpds-py>=0.7.1 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema<5->jupyter-book) (0.10.6)\n",
+      "Requirement already satisfied: uc-micro-py in /usr/local/python/3.10.8/lib/python3.10/site-packages (from linkify-it-py~=2.0.0->jupyter-book) (1.0.2)\n",
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+      "Requirement already satisfied: ipython in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (8.16.1)\n",
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+      "Requirement already satisfied: nbclient in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.7.4)\n",
+      "Requirement already satisfied: myst-parser~=0.18.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (0.18.1)\n",
+      "Requirement already satisfied: nbformat~=5.0 in /home/codespace/.local/lib/python3.10/site-packages (from myst-nb~=0.17.1->jupyter-book) (5.9.2)\n",
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+      "Requirement already satisfied: sphinxcontrib-serializinghtml>=1.1.5 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.1.9)\n",
+      "Requirement already satisfied: sphinxcontrib-qthelp in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.0.6)\n",
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+      "Requirement already satisfied: alabaster<0.8,>=0.7 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (0.7.13)\n",
+      "Requirement already satisfied: imagesize in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (1.4.1)\n",
+      "Requirement already satisfied: requests>=2.5.0 in /home/codespace/.local/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (2.31.0)\n",
+      "Requirement already satisfied: packaging in /home/codespace/.local/lib/python3.10/site-packages (from sphinx<6,>=4->jupyter-book) (23.2)\n",
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+      "Requirement already satisfied: pybtex>=0.24 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (0.24.0)\n",
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+      "Requirement already satisfied: mdit-py-plugins~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.3.5)\n",
+      "Requirement already satisfied: jupyter-client>=6.1.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (8.4.0)\n",
+      "Requirement already satisfied: jupyter-core!=5.0.*,>=4.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (5.4.0)\n",
+      "Requirement already satisfied: traitlets>=5.3 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient->myst-nb~=0.17.1->jupyter-book) (5.11.2)\n",
+      "Requirement already satisfied: fastjsonschema in /home/codespace/.local/lib/python3.10/site-packages (from nbformat~=5.0->myst-nb~=0.17.1->jupyter-book) (2.18.1)\n",
+      "Requirement already satisfied: latexcodec>=1.0.4 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pybtex>=0.24->sphinxcontrib-bibtex<=2.5.0,>=2.2.0->jupyter-book) (2.0.1)\n",
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+      "Requirement already satisfied: prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (3.0.39)\n",
+      "Requirement already satisfied: stack-data in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (0.6.3)\n",
+      "Requirement already satisfied: exceptiongroup in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (1.1.3)\n",
+      "Requirement already satisfied: pexpect>4.3 in /home/codespace/.local/lib/python3.10/site-packages (from ipython->myst-nb~=0.17.1->jupyter-book) (4.8.0)\n",
+      "Requirement already satisfied: parso<0.9.0,>=0.8.3 in /home/codespace/.local/lib/python3.10/site-packages (from jedi>=0.16->ipython->myst-nb~=0.17.1->jupyter-book) (0.8.3)\n",
+      "Requirement already satisfied: python-dateutil>=2.8.2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-client>=6.1.12->nbclient->myst-nb~=0.17.1->jupyter-book) (2.8.2)\n",
+      "Requirement already satisfied: platformdirs>=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-core!=5.0.*,>=4.12->nbclient->myst-nb~=0.17.1->jupyter-book) (3.11.0)\n",
+      "Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py<3.0.0,>=1.0.0->myst-parser~=0.18.0->myst-nb~=0.17.1->jupyter-book) (0.1.2)\n",
+      "Requirement already satisfied: ptyprocess>=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect>4.3->ipython->myst-nb~=0.17.1->jupyter-book) (0.7.0)\n",
+      "Requirement already satisfied: wcwidth in /home/codespace/.local/lib/python3.10/site-packages (from prompt-toolkit!=3.0.37,<3.1.0,>=3.0.30->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.8)\n",
+      "Requirement already satisfied: greenlet!=0.4.17 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sqlalchemy<3,>=1.3.12->jupyter-cache<0.7,>=0.5->myst-nb~=0.17.1->jupyter-book) (3.0.1)\n",
+      "Requirement already satisfied: soupsieve>1.2 in /home/codespace/.local/lib/python3.10/site-packages (from beautifulsoup4->pydata-sphinx-theme>=0.13.3->sphinx-book-theme~=1.0.0->jupyter-book) (2.5)\n",
+      "Requirement already satisfied: executing>=1.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.0.0)\n",
+      "Requirement already satisfied: asttokens>=2.1.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.4.0)\n",
+      "Requirement already satisfied: pure-eval in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.2)\n",
+      "Note: you may need to restart the kernel to use updated packages.\n"
+     ]
+    }
+   ],
    "source": [
     "!python -m pip install --upgrade pip\n",
-    "%pip install jupyter-book"
+    "%pip install jupyter-book\n",
+    "%pip install ghp-import"
    ]
   },
   {
@@ -38,7 +142,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 3,
+   "execution_count": 4,
    "metadata": {},
    "outputs": [
     {
@@ -50,136 +154,123 @@
       "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\n",
       "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\n",
       "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "\u001b[01mmaking output directory... \u001b[39;49;00mdone\n",
-      "[etoc] Changing master_doc to 'README'\n",
+      "[etoc] Changing master_doc to 'intro'\n",
+      "\u001b[01mloading pickled environment... \u001b[39;49;00mdone\n",
       "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
       "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
       "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\n",
       "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 77 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 77 added, 0 changed, 0 removed\n",
+      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 1 source files that are out of date\n",
+      "\u001b[01mupdating environment: \u001b[39;49;00m[config changed ('master_doc')] 78 added, 1 changed, 0 removed\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]  \n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.26 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 1.17 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 1.21 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]     \n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 0.92 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 0.94 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 1.03 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 0.97 seconds [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executed notebook in 1.21 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executing notebook using local CWD [mystnb]      \n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executed notebook in 4.03 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executing notebook using local CWD [mystnb]   \n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 1.23 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executed notebook in 0.89 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 0.87 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 1.25 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/Book.ipynb: Executing notebook using local CWD [mystnb]                          \n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mSECURITY\u001b[39;49;00m                                              \n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb:30008: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Executing notebook failed: CellTimeoutError [mystnb.exec]\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Notebook exception traceback saved in: /workspaces/LeetCode-Solutions/_build/html/reports/Book.err.log [mystnb.exec]\u001b[39;49;00m\n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CODE_OF_CONDUCT.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/SECURITY.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[ 94%] \u001b[32mBook\u001b[39;49;00m                                                   raph\u001b[39;49;00m0m\r"
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 1.33 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/Book.ipynb: Executing notebook using local CWD [mystnb]                          \n"
      ]
     }
    ],
@@ -189,164 +280,28 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "!jupyter-book build /workspaces/LeetCode-Solutions/ --builder pdfhtml"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/html": [
-       "<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>"
-      ],
-      "text/plain": [
-       "<IPython.core.display.HTML object>"
-      ]
-     },
-     "execution_count": 5,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "from IPython.display import HTML\n",
-    "HTML('<a href=\"_build/html/index.html\" target=\"_blank\">Click here to view your Jupyter Book</a>')\n"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 40,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Your book template can be found at\n",
-      "\n",
-      "    mybook/\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book create mybook"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 45,
+   "execution_count": 3,
    "metadata": {},
    "outputs": [
     {
      "name": "stdout",
      "output_type": "stream",
      "text": [
-      "\u001b[32m\u001b[1mRunning Jupyter-Book v0.15.1\u001b[0m\n",
-      "\u001b[34m\u001b[1mSource Folder: \u001b[0m/workspaces/LeetCode-Solutions/mybook\n",
-      "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_config.yml\n",
-      "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/mybook/_build/html\n",
-      "[sphinxcontrib-bibtex] Beware that docutils versions 0.18 and 0.19 (you are running 0.18.1) are known to generate invalid html for citations. If this issue affects you, please use docutils<0.18 (or >=0.20 once released) instead. For more details, see https://sourceforge.net/p/docutils/patches/195/\n",
-      "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
-      "[etoc] Changing master_doc to 'intro'\n",
-      "checking bibtex cache... out of date\n",
-      "parsing bibtex file /workspaces/LeetCode-Solutions/mybook/references.bib... parsed 5 entries\n",
-      "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
-      "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='force', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
-      "Using jupyter-cache at: /workspaces/LeetCode-Solutions/mybook/_build/.jupyter_cache\n",
-      "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 14 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 14 added, 0 changed, 0 removed\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.35 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 0.97 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 0.83 seconds [mystnb]\n",
-      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mintro\u001b[39;49;00m                                                 \n",
-      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
-      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
-      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/mybook/01. Array Hashing/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "\u001b[91m/workspaces/LeetCode-Solutions/mybook/02. Two Pointers/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
-      "done\n",
-      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
-      "\u001b[01mwriting output... \u001b[39;49;00m[100%] \u001b[32mintro\u001b[39;49;00m                                                  \n",
-      "\u001b[01mgenerating indices... \u001b[39;49;00mgenindex done\n",
-      "\u001b[01mwriting additional pages... \u001b[39;49;00msearch done\n",
-      "\u001b[01mcopying static files... \u001b[39;49;00mdone\n",
-      "\u001b[01mcopying extra files... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping search index in English (code: en)... \u001b[39;49;00mdone\n",
-      "\u001b[01mdumping object inventory... \u001b[39;49;00mdone\n",
-      "[etoc] missing index.html written as redirect to 'intro.html'\n",
-      "\u001b[01mbuild succeeded, 2 warnings.\u001b[39;49;00m\n",
-      "\n",
-      "The HTML pages are in mybook/_build/html.\n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n",
-      "Finished generating HTML for book.\n",
-      "Your book's HTML pages are here:\n",
-      "    mybook/_build/html/\n",
-      "You can look at your book by opening this file in a browser:\n",
-      "    mybook/_build/html/index.html\n",
-      "Or paste this line directly into your browser bar:\n",
-      "    file:///workspaces/LeetCode-Solutions/mybook/_build/html/index.html            \n",
-      "\n",
-      "\u001b[92m===============================================================================\u001b[0m\n",
-      "\n"
+      "Requirement already satisfied: ghp-import in /usr/local/python/3.10.8/lib/python3.10/site-packages (2.1.0)\n",
+      "Requirement already satisfied: python-dateutil>=2.8.1 in /home/codespace/.local/lib/python3.10/site-packages (from ghp-import) (2.8.2)\n",
+      "Requirement already satisfied: six>=1.5 in /home/codespace/.local/lib/python3.10/site-packages (from python-dateutil>=2.8.1->ghp-import) (1.16.0)\n",
+      "Note: you may need to restart the kernel to use updated packages.\n"
      ]
     }
    ],
-   "source": [
-    "!jupyter-book build mybook"
-   ]
+   "source": []
   },
   {
    "cell_type": "code",
-   "execution_count": 31,
+   "execution_count": null,
    "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "format: jb-book\n",
-      "root: intro\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "!jupyter-book toc from-project /workspaces/LeetCode-Solutions/mybook/ -f jb-book"
-   ]
+   "outputs": [],
+   "source": []
   }
  ],
  "metadata": {
diff --git a/_sources/intro.md b/_sources/intro.md
new file mode 100644
index 0000000..555837a
--- /dev/null
+++ b/_sources/intro.md
@@ -0,0 +1,23 @@
+## Introduction
+
+Welcome to "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In the dynamic landscape of technical interviews and coding assessments, mastering the art of algorithmic problem-solving is not just a skill—it's a necessity. This book is your guide to conquering the Blind 75, a curated collection of LeetCode challenges meticulously selected to empower you with the knowledge and expertise needed to navigate the complexities of coding interviews.
+
+### Unveiling the Blind 75
+
+The Blind 75 is a carefully chosen set of LeetCode problems, renowned for their frequency in technical interviews across the industry. From fundamental data structures to advanced algorithms, this collection serves as a compass, guiding you through the diverse terrain of coding challenges. Each problem is a stepping stone, designed not only to test your coding prowess but to elevate your problem-solving mindset.
+
+### Building a Strong Foundation
+
+This repository is more than just a list of challenges—it's a hands-on journey to mastery. As you delve into each problem, you'll gain insights into optimal coding practices, algorithmic thinking, and Pythonic solutions. The carefully annotated solutions provide a roadmap, helping you understand not only *how* to solve a problem but also *why* a particular approach works best.
+
+### What to Expect
+
+- **Curated Learning Path:** Progress through a curated set of challenges that cover a spectrum of topics, ensuring a well-rounded understanding of data structures and algorithms.
+  
+- **Practical Application:** Bridge the gap between theoretical knowledge and real-world application as you tackle challenges that mirror the complexities of technical interviews.
+
+- **Pythonic Wisdom:** Learn to harness the power of Python to craft elegant, efficient, and readable code. Each solution is a lesson in leveraging Python's strengths for algorithmic problem-solving.
+
+Whether you are preparing for technical interviews, enhancing your coding skills, or simply seeking to deepen your understanding of Python and algorithms, "Hands-On Problem-Solving in Python" is your companion on this enriching journey. Let's embark on the adventure of mastering the Blind 75 and unlocking the doors to confident, effective problem-solving in Python.
+
+Happy coding!
\ No newline at end of file
diff --git a/genindex.html b/genindex.html
index 13d2f0f..ebd7335 100644
--- a/genindex.html
+++ b/genindex.html
@@ -138,7 +138,7 @@
         <div class="sidebar-primary-item">
   
 
-<a class="navbar-brand logo" href="README.html">
+<a class="navbar-brand logo" href="intro.html">
   
   
   
@@ -153,100 +153,13 @@
         
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
-                <a class="reference internal" href="README.html">
-                    Blind 75 LeetCode Solutions
+                <a class="reference internal" href="intro.html">
+                    Introduction
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
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 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
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+                <h2> Contents </h2>
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+                <ul class="visible nav section-nav flex-column">
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-blind-75">Unveiling the Blind 75</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#building-a-strong-foundation">Building a Strong Foundation</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#what-to-expect">What to Expect</a></li>
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+  <section id="introduction">
+<h1>Introduction<a class="headerlink" href="#introduction" title="Permalink to this heading">#</a></h1>
+<p>Welcome to “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In the dynamic landscape of technical interviews and coding assessments, mastering the art of algorithmic problem-solving is not just a skill—it’s a necessity. This book is your guide to conquering the Blind 75, a curated collection of LeetCode challenges meticulously selected to empower you with the knowledge and expertise needed to navigate the complexities of coding interviews.</p>
+<section id="unveiling-the-blind-75">
+<h2>Unveiling the Blind 75<a class="headerlink" href="#unveiling-the-blind-75" title="Permalink to this heading">#</a></h2>
+<p>The Blind 75 is a carefully chosen set of LeetCode problems, renowned for their frequency in technical interviews across the industry. From fundamental data structures to advanced algorithms, this collection serves as a compass, guiding you through the diverse terrain of coding challenges. Each problem is a stepping stone, designed not only to test your coding prowess but to elevate your problem-solving mindset.</p>
+</section>
+<section id="building-a-strong-foundation">
+<h2>Building a Strong Foundation<a class="headerlink" href="#building-a-strong-foundation" title="Permalink to this heading">#</a></h2>
+<p>This repository is more than just a list of challenges—it’s a hands-on journey to mastery. As you delve into each problem, you’ll gain insights into optimal coding practices, algorithmic thinking, and Pythonic solutions. The carefully annotated solutions provide a roadmap, helping you understand not only <em>how</em> to solve a problem but also <em>why</em> a particular approach works best.</p>
+</section>
+<section id="what-to-expect">
+<h2>What to Expect<a class="headerlink" href="#what-to-expect" title="Permalink to this heading">#</a></h2>
+<ul class="simple">
+<li><p><strong>Curated Learning Path:</strong> Progress through a curated set of challenges that cover a spectrum of topics, ensuring a well-rounded understanding of data structures and algorithms.</p></li>
+<li><p><strong>Practical Application:</strong> Bridge the gap between theoretical knowledge and real-world application as you tackle challenges that mirror the complexities of technical interviews.</p></li>
+<li><p><strong>Pythonic Wisdom:</strong> Learn to harness the power of Python to craft elegant, efficient, and readable code. Each solution is a lesson in leveraging Python’s strengths for algorithmic problem-solving.</p></li>
+</ul>
+<p>Whether you are preparing for technical interviews, enhancing your coding skills, or simply seeking to deepen your understanding of Python and algorithms, “Hands-On Problem-Solving in Python” is your companion on this enriching journey. Let’s embark on the adventure of mastering the Blind 75 and unlocking the doors to confident, effective problem-solving in Python.</p>
+<p>Happy coding!</p>
+</section>
+<div class="toctree-wrapper compound">
+</div>
+</section>
+
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#building-a-strong-foundation">Building a Strong Foundation</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#what-to-expect">What to Expect</a></li>
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diff --git a/search.html b/search.html
index a7412a1..3452898 100644
--- a/search.html
+++ b/search.html
@@ -140,7 +140,7 @@
         <div class="sidebar-primary-item">
   
 
-<a class="navbar-brand logo" href="README.html">
+<a class="navbar-brand logo" href="intro.html">
   
   
   
@@ -155,100 +155,13 @@
         
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
-                <a class="reference internal" href="README.html">
-                    Blind 75 LeetCode Solutions
+                <a class="reference internal" href="intro.html">
+                    Introduction
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             </li>
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         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/README.html">Array &amp; Hashing Solutions - Blind 75 LeetCode Problems</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
-<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/README.html">Two Pointers Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
-<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/README.html">Sliding Window Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
-<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/README.html">Stack Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/README.html">Binary search Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
-<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/README.html">Linked List Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
-<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/README.html">Trees Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/README.html">Trie Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
-<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/README.html">Heap / Priority Queue Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
-<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/README.html">Backtracking Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
-<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/README.html">Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
-<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/README.html">Advanced Graphs Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
-<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
-</ul>
-</li>
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/searchindex.js b/searchindex.js
index 8321071..f6c61a9 100644
--- a/searchindex.js
+++ b/searchindex.js
@@ -1 +1 @@
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14, 23, 26, 28, 30, 31, 32, 33, 34, 35, 36, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 64, 72], "stack": [0, 18, 26, 30, 32, 33, 34, 36, 37, 38, 39, 40, 41, 44, 45, 51, 52, 54, 55, 56, 58, 59, 72], "dure": [0, 10, 24, 30, 31, 33, 34, 36, 39, 40, 41, 44, 45, 49, 54, 55, 59, 61, 72, 73], "proport": [0, 23, 24, 26, 27, 28, 30, 31, 39, 40, 44, 65], "maximum": [0, 1, 4, 5, 6, 9, 13, 14, 15, 16, 20, 30, 31, 35, 40, 41, 42, 44, 45, 48, 51, 56, 58, 59, 61, 63, 64, 66, 68, 75], "depth": [0, 30, 34, 36, 38, 39, 40, 41, 42, 44, 45, 51, 52, 54, 55, 56, 58, 59, 75], "complet": [0, 2, 10, 23, 24, 25, 26, 27, 30, 31, 32, 33, 34, 35, 36, 38, 39, 41, 43, 44, 45, 47, 51, 52, 55, 57], "unbalanc": [0, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41], "all": [0, 1, 2, 3, 4, 6, 9, 10, 11, 13, 14, 16, 18, 20, 21, 24, 25, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 40, 41, 43, 44, 45, 47, 48, 51, 52, 54, 55, 56, 57, 58, 59, 61, 64, 73, 74, 75], "form": [0, 4, 6, 9, 24, 25, 26, 27, 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39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76], "unsort": 1, "arrai": [1, 2, 5, 6, 9, 11, 13, 16, 22, 28, 33, 47, 48, 49, 51, 57, 58, 63, 64, 65, 72], "integ": [1, 2, 3, 5, 9, 11, 14, 15, 18, 20, 21, 23, 33, 37, 39, 47, 48, 49, 51, 54, 59, 63, 64, 65, 71, 75], "num": [1, 2, 3, 5, 11, 20, 21, 48, 49, 63, 64], "length": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 25, 27, 28, 31, 32, 33, 35, 39, 43, 44, 45, 47, 49, 51, 52, 56, 57, 59, 61, 63, 64, 65], "element": [1, 2, 3, 6, 8, 9, 11, 18, 20, 21, 24, 25, 26, 27, 28, 33, 39, 42, 48, 50, 51, 57, 58, 59, 72, 75], "you": [1, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 74, 75, 76], "must": [1, 3, 5, 11, 13, 16, 18, 20, 21, 38, 41, 45, 54, 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13, 15, 16, 23, 24, 27, 38, 39, 40, 41, 44, 47, 48, 49, 56, 59, 63, 64, 73], "effici": [1, 2, 3, 4, 5, 6, 7, 9, 11, 14, 15, 20, 23, 25, 26, 27, 28, 31, 35, 37, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 52, 54, 55, 56, 57, 59, 63, 64], "exist": [1, 2, 18, 27, 31, 33, 35, 37, 38, 40, 43, 44, 52, 56, 61, 74], "keep": [1, 2, 3, 10, 13, 14, 15, 16, 18, 20, 27, 28, 34, 35, 36, 37, 39, 40, 43, 47, 51, 54, 55, 56, 57, 58, 59, 61, 63], "track": [1, 2, 3, 13, 14, 15, 16, 18, 20, 27, 28, 34, 35, 37, 39, 43, 47, 49, 51, 54, 55, 56, 57, 58, 59, 61, 63], "found": [1, 11, 14, 16, 33, 34, 37, 40, 43, 44, 45, 47, 51, 52, 56, 58, 72, 76], "iter": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 20, 23, 24, 25, 26, 27, 28, 31, 33, 35, 36, 37, 43, 44, 45, 47, 49, 51, 52, 54, 56, 57, 58, 59, 61, 63, 64, 65], "through": [1, 2, 3, 4, 5, 6, 7, 10, 11, 13, 14, 15, 16, 18, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 43, 44, 45, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 73], "For": [1, 2, 3, 4, 5, 6, 10, 14, 16, 18, 20, 21, 24, 26, 27, 31, 34, 35, 36, 40, 41, 44, 45, 51, 52, 54, 57, 58, 61, 63, 65, 72, 73], "doesn": [1, 24, 28, 40, 44, 48, 51, 52, 54, 56], "t": [1, 4, 9, 10, 13, 16, 21, 23, 24, 28, 37, 38, 39, 40, 43, 44, 45, 48, 51, 52, 54, 56, 58, 72], "mean": [1, 2, 4, 6, 9, 10, 11, 14, 15, 16, 18, 20, 21, 23, 26, 27, 32, 33, 35, 37, 38, 39, 48, 51, 52, 54, 55, 56, 57, 58, 59, 61, 64], "potenti": [1, 11, 18, 37, 43], "insid": [1, 2, 4, 6, 9, 10, 11, 14, 15, 16, 18, 20, 21, 23, 24, 26, 27, 28, 31, 34, 35, 38, 41, 45, 47, 51, 52, 54, 58, 63], "loop": [1, 2, 3, 4, 5, 6, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 31, 33, 35, 37, 38, 45, 47, 51, 52, 54, 55, 56, 57, 58, 61, 63, 65, 72], "itself": [1, 3, 7, 20, 26, 38, 40], "part": [1, 21, 41, 44, 58, 63], "enter": [1, 4, 11, 20, 23, 26, 27, 31, 35, 37, 47, 51], "continu": [1, 2, 9, 10, 11, 14, 16, 18, 20, 21, 23, 25, 26, 27, 28, 31, 33, 34, 35, 37, 43, 45, 47, 57, 61, 73], "long": [1, 47], "increment": [1, 4, 5, 6, 10, 11, 16, 48, 56, 57, 58, 65], "also": [1, 2, 3, 5, 6, 7, 9, 10, 13, 16, 18, 21, 24, 25, 26, 31, 33, 34, 35, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 57, 58, 59, 61, 72, 73], "account": [1, 14, 44, 73, 74], "updat": [1, 3, 4, 7, 9, 13, 14, 15, 16, 20, 21, 24, 25, 26, 27, 28, 34, 41, 47, 49, 51, 57, 58, 61, 63, 64, 72], "longer": [1, 4, 9], "after": [1, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 21, 23, 24, 25, 26, 27, 28, 31, 33, 35, 37, 39, 43, 44, 45, 47, 51, 52, 55, 57, 58, 59, 61], "move": [1, 9, 10, 11, 15, 16, 23, 24, 25, 26, 27, 37, 38, 43, 44, 48, 51, 52, 56], "outer": [1, 11, 52], "repeat": [1, 2, 17, 20, 24, 54, 56, 72, 75], "process": [1, 2, 6, 7, 9, 11, 18, 20, 21, 24, 26, 27, 28, 31, 33, 35, 36, 37, 39, 43, 44, 48, 51, 54, 56, 57, 58, 59, 61], "result": [1, 3, 4, 5, 6, 7, 9, 10, 11, 18, 20, 21, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 36, 37, 38, 40, 44, 45, 47, 48, 51, 52, 54, 59, 61, 63, 64, 65, 72], "kei": [1, 5, 6, 16, 18, 37, 39, 41, 43, 44, 54, 56, 59, 61, 63, 72, 76], "idea": [1, 37, 63, 74], "By": [1, 7, 15, 34, 37, 49, 74, 75], "do": [1, 4, 10, 13, 15, 16, 18, 21, 23, 24, 25, 27, 34, 35, 37, 39, 47, 48, 51, 55, 56, 63, 65, 74, 76], "find": [1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 21, 22, 23, 24, 29, 33, 34, 35, 37, 38, 42, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 74, 75], "nums1": [1, 2, 20, 21, 63, 64], "100": [1, 6, 25, 27, 32, 35, 36, 42, 48, 54, 63, 64, 72, 75], "200": [1, 59, 60, 72, 75], "nums2": [1, 2, 20, 21, 63, 64], "7": [1, 9, 13, 20, 21, 31, 32, 33, 34, 36, 38, 47, 51, 59, 63, 72], "8": [1, 3, 9, 38, 47, 49, 51, 72], "6": [1, 3, 9, 13, 20, 21, 26, 28, 34, 36, 37, 38, 41, 51, 52, 58, 59, 72], "let": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 61, 63, 64], "provid": [1, 2, 3, 4, 5, 7, 13, 14, 15, 18, 20, 21, 26, 28, 32, 36, 38, 39, 40, 43, 44, 45, 47, 48, 49, 54, 55, 56, 58, 59, 61, 63, 64, 65, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76], "design": [1, 7, 23, 24, 39, 46, 49, 54, 56, 57, 61, 63, 72, 75], "breakdown": [1, 48, 59], "construct": [1, 3, 31, 35, 38, 42, 45, 52, 55, 57, 61, 73, 75], "perform": [1, 2, 4, 6, 13, 14, 15, 16, 18, 21, 25, 27, 28, 31, 33, 34, 35, 37, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 52, 54, 56, 57, 58, 59, 61, 63, 65], "oper": [1, 2, 3, 6, 7, 11, 13, 14, 15, 16, 18, 27, 28, 34, 35, 37, 39, 43, 44, 47, 48, 49, 63, 65], "main": [1, 9, 16, 32, 33, 38, 43, 45, 49, 51, 56, 58, 61, 64, 72, 74], "onli": [1, 5, 9, 10, 11, 13, 15, 18, 23, 24, 26, 28, 34, 35, 38, 40, 41, 43, 47, 49, 52, 54, 55, 56, 59, 63, 64, 65], "mai": [1, 5, 7, 9, 11, 15, 16, 20, 23, 24, 40, 41, 43, 44, 45, 49, 51, 52, 55, 56, 57, 58, 61, 72, 73, 74, 76], "howev": [1, 23, 28, 38, 40, 43, 44, 51, 52, 56, 61, 64], "nest": [1, 11, 45, 51, 52, 56, 72], "its": [1, 3, 5, 6, 7, 14, 16, 18, 25, 26, 27, 28, 31, 32, 33, 34, 35, 36, 38, 39, 40, 41, 43, 45, 47, 48, 49, 54, 55, 56, 57, 58, 59, 61, 64, 67, 68, 69, 70, 71, 75], "remain": [1, 2, 23, 25, 26, 27, 28, 44, 47, 48, 51, 57], "few": [1, 4, 13, 14, 16, 27, 28, 51, 64], "store": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 24, 25, 26, 27, 28, 31, 32, 33, 35, 37, 39, 43, 44, 45, 48, 49, 51, 54, 55, 57, 59, 61, 63, 65], "uniqu": [1, 2, 4, 5, 11, 14, 16, 20, 21, 33, 34, 38, 41, 45, 51, 52, 54, 55, 57, 61, 67, 75], "requir": [1, 2, 3, 4, 9, 10, 11, 16, 28, 35, 43, 44, 45, 47, 48, 49, 51, 54, 57, 65, 72], "depend": [1, 4, 6, 9, 10, 11, 13, 15, 16, 20, 21, 23, 25, 27, 28, 32, 33, 34, 35, 36, 37, 40, 41, 44, 45, 48, 54, 55, 57, 61, 63], "size": [1, 2, 4, 5, 6, 9, 10, 11, 13, 15, 16, 20, 21, 23, 24, 25, 26, 27, 28, 35, 38, 40, 45, 48, 49, 56, 58, 59, 61, 63, 64, 65, 73], "actual": [1, 16, 27, 41, 43, 47, 48, 51, 54, 55, 61, 76], "extend": [1, 5, 13, 15, 27, 33, 34, 35, 38, 44, 45, 48, 61, 63], "certain": [1, 47, 51, 63, 64], "absolut": 1, "e": [1, 5, 16, 18, 21, 23, 25, 26, 31, 33, 34, 37, 39, 43, 45, 47, 48, 51, 52, 54, 55, 57, 58, 61, 65, 72, 73], "g": [1, 5, 21, 25, 33, 45, 47, 54, 55, 65], "less": [1, 9, 10, 11, 15, 20, 21, 38, 40, 41, 43, 45, 47, 48, 51, 59], "than": [1, 2, 5, 9, 10, 11, 13, 15, 16, 20, 21, 27, 34, 37, 38, 40, 41, 43, 45, 47, 48, 51, 52, 54, 57, 59, 65], "k": [1, 2, 6, 8, 11, 13, 14, 15, 21, 24, 26, 27, 29, 37, 38, 45, 47, 49, 52, 63, 72, 75], "instead": [1, 5, 9, 11, 16, 21, 27, 28, 34, 35, 37, 47, 49, 51, 52, 61, 65, 72], "ani": [2, 3, 5, 6, 7, 9, 10, 13, 15, 16, 18, 21, 23, 24, 27, 28, 31, 33, 34, 39, 40, 41, 43, 44, 45, 48, 49, 51, 52, 55, 57, 58, 59, 72, 73, 74, 76], "appear": [2, 24, 25, 27, 33, 34, 55, 61, 73], "least": [2, 48, 49, 51, 58], "twice": [2, 64], "distinct": [2, 6, 14, 16, 21, 49, 51, 56, 58, 65], "containsdupl": 2, "unique_el": 2, "alreadi": [2, 4, 14, 43, 55, 72], "otherwis": [2, 4, 10, 18, 21, 23, 39, 40, 41, 43, 44, 45, 48, 51, 52, 55, 57, 61, 73], "add": [2, 8, 12, 14, 16, 17, 19, 22, 29, 33, 34, 35, 39, 42, 43, 45, 46, 47, 48, 49, 50, 51, 53, 55, 57, 58, 60, 61, 62, 66, 67, 68, 69, 70, 71, 74, 75], "els": [2, 9, 11, 18, 20, 21, 27, 28, 33, 38, 47, 48, 54], "without": [2, 3, 5, 10, 16, 17, 21, 23, 24, 25, 38, 45, 51, 52, 56, 63, 64, 72, 73, 75], "argument": [2, 5, 13, 14, 18, 21, 23, 32, 34, 36, 38, 40, 51, 52, 58], "whether": [2, 4, 15, 40, 41, 43, 44, 52, 55, 56, 57, 58, 61], "condit": [2, 11, 15, 20, 24, 38, 41, 51, 52, 55, 56, 59, 61], "immedi": [2, 4, 43, 55], "ad": [2, 5, 7, 27, 32, 35, 44, 45, 48, 49, 54, 57, 58, 59, 61, 63, 65, 72], "util": [2, 45, 63, 72], "data": [2, 9, 14, 21, 24, 27, 28, 31, 33, 36, 38, 39, 43, 45, 46, 49, 50, 54, 55, 58, 61, 64, 67, 68, 69, 70, 71, 72, 75, 76], "quickli": [2, 49], "detect": [2, 23, 40, 55, 57, 61], "meet": [2, 11, 49, 69, 75], "explain": [2, 24, 25, 26, 27, 43, 44], "earlier": [2, 44, 65], "nums3": [2, 20, 21, 64], "primari": [2, 14, 52], "affect": [2, 6, 44, 48, 72, 73], "averag": [2, 7, 41, 45, 48], "grow": [2, 20, 48], "linearli": [2, 11, 35], "modifi": [2, 3, 5, 7, 9, 11, 13, 14, 15, 18, 20, 24, 25, 26, 31, 33, 34, 35, 37, 39, 41, 45, 47, 49, 51, 52, 54, 55, 56, 57, 59, 61, 63, 64], "paramet": [2, 23, 26, 37, 49, 52, 57], "frequenc": [2, 4, 5, 51], "count": [2, 3, 4, 5, 6, 11, 15, 16, 32, 34, 43, 51, 56, 57, 58, 65, 71, 75], "list": [2, 3, 4, 5, 6, 7, 10, 11, 13, 18, 20, 31, 33, 35, 36, 37, 38, 39, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 72, 74], "more": [2, 7, 8, 12, 17, 19, 22, 25, 26, 29, 38, 39, 41, 42, 45, 46, 47, 48, 50, 52, 53, 56, 57, 58, 60, 61, 62, 66, 67, 68, 69, 70, 71, 72, 74, 75], "along": [2, 32, 63, 64], "answer": [3, 5, 6, 14, 16, 47, 48, 73], "prefix": [3, 44, 45, 46, 61, 75], "suffix": 3, "guarante": [3, 5, 33, 49], "fit": 3, "32": [3, 72], "bit": [3, 39], "divis": [3, 20, 21], "30": [3, 25, 47, 51, 72], "up": [3, 4, 5, 11, 16, 23, 24, 25, 26, 33, 35, 37, 39, 45, 47, 48, 51, 52, 54, 55, 56, 58, 63, 76], "extra": [3, 9, 16, 23, 25, 26, 27, 38, 40, 63, 72], "output": [3, 5, 9, 11, 25, 26, 28, 31, 35, 39, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65, 72], "doe": [3, 6, 9, 10, 11, 13, 15, 18, 20, 21, 25, 27, 28, 34, 38, 43, 44, 45, 48, 49, 52, 55, 57, 58, 59], "productexceptself": 3, "len": [3, 4, 5, 7, 9, 10, 11, 14, 15, 16, 18, 20, 21, 28, 31, 35, 45, 47, 48, 49, 51, 52, 54, 55, 56, 59, 61, 64], "left_product": 3, "right_product": 3, "calcul": [3, 9, 11, 13, 14, 16, 20, 21, 32, 34, 47, 48, 57, 58, 63, 64, 65], "fill": [3, 9, 65], "ones": [3, 49], "A": [3, 10, 23, 26, 32, 33, 34, 38, 39, 40, 41, 43, 45, 51, 52, 55, 57, 58, 72, 73, 74, 76], "index": [3, 11, 14, 15, 20, 21, 23, 24, 33, 37, 39, 51, 52, 54, 72], "includ": [3, 10, 15, 16, 26, 34, 39, 44, 45, 49, 51, 54, 56, 61, 63, 72, 73, 74, 76], "accordingli": [3, 57, 61], "anoth": [3, 11, 37, 39, 42, 57, 58, 61, 65, 72, 75], "revers": [3, 5, 24, 29, 31, 39, 45, 71, 75], "order": [3, 5, 6, 11, 18, 20, 21, 24, 25, 26, 27, 28, 33, 34, 35, 37, 39, 42, 45, 48, 49, 51, 61, 75], "multipli": 3, "correspond": [3, 5, 6, 18, 33, 43, 45, 51, 54, 59], "contain": [3, 4, 5, 6, 7, 8, 11, 12, 15, 16, 18, 20, 21, 24, 25, 26, 27, 28, 31, 33, 34, 35, 38, 39, 40, 41, 44, 47, 48, 51, 54, 57, 58, 59, 61, 64, 67, 68, 69, 70, 71, 72, 75], "desir": [3, 37, 76], "comput": [3, 39, 48, 64], "It": [3, 5, 6, 9, 10, 11, 13, 14, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 36, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "result1": [3, 5, 6, 9, 21, 28, 56], "24": [3, 72], "12": [3, 9, 45, 63, 72], "result2": [3, 5, 6, 9, 21, 28, 56], "first": [3, 4, 5, 6, 13, 23, 24, 25, 26, 27, 31, 32, 33, 34, 35, 36, 37, 39, 41, 44, 45, 49, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "second": [3, 5, 24, 25, 37, 41, 47, 54, 65, 72], "independ": [3, 25, 27, 48, 63], "sequenti": [3, 21, 45, 52], "consum": [3, 10, 14, 15, 27, 38, 52, 55], "respect": [3, 4, 16, 24, 26, 27, 32, 40, 44, 51, 52, 57, 63, 65, 67, 68, 69, 70, 71, 73, 75], "These": [3, 4, 15, 16, 21, 23, 24, 25, 26, 27, 35, 48, 55, 61, 64, 67, 68, 69, 70, 71, 75], "occupi": [3, 24, 28, 65], "toward": [3, 10, 11, 73], "per": [3, 5, 51], "handl": [3, 5, 7, 11, 15, 24, 25, 33, 36, 41, 47, 49, 54, 61, 64, 72], "zero": [3, 11, 14, 26, 51, 56, 70, 75], "should": [3, 5, 7, 10, 20, 21, 23, 26, 27, 31, 39, 41, 43, 44, 47, 48, 52, 54, 55, 56, 57, 61, 65], "place": [3, 5, 24, 26, 52, 64], "modif": [3, 11], "attempt": [3, 16, 76], "directli": [3, 43, 59, 65, 72], "string": [4, 6, 8, 10, 14, 15, 16, 18, 39, 43, 44, 45, 52, 61, 72, 75], "word": [4, 6, 43, 46, 53, 57, 58, 59, 61, 66, 75], "phrase": [4, 6, 10], "rearrang": [4, 6, 27], "letter": [4, 6, 10, 14, 15, 16, 43, 44, 45, 52, 61], "typic": [4, 6, 36, 37, 41, 51, 54, 55], "origin": [4, 6, 7, 10, 20, 24, 26, 39, 45, 51, 52, 54, 55, 57, 74], "consist": [4, 6, 10, 14, 15, 16, 18, 33, 40, 43, 44, 45, 52, 56, 58, 61, 72, 74], "lowercas": [4, 6, 10, 15, 16, 43, 44, 45, 52, 61], "english": [4, 6, 14, 15, 16, 43, 44, 45, 52, 72], "what": [4, 73, 76], "unicod": 4, "charact": [4, 6, 7, 10, 16, 17, 18, 39, 43, 44, 45, 52, 61, 72, 75], "how": [4, 16, 18, 20, 23, 24, 26, 28, 31, 32, 34, 36, 37, 39, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65, 74, 76], "would": [4, 13, 14, 20, 24, 34, 37, 41, 48, 49, 56, 57], "adapt": [4, 20, 21, 41, 73], "your": [4, 5, 13, 21, 33, 39, 47, 48, 52, 55, 58, 63, 67, 68, 69, 70, 71, 72, 74, 75, 76], "isanagram": 4, "str": [4, 6, 7, 39, 44, 52, 56], "bool": [4, 23, 44, 52, 57], "dictionari": [4, 5, 6, 14, 15, 16, 18, 43, 44, 45, 54, 62, 75], "countt": 4, "get": [4, 5, 15, 16, 31, 35, 45, 47, 54, 56], "same": [4, 5, 6, 7, 10, 11, 15, 18, 25, 26, 31, 33, 35, 36, 38, 39, 40, 41, 42, 43, 44, 45, 48, 51, 52, 54, 55, 58, 61, 63, 64, 75], "method": [4, 5, 6, 7, 18, 23, 28, 34, 39, 40, 43, 44, 48, 49, 52, 54, 56, 57, 59, 63, 64, 65], "boolean": [4, 43, 52, 58, 59], "other": [4, 6, 10, 11, 15, 16, 21, 24, 35, 38, 40, 45, 51, 54, 55, 56, 57, 58, 59, 73], 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41, 43, 44, 45, 47, 49, 51, 52, 54, 55, 56, 58, 59, 61], "which": [0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65, 73], "each": [0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 75, 77], "ha": [0, 2, 4, 5, 6, 9, 16, 18, 21, 23, 24, 25, 26, 27, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 43, 44, 45, 47, 48, 51, 52, 54, 55, 56, 57, 58, 63, 64, 76], "s": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 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2, 4, 5, 6, 7, 14, 23, 26, 28, 30, 31, 32, 33, 34, 35, 36, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 64, 72], "stack": [0, 18, 26, 30, 32, 33, 34, 36, 37, 38, 39, 40, 41, 44, 45, 51, 52, 54, 55, 56, 58, 59, 72], "dure": [0, 10, 24, 30, 31, 33, 34, 36, 39, 40, 41, 44, 45, 49, 54, 55, 59, 61, 73], "proport": [0, 23, 24, 26, 27, 28, 30, 31, 39, 40, 44, 65], "maximum": [0, 1, 4, 5, 6, 9, 13, 14, 15, 16, 20, 30, 31, 35, 40, 41, 42, 44, 45, 48, 51, 56, 58, 59, 61, 63, 64, 66, 68, 75], "depth": [0, 30, 34, 36, 38, 39, 40, 41, 42, 44, 45, 51, 52, 54, 55, 56, 58, 59, 75], "complet": [0, 2, 10, 23, 24, 25, 26, 27, 30, 31, 32, 33, 34, 35, 36, 38, 39, 41, 43, 44, 45, 47, 51, 52, 55, 57], "unbalanc": [0, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41], "all": [0, 1, 2, 3, 4, 6, 9, 10, 11, 13, 14, 16, 18, 20, 21, 24, 25, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 40, 41, 43, 44, 45, 47, 48, 51, 52, 54, 55, 56, 57, 58, 59, 61, 64, 73, 74, 75], "form": [0, 4, 6, 9, 24, 25, 26, 27, 30, 45, 51, 52, 55, 56, 57, 61], "singl": [0, 2, 6, 7, 13, 14, 15, 18, 20, 25, 26, 28, 30, 32, 34, 37, 45, 58, 65, 73], "branch": [0, 30, 32, 44, 51, 74], "therefor": [0, 1, 2, 3, 4, 5, 9, 10, 13, 14, 15, 16, 18, 20, 21, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 39, 41, 43, 45, 47, 49, 51, 52, 54, 56, 63, 64, 65], "due": [0, 1, 6, 11, 14, 18, 24, 26, 28, 30, 33, 37, 39, 40, 41, 43, 47, 51, 52, 55, 61], "addit": [0, 1, 2, 3, 4, 8, 9, 10, 11, 12, 13, 15, 17, 19, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 32, 33, 38, 42, 43, 46, 48, 50, 52, 53, 55, 56, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 75, 76], "small": [0, 30, 32, 52, 59], "constant": [0, 1, 4, 6, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 30, 32, 34, 35, 38, 39, 47, 48, 49, 63, 64, 65], "amount": [0, 9, 11, 13, 15, 16, 20, 21, 25, 26, 27, 28, 30, 32, 34, 38, 39, 51, 52, 63, 64], "variabl": [0, 1, 3, 4, 5, 6, 9, 10, 11, 13, 14, 15, 16, 20, 21, 23, 24, 25, 27, 28, 30, 32, 34, 35, 38, 47, 49, 51, 52, 56, 58, 63, 64, 65], "comparison": [0, 4, 10, 11, 13, 14, 27, 28, 30, 39, 61], "within": [0, 1, 3, 6, 10, 13, 14, 15, 16, 21, 24, 30, 34, 37, 39, 40, 41, 45, 47, 48, 51, 54, 58, 73, 76], "signific": [0, 16, 30, 34, 40], "term": [0, 5, 18, 24, 26, 30, 31, 33, 37, 47, 56, 65, 67, 68, 69, 70, 71, 73, 75], "overal": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 14, 16, 18, 21, 24, 25, 28, 30, 31, 33, 34, 35, 39, 40, 41, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65, 73], "summari": [0, 2, 4, 5, 6, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 36, 38, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 58, 61, 63, 64, 65], "medium": [1, 3, 5, 6, 7, 8, 9, 11, 12, 14, 15, 17, 20, 21, 22, 24, 25, 29, 31, 33, 35, 37, 38, 41, 42, 43, 44, 46, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 63, 64, 66, 67, 68, 69, 70, 71, 75], "an": [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76], "unsort": 1, "arrai": [1, 2, 5, 6, 9, 11, 13, 16, 22, 28, 33, 47, 48, 49, 51, 57, 58, 63, 64, 65, 72], "integ": [1, 2, 3, 5, 9, 11, 14, 15, 18, 20, 21, 23, 33, 37, 39, 47, 48, 49, 51, 54, 59, 63, 64, 65, 71, 75], "num": [1, 2, 3, 5, 11, 20, 21, 48, 49, 63, 64], "length": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 25, 27, 28, 31, 32, 33, 35, 39, 43, 44, 45, 47, 49, 51, 52, 56, 57, 59, 61, 63, 64, 65], "element": [1, 2, 3, 6, 8, 9, 11, 18, 20, 21, 24, 25, 26, 27, 28, 33, 39, 42, 48, 50, 51, 57, 58, 59, 72, 75], "you": [1, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77], "must": [1, 3, 5, 11, 13, 16, 18, 20, 21, 38, 41, 45, 54, 55, 57], "algorithm": [1, 3, 5, 7, 9, 11, 14, 15, 16, 20, 21, 23, 24, 34, 38, 39, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65, 67, 68, 69, 70, 71, 75, 77], "run": [1, 3, 9, 10, 11, 14, 15, 16, 20, 27, 38, 47, 63], "5": [1, 2, 3, 4, 5, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 35, 37, 38, 39, 40, 41, 48, 49, 51, 55, 58, 59, 63, 66, 72, 75, 76], "9": [1, 2, 3, 9, 31, 32, 33, 34, 36, 38, 49, 63, 72], "i": [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 16, 18, 20, 21, 23, 25, 26, 28, 31, 33, 34, 35, 37, 39, 43, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "python": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 18, 24, 26, 27, 28, 31, 33, 34, 35, 36, 37, 38, 39, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65, 72, 77], "solv": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 20, 23, 31, 33, 34, 35, 37, 38, 39, 41, 45, 48, 49, 52, 54, 55, 59, 61, 63, 64, 67, 68, 69, 70, 71, 75, 77], "longestconsecut": 1, 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"allow": [1, 2, 7, 13, 15, 16, 23, 24, 27, 38, 39, 40, 41, 44, 47, 48, 49, 56, 59, 63, 64, 73], "effici": [1, 2, 3, 4, 5, 6, 7, 9, 11, 14, 15, 20, 23, 25, 26, 27, 28, 31, 35, 37, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 52, 54, 55, 56, 57, 59, 63, 64, 77], "exist": [1, 2, 18, 27, 31, 33, 35, 37, 38, 40, 43, 44, 52, 56, 61, 74], "keep": [1, 2, 3, 10, 13, 14, 15, 16, 18, 20, 27, 28, 34, 35, 36, 37, 39, 40, 43, 47, 51, 54, 55, 56, 57, 58, 59, 61, 63], "track": [1, 2, 3, 13, 14, 15, 16, 18, 20, 27, 28, 34, 35, 37, 39, 43, 47, 49, 51, 54, 55, 56, 57, 58, 59, 61, 63], "found": [1, 11, 14, 16, 33, 34, 37, 40, 43, 44, 45, 47, 51, 52, 56, 58, 72, 76], "iter": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 20, 23, 24, 25, 26, 27, 28, 31, 33, 35, 36, 37, 43, 44, 45, 47, 49, 51, 52, 54, 56, 57, 58, 59, 61, 63, 64, 65], "through": [1, 2, 3, 4, 5, 6, 7, 10, 11, 13, 14, 15, 16, 18, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 43, 44, 45, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 73, 77], "For": [1, 2, 3, 4, 5, 6, 10, 14, 16, 18, 20, 21, 24, 26, 27, 31, 34, 35, 36, 40, 41, 44, 45, 51, 52, 54, 57, 58, 61, 63, 65, 73], "doesn": [1, 24, 28, 40, 44, 48, 51, 52, 54, 56], "t": [1, 4, 9, 10, 13, 16, 21, 23, 24, 28, 37, 38, 39, 40, 43, 44, 45, 48, 51, 52, 54, 56, 58], "mean": [1, 2, 4, 6, 9, 10, 11, 14, 15, 16, 18, 20, 21, 23, 26, 27, 32, 33, 35, 37, 38, 39, 48, 51, 52, 54, 55, 56, 57, 58, 59, 61, 64], "potenti": [1, 11, 18, 37, 43], "insid": [1, 2, 4, 6, 9, 10, 11, 14, 15, 16, 18, 20, 21, 23, 24, 26, 27, 28, 31, 34, 35, 38, 41, 45, 47, 51, 52, 54, 58, 63], "loop": [1, 2, 3, 4, 5, 6, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 31, 33, 35, 37, 38, 45, 47, 51, 52, 54, 55, 56, 57, 58, 61, 63, 65], "itself": [1, 3, 7, 20, 26, 38, 40], "part": [1, 21, 41, 44, 58, 63], "enter": [1, 4, 11, 20, 23, 26, 27, 31, 35, 37, 47, 51], "continu": [1, 2, 9, 10, 11, 14, 16, 18, 20, 21, 23, 25, 26, 27, 28, 31, 33, 34, 35, 37, 43, 45, 47, 57, 61, 73], "long": [1, 47], "increment": [1, 4, 5, 6, 10, 11, 16, 48, 56, 57, 58, 65], "also": [1, 2, 3, 5, 6, 7, 9, 10, 13, 16, 18, 21, 24, 25, 26, 31, 33, 34, 35, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 57, 58, 59, 61, 73, 77], "account": [1, 14, 44, 73, 74], "updat": [1, 3, 4, 7, 9, 13, 14, 15, 16, 20, 21, 24, 25, 26, 27, 28, 34, 41, 47, 49, 51, 57, 58, 61, 63, 64, 72], "longer": [1, 4, 9], "after": [1, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 21, 23, 24, 25, 26, 27, 28, 31, 33, 35, 37, 39, 43, 44, 45, 47, 51, 52, 55, 57, 58, 59, 61], "move": [1, 9, 10, 11, 15, 16, 23, 24, 25, 26, 27, 37, 38, 43, 44, 48, 51, 52, 56], "outer": [1, 11, 52], "repeat": [1, 2, 17, 20, 24, 54, 56, 72, 75], "process": [1, 2, 6, 7, 9, 11, 18, 20, 21, 24, 26, 27, 28, 31, 33, 35, 36, 37, 39, 43, 44, 48, 51, 54, 56, 57, 58, 59, 61], "result": [1, 3, 4, 5, 6, 7, 9, 10, 11, 18, 20, 21, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 36, 37, 38, 40, 44, 45, 47, 48, 51, 52, 54, 59, 61, 63, 64, 65], "kei": [1, 5, 6, 16, 18, 37, 39, 41, 43, 44, 54, 56, 59, 61, 63, 72, 76], "idea": [1, 37, 63, 74], "By": [1, 7, 15, 34, 37, 49, 74, 75], "do": [1, 4, 10, 13, 15, 16, 18, 21, 23, 24, 25, 27, 34, 35, 37, 39, 47, 48, 51, 55, 56, 63, 65, 74, 76], "find": [1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 21, 22, 23, 24, 29, 33, 34, 35, 37, 38, 42, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 74, 75], "nums1": [1, 2, 20, 21, 63, 64], "100": [1, 6, 25, 27, 32, 35, 36, 42, 48, 54, 63, 64, 75], "200": [1, 59, 60, 72, 75], "nums2": [1, 2, 20, 21, 63, 64], "7": [1, 9, 13, 20, 21, 31, 32, 33, 34, 36, 38, 47, 51, 59, 63, 72], "8": [1, 3, 9, 38, 47, 49, 51, 72], "6": [1, 3, 9, 13, 20, 21, 26, 28, 34, 36, 37, 38, 41, 51, 52, 58, 59, 72], "let": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 61, 63, 64, 77], "provid": [1, 2, 3, 4, 5, 7, 13, 14, 15, 18, 20, 21, 26, 28, 32, 36, 38, 39, 40, 43, 44, 45, 47, 48, 49, 54, 55, 56, 58, 59, 61, 63, 64, 65, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77], "design": [1, 7, 23, 24, 39, 46, 49, 54, 56, 57, 61, 63, 72, 75, 77], "breakdown": [1, 48, 59], "construct": [1, 3, 31, 35, 38, 42, 45, 52, 55, 57, 61, 73, 75], "perform": [1, 2, 4, 6, 13, 14, 15, 16, 18, 21, 25, 27, 28, 31, 33, 34, 35, 37, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 52, 54, 56, 57, 58, 59, 61, 63, 65], "oper": [1, 2, 3, 6, 7, 11, 13, 14, 15, 16, 18, 27, 28, 34, 35, 37, 39, 43, 44, 47, 48, 49, 63, 65], "main": [1, 9, 16, 32, 33, 38, 43, 45, 49, 51, 56, 58, 61, 64, 72, 74], "onli": [1, 5, 9, 10, 11, 13, 15, 18, 23, 24, 26, 28, 34, 35, 38, 40, 41, 43, 47, 49, 52, 54, 55, 56, 59, 63, 64, 65, 77], "mai": [1, 5, 7, 9, 11, 15, 16, 20, 23, 24, 40, 41, 43, 44, 45, 49, 51, 52, 55, 56, 57, 58, 61, 72, 73, 74, 76], "howev": [1, 23, 28, 38, 40, 43, 44, 51, 52, 56, 61, 64], "nest": [1, 11, 45, 51, 52, 56, 72], "its": [1, 3, 5, 6, 7, 14, 16, 18, 25, 26, 27, 28, 31, 32, 33, 34, 35, 36, 38, 39, 40, 41, 43, 45, 47, 48, 49, 54, 55, 56, 57, 58, 59, 61, 64, 67, 68, 69, 70, 71, 75], "remain": [1, 2, 23, 25, 26, 27, 28, 44, 47, 48, 51, 57], "few": [1, 4, 13, 14, 16, 27, 28, 51, 64], "store": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 24, 25, 26, 27, 28, 31, 32, 33, 35, 37, 39, 43, 44, 45, 48, 49, 51, 54, 55, 57, 59, 61, 63, 65], "uniqu": [1, 2, 4, 5, 11, 14, 16, 20, 21, 33, 34, 38, 41, 45, 51, 52, 54, 55, 57, 61, 67, 75], "requir": [1, 2, 3, 4, 9, 10, 11, 16, 28, 35, 43, 44, 45, 47, 48, 49, 51, 54, 57, 65, 72], "depend": [1, 4, 6, 9, 10, 11, 13, 15, 16, 20, 21, 23, 25, 27, 28, 32, 33, 34, 35, 36, 37, 40, 41, 44, 45, 48, 54, 55, 57, 61, 63], "size": [1, 2, 4, 5, 6, 9, 10, 11, 13, 15, 16, 20, 21, 23, 24, 25, 26, 27, 28, 35, 38, 40, 45, 48, 49, 56, 58, 59, 61, 63, 64, 65, 73], "actual": [1, 16, 27, 41, 43, 47, 48, 51, 54, 55, 61, 76], "extend": [1, 5, 13, 15, 27, 33, 34, 35, 38, 44, 45, 48, 61, 63], "certain": [1, 47, 51, 63, 64], "absolut": 1, "e": [1, 5, 16, 18, 21, 23, 25, 26, 31, 33, 34, 37, 39, 43, 45, 47, 48, 51, 52, 54, 55, 57, 58, 61, 65, 73], "g": [1, 5, 21, 25, 33, 45, 47, 54, 55, 65], "less": [1, 9, 10, 11, 15, 20, 21, 38, 40, 41, 43, 45, 47, 48, 51, 59], "than": [1, 2, 5, 9, 10, 11, 13, 15, 16, 20, 21, 27, 34, 37, 38, 40, 41, 43, 45, 47, 48, 51, 52, 54, 57, 59, 65, 77], "k": [1, 2, 6, 8, 11, 13, 14, 15, 21, 24, 26, 27, 29, 37, 38, 45, 47, 49, 52, 63, 72, 75], "instead": [1, 5, 9, 11, 16, 21, 27, 28, 34, 35, 37, 47, 49, 51, 52, 61, 65], "ani": [2, 3, 5, 6, 7, 9, 10, 13, 15, 16, 18, 21, 23, 24, 27, 28, 31, 33, 34, 39, 40, 41, 43, 44, 45, 48, 49, 51, 52, 55, 57, 58, 59, 73, 74, 76], "appear": [2, 24, 25, 27, 33, 34, 55, 61, 73], "least": [2, 48, 49, 51, 58], "twice": [2, 64], "distinct": [2, 6, 14, 16, 21, 49, 51, 56, 58, 65], "containsdupl": 2, "unique_el": 2, "alreadi": [2, 4, 14, 43, 55, 72], "otherwis": [2, 4, 10, 18, 21, 23, 39, 40, 41, 43, 44, 45, 48, 51, 52, 55, 57, 61, 73], "add": [2, 8, 12, 14, 16, 17, 19, 22, 29, 33, 34, 35, 39, 42, 43, 45, 46, 47, 48, 49, 50, 51, 53, 55, 57, 58, 60, 61, 62, 66, 67, 68, 69, 70, 71, 74, 75], "els": [2, 9, 11, 18, 20, 21, 27, 28, 33, 38, 47, 48, 54], "without": [2, 3, 5, 10, 16, 17, 21, 23, 24, 25, 38, 45, 51, 52, 56, 63, 64, 72, 73, 75], "argument": [2, 5, 13, 14, 18, 21, 23, 32, 34, 36, 38, 40, 51, 52, 58], "whether": [2, 4, 15, 40, 41, 43, 44, 52, 55, 56, 57, 58, 61, 77], "condit": [2, 11, 15, 20, 24, 38, 41, 51, 52, 55, 56, 59, 61], "immedi": [2, 4, 43, 55], "ad": [2, 5, 7, 27, 32, 35, 44, 45, 48, 49, 54, 57, 58, 59, 61, 63, 65, 72], "util": [2, 45, 63], "data": [2, 9, 14, 21, 24, 27, 28, 31, 33, 36, 38, 39, 43, 45, 46, 49, 50, 54, 55, 58, 61, 64, 67, 68, 69, 70, 71, 72, 75, 76, 77], "quickli": [2, 49], "detect": [2, 23, 40, 55, 57, 61], "meet": [2, 11, 49, 69, 75], "explain": [2, 24, 25, 26, 27, 43, 44], "earlier": [2, 44, 65], "nums3": [2, 20, 21, 64], "primari": [2, 14, 52], "affect": [2, 6, 44, 48, 73], "averag": [2, 7, 41, 45, 48], "grow": [2, 20, 48], "linearli": [2, 11, 35], "modifi": [2, 3, 5, 7, 9, 11, 13, 14, 15, 18, 20, 24, 25, 26, 31, 33, 34, 35, 37, 39, 41, 45, 47, 49, 51, 52, 54, 55, 56, 57, 59, 61, 63, 64], "paramet": [2, 23, 26, 37, 49, 52, 57], "frequenc": [2, 4, 5, 51, 77], "count": [2, 3, 4, 5, 6, 11, 15, 16, 32, 34, 43, 51, 56, 57, 58, 65, 71, 75], "list": [2, 3, 4, 5, 6, 7, 10, 11, 13, 18, 20, 31, 33, 35, 36, 37, 38, 39, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 72, 74, 77], "more": [2, 7, 8, 12, 17, 19, 22, 25, 26, 29, 38, 39, 41, 42, 45, 46, 47, 48, 50, 52, 53, 56, 57, 58, 60, 61, 62, 66, 67, 68, 69, 70, 71, 74, 75, 77], "along": [2, 32, 63, 64], "answer": [3, 5, 6, 14, 16, 47, 48, 73], "prefix": [3, 44, 45, 46, 61, 75], "suffix": 3, "guarante": [3, 5, 33, 49], "fit": 3, "32": [3, 72], "bit": [3, 39], "divis": [3, 20, 21], "30": [3, 25, 47, 51, 72], "up": [3, 4, 5, 11, 16, 23, 24, 25, 26, 33, 35, 37, 39, 45, 47, 48, 51, 52, 54, 55, 56, 58, 63, 76], "extra": [3, 9, 16, 23, 25, 26, 27, 38, 40, 63], "output": [3, 5, 9, 11, 25, 26, 28, 31, 35, 39, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "doe": [3, 6, 9, 10, 11, 13, 15, 18, 20, 21, 25, 27, 28, 34, 38, 43, 44, 45, 48, 49, 52, 55, 57, 58, 59], "productexceptself": 3, "len": [3, 4, 5, 7, 9, 10, 11, 14, 15, 16, 18, 20, 21, 28, 31, 35, 45, 47, 48, 49, 51, 52, 54, 55, 56, 59, 61, 64], "left_product": 3, "right_product": 3, "calcul": [3, 9, 11, 13, 14, 16, 20, 21, 32, 34, 47, 48, 57, 58, 63, 64, 65], "fill": [3, 9, 65], "ones": [3, 49], "A": [3, 10, 23, 26, 32, 33, 34, 38, 39, 40, 41, 43, 45, 51, 52, 55, 57, 58, 73, 74, 76], "index": [3, 11, 14, 15, 20, 21, 23, 24, 33, 37, 39, 51, 52, 54], "includ": [3, 10, 15, 16, 26, 34, 39, 44, 45, 49, 51, 54, 56, 61, 63, 73, 74, 76], "accordingli": [3, 57, 61], "anoth": [3, 11, 37, 39, 42, 57, 58, 61, 65, 75], "revers": [3, 5, 24, 29, 31, 39, 45, 71, 72, 75], "order": [3, 5, 6, 11, 18, 20, 21, 24, 25, 26, 27, 28, 33, 34, 35, 37, 39, 42, 45, 48, 49, 51, 61, 75], "multipli": 3, "correspond": [3, 5, 6, 18, 33, 43, 45, 51, 54, 59], "contain": [3, 4, 5, 6, 7, 8, 11, 12, 15, 16, 18, 20, 21, 24, 25, 26, 27, 28, 31, 33, 34, 35, 38, 39, 40, 41, 44, 47, 48, 51, 54, 57, 58, 59, 61, 64, 67, 68, 69, 70, 71, 72, 75], "desir": [3, 37, 76], "comput": [3, 39, 48, 64], "It": [3, 5, 6, 9, 10, 11, 13, 14, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 36, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "result1": [3, 5, 6, 9, 21, 28, 56], "24": [3, 72], "12": [3, 9, 45, 63, 72], "result2": [3, 5, 6, 9, 21, 28, 56], "first": [3, 4, 5, 6, 13, 23, 24, 25, 26, 27, 31, 32, 33, 34, 35, 36, 37, 39, 41, 44, 45, 49, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "second": [3, 5, 24, 25, 37, 41, 47, 54, 65, 72], "independ": [3, 25, 27, 48, 63], "sequenti": [3, 21, 45, 52], "consum": [3, 10, 14, 15, 27, 38, 52, 55], "respect": [3, 4, 16, 24, 26, 27, 32, 40, 44, 51, 52, 57, 63, 65, 67, 68, 69, 70, 71, 73, 75], "These": [3, 4, 15, 16, 21, 23, 24, 25, 26, 27, 35, 48, 55, 61, 64, 67, 68, 69, 70, 71, 75], "occupi": [3, 24, 28, 65], "toward": [3, 10, 11, 73], "per": [3, 5, 51], "handl": [3, 5, 7, 11, 15, 24, 25, 33, 36, 41, 47, 49, 54, 61, 64], "zero": [3, 11, 14, 26, 51, 56, 70, 75], "should": [3, 5, 7, 10, 20, 21, 23, 26, 27, 31, 39, 41, 43, 44, 47, 48, 52, 54, 55, 56, 57, 61, 65], "place": [3, 5, 24, 26, 52, 64], "modif": [3, 11], "attempt": [3, 16, 76], "directli": [3, 43, 59, 65], "string": [4, 6, 8, 10, 14, 15, 16, 18, 39, 43, 44, 45, 52, 61, 72, 75], "word": [4, 6, 43, 46, 53, 57, 58, 59, 61, 66, 75], "phrase": [4, 6, 10], "rearrang": [4, 6, 27], "letter": [4, 6, 10, 14, 15, 16, 43, 44, 45, 52, 61], "typic": [4, 6, 36, 37, 41, 51, 54, 55], "origin": [4, 6, 7, 10, 20, 24, 26, 39, 45, 51, 52, 54, 55, 57, 74], "consist": [4, 6, 10, 14, 15, 16, 18, 33, 40, 43, 44, 45, 52, 56, 58, 61, 74], "lowercas": [4, 6, 10, 15, 16, 43, 44, 45, 52, 61], "english": [4, 6, 14, 15, 16, 43, 44, 45, 52], "what": [4, 73, 76], "unicod": 4, "charact": [4, 6, 7, 10, 16, 17, 18, 39, 43, 44, 45, 52, 61, 72, 75], "how": [4, 16, 18, 20, 23, 24, 26, 28, 31, 32, 34, 36, 37, 39, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65, 74, 76, 77], "would": [4, 13, 14, 20, 24, 34, 37, 41, 48, 49, 56, 57], "adapt": [4, 20, 21, 41, 73], "your": [4, 5, 13, 21, 33, 39, 47, 48, 52, 55, 58, 63, 67, 68, 69, 70, 71, 74, 75, 76, 77], "isanagram": 4, "str": [4, 6, 7, 39, 44, 52, 56], "bool": [4, 23, 44, 52, 57], "dictionari": [4, 5, 6, 14, 15, 16, 18, 43, 44, 45, 54, 62, 75], "countt": 4, "get": [4, 5, 15, 16, 31, 35, 45, 47, 54, 56], "same": [4, 5, 6, 7, 10, 11, 15, 18, 25, 26, 31, 33, 35, 36, 38, 39, 40, 41, 42, 43, 44, 45, 48, 51, 52, 54, 55, 58, 61, 63, 64, 75], "method": [4, 5, 6, 7, 18, 23, 28, 34, 39, 40, 43, 44, 48, 49, 52, 54, 56, 57, 59, 63, 64, 65], "boolean": [4, 43, 52, 58, 59], "other": [4, 6, 10, 11, 15, 16, 21, 24, 35, 38, 40, 45, 51, 54, 55, 56, 57, 58, 59, 73], "occurr": [4, 6, 7, 14, 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\ No newline at end of file

From bcf1c22af16b60d64b0b4df9695fcab4921f55b8 Mon Sep 17 00:00:00 2001
From: Mohsen Tabibian <36014419+mohsentabibian@users.noreply.github.com>
Date: Tue, 14 Nov 2023 18:17:14 +0000
Subject: [PATCH 6/7] Update documentation

---
 01. Array Hashing/1. Two Sum.html             | 113 +++-
 .../128. Longest Consecutive Sequence.html    | 113 +++-
 .../217. Contains Duplicate.html              | 113 +++-
 .../238. Product of Array Except Self.html    | 113 +++-
 01. Array Hashing/242. Valid Anagram.html     | 113 +++-
 .../347. Top K Frequent Elements.html         | 113 +++-
 01. Array Hashing/49. Group Anagrams.html     | 113 +++-
 .../659. Encode and Decode Strings.html       | 113 +++-
 01. Array Hashing/README.html                 |  91 ++-
 01. Array Hashing/intro.html                  | 619 +++++++++++++++++
 .../11. Container With Most Water.html        | 113 +++-
 02. Two Pointers/125. Valid Palindrome.html   | 113 +++-
 02. Two Pointers/15. 3Sum.html                | 113 +++-
 02. Two Pointers/README.html                  |  91 ++-
 02. Two Pointers/intro.html                   | 615 +++++++++++++++++
 .../121. Best Time to Buy and Sell Stock.html | 113 +++-
 ...ubstring Without Repeating Characters.html | 113 +++-
 ...ngest Repeating Character Replacement.html | 113 +++-
 .../76. Minimum Window Substring.html         | 113 +++-
 03. Sliding Window/README.html                |  91 ++-
 03. Sliding Window/intro.html                 | 615 +++++++++++++++++
 04. Stack/20. Valid Parentheses.html          | 113 +++-
 04. Stack/README.html                         |  91 ++-
 04. Stack/intro.html                          | 620 +++++++++++++++++
 ... Find Minimum in Rotated Sorted Array.html | 113 +++-
 .../33. Search in Rotated Sorted Array.html   | 113 +++-
 05. Binary Search/README.html                 |  91 ++-
 05. Binary Search/intro.html                  | 627 +++++++++++++++++
 06. Linked List/141. Linked List Cycle.html   | 113 +++-
 06. Linked List/143. Reorder List.html        | 113 +++-
 .../19. Remove Nth Node From End of List.html | 113 +++-
 06. Linked List/206. Reverse Linked List.html | 113 +++-
 .../21. Merge Two Sorted Lists.html           | 113 +++-
 06. Linked List/23. Merge k Sorted Lists.html | 113 +++-
 06. Linked List/README.html                   |  91 ++-
 06. Linked List/intro.html                    | 618 +++++++++++++++++
 07. Trees/100. Same Tree.html                 | 113 +++-
 ...02. Binary Tree Level Order Traversal.html | 113 +++-
 .../104. Maximum Depth of Binary Tree.html    | 113 +++-
 ...e from Preorder and Inorder Traversal.html | 113 +++-
 .../124. Binary Tree Maximum Path Sum.html    | 113 +++-
 .../199. Binary Tree Right Side View.html     | 113 +++-
 07. Trees/226. Invert Binary Tree.html        | 117 +++-
 .../230. Kth Smallest Element in a BST.html   | 113 +++-
 ...mmon Ancestor of a Binary Search Tree.html | 113 +++-
 ...Serialize and Deserialize Binary Tree.html | 113 +++-
 07. Trees/572. Subtree of Another Tree.html   | 113 +++-
 .../98. Validate Binary Search Tree.html      | 113 +++-
 07. Trees/README.html                         |  91 ++-
 07. Trees/intro.html                          | 623 +++++++++++++++++
 .../208. Implement Trie (Prefix Tree).html    | 113 +++-
 ...n Add and Search Words Data Structure.html | 113 +++-
 08. Tries/212. Word Search II.html            | 113 +++-
 08. Tries/README.html                         |  91 ++-
 08. Tries/intro.html                          | 620 +++++++++++++++++
 .../1046. Last Stone Weight.html              | 113 +++-
 .../295. Find Median from Data Stream.html    | 113 +++-
 .../703. Kth Largest Element in a Stream.html | 113 +++-
 09. Heap - Priority Queue/README.html         |  91 ++-
 09. Heap - Priority Queue/intro.html          | 624 +++++++++++++++++
 10. Backtracking/39. Combination Sum.html     | 113 +++-
 10. Backtracking/79. Word Search.html         | 113 +++-
 10. Backtracking/README.html                  |  91 ++-
 10. Backtracking/intro.html                   | 628 ++++++++++++++++++
 11. Graphs/133. Clone Graph.html              | 113 +++-
 11. Graphs/178. Graph Valid Tree.html         | 113 +++-
 11. Graphs/200. Number of Islands.html        | 113 +++-
 11. Graphs/207. Course Schedule.html          | 113 +++-
 ...ted Components in an Undirected Graph.html | 113 +++-
 .../417. Pacific Atlantic Water Flow.html     | 113 +++-
 11. Graphs/README.html                        |  91 ++-
 11. Graphs/intro.html                         | 618 +++++++++++++++++
 .../892. Alien Dictionary.html                | 113 +++-
 12. Advanced Graphs/README.html               |  91 ++-
 12. Advanced Graphs/intro.html                | 621 +++++++++++++++++
 .../198. House Robber.html                    |  91 ++-
 .../213. House Robber II.html                 |  91 ++-
 .../70. Climbing Stairs.html                  |  91 ++-
 13. One-D Dynamic Programming/README.html     |  97 ++-
 14. Two-D Dynamic Programming/README.html     |  91 ++-
 15. Greedy/README.html                        |  91 ++-
 16. Intervals/README.html                     |  91 ++-
 17. Math Geometry/README.html                 |  91 ++-
 18. Bit Manipulation/README.html              |  91 ++-
 Book.html                                     | 350 +++++++---
 CODE_OF_CONDUCT.html                          |  91 ++-
 CONTRIBUTING.html                             |  91 ++-
 README.html                                   |  91 ++-
 SECURITY.html                                 |  91 ++-
 _sources/01. Array Hashing/intro.md           |  33 +
 _sources/02. Two Pointers/intro.md            |  25 +
 _sources/03. Sliding Window/intro.md          |  25 +
 _sources/04. Stack/intro.md                   |  27 +
 _sources/05. Binary Search/intro.md           |  29 +
 _sources/06. Linked List/intro.md             |  31 +
 _sources/07. Trees/intro.md                   |  41 ++
 _sources/08. Tries/intro.md                   |  27 +
 _sources/09. Heap - Priority Queue/intro.md   |  31 +
 _sources/10. Backtracking/intro.md            |  27 +
 _sources/11. Graphs/intro.md                  |  31 +
 _sources/12. Advanced Graphs/intro.md         |  25 +
 _sources/Book.ipynb                           | 169 +++--
 _sources/intro.md                             |  48 +-
 genindex.html                                 |  91 ++-
 intro.html                                    | 173 ++++-
 objects.inv                                   | Bin 2700 -> 2938 bytes
 search.html                                   |  91 ++-
 searchindex.js                                |   2 +-
 108 files changed, 16356 insertions(+), 416 deletions(-)
 create mode 100644 01. Array Hashing/intro.html
 create mode 100644 02. Two Pointers/intro.html
 create mode 100644 03. Sliding Window/intro.html
 create mode 100644 04. Stack/intro.html
 create mode 100644 05. Binary Search/intro.html
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 create mode 100644 07. Trees/intro.html
 create mode 100644 08. Tries/intro.html
 create mode 100644 09. Heap - Priority Queue/intro.html
 create mode 100644 10. Backtracking/intro.html
 create mode 100644 11. Graphs/intro.html
 create mode 100644 12. Advanced Graphs/intro.html
 create mode 100644 _sources/01. Array Hashing/intro.md
 create mode 100644 _sources/02. Two Pointers/intro.md
 create mode 100644 _sources/03. Sliding Window/intro.md
 create mode 100644 _sources/04. Stack/intro.md
 create mode 100644 _sources/05. Binary Search/intro.md
 create mode 100644 _sources/06. Linked List/intro.md
 create mode 100644 _sources/07. Trees/intro.md
 create mode 100644 _sources/08. Tries/intro.md
 create mode 100644 _sources/09. Heap - Priority Queue/intro.md
 create mode 100644 _sources/10. Backtracking/intro.md
 create mode 100644 _sources/11. Graphs/intro.md
 create mode 100644 _sources/12. Advanced Graphs/intro.md

diff --git a/01. Array Hashing/1. Two Sum.html b/01. Array Hashing/1. Two Sum.html
index d7504ca..9c7592f 100644
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+++ b/01. Array Hashing/1. Two Sum.html	
@@ -67,6 +67,8 @@
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     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="128. Longest Consecutive Sequence" href="128.%20Longest%20Consecutive%20Sequence.html" />
+    <link rel="prev" title="Introduction to Array Hashing Challenges" href="intro.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
   <meta name="docsearch:language" content="en"/>
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -540,6 +629,24 @@ <h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-spac
                 <footer class="prev-next-footer">
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+        <p class="prev-next-title">Introduction to Array Hashing Challenges</p>
+      </div>
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+    <a class="right-next"
+       href="128.%20Longest%20Consecutive%20Sequence.html"
+       title="next page">
+      <div class="prev-next-info">
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diff --git a/01. Array Hashing/128. Longest Consecutive Sequence.html b/01. Array Hashing/128. Longest Consecutive Sequence.html
index c7c9681..3c351b8 100644
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     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="217. Contains Duplicate" href="217.%20Contains%20Duplicate.html" />
+    <link rel="prev" title="100: Same Tree" href="1.%20Two%20Sum.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
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         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -531,6 +620,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+        <p class="prev-next-title">100: Same Tree</p>
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+    <a class="right-next"
+       href="217.%20Contains%20Duplicate.html"
+       title="next page">
+      <div class="prev-next-info">
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diff --git a/01. Array Hashing/217. Contains Duplicate.html b/01. Array Hashing/217. Contains Duplicate.html
index ca448d4..4080886 100644
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+    <link rel="next" title="238. Product of Array Except Self" href="238.%20Product%20of%20Array%20Except%20Self.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
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+                    Introduction: Navigating the Coding Cosmos
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -538,6 +627,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/01. Array Hashing/238. Product of Array Except Self.html b/01. Array Hashing/238. Product of Array Except Self.html
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+    <link rel="next" title="242. Valid Anagram" href="242.%20Valid%20Anagram.html" />
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-                    Introduction
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -534,6 +623,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/01. Array Hashing/242. Valid Anagram.html b/01. Array Hashing/242. Valid Anagram.html
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -525,6 +614,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/01. Array Hashing/347. Top K Frequent Elements.html b/01. Array Hashing/347. Top K Frequent Elements.html
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+<li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -538,6 +627,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+    <a class="left-prev"
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+      <i class="fa-solid fa-angle-left"></i>
+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">242. Valid Anagram</p>
+      </div>
+    </a>
+    <a class="right-next"
+       href="49.%20Group%20Anagrams.html"
+       title="next page">
+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">next</p>
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diff --git a/01. Array Hashing/49. Group Anagrams.html b/01. Array Hashing/49. Group Anagrams.html
index fb49bd4..548efb5 100644
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     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="659· Encode and Decode Strings" href="659.%20Encode%20and%20Decode%20Strings.html" />
+    <link rel="prev" title="347. Top K Frequent Elements" href="347.%20Top%20K%20Frequent%20Elements.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
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+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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diff --git a/01. Array Hashing/README.html b/01. Array Hashing/README.html
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+<li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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+    <h1>Introduction to Array Hashing Challenges</h1>
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+                <h2> Contents </h2>
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+                <ul class="visible nav section-nav flex-column">
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-array-hashing-terrain">Navigating the Array Hashing Terrain</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unlocking-pythonic-solutions">Unlocking Pythonic Solutions</a></li>
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+  <section class="tex2jax_ignore mathjax_ignore" id="introduction-to-array-hashing-challenges">
+<h1>Introduction to Array Hashing Challenges<a class="headerlink" href="#introduction-to-array-hashing-challenges" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Array Hashing section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In this segment, we will explore a curated selection of LeetCode problems focused on array hashing, a powerful technique in algorithmic problem-solving.</p>
+<section id="navigating-the-array-hashing-terrain">
+<h2>Navigating the Array Hashing Terrain<a class="headerlink" href="#navigating-the-array-hashing-terrain" title="Permalink to this heading">#</a></h2>
+<p>Array Hashing provides an efficient approach to solving a variety of problems by transforming complex operations into simpler ones through the use of hash functions. The challenges presented in this section will not only sharpen your skills in manipulating arrays but also introduce you to the art of hashing for optimized problem-solving.</p>
+</section>
+<section id="unveiling-the-challenges">
+<h2>Unveiling the Challenges<a class="headerlink" href="#unveiling-the-challenges" title="Permalink to this heading">#</a></h2>
+<ol class="arabic simple">
+<li><p><strong>Two Sum (Problem #1):</strong> Dive into the fundamentals of array manipulation by identifying pairs of elements that sum up to a target value.</p></li>
+<li><p><strong>Longest Consecutive Sequence (Problem #128):</strong> Explore the concept of consecutive elements in an array and devise an algorithm to find the longest sequence.</p></li>
+<li><p><strong>Contains Duplicate (Problem #217):</strong> Investigate the presence of duplicate elements in an array, a common scenario in coding interviews and real-world applications.</p></li>
+<li><p><strong>Product of Array Except Self (Problem #238):</strong> Master the technique of efficiently computing the product of every element in an array except itself.</p></li>
+<li><p><strong>Valid Anagram (Problem #242):</strong> Delve into string manipulation and explore efficient methods to determine if two strings are valid anagrams.</p></li>
+<li><p><strong>Top K Frequent Elements (Problem #347):</strong> Solve the challenge of identifying the top K frequent elements in an array using hash maps and priority queues.</p></li>
+<li><p><strong>Group Anagrams (Problem #49):</strong> Apply array hashing to categorize an array of strings into groups of anagrams, demonstrating the versatility of this technique.</p></li>
+<li><p><strong>Encode and Decode Strings (Problem #659):</strong> Uncover the intricacies of encoding and decoding strings, a problem that draws on array hashing for efficient solutions.</p></li>
+</ol>
+</section>
+<section id="unlocking-pythonic-solutions">
+<h2>Unlocking Pythonic Solutions<a class="headerlink" href="#unlocking-pythonic-solutions" title="Permalink to this heading">#</a></h2>
+<p>As we tackle each challenge, not only will we unravel the intricacies of array hashing, but we’ll also harness the power of Python to craft elegant and efficient solutions. Get ready to embark on a journey of problem-solving that will not only deepen your understanding of arrays and hashing but also enhance your coding skills in Python.</p>
+<p>Let the exploration of Array Hashing challenges begin!</p>
+<p>Happy coding!</p>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-array-hashing-terrain">Navigating the Array Hashing Terrain</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unlocking-pythonic-solutions">Unlocking Pythonic Solutions</a></li>
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diff --git a/02. Two Pointers/11. Container With Most Water.html b/02. Two Pointers/11. Container With Most Water.html
index 6a15259..9d624ba 100644
--- a/02. Two Pointers/11. Container With Most Water.html	
+++ b/02. Two Pointers/11. Container With Most Water.html	
@@ -67,6 +67,8 @@
     <script>DOCUMENTATION_OPTIONS.pagename = '02. Two Pointers/11. Container With Most Water';</script>
     <link rel="index" title="Index" href="../genindex.html" />
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+    <link rel="next" title="125. Valid Palindrome" href="125.%20Valid%20Palindrome.html" />
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   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Two Pointers Technique Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -545,6 +634,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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diff --git a/02. Two Pointers/125. Valid Palindrome.html b/02. Two Pointers/125. Valid Palindrome.html
index 30e9feb..23d25c9 100644
--- a/02. Two Pointers/125. Valid Palindrome.html	
+++ b/02. Two Pointers/125. Valid Palindrome.html	
@@ -67,6 +67,8 @@
     <script>DOCUMENTATION_OPTIONS.pagename = '02. Two Pointers/125. Valid Palindrome';</script>
     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="15. 3Sum" href="15.%203Sum.html" />
+    <link rel="prev" title="11. Container With Most Water" href="11.%20Container%20With%20Most%20Water.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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   </head>
@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Two Pointers Technique Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -545,6 +634,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+    <a class="left-prev"
+       href="11.%20Container%20With%20Most%20Water.html"
+       title="previous page">
+      <i class="fa-solid fa-angle-left"></i>
+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">11. Container With Most Water</p>
+      </div>
+    </a>
+    <a class="right-next"
+       href="15.%203Sum.html"
+       title="next page">
+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">next</p>
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diff --git a/02. Two Pointers/15. 3Sum.html b/02. Two Pointers/15. 3Sum.html
index 556bb00..c0fb8d8 100644
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+++ b/02. Two Pointers/15. 3Sum.html	
@@ -67,6 +67,8 @@
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     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="Introduction to Sliding Window Challenges" href="../03.%20Sliding%20Window/intro.html" />
+    <link rel="prev" title="125. Valid Palindrome" href="125.%20Valid%20Palindrome.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Two Pointers Technique Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -568,6 +657,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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diff --git a/02. Two Pointers/README.html b/02. Two Pointers/README.html
index a9f799d..1ebd112 100644
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+++ b/02. Two Pointers/README.html	
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unleashing-the-efficiency-of-two-pointers">Unleashing the Efficiency of Two Pointers</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exploring-the-challenges">Exploring the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#guiding-principles-and-pythonic-wisdom">Guiding Principles and Pythonic Wisdom</a></li>
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+<h1>Introduction to Two Pointers Technique Challenges<a class="headerlink" href="#introduction-to-two-pointers-technique-challenges" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Two Pointers section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” The Two Pointers technique is a powerful strategy that enhances your ability to efficiently solve problems by maintaining two pointers that traverse a data structure, often an array or a linked list.</p>
+<section id="unleashing-the-efficiency-of-two-pointers">
+<h2>Unleashing the Efficiency of Two Pointers<a class="headerlink" href="#unleashing-the-efficiency-of-two-pointers" title="Permalink to this heading">#</a></h2>
+<p>The Two Pointers technique is a versatile approach known for its effectiveness in optimizing time and space complexity. As we delve into the challenges in this section, you’ll discover how to leverage this technique to navigate through arrays and efficiently address a variety of problems.</p>
+</section>
+<section id="exploring-the-challenges">
+<h2>Exploring the Challenges<a class="headerlink" href="#exploring-the-challenges" title="Permalink to this heading">#</a></h2>
+<ol class="arabic simple">
+<li><p><strong>Container With Most Water (Problem #11):</strong> Immerse yourself in the concept of maximizing the area between two vertical lines in a histogram, showcasing the elegance and efficiency of the Two Pointers technique.</p></li>
+<li><p><strong>Valid Palindrome (Problem #125):</strong> Explore the realm of string manipulation and learn how Two Pointers can be employed to determine whether a given string is a valid palindrome.</p></li>
+<li><p><strong>3Sum (Problem #15):</strong> Embark on a journey to find all unique triplets in an array that sum up to a target value, unraveling the complexities of Three Sum problems with the Two Pointers technique.</p></li>
+</ol>
+</section>
+<section id="guiding-principles-and-pythonic-wisdom">
+<h2>Guiding Principles and Pythonic Wisdom<a class="headerlink" href="#guiding-principles-and-pythonic-wisdom" title="Permalink to this heading">#</a></h2>
+<p>As we tackle each challenge, we’ll not only unravel the intricacies of the Two Pointers technique but also embrace Pythonic solutions that showcase the elegance and readability of Python code. Through practical examples and detailed explanations, you’ll gain a deep understanding of how to apply Two Pointers to real-world coding scenarios.</p>
+<p>Prepare to sharpen your problem-solving skills, enhance your algorithmic thinking, and master the art of using Two Pointers to tackle a diverse array of challenges.</p>
+<p>Let the exploration of Two Pointers challenges commence!</p>
+<p>Happy coding!</p>
+</section>
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+</div>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exploring-the-challenges">Exploring the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#guiding-principles-and-pythonic-wisdom">Guiding Principles and Pythonic Wisdom</a></li>
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diff --git a/03. Sliding Window/121. Best Time to Buy and Sell Stock.html b/03. Sliding Window/121. Best Time to Buy and Sell Stock.html
index f253741..a6cee27 100644
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+    <link rel="next" title="3. Longest Substring Without Repeating Characters" href="3.%20Longest%20Substring%20Without%20Repeating%20Characters.html" />
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -529,6 +618,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/03. Sliding Window/3. Longest Substring Without Repeating Characters.html b/03. Sliding Window/3. Longest Substring Without Repeating Characters.html
index d0f92a1..b5a29ef 100644
--- a/03. Sliding Window/3. Longest Substring Without Repeating Characters.html	
+++ b/03. Sliding Window/3. Longest Substring Without Repeating Characters.html	
@@ -67,6 +67,8 @@
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+    <link rel="next" title="424. Longest Repeating Character Replacement" href="424.%20Longest%20Repeating%20Character%20Replacement.html" />
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         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Sliding Window Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -548,6 +637,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/03. Sliding Window/424. Longest Repeating Character Replacement.html b/03. Sliding Window/424. Longest Repeating Character Replacement.html
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--- a/03. Sliding Window/424. Longest Repeating Character Replacement.html	
+++ b/03. Sliding Window/424. Longest Repeating Character Replacement.html	
@@ -67,6 +67,8 @@
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+    <link rel="next" title="76. Minimum Window Substring" href="76.%20Minimum%20Window%20Substring.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
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-        <ul class="nav bd-sidenav">
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+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Sliding Window Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -534,6 +623,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+       title="previous page">
+      <i class="fa-solid fa-angle-left"></i>
+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">3. Longest Substring Without Repeating Characters</p>
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+       title="next page">
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diff --git a/03. Sliding Window/76. Minimum Window Substring.html b/03. Sliding Window/76. Minimum Window Substring.html
index 3b30aa6..51218d8 100644
--- a/03. Sliding Window/76. Minimum Window Substring.html	
+++ b/03. Sliding Window/76. Minimum Window Substring.html	
@@ -67,6 +67,8 @@
     <script>DOCUMENTATION_OPTIONS.pagename = '03. Sliding Window/76. Minimum Window Substring';</script>
     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="Introduction to Stack Challenges: Unveiling Valid Parentheses" href="../04.%20Stack/intro.html" />
+    <link rel="prev" title="424. Longest Repeating Character Replacement" href="424.%20Longest%20Repeating%20Character%20Replacement.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Sliding Window Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -580,6 +669,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/03. Sliding Window/README.html b/03. Sliding Window/README.html
index 397bb7f..ab7d86c 100644
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+++ b/03. Sliding Window/README.html	
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
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         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/03. Sliding Window/intro.html b/03. Sliding Window/intro.html
new file mode 100644
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--- /dev/null
+++ b/03. Sliding Window/intro.html	
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+
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+<!DOCTYPE html>
+
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+    <meta charset="utf-8" />
+    <meta name="viewport" content="width=device-width, initial-scale=1.0" /><meta name="generator" content="Docutils 0.18.1: http://docutils.sourceforge.net/" />
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+    <title>Introduction to Sliding Window Challenges &#8212; Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges</title>
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+  <script data-cfasync="false">
+    document.documentElement.dataset.mode = localStorage.getItem("mode") || "";
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-sliding-window-approach">Navigating the Sliding Window Approach</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-efficiency-unleashed">Pythonic Efficiency Unleashed</a></li>
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+<h1>Introduction to Sliding Window Challenges<a class="headerlink" href="#introduction-to-sliding-window-challenges" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Sliding Window section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In this chapter, we will delve into the powerful technique of Sliding Window, a strategy designed to efficiently solve problems by maintaining a dynamic window over a sequence.</p>
+<section id="navigating-the-sliding-window-approach">
+<h2>Navigating the Sliding Window Approach<a class="headerlink" href="#navigating-the-sliding-window-approach" title="Permalink to this heading">#</a></h2>
+<p>The Sliding Window technique is particularly effective for tackling problems that involve arrays or strings by efficiently processing a continuous subarray or substring. As we navigate through the challenges in this section, you’ll witness how this approach optimizes both time and space complexity, making it an invaluable tool in your problem-solving arsenal.</p>
+</section>
+<section id="unveiling-the-challenges">
+<h2>Unveiling the Challenges<a class="headerlink" href="#unveiling-the-challenges" title="Permalink to this heading">#</a></h2>
+<ol class="arabic simple">
+<li><p><strong>Best Time to Buy and Sell Stock (Problem #121):</strong> Explore the dynamic world of stock prices and optimize your buying and selling strategy to maximize profit.</p></li>
+<li><p><strong>Longest Substring Without Repeating Characters (Problem #3):</strong> Dive into string manipulation as you discover the longest substring without any repeating characters using the Sliding Window technique.</p></li>
+<li><p><strong>Longest Repeating Character Replacement (Problem #424):</strong> Investigate the art of replacing characters in a string to create the longest substring with the same character, unveiling the versatility of the Sliding Window approach.</p></li>
+<li><p><strong>Minimum Window Substring (Problem #76):</strong> Tackle the challenge of finding the minimum window in a string that contains all characters of another string, showcasing the adaptability of Sliding Window for string-related problems.</p></li>
+</ol>
+</section>
+<section id="pythonic-efficiency-unleashed">
+<h2>Pythonic Efficiency Unleashed<a class="headerlink" href="#pythonic-efficiency-unleashed" title="Permalink to this heading">#</a></h2>
+<p>As we unravel each challenge, we’ll not only grasp the intricacies of the Sliding Window technique but also unleash the power of Python to implement elegant and efficient solutions. Get ready to witness how this dynamic approach can transform complex problems into manageable and optimized solutions.</p>
+<p>Let’s embark on the exploration of Sliding Window challenges, where dynamic problem-solving meets the efficiency of the sliding window!</p>
+<p>Happy coding!</p>
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diff --git a/04. Stack/20. Valid Parentheses.html b/04. Stack/20. Valid Parentheses.html
index 7e450f3..97a1a19 100644
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+++ b/04. Stack/20. Valid Parentheses.html	
@@ -67,6 +67,8 @@
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -542,6 +631,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/04. Stack/README.html b/04. Stack/README.html
index e7e06b5..03efbc1 100644
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+++ b/04. Stack/README.html	
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/04. Stack/intro.html b/04. Stack/intro.html
new file mode 100644
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-stack-landscape">Navigating the Stack Landscape</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenge-valid-parentheses-problem-20">Unveiling the Challenge: Valid Parentheses (Problem #20)</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#what-to-expect">What to Expect</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unlocking-the-valid-parentheses-challenge">Unlocking the Valid Parentheses Challenge</a></li>
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+<h1>Introduction to Stack Challenges: Unveiling Valid Parentheses<a class="headerlink" href="#introduction-to-stack-challenges-unveiling-valid-parentheses" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Stack section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” This segment is dedicated to exploring the versatility and efficiency of the stack data structure, a fundamental tool in the arsenal of every proficient coder.</p>
+<section id="navigating-the-stack-landscape">
+<h2>Navigating the Stack Landscape<a class="headerlink" href="#navigating-the-stack-landscape" title="Permalink to this heading">#</a></h2>
+<p>In this journey, we will unravel the power of stacks, a data structure that follows the Last-In, First-Out (LIFO) principle. The challenges presented in this section will not only enhance your understanding of stacks but also showcase their applicability in solving a variety of real-world problems.</p>
+</section>
+<section id="unveiling-the-challenge-valid-parentheses-problem-20">
+<h2>Unveiling the Challenge: Valid Parentheses (Problem #20)<a class="headerlink" href="#unveiling-the-challenge-valid-parentheses-problem-20" title="Permalink to this heading">#</a></h2>
+<p>Our focus for this section is on the “Valid Parentheses” problem. As you delve into this challenge, you’ll encounter a scenario commonly faced in coding interviews and real-world applications: determining the validity of a string containing various types of parentheses—curly braces, square brackets, and parentheses.</p>
+</section>
+<section id="what-to-expect">
+<h2>What to Expect<a class="headerlink" href="#what-to-expect" title="Permalink to this heading">#</a></h2>
+<ul class="simple">
+<li><p><strong>Fundamental Stack Operations:</strong> Dive into the core operations of stacks, such as push and pop, to efficiently solve problems that involve nested structures.</p></li>
+<li><p><strong>Balancing Act:</strong> Explore the concept of balancing parentheses and how stacks provide an elegant solution to ensure the correct order of opening and closing brackets.</p></li>
+<li><p><strong>Pythonic Solutions:</strong> Leverage the strengths of Python to implement clean and concise code that showcases the simplicity and power of the stack data structure.</p></li>
+</ul>
+</section>
+<section id="unlocking-the-valid-parentheses-challenge">
+<h2>Unlocking the Valid Parentheses Challenge<a class="headerlink" href="#unlocking-the-valid-parentheses-challenge" title="Permalink to this heading">#</a></h2>
+<p>As we embark on the journey of solving the “Valid Parentheses” problem, you’ll not only gain a deeper understanding of stack operations but also witness how this seemingly simple data structure can be a robust tool for solving complex problems.</p>
+<p>Prepare to strengthen your foundation in stack-based problem-solving and discover the elegance of Pythonic solutions. Let’s unravel the mysteries of stacks together!</p>
+<p>Happy coding!</p>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenge-valid-parentheses-problem-20">Unveiling the Challenge: Valid Parentheses (Problem #20)</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#what-to-expect">What to Expect</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unlocking-the-valid-parentheses-challenge">Unlocking the Valid Parentheses Challenge</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
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+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Binary Search Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
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+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
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+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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diff --git a/05. Binary Search/README.html b/05. Binary Search/README.html
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
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+                <ul class="visible nav section-nav flex-column">
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-binary-search-landscape">Navigating the Binary Search Landscape</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a><ul class="nav section-nav flex-column">
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+  <section class="tex2jax_ignore mathjax_ignore" id="introduction-to-binary-search-challenges">
+<h1>Introduction to Binary Search Challenges<a class="headerlink" href="#introduction-to-binary-search-challenges" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Binary Search section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In this segment, we’ll unravel the power of binary search, a fundamental algorithmic technique that is both elegant and efficient. The challenges presented here will sharpen your skills in applying binary search to unique problem scenarios.</p>
+<section id="navigating-the-binary-search-landscape">
+<h2>Navigating the Binary Search Landscape<a class="headerlink" href="#navigating-the-binary-search-landscape" title="Permalink to this heading">#</a></h2>
+<p>Binary search is a divide-and-conquer algorithm that plays a crucial role in solving problems efficiently, especially in scenarios where the input is sorted or can be manipulated to exhibit a sorted structure. As we delve into this section, we’ll explore the nuances of binary search through hands-on problem-solving.</p>
+</section>
+<section id="unveiling-the-challenges">
+<h2>Unveiling the Challenges<a class="headerlink" href="#unveiling-the-challenges" title="Permalink to this heading">#</a></h2>
+<ol class="arabic simple">
+<li><p><strong>Find Minimum in Rotated Sorted Array (Problem #153):</strong> Encounter the intricacies of searching for the minimum element in a rotated and sorted array. This challenge will test your ability to adapt binary search to unconventional data structures.</p></li>
+<li><p><strong>Search in Rotated Sorted Array (Problem #33):</strong> Embark on a journey to efficiently locate a target element in a rotated and sorted array. This problem showcases the versatility of binary search in scenarios beyond the traditional sorted arrays.</p></li>
+</ol>
+<section id="pythonic-solutions-await">
+<h3>Pythonic Solutions Await<a class="headerlink" href="#pythonic-solutions-await" title="Permalink to this heading">#</a></h3>
+<p>As we immerse ourselves in these challenges, the focus goes beyond mere problem-solving. We’ll leverage the Python programming language to craft elegant, readable, and efficient solutions. Each problem serves as a canvas for Pythonic coding techniques, showcasing how the language’s features can be harnessed for optimal algorithmic solutions.</p>
+</section>
+</section>
+<section id="what-to-expect">
+<h2>What to Expect<a class="headerlink" href="#what-to-expect" title="Permalink to this heading">#</a></h2>
+<ul class="simple">
+<li><p><strong>Algorithmic Mastery:</strong> Develop a deep understanding of binary search and its application in diverse problem scenarios.</p></li>
+<li><p><strong>Adaptability:</strong> Learn to adapt binary search to unconventional data structures, broadening your problem-solving toolkit.</p></li>
+<li><p><strong>Pythonic Coding:</strong> Immerse yourself in Pythonic coding practices, gaining insights into writing clean and effective code.</p></li>
+</ul>
+<p>Let the exploration of Binary Search challenges commence! Whether you’re a seasoned coder or just beginning your algorithmic journey, these challenges will provide a solid foundation in the art of binary search.</p>
+<p>Happy coding!</p>
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diff --git a/06. Linked List/141. Linked List Cycle.html b/06. Linked List/141. Linked List Cycle.html
index 8d40ea8..1d137ab 100644
--- a/06. Linked List/141. Linked List Cycle.html	
+++ b/06. Linked List/141. Linked List Cycle.html	
@@ -67,6 +67,8 @@
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     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="143. Reorder List" href="143.%20Reorder%20List.html" />
+    <link rel="prev" title="Introduction to Linked List Mastery" href="intro.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Linked List Mastery</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -575,6 +664,24 @@ <h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this
                 <footer class="prev-next-footer">
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+       title="next page">
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diff --git a/06. Linked List/143. Reorder List.html b/06. Linked List/143. Reorder List.html
index 2a8df5b..ceb88f4 100644
--- a/06. Linked List/143. Reorder List.html	
+++ b/06. Linked List/143. Reorder List.html	
@@ -67,6 +67,8 @@
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     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="19. Remove Nth Node From End of List" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html" />
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   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Linked List Mastery</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -617,6 +706,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+      <div class="prev-next-info">
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+        <p class="prev-next-title">141. Linked List Cycle</p>
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diff --git a/06. Linked List/19. Remove Nth Node From End of List.html b/06. Linked List/19. Remove Nth Node From End of List.html
index 5203afc..4a91bff 100644
--- a/06. Linked List/19. Remove Nth Node From End of List.html	
+++ b/06. Linked List/19. Remove Nth Node From End of List.html	
@@ -67,6 +67,8 @@
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     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="206. Reverse Linked List" href="206.%20Reverse%20Linked%20List.html" />
+    <link rel="prev" title="143. Reorder List" href="143.%20Reorder%20List.html" />
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Linked List Mastery</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -595,6 +684,24 @@ <h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this
                 <footer class="prev-next-footer">
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+      <i class="fa-solid fa-angle-left"></i>
+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">143. Reorder List</p>
+      </div>
+    </a>
+    <a class="right-next"
+       href="206.%20Reverse%20Linked%20List.html"
+       title="next page">
+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">next</p>
+        <p class="prev-next-title">206. Reverse Linked List</p>
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diff --git a/06. Linked List/206. Reverse Linked List.html b/06. Linked List/206. Reverse Linked List.html
index 6c1d33b..7cc900c 100644
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+++ b/06. Linked List/206. Reverse Linked List.html	
@@ -67,6 +67,8 @@
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     <link rel="index" title="Index" href="../genindex.html" />
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+    <link rel="next" title="21. Merge Two Sorted Lists" href="21.%20Merge%20Two%20Sorted%20Lists.html" />
+    <link rel="prev" title="19. Remove Nth Node From End of List" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html" />
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
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-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
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         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Linked List Mastery</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -650,6 +739,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+        <p class="prev-next-title">19. Remove Nth Node From End of List</p>
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diff --git a/06. Linked List/21. Merge Two Sorted Lists.html b/06. Linked List/21. Merge Two Sorted Lists.html
index 4248338..ac740d1 100644
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+++ b/06. Linked List/21. Merge Two Sorted Lists.html	
@@ -67,6 +67,8 @@
     <script>DOCUMENTATION_OPTIONS.pagename = '06. Linked List/21. Merge Two Sorted Lists';</script>
     <link rel="index" title="Index" href="../genindex.html" />
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+    <link rel="next" title="23. Merge k Sorted Lists" href="23.%20Merge%20k%20Sorted%20Lists.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
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-        <ul class="nav bd-sidenav">
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+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Linked List Mastery</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -574,6 +663,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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+      <i class="fa-solid fa-angle-left"></i>
+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">206. Reverse Linked List</p>
+      </div>
+    </a>
+    <a class="right-next"
+       href="23.%20Merge%20k%20Sorted%20Lists.html"
+       title="next page">
+      <div class="prev-next-info">
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diff --git a/06. Linked List/23. Merge k Sorted Lists.html b/06. Linked List/23. Merge k Sorted Lists.html
index 40bbb11..5c3da21 100644
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+    <link rel="next" title="Introduction to Tree Challenges" href="../07.%20Trees/intro.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Linked List Mastery</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -603,6 +692,24 @@ <h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this
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+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">21. Merge Two Sorted Lists</p>
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diff --git a/06. Linked List/README.html b/06. Linked List/README.html
index 2b2b62f..b3e9426 100644
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@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/06. Linked List/intro.html b/06. Linked List/intro.html
new file mode 100644
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--- /dev/null
+++ b/06. Linked List/intro.html	
@@ -0,0 +1,618 @@
+
+
+<!DOCTYPE html>
+
+
+<html lang="en" >
+
+  <head>
+    <meta charset="utf-8" />
+    <meta name="viewport" content="width=device-width, initial-scale=1.0" /><meta name="generator" content="Docutils 0.18.1: http://docutils.sourceforge.net/" />
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+    <h1>Introduction to Linked List Mastery</h1>
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+                <ul class="visible nav section-nav flex-column">
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-linked-list-landscape">Navigating the Linked List Landscape</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-solutions-unleashed">Pythonic Solutions Unleashed</a></li>
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+  <section class="tex2jax_ignore mathjax_ignore" id="introduction-to-linked-list-mastery">
+<h1>Introduction to Linked List Mastery<a class="headerlink" href="#introduction-to-linked-list-mastery" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Linked List section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In this segment, we embark on a journey through the intricacies of linked lists — a fundamental data structure in computer science and a cornerstone of algorithmic problem-solving.</p>
+<section id="navigating-the-linked-list-landscape">
+<h2>Navigating the Linked List Landscape<a class="headerlink" href="#navigating-the-linked-list-landscape" title="Permalink to this heading">#</a></h2>
+<p>Linked lists offer a dynamic and flexible way to organize and manipulate data, making them a critical asset in the programmer’s toolkit. The challenges presented in this section are crafted to sharpen your understanding of linked list operations and empower you with the skills to tackle complex problems efficiently.</p>
+</section>
+<section id="unveiling-the-challenges">
+<h2>Unveiling the Challenges<a class="headerlink" href="#unveiling-the-challenges" title="Permalink to this heading">#</a></h2>
+<ol class="arabic simple">
+<li><p><strong>Linked List Cycle (Problem #141):</strong> Delve into the exploration of cycles within linked lists, a common challenge in detecting loops and optimizing algorithmic solutions.</p></li>
+<li><p><strong>Reorder List (Problem #143):</strong> Uncover the art of reordering linked list elements, an essential skill that finds applications in various real-world scenarios.</p></li>
+<li><p><strong>Remove Nth Node From End of List (Problem #19):</strong> Tackle the challenge of efficiently removing the nth node from the end of a linked list, mastering traversal and pointer manipulation.</p></li>
+<li><p><strong>Reverse Linked List (Problem #206):</strong> Explore the reversal of linked lists, a fundamental operation with applications in numerous algorithmic problems.</p></li>
+<li><p><strong>Merge Two Sorted Lists (Problem #21):</strong> Combine the power of linked list manipulation and sorting algorithms to merge two sorted lists into a single, ordered list.</p></li>
+<li><p><strong>Merge k Sorted Lists (Problem #23):</strong> Rise to the challenge of merging k sorted linked lists, showcasing your ability to handle complexity in data structures.</p></li>
+</ol>
+</section>
+<section id="pythonic-solutions-unleashed">
+<h2>Pythonic Solutions Unleashed<a class="headerlink" href="#pythonic-solutions-unleashed" title="Permalink to this heading">#</a></h2>
+<p>As we navigate through each challenge, we not only unravel the intricacies of linked list operations but also harness the expressive power of Python to create elegant and efficient solutions. Prepare to witness the marriage of algorithmic thinking and Pythonic coding in a seamless blend.</p>
+<p>Whether you are a coding enthusiast, a job seeker preparing for technical interviews, or a seasoned programmer seeking to deepen your understanding, the Linked List challenges in this section are designed to elevate your problem-solving skills to new heights.</p>
+<p>Let’s dive into the world of linked lists and master the art of problem-solving in Python!</p>
+<p>Happy coding!</p>
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+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -545,6 +634,24 @@ <h2>In summary:<a class="headerlink" href="#in-summary" title="Permalink to this
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+        <p class="prev-next-title">Introduction to Tree Challenges</p>
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diff --git a/07. Trees/102. Binary Tree Level Order Traversal.html b/07. Trees/102. Binary Tree Level Order Traversal.html
index dc82b4f..8605bfc 100644
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+    <link rel="next" title="104. Maximum Depth of Binary Tree" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
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-                    Introduction
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -584,6 +673,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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+        <p class="prev-next-title">100: Same Tree</p>
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diff --git a/07. Trees/104. Maximum Depth of Binary Tree.html b/07. Trees/104. Maximum Depth of Binary Tree.html
index 2948025..0e8b8fd 100644
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+++ b/07. Trees/104. Maximum Depth of Binary Tree.html	
@@ -67,6 +67,8 @@
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -545,6 +634,24 @@ <h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this
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+    <link rel="next" title="124. Binary Tree Maximum Path Sum" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html" />
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             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -618,6 +707,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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+        <p class="prev-next-title">104. Maximum Depth of Binary Tree</p>
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diff --git a/07. Trees/124. Binary Tree Maximum Path Sum.html b/07. Trees/124. Binary Tree Maximum Path Sum.html
index 826864a..0d82697 100644
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     <link rel="index" title="Index" href="../genindex.html" />
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+    <link rel="next" title="199. Binary Tree Right Side View" href="199.%20Binary%20Tree%20Right%20Side%20View.html" />
+    <link rel="prev" title="105. Construct Binary Tree from Preorder and Inorder Traversal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -575,6 +664,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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+        <p class="prev-next-title">105. Construct Binary Tree from Preorder and Inorder Traversal</p>
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diff --git a/07. Trees/199. Binary Tree Right Side View.html b/07. Trees/199. Binary Tree Right Side View.html
index 89095fd..7d9d337 100644
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+    <link rel="next" title="226: Invert Binary Tree" href="226.%20Invert%20Binary%20Tree.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -566,6 +655,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/07. Trees/226. Invert Binary Tree.html b/07. Trees/226. Invert Binary Tree.html
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+++ b/07. Trees/226. Invert Binary Tree.html	
@@ -67,6 +67,8 @@
     <script>DOCUMENTATION_OPTIONS.pagename = '07. Trees/226. Invert Binary Tree';</script>
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+    <link rel="next" title="230. Kth Smallest Element in a BST" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -448,7 +537,7 @@ <h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this
 </div>
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Inverted Tree (Example 1): &lt;__main__.TreeNode object at 0x7f2ef0355930&gt;
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Inverted Tree (Example 2): &lt;__main__.TreeNode object at 0x7f9db83bc2e0&gt;
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Inverted Tree (Example 2): &lt;__main__.TreeNode object at 0x7f2ef0355c00&gt;
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                 <footer class="prev-next-footer">
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+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">199. Binary Tree Right Side View</p>
+      </div>
+    </a>
+    <a class="right-next"
+       href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html"
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diff --git a/07. Trees/230. Kth Smallest Element in a BST.html b/07. Trees/230. Kth Smallest Element in a BST.html
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+    <link rel="next" title="235. Lowest Common Ancestor of a Binary Search Tree" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html" />
+    <link rel="prev" title="226: Invert Binary Tree" href="226.%20Invert%20Binary%20Tree.html" />
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
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+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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diff --git a/07. Trees/297. Serialize and Deserialize Binary Tree.html b/07. Trees/297. Serialize and Deserialize Binary Tree.html
index 046eff8..544e7e8 100644
--- a/07. Trees/297. Serialize and Deserialize Binary Tree.html	
+++ b/07. Trees/297. Serialize and Deserialize Binary Tree.html	
@@ -67,6 +67,8 @@
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     <link rel="index" title="Index" href="../genindex.html" />
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+    <link rel="next" title="572: Subtree of Another Tree" href="572.%20Subtree%20of%20Another%20Tree.html" />
+    <link rel="prev" title="235. Lowest Common Ancestor of a Binary Search Tree" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html" />
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
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         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -655,6 +744,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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diff --git a/07. Trees/572. Subtree of Another Tree.html b/07. Trees/572. Subtree of Another Tree.html
index 17cdfcb..6418426 100644
--- a/07. Trees/572. Subtree of Another Tree.html	
+++ b/07. Trees/572. Subtree of Another Tree.html	
@@ -67,6 +67,8 @@
     <script>DOCUMENTATION_OPTIONS.pagename = '07. Trees/572. Subtree of Another Tree';</script>
     <link rel="index" title="Index" href="../genindex.html" />
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+    <link rel="next" title="98. Validate Binary Search Tree" href="98.%20Validate%20Binary%20Search%20Tree.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
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             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -608,6 +697,24 @@ <h2>Test cases<a class="headerlink" href="#test-cases" title="Permalink to this
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diff --git a/07. Trees/98. Validate Binary Search Tree.html b/07. Trees/98. Validate Binary Search Tree.html
index 3a458e0..ff3f893 100644
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+    <link rel="next" title="Unveiling the Power of Tries: A Journey in Prefix Trees" href="../08.%20Tries/intro.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -557,6 +646,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">572: Subtree of Another Tree</p>
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diff --git a/07. Trees/README.html b/07. Trees/README.html
index c0b53fe..87ca37a 100644
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+++ b/07. Trees/README.html	
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/07. Trees/intro.html b/07. Trees/intro.html
new file mode 100644
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@@ -0,0 +1,623 @@
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+    <meta name="viewport" content="width=device-width, initial-scale=1.0" /><meta name="generator" content="Docutils 0.18.1: http://docutils.sourceforge.net/" />
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-arboreal-adventures">Unveiling the Arboreal Adventures</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-canopy-exploration">Pythonic Canopy Exploration</a></li>
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+  <section class="tex2jax_ignore mathjax_ignore" id="introduction-to-tree-challenges">
+<h1>Introduction to Tree Challenges<a class="headerlink" href="#introduction-to-tree-challenges" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Trees section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” This segment delves into the fascinating realm of trees, exploring a diverse set of problems that will deepen your understanding of tree structures and sharpen your skills in traversing and manipulating them.</p>
+<section id="navigating-the-lush-canopy-of-challenges">
+<h2>Navigating the Lush Canopy of Challenges<a class="headerlink" href="#navigating-the-lush-canopy-of-challenges" title="Permalink to this heading">#</a></h2>
+<p>Trees are a fundamental data structure in computer science, and their applications extend far and wide. The challenges in this section are designed to guide you through various aspects of tree manipulation, from basic operations to advanced traversal techniques and beyond.</p>
+</section>
+<section id="unveiling-the-arboreal-adventures">
+<h2>Unveiling the Arboreal Adventures<a class="headerlink" href="#unveiling-the-arboreal-adventures" title="Permalink to this heading">#</a></h2>
+<ol class="arabic simple">
+<li><p><strong>Same Tree (Problem #100):</strong> Compare two binary trees and determine if they are structurally identical.</p></li>
+<li><p><strong>Binary Tree Level Order Traversal (Problem #102):</strong> Embark on a journey through the levels of a binary tree, unraveling its secrets in an ordered traversal.</p></li>
+<li><p><strong>Maximum Depth of Binary Tree (Problem #104):</strong> Measure the depth of a binary tree, a fundamental metric in understanding its structure.</p></li>
+<li><p><strong>Construct Binary Tree from Preorder and Inorder Traversal (Problem #105):</strong> Unravel the puzzle of reconstructing a binary tree from its preorder and inorder traversals.</p></li>
+<li><p><strong>Binary Tree Maximum Path Sum (Problem #124):</strong> Navigate the complex web of paths in a binary tree and find the maximum sum achievable.</p></li>
+<li><p><strong>Binary Tree Right Side View (Problem #199):</strong> Gain a unique perspective on a binary tree by capturing the view from its right side.</p></li>
+<li><p><strong>Invert Binary Tree (Problem #226):</strong> Explore the transformative power of inverting a binary tree, changing its structure in a profound way.</p></li>
+<li><p><strong>Kth Smallest Element in a BST (Problem #230):</strong> Traverse a binary search tree to find the Kth smallest element, mastering the art of efficient search.</p></li>
+<li><p><strong>Lowest Common Ancestor of a Binary Search Tree (Problem #235):</strong> Discover the common ground in a binary search tree by finding the lowest common ancestor.</p></li>
+<li><p><strong>Serialize and Deserialize Binary Tree (Problem #297):</strong> Encode and decode the intricate structure of a binary tree for efficient storage and retrieval.</p></li>
+<li><p><strong>Subtree of Another Tree (Problem #572):</strong> Investigate whether one tree is a subtree of another, unveiling the subtleties of tree comparison.</p></li>
+<li><p><strong>Validate Binary Search Tree (Problem #98):</strong> Examine the properties of a binary tree to determine if it is a valid binary search tree.</p></li>
+</ol>
+</section>
+<section id="pythonic-canopy-exploration">
+<h2>Pythonic Canopy Exploration<a class="headerlink" href="#pythonic-canopy-exploration" title="Permalink to this heading">#</a></h2>
+<p>As we journey through these arboreal challenges, we’ll not only uncover the intricacies of tree structures and algorithms but also leverage the power of Python to craft elegant and efficient solutions. Get ready to traverse the lush canopy of tree challenges and elevate your problem-solving skills.</p>
+<p>Let the exploration of Tree challenges begin!</p>
+<p>Happy coding!</p>
+</section>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-lush-canopy-of-challenges">Navigating the Lush Canopy of Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-arboreal-adventures">Unveiling the Arboreal Adventures</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-canopy-exploration">Pythonic Canopy Exploration</a></li>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -608,6 +697,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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+        <p class="prev-next-title">Unveiling the Power of Tries: A Journey in Prefix Trees</p>
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diff --git a/08. Tries/211. Design Add and Search Words Data Structure.html b/08. Tries/211. Design Add and Search Words Data Structure.html
index 742f21b..e5710db 100644
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+++ b/08. Tries/211. Design Add and Search Words Data Structure.html	
@@ -67,6 +67,8 @@
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+    <link rel="next" title="212. Word Search II" href="212.%20Word%20Search%20II.html" />
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -622,6 +711,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/08. Tries/212. Word Search II.html b/08. Tries/212. Word Search II.html
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+++ b/08. Tries/212. Word Search II.html	
@@ -67,6 +67,8 @@
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     <link rel="index" title="Index" href="../genindex.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -652,6 +741,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/08. Tries/README.html b/08. Tries/README.html
index 6b5457f..45ee555 100644
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+++ b/08. Tries/README.html	
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
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+<li class="toctree-l1 current active has-children"><a class="current reference internal" href="#">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
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+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
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+    <h1>Unveiling the Power of Tries: A Journey in Prefix Trees</h1>
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+                <ul class="visible nav section-nav flex-column">
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#understanding-the-essence-of-tries">Understanding the Essence of Tries</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-chapters">Navigating the Chapters</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-wisdom-unleashed">Pythonic Wisdom Unleashed</a></li>
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+<h1>Unveiling the Power of Tries: A Journey in Prefix Trees<a class="headerlink" href="#unveiling-the-power-of-tries-a-journey-in-prefix-trees" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Tries section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In this segment, we embark on a fascinating exploration of Tries, also known as Prefix Trees, a powerful data structure that excels in handling strings and words.</p>
+<section id="understanding-the-essence-of-tries">
+<h2>Understanding the Essence of Tries<a class="headerlink" href="#understanding-the-essence-of-tries" title="Permalink to this heading">#</a></h2>
+<p>Tries stand as a testament to the elegance of data structure design, particularly when it comes to storing and retrieving words efficiently. This section delves into the intricacies of Tries, unraveling their structure and demonstrating their application in various problem-solving scenarios.</p>
+</section>
+<section id="navigating-the-chapters">
+<h2>Navigating the Chapters<a class="headerlink" href="#navigating-the-chapters" title="Permalink to this heading">#</a></h2>
+<ol class="arabic simple">
+<li><p><strong>Implement Trie (Prefix Tree) (Problem #208):</strong> Lay the foundation by mastering the implementation of a Trie, a versatile data structure that stores a dynamic set of strings, making it an invaluable tool in a coder’s arsenal.</p></li>
+<li><p><strong>Design Add and Search Words Data Structure (Problem #211):</strong> Explore the versatility of Tries in designing a data structure that supports the addition and search of words with the flexibility to handle wildcard characters.</p></li>
+<li><p><strong>Word Search II (Problem #212):</strong> Elevate your problem-solving skills as we tackle a more advanced challenge involving Tries, where we embark on a word search adventure through a 2D board.</p></li>
+</ol>
+</section>
+<section id="pythonic-wisdom-unleashed">
+<h2>Pythonic Wisdom Unleashed<a class="headerlink" href="#pythonic-wisdom-unleashed" title="Permalink to this heading">#</a></h2>
+<p>Throughout our journey, Python will be our trusty companion. We’ll leverage the expressive nature of the language to craft elegant and efficient solutions to the challenges posed by Tries. Whether you are new to Tries or seeking to deepen your understanding, this section provides a comprehensive guide to mastering this fundamental data structure.</p>
+</section>
+<section id="your-journey-begins">
+<h2>Your Journey Begins<a class="headerlink" href="#your-journey-begins" title="Permalink to this heading">#</a></h2>
+<p>Get ready to traverse the nodes and edges of Tries, exploring their potential in solving a diverse range of problems. As we unravel each challenge, you’ll not only enhance your problem-solving prowess but also gain a deeper appreciation for the beauty of Tries.</p>
+<p>Let the exploration of Tries commence!</p>
+<p>Happy coding!</p>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#understanding-the-essence-of-tries">Understanding the Essence of Tries</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-chapters">Navigating the Chapters</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-wisdom-unleashed">Pythonic Wisdom Unleashed</a></li>
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diff --git a/09. Heap - Priority Queue/1046. Last Stone Weight.html b/09. Heap - Priority Queue/1046. Last Stone Weight.html
index 1c561df..6f2b8b2 100644
--- a/09. Heap - Priority Queue/1046. Last Stone Weight.html	
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Unveiling the Power of Heaps and Priority Queues</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -579,6 +668,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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+    <link rel="next" title="703. Kth Largest Element in a Stream" href="703.%20Kth%20Largest%20Element%20in%20a%20Stream.html" />
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-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Unveiling the Power of Heaps and Priority Queues</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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@@ -574,6 +663,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
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+<li class="toctree-l2"><a class="reference internal" href="295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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diff --git a/09. Heap - Priority Queue/README.html b/09. Heap - Priority Queue/README.html
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
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+    <h1>Unveiling the Power of Heaps and Priority Queues</h1>
+    <!-- Table of contents -->
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+            <nav aria-label="Page">
+                <ul class="visible nav section-nav flex-column">
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-heap-landscape">Navigating the Heap Landscape</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-mastery">Pythonic Mastery</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#what-to-expect">What to Expect</a></li>
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+  <section class="tex2jax_ignore mathjax_ignore" id="unveiling-the-power-of-heaps-and-priority-queues">
+<h1>Unveiling the Power of Heaps and Priority Queues<a class="headerlink" href="#unveiling-the-power-of-heaps-and-priority-queues" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Heap and Priority Queue section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In this chapter, we will delve into the world of heaps and priority queues—a realm where efficient data manipulation and retrieval are key.</p>
+<section id="navigating-the-heap-landscape">
+<h2>Navigating the Heap Landscape<a class="headerlink" href="#navigating-the-heap-landscape" title="Permalink to this heading">#</a></h2>
+<p>Heaps and Priority Queues play a pivotal role in algorithmic problem-solving, providing swift solutions to problems that involve maintaining a dynamic set of elements with varying priorities. This section introduces you to the fundamental concepts and challenges associated with heaps, guiding you through their application in real-world scenarios.</p>
+</section>
+<section id="unveiling-the-challenges">
+<h2>Unveiling the Challenges<a class="headerlink" href="#unveiling-the-challenges" title="Permalink to this heading">#</a></h2>
+<ol class="arabic simple">
+<li><p><strong>Last Stone Weight (Problem #1046):</strong> Dive into the dynamic world of stone smashing, where a priority queue becomes the tool of choice for determining the last remaining stone’s weight.</p></li>
+<li><p><strong>Find Median from Data Stream (Problem #295):</strong> Explore the intricacies of streaming data and discover how heaps facilitate the efficient calculation of medians in real-time.</p></li>
+<li><p><strong>Kth Largest Element in a Stream (Problem #703):</strong> Uncover the power of heaps in maintaining and updating a dynamic stream of elements, efficiently identifying the Kth largest element.</p></li>
+</ol>
+</section>
+<section id="pythonic-mastery">
+<h2>Pythonic Mastery<a class="headerlink" href="#pythonic-mastery" title="Permalink to this heading">#</a></h2>
+<p>As we journey through each challenge, we will not only unravel the intricacies of heaps and priority queues but also wield the elegance of Python to craft solutions that are both efficient and readable. From understanding heap operations to mastering priority queue nuances, this section is designed to sharpen your problem-solving skills in Python.</p>
+</section>
+<section id="what-to-expect">
+<h2>What to Expect<a class="headerlink" href="#what-to-expect" title="Permalink to this heading">#</a></h2>
+<ul class="simple">
+<li><p><strong>Foundational Concepts:</strong> Grasp the core principles of heaps and priority queues, laying the groundwork for efficient algorithmic solutions.</p></li>
+<li><p><strong>Real-world Applications:</strong> Explore challenges that mirror scenarios in which heaps excel, from dynamic data streams to problem domains requiring prioritized element retrieval.</p></li>
+<li><p><strong>Pythonic Elegance:</strong> Learn to implement solutions that leverage Python’s strengths, making your code not only functional but also a pleasure to read and maintain.</p></li>
+</ul>
+<p>Join us on this exploration of Heap and Priority Queue challenges, where efficiency meets elegance, and Python becomes your ally in conquering the complexities of algorithmic problem-solving.</p>
+<p>Happy coding!</p>
+</section>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-heap-landscape">Navigating the Heap Landscape</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-mastery">Pythonic Mastery</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#what-to-expect">What to Expect</a></li>
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diff --git a/10. Backtracking/39. Combination Sum.html b/10. Backtracking/39. Combination Sum.html
index 221bace..6fcf36d 100644
--- a/10. Backtracking/39. Combination Sum.html	
+++ b/10. Backtracking/39. Combination Sum.html	
@@ -67,6 +67,8 @@
     <script>DOCUMENTATION_OPTIONS.pagename = '10. Backtracking/39. Combination Sum';</script>
     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="79. Word Search" href="79.%20Word%20Search.html" />
+    <link rel="prev" title="Introduction to Backtracking Challenges" href="intro.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Backtracking Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -589,6 +678,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+        <p class="prev-next-title">Introduction to Backtracking Challenges</p>
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diff --git a/10. Backtracking/79. Word Search.html b/10. Backtracking/79. Word Search.html
index e944b35..d94b635 100644
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+++ b/10. Backtracking/79. Word Search.html	
@@ -67,6 +67,8 @@
     <script>DOCUMENTATION_OPTIONS.pagename = '10. Backtracking/79. Word Search';</script>
     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="Introduction to Graph Challenges" href="../11.%20Graphs/intro.html" />
+    <link rel="prev" title="39. Combination Sum" href="39.%20Combination%20Sum.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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@@ -157,12 +159,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Backtracking Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -616,6 +705,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+      <i class="fa-solid fa-angle-left"></i>
+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">39. Combination Sum</p>
+      </div>
+    </a>
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+       href="../11.%20Graphs/intro.html"
+       title="next page">
+      <div class="prev-next-info">
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diff --git a/10. Backtracking/README.html b/10. Backtracking/README.html
index 1549ca4..c2d8617 100644
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+++ b/10. Backtracking/README.html	
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unraveling-the-backtracking-paradigm">Unraveling the Backtracking Paradigm</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#discover-the-challenges">Discover the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-readme">Navigating the README</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-problem-solving">Pythonic Problem-Solving</a><ul class="nav section-nav flex-column">
+<li class="toc-h3 nav-item toc-entry"><a class="reference internal nav-link" href="#let-the-backtracking-journey-begin">Let the Backtracking Journey Begin</a></li>
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+  <section class="tex2jax_ignore mathjax_ignore" id="introduction-to-backtracking-challenges">
+<h1>Introduction to Backtracking Challenges<a class="headerlink" href="#introduction-to-backtracking-challenges" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Backtracking section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” Backtracking is a powerful technique employed in algorithmic problem-solving, enabling us to explore and find solutions to combinatorial problems efficiently.</p>
+<section id="unraveling-the-backtracking-paradigm">
+<h2>Unraveling the Backtracking Paradigm<a class="headerlink" href="#unraveling-the-backtracking-paradigm" title="Permalink to this heading">#</a></h2>
+<p>Backtracking is a systematic way of searching through all possible solutions to a problem and systematically discarding those that do not satisfy the problem constraints. In this section, we will delve into the intricacies of backtracking and apply it to two fascinating challenges.</p>
+</section>
+<section id="discover-the-challenges">
+<h2>Discover the Challenges<a class="headerlink" href="#discover-the-challenges" title="Permalink to this heading">#</a></h2>
+<ol class="arabic simple">
+<li><p><strong>Combination Sum (Problem #39):</strong> Explore the world of combinations as we navigate through the possibilities of summing up to a target value. This challenge will sharpen your ability to generate combinations and filter out the valid ones.</p></li>
+<li><p><strong>Word Search (Problem #79):</strong> Embark on a journey through a grid of letters, uncovering words hidden within. This challenge not only tests your backtracking skills but also introduces you to the realm of searching for words efficiently.</p></li>
+</ol>
+</section>
+<section id="navigating-the-readme">
+<h2>Navigating the README<a class="headerlink" href="#navigating-the-readme" title="Permalink to this heading">#</a></h2>
+<p>Before we dive into the challenges, take a moment to explore the README file. It serves as a guide, offering insights into the core concepts of backtracking, tips for approaching related problems, and Pythonic solutions for a better understanding.</p>
+</section>
+<section id="pythonic-problem-solving">
+<h2>Pythonic Problem-Solving<a class="headerlink" href="#pythonic-problem-solving" title="Permalink to this heading">#</a></h2>
+<p>As we venture through each challenge, we’ll not only unravel the mysteries of backtracking but also leverage the capabilities of Python to create elegant, concise, and efficient solutions. Get ready to enhance your problem-solving toolkit and deepen your appreciation for the versatility of backtracking.</p>
+<section id="let-the-backtracking-journey-begin">
+<h3>Let the Backtracking Journey Begin<a class="headerlink" href="#let-the-backtracking-journey-begin" title="Permalink to this heading">#</a></h3>
+<p>Are you ready to navigate the intricacies of backtracking and unlock solutions to challenging problems? Grab your Python interpreter, open your code editor, and let’s embark on a hands-on journey through the Backtracking challenges.</p>
+<p>Happy coding!</p>
+</section>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unraveling-the-backtracking-paradigm">Unraveling the Backtracking Paradigm</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#discover-the-challenges">Discover the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-readme">Navigating the README</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-problem-solving">Pythonic Problem-Solving</a><ul class="nav section-nav flex-column">
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Graph Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -627,6 +716,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">Introduction to Graph Challenges</p>
+      </div>
+    </a>
+    <a class="right-next"
+       href="178.%20Graph%20Valid%20Tree.html"
+       title="next page">
+      <div class="prev-next-info">
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diff --git a/11. Graphs/178. Graph Valid Tree.html b/11. Graphs/178. Graph Valid Tree.html
index 3143a84..6b5605c 100644
--- a/11. Graphs/178. Graph Valid Tree.html	
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     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
+    <link rel="next" title="200. Number of Islands" href="200.%20Number%20of%20Islands.html" />
+    <link rel="prev" title="133. Clone Graph" href="133.%20Clone%20Graph.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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@@ -155,12 +157,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Graph Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -565,6 +654,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+      <div class="prev-next-info">
+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">133. Clone Graph</p>
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+    </a>
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+       href="200.%20Number%20of%20Islands.html"
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diff --git a/11. Graphs/200. Number of Islands.html b/11. Graphs/200. Number of Islands.html
index 357214b..88f2ade 100644
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     <link rel="index" title="Index" href="../genindex.html" />
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+    <link rel="next" title="207. Course Schedule" href="207.%20Course%20Schedule.html" />
+    <link rel="prev" title="178. Graph Valid Tree" href="178.%20Graph%20Valid%20Tree.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
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-                    Introduction
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Graph Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -560,6 +649,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
                 <footer class="prev-next-footer">
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+        <p class="prev-next-title">178. Graph Valid Tree</p>
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diff --git a/11. Graphs/207. Course Schedule.html b/11. Graphs/207. Course Schedule.html
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+++ b/11. Graphs/207. Course Schedule.html	
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     <link rel="index" title="Index" href="../genindex.html" />
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+    <link rel="next" title="3651. Number of Connected Components in an Undirected Graph" href="3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Graph Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -563,6 +652,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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+        <p class="prev-next-title">200. Number of Islands</p>
+      </div>
+    </a>
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+       href="3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html"
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+    <link rel="next" title="417. Pacific Atlantic Water Flow" href="417.%20Pacific%20Atlantic%20Water%20Flow.html" />
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
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             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Graph Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -550,6 +639,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
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-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
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             </li>
         </ul>
-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Graph Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2"><a class="reference internal" href="133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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@@ -573,6 +662,24 @@ <h2>Challenging Exercises:<a class="headerlink" href="#challenging-exercises" ti
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diff --git a/11. Graphs/README.html b/11. Graphs/README.html
index d3926dd..c72ee0e 100644
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
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+                    Introduction: Navigating the Coding Cosmos
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/11. Graphs/intro.html b/11. Graphs/intro.html
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
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+                <h2> Contents </h2>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-graph-landscape">Navigating the Graph Landscape</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-exploration">Pythonic Exploration</a></li>
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+  <section class="tex2jax_ignore mathjax_ignore" id="introduction-to-graph-challenges">
+<h1>Introduction to Graph Challenges<a class="headerlink" href="#introduction-to-graph-challenges" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Graphs section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In this segment, we dive into the fascinating world of graphs, a fundamental data structure with applications ranging from social networks to network routing algorithms.</p>
+<section id="navigating-the-graph-landscape">
+<h2>Navigating the Graph Landscape<a class="headerlink" href="#navigating-the-graph-landscape" title="Permalink to this heading">#</a></h2>
+<p>Graphs serve as a powerful model for representing relationships and connections between entities. The challenges presented in this section will not only test your understanding of graph theory but also equip you with the skills needed to tackle real-world problems involving interconnected data.</p>
+</section>
+<section id="unveiling-the-challenges">
+<h2>Unveiling the Challenges<a class="headerlink" href="#unveiling-the-challenges" title="Permalink to this heading">#</a></h2>
+<ol class="arabic simple">
+<li><p><strong>Clone Graph (Problem #133):</strong> Embark on the journey of graph cloning, a task that involves creating an identical copy of a connected graph while preserving its structure.</p></li>
+<li><p><strong>Graph Valid Tree (Problem #178):</strong> Explore the concept of a valid tree within a graph, addressing the conditions that define a tree structure and applying them to ensure validity.</p></li>
+<li><p><strong>Number of Islands (Problem #200):</strong> Grapple with the challenge of counting the number of islands in a matrix, navigating through interconnected regions in a sea of elements.</p></li>
+<li><p><strong>Course Schedule (Problem #207):</strong> Delve into the world of course scheduling, solving a problem that explores the dependencies between courses in an academic curriculum.</p></li>
+<li><p><strong>Number of Connected Components in an Undirected Graph (Problem #3651):</strong> Uncover the dynamics of connected components in undirected graphs, a crucial concept in network analysis and data connectivity.</p></li>
+<li><p><strong>Pacific Atlantic Water Flow (Problem #417):</strong> Tackle a geographical challenge involving the flow of water in the Pacific and Atlantic Oceans, demonstrating the practical applications of graph theory.</p></li>
+</ol>
+</section>
+<section id="pythonic-exploration">
+<h2>Pythonic Exploration<a class="headerlink" href="#pythonic-exploration" title="Permalink to this heading">#</a></h2>
+<p>As we navigate through each graph-related challenge, we’ll not only unravel the complexities of graph theory but also harness the power of Python to implement elegant and efficient solutions. Get ready to embark on a journey of problem-solving that will not only deepen your understanding of graphs but also enhance your coding skills in Python.</p>
+<p>Whether you are a coding enthusiast, a student exploring graph theory, or a professional honing your problem-solving skills, this section is designed to provide you with a solid foundation in graph-based algorithms.</p>
+<p>Let the exploration of Graph challenges begin!</p>
+<p>Happy coding!</p>
+</section>
+<div class="toctree-wrapper compound">
+</div>
+</section>
+
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+        <p class="prev-next-subtitle">previous</p>
+        <p class="prev-next-title">79. Word Search</p>
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+    <ul class="visible nav section-nav flex-column">
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-graph-landscape">Navigating the Graph Landscape</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-exploration">Pythonic Exploration</a></li>
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diff --git a/12. Advanced Graphs/892. Alien Dictionary.html b/12. Advanced Graphs/892. Alien Dictionary.html
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
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-                    Introduction
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-        <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+        <ul class="current nav bd-sidenav">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
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+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l2 current active"><a class="current reference internal" href="#">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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diff --git a/12. Advanced Graphs/README.html b/12. Advanced Graphs/README.html
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
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+<li class="toctree-l1 current active has-children"><a class="current reference internal" href="#">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+    <h1>Introduction to Advanced Graphs: Navigating the Alien Dictionary</h1>
+    <!-- Table of contents -->
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+                <h2> Contents </h2>
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+                <ul class="visible nav section-nav flex-column">
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-complexity">Unveiling the Complexity</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#exploring-the-alien-dictionary">Exploring the Alien Dictionary</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-sections">Navigating Sections</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-exploration">Pythonic Exploration</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#what-lies-ahead">What Lies Ahead</a></li>
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+  <section class="tex2jax_ignore mathjax_ignore" id="introduction-to-advanced-graphs-navigating-the-alien-dictionary">
+<h1>Introduction to Advanced Graphs: Navigating the Alien Dictionary<a class="headerlink" href="#introduction-to-advanced-graphs-navigating-the-alien-dictionary" title="Permalink to this heading">#</a></h1>
+<p>Welcome to the Advanced Graphs section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In this segment, we delve into the fascinating world of graph theory, tackling intricate problems that demand a deep understanding of graph traversal and manipulation.</p>
+<section id="unveiling-the-complexity">
+<h2>Unveiling the Complexity<a class="headerlink" href="#unveiling-the-complexity" title="Permalink to this heading">#</a></h2>
+<p>Graphs, with their interconnected nodes and edges, provide a rich landscape for problem-solving. As we venture into the Advanced Graphs section, we confront challenges that push the boundaries of conventional graph algorithms. Our focus goes beyond basic graph traversal, exploring advanced scenarios where nuanced strategies are required.</p>
+</section>
+<section id="exploring-the-alien-dictionary">
+<h2>Exploring the Alien Dictionary<a class="headerlink" href="#exploring-the-alien-dictionary" title="Permalink to this heading">#</a></h2>
+<p><strong>892. Alien Dictionary:</strong> Among the celestial challenges in this section, we encounter the enigmatic “Alien Dictionary.” This problem propels us into the realm of lexicographical ordering in an extraterrestrial language. As we decipher the rules governing the alien script, we will harness graph theory to unravel the order of the alphabet and construct a dictionary unique to this cosmic language.</p>
+</section>
+<section id="navigating-sections">
+<h2>Navigating Sections<a class="headerlink" href="#navigating-sections" title="Permalink to this heading">#</a></h2>
+<p>This section is not just about solving problems but also about understanding the underlying principles of advanced graph theory. As we progress, each problem becomes a stepping stone, preparing you for the intricacies of real-world applications and coding interviews.</p>
+</section>
+<section id="pythonic-exploration">
+<h2>Pythonic Exploration<a class="headerlink" href="#pythonic-exploration" title="Permalink to this heading">#</a></h2>
+<p>Just as constellations form patterns in the night sky, Pythonic solutions emerge from the code. Throughout this journey, we’ll leverage the elegance and power of Python to craft solutions that not only solve the challenges at hand but also showcase the beauty of efficient and readable code.</p>
+</section>
+<section id="what-lies-ahead">
+<h2>What Lies Ahead<a class="headerlink" href="#what-lies-ahead" title="Permalink to this heading">#</a></h2>
+<p>From alien languages to intricate graph scenarios, the challenges in this section are designed to stretch your problem-solving muscles. Embrace the complexity, explore the uncharted territories of advanced graphs, and let the mysteries of the “Alien Dictionary” be the guide on this cosmic coding adventure.</p>
+<p>Happy coding!</p>
+</section>
+<div class="toctree-wrapper compound">
+</div>
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
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+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
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+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
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diff --git a/13. One-D Dynamic Programming/213. House Robber II.html b/13. One-D Dynamic Programming/213. House Robber II.html
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--- a/13. One-D Dynamic Programming/213. House Robber II.html	
+++ b/13. One-D Dynamic Programming/213. House Robber II.html	
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         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
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         <ul class="current nav bd-sidenav">
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+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
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+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
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+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
 <li class="toctree-l2"><a class="reference internal" href="70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2 current active"><a class="current reference internal" href="#">House Robber II</a></li>
diff --git a/13. One-D Dynamic Programming/70. Climbing Stairs.html b/13. One-D Dynamic Programming/70. Climbing Stairs.html
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+++ b/13. One-D Dynamic Programming/70. Climbing Stairs.html	
@@ -157,12 +157,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 current active has-children"><a class="reference internal" href="README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
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 <li class="toctree-l2"><a class="reference internal" href="198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/13. One-D Dynamic Programming/README.html b/13. One-D Dynamic Programming/README.html
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--- a/13. One-D Dynamic Programming/README.html	
+++ b/13. One-D Dynamic Programming/README.html	
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     <link rel="search" title="Search" href="../search.html" />
     <link rel="next" title="70. Climbing Stairs" href="70.%20Climbing%20Stairs.html" />
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+    <link rel="prev" title="892. Alien Dictionary" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html" />
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@@ -157,12 +157,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
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         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 current active has-children"><a class="current reference internal" href="#">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 current active has-children"><a class="current reference internal" href="#">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -467,12 +554,12 @@ <h2>Resources<a class="headerlink" href="#resources" title="Permalink to this he
                   <!-- Previous / next buttons -->
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diff --git a/14. Two-D Dynamic Programming/README.html b/14. Two-D Dynamic Programming/README.html
index e736b8c..8b5be45 100644
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@@ -157,12 +157,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
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-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/15. Greedy/README.html b/15. Greedy/README.html
index df90a1d..2cead9a 100644
--- a/15. Greedy/README.html	
+++ b/15. Greedy/README.html	
@@ -157,12 +157,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/16. Intervals/README.html b/16. Intervals/README.html
index 8e99b0a..c31f5c2 100644
--- a/16. Intervals/README.html	
+++ b/16. Intervals/README.html	
@@ -157,12 +157,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/17. Math Geometry/README.html b/17. Math Geometry/README.html
index 3dcca5e..0d567da 100644
--- a/17. Math Geometry/README.html	
+++ b/17. Math Geometry/README.html	
@@ -157,12 +157,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/18. Bit Manipulation/README.html b/18. Bit Manipulation/README.html
index f78b2bf..4bf5238 100644
--- a/18. Bit Manipulation/README.html	
+++ b/18. Bit Manipulation/README.html	
@@ -156,12 +156,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="../intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="../11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="../12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/Book.html b/Book.html
index 858c3b4..9122639 100644
--- a/Book.html
+++ b/Book.html
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -380,6 +467,7 @@ <h2> Contents </h2>
 <div class="cell_input docutils container">
 <div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="o">!</span>python<span class="w"> </span>-m<span class="w"> </span>pip<span class="w"> </span>install<span class="w"> </span>--upgrade<span class="w"> </span>pip
 <span class="o">%</span><span class="k">pip</span> install jupyter-book
+<span class="o">%</span><span class="k">pip</span> install ghp-import
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 Requirement already satisfied: pybtex-docutils&gt;=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinxcontrib-bibtex&lt;=2.5.0,&gt;=2.2.0-&gt;jupyter-book) (1.0.3)
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+Requirement already satisfied: setuptools in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx_togglebutton-&gt;jupyter-book) (68.0.0)
 Requirement already satisfied: wheel in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx_togglebutton-&gt;jupyter-book) (0.41.3)
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 Requirement already satisfied: tabulate in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-cache&lt;0.7,&gt;=0.5-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.9.0)
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+Requirement already satisfied: markdown-it-py&lt;3.0.0,&gt;=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.2.0)
 Requirement already satisfied: mdit-py-plugins~=0.3.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.3.5)
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+Requirement already satisfied: jupyter-client&gt;=6.1.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (8.4.0)
 Requirement already satisfied: jupyter-core!=5.0.*,&gt;=4.12 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (5.4.0)
 Requirement already satisfied: traitlets&gt;=5.3 in /home/codespace/.local/lib/python3.10/site-packages (from nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (5.11.2)
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+Requirement already satisfied: accessible-pygments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pydata-sphinx-theme&gt;=0.13.3-&gt;sphinx-book-theme~=1.0.0-&gt;jupyter-book) (0.0.4)
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 Requirement already satisfied: idna&lt;4,&gt;=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (3.4)
 Requirement already satisfied: urllib3&lt;3,&gt;=1.21.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (1.26.18)
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-Requirement already satisfied: comm&gt;=0.1.1 in /home/codespace/.local/lib/python3.10/site-packages (from ipykernel-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.1.4)
+</pre></div>
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: python-dateutil&gt;=2.8.2 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-client&gt;=6.1.12-&gt;nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.8.2)
+Requirement already satisfied: platformdirs&gt;=2.5 in /home/codespace/.local/lib/python3.10/site-packages (from jupyter-core!=5.0.*,&gt;=4.12-&gt;nbclient-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (3.11.0)
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: mdurl~=0.1 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from markdown-it-py&lt;3.0.0,&gt;=1.0.0-&gt;myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.1.2)
+Requirement already satisfied: ptyprocess&gt;=0.5 in /home/codespace/.local/lib/python3.10/site-packages (from pexpect&gt;4.3-&gt;ipython-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (0.7.0)
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: ghp-import in /usr/local/python/3.10.8/lib/python3.10/site-packages (2.1.0)
+Requirement already satisfied: python-dateutil&gt;=2.8.1 in /home/codespace/.local/lib/python3.10/site-packages (from ghp-import) (2.8.2)
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+</pre></div>
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@@ -587,195 +677,275 @@ <h2> Contents </h2>
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-<span class=" -Color -Color-Bold">myst v0.18.1:</span> MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=[&#39;colon_fence&#39;, &#39;dollarmath&#39;, &#39;linkify&#39;, &#39;substitution&#39;, &#39;tasklist&#39;], disable_syntax=[], all_links_external=False, url_schemes=[&#39;mailto&#39;, &#39;http&#39;, &#39;https&#39;], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=(&#39;{&#39;, &#39;}&#39;), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes=&#39;tex2jax_process|mathjax_process|math|output_area&#39;)
-<span class=" -Color -Color-Bold">myst-nb v0.17.2:</span> NbParserConfig(custom_formats={}, metadata_key=&#39;mystnb&#39;, cell_metadata_key=&#39;mystnb&#39;, kernel_rgx_aliases={}, execution_mode=&#39;auto&#39;, execution_cache_path=&#39;&#39;, execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin=&#39;default&#39;, remove_code_source=False, remove_code_outputs=False, code_prompt_show=&#39;Show code cell {type}&#39;, code_prompt_hide=&#39;Hide code cell {type}&#39;, number_source_lines=False, output_stderr=&#39;show&#39;, render_text_lexer=&#39;myst-ansi&#39;, render_error_lexer=&#39;ipythontb&#39;, render_image_options={}, render_figure_options={}, render_markdown_format=&#39;commonmark&#39;, output_folder=&#39;build&#39;, append_css=True, metadata_to_fm=False)
-Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">building [mo]: </span>targets for 0 po files that are out of date
-<span class=" -Color -Color-Bold">building [html]: </span>targets for 1 source files that are out of date
-<span class=" -Color -Color-Bold">updating environment: </span>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">myst v0.18.1:</span> MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=[&#39;colon_fence&#39;, &#39;dollarmath&#39;, &#39;linkify&#39;, &#39;substitution&#39;, &#39;tasklist&#39;], disable_syntax=[], all_links_external=False, url_schemes=[&#39;mailto&#39;, &#39;http&#39;, &#39;https&#39;], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=(&#39;{&#39;, &#39;}&#39;), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes=&#39;tex2jax_process|mathjax_process|math|output_area&#39;)
+<span class=" -Color -Color-Bold">myst-nb v0.17.2:</span> NbParserConfig(custom_formats={}, metadata_key=&#39;mystnb&#39;, cell_metadata_key=&#39;mystnb&#39;, kernel_rgx_aliases={}, execution_mode=&#39;auto&#39;, execution_cache_path=&#39;&#39;, execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin=&#39;default&#39;, remove_code_source=False, remove_code_outputs=False, code_prompt_show=&#39;Show code cell {type}&#39;, code_prompt_hide=&#39;Hide code cell {type}&#39;, number_source_lines=False, output_stderr=&#39;show&#39;, render_text_lexer=&#39;myst-ansi&#39;, render_error_lexer=&#39;ipythontb&#39;, render_image_options={}, render_figure_options={}, render_markdown_format=&#39;commonmark&#39;, output_folder=&#39;build&#39;, append_css=True, metadata_to_fm=False)
+Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache
+<span class=" -Color -Color-Bold">building [mo]: </span>targets for 0 po files that are out of date
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 18%] <span class=" -Color -Color-Magenta">03. Sliding Window/3. Longest Substring Without Repeating Characters</span>
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 1.15 seconds [mystnb]
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 0.94 seconds [mystnb]
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 1.13 seconds [mystnb]
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 4.02 seconds [mystnb]
+<span class=" -Color -Color-Bold">reading sources... </span>[ 22%] <span class=" -Color -Color-Magenta">03. Sliding Window/README</span>                             
+<span class=" -Color -Color-Bold">reading sources... </span>[ 23%] <span class=" -Color -Color-Magenta">03. Sliding Window/intro</span>                              
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-<span class=" -Color -Color-Bold">reading sources... </span>[ 24%] <span class=" -Color -Color-Magenta">04. Stack/20. Valid Parentheses</span>                       
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 24%] <span class=" -Color -Color-Magenta">04. Stack/20. Valid Parentheses</span>                       
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 1.10 seconds [mystnb]
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+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 26%] <span class=" -Color -Color-Magenta">04. Stack/intro</span>                                       
+<span class=" -Color -Color-Bold">reading sources... </span>[ 27%] <span class=" -Color -Color-Magenta">05. Binary Search/153. Find Minimum in Rotated Sorted Array</span>
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 </pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 1.21 seconds [mystnb]
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 1.12 seconds [mystnb]
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-<span class=" -Color -Color-Bold">reading sources... </span>[ 29%] <span class=" -Color -Color-Magenta">05. Binary Search/README</span>                              
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.00 seconds [mystnb]
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-/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 30%] <span class=" -Color -Color-Magenta">05. Binary Search/README</span>                              
+<span class=" -Color -Color-Bold">reading sources... </span>[ 31%] <span class=" -Color -Color-Magenta">05. Binary Search/intro</span>                               
+<span class=" -Color -Color-Bold">reading sources... </span>[ 32%] <span class=" -Color -Color-Magenta">06. Linked List/141. Linked List Cycle</span>                
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]
 </pre></div>
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 0.97 seconds [mystnb]
+<span class=" -Color -Color-Bold">reading sources... </span>[ 33%] <span class=" -Color -Color-Magenta">06. Linked List/143. Reorder List</span>                     
 /workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]
 </pre></div>
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executed notebook in 0.88 seconds [mystnb]
 </pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 33%] <span class=" -Color -Color-Magenta">06. Linked List/19. Remove Nth Node From End of List</span>  
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 34%] <span class=" -Color -Color-Magenta">06. Linked List/19. Remove Nth Node From End of List</span>  
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 </pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 1.05 seconds [mystnb]
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-/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 1.24 seconds [mystnb]
+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 35%] <span class=" -Color -Color-Magenta">06. Linked List/206. Reverse Linked List</span>              
+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]
 </pre></div>
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 1.15 seconds [mystnb]
 </pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 35%] <span class=" -Color -Color-Magenta">06. Linked List/21. Merge Two Sorted Lists</span>            
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 36%] <span class=" -Color -Color-Magenta">06. Linked List/21. Merge Two Sorted Lists</span>            
 /workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]
 </pre></div>
 </div>
 <div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>^C
-Exception ignored in: &lt;function WeakSet.__init__.&lt;locals&gt;._remove at 0x7f861ec2d2d0&gt;
+/workspaces/LeetCode-Solutions/04. Stack/intro.md:5: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]
 Traceback (most recent call last):
-  File &quot;/usr/local/python/3.10.8/lib/python3.10/_weakrefset.py&quot;, line 39, in _remove
-    def _remove(item, selfref=ref(self)):
-KeyboardInterrupt: 
-<span class=" -Color -Color-Bold">reading sources... </span>[ 37%] <span class=" -Color -Color-Magenta">06. Linked List/23. Merge k Sorted Lists</span>              
-/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 689, in async_execute
+    async with self.async_setup_kernel(**kwargs):
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/contextlib.py&quot;, line 199, in __aenter__
+    return await anext(self.gen)
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 647, in async_setup_kernel
+    await self.async_start_new_kernel_client()
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 562, in async_start_new_kernel_client
+    await ensure_async(self.kc.wait_for_ready(timeout=self.startup_timeout))
+  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py&quot;, line 189, in ensure_async
+    result = await obj
+  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/client.py&quot;, line 212, in _async_wait_for_ready
+    msg = await ensure_async(self.iopub_channel.get_msg(timeout=0.2))
+  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py&quot;, line 189, in ensure_async
+    result = await obj
+  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_client/channels.py&quot;, line 310, in get_msg
+    ready = await self.socket.poll(timeout)
+asyncio.exceptions.CancelledError
+
+During handling of the above exception, another exception occurred:
+
+Traceback (most recent call last):
+  File &quot;/home/codespace/.python/current/bin/jupyter-book&quot;, line 8, in &lt;module&gt;
+    sys.exit(main())
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1157, in __call__
+    return self.main(*args, **kwargs)
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1078, in main
+    rv = self.invoke(ctx)
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1688, in invoke
+    return _process_result(sub_ctx.command.invoke(sub_ctx))
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 1434, in invoke
+    return ctx.invoke(self.callback, **ctx.params)
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/click/core.py&quot;, line 783, in invoke
+    return __callback(*args, **kwargs)
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/cli/main.py&quot;, line 301, in build
+    result = build_sphinx(
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_book/sphinx.py&quot;, line 171, in build_sphinx
+    app.build(force_all, filenames)
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/application.py&quot;, line 329, in build
+    self.builder.build_update()
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 288, in build_update
+    self.build(to_build,
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 302, in build
+    updated_docnames = set(self.read())
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 409, in read
+    self._read_serial(docnames)
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 430, in _read_serial
+    self.read_doc(docname)
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/builders/__init__.py&quot;, line 483, in read_doc
+    publisher.publish()
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/core.py&quot;, line 217, in publish
+    self.document = self.reader.read(self.source, self.parser,
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/sphinx/io.py&quot;, line 103, in read
+    self.parse()
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/docutils/readers/__init__.py&quot;, line 78, in parse
+    self.parser.parse(self.input, document)
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/sphinx_.py&quot;, line 150, in parse
+    with create_client(
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/base.py&quot;, line 83, in __enter__
+    self.start_client()
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/myst_nb/core/execute/direct.py&quot;, line 40, in start_client
+    result = single_nb_execution(
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/jupyter_cache/executors/utils.py&quot;, line 58, in single_nb_execution
+    executenb(
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/site-packages/nbclient/client.py&quot;, line 1305, in execute
+    return NotebookClient(nb=nb, resources=resources, km=km, **kwargs).execute()
+  File &quot;/home/codespace/.local/lib/python3.10/site-packages/jupyter_core/utils/__init__.py&quot;, line 173, in wrapped
+    return loop.run_until_complete(inner)
+  File &quot;/usr/local/python/3.10.8/lib/python3.10/asyncio/base_events.py&quot;, line 649, in run_until_complete
+    return future.result()
+asyncio.exceptions.CancelledError
 </pre></div>
 </div>
 </div>
 </div>
 <div class="cell docutils container">
 <div class="cell_input docutils container">
-<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="o">%</span><span class="k">pip</span> install ghp-import
+<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="o">!</span>ghp-import<span class="w"> </span>-n<span class="w"> </span>-p<span class="w"> </span>-f<span class="w"> </span>_build/html
 </pre></div>
 </div>
 </div>
 <div class="cell_output docutils container">
-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: ghp-import in /usr/local/python/3.10.8/lib/python3.10/site-packages (2.1.0)
-Requirement already satisfied: python-dateutil&gt;=2.8.1 in /home/codespace/.local/lib/python3.10/site-packages (from ghp-import) (2.8.2)
-Requirement already satisfied: six&gt;=1.5 in /home/codespace/.local/lib/python3.10/site-packages (from python-dateutil&gt;=2.8.1-&gt;ghp-import) (1.16.0)
-Note: you may need to restart the kernel to use updated packages.
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Enumerating objects: 194, done.
+Counting objects: 100% (194/194), done.
+Delta compression using up to 2 threads
+Compressing objects: 100% (23/23), done.
+Writing objects: 100% (105/105), 219.17 KiB | 73.06 MiB/s, done.
+Total 105 (delta 85), reused 97 (delta 77), pack-reused 0
+remote: Resolving deltas: 100% (85/85), completed with 8 local objects.
+To https://github.com/mohsentabibian/LeetCode-Solutions
+   9b1f6d2..f40f6c9  gh-pages -&gt; gh-pages
 </pre></div>
 </div>
 </div>
diff --git a/CODE_OF_CONDUCT.html b/CODE_OF_CONDUCT.html
index 9cad015..2b3d29e 100644
--- a/CODE_OF_CONDUCT.html
+++ b/CODE_OF_CONDUCT.html
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/CONTRIBUTING.html b/CONTRIBUTING.html
index 1a3aa7a..d64348c 100644
--- a/CONTRIBUTING.html
+++ b/CONTRIBUTING.html
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/README.html b/README.html
index c137f96..ee74290 100644
--- a/README.html
+++ b/README.html
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/SECURITY.html b/SECURITY.html
index 37fcc97..4a40660 100644
--- a/SECURITY.html
+++ b/SECURITY.html
@@ -155,12 +155,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/_sources/01. Array Hashing/intro.md b/_sources/01. Array Hashing/intro.md
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+# Introduction to Array Hashing Challenges
+
+Welcome to the Array Hashing section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In this segment, we will explore a curated selection of LeetCode problems focused on array hashing, a powerful technique in algorithmic problem-solving.
+
+## Navigating the Array Hashing Terrain
+
+Array Hashing provides an efficient approach to solving a variety of problems by transforming complex operations into simpler ones through the use of hash functions. The challenges presented in this section will not only sharpen your skills in manipulating arrays but also introduce you to the art of hashing for optimized problem-solving.
+
+## Unveiling the Challenges
+
+1. **Two Sum (Problem #1):** Dive into the fundamentals of array manipulation by identifying pairs of elements that sum up to a target value.
+
+2. **Longest Consecutive Sequence (Problem #128):** Explore the concept of consecutive elements in an array and devise an algorithm to find the longest sequence.
+
+3. **Contains Duplicate (Problem #217):** Investigate the presence of duplicate elements in an array, a common scenario in coding interviews and real-world applications.
+
+4. **Product of Array Except Self (Problem #238):** Master the technique of efficiently computing the product of every element in an array except itself.
+
+5. **Valid Anagram (Problem #242):** Delve into string manipulation and explore efficient methods to determine if two strings are valid anagrams.
+
+6. **Top K Frequent Elements (Problem #347):** Solve the challenge of identifying the top K frequent elements in an array using hash maps and priority queues.
+
+7. **Group Anagrams (Problem #49):** Apply array hashing to categorize an array of strings into groups of anagrams, demonstrating the versatility of this technique.
+
+8. **Encode and Decode Strings (Problem #659):** Uncover the intricacies of encoding and decoding strings, a problem that draws on array hashing for efficient solutions.
+
+## Unlocking Pythonic Solutions
+
+As we tackle each challenge, not only will we unravel the intricacies of array hashing, but we'll also harness the power of Python to craft elegant and efficient solutions. Get ready to embark on a journey of problem-solving that will not only deepen your understanding of arrays and hashing but also enhance your coding skills in Python.
+
+Let the exploration of Array Hashing challenges begin!
+
+Happy coding!
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diff --git a/_sources/02. Two Pointers/intro.md b/_sources/02. Two Pointers/intro.md
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+# Introduction to Two Pointers Technique Challenges
+
+Welcome to the Two Pointers section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." The Two Pointers technique is a powerful strategy that enhances your ability to efficiently solve problems by maintaining two pointers that traverse a data structure, often an array or a linked list.
+
+## Unleashing the Efficiency of Two Pointers
+
+The Two Pointers technique is a versatile approach known for its effectiveness in optimizing time and space complexity. As we delve into the challenges in this section, you'll discover how to leverage this technique to navigate through arrays and efficiently address a variety of problems.
+
+## Exploring the Challenges
+
+1. **Container With Most Water (Problem #11):** Immerse yourself in the concept of maximizing the area between two vertical lines in a histogram, showcasing the elegance and efficiency of the Two Pointers technique.
+
+2. **Valid Palindrome (Problem #125):** Explore the realm of string manipulation and learn how Two Pointers can be employed to determine whether a given string is a valid palindrome.
+
+3. **3Sum (Problem #15):** Embark on a journey to find all unique triplets in an array that sum up to a target value, unraveling the complexities of Three Sum problems with the Two Pointers technique.
+
+## Guiding Principles and Pythonic Wisdom
+
+As we tackle each challenge, we'll not only unravel the intricacies of the Two Pointers technique but also embrace Pythonic solutions that showcase the elegance and readability of Python code. Through practical examples and detailed explanations, you'll gain a deep understanding of how to apply Two Pointers to real-world coding scenarios.
+
+Prepare to sharpen your problem-solving skills, enhance your algorithmic thinking, and master the art of using Two Pointers to tackle a diverse array of challenges.
+
+Let the exploration of Two Pointers challenges commence!
+
+Happy coding!
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diff --git a/_sources/03. Sliding Window/intro.md b/_sources/03. Sliding Window/intro.md
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+# Introduction to Sliding Window Challenges
+
+Welcome to the Sliding Window section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In this chapter, we will delve into the powerful technique of Sliding Window, a strategy designed to efficiently solve problems by maintaining a dynamic window over a sequence.
+
+## Navigating the Sliding Window Approach
+
+The Sliding Window technique is particularly effective for tackling problems that involve arrays or strings by efficiently processing a continuous subarray or substring. As we navigate through the challenges in this section, you'll witness how this approach optimizes both time and space complexity, making it an invaluable tool in your problem-solving arsenal.
+
+## Unveiling the Challenges
+
+1. **Best Time to Buy and Sell Stock (Problem #121):** Explore the dynamic world of stock prices and optimize your buying and selling strategy to maximize profit.
+
+2. **Longest Substring Without Repeating Characters (Problem #3):** Dive into string manipulation as you discover the longest substring without any repeating characters using the Sliding Window technique.
+
+3. **Longest Repeating Character Replacement (Problem #424):** Investigate the art of replacing characters in a string to create the longest substring with the same character, unveiling the versatility of the Sliding Window approach.
+
+4. **Minimum Window Substring (Problem #76):** Tackle the challenge of finding the minimum window in a string that contains all characters of another string, showcasing the adaptability of Sliding Window for string-related problems.
+
+## Pythonic Efficiency Unleashed
+
+As we unravel each challenge, we'll not only grasp the intricacies of the Sliding Window technique but also unleash the power of Python to implement elegant and efficient solutions. Get ready to witness how this dynamic approach can transform complex problems into manageable and optimized solutions.
+
+Let's embark on the exploration of Sliding Window challenges, where dynamic problem-solving meets the efficiency of the sliding window!
+
+Happy coding!
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diff --git a/_sources/04. Stack/intro.md b/_sources/04. Stack/intro.md
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+# Introduction to Stack Challenges: Unveiling Valid Parentheses
+
+Welcome to the Stack section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." This segment is dedicated to exploring the versatility and efficiency of the stack data structure, a fundamental tool in the arsenal of every proficient coder.
+
+### Navigating the Stack Landscape
+
+In this journey, we will unravel the power of stacks, a data structure that follows the Last-In, First-Out (LIFO) principle. The challenges presented in this section will not only enhance your understanding of stacks but also showcase their applicability in solving a variety of real-world problems.
+
+## Unveiling the Challenge: Valid Parentheses (Problem #20)
+
+Our focus for this section is on the "Valid Parentheses" problem. As you delve into this challenge, you'll encounter a scenario commonly faced in coding interviews and real-world applications: determining the validity of a string containing various types of parentheses—curly braces, square brackets, and parentheses.
+
+## What to Expect
+
+- **Fundamental Stack Operations:** Dive into the core operations of stacks, such as push and pop, to efficiently solve problems that involve nested structures.
+
+- **Balancing Act:** Explore the concept of balancing parentheses and how stacks provide an elegant solution to ensure the correct order of opening and closing brackets.
+
+- **Pythonic Solutions:** Leverage the strengths of Python to implement clean and concise code that showcases the simplicity and power of the stack data structure.
+
+## Unlocking the Valid Parentheses Challenge
+
+As we embark on the journey of solving the "Valid Parentheses" problem, you'll not only gain a deeper understanding of stack operations but also witness how this seemingly simple data structure can be a robust tool for solving complex problems.
+
+Prepare to strengthen your foundation in stack-based problem-solving and discover the elegance of Pythonic solutions. Let's unravel the mysteries of stacks together!
+
+Happy coding!
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diff --git a/_sources/05. Binary Search/intro.md b/_sources/05. Binary Search/intro.md
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+# Introduction to Binary Search Challenges
+
+Welcome to the Binary Search section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In this segment, we'll unravel the power of binary search, a fundamental algorithmic technique that is both elegant and efficient. The challenges presented here will sharpen your skills in applying binary search to unique problem scenarios.
+
+## Navigating the Binary Search Landscape
+
+Binary search is a divide-and-conquer algorithm that plays a crucial role in solving problems efficiently, especially in scenarios where the input is sorted or can be manipulated to exhibit a sorted structure. As we delve into this section, we'll explore the nuances of binary search through hands-on problem-solving.
+
+## Unveiling the Challenges
+
+1. **Find Minimum in Rotated Sorted Array (Problem #153):** Encounter the intricacies of searching for the minimum element in a rotated and sorted array. This challenge will test your ability to adapt binary search to unconventional data structures.
+
+2. **Search in Rotated Sorted Array (Problem #33):** Embark on a journey to efficiently locate a target element in a rotated and sorted array. This problem showcases the versatility of binary search in scenarios beyond the traditional sorted arrays.
+
+### Pythonic Solutions Await
+
+As we immerse ourselves in these challenges, the focus goes beyond mere problem-solving. We'll leverage the Python programming language to craft elegant, readable, and efficient solutions. Each problem serves as a canvas for Pythonic coding techniques, showcasing how the language's features can be harnessed for optimal algorithmic solutions.
+
+## What to Expect
+
+- **Algorithmic Mastery:** Develop a deep understanding of binary search and its application in diverse problem scenarios.
+
+- **Adaptability:** Learn to adapt binary search to unconventional data structures, broadening your problem-solving toolkit.
+
+- **Pythonic Coding:** Immerse yourself in Pythonic coding practices, gaining insights into writing clean and effective code.
+
+Let the exploration of Binary Search challenges commence! Whether you're a seasoned coder or just beginning your algorithmic journey, these challenges will provide a solid foundation in the art of binary search.
+
+Happy coding!
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diff --git a/_sources/06. Linked List/intro.md b/_sources/06. Linked List/intro.md
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+# Introduction to Linked List Mastery
+
+Welcome to the Linked List section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In this segment, we embark on a journey through the intricacies of linked lists — a fundamental data structure in computer science and a cornerstone of algorithmic problem-solving.
+
+## Navigating the Linked List Landscape
+
+Linked lists offer a dynamic and flexible way to organize and manipulate data, making them a critical asset in the programmer's toolkit. The challenges presented in this section are crafted to sharpen your understanding of linked list operations and empower you with the skills to tackle complex problems efficiently.
+
+## Unveiling the Challenges
+
+1. **Linked List Cycle (Problem #141):** Delve into the exploration of cycles within linked lists, a common challenge in detecting loops and optimizing algorithmic solutions.
+
+2. **Reorder List (Problem #143):** Uncover the art of reordering linked list elements, an essential skill that finds applications in various real-world scenarios.
+
+3. **Remove Nth Node From End of List (Problem #19):** Tackle the challenge of efficiently removing the nth node from the end of a linked list, mastering traversal and pointer manipulation.
+
+4. **Reverse Linked List (Problem #206):** Explore the reversal of linked lists, a fundamental operation with applications in numerous algorithmic problems.
+
+5. **Merge Two Sorted Lists (Problem #21):** Combine the power of linked list manipulation and sorting algorithms to merge two sorted lists into a single, ordered list.
+
+6. **Merge k Sorted Lists (Problem #23):** Rise to the challenge of merging k sorted linked lists, showcasing your ability to handle complexity in data structures.
+
+## Pythonic Solutions Unleashed
+
+As we navigate through each challenge, we not only unravel the intricacies of linked list operations but also harness the expressive power of Python to create elegant and efficient solutions. Prepare to witness the marriage of algorithmic thinking and Pythonic coding in a seamless blend.
+
+Whether you are a coding enthusiast, a job seeker preparing for technical interviews, or a seasoned programmer seeking to deepen your understanding, the Linked List challenges in this section are designed to elevate your problem-solving skills to new heights.
+
+Let's dive into the world of linked lists and master the art of problem-solving in Python!
+
+Happy coding!
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diff --git a/_sources/07. Trees/intro.md b/_sources/07. Trees/intro.md
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+# Introduction to Tree Challenges
+
+Welcome to the Trees section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." This segment delves into the fascinating realm of trees, exploring a diverse set of problems that will deepen your understanding of tree structures and sharpen your skills in traversing and manipulating them.
+
+## Navigating the Lush Canopy of Challenges
+
+Trees are a fundamental data structure in computer science, and their applications extend far and wide. The challenges in this section are designed to guide you through various aspects of tree manipulation, from basic operations to advanced traversal techniques and beyond.
+
+## Unveiling the Arboreal Adventures
+
+1. **Same Tree (Problem #100):** Compare two binary trees and determine if they are structurally identical.
+
+2. **Binary Tree Level Order Traversal (Problem #102):** Embark on a journey through the levels of a binary tree, unraveling its secrets in an ordered traversal.
+
+3. **Maximum Depth of Binary Tree (Problem #104):** Measure the depth of a binary tree, a fundamental metric in understanding its structure.
+
+4. **Construct Binary Tree from Preorder and Inorder Traversal (Problem #105):** Unravel the puzzle of reconstructing a binary tree from its preorder and inorder traversals.
+
+5. **Binary Tree Maximum Path Sum (Problem #124):** Navigate the complex web of paths in a binary tree and find the maximum sum achievable.
+
+6. **Binary Tree Right Side View (Problem #199):** Gain a unique perspective on a binary tree by capturing the view from its right side.
+
+7. **Invert Binary Tree (Problem #226):** Explore the transformative power of inverting a binary tree, changing its structure in a profound way.
+
+8. **Kth Smallest Element in a BST (Problem #230):** Traverse a binary search tree to find the Kth smallest element, mastering the art of efficient search.
+
+9. **Lowest Common Ancestor of a Binary Search Tree (Problem #235):** Discover the common ground in a binary search tree by finding the lowest common ancestor.
+
+10. **Serialize and Deserialize Binary Tree (Problem #297):** Encode and decode the intricate structure of a binary tree for efficient storage and retrieval.
+
+11. **Subtree of Another Tree (Problem #572):** Investigate whether one tree is a subtree of another, unveiling the subtleties of tree comparison.
+
+12. **Validate Binary Search Tree (Problem #98):** Examine the properties of a binary tree to determine if it is a valid binary search tree.
+
+## Pythonic Canopy Exploration
+
+As we journey through these arboreal challenges, we'll not only uncover the intricacies of tree structures and algorithms but also leverage the power of Python to craft elegant and efficient solutions. Get ready to traverse the lush canopy of tree challenges and elevate your problem-solving skills.
+
+Let the exploration of Tree challenges begin!
+
+Happy coding!
\ No newline at end of file
diff --git a/_sources/08. Tries/intro.md b/_sources/08. Tries/intro.md
new file mode 100644
index 0000000..3aea3e5
--- /dev/null
+++ b/_sources/08. Tries/intro.md	
@@ -0,0 +1,27 @@
+# Unveiling the Power of Tries: A Journey in Prefix Trees
+
+Welcome to the Tries section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In this segment, we embark on a fascinating exploration of Tries, also known as Prefix Trees, a powerful data structure that excels in handling strings and words.
+
+## Understanding the Essence of Tries
+
+Tries stand as a testament to the elegance of data structure design, particularly when it comes to storing and retrieving words efficiently. This section delves into the intricacies of Tries, unraveling their structure and demonstrating their application in various problem-solving scenarios.
+
+## Navigating the Chapters
+
+1. **Implement Trie (Prefix Tree) (Problem #208):** Lay the foundation by mastering the implementation of a Trie, a versatile data structure that stores a dynamic set of strings, making it an invaluable tool in a coder's arsenal.
+
+2. **Design Add and Search Words Data Structure (Problem #211):** Explore the versatility of Tries in designing a data structure that supports the addition and search of words with the flexibility to handle wildcard characters.
+
+3. **Word Search II (Problem #212):** Elevate your problem-solving skills as we tackle a more advanced challenge involving Tries, where we embark on a word search adventure through a 2D board.
+
+## Pythonic Wisdom Unleashed
+
+Throughout our journey, Python will be our trusty companion. We'll leverage the expressive nature of the language to craft elegant and efficient solutions to the challenges posed by Tries. Whether you are new to Tries or seeking to deepen your understanding, this section provides a comprehensive guide to mastering this fundamental data structure.
+
+## Your Journey Begins
+
+Get ready to traverse the nodes and edges of Tries, exploring their potential in solving a diverse range of problems. As we unravel each challenge, you'll not only enhance your problem-solving prowess but also gain a deeper appreciation for the beauty of Tries.
+
+Let the exploration of Tries commence!
+
+Happy coding!
\ No newline at end of file
diff --git a/_sources/09. Heap - Priority Queue/intro.md b/_sources/09. Heap - Priority Queue/intro.md
new file mode 100644
index 0000000..b2c02f7
--- /dev/null
+++ b/_sources/09. Heap - Priority Queue/intro.md	
@@ -0,0 +1,31 @@
+# Unveiling the Power of Heaps and Priority Queues
+
+Welcome to the Heap and Priority Queue section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In this chapter, we will delve into the world of heaps and priority queues—a realm where efficient data manipulation and retrieval are key.
+
+## Navigating the Heap Landscape
+
+Heaps and Priority Queues play a pivotal role in algorithmic problem-solving, providing swift solutions to problems that involve maintaining a dynamic set of elements with varying priorities. This section introduces you to the fundamental concepts and challenges associated with heaps, guiding you through their application in real-world scenarios.
+
+## Unveiling the Challenges
+
+1. **Last Stone Weight (Problem #1046):** Dive into the dynamic world of stone smashing, where a priority queue becomes the tool of choice for determining the last remaining stone's weight.
+
+2. **Find Median from Data Stream (Problem #295):** Explore the intricacies of streaming data and discover how heaps facilitate the efficient calculation of medians in real-time.
+
+3. **Kth Largest Element in a Stream (Problem #703):** Uncover the power of heaps in maintaining and updating a dynamic stream of elements, efficiently identifying the Kth largest element.
+
+## Pythonic Mastery
+
+As we journey through each challenge, we will not only unravel the intricacies of heaps and priority queues but also wield the elegance of Python to craft solutions that are both efficient and readable. From understanding heap operations to mastering priority queue nuances, this section is designed to sharpen your problem-solving skills in Python.
+
+## What to Expect
+
+- **Foundational Concepts:** Grasp the core principles of heaps and priority queues, laying the groundwork for efficient algorithmic solutions.
+
+- **Real-world Applications:** Explore challenges that mirror scenarios in which heaps excel, from dynamic data streams to problem domains requiring prioritized element retrieval.
+
+- **Pythonic Elegance:** Learn to implement solutions that leverage Python's strengths, making your code not only functional but also a pleasure to read and maintain.
+
+Join us on this exploration of Heap and Priority Queue challenges, where efficiency meets elegance, and Python becomes your ally in conquering the complexities of algorithmic problem-solving.
+
+Happy coding!
\ No newline at end of file
diff --git a/_sources/10. Backtracking/intro.md b/_sources/10. Backtracking/intro.md
new file mode 100644
index 0000000..0421341
--- /dev/null
+++ b/_sources/10. Backtracking/intro.md	
@@ -0,0 +1,27 @@
+# Introduction to Backtracking Challenges
+
+Welcome to the Backtracking section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." Backtracking is a powerful technique employed in algorithmic problem-solving, enabling us to explore and find solutions to combinatorial problems efficiently.
+
+## Unraveling the Backtracking Paradigm
+
+Backtracking is a systematic way of searching through all possible solutions to a problem and systematically discarding those that do not satisfy the problem constraints. In this section, we will delve into the intricacies of backtracking and apply it to two fascinating challenges.
+
+## Discover the Challenges
+
+1. **Combination Sum (Problem #39):** Explore the world of combinations as we navigate through the possibilities of summing up to a target value. This challenge will sharpen your ability to generate combinations and filter out the valid ones.
+
+2. **Word Search (Problem #79):** Embark on a journey through a grid of letters, uncovering words hidden within. This challenge not only tests your backtracking skills but also introduces you to the realm of searching for words efficiently.
+
+## Navigating the README
+
+Before we dive into the challenges, take a moment to explore the README file. It serves as a guide, offering insights into the core concepts of backtracking, tips for approaching related problems, and Pythonic solutions for a better understanding.
+
+## Pythonic Problem-Solving
+
+As we venture through each challenge, we'll not only unravel the mysteries of backtracking but also leverage the capabilities of Python to create elegant, concise, and efficient solutions. Get ready to enhance your problem-solving toolkit and deepen your appreciation for the versatility of backtracking.
+
+### Let the Backtracking Journey Begin
+
+Are you ready to navigate the intricacies of backtracking and unlock solutions to challenging problems? Grab your Python interpreter, open your code editor, and let's embark on a hands-on journey through the Backtracking challenges.
+
+Happy coding!
\ No newline at end of file
diff --git a/_sources/11. Graphs/intro.md b/_sources/11. Graphs/intro.md
new file mode 100644
index 0000000..813fd6a
--- /dev/null
+++ b/_sources/11. Graphs/intro.md	
@@ -0,0 +1,31 @@
+# Introduction to Graph Challenges
+
+Welcome to the Graphs section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In this segment, we dive into the fascinating world of graphs, a fundamental data structure with applications ranging from social networks to network routing algorithms.
+
+## Navigating the Graph Landscape
+
+Graphs serve as a powerful model for representing relationships and connections between entities. The challenges presented in this section will not only test your understanding of graph theory but also equip you with the skills needed to tackle real-world problems involving interconnected data.
+
+## Unveiling the Challenges
+
+1. **Clone Graph (Problem #133):** Embark on the journey of graph cloning, a task that involves creating an identical copy of a connected graph while preserving its structure.
+
+2. **Graph Valid Tree (Problem #178):** Explore the concept of a valid tree within a graph, addressing the conditions that define a tree structure and applying them to ensure validity.
+
+3. **Number of Islands (Problem #200):** Grapple with the challenge of counting the number of islands in a matrix, navigating through interconnected regions in a sea of elements.
+
+4. **Course Schedule (Problem #207):** Delve into the world of course scheduling, solving a problem that explores the dependencies between courses in an academic curriculum.
+
+5. **Number of Connected Components in an Undirected Graph (Problem #3651):** Uncover the dynamics of connected components in undirected graphs, a crucial concept in network analysis and data connectivity.
+
+6. **Pacific Atlantic Water Flow (Problem #417):** Tackle a geographical challenge involving the flow of water in the Pacific and Atlantic Oceans, demonstrating the practical applications of graph theory.
+
+## Pythonic Exploration
+
+As we navigate through each graph-related challenge, we'll not only unravel the complexities of graph theory but also harness the power of Python to implement elegant and efficient solutions. Get ready to embark on a journey of problem-solving that will not only deepen your understanding of graphs but also enhance your coding skills in Python.
+
+Whether you are a coding enthusiast, a student exploring graph theory, or a professional honing your problem-solving skills, this section is designed to provide you with a solid foundation in graph-based algorithms.
+
+Let the exploration of Graph challenges begin!
+
+Happy coding!
\ No newline at end of file
diff --git a/_sources/12. Advanced Graphs/intro.md b/_sources/12. Advanced Graphs/intro.md
new file mode 100644
index 0000000..f7fabf8
--- /dev/null
+++ b/_sources/12. Advanced Graphs/intro.md	
@@ -0,0 +1,25 @@
+# Introduction to Advanced Graphs: Navigating the Alien Dictionary
+
+Welcome to the Advanced Graphs section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In this segment, we delve into the fascinating world of graph theory, tackling intricate problems that demand a deep understanding of graph traversal and manipulation.
+
+## Unveiling the Complexity
+
+Graphs, with their interconnected nodes and edges, provide a rich landscape for problem-solving. As we venture into the Advanced Graphs section, we confront challenges that push the boundaries of conventional graph algorithms. Our focus goes beyond basic graph traversal, exploring advanced scenarios where nuanced strategies are required.
+
+## Exploring the Alien Dictionary
+
+**892. Alien Dictionary:** Among the celestial challenges in this section, we encounter the enigmatic "Alien Dictionary." This problem propels us into the realm of lexicographical ordering in an extraterrestrial language. As we decipher the rules governing the alien script, we will harness graph theory to unravel the order of the alphabet and construct a dictionary unique to this cosmic language.
+
+## Navigating Sections
+
+This section is not just about solving problems but also about understanding the underlying principles of advanced graph theory. As we progress, each problem becomes a stepping stone, preparing you for the intricacies of real-world applications and coding interviews.
+
+## Pythonic Exploration
+
+Just as constellations form patterns in the night sky, Pythonic solutions emerge from the code. Throughout this journey, we'll leverage the elegance and power of Python to craft solutions that not only solve the challenges at hand but also showcase the beauty of efficient and readable code.
+
+## What Lies Ahead
+
+From alien languages to intricate graph scenarios, the challenges in this section are designed to stretch your problem-solving muscles. Embrace the complexity, explore the uncharted territories of advanced graphs, and let the mysteries of the "Alien Dictionary" be the guide on this cosmic coding adventure.
+
+Happy coding!
\ No newline at end of file
diff --git a/_sources/Book.ipynb b/_sources/Book.ipynb
index 59960cb..64a1214 100644
--- a/_sources/Book.ipynb
+++ b/_sources/Book.ipynb
@@ -142,7 +142,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 4,
+   "execution_count": 1,
    "metadata": {},
    "outputs": [
     {
@@ -160,117 +160,144 @@
       "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
       "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\n",
       "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 1 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m[config changed ('master_doc')] 78 added, 1 changed, 0 removed\n",
+      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 0 source files that are out of date\n",
+      "\u001b[01mupdating environment: \u001b[39;49;00m12 added, 78 changed, 0 removed\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]  \n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.21 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/128. Longest Consecutive Sequence.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/217. Contains Duplicate.ipynb: Executed notebook in 1.22 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/238. Product of Array Except Self.ipynb: Executed notebook in 1.27 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/242. Valid Anagram.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/347. Top K Frequent Elements.ipynb: Executed notebook in 1.20 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 1.17 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/49. Group Anagrams.ipynb: Executed notebook in 1.35 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/659. Encode and Decode Strings.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/11. Container With Most Water.ipynb: Executed notebook in 1.00 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 1.21 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/125. Valid Palindrome.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executing notebook using local CWD [mystnb]     \n",
-      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/02. Two Pointers/15. 3Sum.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/03. Sliding Window/121. Best Time to Buy and Sell Stock.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/3. Longest Substring Without Repeating Characters.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 0.94 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/424. Longest Repeating Character Replacement.ipynb: Executed notebook in 0.89 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/03. Sliding Window/76. Minimum Window Substring.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/04. Stack/20. Valid Parentheses.ipynb: Executed notebook in 0.95 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 0.97 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/05. Binary Search/153. Find Minimum in Rotated Sorted Array.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/05. Binary Search/33. Search in Rotated Sorted Array.ipynb: Executed notebook in 0.83 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executed notebook in 1.21 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executed notebook in 0.98 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executing notebook using local CWD [mystnb]      \n",
-      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executed notebook in 0.84 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executed notebook in 1.38 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executed notebook in 1.03 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/124. Binary Tree Maximum Path Sum.ipynb: Executed notebook in 1.16 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executed notebook in 1.14 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executed notebook in 4.02 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executed notebook in 1.02 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executed notebook in 0.95 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executed notebook in 0.94 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executing notebook using local CWD [mystnb] \n",
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+      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executed notebook in 1.03 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 1.15 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executed notebook in 1.17 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executing notebook using local CWD [mystnb]   \n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 1.23 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 0.92 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 1.27 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.12 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.25 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 1.03 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 1.25 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 1.18 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 1.03 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 1.10 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 1.33 seconds [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/Book.ipynb: Executing notebook using local CWD [mystnb]                          \n"
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/Book.ipynb: Executing notebook using local CWD [mystnb]                          \n",
+      "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mintro\u001b[39;49;00m                                                 \n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/04. Stack/intro.md:5: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb:30008: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Executing notebook failed: CellTimeoutError [mystnb.exec]\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: Notebook exception traceback saved in: /workspaces/LeetCode-Solutions/_build/html/reports/Book.err.log [mystnb.exec]\u001b[39;49;00m\n",
+      "\u001b[01mlooking for now-outdated files... \u001b[39;49;00mnone found\n",
+      "\u001b[01mpickling environment... \u001b[39;49;00mdone\n",
+      "\u001b[01mchecking consistency... \u001b[39;49;00m\u001b[91m/workspaces/LeetCode-Solutions/01. Array Hashing/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/02. Two Pointers/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/03. Sliding Window/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/04. Stack/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/05. Binary Search/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/06. Linked List/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/07. Trees/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/08. Tries/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/10. Backtracking/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/11. Graphs/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/12. Advanced Graphs/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/Book.ipynb: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/CODE_OF_CONDUCT.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/CONTRIBUTING.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/README.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "\u001b[91m/workspaces/LeetCode-Solutions/SECURITY.md: WARNING: document isn't included in any toctree\u001b[39;49;00m\n",
+      "done\n",
+      "\u001b[01mpreparing documents... \u001b[39;49;00mdone\n",
+      "\u001b[01mwriting output... \u001b[39;49;00m[ 94%] \u001b[32mBook\u001b[39;49;00m                                                   raph\u001b[39;49;00m0m\r"
      ]
     }
    ],
@@ -280,28 +307,28 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 3,
+   "execution_count": null,
    "metadata": {},
    "outputs": [
     {
      "name": "stdout",
      "output_type": "stream",
      "text": [
-      "Requirement already satisfied: ghp-import in /usr/local/python/3.10.8/lib/python3.10/site-packages (2.1.0)\n",
-      "Requirement already satisfied: python-dateutil>=2.8.1 in /home/codespace/.local/lib/python3.10/site-packages (from ghp-import) (2.8.2)\n",
-      "Requirement already satisfied: six>=1.5 in /home/codespace/.local/lib/python3.10/site-packages (from python-dateutil>=2.8.1->ghp-import) (1.16.0)\n",
-      "Note: you may need to restart the kernel to use updated packages.\n"
+      "Enumerating objects: 194, done.\n",
+      "Counting objects: 100% (194/194), done.\n",
+      "Delta compression using up to 2 threads\n",
+      "Compressing objects: 100% (23/23), done.\n",
+      "Writing objects: 100% (105/105), 219.17 KiB | 73.06 MiB/s, done.\n",
+      "Total 105 (delta 85), reused 97 (delta 77), pack-reused 0\n",
+      "remote: Resolving deltas: 100% (85/85), completed with 8 local objects.\u001b[K\n",
+      "To https://github.com/mohsentabibian/LeetCode-Solutions\n",
+      "   9b1f6d2..f40f6c9  gh-pages -> gh-pages\n"
      ]
     }
    ],
-   "source": []
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
+   "source": [
+    "!ghp-import -n -p -f _build/html"
+   ]
   }
  ],
  "metadata": {
diff --git a/_sources/intro.md b/_sources/intro.md
index 555837a..eebac5d 100644
--- a/_sources/intro.md
+++ b/_sources/intro.md
@@ -1,23 +1,47 @@
-## Introduction
+# Introduction: Navigating the Coding Cosmos
 
-Welcome to "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In the dynamic landscape of technical interviews and coding assessments, mastering the art of algorithmic problem-solving is not just a skill—it's a necessity. This book is your guide to conquering the Blind 75, a curated collection of LeetCode challenges meticulously selected to empower you with the knowledge and expertise needed to navigate the complexities of coding interviews.
+Greetings, coding adventurer! Welcome to the expansive odyssey that is "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In this vast tome, we embark on a transformative journey that transcends traditional problem-solving, delving into the intricate dance of algorithms, the elegance of Pythonic solutions, and the practical application of coding wisdom.
 
-### Unveiling the Blind 75
+## Unveiling the Coding Cosmos: Foundations of Array Mastery
 
-The Blind 75 is a carefully chosen set of LeetCode problems, renowned for their frequency in technical interviews across the industry. From fundamental data structures to advanced algorithms, this collection serves as a compass, guiding you through the diverse terrain of coding challenges. Each problem is a stepping stone, designed not only to test your coding prowess but to elevate your problem-solving mindset.
+Our cosmic journey commences with the fundamental realm of "Array Manipulation." In this foundational section, we delve into the very essence of arrays—the building blocks of many algorithms. Understanding arrays isn't merely about memorizing syntax; it's about cultivating an intuition for array manipulation that transcends specific challenges. We explore not only the "how" but the "why" behind array manipulation, unraveling the logic that makes Pythonic solutions both powerful and elegant. Each problem is a microcosm, illustrating the nuanced strategies required for efficient array traversal, manipulation, and transformation.
 
-### Building a Strong Foundation
+As we navigate through the cosmos of arrays, it's essential to recognize that arrays are not static entities; they are dynamic landscapes that, when navigated skillfully, reveal hidden patterns and efficient solutions. Consider the "Product of Array Except Self" challenge as an example—beyond providing an answer, we explore the underlying mathematical principles, unlocking a broader understanding of the problem's intricacies.
 
-This repository is more than just a list of challenges—it's a hands-on journey to mastery. As you delve into each problem, you'll gain insights into optimal coding practices, algorithmic thinking, and Pythonic solutions. The carefully annotated solutions provide a roadmap, helping you understand not only *how* to solve a problem but also *why* a particular approach works best.
+## Navigating Graphical Constellations: Advanced Graph Theory
 
-### What to Expect
+Journeying deeper into the coding cosmos, we set our sights on the celestial realm of "Advanced Graphs." In this section, the challenges are like constellations, forming intricate patterns that guide us through the complexities of graph theory. The "Alien Dictionary" challenge, for instance, isn't just a problem to be solved; it's a portal to understanding the topological order, a gateway to deciphering language hierarchies in the cosmic lexicon.
 
-- **Curated Learning Path:** Progress through a curated set of challenges that cover a spectrum of topics, ensuring a well-rounded understanding of data structures and algorithms.
-  
-- **Practical Application:** Bridge the gap between theoretical knowledge and real-world application as you tackle challenges that mirror the complexities of technical interviews.
+Graphs, like constellations in the night sky, may seem daunting at first, but as we explore their interconnected nodes and edges, we begin to discern the hidden order. We discuss not only the algorithmic solutions to graph challenges but also the real-world applications of graph theory. Understanding how to represent and manipulate relationships between entities is a skill that extends beyond coding interviews, proving invaluable in scenarios ranging from social network analysis to optimizing network infrastructures.
 
-- **Pythonic Wisdom:** Learn to harness the power of Python to craft elegant, efficient, and readable code. Each solution is a lesson in leveraging Python's strengths for algorithmic problem-solving.
+## The Apex: Mastering the Blind 75 Challenges
 
-Whether you are preparing for technical interviews, enhancing your coding skills, or simply seeking to deepen your understanding of Python and algorithms, "Hands-On Problem-Solving in Python" is your companion on this enriching journey. Let's embark on the adventure of mastering the Blind 75 and unlocking the doors to confident, effective problem-solving in Python.
+Our cosmic vessel propels us to the highest peaks of algorithmic prowess—the summit known as the Blind 75. Curated to mirror the challenges encountered in the revered halls of technical interviews, these problems serve as beacons illuminating the path to coding excellence. This section isn't merely a collection of challenges; it's a training ground for honing problem-solving skills to a razor-sharp edge.
+
+As we ascend, each challenge becomes a chapter in a narrative of skill development. Take "Longest Consecutive Sequence," for instance; it isn't just about finding the longest sequence but understanding how to traverse and manipulate sequences efficiently. Likewise, "Encode and Decode Strings" is more than encoding and decoding; it's an exploration of efficient data representation and transmission—a skill crucial in real-world scenarios, from data compression to network communication.
+
+## Pythonic Symphony: Art and Language Unite
+
+In the vast coding cosmos, Python is not just a language; it's a symphony. Our solutions aren't mere code; they are compositions that resonate with the clarity and elegance that Python provides. As we traverse the cosmic landscape, we explore the nuances of Pythonic coding—leveraging list comprehensions, exploiting the power of generators, and embracing the dynamic nature of the language.
+
+Pythonic solutions are not a luxury but a necessity. The art of crafting Pythonic code is about more than adhering to a set of conventions; it's about understanding the language's philosophy and utilizing its features to create code that is not only functional but a joy to read. A well-crafted Pythonic solution is a testament to the coder's understanding of the language's idioms and the ability to leverage them to their advantage.
+
+## Bridging Theory to Real-World Challenges
+
+This guide isn't a mere compendium of challenges; it's a bridge connecting theoretical knowledge to real-world challenges. Each chapter prepares you not only for the academic rigors of coding interviews but also for the dynamic challenges you'll encounter in the field. The Blind 75, in particular, serves as a beacon, illuminating the path to coding excellence. We delve into not just how to solve a problem but why a specific approach is chosen—bridging the gap between theoretical understanding and practical application.
+
+In the real-world coding cosmos, challenges are not neatly packaged LeetCode problems; they are dynamic, evolving scenarios that require adaptability and a deep understanding of underlying principles. The Blind 75 serves as a microcosm, preparing you for the unpredictability and complexity of real-world coding challenges.
+
+## A Holistic Coding Odyssey: Beyond the Challenges
+
+As we progress through the cosmic expanse of this book, it's crucial to recognize that the journey extends beyond the challenges laid before you. It's a holistic odyssey—an exploration of languages, algorithms, and the art of problem-solving. Each challenge is not just a problem to be solved; it's an opportunity to develop a mindset—a mindset of a skilled problem-solver ready to tackle any coding puzzle that comes your way.
+
+In each chapter, we provide insights into the broader context of the challenges, discussing their relevance in real-world scenarios and how the skills you acquire are transferable to a variety of coding endeavors. We emphasize not just the technical aspects of coding but the problem-solving mindset—a mindset that transcends specific challenges and prepares you for the unpredictable nature of the coding cosmos.
+
+As you immerse yourself in the challenges, you're not just learning to solve problems; you're honing the mindset of a skilled problem-solver, ready to tackle any coding puzzle that comes your way. This holistic approach ensures that you're not just a proficient coder but a versatile problem-solver—a coder equipped to navigate the dynamic and ever-evolving landscape of the coding cosmos.
+
+## Beyond the Pages: A Cosmic Coding Adventure
+
+So, fellow coder, embark on this cosmic journey with pen, paper, and Python at your side. Let the coding cosmos be your guide, and may your endeavors be met with the joy of discovery. Beyond the pages of this book, a cosmic coding adventure awaits—one that extends into the realms of real-world challenges, continuous learning, and the ever-expanding frontiers of coding excellence.
 
 Happy coding!
\ No newline at end of file
diff --git a/genindex.html b/genindex.html
index ebd7335..5e8acc9 100644
--- a/genindex.html
+++ b/genindex.html
@@ -154,12 +154,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
diff --git a/intro.html b/intro.html
index 1cffec9..f39fbcc 100644
--- a/intro.html
+++ b/intro.html
@@ -9,7 +9,7 @@
     <meta charset="utf-8" />
     <meta name="viewport" content="width=device-width, initial-scale=1.0" /><meta name="generator" content="Docutils 0.18.1: http://docutils.sourceforge.net/" />
 
-    <title>Introduction &#8212; Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges</title>
+    <title>Introduction: Navigating the Coding Cosmos &#8212; Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges</title>
   
   
   
@@ -65,7 +65,7 @@
     <script>DOCUMENTATION_OPTIONS.pagename = 'intro';</script>
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     <link rel="search" title="Search" href="search.html" />
-    <link rel="next" title="1-D Dynamic Programming Problems - Blind 75 LeetCode" href="13.%20One-D%20Dynamic%20Programming/README.html" />
+    <link rel="next" title="Introduction to Array Hashing Challenges" href="01.%20Array%20Hashing/intro.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
   <meta name="docsearch:language" content="en"/>
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@@ -156,12 +156,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1 current active">
                 <a class="reference internal" href="#">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
@@ -355,7 +442,7 @@
               
 
 <div id="jb-print-docs-body" class="onlyprint">
-    <h1>Introduction</h1>
+    <h1>Introduction: Navigating the Coding Cosmos</h1>
     <!-- Table of contents -->
     <div id="print-main-content">
         <div id="jb-print-toc">
@@ -365,9 +452,13 @@ <h2> Contents </h2>
             </div>
             <nav aria-label="Page">
                 <ul class="visible nav section-nav flex-column">
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-blind-75">Unveiling the Blind 75</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#building-a-strong-foundation">Building a Strong Foundation</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#what-to-expect">What to Expect</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-coding-cosmos-foundations-of-array-mastery">Unveiling the Coding Cosmos: Foundations of Array Mastery</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-graphical-constellations-advanced-graph-theory">Navigating Graphical Constellations: Advanced Graph Theory</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#the-apex-mastering-the-blind-75-challenges">The Apex: Mastering the Blind 75 Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-symphony-art-and-language-unite">Pythonic Symphony: Art and Language Unite</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#bridging-theory-to-real-world-challenges">Bridging Theory to Real-World Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#a-holistic-coding-odyssey-beyond-the-challenges">A Holistic Coding Odyssey: Beyond the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#beyond-the-pages-a-cosmic-coding-adventure">Beyond the Pages: A Cosmic Coding Adventure</a></li>
 </ul>
             </nav>
         </div>
@@ -379,25 +470,43 @@ <h2> Contents </h2>
 <div id="searchbox"></div>
                 <article class="bd-article" role="main">
                   
-  <section id="introduction">
-<h1>Introduction<a class="headerlink" href="#introduction" title="Permalink to this heading">#</a></h1>
-<p>Welcome to “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In the dynamic landscape of technical interviews and coding assessments, mastering the art of algorithmic problem-solving is not just a skill—it’s a necessity. This book is your guide to conquering the Blind 75, a curated collection of LeetCode challenges meticulously selected to empower you with the knowledge and expertise needed to navigate the complexities of coding interviews.</p>
-<section id="unveiling-the-blind-75">
-<h2>Unveiling the Blind 75<a class="headerlink" href="#unveiling-the-blind-75" title="Permalink to this heading">#</a></h2>
-<p>The Blind 75 is a carefully chosen set of LeetCode problems, renowned for their frequency in technical interviews across the industry. From fundamental data structures to advanced algorithms, this collection serves as a compass, guiding you through the diverse terrain of coding challenges. Each problem is a stepping stone, designed not only to test your coding prowess but to elevate your problem-solving mindset.</p>
+  <section class="tex2jax_ignore mathjax_ignore" id="introduction-navigating-the-coding-cosmos">
+<h1>Introduction: Navigating the Coding Cosmos<a class="headerlink" href="#introduction-navigating-the-coding-cosmos" title="Permalink to this heading">#</a></h1>
+<p>Greetings, coding adventurer! Welcome to the expansive odyssey that is “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In this vast tome, we embark on a transformative journey that transcends traditional problem-solving, delving into the intricate dance of algorithms, the elegance of Pythonic solutions, and the practical application of coding wisdom.</p>
+<section id="unveiling-the-coding-cosmos-foundations-of-array-mastery">
+<h2>Unveiling the Coding Cosmos: Foundations of Array Mastery<a class="headerlink" href="#unveiling-the-coding-cosmos-foundations-of-array-mastery" title="Permalink to this heading">#</a></h2>
+<p>Our cosmic journey commences with the fundamental realm of “Array Manipulation.” In this foundational section, we delve into the very essence of arrays—the building blocks of many algorithms. Understanding arrays isn’t merely about memorizing syntax; it’s about cultivating an intuition for array manipulation that transcends specific challenges. We explore not only the “how” but the “why” behind array manipulation, unraveling the logic that makes Pythonic solutions both powerful and elegant. Each problem is a microcosm, illustrating the nuanced strategies required for efficient array traversal, manipulation, and transformation.</p>
+<p>As we navigate through the cosmos of arrays, it’s essential to recognize that arrays are not static entities; they are dynamic landscapes that, when navigated skillfully, reveal hidden patterns and efficient solutions. Consider the “Product of Array Except Self” challenge as an example—beyond providing an answer, we explore the underlying mathematical principles, unlocking a broader understanding of the problem’s intricacies.</p>
 </section>
-<section id="building-a-strong-foundation">
-<h2>Building a Strong Foundation<a class="headerlink" href="#building-a-strong-foundation" title="Permalink to this heading">#</a></h2>
-<p>This repository is more than just a list of challenges—it’s a hands-on journey to mastery. As you delve into each problem, you’ll gain insights into optimal coding practices, algorithmic thinking, and Pythonic solutions. The carefully annotated solutions provide a roadmap, helping you understand not only <em>how</em> to solve a problem but also <em>why</em> a particular approach works best.</p>
+<section id="navigating-graphical-constellations-advanced-graph-theory">
+<h2>Navigating Graphical Constellations: Advanced Graph Theory<a class="headerlink" href="#navigating-graphical-constellations-advanced-graph-theory" title="Permalink to this heading">#</a></h2>
+<p>Journeying deeper into the coding cosmos, we set our sights on the celestial realm of “Advanced Graphs.” In this section, the challenges are like constellations, forming intricate patterns that guide us through the complexities of graph theory. The “Alien Dictionary” challenge, for instance, isn’t just a problem to be solved; it’s a portal to understanding the topological order, a gateway to deciphering language hierarchies in the cosmic lexicon.</p>
+<p>Graphs, like constellations in the night sky, may seem daunting at first, but as we explore their interconnected nodes and edges, we begin to discern the hidden order. We discuss not only the algorithmic solutions to graph challenges but also the real-world applications of graph theory. Understanding how to represent and manipulate relationships between entities is a skill that extends beyond coding interviews, proving invaluable in scenarios ranging from social network analysis to optimizing network infrastructures.</p>
 </section>
-<section id="what-to-expect">
-<h2>What to Expect<a class="headerlink" href="#what-to-expect" title="Permalink to this heading">#</a></h2>
-<ul class="simple">
-<li><p><strong>Curated Learning Path:</strong> Progress through a curated set of challenges that cover a spectrum of topics, ensuring a well-rounded understanding of data structures and algorithms.</p></li>
-<li><p><strong>Practical Application:</strong> Bridge the gap between theoretical knowledge and real-world application as you tackle challenges that mirror the complexities of technical interviews.</p></li>
-<li><p><strong>Pythonic Wisdom:</strong> Learn to harness the power of Python to craft elegant, efficient, and readable code. Each solution is a lesson in leveraging Python’s strengths for algorithmic problem-solving.</p></li>
-</ul>
-<p>Whether you are preparing for technical interviews, enhancing your coding skills, or simply seeking to deepen your understanding of Python and algorithms, “Hands-On Problem-Solving in Python” is your companion on this enriching journey. Let’s embark on the adventure of mastering the Blind 75 and unlocking the doors to confident, effective problem-solving in Python.</p>
+<section id="the-apex-mastering-the-blind-75-challenges">
+<h2>The Apex: Mastering the Blind 75 Challenges<a class="headerlink" href="#the-apex-mastering-the-blind-75-challenges" title="Permalink to this heading">#</a></h2>
+<p>Our cosmic vessel propels us to the highest peaks of algorithmic prowess—the summit known as the Blind 75. Curated to mirror the challenges encountered in the revered halls of technical interviews, these problems serve as beacons illuminating the path to coding excellence. This section isn’t merely a collection of challenges; it’s a training ground for honing problem-solving skills to a razor-sharp edge.</p>
+<p>As we ascend, each challenge becomes a chapter in a narrative of skill development. Take “Longest Consecutive Sequence,” for instance; it isn’t just about finding the longest sequence but understanding how to traverse and manipulate sequences efficiently. Likewise, “Encode and Decode Strings” is more than encoding and decoding; it’s an exploration of efficient data representation and transmission—a skill crucial in real-world scenarios, from data compression to network communication.</p>
+</section>
+<section id="pythonic-symphony-art-and-language-unite">
+<h2>Pythonic Symphony: Art and Language Unite<a class="headerlink" href="#pythonic-symphony-art-and-language-unite" title="Permalink to this heading">#</a></h2>
+<p>In the vast coding cosmos, Python is not just a language; it’s a symphony. Our solutions aren’t mere code; they are compositions that resonate with the clarity and elegance that Python provides. As we traverse the cosmic landscape, we explore the nuances of Pythonic coding—leveraging list comprehensions, exploiting the power of generators, and embracing the dynamic nature of the language.</p>
+<p>Pythonic solutions are not a luxury but a necessity. The art of crafting Pythonic code is about more than adhering to a set of conventions; it’s about understanding the language’s philosophy and utilizing its features to create code that is not only functional but a joy to read. A well-crafted Pythonic solution is a testament to the coder’s understanding of the language’s idioms and the ability to leverage them to their advantage.</p>
+</section>
+<section id="bridging-theory-to-real-world-challenges">
+<h2>Bridging Theory to Real-World Challenges<a class="headerlink" href="#bridging-theory-to-real-world-challenges" title="Permalink to this heading">#</a></h2>
+<p>This guide isn’t a mere compendium of challenges; it’s a bridge connecting theoretical knowledge to real-world challenges. Each chapter prepares you not only for the academic rigors of coding interviews but also for the dynamic challenges you’ll encounter in the field. The Blind 75, in particular, serves as a beacon, illuminating the path to coding excellence. We delve into not just how to solve a problem but why a specific approach is chosen—bridging the gap between theoretical understanding and practical application.</p>
+<p>In the real-world coding cosmos, challenges are not neatly packaged LeetCode problems; they are dynamic, evolving scenarios that require adaptability and a deep understanding of underlying principles. The Blind 75 serves as a microcosm, preparing you for the unpredictability and complexity of real-world coding challenges.</p>
+</section>
+<section id="a-holistic-coding-odyssey-beyond-the-challenges">
+<h2>A Holistic Coding Odyssey: Beyond the Challenges<a class="headerlink" href="#a-holistic-coding-odyssey-beyond-the-challenges" title="Permalink to this heading">#</a></h2>
+<p>As we progress through the cosmic expanse of this book, it’s crucial to recognize that the journey extends beyond the challenges laid before you. It’s a holistic odyssey—an exploration of languages, algorithms, and the art of problem-solving. Each challenge is not just a problem to be solved; it’s an opportunity to develop a mindset—a mindset of a skilled problem-solver ready to tackle any coding puzzle that comes your way.</p>
+<p>In each chapter, we provide insights into the broader context of the challenges, discussing their relevance in real-world scenarios and how the skills you acquire are transferable to a variety of coding endeavors. We emphasize not just the technical aspects of coding but the problem-solving mindset—a mindset that transcends specific challenges and prepares you for the unpredictable nature of the coding cosmos.</p>
+<p>As you immerse yourself in the challenges, you’re not just learning to solve problems; you’re honing the mindset of a skilled problem-solver, ready to tackle any coding puzzle that comes your way. This holistic approach ensures that you’re not just a proficient coder but a versatile problem-solver—a coder equipped to navigate the dynamic and ever-evolving landscape of the coding cosmos.</p>
+</section>
+<section id="beyond-the-pages-a-cosmic-coding-adventure">
+<h2>Beyond the Pages: A Cosmic Coding Adventure<a class="headerlink" href="#beyond-the-pages-a-cosmic-coding-adventure" title="Permalink to this heading">#</a></h2>
+<p>So, fellow coder, embark on this cosmic journey with pen, paper, and Python at your side. Let the coding cosmos be your guide, and may your endeavors be met with the joy of discovery. Beyond the pages of this book, a cosmic coding adventure awaits—one that extends into the realms of real-world challenges, continuous learning, and the ever-expanding frontiers of coding excellence.</p>
 <p>Happy coding!</p>
 </section>
 <div class="toctree-wrapper compound">
@@ -435,11 +544,11 @@ <h2>What to Expect<a class="headerlink" href="#what-to-expect" title="Permalink
                   <!-- Previous / next buttons -->
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+       href="01.%20Array%20Hashing/intro.html"
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       <div class="prev-next-info">
         <p class="prev-next-subtitle">next</p>
-        <p class="prev-next-title">1-D Dynamic Programming Problems - Blind 75 LeetCode</p>
+        <p class="prev-next-title">Introduction to Array Hashing Challenges</p>
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@@ -458,9 +567,13 @@ <h2>What to Expect<a class="headerlink" href="#what-to-expect" title="Permalink
   </div>
   <nav class="bd-toc-nav page-toc">
     <ul class="visible nav section-nav flex-column">
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-blind-75">Unveiling the Blind 75</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#building-a-strong-foundation">Building a Strong Foundation</a></li>
-<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#what-to-expect">What to Expect</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-coding-cosmos-foundations-of-array-mastery">Unveiling the Coding Cosmos: Foundations of Array Mastery</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-graphical-constellations-advanced-graph-theory">Navigating Graphical Constellations: Advanced Graph Theory</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#the-apex-mastering-the-blind-75-challenges">The Apex: Mastering the Blind 75 Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#pythonic-symphony-art-and-language-unite">Pythonic Symphony: Art and Language Unite</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#bridging-theory-to-real-world-challenges">Bridging Theory to Real-World Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#a-holistic-coding-odyssey-beyond-the-challenges">A Holistic Coding Odyssey: Beyond the Challenges</a></li>
+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#beyond-the-pages-a-cosmic-coding-adventure">Beyond the Pages: A Cosmic Coding Adventure</a></li>
 </ul>
   </nav></div>
 
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diff --git a/search.html b/search.html
index 3452898..cf18f55 100644
--- a/search.html
+++ b/search.html
@@ -156,12 +156,99 @@
         <ul class="nav bd-sidenav bd-sidenav__home-link">
             <li class="toctree-l1">
                 <a class="reference internal" href="intro.html">
-                    Introduction
+                    Introduction: Navigating the Coding Cosmos
                 </a>
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/238.%20Product%20of%20Array%20Except%20Self.html">238. Product of Array Except Self</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/242.%20Valid%20Anagram.html">242. Valid Anagram</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/347.%20Top%20K%20Frequent%20Elements.html">347. Top K Frequent Elements</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/49.%20Group%20Anagrams.html">49. Group Anagrams</a></li>
+<li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/659.%20Encode%20and%20Decode%20Strings.html">659· Encode and Decode Strings</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="02.%20Two%20Pointers/intro.html">Introduction to Two Pointers Technique Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-2" name="toctree-checkbox-2" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-2"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/11.%20Container%20With%20Most%20Water.html">11. Container With Most Water</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/125.%20Valid%20Palindrome.html">125. Valid Palindrome</a></li>
+<li class="toctree-l2"><a class="reference internal" href="02.%20Two%20Pointers/15.%203Sum.html">15. 3Sum</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="03.%20Sliding%20Window/intro.html">Introduction to Sliding Window Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-3" name="toctree-checkbox-3" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-3"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/121.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock.html">121. Best Time to Buy and Sell Stock</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/3.%20Longest%20Substring%20Without%20Repeating%20Characters.html">3. Longest Substring Without Repeating Characters</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/424.%20Longest%20Repeating%20Character%20Replacement.html">424. Longest Repeating Character Replacement</a></li>
+<li class="toctree-l2"><a class="reference internal" href="03.%20Sliding%20Window/76.%20Minimum%20Window%20Substring.html">76. Minimum Window Substring</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="04.%20Stack/intro.html">Introduction to Stack Challenges: Unveiling Valid Parentheses</a><input class="toctree-checkbox" id="toctree-checkbox-4" name="toctree-checkbox-4" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-4"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="04.%20Stack/20.%20Valid%20Parentheses.html">20. Valid Parentheses</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="05.%20Binary%20Search/intro.html">Introduction to Binary Search Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-5" name="toctree-checkbox-5" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-5"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/153.%20Find%20Minimum%20in%20Rotated%20Sorted%20Array.html">153. Find Minimum in Rotated Sorted Array</a></li>
+<li class="toctree-l2"><a class="reference internal" href="05.%20Binary%20Search/33.%20Search%20in%20Rotated%20Sorted%20Array.html">33. Search in Rotated Sorted Array</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="06.%20Linked%20List/intro.html">Introduction to Linked List Mastery</a><input class="toctree-checkbox" id="toctree-checkbox-6" name="toctree-checkbox-6" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-6"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/141.%20Linked%20List%20Cycle.html">141. Linked List Cycle</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/143.%20Reorder%20List.html">143. Reorder List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/19.%20Remove%20Nth%20Node%20From%20End%20of%20List.html">19. Remove Nth Node From End of List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/206.%20Reverse%20Linked%20List.html">206. Reverse Linked List</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/21.%20Merge%20Two%20Sorted%20Lists.html">21. Merge Two Sorted Lists</a></li>
+<li class="toctree-l2"><a class="reference internal" href="06.%20Linked%20List/23.%20Merge%20k%20Sorted%20Lists.html">23. Merge k Sorted Lists</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="07.%20Trees/intro.html">Introduction to Tree Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-7" name="toctree-checkbox-7" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-7"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/100.%20Same%20Tree.html">100: Same Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/102.%20Binary%20Tree%20Level%20Order%20Traversal.html">102. Binary Tree Level Order Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/104.%20Maximum%20Depth%20of%20Binary%20Tree.html">104. Maximum Depth of Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/105.%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/124.%20Binary%20Tree%20Maximum%20Path%20Sum.html">124. Binary Tree Maximum Path Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/199.%20Binary%20Tree%20Right%20Side%20View.html">199. Binary Tree Right Side View</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/226.%20Invert%20Binary%20Tree.html">226: Invert Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/230.%20Kth%20Smallest%20Element%20in%20a%20BST.html">230. Kth Smallest Element in a BST</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/235.%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree.html">235. Lowest Common Ancestor of a Binary Search Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.html">297. Serialize and Deserialize Binary Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/572.%20Subtree%20of%20Another%20Tree.html">572: Subtree of Another Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="07.%20Trees/98.%20Validate%20Binary%20Search%20Tree.html">98. Validate Binary Search Tree</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="08.%20Tries/intro.html">Unveiling the Power of Tries: A Journey in Prefix Trees</a><input class="toctree-checkbox" id="toctree-checkbox-8" name="toctree-checkbox-8" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-8"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/208.%20Implement%20Trie%20%28Prefix%20Tree%29.html">208. Implement Trie (Prefix Tree)</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/211.%20Design%20Add%20and%20Search%20Words%20Data%20Structure.html">211. Design Add and Search Words Data Structure</a></li>
+<li class="toctree-l2"><a class="reference internal" href="08.%20Tries/212.%20Word%20Search%20II.html">212. Word Search II</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/intro.html">Unveiling the Power of Heaps and Priority Queues</a><input class="toctree-checkbox" id="toctree-checkbox-9" name="toctree-checkbox-9" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-9"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/1046.%20Last%20Stone%20Weight.html">1046. Last Stone Weight</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/295.%20Find%20Median%20from%20Data%20Stream.html">295. Find Median from Data Stream</a></li>
+<li class="toctree-l2"><a class="reference internal" href="09.%20Heap%20-%20Priority%20Queue/703.%20Kth%20Largest%20Element%20in%20a%20Stream.html">703. Kth Largest Element in a Stream</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="10.%20Backtracking/intro.html">Introduction to Backtracking Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-10" name="toctree-checkbox-10" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-10"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/39.%20Combination%20Sum.html">39. Combination Sum</a></li>
+<li class="toctree-l2"><a class="reference internal" href="10.%20Backtracking/79.%20Word%20Search.html">79. Word Search</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="11.%20Graphs/intro.html">Introduction to Graph Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-11" name="toctree-checkbox-11" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-11"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/133.%20Clone%20Graph.html">133. Clone Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/178.%20Graph%20Valid%20Tree.html">178. Graph Valid Tree</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/200.%20Number%20of%20Islands.html">200. Number of Islands</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/207.%20Course%20Schedule.html">207. Course Schedule</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/3651.%20Number%20of%20Connected%20Components%20in%20an%20Undirected%20Graph.html">3651. Number of Connected Components in an Undirected Graph</a></li>
+<li class="toctree-l2"><a class="reference internal" href="11.%20Graphs/417.%20Pacific%20Atlantic%20Water%20Flow.html">417. Pacific Atlantic Water Flow</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="12.%20Advanced%20Graphs/intro.html">Introduction to Advanced Graphs: Navigating the Alien Dictionary</a><input class="toctree-checkbox" id="toctree-checkbox-12" name="toctree-checkbox-12" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-12"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l2"><a class="reference internal" href="12.%20Advanced%20Graphs/892.%20Alien%20Dictionary.html">892. Alien Dictionary</a></li>
+</ul>
+</li>
+<li class="toctree-l1 has-children"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/README.html">1-D Dynamic Programming Problems - Blind 75 LeetCode</a><input class="toctree-checkbox" id="toctree-checkbox-13" name="toctree-checkbox-13" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-13"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/70.%20Climbing%20Stairs.html">70. Climbing Stairs</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/198.%20House%20Robber.html">House Robber</a></li>
 <li class="toctree-l2"><a class="reference internal" href="13.%20One-D%20Dynamic%20Programming/213.%20House%20Robber%20II.html">House Robber II</a></li>
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index f6c61a9..6171ed8 100644
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48, 49, 51, 61, 75], "multipli": 3, "correspond": [3, 5, 6, 18, 33, 43, 45, 51, 54, 59], "contain": [3, 4, 5, 6, 7, 8, 11, 12, 15, 16, 18, 20, 21, 24, 25, 26, 27, 28, 31, 33, 34, 35, 38, 39, 40, 41, 44, 47, 48, 51, 54, 57, 58, 59, 61, 64, 67, 68, 69, 70, 71, 72, 75], "desir": [3, 37, 76], "comput": [3, 39, 48, 64], "It": [3, 5, 6, 9, 10, 11, 13, 14, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 36, 38, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "result1": [3, 5, 6, 9, 21, 28, 56], "24": [3, 72], "12": [3, 9, 45, 63, 72], "result2": [3, 5, 6, 9, 21, 28, 56], "first": [3, 4, 5, 6, 13, 23, 24, 25, 26, 27, 31, 32, 33, 34, 35, 36, 37, 39, 41, 44, 45, 49, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64, 65], "second": [3, 5, 24, 25, 37, 41, 47, 54, 65, 72], "independ": [3, 25, 27, 48, 63], "sequenti": [3, 21, 45, 52], "consum": [3, 10, 14, 15, 27, 38, 52, 55], "respect": [3, 4, 16, 24, 26, 27, 32, 40, 44, 51, 52, 57, 63, 65, 67, 68, 69, 70, 71, 73, 75], "These": 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66, 67, 68, 69, 70, 71, 73, 75], "websit": [8, 12, 17, 19, 22, 29, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 75], "forum": [8, 12, 17, 19, 22, 29, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 75], "hint": [8, 12, 15, 17, 19, 22, 29, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 75], "happi": [8, 12, 17, 19, 22, 29, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 75, 77], "height": [9, 32, 33, 34, 36, 37, 38, 39, 40, 41, 59], "There": [9, 21, 23, 39, 43, 44, 48, 54, 55, 57, 58, 59, 61, 65, 74], "vertic": [9, 45, 52, 54, 56, 57, 61], "drawn": 9, "endpoint": 9, "th": [9, 13, 25, 37, 47, 63], "x": [9, 39, 45, 47, 52, 56, 59, 61, 74], "axi": 9, "notic": [9, 11, 20], "slant": 9, "maxarea": 9, "area": 9, "max_area": 9, "pointer": [9, 10, 11, 16, 20, 21, 23, 24, 25, 26, 27, 28, 72], "until": [9, 10, 11, 15, 20, 21, 24, 25, 26, 27, 28, 31, 33, 35, 37, 38, 43, 47, 56, 57, 61, 76], "h": [9, 32, 34, 36, 37, 38, 39, 45], "min": [9, 13, 14, 48, 49, 59, 61], "width": [9, 35], "distanc": 9, "shorter": 9, "inward": 9, "possibl": [9, 11, 13, 21, 34, 37, 40, 44, 45, 51, 52, 56, 57, 59, 61, 65, 74], "systemat": 9, "hold": [9, 15, 23], "point": [9, 11, 20, 21, 23, 24, 25, 26, 27, 34, 39, 43, 54], "taller": 9, "won": 9, "increas": [9, 11, 38, 66, 75], "decreas": [9, 11, 27], "smaller": [9, 11, 16, 27, 33, 34, 38, 48, 49, 51, 55, 57, 65], "height1": 9, "height2": 9, "height3": 9, "45": [9, 65], "approach": [9, 15, 26, 28, 33, 34, 37, 39, 40, 41, 45, 49, 51, 56, 63, 77], "like": [9, 11, 15, 26, 34, 35, 36, 38, 39, 41, 43, 45, 55, 56, 73], "whose": 9, "consumpt": 9, "As": [9, 24, 27, 37, 77], "variant": [9, 35], "gap": [9, 77], "ll": [9, 77], "volum": 9, "uppercas": [10, 15, 16, 52], "non": [10, 27, 34, 36, 61, 69, 75], "alphanumer": 10, "read": 10, "forward": [10, 74], "backward": 10, "printabl": 10, "ispalindrom": 10, "lower": [10, 13], "cleaned_": 10, "join": [10, 39, 61], "isalnum": 10, "clean": [10, 33], "don": [10, 21, 28, 39, 43], "insensit": 10, "digit": [10, 14], "punctuat": 10, "closer": [10, 36, 37], "decrement": [10, 11, 15, 16, 37, 57], "mismatch": [10, 40], "man": 10, "plan": [10, 63, 64], "canal": 10, "panama": 10, "output1": 10, "race": [10, 73], "output2": 10, "input3": 10, "output3": 10, "comprehens": 10, "half": [10, 20, 21, 24, 48], "still": [10, 15, 16, 20, 23, 31, 37, 41, 44, 49, 52], "primarili": [10, 15, 24, 40, 41, 44, 47, 64], "version": [10, 56, 73], "larg": [10, 38, 40, 47, 49], "addition": [10, 14, 25, 33, 34, 40, 52, 56, 58, 61], "babad": 10, "bab": 10, "aba": 10, "limit": [10, 45, 65], "allowed_char": 10, "ignor": [10, 18, 72], "triplet": 11, "j": [11, 28, 45, 52, 54, 56, 59, 61], "3000": [11, 33], "threesum": 11, "ascend": [11, 20, 21, 28, 37], "skip": [11, 25, 51, 63, 64], "elif": [11, 18, 21, 38, 44, 48, 64], "neg": [11, 34, 41, 51], "Then": [11, 13, 48, 59, 61], "larger": [11, 16, 38, 48, 52], "been": [11, 16, 18, 24, 31, 33, 45, 54, 55, 57, 58, 76], "contribut": [11, 21, 24, 35, 47, 52, 55, 65, 73], "roughli": [11, 35], "mani": [11, 16, 44, 51, 56, 58, 61, 65], "criteria": 11, "rather": [11, 20, 54], "gener": [11, 16, 18, 24, 48, 51, 52, 65, 74], "target": [11, 21, 34, 38, 51], "satisfi": [11, 41, 55, 59, 61, 72], "themselv": [11, 24], "some": [11, 15, 18, 23, 38, 39], "125": [12, 72, 75], "15": [12, 20, 31, 32, 33, 34, 52, 72, 75], "3sum": [12, 72, 75], "11": [12, 20, 72, 75], "With": [12, 13, 72, 75], "water": [12, 56, 60, 72, 75], "solut": [12, 17, 19, 22, 23, 27, 28, 29, 36, 40, 42, 46, 50, 53, 60, 62, 66, 67, 68, 69, 70, 71, 72, 74, 77], "price": 13, "dai": [13, 76], "maxim": [13, 63], "profit": 13, "choos": [13, 15, 34, 47, 64], "futur": 13, "transact": 13, "maxprofit": 13, "min_pric": 13, "max_profit": 13, "encount": [13, 14, 18, 34, 55, 56, 74], "made": [13, 27, 28, 33, 40, 43, 44, 48, 49, 51], "seen": [13, 14, 49], "far": [13, 14, 16, 37, 48, 49], "haven": [13, 37, 45], "lowest": [13, 42, 75], "sold": 13, "assum": [13, 33, 44, 54, 55, 56, 58, 61], "bought": 13, "subtract": [13, 51], "previous": [13, 43, 44], "prices1": 13, "prices2": 13, "examin": 13, "rel": [13, 28, 38, 59], "regardless": [13, 15, 23, 24, 26, 27, 38, 63, 64, 73], "memori": [13, 23, 26, 27, 28, 38, 39, 44, 52, 54, 56], "usag": [13, 23, 25, 33, 34, 37, 39, 41, 43, 44, 47, 48, 49, 52, 59, 63], "again": [13, 15, 21, 23, 43], "fee": 13, "introduc": [13, 74], "symbol": 14, "length_of_longest_substr": 14, "last": [14, 21, 23, 24, 26, 31, 43, 50, 63, 64, 72], "char_index": 14, "necessari": [14, 44, 48, 49, 52], "wa": [14, 16, 20, 26, 43, 61, 73], "new": [14, 23, 24, 25, 26, 27, 28, 31, 33, 34, 35, 38, 39, 43, 44, 45, 47, 49, 54, 58, 61, 63, 74], "fact": 14, "abcabcbb": 14, "bbbbb": 14, "pwwkew": 14, "lookup": 14, "arithmet": [14, 18], "abca": 14, "abcd": 14, "chang": [15, 24, 26, 45, 49, 56, 66, 72, 74, 75], "time": [15, 16, 17, 18, 20, 21, 23, 24, 25, 26, 27, 28, 32, 36, 40, 72, 73, 74, 75], "characterreplac": 15, "max_count": 15, "slide": [15, 16, 48, 72], "window": [15, 48, 72], "char_count": 15, "plu": [15, 34], "minu": [15, 52], "exceed": 15, "shrink": [15, 16], "side": [15, 36], "alwai": [15, 49, 54], "s1": 15, "abab": 15, "k1": [15, 37], "s2": 15, "aababba": 15, "k2": [15, 37], "space": [15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 32, 36, 40, 73, 74], "complex": [15, 16, 18, 20, 21, 23, 24, 25, 26, 27, 28, 32, 36, 40, 74, 77], "sever": [15, 39, 74], "inform": [15, 26, 55, 61, 73, 75, 76], "variat": [15, 16, 20, 21, 47, 63, 64], "testcas": 16, "min_window_substr": 16, "t_dict": 16, "current_window_dict": 16, "popul": [16, 49, 57, 58], "char": [16, 18, 43, 44, 45, 61], "min_len": 16, "float": [16, 34, 41, 48], "inf": [16, 34, 41], "min_window": 16, "required_char": 16, "found_char": 16, "window_s": 16, "left_char": 16, "further": [16, 26, 56, 61, 74, 76], "adobecodebanc": 16, "abc": [16, 52], "banc": 16, "aa": 16, "smallest": [16, 20, 21, 42, 48, 49, 61, 75], "No": [16, 73], "121": [17, 72, 75], "best": [17, 32, 34, 37, 72, 73, 75, 77], "bui": [17, 72, 75], "And": [17, 42, 46, 75], "sell": [17, 72, 75], "stock": [17, 72, 75], "424": [17, 72, 75], "76": [17, 72, 75], "bracket": 18, "close": [18, 37, 41, 55], "correct": [18, 34, 35, 47, 48, 52, 59, 61], "isvalid": 18, "map": [18, 54, 56], "bracket_map": 18, "push": [18, 37, 47, 74], "onto": [18, 37], "pop": [18, 31, 33, 35, 37], "dummi": [18, 23, 25, 27, 28], "top_el": 18, "unmatch": 18, "recent": [18, 72], "error": 18, "properli": 18, "bottom": [18, 31, 34, 35, 45, 59], "demonstr": [18, 34, 39, 43, 47, 51, 55, 58, 59, 63, 64, 73], "mainli": [18, 44, 45], "express": [18, 51, 73], "combin": [18, 24, 27, 33, 35, 53, 65, 75], "20": [19, 31, 32, 33, 34, 72, 75], "parenthes": [19, 72, 75], "suppos": [20, 45, 47], "might": [20, 21, 33, 48], "becom": [20, 21, 25, 26, 27, 31, 37, 38, 51, 57], "5000": [20, 21, 26, 57], "findmin": 20, "mid": [20, 21], "rightmost": [20, 35], "break": [20, 24, 31, 33, 40, 61, 65, 66, 75], "converg": 20, "At": [20, 34, 39, 43, 44, 45, 47, 48, 49], "search": [20, 33, 37, 39, 40, 42, 43, 46, 49, 51, 53, 54, 55, 56, 58, 59, 72], "middl": [20, 21, 24, 48], "elimin": [20, 38, 51], "met": [20, 24, 38, 52, 55, 56], "impli": 20, "13": [20, 72], "17": [20, 72], "emploi": 20, "divid": [20, 21, 28, 33, 48, 59], "halv": [20, 24, 48], "logarithm": [20, 38, 48], "reduc": [20, 28, 43, 51, 56, 59], "henc": 20, "kth": [20, 42, 47, 48, 50, 75], "extens": [20, 34, 38, 48, 54, 72], "prior": 21, "pass": [21, 23, 25, 26, 34, 49, 52], "unknown": [21, 61], "pivot": 21, "runtim": 21, "li": 21, "reset": 21, "chosen": [21, 51, 77], "target1": [21, 51], "target2": [21, 51], "target3": [21, 51], "veri": [21, 40, 49], "scale": [21, 35, 40, 64], "circularli": 21, "circular": [21, 57, 63, 64], "manner": [21, 27, 32, 37], "maintain": [21, 41, 43, 45, 47, 48, 49, 54, 59, 63], "153": [22, 72, 75], "rotat": [22, 26, 70, 72, 75], "33": [22, 72, 75], "head": [23, 24, 25, 26, 27], "reach": [23, 24, 25, 26, 32, 37, 38, 43, 44, 51, 55, 58, 59, 65], "intern": 23, "po": [23, 72], "denot": [23, 59], "tail": [23, 26, 28], "connect": [23, 25, 34, 39, 54, 55, 56, 59, 60, 63, 64, 75], "singli": [23, 24, 26, 27, 28], "listnod": [23, 24, 25, 26, 27, 28], "hascycl": 23, "slow": [23, 24], "fast": [23, 24, 49], "attribut": [23, 24, 25, 27, 32, 33, 34, 38, 39, 43, 44], "eventu": [23, 38], "catch": 23, "helper": [23, 24, 25, 26, 27, 34, 40, 41, 44, 51, 52, 58, 64], "create_linked_list_with_cycl": 23, "cycle_nod": 23, "enumer": [23, 54], "purpos": [23, 24, 26, 27, 67, 68, 69, 70, 71, 75, 76], "simplifi": [23, 24, 25, 26, 27, 28, 58, 59], "creation": [23, 24, 47], "specifi": [23, 24, 44, 49, 52, 61, 73], "simul": [23, 37, 47], "easili": [23, 24, 27, 39], "particularli": [23, 43], "abil": [23, 43, 55], "reason": [23, 36, 38, 44, 56, 73], "speed": 23, "approxim": [23, 31, 36, 41, 47], "l_0": 24, "l_1": 24, "l_": 24, "l_n": 24, "l_2": 24, "1000": [24, 31, 34, 39, 40, 47, 64], "reorderlist": 24, "first_half": 24, "second_half": 24, "prev": [24, 26, 63, 64], "next_nod": [24, 26], "merg": [24, 29, 37, 69, 72, 74, 75], "altern": [24, 35], "next_p1": 24, "next_p2": 24, "pattern": [24, 27, 31, 73], "describ": [24, 47, 54, 64, 76], "statement": [24, 43, 47, 51, 61, 64, 65], "detail": [24, 51, 58, 59, 61, 74, 76], "direct": [24, 26, 52, 56, 57, 58, 61], "now": [24, 25, 33, 43, 47], "adjust": 24, "printlinkedlist": [24, 27], "debug": [24, 27], "displai": [24, 27], "regular": [24, 27], "visual": [24, 27, 55], "head1": [24, 25, 26], "head2": [24, 25, 26], "compon": [24, 54, 59, 60, 75], "manipul": [24, 43, 47, 49], "segment": 24, "odd": [24, 31, 48], "come": [24, 39, 52, 61], "sz": 25, "removenthfromend": 25, "advanc": [25, 26, 73, 77], "_": [25, 31, 58, 59], "simultan": [25, 63], "individu": [25, 40, 56, 67, 68, 69, 70, 71, 73, 75], "ahead": 25, "createlinkedlist": [25, 26], "converttolist": 25, "familiar": [25, 74], "n1": [25, 55, 58], "new_head1": 25, "n2": [25, 55, 58], "new_head2": 25, "head3": [25, 26], "n3": 25, "new_head3": 25, "certainli": [25, 64], "isol": [25, 54], "reverselistit": 26, "reverselistrecurs": 26, "new_head": 26, "cycl": [26, 29, 54, 55, 57, 61, 72, 75], "linkedlisttolist": 26, "newli": [26, 27], "reversed_head1": 26, "reversed_head2": 26, "reversed_head3": 26, "about": [26, 55, 56, 57, 73], "state": 26, "list1": 27, "list2": 27, "splice": 27, "50": 27, "mergetwolist": 27, "serv": [27, 43, 54, 77], "placehold": 27, "attach": 27, "essenti": [27, 36, 38, 57, 65], "merged_list": 27, "why": [27, 35, 49, 73, 77], "temporari": [27, 63], "manag": 27, "alloc": [27, 44], "zigzag": [27, 31, 35], "500": 28, "exce": [28, 49, 51], "mergeklist": 28, "mergedlist": 28, "l1": 28, "l2": 28, "mergelist": 28, "conquer": [28, 77], "high": [28, 58], "level": [28, 32, 33, 35, 42, 44, 58, 59, 72, 73, 75], "overview": [28, 51], "strategi": [28, 63, 64], "lists1": 28, "lists2": 28, "lists3": 28, "fashion": 28, "across": [28, 39, 43, 77], "low": [28, 49], "overhead": [28, 32, 44], "206": [29, 72, 75], "21": [29, 72, 75], "143": [29, 72, 75], "reorder": [29, 72, 75], "19": [29, 72, 75], "nth": [29, 72, 75], "141": [29, 72, 75], "23": [29, 72, 75], "2000": [31, 35, 40, 43, 57], "levelord": 31, "bf": [31, 55, 56], "level_valu": 31, "level_s": [31, 35], "dequeu": [31, 33, 35, 57], "front": [31, 33, 57], "enqueu": [31, 33, 35], "children": [31, 33, 35, 37, 39, 43, 44, 45], "object": [31, 33, 36, 39, 43, 44, 48, 49, 56], "breadth": [31, 54, 55, 56], "balanc": [31, 32, 34, 36, 37, 38, 41, 48], "wors": 31, "invert": [32, 42, 75], "path": [32, 42, 45, 51, 52, 55, 58, 67, 72, 75, 77], "farthest": 32, "leaf": [32, 34, 37, 39], "maxdepth": 32, "left_depth": 32, "right_depth": 32, "understand": [32, 74, 76, 77], "three": [32, 33, 38, 39, 51, 52, 64, 65], "well": [32, 38, 40, 44, 48, 56, 65, 73, 77], "explor": [32, 34, 44, 45, 51, 52, 54, 55, 56, 58, 59], "skew": [32, 33, 36, 41], "deep": [32, 33, 54, 56, 59], "perfectli": [32, 37], "vari": [32, 37, 41], "shape": [32, 41], "buildtre": 33, "root_val": 33, "root_idx_inord": 33, "appropri": [33, 48, 73], "sublist": 33, "slice": 33, "subproblem": [33, 65], "treetolist": 33, "trail": 33, "b": [33, 35, 44, 45, 52, 58, 61, 74], "represent": [33, 39, 57, 61], "reflect": [33, 61], "highli": [33, 37], "postord": 33, "reconstruct": [33, 39], "adjac": [34, 45, 52, 54, 55, 56, 57, 58, 59, 61, 63, 64], "edg": [34, 54, 55, 56, 57, 58, 59, 61], "choic": 34, "option": 34, "global": 34, "ultim": 34, "aim": [34, 47, 59, 64], "maxpathsum": 34, "max_path_sum_help": 34, "left_max": 34, "right_max": 34, "max_sum": 34, "infin": [34, 41], "importantli": 34, "root1": [34, 35, 36, 37, 41], "root2": [34, 35, 36, 37, 38, 41], "42": 34, "anywher": 34, "imagin": 35, "yourself": [35, 39], "stand": 35, "rightsideview": 35, "d": [35, 45, 52], "subsequ": [35, 66, 67, 75], "obtain": [35, 39, 45], "visibl": [35, 73], "auxiliari": [35, 36, 51, 54], "leftmost": [35, 37], "inverttre": 36, "swap": [36, 38], "noth": 36, "inverted_root1": 36, "__main__": 36, "0x7faa5c445ab0": [], "inverted_root2": 36, "0x7faa5c444310": [], "root3": 36, "inverted_root3": 36, "occur": [36, 43, 51, 56], "full": 36, "practic": [36, 37, 38, 40, 41, 43, 44, 45, 51, 55, 56, 57, 77], "often": [36, 37, 41, 51, 56, 57], "mind": [36, 40, 56, 61], "task": [36, 39, 43, 45, 52, 55, 58, 59, 63], "mimic": 36, "tree": [37, 46, 60], "kthsmallest": [37, 49], "yet": [37, 58], "concis": [37, 76], "resembl": 37, "moder": [37, 40, 73], "usual": [37, 55], "bst": [38, 41, 42, 75], "lca": 38, "wikipedia": 38, "descend": [38, 40], "lowestcommonancestor": 38, "inclus": [38, 73], "simpl": [38, 39], "logic": [38, 43], "neither": 38, "advantag": [38, 76], "properti": [38, 41, 49, 55], "narrow": 38, "lead": [38, 43, 51, 73], "quick": [38, 48], "Its": 38, "suitabl": [38, 48], "file": [39, 72], "buffer": 39, "transmit": 39, "later": [39, 44], "environ": [39, 72, 73, 76], "restrict": [39, 64], "clarif": 39, "necessarili": 39, "creativ": 39, "tthe": 39, "byte": 39, "persist": 39, "context": [39, 51, 55], "recreat": 39, "flexibl": 39, "accur": [39, 54], "unambigu": 39, 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60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 84], "problem": [0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 17, 18, 20, 21, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 64, 65, 66, 67, 68, 69, 71, 72, 74, 75, 76, 77, 86, 89], "To": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 84], "see": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 85], "leetcod": [0, 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 14, 15, 16, 17, 18, 20, 21, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 64, 66, 67, 69, 71, 74, 75, 76, 77, 84, 86, 88, 89], "click": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 84, 86], "here": [0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 19, 21, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 72, 73, 75, 76, 77, 78, 79, 80, 81, 82, 83, 86, 87], "given": [0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 14, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77], "root": [0, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 57], "two": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 16, 18, 24, 25, 28, 29, 30, 31, 33, 34, 35, 36, 37, 39, 40, 43, 44, 46, 49, 50, 51, 55, 56, 57, 60, 61, 63, 65, 67, 68, 69, 72, 75, 76, 77, 83, 84], "binari": [0, 24, 25, 36, 43, 46, 48, 49, 66, 84], "p": [0, 11, 36, 44, 50, 84], "q": [0, 36, 44, 68], "write": [0, 1, 2, 3, 16, 17, 21, 24, 25, 27, 31, 36, 37, 65, 68, 72, 84], "function": [0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 36, 37, 38, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 59, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 89], "check": [0, 1, 2, 4, 10, 11, 12, 15, 16, 17, 18, 21, 24, 28, 29, 30, 31, 32, 33, 36, 37, 39, 40, 41, 44, 46, 47, 50, 51, 52, 56, 57, 61, 64, 65, 66, 67, 68, 69, 72, 76, 79, 80, 81, 82, 83, 87], "thei": [0, 1, 4, 11, 17, 21, 25, 28, 29, 30, 31, 32, 36, 37, 39, 41, 46, 49, 50, 52, 55, 56, 60, 65, 75, 76, 85, 88, 89], "ar": [0, 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 71, 72, 74, 75, 76, 77, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89], "consid": [0, 1, 4, 11, 12, 15, 16, 25, 36, 40, 43, 45, 46, 47, 56, 60, 61, 65, 67, 69, 72, 75, 76, 77, 85, 86, 89], "structur": [0, 2, 10, 14, 16, 23, 25, 27, 29, 32, 33, 35, 36, 37, 38, 39, 40, 42, 44, 45, 46, 49, 50, 52, 53, 54, 56, 57, 64, 65, 68, 71, 72, 76, 79, 80, 81, 82, 83, 86, 87], "ident": [0, 36, 45, 49, 71, 85], "node": [0, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 54, 64, 65, 67, 68, 74, 84, 87, 89], "have": [0, 1, 4, 5, 6, 7, 8, 10, 12, 13, 16, 17, 18, 19, 21, 22, 25, 26, 28, 29, 34, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 64, 65, 67, 68, 70, 72, 73, 75, 76, 78, 79, 80, 81, 82, 83, 85, 86, 87, 88], "valu": [0, 2, 5, 6, 9, 12, 14, 15, 16, 18, 21, 25, 28, 29, 30, 31, 32, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 55, 56, 57, 60, 61, 63, 64, 68, 69, 75, 77], "constraint": [0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 63, 64, 65, 66, 67, 69, 72, 75, 76, 77], "The": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 75, 76, 77, 78, 79, 80, 81, 82, 83, 85, 87], "number": [0, 1, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 67, 69, 70, 71, 72, 75, 76, 77, 83, 84, 87], "both": [0, 4, 5, 7, 10, 12, 17, 20, 27, 28, 29, 30, 31, 32, 36, 40, 43, 44, 45, 46, 47, 51, 55, 56, 57, 59, 64, 68, 69, 75, 76, 77, 89], "rang": [0, 3, 4, 5, 12, 16, 17, 18, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 40, 41, 42, 44, 45, 46, 47, 52, 54, 56, 60, 61, 64, 65, 66, 67, 68, 69, 71, 72, 77, 79, 80, 81, 82, 83, 87, 89], "0": [0, 1, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 84, 85], "10": [0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 15, 16, 17, 18, 21, 25, 28, 29, 33, 36, 38, 40, 43, 44, 45, 46, 47, 50, 51, 52, 56, 57, 69, 84], "4": [0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 15, 16, 17, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 38, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 61, 64, 65, 66, 68, 69, 75, 76, 84], "val": [0, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 57, 64], "definit": [0, 28, 31, 33, 36, 38, 39, 42, 44, 45, 46, 64], "class": [0, 4, 6, 7, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 56, 57, 61, 64, 66, 67, 69, 75, 76, 77, 85], "treenod": [0, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47], "def": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77], "__init__": [0, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 56, 57, 64, 84], "self": [0, 7, 8, 9, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 56, 57, 61, 64, 66, 67, 69, 75, 76, 77, 84, 87, 89], "left": [0, 3, 4, 10, 11, 12, 17, 18, 21, 24, 25, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 51, 52, 55, 61, 66, 69], "none": [0, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 56, 64, 84], "right": [0, 3, 10, 11, 12, 17, 18, 24, 25, 31, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 47, 49, 52, 61, 66, 69, 85], "initi": [0, 1, 3, 4, 6, 10, 11, 12, 15, 16, 17, 18, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 39, 40, 41, 43, 45, 47, 50, 51, 55, 56, 57, 60, 61, 64, 66, 67, 68, 69, 72, 75, 77], "child": [0, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 50, 51, 52], "issametre": [0, 36, 46], "base": [0, 4, 5, 6, 16, 23, 29, 31, 36, 38, 40, 42, 44, 47, 50, 51, 56, 57, 65, 71, 72, 75, 76, 77, 84], "If": [0, 1, 2, 4, 5, 6, 8, 11, 12, 13, 15, 16, 17, 18, 19, 21, 22, 24, 25, 26, 28, 29, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 72, 73, 75, 77, 78, 79, 80, 81, 82, 83, 86, 87, 88], "return": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 84], "true": [0, 2, 4, 11, 21, 24, 28, 36, 46, 47, 50, 51, 52, 61, 65, 67, 68, 69, 84], "either": [0, 11, 18, 29, 31, 32, 36, 40, 43, 44, 46, 61, 65, 66, 69, 75, 76, 77], "differ": [0, 1, 4, 5, 6, 10, 15, 17, 21, 28, 29, 36, 40, 45, 46, 56, 60, 68, 72, 75, 76, 77, 85], "fals": [0, 2, 4, 11, 21, 28, 36, 46, 47, 50, 51, 52, 61, 65, 67, 68, 69, 84], "current": [0, 1, 3, 5, 6, 7, 10, 12, 15, 16, 17, 18, 24, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 51, 52, 56, 60, 61, 66, 68, 69, 75, 76, 84], "equal": [0, 1, 3, 4, 5, 11, 16, 17, 24, 25, 28, 32, 36, 40, 42, 43, 46, 47, 51, 55, 56, 60, 61, 65, 69], "recurs": [0, 31, 36, 38, 39, 40, 42, 44, 45, 46, 47, 51, 52, 60, 61, 64, 65, 66, 68, 69], "subtre": [0, 36, 38, 39, 40, 42, 43, 44, 45, 47, 48, 49, 87], "entir": [0, 10, 21, 30, 36, 39, 43, 44, 46, 50, 52, 57, 64, 66, 69], "In": [0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 15, 16, 17, 18, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77, 87, 89], "thi": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 20, 21, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 71, 72, 74, 75, 76, 77, 79, 80, 81, 82, 83, 85, 86, 87, 88, 89], "code": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84], "we": [0, 1, 2, 6, 7, 9, 10, 12, 14, 15, 16, 17, 18, 20, 21, 23, 25, 27, 29, 30, 32, 33, 35, 36, 37, 38, 40, 41, 42, 43, 44, 46, 47, 49, 51, 52, 54, 55, 57, 59, 60, 61, 63, 64, 65, 67, 68, 69, 71, 72, 74, 77, 85, 86, 87, 88, 89], "defin": [0, 1, 4, 7, 16, 21, 28, 30, 32, 33, 36, 37, 38, 40, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 71, 76, 77], "repres": [0, 1, 5, 6, 10, 15, 16, 17, 18, 24, 25, 28, 29, 30, 32, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 57, 61, 64, 65, 66, 67, 68, 69, 71, 72, 75, 76, 77, 85, 89], "us": [0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 17, 18, 20, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 72, 74, 75, 76, 77, 84, 85, 88, 89], "travers": [0, 5, 10, 14, 28, 29, 30, 31, 32, 35, 36, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 54, 64, 65, 66, 69, 74, 87, 89], "compar": [0, 1, 4, 11, 18, 21, 25, 32, 33, 36, 44, 46, 49, 56, 57, 60, 65, 69, 72, 79, 80, 81, 82, 83, 87], "start": [0, 1, 5, 7, 10, 11, 12, 15, 16, 17, 18, 21, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 46, 47, 50, 51, 52, 55, 57, 60, 61, 64, 65, 66, 68, 69, 72], "which": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 59, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 85], "each": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 87, 89], "ha": [0, 2, 4, 5, 6, 10, 18, 21, 25, 28, 29, 30, 31, 32, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 50, 51, 52, 55, 56, 60, 61, 64, 65, 66, 67, 68, 75, 76, 88], "s": [0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 48, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 72, 73, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 89], "help": [0, 8, 12, 13, 16, 19, 21, 22, 24, 26, 32, 34, 36, 48, 53, 58, 60, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87, 88], "creat": [0, 1, 2, 3, 4, 5, 6, 7, 8, 11, 13, 16, 19, 20, 21, 22, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 48, 50, 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12, 15, 16, 17, 18, 21, 23, 32, 33, 35, 40, 41, 43, 45, 49, 50, 51, 55, 56, 57, 59, 75, 77], "main": [1, 10, 18, 38, 39, 44, 50, 52, 57, 60, 66, 68, 72, 76, 84, 86], "onli": [1, 5, 9, 10, 11, 12, 14, 15, 17, 20, 21, 23, 28, 29, 31, 33, 35, 40, 41, 44, 46, 47, 49, 50, 54, 55, 57, 59, 61, 63, 64, 65, 66, 69, 71, 74, 75, 76, 77, 89], "mai": [1, 5, 7, 10, 12, 17, 18, 24, 28, 29, 46, 47, 50, 51, 52, 57, 60, 61, 65, 66, 67, 68, 72, 84, 85, 86, 88, 89], "howev": [1, 28, 33, 44, 46, 50, 51, 60, 61, 66, 72, 76], "nest": [1, 12, 23, 52, 60, 61, 66, 84], "its": [1, 3, 5, 6, 7, 14, 16, 18, 21, 27, 30, 31, 32, 33, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 49, 50, 52, 55, 56, 57, 64, 65, 66, 67, 68, 69, 71, 72, 76, 79, 80, 81, 82, 83, 87, 89], "remain": [1, 2, 28, 30, 31, 32, 33, 51, 55, 56, 59, 60, 67], "few": [1, 4, 15, 16, 18, 32, 33, 60, 76], "store": [1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 29, 30, 31, 32, 33, 37, 38, 39, 41, 43, 45, 50, 51, 52, 54, 56, 57, 60, 64, 65, 67, 69, 72, 75, 77], "uniqu": [1, 2, 4, 5, 12, 14, 16, 18, 24, 25, 27, 39, 40, 44, 47, 49, 52, 60, 61, 64, 65, 67, 72, 74, 79, 87], "requir": [1, 2, 3, 4, 10, 11, 12, 18, 33, 41, 50, 51, 52, 55, 56, 57, 59, 60, 64, 67, 74, 77, 84, 89], "depend": [1, 4, 6, 10, 11, 12, 15, 17, 18, 24, 25, 28, 30, 32, 33, 38, 39, 40, 41, 42, 43, 46, 47, 51, 52, 56, 64, 65, 67, 71, 72, 75], "size": [1, 2, 4, 5, 6, 10, 11, 12, 15, 17, 18, 24, 25, 28, 29, 30, 31, 32, 33, 41, 44, 46, 52, 56, 57, 66, 68, 69, 72, 75, 76, 77, 85], "actual": [1, 18, 32, 47, 50, 55, 56, 60, 64, 65, 72, 88], "extend": [1, 5, 15, 17, 32, 39, 40, 41, 44, 49, 51, 52, 56, 72, 75, 89], "certain": [1, 55, 60, 75, 76], "absolut": 1, "e": [1, 5, 18, 21, 25, 28, 30, 31, 37, 39, 40, 43, 45, 50, 52, 55, 56, 60, 61, 64, 65, 67, 68, 72, 77, 85], "g": [1, 5, 25, 30, 39, 52, 55, 64, 65, 77], "less": [1, 10, 11, 12, 17, 24, 25, 44, 46, 47, 50, 52, 55, 56, 60, 69], "than": [1, 2, 5, 10, 11, 12, 15, 17, 18, 24, 25, 32, 40, 43, 44, 46, 47, 50, 52, 55, 56, 60, 61, 64, 67, 69, 77, 89], "k": [1, 2, 6, 8, 9, 12, 15, 16, 17, 25, 29, 31, 32, 34, 35, 43, 44, 52, 55, 57, 61, 75, 84, 87], "instead": [1, 5, 10, 12, 18, 25, 32, 33, 40, 41, 43, 55, 57, 60, 61, 72, 77], "ani": [2, 3, 5, 6, 7, 10, 11, 15, 17, 18, 20, 21, 25, 28, 29, 32, 33, 37, 39, 40, 45, 46, 47, 50, 51, 52, 56, 57, 60, 61, 65, 67, 68, 69, 85, 86, 88, 89], "appear": [2, 29, 30, 32, 39, 40, 65, 72, 85], "least": [2, 56, 57, 60, 68], "twice": [2, 76], "distinct": [2, 6, 16, 18, 25, 57, 60, 66, 68, 77], "containsdupl": 2, "unique_el": 2, "alreadi": [2, 4, 16, 50, 65, 84], "otherwis": [2, 4, 11, 21, 25, 28, 45, 46, 47, 50, 51, 52, 56, 60, 61, 65, 67, 72, 85], "add": [2, 8, 13, 16, 18, 19, 22, 26, 34, 39, 40, 41, 45, 48, 50, 52, 53, 54, 55, 56, 57, 58, 60, 62, 65, 67, 68, 70, 72, 73, 78, 79, 80, 81, 82, 83, 86, 87], "els": [2, 10, 12, 21, 24, 25, 32, 33, 39, 44, 55, 56, 64], "without": [2, 3, 5, 11, 18, 19, 20, 25, 28, 29, 30, 44, 52, 60, 61, 66, 75, 76, 84, 85, 87], "argument": [2, 5, 15, 16, 21, 25, 28, 38, 40, 42, 44, 46, 60, 61, 68], "whether": [2, 4, 14, 17, 27, 35, 46, 47, 49, 50, 51, 54, 61, 65, 66, 67, 68, 71, 72], "condit": [2, 12, 17, 24, 29, 44, 47, 60, 61, 65, 66, 69, 71, 72], "immedi": [2, 4, 50, 65], "ad": [2, 5, 7, 32, 38, 41, 51, 52, 56, 57, 64, 67, 68, 69, 72, 75, 77, 84], "util": [2, 52, 75, 84, 89], "data": [2, 10, 14, 16, 23, 25, 27, 29, 32, 33, 35, 37, 39, 42, 44, 45, 49, 50, 52, 53, 54, 57, 58, 59, 64, 65, 68, 71, 72, 76, 79, 80, 81, 82, 83, 84, 87, 88, 89], "quickli": [2, 57], "detect": [2, 28, 35, 46, 65, 67, 72], "meet": [2, 12, 20, 57, 59, 81, 87], "explain": [2, 29, 30, 31, 32, 50, 51], "earlier": [2, 51, 77], "nums3": [2, 24, 25, 76], "primari": [2, 16, 61], "affect": [2, 6, 51, 56, 85], "averag": [2, 7, 47, 52, 56], "grow": [2, 24, 56], "linearli": [2, 12, 41], "modifi": [2, 3, 5, 7, 10, 12, 15, 16, 17, 21, 24, 29, 30, 31, 37, 39, 40, 41, 43, 45, 47, 52, 55, 57, 60, 61, 64, 65, 66, 67, 69, 72, 75, 76], "paramet": [2, 28, 31, 43, 57, 61, 67], "frequenc": [2, 4, 5, 60], "count": [2, 3, 4, 5, 6, 12, 17, 18, 38, 40, 50, 60, 66, 67, 68, 71, 77, 83, 84, 87], "list": [2, 3, 4, 5, 6, 7, 11, 12, 14, 15, 21, 24, 37, 39, 41, 42, 43, 44, 45, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 84, 86, 89], "more": [2, 7, 8, 13, 19, 22, 26, 30, 31, 34, 44, 45, 47, 48, 52, 53, 54, 55, 56, 58, 61, 62, 66, 67, 68, 70, 72, 73, 78, 79, 80, 81, 82, 83, 86, 87, 89], "along": [2, 38, 75, 76], "answer": [3, 5, 6, 16, 18, 55, 56, 85, 89], "prefix": [3, 51, 52, 53, 72, 87], "suffix": 3, "guarante": [3, 5, 39, 57], "fit": 3, "32": [3, 84], "bit": [3, 45], "divis": [3, 24, 25], "30": [3, 30, 55, 60, 84], "up": [3, 4, 5, 9, 12, 14, 18, 28, 29, 30, 31, 39, 41, 43, 45, 52, 55, 56, 60, 61, 63, 64, 65, 66, 68, 75, 84, 88], "extra": [3, 10, 18, 28, 30, 31, 32, 44, 46, 75], "output": [3, 5, 10, 12, 30, 31, 33, 37, 41, 45, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77], "doe": [3, 6, 10, 11, 12, 15, 17, 21, 24, 25, 30, 32, 33, 40, 44, 50, 51, 52, 56, 57, 61, 65, 67, 68, 69], "productexceptself": 3, "len": [3, 4, 5, 7, 10, 11, 12, 16, 17, 18, 21, 24, 25, 33, 37, 41, 52, 55, 56, 57, 60, 61, 64, 65, 66, 69, 72, 76], "left_product": 3, "right_product": 3, "calcul": [3, 10, 12, 15, 16, 18, 24, 25, 38, 40, 55, 56, 59, 67, 68, 75, 76, 77], "fill": [3, 10, 77], "ones": [3, 9, 57, 63], "A": [3, 11, 28, 31, 38, 39, 40, 44, 45, 46, 47, 50, 52, 60, 61, 65, 67, 68, 85, 86, 88], "index": [3, 12, 16, 17, 24, 25, 28, 29, 39, 43, 45, 60, 61, 64], "includ": [3, 11, 17, 18, 31, 40, 45, 51, 52, 57, 60, 64, 66, 72, 75, 85, 86, 88], "accordingli": [3, 67, 72], "anoth": [3, 12, 20, 43, 45, 48, 49, 67, 68, 72, 77, 84, 87], "revers": [3, 5, 29, 34, 35, 37, 45, 52, 83, 84, 87], "order": [3, 5, 6, 12, 21, 23, 24, 25, 29, 30, 31, 32, 33, 35, 39, 40, 41, 43, 45, 48, 49, 52, 56, 57, 60, 72, 74, 87, 89], "multipli": 3, "correspond": [3, 5, 6, 21, 39, 50, 52, 60, 64, 69], "contain": [3, 4, 5, 6, 7, 8, 9, 12, 13, 14, 17, 18, 20, 21, 23, 24, 25, 29, 30, 31, 32, 33, 37, 39, 40, 41, 44, 45, 46, 47, 51, 55, 56, 60, 64, 67, 68, 69, 72, 76, 79, 80, 81, 82, 83, 84, 87], "desir": [3, 43, 88], "comput": [3, 9, 35, 45, 49, 56, 76], "It": [3, 5, 6, 10, 11, 12, 15, 16, 28, 29, 30, 31, 32, 33, 37, 38, 39, 40, 42, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 63, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 89], "result1": [3, 5, 6, 10, 25, 33, 66], "24": [3, 84], "12": [3, 10, 52, 75, 84], "result2": [3, 5, 6, 10, 25, 33, 66], "first": [3, 4, 5, 6, 15, 23, 28, 29, 30, 31, 32, 37, 38, 39, 40, 41, 42, 43, 45, 47, 51, 52, 57, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 89], "second": [3, 5, 29, 30, 43, 47, 55, 64, 77, 84], "independ": [3, 30, 32, 56, 75], "sequenti": [3, 25, 52, 61], "consum": [3, 11, 16, 17, 32, 44, 61, 65], "respect": [3, 4, 18, 29, 31, 32, 38, 46, 51, 60, 61, 67, 75, 77, 79, 80, 81, 82, 83, 85, 87], "These": [3, 4, 17, 18, 25, 28, 29, 30, 31, 32, 41, 56, 65, 72, 76, 79, 80, 81, 82, 83, 87], "occupi": [3, 29, 33, 77], "toward": [3, 11, 12, 85], "per": [3, 5, 60], "handl": [3, 5, 7, 12, 17, 29, 30, 35, 39, 42, 47, 54, 55, 57, 64, 72, 76, 84], "zero": [3, 12, 16, 31, 60, 66, 82, 87], "should": [3, 5, 7, 11, 24, 25, 28, 31, 32, 37, 45, 47, 50, 51, 55, 56, 61, 64, 65, 66, 67, 72, 77], "place": [3, 5, 29, 31, 61, 76], "modif": [3, 12], "attempt": [3, 18, 88], "directli": [3, 50, 69, 77], "string": [4, 6, 8, 9, 11, 14, 16, 17, 18, 20, 21, 23, 45, 50, 51, 52, 54, 61, 72, 84, 87, 89], "word": [4, 6, 50, 53, 54, 62, 63, 67, 68, 69, 72, 78, 87], "phrase": [4, 6, 11], "rearrang": [4, 6, 32], "letter": [4, 6, 11, 16, 17, 18, 50, 51, 52, 61, 63, 72], "typic": [4, 6, 42, 43, 47, 60, 64, 65], "origin": [4, 6, 7, 11, 24, 29, 31, 45, 52, 60, 61, 64, 65, 67, 86], "consist": [4, 6, 11, 16, 17, 18, 21, 39, 46, 50, 51, 52, 61, 66, 68, 72, 86], "lowercas": [4, 6, 11, 17, 18, 50, 51, 52, 61, 72], "english": [4, 6, 16, 17, 18, 50, 51, 52, 61], "what": [4, 85, 88], "unicod": 4, "charact": [4, 6, 7, 11, 18, 19, 20, 21, 45, 50, 51, 52, 54, 61, 72, 84, 87], "how": [4, 14, 18, 20, 21, 23, 24, 27, 28, 29, 31, 33, 37, 38, 40, 42, 43, 45, 47, 50, 51, 52, 55, 56, 57, 59, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 86, 88, 89], "would": [4, 15, 16, 24, 29, 40, 43, 47, 56, 57, 66, 67], "adapt": [4, 20, 24, 25, 27, 47, 85, 89], "your": [4, 5, 9, 14, 15, 20, 23, 25, 27, 35, 39, 45, 49, 55, 56, 59, 61, 63, 65, 68, 71, 74, 75, 79, 80, 81, 82, 83, 86, 87, 88, 89], "isanagram": 4, "str": [4, 6, 7, 45, 51, 61, 66], "bool": [4, 28, 51, 61, 67], "dictionari": [4, 5, 6, 16, 17, 18, 21, 50, 51, 52, 64, 73, 87, 89], "countt": 4, "get": [4, 5, 9, 17, 18, 20, 37, 41, 49, 52, 54, 55, 63, 64, 66, 71], "same": [4, 5, 6, 7, 11, 12, 17, 20, 21, 30, 31, 37, 39, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 56, 60, 61, 64, 65, 68, 72, 75, 76, 87], "method": [4, 5, 6, 7, 9, 21, 28, 33, 40, 45, 46, 50, 51, 56, 57, 61, 64, 66, 67, 69, 75, 76, 77], "boolean": [4, 50, 61, 68, 69], "other": [4, 6, 11, 12, 17, 18, 25, 29, 41, 44, 46, 52, 60, 64, 65, 66, 67, 68, 69, 85], "occurr": [4, 6, 7, 16, 18, 60], "line": [4, 5, 10, 14, 15, 39, 61, 84], "similarli": 4, "wai": [4, 10, 16, 18, 31, 35, 45, 49, 63, 67, 72, 77, 78, 85, 87, 89], "indic": [4, 7, 10, 11, 16, 24, 25, 28, 30, 31, 37, 39, 43, 47, 50, 51, 55, 61, 65, 66, 67, 68], "match": [4, 11, 18, 21, 28, 44, 46, 51, 52, 61, 64], "nagaram": 4, "rat": 4, "car": [4, 11], "scenario": [4, 9, 14, 15, 23, 27, 35, 39, 42, 43, 50, 54, 56, 59, 60, 65, 69, 74, 75, 76, 77, 89], "linear": [4, 15, 16, 17, 31, 32, 38, 45, 51, 66, 75, 76, 77], "simpli": [4, 16, 21, 56], "bookkeep": 4, "significantli": [4, 6, 29, 55, 56, 61, 77], "impact": [4, 18, 29, 85], "chain": 4, "longest": [4, 8, 9, 11, 19, 20, 38, 52, 78, 79, 84, 87, 89], "previou": [4, 5, 12, 30, 75, 76, 77], "bat": [4, 6], "tab": [4, 6], "cat": 4, "tac": [4, 6], "dog": [4, 6], "minimum": [4, 10, 15, 19, 20, 25, 26, 27, 56, 65, 77, 84, 87], "delet": [4, 43, 51], "make": [4, 10, 15, 20, 28, 30, 31, 32, 35, 40, 41, 44, 52, 54, 56, 57, 59, 61, 65, 69, 77, 85, 86, 89], "them": [4, 6, 11, 12, 18, 21, 30, 31, 33, 35, 40, 41, 42, 46, 49, 55, 56, 57, 61, 65, 66, 67, 68, 71, 72, 75, 89], "hello": 4, "billion": 4, "remov": [4, 11, 17, 18, 34, 35, 37, 39, 51, 55, 57, 67, 72, 84, 85, 87], "heo": 4, "billi": 4, "lo": 4, "most": [5, 13, 14, 15, 17, 21, 40, 43, 45, 50, 51, 52, 55, 56, 57, 61, 65, 67, 69, 75, 84, 87], "better": [5, 8, 13, 19, 22, 26, 34, 40, 48, 53, 58, 62, 63, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "log": [5, 12, 24, 25, 33, 37, 38, 40, 42, 43, 44, 47, 55, 56, 57], "topkfrequ": 5, "freq": 5, "accord": [5, 29, 44, 72], "c": [5, 6, 11, 39, 41, 52, 61, 69, 84], "item": 5, "append": [5, 6, 7, 12, 21, 28, 29, 30, 31, 32, 33, 37, 39, 41, 43, 45, 57, 60, 64, 65, 67, 68, 69, 72], "re": [5, 27, 57, 89], "end": [5, 7, 10, 11, 12, 16, 17, 18, 24, 25, 28, 29, 31, 32, 33, 34, 35, 38, 40, 50, 51, 52, 55, 68, 75, 84, 87], "higher": [5, 15, 61], "similar": [5, 7, 32, 39, 50], "one": [5, 7, 10, 11, 15, 17, 25, 28, 29, 30, 31, 32, 33, 42, 43, 44, 45, 46, 49, 55, 56, 60, 61, 64, 66, 68, 72, 75, 76, 77, 89], "greater": [5, 10, 11, 12, 15, 17, 24, 25, 40, 43, 44, 46, 47, 56, 77], "retriev": [5, 49, 50, 54, 55, 56, 59], "default": [5, 6, 72, 84], "abov": [5, 17, 29, 44, 51, 65, 69, 77, 84], "pair": [5, 9, 10, 18, 21, 33, 40, 65, 67, 68, 72], "bucket": 5, "group": [5, 8, 9, 31, 66, 68, 84, 87], "highest": [5, 89], "go": [5, 25, 28, 38, 39, 68, 69, 86], "down": [5, 29, 31, 37, 38, 39, 44, 46, 52, 61, 66, 77], "note": [5, 12, 28, 40, 51, 57, 66, 84, 86], "typo": 5, "exit": [5, 24, 25, 32, 84], "sort": [5, 12, 26, 27, 34, 35, 43, 57, 60, 67, 72, 84, 85, 87], "being": [5, 18, 25, 40, 46, 50, 51, 77], "name": [5, 8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 69, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "third": 5, "domin": [5, 11, 12, 18, 25, 29, 32, 37, 41, 45, 47, 52, 55, 60, 61, 69, 72], "factor": [5, 12, 18, 25, 32, 37, 41, 45, 51, 60], "105": [5, 48, 49, 84, 87], "optim": [5, 7, 9, 10, 14, 17, 18, 20, 27, 35, 37, 43, 45, 50, 52, 56, 60, 61, 66, 67, 69, 75, 76, 89], "ensur": [5, 7, 10, 11, 12, 16, 17, 23, 30, 39, 40, 41, 44, 45, 47, 51, 52, 56, 57, 61, 64, 65, 69, 71, 72, 86, 89], "klogk": 5, "prioriti": [5, 9, 55, 56], "queue": [5, 9, 37, 39, 41, 42, 44, 55, 56, 67, 72], "heap": [5, 33, 55, 56, 57], "achiev": [5, 7, 15, 17, 31, 42, 49, 60, 64, 75], "duplic": [5, 8, 9, 12, 18, 24, 39, 47, 52, 60, 64, 65, 84, 87], "multipl": [5, 6, 7, 15, 29, 44, 47, 51, 55, 68, 72], "among": [5, 40, 50, 52, 72, 74], "togeth": [6, 7, 10, 23, 30, 31, 32, 41, 55, 65, 72, 86], "collect": [6, 8, 13, 19, 22, 26, 34, 37, 48, 52, 53, 55, 58, 61, 62, 64, 67, 70, 72, 73, 78, 79, 80, 81, 82, 83, 87, 89], "import": [6, 12, 41, 46, 55, 56, 57, 61, 66, 67, 69, 72, 75, 76, 84], "defaultdict": [6, 67, 72], "groupanagram": 6, "z": [6, 72], "26": [6, 51, 84], "ord": 6, "tupl": 6, "convert": [6, 11, 29, 30, 31, 32, 39, 45, 52, 55], "associ": [6, 28, 31, 59, 68], "step": [6, 11, 12, 16, 18, 24, 28, 29, 31, 32, 33, 37, 38, 39, 41, 44, 45, 50, 51, 52, 55, 57, 60, 64, 66, 69, 72, 74, 77, 88], "specif": [6, 29, 45, 50, 51, 52, 55, 61, 65, 67, 72, 77, 79, 80, 81, 82, 83, 84, 87, 89], "ascii": [6, 11], "under": [6, 75], "constructor": [6, 50, 51], "inner": [6, 61, 84], "strs1": 6, "eat": [6, 52], "tea": 6, "tan": 6, "ate": 6, "nat": 6, "strs2": 6, "strs3": 6, "result3": [6, 10, 25, 33], "instanc": [6, 7, 39, 46, 57, 61, 64, 66, 76, 85, 89], "done": [6, 7, 11, 12, 29, 37, 40, 60, 61, 72, 84], "belong": [6, 68], "fix": [6, 15, 24, 32, 86, 88], "effect": [6, 14, 17, 20, 24, 27, 28, 29, 31, 33, 37, 39, 40, 41, 42, 46, 50, 52], "palindrom": [6, 13, 14, 78, 84, 87], "act": [6, 23, 85], "largest": [6, 33, 43, 56, 58, 59, 66, 68], "lintcod": [7, 65, 68, 72], "sent": 7, "over": [7, 18, 20, 30, 45, 56, 57, 60], "network": [7, 45, 71, 89], "back": [7, 30, 31, 45, 55], "pleas": [7, 45, 79, 80, 81, 82, 83, 86, 87, 88], "implement": [7, 12, 20, 23, 24, 31, 39, 40, 42, 45, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 64, 65, 66, 67, 68, 69, 71, 72, 75, 76, 77, 86, 87], "codec": [7, 45], "type": [7, 21, 23, 57, 61, 65, 66, 67, 69, 75, 76, 84], "rtype": 7, "replac": [7, 19, 20, 33, 61, 84, 87], "encapsul": [7, 61, 66], "involv": [7, 20, 23, 29, 30, 32, 40, 43, 50, 54, 55, 56, 59, 61, 65, 66, 69, 71, 72, 75, 85, 88], "concaten": [7, 52], "delimit": 7, "avoid": [7, 12, 21, 31, 46, 52, 60, 64, 65, 72, 75, 76, 85], "conflict": 7, "split": [7, 29, 45], "begin": [7, 9, 10, 11, 16, 17, 24, 27, 28, 29, 30, 32, 49, 71, 89], "repeatedli": [7, 33, 43, 72], "extract": [7, 39], "substr": [7, 11, 17, 19, 20, 78, 84, 87], "between": [7, 10, 14, 17, 24, 31, 41, 44, 51, 64, 65, 67, 68, 71, 72, 76, 89], "posit": [7, 12, 17, 21, 28, 30, 32, 39, 45, 47, 55, 61, 85], "befor": [7, 15, 29, 37, 40, 44, 50, 52, 56, 63, 64, 67, 72, 75, 86, 89], "revert": 7, "escap": 7, "preserv": [7, 64, 71], "content": [7, 29, 32], "even": [7, 29, 37, 44, 56, 86], "colon": 7, "input1": [7, 11], "lint": 7, "love": 7, "encoded1": 7, "decoded1": 7, 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50, 52, 60, 61, 63, 70, 71, 72, 84, 87], "anagram": [8, 9, 84, 87], "sum": [8, 9, 12, 14, 33, 48, 49, 62, 63, 69, 77, 83, 84, 87], "49": [8, 9, 10, 84, 87], "347": [8, 9, 84, 87], "top": [8, 9, 21, 37, 41, 52, 56, 69, 77, 84, 86, 87], "frequent": [8, 9, 43, 79, 80, 81, 82, 83, 84, 87], "238": [8, 9, 84, 87], "product": [8, 9, 78, 84, 87, 89], "except": [8, 9, 84, 87, 89], "659": [8, 9, 84, 87], "encod": [8, 9, 45, 49, 84, 87, 89], "decod": [8, 9, 49, 78, 84, 87, 89], "128": [8, 9, 84, 87], "consecut": [8, 9, 76, 84, 87, 89], "sequenc": [8, 9, 20, 40, 45, 52, 56, 77, 84, 87, 89], "hard": [8, 18, 19, 33, 34, 40, 45, 48, 52, 53, 56, 58, 72, 73, 81, 87], "want": [8, 13, 15, 16, 19, 22, 25, 26, 30, 34, 43, 44, 48, 53, 58, 62, 67, 70, 73, 78, 79, 80, 81, 82, 83, 87], "feel": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "free": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 85, 87], "pull": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "request": [8, 13, 19, 22, 26, 34, 48, 53, 56, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87], "open": [8, 13, 19, 21, 22, 23, 26, 34, 48, 53, 58, 62, 63, 70, 73, 78, 79, 80, 81, 82, 83, 85, 86, 87], "issu": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 85, 86, 87, 88], "discuss": [8, 13, 19, 22, 26, 28, 34, 42, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87, 89], "improv": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "offici": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 85, 87], "websit": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "forum": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "hint": [8, 13, 17, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "happi": [8, 9, 13, 14, 19, 20, 22, 23, 26, 27, 34, 35, 48, 49, 53, 54, 58, 59, 62, 63, 70, 71, 73, 74, 78, 79, 80, 81, 82, 83, 87, 89], "height": [10, 35, 38, 39, 40, 42, 43, 44, 45, 46, 47, 69], "There": [10, 25, 28, 45, 50, 51, 56, 64, 65, 67, 68, 69, 72, 77, 86], "vertic": [10, 14, 52, 61, 64, 66, 67, 72], "drawn": 10, "endpoint": 10, "th": [10, 15, 30, 43, 55, 75], "x": [10, 45, 52, 55, 61, 66, 69, 72, 86], "axi": 10, "notic": [10, 12, 24], "slant": 10, "maxarea": 10, "area": [10, 14], "max_area": 10, "pointer": [10, 11, 12, 18, 24, 25, 28, 29, 30, 31, 32, 33, 35, 84], "until": [10, 11, 12, 17, 24, 25, 29, 30, 31, 32, 33, 37, 39, 41, 43, 44, 50, 55, 66, 67, 72, 88], "h": [10, 38, 40, 42, 43, 44, 45, 52], "min": [10, 15, 16, 56, 57, 69, 72], "width": [10, 41], "distanc": 10, "shorter": 10, "inward": 10, "possibl": [10, 12, 15, 25, 40, 43, 46, 51, 52, 60, 61, 63, 66, 67, 69, 72, 77, 86], "systemat": [10, 63], "hold": [10, 17, 28], "point": [10, 12, 24, 25, 28, 29, 30, 31, 32, 40, 45, 50, 64], "taller": 10, "won": 10, "increas": [10, 12, 44, 78, 84, 87], "decreas": [10, 12, 32], "smaller": 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11, "output1": 11, "race": [11, 85], "output2": 11, "input3": 11, "output3": 11, "comprehens": [11, 54, 89], "half": [11, 24, 25, 29, 56], "still": [11, 17, 18, 24, 28, 37, 43, 47, 51, 57, 61], "primarili": [11, 17, 29, 46, 47, 51, 55, 76], "version": [11, 66, 85], "larg": [11, 44, 46, 55, 57], "addition": [11, 16, 30, 39, 40, 46, 61, 66, 68, 72], "babad": 11, "bab": 11, "aba": 11, "limit": [11, 52, 77], "allowed_char": 11, "ignor": [11, 21], "triplet": [12, 14], "j": [12, 33, 52, 61, 64, 66, 69, 72], "3000": [12, 39], "threesum": 12, "ascend": [12, 24, 25, 33, 43, 89], "skip": [12, 30, 60, 75, 76], "elif": [12, 21, 25, 44, 51, 56, 76], "neg": [12, 40, 47, 60], "Then": [12, 15, 56, 69, 72], "larger": [12, 18, 44, 56, 61], "been": [12, 18, 21, 29, 37, 39, 52, 64, 65, 67, 68, 88], "contribut": [12, 25, 29, 41, 55, 61, 65, 77, 85], "roughli": [12, 41], "mani": [12, 18, 51, 60, 66, 68, 72, 77, 89], "criteria": 12, "rather": [12, 24, 64], "gener": [12, 18, 21, 29, 56, 60, 61, 63, 77, 86, 89], "target": [9, 12, 14, 25, 27, 40, 44, 60, 63], "satisfi": [12, 47, 63, 65, 69, 72, 84], "themselv": [12, 29], "some": [12, 17, 21, 28, 44, 45], "125": [13, 14, 84, 87], "15": [13, 14, 24, 37, 38, 39, 40, 61, 84, 87], "3sum": [13, 14, 84, 87], "11": [13, 14, 24, 84, 87], "With": [13, 14, 15, 84, 87], "water": [13, 14, 66, 70, 71, 84, 87], "solut": [13, 14, 19, 20, 22, 23, 26, 28, 32, 33, 34, 42, 46, 48, 49, 53, 54, 58, 59, 62, 63, 70, 71, 73, 74, 78, 79, 80, 81, 82, 83, 84, 86, 89], "price": [15, 20], "dai": [15, 88], "maxim": [14, 15, 20, 75], "profit": [15, 20], "choos": [15, 17, 40, 55, 76], "futur": [15, 84], "transact": 15, "maxprofit": 15, "min_pric": 15, "max_profit": 15, "encount": [15, 16, 21, 23, 27, 40, 65, 66, 74, 86, 89], "made": [15, 32, 33, 39, 46, 50, 51, 56, 57, 60], "seen": [15, 16, 57], "far": [15, 16, 18, 43, 49, 56, 57], "haven": [15, 43, 52], "lowest": [15, 48, 49, 87], "sold": 15, "assum": [15, 39, 51, 64, 65, 66, 68, 72], "bought": 15, "subtract": [15, 60], "previous": [15, 50, 51], "prices1": 15, "prices2": 15, "examin": [15, 49], "rel": [15, 33, 44, 69], "regardless": [15, 17, 28, 29, 31, 32, 44, 75, 76, 85], "memori": [15, 28, 31, 32, 33, 44, 45, 51, 61, 64, 66], "usag": [15, 28, 30, 39, 40, 43, 45, 47, 50, 51, 55, 56, 57, 61, 69, 75], "again": [15, 17, 25, 28, 50], "fee": 15, "introduc": [9, 15, 59, 63, 86], "symbol": 16, "length_of_longest_substr": 16, "last": [16, 23, 25, 28, 29, 31, 37, 50, 58, 59, 75, 76, 84], "char_index": 16, "necessari": [16, 51, 56, 57, 61], "wa": [16, 18, 24, 31, 50, 72, 85], "new": [16, 28, 29, 30, 31, 32, 33, 35, 37, 39, 40, 41, 44, 45, 50, 51, 52, 54, 55, 57, 64, 68, 72, 75, 86], "fact": 16, "abcabcbb": 16, "bbbbb": 16, "pwwkew": 16, "lookup": 16, "arithmet": [16, 21], "abca": 16, "abcd": 16, "chang": [17, 29, 31, 49, 52, 57, 66, 78, 84, 86, 87], "time": [14, 17, 18, 19, 20, 21, 24, 25, 28, 29, 30, 31, 32, 33, 38, 42, 46, 59, 84, 85, 86, 87], "characterreplac": 17, "max_count": 17, "slide": [17, 18, 56, 84], "window": [17, 56, 84], "char_count": 17, "plu": [17, 40], "minu": [17, 61], "exceed": 17, "shrink": [17, 18], "side": [17, 42, 49, 89], "alwai": [17, 57, 64], "s1": 17, "abab": 17, "k1": [17, 43], "s2": 17, "aababba": 17, "k2": [17, 43], "space": [14, 17, 18, 20, 21, 24, 25, 28, 29, 30, 31, 32, 33, 38, 42, 46, 85, 86], "complex": [9, 14, 17, 18, 20, 21, 23, 24, 25, 28, 29, 30, 31, 32, 33, 35, 38, 42, 46, 49, 59, 71, 86, 89], "sever": [17, 45, 86], "inform": [17, 31, 65, 72, 85, 87, 88], "variat": [17, 18, 24, 25, 55, 75, 76], "testcas": 18, "min_window_substr": 18, "t_dict": 18, "current_window_dict": 18, "popul": [18, 57, 67, 68], "char": [18, 21, 50, 51, 52, 72], "min_len": 18, "float": [18, 40, 47, 56], "inf": [18, 40, 47], "min_window": 18, "required_char": 18, "found_char": 18, "window_s": 18, "left_char": 18, "further": [18, 31, 66, 72, 86, 88], "adobecodebanc": 18, "abc": [18, 61], "banc": 18, "aa": 18, "smallest": [18, 24, 25, 48, 49, 56, 57, 72, 87], "No": [18, 85], "121": [19, 20, 84, 87], "best": [19, 20, 38, 40, 43, 84, 85, 87], "bui": [19, 20, 84, 87], "And": [19, 48, 53, 87], "sell": [19, 20, 84, 87], "stock": [19, 20, 84, 87], "424": [19, 20, 84, 87], "76": [19, 20, 84, 87], "bracket": [21, 23], "close": [21, 23, 43, 47, 65], "correct": [21, 23, 40, 41, 55, 56, 61, 69, 72], "isvalid": 21, "map": [9, 21, 64, 66], "bracket_map": 21, "push": [21, 23, 43, 55, 74, 86], "onto": [21, 43], "pop": [21, 23, 37, 39, 41, 43], "dummi": [21, 28, 30, 32, 33], "top_el": 21, "unmatch": 21, "recent": [21, 84], "error": 21, "properli": 21, "bottom": [21, 37, 40, 41, 52, 69], "demonstr": [9, 21, 40, 45, 50, 54, 55, 60, 65, 68, 69, 71, 75, 76, 85], "mainli": [21, 51, 52], "express": [21, 35, 54, 60, 85], "combin": [21, 29, 32, 35, 39, 41, 62, 63, 77, 87], "20": [22, 37, 38, 39, 40, 84, 87], "parenthes": [22, 84, 87], "suppos": [24, 52, 55], "might": [24, 25, 39, 56], "becom": [24, 25, 30, 31, 32, 37, 43, 44, 59, 60, 67, 74, 89], "5000": [24, 25, 31, 67], "findmin": 24, "mid": [24, 25], "rightmost": [24, 41], "break": [24, 29, 37, 39, 46, 72, 77, 78, 87], "converg": 24, "At": [24, 40, 45, 50, 51, 52, 55, 56, 57], "search": [24, 39, 43, 45, 46, 48, 49, 50, 53, 54, 57, 60, 62, 63, 64, 65, 66, 68, 69, 84], "middl": [24, 25, 29, 56], "elimin": [24, 44, 60], "met": [24, 29, 44, 61, 65, 66, 89], "impli": 24, "13": [24, 84], "17": [24, 84], "emploi": [14, 24, 63], "divid": [24, 25, 27, 33, 39, 56, 69], "halv": [24, 29, 56], "logarithm": [24, 44, 56], "reduc": [24, 33, 50, 60, 66, 69], "henc": 24, "kth": [24, 48, 49, 55, 56, 58, 59, 87], "extens": [24, 40, 44, 56, 64, 84], "prior": 25, "pass": [25, 28, 30, 31, 40, 57, 61], "unknown": [25, 72], "pivot": [25, 59], "runtim": 25, "li": 25, "reset": 25, "chosen": [25, 60, 89], "target1": [25, 60], "target2": [25, 60], "target3": [25, 60], "veri": [25, 46, 57, 89], "scale": [25, 41, 46, 76], "circularli": 25, "circular": [25, 67, 75, 76], "manner": [25, 32, 38, 43], "maintain": [14, 20, 25, 47, 50, 52, 55, 56, 57, 59, 64, 69, 75], "153": [26, 27, 84, 87], "rotat": [26, 27, 31, 82, 84, 87], "33": [26, 27, 84, 87], "head": [28, 29, 30, 31, 32], "reach": [28, 29, 30, 31, 38, 43, 44, 50, 51, 60, 65, 68, 69, 77], "intern": 28, "po": [28, 84], "denot": [28, 69], "tail": [28, 31, 33], "connect": [28, 30, 40, 45, 64, 65, 66, 69, 70, 71, 75, 76, 87, 89], "singli": [28, 29, 31, 32, 33], "listnod": [28, 29, 30, 31, 32, 33], "hascycl": 28, "slow": [28, 29], "fast": [28, 29, 57], "attribut": [28, 29, 30, 32, 38, 39, 40, 44, 45, 50, 51], "eventu": [28, 44], "catch": 28, "helper": [28, 29, 30, 31, 32, 40, 46, 47, 51, 60, 61, 68, 76], "create_linked_list_with_cycl": 28, "cycle_nod": 28, "enumer": [28, 64, 84], "purpos": [28, 29, 31, 32, 79, 80, 81, 82, 83, 87, 88], "simplifi": [28, 29, 30, 31, 32, 33, 68, 69], "creation": [28, 29, 55], "specifi": [28, 29, 51, 57, 61, 72, 85], "simul": [28, 43, 55], "easili": [28, 29, 32, 45], "particularli": [20, 28, 50, 54], "abil": [14, 27, 28, 35, 50, 63, 65, 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13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87, 89], "75": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87, 89], "leetcod": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "problem": [8, 13, 19, 22, 23, 26, 34, 48, 53, 58, 62, 63, 70, 73, 78, 79, 80, 81, 82, 83, 87], "contribut": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "resourc": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "11": 10, "With": 10, "most": 10, "water": [10, 69], "125": 11, "palindrom": 11, "15": 12, "3sum": 12, "two": [13, 14, 32, 87], "pointer": [13, 14, 87], "121": 15, "best": 15, "bui": 15, "sell": 15, "stock": 15, "3": [16, 85], "substr": [16, 18], "without": 16, "repeat": [16, 17], "charact": [16, 17], "424": 17, "replac": 17, "76": 18, "minimum": [18, 24], "window": [18, 19, 20, 87], "slide": [19, 20, 87], "20": [21, 23], "parenthes": [21, 23], "stack": [22, 23, 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From bfaafaf1419cdf98411de2e2beed80d919e5c31d Mon Sep 17 00:00:00 2001
From: Mohsen Tabibian <36014419+mohsentabibian@users.noreply.github.com>
Date: Tue, 14 Nov 2023 20:35:38 +0000
Subject: [PATCH 7/7] Update documentation

---
 01. Array Hashing/1. Two Sum.html             |   6 +-
 .../128. Longest Consecutive Sequence.html    |   2 +-
 .../217. Contains Duplicate.html              |   2 +-
 .../238. Product of Array Except Self.html    |   2 +-
 01. Array Hashing/242. Valid Anagram.html     |   2 +-
 .../347. Top K Frequent Elements.html         |   2 +-
 01. Array Hashing/49. Group Anagrams.html     |   2 +-
 .../659. Encode and Decode Strings.html       |   2 +-
 01. Array Hashing/README.html                 |   2 +-
 01. Array Hashing/intro.html                  |  15 +-
 .../11. Container With Most Water.html        |   2 +-
 02. Two Pointers/125. Valid Palindrome.html   |   2 +-
 02. Two Pointers/15. 3Sum.html                |   2 +-
 02. Two Pointers/README.html                  |   2 +-
 02. Two Pointers/intro.html                   |   2 +-
 .../121. Best Time to Buy and Sell Stock.html |   2 +-
 ...ubstring Without Repeating Characters.html |   2 +-
 ...ngest Repeating Character Replacement.html |   2 +-
 .../76. Minimum Window Substring.html         |   2 +-
 03. Sliding Window/README.html                |   2 +-
 03. Sliding Window/intro.html                 |   2 +-
 04. Stack/20. Valid Parentheses.html          |   2 +-
 04. Stack/README.html                         |   2 +-
 04. Stack/intro.html                          |   2 +-
 ... Find Minimum in Rotated Sorted Array.html |   2 +-
 .../33. Search in Rotated Sorted Array.html   |   2 +-
 05. Binary Search/README.html                 |   2 +-
 05. Binary Search/intro.html                  |   2 +-
 06. Linked List/141. Linked List Cycle.html   |   2 +-
 06. Linked List/143. Reorder List.html        |   2 +-
 .../19. Remove Nth Node From End of List.html |   2 +-
 06. Linked List/206. Reverse Linked List.html |   2 +-
 .../21. Merge Two Sorted Lists.html           |   2 +-
 06. Linked List/23. Merge k Sorted Lists.html |   2 +-
 06. Linked List/README.html                   |   2 +-
 06. Linked List/intro.html                    |   2 +-
 07. Trees/100. Same Tree.html                 |   2 +-
 ...02. Binary Tree Level Order Traversal.html |   2 +-
 .../104. Maximum Depth of Binary Tree.html    |   2 +-
 ...e from Preorder and Inorder Traversal.html |   2 +-
 .../124. Binary Tree Maximum Path Sum.html    |   2 +-
 .../199. Binary Tree Right Side View.html     |   2 +-
 07. Trees/226. Invert Binary Tree.html        |   6 +-
 .../230. Kth Smallest Element in a BST.html   |   2 +-
 ...mmon Ancestor of a Binary Search Tree.html |   2 +-
 ...Serialize and Deserialize Binary Tree.html |   2 +-
 07. Trees/572. Subtree of Another Tree.html   |   2 +-
 .../98. Validate Binary Search Tree.html      |   2 +-
 07. Trees/README.html                         |   2 +-
 07. Trees/intro.html                          |   2 +-
 .../208. Implement Trie (Prefix Tree).html    |   2 +-
 ...n Add and Search Words Data Structure.html |   2 +-
 08. Tries/212. Word Search II.html            |   2 +-
 08. Tries/README.html                         |   2 +-
 08. Tries/intro.html                          |   2 +-
 .../1046. Last Stone Weight.html              |   2 +-
 .../295. Find Median from Data Stream.html    |   2 +-
 .../703. Kth Largest Element in a Stream.html |   2 +-
 09. Heap - Priority Queue/README.html         |   2 +-
 09. Heap - Priority Queue/intro.html          |   2 +-
 10. Backtracking/39. Combination Sum.html     |   2 +-
 10. Backtracking/79. Word Search.html         |   2 +-
 10. Backtracking/README.html                  |   2 +-
 10. Backtracking/intro.html                   |   2 +-
 11. Graphs/133. Clone Graph.html              |   2 +-
 11. Graphs/178. Graph Valid Tree.html         |   2 +-
 11. Graphs/200. Number of Islands.html        |   2 +-
 11. Graphs/207. Course Schedule.html          |   2 +-
 ...ted Components in an Undirected Graph.html |   2 +-
 .../417. Pacific Atlantic Water Flow.html     |   2 +-
 11. Graphs/README.html                        |   2 +-
 11. Graphs/intro.html                         |   2 +-
 .../892. Alien Dictionary.html                |   2 +-
 12. Advanced Graphs/README.html               |   2 +-
 12. Advanced Graphs/intro.html                |   2 +-
 .../198. House Robber.html                    |   2 +-
 .../213. House Robber II.html                 |   2 +-
 .../70. Climbing Stairs.html                  |   2 +-
 13. One-D Dynamic Programming/README.html     |   2 +-
 14. Two-D Dynamic Programming/README.html     |   2 +-
 15. Greedy/README.html                        |   2 +-
 16. Intervals/README.html                     |   2 +-
 17. Math Geometry/README.html                 |   2 +-
 18. Bit Manipulation/README.html              |   2 +-
 Book.html                                     | 168 ++++++++++--------
 CODE_OF_CONDUCT.html                          |   2 +-
 CONTRIBUTING.html                             |   2 +-
 README.html                                   |   2 +-
 SECURITY.html                                 |   2 +-
 _sources/01. Array Hashing/intro.md           |   4 +-
 _sources/Book.ipynb                           | 136 +++++++-------
 genindex.html                                 |   2 +-
 index.html                                    |   2 +-
 intro.html                                    |   6 +-
 objects.inv                                   | Bin 2938 -> 2945 bytes
 search.html                                   |   2 +-
 searchindex.js                                |   2 +-
 97 files changed, 273 insertions(+), 246 deletions(-)

diff --git a/01. Array Hashing/1. Two Sum.html b/01. Array Hashing/1. Two Sum.html
index 9c7592f..a95ac97 100644
--- a/01. Array Hashing/1. Two Sum.html	
+++ b/01. Array Hashing/1. Two Sum.html	
@@ -68,7 +68,7 @@
     <link rel="index" title="Index" href="../genindex.html" />
     <link rel="search" title="Search" href="../search.html" />
     <link rel="next" title="128. Longest Consecutive Sequence" href="128.%20Longest%20Consecutive%20Sequence.html" />
-    <link rel="prev" title="Introduction to Array Hashing Challenges" href="intro.html" />
+    <link rel="prev" title="Chapter 1: Array Hashing" href="intro.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
   <meta name="docsearch:language" content="en"/>
   </head>
@@ -164,7 +164,7 @@
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Chapter 1: Array Hashing</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
 <li class="toctree-l2 current active"><a class="current reference internal" href="#">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
@@ -635,7 +635,7 @@ <h2>Time and Space Complexity Analysis<a class="headerlink" href="#time-and-spac
       <i class="fa-solid fa-angle-left"></i>
       <div class="prev-next-info">
         <p class="prev-next-subtitle">previous</p>
-        <p class="prev-next-title">Introduction to Array Hashing Challenges</p>
+        <p class="prev-next-title">Chapter 1: Array Hashing</p>
       </div>
     </a>
     <a class="right-next"
diff --git a/01. Array Hashing/128. Longest Consecutive Sequence.html b/01. Array Hashing/128. Longest Consecutive Sequence.html
index 3c351b8..f909eb0 100644
--- a/01. Array Hashing/128. Longest Consecutive Sequence.html	
+++ b/01. Array Hashing/128. Longest Consecutive Sequence.html	
@@ -164,7 +164,7 @@
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Chapter 1: Array Hashing</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
 <li class="toctree-l2"><a class="reference internal" href="1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2 current active"><a class="current reference internal" href="#">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/01. Array Hashing/217. Contains Duplicate.html b/01. Array Hashing/217. Contains Duplicate.html
index 4080886..70a47e1 100644
--- a/01. Array Hashing/217. Contains Duplicate.html	
+++ b/01. Array Hashing/217. Contains Duplicate.html	
@@ -164,7 +164,7 @@
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Chapter 1: Array Hashing</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
 <li class="toctree-l2"><a class="reference internal" href="1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2 current active"><a class="current reference internal" href="#">217. Contains Duplicate</a></li>
diff --git a/01. Array Hashing/238. Product of Array Except Self.html b/01. Array Hashing/238. Product of Array Except Self.html
index 024a943..86b597e 100644
--- a/01. Array Hashing/238. Product of Array Except Self.html	
+++ b/01. Array Hashing/238. Product of Array Except Self.html	
@@ -164,7 +164,7 @@
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Chapter 1: Array Hashing</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
 <li class="toctree-l2"><a class="reference internal" href="1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/01. Array Hashing/242. Valid Anagram.html b/01. Array Hashing/242. Valid Anagram.html
index 5dcfd44..75d1f01 100644
--- a/01. Array Hashing/242. Valid Anagram.html	
+++ b/01. Array Hashing/242. Valid Anagram.html	
@@ -164,7 +164,7 @@
             </li>
         </ul>
         <ul class="current nav bd-sidenav">
-<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Introduction to Array Hashing Challenges</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
+<li class="toctree-l1 current active has-children"><a class="reference internal" href="intro.html">Chapter 1: Array Hashing</a><input checked="" class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul class="current">
 <li class="toctree-l2"><a class="reference internal" href="1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/01. Array Hashing/347. Top K Frequent Elements.html b/01. Array Hashing/347. Top K Frequent Elements.html
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 <li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-array-hashing-terrain">Navigating the Array Hashing Terrain</a></li>
 <li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a></li>
 <li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unlocking-pythonic-solutions">Unlocking Pythonic Solutions</a></li>
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-<h1>Introduction to Array Hashing Challenges<a class="headerlink" href="#introduction-to-array-hashing-challenges" title="Permalink to this heading">#</a></h1>
+  <section class="tex2jax_ignore mathjax_ignore" id="chapter-1-array-hashing">
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+<section id="introduction-to-array-hashing-challenges">
+<h2>Introduction to Array Hashing Challenges<a class="headerlink" href="#introduction-to-array-hashing-challenges" title="Permalink to this heading">#</a></h2>
 <p>Welcome to the Array Hashing section of “Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges.” In this segment, we will explore a curated selection of LeetCode problems focused on array hashing, a powerful technique in algorithmic problem-solving.</p>
+</section>
 <section id="navigating-the-array-hashing-terrain">
 <h2>Navigating the Array Hashing Terrain<a class="headerlink" href="#navigating-the-array-hashing-terrain" title="Permalink to this heading">#</a></h2>
 <p>Array Hashing provides an efficient approach to solving a variety of problems by transforming complex operations into simpler ones through the use of hash functions. The challenges presented in this section will not only sharpen your skills in manipulating arrays but also introduce you to the art of hashing for optimized problem-solving.</p>
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+<li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#introduction-to-array-hashing-challenges">Introduction to Array Hashing Challenges</a></li>
 <li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#navigating-the-array-hashing-terrain">Navigating the Array Hashing Terrain</a></li>
 <li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unveiling-the-challenges">Unveiling the Challenges</a></li>
 <li class="toc-h2 nav-item toc-entry"><a class="reference internal nav-link" href="#unlocking-pythonic-solutions">Unlocking Pythonic Solutions</a></li>
diff --git a/02. Two Pointers/11. Container With Most Water.html b/02. Two Pointers/11. Container With Most Water.html
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 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/14. Two-D Dynamic Programming/README.html b/14. Two-D Dynamic Programming/README.html
index 8b5be45..092b669 100644
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+++ b/14. Two-D Dynamic Programming/README.html	
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Chapter 1: Array Hashing</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/15. Greedy/README.html b/15. Greedy/README.html
index 2cead9a..8af4c04 100644
--- a/15. Greedy/README.html	
+++ b/15. Greedy/README.html	
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Chapter 1: Array Hashing</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/16. Intervals/README.html b/16. Intervals/README.html
index c31f5c2..af17f27 100644
--- a/16. Intervals/README.html	
+++ b/16. Intervals/README.html	
@@ -162,7 +162,7 @@
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 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/17. Math Geometry/README.html b/17. Math Geometry/README.html
index 0d567da..b5db50a 100644
--- a/17. Math Geometry/README.html	
+++ b/17. Math Geometry/README.html	
@@ -162,7 +162,7 @@
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 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/18. Bit Manipulation/README.html b/18. Bit Manipulation/README.html
index 4bf5238..1adfd74 100644
--- a/18. Bit Manipulation/README.html	
+++ b/18. Bit Manipulation/README.html	
@@ -161,7 +161,7 @@
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+<li class="toctree-l1 has-children"><a class="reference internal" href="../01.%20Array%20Hashing/intro.html">Chapter 1: Array Hashing</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="../01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/Book.html b/Book.html
index 9122639..2fc06b3 100644
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+++ b/Book.html
@@ -160,7 +160,7 @@
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Chapter 1: Array Hashing</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
@@ -496,14 +496,16 @@ <h2> Contents </h2>
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 Requirement already satisfied: sphinx-multitoc-numbering~=0.1.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from jupyter-book) (0.1.3)
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+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: attrs&gt;=22.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from jsonschema&lt;5-&gt;jupyter-book) (23.1.0)
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+Requirement already satisfied: uc-micro-py in /usr/local/python/3.10.8/lib/python3.10/site-packages (from linkify-it-py~=2.0.0-&gt;jupyter-book) (1.0.2)
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: uc-micro-py in /usr/local/python/3.10.8/lib/python3.10/site-packages (from linkify-it-py~=2.0.0-&gt;jupyter-book) (1.0.2)
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: importlib_metadata in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-nb~=0.17.1-&gt;jupyter-book) (6.8.0)
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@@ -527,18 +529,24 @@ <h2> Contents </h2>
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+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: pydata-sphinx-theme&gt;=0.13.3 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx-book-theme~=1.0.0-&gt;jupyter-book) (0.14.3)
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+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: setuptools in /usr/local/python/3.10.8/lib/python3.10/site-packages (from sphinx_togglebutton-&gt;jupyter-book) (68.0.0)
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+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: markdown-it-py&lt;3.0.0,&gt;=1.0.0 in /usr/local/python/3.10.8/lib/python3.10/site-packages (from myst-parser~=0.18.0-&gt;myst-nb~=0.17.1-&gt;jupyter-book) (2.2.0)
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+Requirement already satisfied: beautifulsoup4 in /home/codespace/.local/lib/python3.10/site-packages (from pydata-sphinx-theme&gt;=0.13.3-&gt;sphinx-book-theme~=1.0.0-&gt;jupyter-book) (4.12.2)
+Requirement already satisfied: accessible-pygments in /usr/local/python/3.10.8/lib/python3.10/site-packages (from pydata-sphinx-theme&gt;=0.13.3-&gt;sphinx-book-theme~=1.0.0-&gt;jupyter-book) (0.0.4)
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Requirement already satisfied: charset-normalizer&lt;4,&gt;=2 in /home/codespace/.local/lib/python3.10/site-packages (from requests&gt;=2.5.0-&gt;sphinx&lt;6,&gt;=4-&gt;jupyter-book) (3.3.0)
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@@ -676,168 +684,167 @@ <h2> Contents </h2>
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 <div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>[etoc] Changing master_doc to &#39;intro&#39;
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 Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache
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+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/143. Reorder List.ipynb: Executed notebook in 1.07 seconds [mystnb]
+<span class=" -Color -Color-Bold">reading sources... </span>[ 34%] <span class=" -Color -Color-Magenta">06. Linked List/19. Remove Nth Node From End of List</span>  
 </pre></div>
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-/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executing notebook using local CWD [mystnb]
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executing notebook using local CWD [mystnb]
 </pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 1.24 seconds [mystnb]
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 1.16 seconds [mystnb]
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 <div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]
 </pre></div>
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 1.15 seconds [mystnb]
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 0.88 seconds [mystnb]
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 /workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]
 </pre></div>
 </div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executed notebook in 1.06 seconds [mystnb]
+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span><span class=" -Color -Color-Bold">reading sources... </span>[ 37%] <span class=" -Color -Color-Magenta">06. Linked List/23. Merge k Sorted Lists</span>              
+</pre></div>
+</div>
+<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]
+</pre></div>
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 <div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>^C
 /workspaces/LeetCode-Solutions/04. Stack/intro.md:5: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]
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-<div class="output stream highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>Enumerating objects: 194, done.
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-remote: Resolving deltas: 100% (85/85), completed with 8 local objects.
+Compressing objects: 100% (34/34), done.
+Writing objects: 100% (141/141), 301.49 KiB | 75.37 MiB/s, done.
+Total 141 (delta 117), reused 126 (delta 102), pack-reused 0
+remote: Resolving deltas: 100% (117/117), completed with 15 local objects.
 To https://github.com/mohsentabibian/LeetCode-Solutions
-   9b1f6d2..f40f6c9  gh-pages -&gt; gh-pages
+   f40f6c9..bcf1c22  gh-pages -&gt; gh-pages
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diff --git a/CODE_OF_CONDUCT.html b/CODE_OF_CONDUCT.html
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             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Chapter 1: Array Hashing</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/CONTRIBUTING.html b/CONTRIBUTING.html
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             </li>
         </ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Chapter 1: Array Hashing</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/README.html b/README.html
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             </li>
         </ul>
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-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Chapter 1: Array Hashing</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/SECURITY.html b/SECURITY.html
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             </li>
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+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Chapter 1: Array Hashing</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/_sources/01. Array Hashing/intro.md b/_sources/01. Array Hashing/intro.md
index 9b6c3a2..210856e 100644
--- a/_sources/01. Array Hashing/intro.md	
+++ b/_sources/01. Array Hashing/intro.md	
@@ -1,4 +1,6 @@
-# Introduction to Array Hashing Challenges
+# Chapter 1: Array Hashing
+
+## Introduction to Array Hashing Challenges
 
 Welcome to the Array Hashing section of "Hands-On Problem-Solving in Python: Mastering the Blind 75 LeetCode Challenges." In this segment, we will explore a curated selection of LeetCode problems focused on array hashing, a powerful technique in algorithmic problem-solving.
 
diff --git a/_sources/Book.ipynb b/_sources/Book.ipynb
index 64a1214..fcb22ba 100644
--- a/_sources/Book.ipynb
+++ b/_sources/Book.ipynb
@@ -2,7 +2,7 @@
  "cells": [
   {
    "cell_type": "code",
-   "execution_count": 2,
+   "execution_count": 3,
    "metadata": {},
    "outputs": [
     {
@@ -104,6 +104,10 @@
       "Requirement already satisfied: executing>=1.2.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.0.0)\n",
       "Requirement already satisfied: asttokens>=2.1.0 in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (2.4.0)\n",
       "Requirement already satisfied: pure-eval in /home/codespace/.local/lib/python3.10/site-packages (from stack-data->ipython->myst-nb~=0.17.1->jupyter-book) (0.2.2)\n",
+      "Note: you may need to restart the kernel to use updated packages.\n",
+      "Requirement already satisfied: ghp-import in /usr/local/python/3.10.8/lib/python3.10/site-packages (2.1.0)\n",
+      "Requirement already satisfied: python-dateutil>=2.8.1 in /home/codespace/.local/lib/python3.10/site-packages (from ghp-import) (2.8.2)\n",
+      "Requirement already satisfied: six>=1.5 in /home/codespace/.local/lib/python3.10/site-packages (from python-dateutil>=2.8.1->ghp-import) (1.16.0)\n",
       "Note: you may need to restart the kernel to use updated packages.\n"
      ]
     }
@@ -142,7 +146,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 1,
+   "execution_count": 4,
    "metadata": {},
    "outputs": [
     {
@@ -154,122 +158,122 @@
       "\u001b[34m\u001b[1mConfig Path: \u001b[0m/workspaces/LeetCode-Solutions/_config.yml\n",
       "\u001b[34m\u001b[1mOutput Path: \u001b[0m/workspaces/LeetCode-Solutions/_build/html\n",
       "\u001b[01mRunning Sphinx v5.0.2\u001b[39;49;00m\n",
+      "\u001b[01mmaking output directory... \u001b[39;49;00mdone\n",
       "[etoc] Changing master_doc to 'intro'\n",
-      "\u001b[01mloading pickled environment... \u001b[39;49;00mdone\n",
       "\u001b[01mmyst v0.18.1:\u001b[39;49;00m MdParserConfig(commonmark_only=False, gfm_only=False, enable_extensions=['colon_fence', 'dollarmath', 'linkify', 'substitution', 'tasklist'], disable_syntax=[], all_links_external=False, url_schemes=['mailto', 'http', 'https'], ref_domains=None, highlight_code_blocks=True, number_code_blocks=[], title_to_header=False, heading_anchors=None, heading_slug_func=None, footnote_transition=True, words_per_minute=200, sub_delimiters=('{', '}'), linkify_fuzzy_links=True, dmath_allow_labels=True, dmath_allow_space=True, dmath_allow_digits=True, dmath_double_inline=False, update_mathjax=True, mathjax_classes='tex2jax_process|mathjax_process|math|output_area')\n",
       "\u001b[01mmyst-nb v0.17.2:\u001b[39;49;00m NbParserConfig(custom_formats={}, metadata_key='mystnb', cell_metadata_key='mystnb', kernel_rgx_aliases={}, execution_mode='auto', execution_cache_path='', execution_excludepatterns=[], execution_timeout=30, execution_in_temp=False, execution_allow_errors=False, execution_raise_on_error=False, execution_show_tb=False, merge_streams=False, render_plugin='default', remove_code_source=False, remove_code_outputs=False, code_prompt_show='Show code cell {type}', code_prompt_hide='Hide code cell {type}', number_source_lines=False, output_stderr='show', render_text_lexer='myst-ansi', render_error_lexer='ipythontb', render_image_options={}, render_figure_options={}, render_markdown_format='commonmark', output_folder='build', append_css=True, metadata_to_fm=False)\n",
       "Using jupyter-cache at: /workspaces/LeetCode-Solutions/_build/.jupyter_cache\n",
       "\u001b[01mbuilding [mo]: \u001b[39;49;00mtargets for 0 po files that are out of date\n",
-      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 0 source files that are out of date\n",
-      "\u001b[01mupdating environment: \u001b[39;49;00m12 added, 78 changed, 0 removed\n",
+      "\u001b[01mbuilding [html]: \u001b[39;49;00mtargets for 90 source files that are out of date\n",
+      "\u001b[01mupdating environment: \u001b[39;49;00m[new config] 90 added, 0 changed, 0 removed\n",
       "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executing notebook using local CWD [mystnb]  \n",
-      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.21 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/01. Array Hashing/1. Two Sum.ipynb: Executed notebook in 1.24 seconds [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/06. Linked List/141. Linked List Cycle.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/06. Linked List/19. Remove Nth Node From End of List.ipynb: Executed notebook in 0.96 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/06. Linked List/206. Reverse Linked List.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/06. Linked List/21. Merge Two Sorted Lists.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/06. Linked List/23. Merge k Sorted Lists.ipynb: Executed notebook in 0.92 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executing notebook using local CWD [mystnb]      \n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/100. Same Tree.ipynb: Executed notebook in 1.27 seconds [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/102. Binary Tree Level Order Traversal.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/104. Maximum Depth of Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/105. Construct Binary Tree from Preorder and Inorder Traversal.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/199. Binary Tree Right Side View.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/226. Invert Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/230. Kth Smallest Element in a BST.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/235. Lowest Common Ancestor of a Binary Search Tree.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/297. Serialize and Deserialize Binary Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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       "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executed notebook in 0.95 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/07. Trees/572. Subtree of Another Tree.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/07. Trees/98. Validate Binary Search Tree.ipynb: Executed notebook in 0.86 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/08. Tries/208. Implement Trie (Prefix Tree).ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/08. Tries/211. Design Add and Search Words Data Structure.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executing notebook using local CWD [mystnb] \n",
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+      "/workspaces/LeetCode-Solutions/08. Tries/212. Word Search II.ipynb: Executed notebook in 0.94 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/1046. Last Stone Weight.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 0.90 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/295. Find Median from Data Stream.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/09. Heap - Priority Queue/703. Kth Largest Element in a Stream.ipynb: Executed notebook in 1.03 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executing notebook using local CWD [mystnb]\n",
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+      "/workspaces/LeetCode-Solutions/10. Backtracking/39. Combination Sum.ipynb: Executed notebook in 0.93 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executed notebook in 1.04 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/10. Backtracking/79. Word Search.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executing notebook using local CWD [mystnb]   \n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 0.92 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/133. Clone Graph.ipynb: Executed notebook in 1.09 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executed notebook in 1.06 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/178. Graph Valid Tree.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 1.27 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/200. Number of Islands.ipynb: Executed notebook in 1.01 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.25 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/207. Course Schedule.ipynb: Executed notebook in 1.20 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 1.03 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/3651. Number of Connected Components in an Undirected Graph.ipynb: Executed notebook in 1.07 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 1.11 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/11. Graphs/417. Pacific Atlantic Water Flow.ipynb: Executed notebook in 1.05 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 1.18 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/12. Advanced Graphs/892. Alien Dictionary.ipynb: Executed notebook in 0.88 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 1.03 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/198. House Robber.ipynb: Executed notebook in 0.89 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 1.13 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/213. House Robber II.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executing notebook using local CWD [mystnb]\n",
-      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 0.99 seconds [mystnb]\n",
+      "/workspaces/LeetCode-Solutions/13. One-D Dynamic Programming/70. Climbing Stairs.ipynb: Executed notebook in 1.08 seconds [mystnb]\n",
       "/workspaces/LeetCode-Solutions/Book.ipynb: Executing notebook using local CWD [mystnb]                          \n",
       "\u001b[01mreading sources... \u001b[39;49;00m[100%] \u001b[35mintro\u001b[39;49;00m                                                 \n",
       "\u001b[91m/workspaces/LeetCode-Solutions/04. Stack/intro.md:5: WARNING: Non-consecutive header level increase; H1 to H3 [myst.header]\u001b[39;49;00m\n",
@@ -307,22 +311,22 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 2,
    "metadata": {},
    "outputs": [
     {
      "name": "stdout",
      "output_type": "stream",
      "text": [
-      "Enumerating objects: 194, done.\n",
-      "Counting objects: 100% (194/194), done.\n",
+      "Enumerating objects: 241, done.\n",
+      "Counting objects: 100% (241/241), done.\n",
       "Delta compression using up to 2 threads\n",
-      "Compressing objects: 100% (23/23), done.\n",
-      "Writing objects: 100% (105/105), 219.17 KiB | 73.06 MiB/s, done.\n",
-      "Total 105 (delta 85), reused 97 (delta 77), pack-reused 0\n",
-      "remote: Resolving deltas: 100% (85/85), completed with 8 local objects.\u001b[K\n",
+      "Compressing objects: 100% (34/34), done.\n",
+      "Writing objects: 100% (141/141), 301.49 KiB | 75.37 MiB/s, done.\n",
+      "Total 141 (delta 117), reused 126 (delta 102), pack-reused 0\n",
+      "remote: Resolving deltas: 100% (117/117), completed with 15 local objects.\u001b[K\n",
       "To https://github.com/mohsentabibian/LeetCode-Solutions\n",
-      "   9b1f6d2..f40f6c9  gh-pages -> gh-pages\n"
+      "   f40f6c9..bcf1c22  gh-pages -> gh-pages\n"
      ]
     }
    ],
diff --git a/genindex.html b/genindex.html
index 5e8acc9..dfc4276 100644
--- a/genindex.html
+++ b/genindex.html
@@ -159,7 +159,7 @@
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Chapter 1: Array Hashing</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
diff --git a/index.html b/index.html
index 960bb69..3157386 100644
--- a/index.html
+++ b/index.html
@@ -1 +1 @@
-<meta http-equiv="Refresh" content="0; url=README.html" />
+<meta http-equiv="Refresh" content="0; url=intro.html" />
diff --git a/intro.html b/intro.html
index f39fbcc..34763ab 100644
--- a/intro.html
+++ b/intro.html
@@ -65,7 +65,7 @@
     <script>DOCUMENTATION_OPTIONS.pagename = 'intro';</script>
     <link rel="index" title="Index" href="genindex.html" />
     <link rel="search" title="Search" href="search.html" />
-    <link rel="next" title="Introduction to Array Hashing Challenges" href="01.%20Array%20Hashing/intro.html" />
+    <link rel="next" title="Chapter 1: Array Hashing" href="01.%20Array%20Hashing/intro.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
   <meta name="docsearch:language" content="en"/>
   </head>
@@ -161,7 +161,7 @@
             </li>
         </ul>
         <ul class="nav bd-sidenav">
-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
+<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Chapter 1: Array Hashing</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/217.%20Contains%20Duplicate.html">217. Contains Duplicate</a></li>
@@ -548,7 +548,7 @@ <h2>Beyond the Pages: A Cosmic Coding Adventure<a class="headerlink" href="#beyo
        title="next page">
       <div class="prev-next-info">
         <p class="prev-next-subtitle">next</p>
-        <p class="prev-next-title">Introduction to Array Hashing Challenges</p>
+        <p class="prev-next-title">Chapter 1: Array Hashing</p>
       </div>
       <i class="fa-solid fa-angle-right"></i>
     </a>
diff --git a/objects.inv b/objects.inv
index fa6b5e9479ca14fe20743bceb1cf5baadb16c3c4..9347a9642a435c8cadd1e4c46681e13741eef37d 100644
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-<li class="toctree-l1 has-children"><a class="reference internal" href="01.%20Array%20Hashing/intro.html">Introduction to Array Hashing Challenges</a><input class="toctree-checkbox" id="toctree-checkbox-1" name="toctree-checkbox-1" type="checkbox"/><label class="toctree-toggle" for="toctree-checkbox-1"><i class="fa-solid fa-chevron-down"></i></label><ul>
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 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/1.%20Two%20Sum.html">100: Same Tree</a></li>
 <li class="toctree-l2"><a class="reference internal" href="01.%20Array%20Hashing/128.%20Longest%20Consecutive%20Sequence.html">128. Longest Consecutive Sequence</a></li>
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Binary Tree Level Order Traversal", "104. Maximum Depth of Binary Tree", "105. Construct Binary Tree from Preorder and Inorder Traversal", "124. Binary Tree Maximum Path Sum", "199. Binary Tree Right Side View", "226: Invert Binary Tree", "230. Kth Smallest Element in a BST", "235. Lowest Common Ancestor of a Binary Search Tree", "297. Serialize and Deserialize Binary Tree", "572: Subtree of Another Tree", "98. Validate Binary Search Tree", "Trees Problems - Blind 75 LeetCode", "Introduction to Tree Challenges", "208. Implement Trie (Prefix Tree)", "211. Design Add and Search Words Data Structure", "212. Word Search II", "Trie Problems - Blind 75 LeetCode", "Unveiling the Power of Tries: A Journey in Prefix Trees", "1046. Last Stone Weight", "295. Find Median from Data Stream", "703. Kth Largest Element in a Stream", "Heap / Priority Queue Problems - Blind 75 LeetCode", "Unveiling the Power of Heaps and Priority Queues", "39. Combination Sum", "79. Word Search", "Backtracking Problems - Blind 75 LeetCode", "Introduction to Backtracking Challenges", "133. Clone Graph", "178. Graph Valid Tree", "200. Number of Islands", "207. Course Schedule", "3651. Number of Connected Components in an Undirected Graph", "417. Pacific Atlantic Water Flow", "Graphs Problems - Blind 75 LeetCode", "Introduction to Graph Challenges", "892. Alien Dictionary", "Advanced Graphs Problems - Blind 75 LeetCode", "Introduction to Advanced Graphs: Navigating the Alien Dictionary", "House Robber", "House Robber II", "70. Climbing Stairs", "1-D Dynamic Programming Problems - Blind 75 LeetCode", "Blind 75 LeetCode Problems", "Blind 75 LeetCode Problems", "Blind 75 LeetCode Problems", "Blind 75 LeetCode Problems", "Blind 75 LeetCode Problems", "&lt;no title&gt;", "Contributor Covenant Code of Conduct", "Contributing Guidelines", "Blind 75 LeetCode Solutions", "Security Policy", "Introduction: Navigating the Coding Cosmos"], "terms": {"difficulti": [0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 19, 21, 22, 24, 25, 26, 28, 29, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 72, 73, 75, 76, 77, 78, 79, 80, 81, 82, 83, 87], "easi": [0, 2, 4, 8, 11, 13, 15, 19, 21, 22, 28, 31, 32, 34, 36, 38, 42, 46, 48, 55, 57, 58, 77, 78, 80, 81, 83, 87], "link": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 14, 15, 16, 17, 18, 21, 24, 25, 29, 30, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 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24, 25, 37, 38, 39, 40, 42, 44, 55, 60, 69, 75, 84], "8": [1, 3, 10, 44, 55, 57, 60, 84], "6": [1, 3, 10, 15, 24, 25, 31, 33, 40, 42, 43, 44, 47, 60, 61, 68, 69, 84], "let": [1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 17, 18, 20, 21, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 54, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 71, 72, 74, 75, 76, 89], "provid": [1, 2, 3, 4, 5, 7, 9, 15, 16, 17, 21, 23, 24, 25, 27, 31, 33, 38, 42, 44, 45, 46, 50, 51, 52, 54, 55, 56, 57, 59, 64, 65, 66, 68, 69, 71, 72, 74, 75, 76, 77, 79, 80, 81, 82, 83, 85, 86, 87, 88, 89], "design": [1, 7, 20, 28, 29, 35, 45, 49, 53, 54, 57, 59, 64, 66, 67, 71, 72, 74, 75, 84, 87], "breakdown": [1, 56, 69], "construct": [1, 3, 37, 41, 44, 48, 49, 52, 61, 65, 67, 72, 74, 85, 87], "perform": [1, 2, 4, 6, 15, 16, 17, 18, 21, 25, 30, 32, 33, 37, 39, 40, 41, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 61, 64, 66, 67, 68, 69, 72, 75, 77], "oper": [1, 2, 3, 6, 7, 9, 12, 15, 16, 17, 18, 21, 23, 32, 33, 35, 40, 41, 43, 45, 49, 50, 51, 55, 56, 57, 59, 75, 77], "main": [1, 10, 18, 38, 39, 44, 50, 52, 57, 60, 66, 68, 72, 76, 84, 86], "onli": [1, 5, 9, 10, 11, 12, 14, 15, 17, 20, 21, 23, 28, 29, 31, 33, 35, 40, 41, 44, 46, 47, 49, 50, 54, 55, 57, 59, 61, 63, 64, 65, 66, 69, 71, 74, 75, 76, 77, 89], "mai": [1, 5, 7, 10, 12, 17, 18, 24, 28, 29, 46, 47, 50, 51, 52, 57, 60, 61, 65, 66, 67, 68, 72, 84, 85, 86, 88, 89], "howev": [1, 28, 33, 44, 46, 50, 51, 60, 61, 66, 72, 76], "nest": [1, 12, 23, 52, 60, 61, 66, 84], "its": [1, 3, 5, 6, 7, 14, 16, 18, 21, 27, 30, 31, 32, 33, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 49, 50, 52, 55, 56, 57, 64, 65, 66, 67, 68, 69, 71, 72, 76, 79, 80, 81, 82, 83, 87, 89], "remain": [1, 2, 28, 30, 31, 32, 33, 51, 55, 56, 59, 60, 67], "few": [1, 4, 15, 16, 18, 32, 33, 60, 76], "store": [1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 29, 30, 31, 32, 33, 37, 38, 39, 41, 43, 45, 50, 51, 52, 54, 56, 57, 60, 64, 65, 67, 69, 72, 75, 77], "uniqu": [1, 2, 4, 5, 12, 14, 16, 18, 24, 25, 27, 39, 40, 44, 47, 49, 52, 60, 61, 64, 65, 67, 72, 74, 79, 87], "requir": [1, 2, 3, 4, 10, 11, 12, 18, 33, 41, 50, 51, 52, 55, 56, 57, 59, 60, 64, 67, 74, 77, 84, 89], "depend": [1, 4, 6, 10, 11, 12, 15, 17, 18, 24, 25, 28, 30, 32, 33, 38, 39, 40, 41, 42, 43, 46, 47, 51, 52, 56, 64, 65, 67, 71, 72, 75], "size": [1, 2, 4, 5, 6, 10, 11, 12, 15, 17, 18, 24, 25, 28, 29, 30, 31, 32, 33, 41, 44, 46, 52, 56, 57, 66, 68, 69, 72, 75, 76, 77, 85], "actual": [1, 18, 32, 47, 50, 55, 56, 60, 64, 65, 72, 88], "extend": [1, 5, 15, 17, 32, 39, 40, 41, 44, 49, 51, 52, 56, 72, 75, 89], "certain": [1, 55, 60, 75, 76], "absolut": 1, "e": [1, 5, 18, 21, 25, 28, 30, 31, 37, 39, 40, 43, 45, 50, 52, 55, 56, 60, 61, 64, 65, 67, 68, 72, 77, 85], "g": [1, 5, 25, 30, 39, 52, 55, 64, 65, 77], "less": [1, 10, 11, 12, 17, 24, 25, 44, 46, 47, 50, 52, 55, 56, 60, 69], "than": [1, 2, 5, 10, 11, 12, 15, 17, 18, 24, 25, 32, 40, 43, 44, 46, 47, 50, 52, 55, 56, 60, 61, 64, 67, 69, 77, 89], "k": [1, 2, 6, 8, 9, 12, 15, 16, 17, 25, 29, 31, 32, 34, 35, 43, 44, 52, 55, 57, 61, 75, 84, 87], "instead": [1, 5, 10, 12, 18, 25, 32, 33, 40, 41, 43, 55, 57, 60, 61, 72, 77], "ani": [2, 3, 5, 6, 7, 10, 11, 15, 17, 18, 20, 21, 25, 28, 29, 32, 33, 37, 39, 40, 45, 46, 47, 50, 51, 52, 56, 57, 60, 61, 65, 67, 68, 69, 85, 86, 88, 89], "appear": [2, 29, 30, 32, 39, 40, 65, 72, 85], "least": [2, 56, 57, 60, 68], "twice": [2, 76], "distinct": [2, 6, 16, 18, 25, 57, 60, 66, 68, 77], "containsdupl": 2, "unique_el": 2, "alreadi": [2, 4, 16, 50, 65, 84], "otherwis": [2, 4, 11, 21, 25, 28, 45, 46, 47, 50, 51, 52, 56, 60, 61, 65, 67, 72, 85], "add": [2, 8, 13, 16, 18, 19, 22, 26, 34, 39, 40, 41, 45, 48, 50, 52, 53, 54, 55, 56, 57, 58, 60, 62, 65, 67, 68, 70, 72, 73, 78, 79, 80, 81, 82, 83, 86, 87], "els": [2, 10, 12, 21, 24, 25, 32, 33, 39, 44, 55, 56, 64], "without": [2, 3, 5, 11, 18, 19, 20, 25, 28, 29, 30, 44, 52, 60, 61, 66, 75, 76, 84, 85, 87], "argument": [2, 5, 15, 16, 21, 25, 28, 38, 40, 42, 44, 46, 60, 61, 68], "whether": [2, 4, 14, 17, 27, 35, 46, 47, 49, 50, 51, 54, 61, 65, 66, 67, 68, 71, 72], "condit": [2, 12, 17, 24, 29, 44, 47, 60, 61, 65, 66, 69, 71, 72], "immedi": [2, 4, 50, 65], "ad": [2, 5, 7, 32, 38, 41, 51, 52, 56, 57, 64, 67, 68, 69, 72, 75, 77, 84], "util": [2, 52, 75, 84, 89], "data": [2, 10, 14, 16, 23, 25, 27, 29, 32, 33, 35, 37, 39, 42, 44, 45, 49, 50, 52, 53, 54, 57, 58, 59, 64, 65, 68, 71, 72, 76, 79, 80, 81, 82, 83, 84, 87, 88, 89], "quickli": [2, 57], "detect": [2, 28, 35, 46, 65, 67, 72], "meet": [2, 12, 20, 57, 59, 81, 87], "explain": [2, 29, 30, 31, 32, 50, 51], "earlier": [2, 51, 77], "nums3": [2, 24, 25, 76], "primari": [2, 16, 61], "affect": [2, 6, 51, 56, 85], "averag": [2, 7, 47, 52, 56], "grow": [2, 24, 56], "linearli": [2, 12, 41], "modifi": [2, 3, 5, 7, 10, 12, 15, 16, 17, 21, 24, 29, 30, 31, 37, 39, 40, 41, 43, 45, 47, 52, 55, 57, 60, 61, 64, 65, 66, 67, 69, 72, 75, 76], "paramet": [2, 28, 31, 43, 57, 61, 67], "frequenc": [2, 4, 5, 60], "count": [2, 3, 4, 5, 6, 12, 17, 18, 38, 40, 50, 60, 66, 67, 68, 71, 77, 83, 84, 87], "list": [2, 3, 4, 5, 6, 7, 11, 12, 14, 15, 21, 24, 37, 39, 41, 42, 43, 44, 45, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 84, 86, 89], "more": [2, 7, 8, 13, 19, 22, 26, 30, 31, 34, 44, 45, 47, 48, 52, 53, 54, 55, 56, 58, 61, 62, 66, 67, 68, 70, 72, 73, 78, 79, 80, 81, 82, 83, 86, 87, 89], "along": [2, 38, 75, 76], "answer": [3, 5, 6, 16, 18, 55, 56, 85, 89], "prefix": [3, 51, 52, 53, 72, 87], "suffix": 3, "guarante": [3, 5, 39, 57], "fit": 3, "32": [3, 84], "bit": [3, 45], "divis": [3, 24, 25], "30": [3, 30, 55, 60, 84], "up": [3, 4, 5, 9, 12, 14, 18, 28, 29, 30, 31, 39, 41, 43, 45, 52, 55, 56, 60, 61, 63, 64, 65, 66, 68, 75, 84, 88], "extra": [3, 10, 18, 28, 30, 31, 32, 44, 46, 75], "output": [3, 5, 10, 12, 30, 31, 33, 37, 41, 45, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77], "doe": [3, 6, 10, 11, 12, 15, 17, 21, 24, 25, 30, 32, 33, 40, 44, 50, 51, 52, 56, 57, 61, 65, 67, 68, 69], "productexceptself": 3, "len": [3, 4, 5, 7, 10, 11, 12, 16, 17, 18, 21, 24, 25, 33, 37, 41, 52, 55, 56, 57, 60, 61, 64, 65, 66, 69, 72, 76], "left_product": 3, "right_product": 3, "calcul": [3, 10, 12, 15, 16, 18, 24, 25, 38, 40, 55, 56, 59, 67, 68, 75, 76, 77], "fill": [3, 10, 77], "ones": [3, 9, 57, 63], "A": [3, 11, 28, 31, 38, 39, 40, 44, 45, 46, 47, 50, 52, 60, 61, 65, 67, 68, 85, 86, 88], "index": [3, 12, 16, 17, 24, 25, 28, 29, 39, 43, 45, 60, 61, 64], "includ": [3, 11, 17, 18, 31, 40, 45, 51, 52, 57, 60, 64, 66, 72, 75, 85, 86, 88], "accordingli": [3, 67, 72], "anoth": [3, 12, 20, 43, 45, 48, 49, 67, 68, 72, 77, 84, 87], "revers": [3, 5, 29, 34, 35, 37, 45, 52, 83, 84, 87], "order": [3, 5, 6, 12, 21, 23, 24, 25, 29, 30, 31, 32, 33, 35, 39, 40, 41, 43, 45, 48, 49, 52, 56, 57, 60, 72, 74, 87, 89], "multipli": 3, "correspond": [3, 5, 6, 21, 39, 50, 52, 60, 64, 69], "contain": [3, 4, 5, 6, 7, 8, 9, 12, 13, 14, 17, 18, 20, 21, 23, 24, 25, 29, 30, 31, 32, 33, 37, 39, 40, 41, 44, 45, 46, 47, 51, 55, 56, 60, 64, 67, 68, 69, 72, 76, 79, 80, 81, 82, 83, 84, 87], "desir": [3, 43, 88], "comput": [3, 9, 35, 45, 49, 56, 76], "It": [3, 5, 6, 10, 11, 12, 15, 16, 28, 29, 30, 31, 32, 33, 37, 38, 39, 40, 42, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 63, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 89], "result1": [3, 5, 6, 10, 25, 33, 66], "24": [3, 84], "12": [3, 10, 52, 75, 84], "result2": [3, 5, 6, 10, 25, 33, 66], "first": [3, 4, 5, 6, 15, 23, 28, 29, 30, 31, 32, 37, 38, 39, 40, 41, 42, 43, 45, 47, 51, 52, 57, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 89], "second": [3, 5, 29, 30, 43, 47, 55, 64, 77, 84], "independ": [3, 30, 32, 56, 75], "sequenti": [3, 25, 52, 61], "consum": [3, 11, 16, 17, 32, 44, 61, 65], "respect": [3, 4, 18, 29, 31, 32, 38, 46, 51, 60, 61, 67, 75, 77, 79, 80, 81, 82, 83, 85, 87], "These": [3, 4, 17, 18, 25, 28, 29, 30, 31, 32, 41, 56, 65, 72, 76, 79, 80, 81, 82, 83, 87], "occupi": [3, 29, 33, 77], "toward": [3, 11, 12, 85], "per": [3, 5, 60], "handl": [3, 5, 7, 12, 17, 29, 30, 35, 39, 42, 47, 54, 55, 57, 64, 72, 76, 84], "zero": [3, 12, 16, 31, 60, 66, 82, 87], "should": [3, 5, 7, 11, 24, 25, 28, 31, 32, 37, 45, 47, 50, 51, 55, 56, 61, 64, 65, 66, 67, 72, 77], "place": [3, 5, 29, 31, 61, 76], "modif": [3, 12], "attempt": [3, 18, 88], "directli": [3, 50, 69, 77], "string": [4, 6, 8, 9, 11, 14, 16, 17, 18, 20, 21, 23, 45, 50, 51, 52, 54, 61, 72, 84, 87, 89], "word": [4, 6, 50, 53, 54, 62, 63, 67, 68, 69, 72, 78, 87], "phrase": [4, 6, 11], "rearrang": [4, 6, 32], "letter": [4, 6, 11, 16, 17, 18, 50, 51, 52, 61, 63, 72], "typic": [4, 6, 42, 43, 47, 60, 64, 65], "origin": [4, 6, 7, 11, 24, 29, 31, 45, 52, 60, 61, 64, 65, 67, 86], "consist": [4, 6, 11, 16, 17, 18, 21, 39, 46, 50, 51, 52, 61, 66, 68, 72, 86], "lowercas": [4, 6, 11, 17, 18, 50, 51, 52, 61, 72], "english": [4, 6, 16, 17, 18, 50, 51, 52, 61], "what": [4, 85, 88], "unicod": 4, "charact": [4, 6, 7, 11, 18, 19, 20, 21, 45, 50, 51, 52, 54, 61, 72, 84, 87], "how": [4, 14, 18, 20, 21, 23, 24, 27, 28, 29, 31, 33, 37, 38, 40, 42, 43, 45, 47, 50, 51, 52, 55, 56, 57, 59, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 86, 88, 89], "would": [4, 15, 16, 24, 29, 40, 43, 47, 56, 57, 66, 67], "adapt": [4, 20, 24, 25, 27, 47, 85, 89], "your": [4, 5, 9, 14, 15, 20, 23, 25, 27, 35, 39, 45, 49, 55, 56, 59, 61, 63, 65, 68, 71, 74, 75, 79, 80, 81, 82, 83, 86, 87, 88, 89], "isanagram": 4, "str": [4, 6, 7, 45, 51, 61, 66], "bool": [4, 28, 51, 61, 67], "dictionari": [4, 5, 6, 16, 17, 18, 21, 50, 51, 52, 64, 73, 87, 89], "countt": 4, "get": [4, 5, 9, 17, 18, 20, 37, 41, 49, 52, 54, 55, 63, 64, 66, 71], "same": [4, 5, 6, 7, 11, 12, 17, 20, 21, 30, 31, 37, 39, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 56, 60, 61, 64, 65, 68, 72, 75, 76, 87], "method": [4, 5, 6, 7, 9, 21, 28, 33, 40, 45, 46, 50, 51, 56, 57, 61, 64, 66, 67, 69, 75, 76, 77], "boolean": [4, 50, 61, 68, 69], "other": [4, 6, 11, 12, 17, 18, 25, 29, 41, 44, 46, 52, 60, 64, 65, 66, 67, 68, 69, 85], "occurr": [4, 6, 7, 16, 18, 60], "line": [4, 5, 10, 14, 15, 39, 61, 84], "similarli": 4, "wai": [4, 10, 16, 18, 31, 35, 45, 49, 63, 67, 72, 77, 78, 85, 87, 89], "indic": [4, 7, 10, 11, 16, 24, 25, 28, 30, 31, 37, 39, 43, 47, 50, 51, 55, 61, 65, 66, 67, 68], "match": [4, 11, 18, 21, 28, 44, 46, 51, 52, 61, 64], "nagaram": 4, "rat": 4, "car": [4, 11], "scenario": [4, 9, 14, 15, 23, 27, 35, 39, 42, 43, 50, 54, 56, 59, 60, 65, 69, 74, 75, 76, 77, 89], "linear": [4, 15, 16, 17, 31, 32, 38, 45, 51, 66, 75, 76, 77], "simpli": [4, 16, 21, 56], "bookkeep": 4, "significantli": [4, 6, 29, 55, 56, 61, 77], "impact": [4, 18, 29, 85], "chain": 4, "longest": [4, 8, 9, 11, 19, 20, 38, 52, 78, 79, 84, 87, 89], "previou": [4, 5, 12, 30, 75, 76, 77], "bat": [4, 6], "tab": [4, 6], "cat": 4, "tac": [4, 6], "dog": [4, 6], "minimum": [4, 10, 15, 19, 20, 25, 26, 27, 56, 65, 77, 84, 87], "delet": [4, 43, 51], "make": [4, 10, 15, 20, 28, 30, 31, 32, 35, 40, 41, 44, 52, 54, 56, 57, 59, 61, 65, 69, 77, 85, 86, 89], "them": [4, 6, 11, 12, 18, 21, 30, 31, 33, 35, 40, 41, 42, 46, 49, 55, 56, 57, 61, 65, 66, 67, 68, 71, 72, 75, 89], "hello": 4, "billion": 4, "remov": [4, 11, 17, 18, 34, 35, 37, 39, 51, 55, 57, 67, 72, 84, 85, 87], "heo": 4, "billi": 4, "lo": 4, "most": [5, 13, 14, 15, 17, 21, 40, 43, 45, 50, 51, 52, 55, 56, 57, 61, 65, 67, 69, 75, 84, 87], "better": [5, 8, 13, 19, 22, 26, 34, 40, 48, 53, 58, 62, 63, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "log": [5, 12, 24, 25, 33, 37, 38, 40, 42, 43, 44, 47, 55, 56, 57], "topkfrequ": 5, "freq": 5, "accord": [5, 29, 44, 72], "c": [5, 6, 11, 39, 41, 52, 61, 69, 84], "item": 5, "append": [5, 6, 7, 12, 21, 28, 29, 30, 31, 32, 33, 37, 39, 41, 43, 45, 57, 60, 64, 65, 67, 68, 69, 72], "re": [5, 27, 57, 89], "end": [5, 7, 10, 11, 12, 16, 17, 18, 24, 25, 28, 29, 31, 32, 33, 34, 35, 38, 40, 50, 51, 52, 55, 68, 75, 84, 87], "higher": [5, 15, 61], "similar": [5, 7, 32, 39, 50], "one": [5, 7, 10, 11, 15, 17, 25, 28, 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35, 37, 43, 45, 50, 52, 56, 60, 61, 66, 67, 69, 75, 76, 89], "ensur": [5, 7, 10, 11, 12, 16, 17, 23, 30, 39, 40, 41, 44, 45, 47, 51, 52, 56, 57, 61, 64, 65, 69, 71, 72, 86, 89], "klogk": 5, "prioriti": [5, 9, 55, 56], "queue": [5, 9, 37, 39, 41, 42, 44, 55, 56, 67, 72], "heap": [5, 33, 55, 56, 57], "achiev": [5, 7, 15, 17, 31, 42, 49, 60, 64, 75], "duplic": [5, 8, 9, 12, 18, 24, 39, 47, 52, 60, 64, 65, 84, 87], "multipl": [5, 6, 7, 15, 29, 44, 47, 51, 55, 68, 72], "among": [5, 40, 50, 52, 72, 74], "togeth": [6, 7, 10, 23, 30, 31, 32, 41, 55, 65, 72, 86], "collect": [6, 8, 13, 19, 22, 26, 34, 37, 48, 52, 53, 55, 58, 61, 62, 64, 67, 70, 72, 73, 78, 79, 80, 81, 82, 83, 87, 89], "import": [6, 12, 41, 46, 55, 56, 57, 61, 66, 67, 69, 72, 75, 76, 84], "defaultdict": [6, 67, 72], "groupanagram": 6, "z": [6, 72], "26": [6, 51, 84], "ord": 6, "tupl": 6, "convert": [6, 11, 29, 30, 31, 32, 39, 45, 52, 55], "associ": [6, 28, 31, 59, 68], "step": [6, 11, 12, 16, 18, 24, 28, 29, 31, 32, 33, 37, 38, 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67, 68, 69, 71, 72, 75, 76, 77, 86, 87], "codec": [7, 45], "type": [7, 21, 23, 57, 61, 65, 66, 67, 69, 75, 76, 84], "rtype": 7, "replac": [7, 19, 20, 33, 61, 84, 87], "encapsul": [7, 61, 66], "involv": [7, 20, 23, 29, 30, 32, 40, 43, 50, 54, 55, 56, 59, 61, 65, 66, 69, 71, 72, 75, 85, 88], "concaten": [7, 52], "delimit": 7, "avoid": [7, 12, 21, 31, 46, 52, 60, 64, 65, 72, 75, 76, 85], "conflict": 7, "split": [7, 29, 45], "begin": [7, 9, 10, 11, 16, 17, 24, 27, 28, 29, 30, 32, 49, 71, 89], "repeatedli": [7, 33, 43, 72], "extract": [7, 39], "substr": [7, 11, 17, 19, 20, 78, 84, 87], "between": [7, 10, 14, 17, 24, 31, 41, 44, 51, 64, 65, 67, 68, 71, 72, 76, 89], "posit": [7, 12, 17, 21, 28, 30, 32, 39, 45, 47, 55, 61, 85], "befor": [7, 15, 29, 37, 40, 44, 50, 52, 56, 63, 64, 67, 72, 75, 86, 89], "revert": 7, "escap": 7, "preserv": [7, 64, 71], "content": [7, 29, 32], "even": [7, 29, 37, 44, 56, 86], "colon": 7, "input1": [7, 11], "lint": 7, "love": 7, "encoded1": 7, "decoded1": 7, 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50, 52, 60, 61, 63, 70, 71, 72, 84, 87], "anagram": [8, 9, 84, 87], "sum": [8, 9, 12, 14, 33, 48, 49, 62, 63, 69, 77, 83, 84, 87], "49": [8, 9, 10, 84, 87], "347": [8, 9, 84, 87], "top": [8, 9, 21, 37, 41, 52, 56, 69, 77, 84, 86, 87], "frequent": [8, 9, 43, 79, 80, 81, 82, 83, 84, 87], "238": [8, 9, 84, 87], "product": [8, 9, 78, 84, 87, 89], "except": [8, 9, 84, 87, 89], "659": [8, 9, 84, 87], "encod": [8, 9, 45, 49, 84, 87, 89], "decod": [8, 9, 49, 78, 84, 87, 89], "128": [8, 9, 84, 87], "consecut": [8, 9, 76, 84, 87, 89], "sequenc": [8, 9, 20, 40, 45, 52, 56, 77, 84, 87, 89], "hard": [8, 18, 19, 33, 34, 40, 45, 48, 52, 53, 56, 58, 72, 73, 81, 87], "want": [8, 13, 15, 16, 19, 22, 25, 26, 30, 34, 43, 44, 48, 53, 58, 62, 67, 70, 73, 78, 79, 80, 81, 82, 83, 87], "feel": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "free": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 85, 87], "pull": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "request": [8, 13, 19, 22, 26, 34, 48, 53, 56, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87], "open": [8, 13, 19, 21, 22, 23, 26, 34, 48, 53, 58, 62, 63, 70, 73, 78, 79, 80, 81, 82, 83, 85, 86, 87], "issu": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 85, 86, 87, 88], "discuss": [8, 13, 19, 22, 26, 28, 34, 42, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87, 89], "improv": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "offici": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 85, 87], "websit": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "forum": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "hint": [8, 13, 17, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "happi": [8, 9, 13, 14, 19, 20, 22, 23, 26, 27, 34, 35, 48, 49, 53, 54, 58, 59, 62, 63, 70, 71, 73, 74, 78, 79, 80, 81, 82, 83, 87, 89], "height": [10, 35, 38, 39, 40, 42, 43, 44, 45, 46, 47, 69], "There": [10, 25, 28, 45, 50, 51, 56, 64, 65, 67, 68, 69, 72, 77, 86], "vertic": [10, 14, 52, 61, 64, 66, 67, 72], "drawn": 10, "endpoint": 10, "th": [10, 15, 30, 43, 55, 75], "x": [10, 45, 52, 55, 61, 66, 69, 72, 86], "axi": 10, "notic": [10, 12, 24], "slant": 10, "maxarea": 10, "area": [10, 14], "max_area": 10, "pointer": [10, 11, 12, 18, 24, 25, 28, 29, 30, 31, 32, 33, 35, 84], "until": [10, 11, 12, 17, 24, 25, 29, 30, 31, 32, 33, 37, 39, 41, 43, 44, 50, 55, 66, 67, 72, 88], "h": [10, 38, 40, 42, 43, 44, 45, 52], "min": [10, 15, 16, 56, 57, 69, 72], "width": [10, 41], "distanc": 10, "shorter": 10, "inward": 10, "possibl": [10, 12, 15, 25, 40, 43, 46, 51, 52, 60, 61, 63, 66, 67, 69, 72, 77, 86], "systemat": [10, 63], "hold": [10, 17, 28], "point": [10, 12, 24, 25, 28, 29, 30, 31, 32, 40, 45, 50, 64], "taller": 10, "won": 10, "increas": [10, 12, 44, 78, 84, 87], "decreas": [10, 12, 32], "smaller": 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11, "output1": 11, "race": [11, 85], "output2": 11, "input3": 11, "output3": 11, "comprehens": [11, 54, 89], "half": [11, 24, 25, 29, 56], "still": [11, 17, 18, 24, 28, 37, 43, 47, 51, 57, 61], "primarili": [11, 17, 29, 46, 47, 51, 55, 76], "version": [11, 66, 85], "larg": [11, 44, 46, 55, 57], "addition": [11, 16, 30, 39, 40, 46, 61, 66, 68, 72], "babad": 11, "bab": 11, "aba": 11, "limit": [11, 52, 77], "allowed_char": 11, "ignor": [11, 21], "triplet": [12, 14], "j": [12, 33, 52, 61, 64, 66, 69, 72], "3000": [12, 39], "threesum": 12, "ascend": [12, 24, 25, 33, 43, 89], "skip": [12, 30, 60, 75, 76], "elif": [12, 21, 25, 44, 51, 56, 76], "neg": [12, 40, 47, 60], "Then": [12, 15, 56, 69, 72], "larger": [12, 18, 44, 56, 61], "been": [12, 18, 21, 29, 37, 39, 52, 64, 65, 67, 68, 88], "contribut": [12, 25, 29, 41, 55, 61, 65, 77, 85], "roughli": [12, 41], "mani": [12, 18, 51, 60, 66, 68, 72, 77, 89], "criteria": 12, "rather": [12, 24, 64], "gener": [12, 18, 21, 29, 56, 60, 61, 63, 77, 86, 89], "target": [9, 12, 14, 25, 27, 40, 44, 60, 63], "satisfi": [12, 47, 63, 65, 69, 72, 84], "themselv": [12, 29], "some": [12, 17, 21, 28, 44, 45], "125": [13, 14, 84, 87], "15": [13, 14, 24, 37, 38, 39, 40, 61, 84, 87], "3sum": [13, 14, 84, 87], "11": [13, 14, 24, 84, 87], "With": [13, 14, 15, 84, 87], "water": [13, 14, 66, 70, 71, 84, 87], "solut": [13, 14, 19, 20, 22, 23, 26, 28, 32, 33, 34, 42, 46, 48, 49, 53, 54, 58, 59, 62, 63, 70, 71, 73, 74, 78, 79, 80, 81, 82, 83, 84, 86, 89], "price": [15, 20], "dai": [15, 88], "maxim": [14, 15, 20, 75], "profit": [15, 20], "choos": [15, 17, 40, 55, 76], "futur": [15, 84], "transact": 15, "maxprofit": 15, "min_pric": 15, "max_profit": 15, "encount": [15, 16, 21, 23, 27, 40, 65, 66, 74, 86, 89], "made": [15, 32, 33, 39, 46, 50, 51, 56, 57, 60], "seen": [15, 16, 57], "far": [15, 16, 18, 43, 49, 56, 57], "haven": [15, 43, 52], "lowest": [15, 48, 49, 87], "sold": 15, "assum": [15, 39, 51, 64, 65, 66, 68, 72], "bought": 15, "subtract": [15, 60], "previous": [15, 50, 51], "prices1": 15, "prices2": 15, "examin": [15, 49], "rel": [15, 33, 44, 69], "regardless": [15, 17, 28, 29, 31, 32, 44, 75, 76, 85], "memori": [15, 28, 31, 32, 33, 44, 45, 51, 61, 64, 66], "usag": [15, 28, 30, 39, 40, 43, 45, 47, 50, 51, 55, 56, 57, 61, 69, 75], "again": [15, 17, 25, 28, 50], "fee": 15, "introduc": [9, 15, 59, 63, 86], "symbol": 16, "length_of_longest_substr": 16, "last": [16, 23, 25, 28, 29, 31, 37, 50, 58, 59, 75, 76, 84], "char_index": 16, "necessari": [16, 51, 56, 57, 61], "wa": [16, 18, 24, 31, 50, 72, 85], "new": [16, 28, 29, 30, 31, 32, 33, 35, 37, 39, 40, 41, 44, 45, 50, 51, 52, 54, 55, 57, 64, 68, 72, 75, 86], "fact": 16, "abcabcbb": 16, "bbbbb": 16, "pwwkew": 16, "lookup": 16, "arithmet": [16, 21], "abca": 16, "abcd": 16, "chang": [17, 29, 31, 49, 52, 57, 66, 78, 84, 86, 87], "time": [14, 17, 18, 19, 20, 21, 24, 25, 28, 29, 30, 31, 32, 33, 38, 42, 46, 59, 84, 85, 86, 87], "characterreplac": 17, "max_count": 17, "slide": [17, 18, 56, 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"121": [19, 20, 84, 87], "best": [19, 20, 38, 40, 43, 84, 85, 87], "bui": [19, 20, 84, 87], "And": [19, 48, 53, 87], "sell": [19, 20, 84, 87], "stock": [19, 20, 84, 87], "424": [19, 20, 84, 87], "76": [19, 20, 84, 87], "bracket": [21, 23], "close": [21, 23, 43, 47, 65], "correct": [21, 23, 40, 41, 55, 56, 61, 69, 72], "isvalid": 21, "map": [9, 21, 64, 66], "bracket_map": 21, "push": [21, 23, 43, 55, 74, 86], "onto": [21, 43], "pop": [21, 23, 37, 39, 41, 43], "dummi": [21, 28, 30, 32, 33], "top_el": 21, "unmatch": 21, "recent": [21, 84], "error": 21, "properli": 21, "bottom": [21, 37, 40, 41, 52, 69], "demonstr": [9, 21, 40, 45, 50, 54, 55, 60, 65, 68, 69, 71, 75, 76, 85], "mainli": [21, 51, 52], "express": [21, 35, 54, 60, 85], "combin": [21, 29, 32, 35, 39, 41, 62, 63, 77, 87], "20": [22, 37, 38, 39, 40, 84, 87], "parenthes": [22, 84, 87], "suppos": [24, 52, 55], "might": [24, 25, 39, 56], "becom": [24, 25, 30, 31, 32, 37, 43, 44, 59, 60, 67, 74, 89], "5000": [24, 25, 31, 67], "findmin": 24, "mid": [24, 25], "rightmost": [24, 41], "break": [24, 29, 37, 39, 46, 72, 77, 78, 87], "converg": 24, "At": [24, 40, 45, 50, 51, 52, 55, 56, 57], "search": [24, 39, 43, 45, 46, 48, 49, 50, 53, 54, 57, 60, 62, 63, 64, 65, 66, 68, 69, 84], "middl": [24, 25, 29, 56], "elimin": [24, 44, 60], "met": [24, 29, 44, 61, 65, 66, 89], "impli": 24, "13": [24, 84], "17": [24, 84], "emploi": [14, 24, 63], "divid": [24, 25, 27, 33, 39, 56, 69], "halv": [24, 29, 56], "logarithm": [24, 44, 56], "reduc": [24, 33, 50, 60, 66, 69], "henc": 24, "kth": [24, 48, 49, 55, 56, 58, 59, 87], "extens": [24, 40, 44, 56, 64, 84], "prior": 25, "pass": [25, 28, 30, 31, 40, 57, 61], "unknown": [25, 72], "pivot": [25, 59], "runtim": 25, "li": 25, "reset": 25, "chosen": [25, 60, 89], "target1": [25, 60], "target2": [25, 60], "target3": [25, 60], "veri": [25, 46, 57, 89], "scale": [25, 41, 46, 76], "circularli": 25, "circular": [25, 67, 75, 76], "manner": [25, 32, 38, 43], "maintain": [14, 20, 25, 47, 50, 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74, 75, 76, 77, 84, 86, 88, 89], "click": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 84, 86], "here": [0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 19, 21, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 72, 73, 75, 76, 77, 78, 79, 80, 81, 82, 83, 86, 87], "given": [0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 14, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77], "root": [0, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 57], "two": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 16, 18, 24, 25, 28, 29, 30, 31, 33, 34, 35, 36, 37, 39, 40, 43, 44, 46, 49, 50, 51, 55, 56, 57, 60, 61, 63, 65, 67, 68, 69, 72, 75, 76, 77, 83, 84], "binari": [0, 24, 25, 36, 43, 46, 48, 49, 66, 84], "p": [0, 11, 36, 44, 50, 84], "q": [0, 36, 44, 68], "write": [0, 1, 2, 3, 16, 17, 21, 24, 25, 27, 31, 36, 37, 65, 68, 72, 84], "function": [0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 36, 37, 38, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 59, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 89], "check": [0, 1, 2, 4, 10, 11, 12, 15, 16, 17, 18, 21, 24, 28, 29, 30, 31, 32, 33, 36, 37, 39, 40, 41, 44, 46, 47, 50, 51, 52, 56, 57, 61, 64, 65, 66, 67, 68, 69, 72, 76, 79, 80, 81, 82, 83, 87], "thei": [0, 1, 4, 11, 17, 21, 25, 28, 29, 30, 31, 32, 36, 37, 39, 41, 46, 49, 50, 52, 55, 56, 60, 65, 75, 76, 85, 88, 89], "ar": [0, 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 71, 72, 74, 75, 76, 77, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89], "consid": [0, 1, 4, 11, 12, 15, 16, 25, 36, 40, 43, 45, 46, 47, 56, 60, 61, 65, 67, 69, 72, 75, 76, 77, 85, 86, 89], "structur": [0, 2, 10, 14, 16, 23, 25, 27, 29, 32, 33, 35, 36, 37, 38, 39, 40, 42, 44, 45, 46, 49, 50, 52, 53, 54, 56, 57, 64, 65, 68, 71, 72, 76, 79, 80, 81, 82, 83, 86, 87], "ident": [0, 36, 45, 49, 71, 85], "node": [0, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 54, 64, 65, 67, 68, 74, 84, 87, 89], "have": [0, 1, 4, 5, 6, 7, 8, 10, 12, 13, 16, 17, 18, 19, 21, 22, 25, 26, 28, 29, 34, 36, 37, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 64, 65, 67, 68, 70, 72, 73, 75, 76, 78, 79, 80, 81, 82, 83, 85, 86, 87, 88], "valu": [0, 2, 5, 6, 9, 12, 14, 15, 16, 18, 21, 25, 28, 29, 30, 31, 32, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 55, 56, 57, 60, 61, 63, 64, 68, 69, 75, 77], "constraint": [0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 63, 64, 65, 66, 67, 69, 72, 75, 76, 77], "The": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 75, 76, 77, 78, 79, 80, 81, 82, 83, 85, 87], "number": [0, 1, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 67, 69, 70, 71, 72, 75, 76, 77, 83, 84, 87], "both": [0, 4, 5, 7, 10, 12, 17, 20, 27, 28, 29, 30, 31, 32, 36, 40, 43, 44, 45, 46, 47, 51, 55, 56, 57, 59, 64, 68, 69, 75, 76, 77, 89], "rang": [0, 3, 4, 5, 12, 16, 17, 18, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 40, 41, 42, 44, 45, 46, 47, 52, 54, 56, 60, 61, 64, 65, 66, 67, 68, 69, 71, 72, 77, 79, 80, 81, 82, 83, 87, 89], "0": [0, 1, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 84, 85], "10": [0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 15, 16, 17, 18, 21, 25, 28, 29, 33, 36, 38, 40, 43, 44, 45, 46, 47, 50, 51, 52, 56, 57, 69, 84], "4": [0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 15, 16, 17, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 38, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 61, 64, 65, 66, 68, 69, 75, 76, 84], "val": [0, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 57, 64], "definit": [0, 28, 31, 33, 36, 38, 39, 42, 44, 45, 46, 64], "class": [0, 4, 6, 7, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 56, 57, 61, 64, 66, 67, 69, 75, 76, 77, 85], "treenod": [0, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47], "def": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77], "__init__": [0, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 56, 57, 64, 84], "self": [0, 7, 8, 9, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 56, 57, 61, 64, 66, 67, 69, 75, 76, 77, 84, 87, 89], "left": [0, 3, 4, 10, 11, 12, 17, 18, 21, 24, 25, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 51, 52, 55, 61, 66, 69], "none": [0, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 56, 64, 84], "right": [0, 3, 10, 11, 12, 17, 18, 24, 25, 31, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 47, 49, 52, 61, 66, 69, 85], "initi": [0, 1, 3, 4, 6, 10, 11, 12, 15, 16, 17, 18, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 39, 40, 41, 43, 45, 47, 50, 51, 55, 56, 57, 60, 61, 64, 66, 67, 68, 69, 72, 75, 77], "child": [0, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 50, 51, 52], "issametre": [0, 36, 46], "base": [0, 4, 5, 6, 16, 23, 29, 31, 36, 38, 40, 42, 44, 47, 50, 51, 56, 57, 65, 71, 72, 75, 76, 77, 84], "If": [0, 1, 2, 4, 5, 6, 8, 11, 12, 13, 15, 16, 17, 18, 19, 21, 22, 24, 25, 26, 28, 29, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 72, 73, 75, 77, 78, 79, 80, 81, 82, 83, 86, 87, 88], "return": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 84], "true": [0, 2, 4, 11, 21, 24, 28, 36, 46, 47, 50, 51, 52, 61, 65, 67, 68, 69, 84], "either": [0, 11, 18, 29, 31, 32, 36, 40, 43, 44, 46, 61, 65, 66, 69, 75, 76, 77], "differ": [0, 1, 4, 5, 6, 10, 15, 17, 21, 28, 29, 36, 40, 45, 46, 56, 60, 68, 72, 75, 76, 77, 85], "fals": [0, 2, 4, 11, 21, 28, 36, 46, 47, 50, 51, 52, 61, 65, 67, 68, 69, 84], "current": [0, 1, 3, 5, 6, 7, 10, 12, 15, 16, 17, 18, 24, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 51, 52, 56, 60, 61, 66, 68, 69, 75, 76, 84], "equal": [0, 1, 3, 4, 5, 11, 16, 17, 24, 25, 28, 32, 36, 40, 42, 43, 46, 47, 51, 55, 56, 60, 61, 65, 69], "recurs": [0, 31, 36, 38, 39, 40, 42, 44, 45, 46, 47, 51, 52, 60, 61, 64, 65, 66, 68, 69], "subtre": [0, 36, 38, 39, 40, 42, 43, 44, 45, 47, 48, 49, 87], "entir": [0, 10, 21, 30, 36, 39, 43, 44, 46, 50, 52, 57, 64, 66, 69], "In": [0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 15, 16, 17, 18, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77, 87, 89], "thi": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 20, 21, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 71, 72, 74, 75, 76, 77, 79, 80, 81, 82, 83, 85, 86, 87, 88, 89], "code": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84], "we": [0, 1, 2, 6, 7, 9, 10, 12, 14, 15, 16, 17, 18, 20, 21, 23, 25, 27, 29, 30, 32, 33, 35, 36, 37, 38, 40, 41, 42, 43, 44, 46, 47, 49, 51, 52, 54, 55, 57, 59, 60, 61, 63, 64, 65, 67, 68, 69, 71, 72, 74, 77, 85, 86, 87, 88, 89], "defin": [0, 1, 4, 7, 16, 21, 28, 30, 32, 33, 36, 37, 38, 40, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 71, 76, 77], "repres": [0, 1, 5, 6, 10, 15, 16, 17, 18, 24, 25, 28, 29, 30, 32, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 57, 61, 64, 65, 66, 67, 68, 69, 71, 72, 75, 76, 77, 85, 89], "us": [0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 17, 18, 20, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 72, 74, 75, 76, 77, 84, 85, 88, 89], "travers": [0, 5, 10, 14, 28, 29, 30, 31, 32, 35, 36, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 54, 64, 65, 66, 69, 74, 87, 89], "compar": [0, 1, 4, 11, 18, 21, 25, 32, 33, 36, 44, 46, 49, 56, 57, 60, 65, 69, 72, 79, 80, 81, 82, 83, 87], "start": [0, 1, 5, 7, 10, 11, 12, 15, 16, 17, 18, 21, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 46, 47, 50, 51, 52, 55, 57, 60, 61, 64, 65, 66, 68, 69, 72], "which": [0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 15, 16, 17, 18, 21, 24, 25, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 59, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 85], "each": [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 87, 89], "ha": [0, 2, 4, 5, 6, 10, 18, 21, 25, 28, 29, 30, 31, 32, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 50, 51, 52, 55, 56, 60, 61, 64, 65, 66, 67, 68, 75, 76, 88], "s": [0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 48, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 72, 73, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 89], "help": [0, 8, 12, 13, 16, 19, 21, 22, 24, 26, 32, 34, 36, 48, 53, 58, 60, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87, 88], "creat": [0, 1, 2, 3, 4, 5, 6, 7, 8, 11, 13, 16, 19, 20, 21, 22, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 76, 77, 78, 79, 80, 81, 82, 83, 86, 87, 88, 89], "work": [0, 21, 25, 28, 29, 30, 31, 33, 36, 38, 45, 46, 50, 51, 55, 56, 57, 60, 61, 68, 86, 88], "next": [0, 1, 2, 7, 11, 12, 15, 16, 28, 29, 30, 31, 32, 33, 36, 39, 40, 41, 51, 60, 66, 67, 75], "when": [0, 11, 12, 16, 18, 24, 28, 29, 31, 36, 41, 42, 43, 46, 51, 52, 54, 56, 57, 60, 61, 65, 66, 68, 79, 80, 81, 82, 83, 85, 86, 87, 89], "so": [0, 1, 5, 6, 7, 11, 12, 15, 16, 18, 24, 25, 28, 29, 31, 33, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 51, 55, 56, 57, 60, 61, 64, 66, 67, 68, 69, 76, 89], "cannot": [0, 4, 15, 29, 36, 44, 60, 61, 65, 75, 76], "becaus": [0, 1, 2, 3, 4, 7, 10, 11, 12, 15, 16, 18, 21, 24, 28, 30, 31, 36, 37, 38, 39, 41, 42, 43, 44, 46, 50, 51, 56, 57, 60, 61, 64, 65, 69, 76, 77], "final": [0, 1, 3, 4, 6, 7, 10, 12, 15, 17, 18, 21, 28, 29, 30, 31, 32, 33, 36, 37, 38, 39, 40, 41, 42, 43, 50, 51, 52, 55, 57, 60, 64, 65, 66, 69, 72, 75, 76, 77], 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64, 67, 69, 77, 89], "k": [1, 2, 6, 8, 9, 12, 15, 16, 17, 25, 29, 31, 32, 34, 35, 43, 44, 52, 55, 57, 61, 75, 84, 87], "instead": [1, 5, 10, 12, 18, 25, 32, 33, 40, 41, 43, 55, 57, 60, 61, 72, 77], "ani": [2, 3, 5, 6, 7, 10, 11, 15, 17, 18, 20, 21, 25, 28, 29, 32, 33, 37, 39, 40, 45, 46, 47, 50, 51, 52, 56, 57, 60, 61, 65, 67, 68, 69, 85, 86, 88, 89], "appear": [2, 29, 30, 32, 39, 40, 65, 72, 85], "least": [2, 56, 57, 60, 68], "twice": [2, 76], "distinct": [2, 6, 16, 18, 25, 57, 60, 66, 68, 77], "containsdupl": 2, "unique_el": 2, "alreadi": [2, 4, 16, 50, 65, 84], "otherwis": [2, 4, 11, 21, 25, 28, 45, 46, 47, 50, 51, 52, 56, 60, 61, 65, 67, 72, 85], "add": [2, 8, 13, 16, 18, 19, 22, 26, 34, 39, 40, 41, 45, 48, 50, 52, 53, 54, 55, 56, 57, 58, 60, 62, 65, 67, 68, 70, 72, 73, 78, 79, 80, 81, 82, 83, 86, 87], "els": [2, 10, 12, 21, 24, 25, 32, 33, 39, 44, 55, 56, 64], "without": [2, 3, 5, 11, 18, 19, 20, 25, 28, 29, 30, 44, 52, 60, 61, 66, 75, 76, 84, 85, 87], "argument": [2, 5, 15, 16, 21, 25, 28, 38, 40, 42, 44, 46, 60, 61, 68], "whether": [2, 4, 14, 17, 27, 35, 46, 47, 49, 50, 51, 54, 61, 65, 66, 67, 68, 71, 72], "condit": [2, 12, 17, 24, 29, 44, 47, 60, 61, 65, 66, 69, 71, 72], "immedi": [2, 4, 50, 65], "ad": [2, 5, 7, 32, 38, 41, 51, 52, 56, 57, 64, 67, 68, 69, 72, 75, 77, 84], "util": [2, 52, 75, 84, 89], "data": [2, 10, 14, 16, 23, 25, 27, 29, 32, 33, 35, 37, 39, 42, 44, 45, 49, 50, 52, 53, 54, 57, 58, 59, 64, 65, 68, 71, 72, 76, 79, 80, 81, 82, 83, 84, 87, 88, 89], "quickli": [2, 57], "detect": [2, 28, 35, 46, 65, 67, 72], "meet": [2, 12, 20, 57, 59, 81, 87], "explain": [2, 29, 30, 31, 32, 50, 51], "earlier": [2, 51, 77], "nums3": [2, 24, 25, 76], "primari": [2, 16, 61], "affect": [2, 6, 51, 56, 85], "averag": [2, 7, 47, 52, 56], "grow": [2, 24, 56], "linearli": [2, 12, 41], "modifi": [2, 3, 5, 7, 10, 12, 15, 16, 17, 21, 24, 29, 30, 31, 37, 39, 40, 41, 43, 45, 47, 52, 55, 57, 60, 61, 64, 65, 66, 67, 69, 72, 75, 76], "paramet": [2, 28, 31, 43, 57, 61, 67], "frequenc": [2, 4, 5, 60], "count": [2, 3, 4, 5, 6, 12, 17, 18, 38, 40, 50, 60, 66, 67, 68, 71, 77, 83, 84, 87], "list": [2, 3, 4, 5, 6, 7, 11, 12, 14, 15, 21, 24, 37, 39, 41, 42, 43, 44, 45, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 84, 86, 89], "more": [2, 7, 8, 13, 19, 22, 26, 30, 31, 34, 44, 45, 47, 48, 52, 53, 54, 55, 56, 58, 61, 62, 66, 67, 68, 70, 72, 73, 78, 79, 80, 81, 82, 83, 86, 87, 89], "along": [2, 38, 75, 76], "answer": [3, 5, 6, 16, 18, 55, 56, 85, 89], "prefix": [3, 51, 52, 53, 72, 87], "suffix": 3, "guarante": [3, 5, 39, 57], "fit": 3, "32": [3, 84], "bit": [3, 45], "divis": [3, 24, 25], "30": [3, 30, 55, 60, 84], "up": [3, 4, 5, 9, 12, 14, 18, 28, 29, 30, 31, 39, 41, 43, 45, 52, 55, 56, 60, 61, 63, 64, 65, 66, 68, 75, 84, 88], "extra": [3, 10, 18, 28, 30, 31, 32, 44, 46, 75], "output": [3, 5, 10, 12, 30, 31, 33, 37, 41, 45, 50, 51, 52, 55, 56, 57, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77], "doe": [3, 6, 10, 11, 12, 15, 17, 21, 24, 25, 30, 32, 33, 40, 44, 50, 51, 52, 56, 57, 61, 65, 67, 68, 69], "productexceptself": 3, "len": [3, 4, 5, 7, 10, 11, 12, 16, 17, 18, 21, 24, 25, 33, 37, 41, 52, 55, 56, 57, 60, 61, 64, 65, 66, 69, 72, 76], "left_product": 3, "right_product": 3, "calcul": [3, 10, 12, 15, 16, 18, 24, 25, 38, 40, 55, 56, 59, 67, 68, 75, 76, 77], "fill": [3, 10, 77], "ones": [3, 9, 57, 63], "A": [3, 11, 28, 31, 38, 39, 40, 44, 45, 46, 47, 50, 52, 60, 61, 65, 67, 68, 85, 86, 88], "index": [3, 12, 16, 17, 24, 25, 28, 29, 39, 43, 45, 60, 61, 64], "includ": [3, 11, 17, 18, 31, 40, 45, 51, 52, 57, 60, 64, 66, 72, 75, 85, 86, 88], "accordingli": [3, 67, 72], "anoth": [3, 12, 20, 43, 45, 48, 49, 67, 68, 72, 77, 84, 87], "revers": [3, 5, 29, 34, 35, 37, 45, 52, 83, 84, 87], "order": [3, 5, 6, 12, 21, 23, 24, 25, 29, 30, 31, 32, 33, 35, 39, 40, 41, 43, 45, 48, 49, 52, 56, 57, 60, 72, 74, 87, 89], "multipli": 3, "correspond": [3, 5, 6, 21, 39, 50, 52, 60, 64, 69], "contain": [3, 4, 5, 6, 7, 8, 9, 12, 13, 14, 17, 18, 20, 21, 23, 24, 25, 29, 30, 31, 32, 33, 37, 39, 40, 41, 44, 45, 46, 47, 51, 55, 56, 60, 64, 67, 68, 69, 72, 76, 79, 80, 81, 82, 83, 84, 87], "desir": [3, 43, 88], "comput": [3, 9, 35, 45, 49, 56, 76], "It": [3, 5, 6, 10, 11, 12, 15, 16, 28, 29, 30, 31, 32, 33, 37, 38, 39, 40, 42, 44, 45, 46, 47, 50, 51, 52, 55, 56, 57, 60, 61, 63, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 89], "result1": [3, 5, 6, 10, 25, 33, 66], "24": [3, 84], "12": [3, 10, 52, 75, 84], "result2": [3, 5, 6, 10, 25, 33, 66], "first": [3, 4, 5, 6, 15, 23, 28, 29, 30, 31, 32, 37, 38, 39, 40, 41, 42, 43, 45, 47, 51, 52, 57, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 89], "second": [3, 5, 29, 30, 43, 47, 55, 64, 77, 84], "independ": [3, 30, 32, 56, 75], "sequenti": [3, 25, 52, 61], "consum": [3, 11, 16, 17, 32, 44, 61, 65], "respect": [3, 4, 18, 29, 31, 32, 38, 46, 51, 60, 61, 67, 75, 77, 79, 80, 81, 82, 83, 85, 87], "These": [3, 4, 17, 18, 25, 28, 29, 30, 31, 32, 41, 56, 65, 72, 76, 79, 80, 81, 82, 83, 87], "occupi": [3, 29, 33, 77], "toward": [3, 11, 12, 85], "per": [3, 5, 60], "handl": [3, 5, 7, 12, 17, 29, 30, 35, 39, 42, 47, 54, 55, 57, 64, 72, 76, 84], "zero": [3, 12, 16, 31, 60, 66, 82, 87], "should": [3, 5, 7, 11, 24, 25, 28, 31, 32, 37, 45, 47, 50, 51, 55, 56, 61, 64, 65, 66, 67, 72, 77], "place": [3, 5, 29, 31, 61, 76], "modif": [3, 12], "attempt": [3, 18, 88], "directli": [3, 50, 69, 77], "string": [4, 6, 8, 9, 11, 14, 16, 17, 18, 20, 21, 23, 45, 50, 51, 52, 54, 61, 72, 84, 87, 89], "word": [4, 6, 50, 53, 54, 62, 63, 67, 68, 69, 72, 78, 87], "phrase": [4, 6, 11], "rearrang": [4, 6, 32], "letter": [4, 6, 11, 16, 17, 18, 50, 51, 52, 61, 63, 72], "typic": [4, 6, 42, 43, 47, 60, 64, 65], "origin": [4, 6, 7, 11, 24, 29, 31, 45, 52, 60, 61, 64, 65, 67, 86], "consist": [4, 6, 11, 16, 17, 18, 21, 39, 46, 50, 51, 52, 61, 66, 68, 72, 86], "lowercas": [4, 6, 11, 17, 18, 50, 51, 52, 61, 72], "english": [4, 6, 16, 17, 18, 50, 51, 52, 61], "what": [4, 85, 88], "unicod": 4, "charact": [4, 6, 7, 11, 18, 19, 20, 21, 45, 50, 51, 52, 54, 61, 72, 84, 87], "how": [4, 14, 18, 20, 21, 23, 24, 27, 28, 29, 31, 33, 37, 38, 40, 42, 43, 45, 47, 50, 51, 52, 55, 56, 57, 59, 60, 61, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 86, 88, 89], "would": [4, 15, 16, 24, 29, 40, 43, 47, 56, 57, 66, 67], "adapt": [4, 20, 24, 25, 27, 47, 85, 89], "your": [4, 5, 9, 14, 15, 20, 23, 25, 27, 35, 39, 45, 49, 55, 56, 59, 61, 63, 65, 68, 71, 74, 75, 79, 80, 81, 82, 83, 86, 87, 88, 89], "isanagram": 4, "str": [4, 6, 7, 45, 51, 61, 66], "bool": [4, 28, 51, 61, 67], "dictionari": [4, 5, 6, 16, 17, 18, 21, 50, 51, 52, 64, 73, 87, 89], "countt": 4, "get": [4, 5, 9, 17, 18, 20, 37, 41, 49, 52, 54, 55, 63, 64, 66, 71], "same": [4, 5, 6, 7, 11, 12, 17, 20, 21, 30, 31, 37, 39, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 56, 60, 61, 64, 65, 68, 72, 75, 76, 87], "method": [4, 5, 6, 7, 9, 21, 28, 33, 40, 45, 46, 50, 51, 56, 57, 61, 64, 66, 67, 69, 75, 76, 77], "boolean": [4, 50, 61, 68, 69], "other": [4, 6, 11, 12, 17, 18, 25, 29, 41, 44, 46, 52, 60, 64, 65, 66, 67, 68, 69, 85], "occurr": [4, 6, 7, 16, 18, 60], "line": [4, 5, 10, 14, 15, 39, 61, 84], "similarli": 4, "wai": [4, 10, 16, 18, 31, 35, 45, 49, 63, 67, 72, 77, 78, 85, 87, 89], "indic": [4, 7, 10, 11, 16, 24, 25, 28, 30, 31, 37, 39, 43, 47, 50, 51, 55, 61, 65, 66, 67, 68], "match": [4, 11, 18, 21, 28, 44, 46, 51, 52, 61, 64], "nagaram": 4, "rat": 4, "car": [4, 11], "scenario": [4, 9, 14, 15, 23, 27, 35, 39, 42, 43, 50, 54, 56, 59, 60, 65, 69, 74, 75, 76, 77, 89], "linear": [4, 15, 16, 17, 31, 32, 38, 45, 51, 66, 75, 76, 77], "simpli": [4, 16, 21, 56], "bookkeep": 4, "significantli": [4, 6, 29, 55, 56, 61, 77], "impact": [4, 18, 29, 85], "chain": 4, "longest": [4, 8, 9, 11, 19, 20, 38, 52, 78, 79, 84, 87, 89], "previou": [4, 5, 12, 30, 75, 76, 77], "bat": [4, 6], "tab": [4, 6], "cat": 4, "tac": [4, 6], "dog": [4, 6], "minimum": [4, 10, 15, 19, 20, 25, 26, 27, 56, 65, 77, 84, 87], "delet": [4, 43, 51], "make": [4, 10, 15, 20, 28, 30, 31, 32, 35, 40, 41, 44, 52, 54, 56, 57, 59, 61, 65, 69, 77, 85, 86, 89], "them": [4, 6, 11, 12, 18, 21, 30, 31, 33, 35, 40, 41, 42, 46, 49, 55, 56, 57, 61, 65, 66, 67, 68, 71, 72, 75, 89], "hello": 4, "billion": 4, "remov": [4, 11, 17, 18, 34, 35, 37, 39, 51, 55, 57, 67, 72, 84, 85, 87], "heo": 4, "billi": 4, "lo": 4, "most": [5, 13, 14, 15, 17, 21, 40, 43, 45, 50, 51, 52, 55, 56, 57, 61, 65, 67, 69, 75, 84, 87], "better": [5, 8, 13, 19, 22, 26, 34, 40, 48, 53, 58, 62, 63, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "log": [5, 12, 24, 25, 33, 37, 38, 40, 42, 43, 44, 47, 55, 56, 57], "topkfrequ": 5, "freq": 5, "accord": [5, 29, 44, 72], "c": [5, 6, 11, 39, 41, 52, 61, 69, 84], "item": 5, "append": [5, 6, 7, 12, 21, 28, 29, 30, 31, 32, 33, 37, 39, 41, 43, 45, 57, 60, 64, 65, 67, 68, 69, 72], "re": [5, 27, 57, 89], "end": [5, 7, 10, 11, 12, 16, 17, 18, 24, 25, 28, 29, 31, 32, 33, 34, 35, 38, 40, 50, 51, 52, 55, 68, 75, 84, 87], "higher": [5, 15, 61], "similar": [5, 7, 32, 39, 50], "one": [5, 7, 10, 11, 15, 17, 25, 28, 29, 30, 31, 32, 33, 42, 43, 44, 45, 46, 49, 55, 56, 60, 61, 64, 66, 68, 72, 75, 76, 77, 89], "greater": [5, 10, 11, 12, 15, 17, 24, 25, 40, 43, 44, 46, 47, 56, 77], "retriev": [5, 49, 50, 54, 55, 56, 59], "default": [5, 6, 72, 84], "abov": [5, 17, 29, 44, 51, 65, 69, 77, 84], "pair": [5, 9, 10, 18, 21, 33, 40, 65, 67, 68, 72], "bucket": 5, "group": [5, 8, 9, 31, 66, 68, 84, 87], "highest": [5, 89], "go": [5, 25, 28, 38, 39, 68, 69, 86], "down": [5, 29, 31, 37, 38, 39, 44, 46, 52, 61, 66, 77], "note": [5, 12, 28, 40, 51, 57, 66, 84, 86], "typo": 5, "exit": [5, 24, 25, 32, 84], "sort": [5, 12, 26, 27, 34, 35, 43, 57, 60, 67, 72, 84, 85, 87], "being": [5, 18, 25, 40, 46, 50, 51, 77], "name": [5, 8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 69, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "third": 5, "domin": [5, 11, 12, 18, 25, 29, 32, 37, 41, 45, 47, 52, 55, 60, 61, 69, 72], "factor": [5, 12, 18, 25, 32, 37, 41, 45, 51, 60], "105": [5, 48, 49, 87], "optim": [5, 7, 9, 10, 14, 17, 18, 20, 27, 35, 37, 43, 45, 50, 52, 56, 60, 61, 66, 67, 69, 75, 76, 89], "ensur": [5, 7, 10, 11, 12, 16, 17, 23, 30, 39, 40, 41, 44, 45, 47, 51, 52, 56, 57, 61, 64, 65, 69, 71, 72, 86, 89], "klogk": 5, "prioriti": [5, 9, 55, 56], "queue": [5, 9, 37, 39, 41, 42, 44, 55, 56, 67, 72], "heap": [5, 33, 55, 56, 57], "achiev": [5, 7, 15, 17, 31, 42, 49, 60, 64, 75], "duplic": [5, 8, 9, 12, 18, 24, 39, 47, 52, 60, 64, 65, 84, 87], "multipl": [5, 6, 7, 15, 29, 44, 47, 51, 55, 68, 72], "among": [5, 40, 50, 52, 72, 74], "togeth": [6, 7, 10, 23, 30, 31, 32, 41, 55, 65, 72, 86], "collect": [6, 8, 13, 19, 22, 26, 34, 37, 48, 52, 53, 55, 58, 61, 62, 64, 67, 70, 72, 73, 78, 79, 80, 81, 82, 83, 87, 89], "import": [6, 12, 41, 46, 55, 56, 57, 61, 66, 67, 69, 72, 75, 76, 84], "defaultdict": [6, 67, 72], "groupanagram": 6, "z": [6, 72], "26": [6, 51, 84], "ord": 6, "tupl": 6, "convert": [6, 11, 29, 30, 31, 32, 39, 45, 52, 55], "associ": [6, 28, 31, 59, 68], "step": [6, 11, 12, 16, 18, 24, 28, 29, 31, 32, 33, 37, 38, 39, 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68, 69, 71, 72, 75, 76, 77, 86, 87], "codec": [7, 45], "type": [7, 21, 23, 57, 61, 65, 66, 67, 69, 75, 76, 84], "rtype": 7, "replac": [7, 19, 20, 33, 61, 84, 87], "encapsul": [7, 61, 66], "involv": [7, 20, 23, 29, 30, 32, 40, 43, 50, 54, 55, 56, 59, 61, 65, 66, 69, 71, 72, 75, 85, 88], "concaten": [7, 52], "delimit": 7, "avoid": [7, 12, 21, 31, 46, 52, 60, 64, 65, 72, 75, 76, 85], "conflict": 7, "split": [7, 29, 45], "begin": [7, 9, 10, 11, 16, 17, 24, 27, 28, 29, 30, 32, 49, 71, 89], "repeatedli": [7, 33, 43, 72], "extract": [7, 39], "substr": [7, 11, 17, 19, 20, 78, 84, 87], "between": [7, 10, 14, 17, 24, 31, 41, 44, 51, 64, 65, 67, 68, 71, 72, 76, 89], "posit": [7, 12, 17, 21, 28, 30, 32, 39, 45, 47, 55, 61, 85], "befor": [7, 15, 29, 37, 40, 44, 50, 52, 56, 63, 64, 67, 72, 75, 86, 89], "revert": 7, "escap": 7, "preserv": [7, 64, 71], "content": [7, 29, 32], "even": [7, 29, 37, 44, 56, 86], "colon": 7, "input1": [7, 11], "lint": 7, "love": 7, "encoded1": 7, "decoded1": 7, "input2": 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61, 63, 70, 71, 72, 84, 87], "anagram": [8, 9, 84, 87], "sum": [8, 9, 12, 14, 33, 48, 49, 62, 63, 69, 77, 83, 84, 87], "49": [8, 9, 10, 84, 87], "347": [8, 9, 84, 87], "top": [8, 9, 21, 37, 41, 52, 56, 69, 77, 84, 86, 87], "frequent": [8, 9, 43, 79, 80, 81, 82, 83, 84, 87], "238": [8, 9, 84, 87], "product": [8, 9, 78, 84, 87, 89], "except": [8, 9, 84, 87, 89], "659": [8, 9, 84, 87], "encod": [8, 9, 45, 49, 84, 87, 89], "decod": [8, 9, 49, 78, 84, 87, 89], "128": [8, 9, 84, 87], "consecut": [8, 9, 76, 84, 87, 89], "sequenc": [8, 9, 20, 40, 45, 52, 56, 77, 84, 87, 89], "hard": [8, 18, 19, 33, 34, 40, 45, 48, 52, 53, 56, 58, 72, 73, 81, 87], "want": [8, 13, 15, 16, 19, 22, 25, 26, 30, 34, 43, 44, 48, 53, 58, 62, 67, 70, 73, 78, 79, 80, 81, 82, 83, 87], "feel": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "free": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 85, 87], "pull": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "request": [8, 13, 19, 22, 26, 34, 48, 53, 56, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87], "open": [8, 13, 19, 21, 22, 23, 26, 34, 48, 53, 58, 62, 63, 70, 73, 78, 79, 80, 81, 82, 83, 85, 86, 87], "issu": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 85, 86, 87, 88], "discuss": [8, 13, 19, 22, 26, 28, 34, 42, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87, 89], "improv": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 86, 87], "offici": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 85, 87], "websit": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "forum": [8, 13, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "hint": [8, 13, 17, 19, 22, 26, 34, 48, 53, 58, 62, 70, 73, 78, 79, 80, 81, 82, 83, 87], "happi": [8, 9, 13, 14, 19, 20, 22, 23, 26, 27, 34, 35, 48, 49, 53, 54, 58, 59, 62, 63, 70, 71, 73, 74, 78, 79, 80, 81, 82, 83, 87, 89], "welcom": [9, 14, 20, 23, 27, 35, 49, 54, 59, 63, 71, 74, 85, 86, 89], "section": [9, 14, 20, 23, 27, 35, 49, 54, 59, 63, 71, 89], "hand": [9, 14, 20, 23, 27, 35, 49, 50, 54, 59, 63, 64, 68, 69, 71, 72, 74, 89], "On": [9, 14, 20, 23, 27, 35, 49, 54, 55, 59, 63, 71, 74, 89], "master": [9, 14, 20, 23, 27, 35, 49, 54, 59, 63, 71, 74], "blind": [9, 14, 20, 23, 27, 35, 49, 54, 59, 63, 71, 74, 86, 88], "75": [9, 14, 20, 23, 27, 35, 49, 54, 59, 63, 71, 74, 84, 86, 88], "segment": [9, 23, 27, 29, 35, 49, 54, 71, 74], "explor": [9, 20, 23, 27, 35, 38, 40, 51, 52, 54, 59, 60, 61, 63, 64, 65, 66, 68, 69, 89], "curat": [9, 79, 80, 81, 82, 83, 87, 89], "select": [9, 55, 60], "focus": [9, 85], "power": [9, 14, 20, 23, 27, 35, 49, 63, 71, 74, 89], "techniqu": [9, 20, 27, 45, 49, 63, 66], "approach": [9, 10, 14, 17, 31, 33, 39, 40, 43, 45, 46, 47, 52, 57, 60, 63, 66, 75, 89], "varieti": [9, 14, 23, 89], "transform": [9, 20, 49, 89], "complex": [9, 14, 17, 18, 20, 21, 23, 24, 25, 28, 29, 30, 31, 32, 33, 35, 38, 42, 46, 49, 59, 71, 86, 89], "simpler": 9, "sharpen": [9, 14, 27, 35, 49, 59, 63], "skill": [9, 14, 27, 35, 49, 54, 59, 63, 71, 89], "manipul": [9, 14, 20, 27, 29, 35, 49, 50, 55, 57, 59, 74, 89], "introduc": [9, 15, 59, 63, 86], "art": [9, 14, 20, 27, 35, 49], "dive": [9, 20, 23, 35, 59, 63, 71], "fundament": [9, 23, 27, 35, 49, 54, 59, 71, 89], "identifi": [9, 59, 65, 66, 67, 68, 69], "target": [9, 12, 14, 25, 27, 40, 44, 60, 63], "concept": [9, 14, 23, 59, 63, 65, 71], "devis": 9, "investig": [9, 20, 49, 85], "presenc": [9, 50, 67], "common": [9, 35, 45, 48, 49, 50, 52, 79, 85, 87], "interview": [9, 23, 35, 74, 79, 80, 81, 82, 83, 87, 89], "real": [9, 14, 23, 35, 59, 71, 74], "world": [9, 14, 20, 23, 35, 59, 63, 71, 74], "applic": [9, 23, 27, 35, 49, 50, 54, 59, 71, 74, 84, 89], "delv": [9, 14, 20, 23, 27, 35, 49, 54, 59, 63, 71, 74, 89], "map": [9, 21, 64, 66], "appli": [9, 14, 27, 55, 63, 71, 76, 85], "categor": 9, "demonstr": [9, 21, 40, 45, 50, 54, 55, 60, 65, 68, 69, 71, 75, 76, 85], "versatil": [9, 14, 20, 23, 27, 54, 63, 89], "uncov": [9, 35, 49, 59, 63, 71], "intricaci": [9, 14, 20, 27, 35, 49, 54, 59, 63, 74, 89], "draw": 9, "As": [9, 10, 14, 20, 23, 27, 29, 32, 35, 43, 49, 54, 59, 63, 71, 74, 89], "tackl": [9, 14, 20, 35, 54, 71, 74, 89], "unravel": [9, 14, 20, 23, 27, 35, 49, 54, 59, 71, 74, 89], "ll": [9, 10, 14, 20, 23, 27, 49, 54, 63, 71, 74, 89], "har": [9, 27, 35, 71, 74], "craft": [9, 27, 35, 49, 54, 59, 74, 89], "eleg": [9, 14, 20, 23, 27, 35, 49, 54, 59, 63, 71, 74, 89], "readi": [9, 20, 49, 54, 63, 71, 84, 89], "embark": [9, 14, 20, 23, 27, 35, 49, 54, 63, 71, 89], "journei": [9, 14, 23, 27, 35, 49, 59, 71, 74, 89], "deepen": [9, 35, 49, 54, 63, 71], "understand": [9, 14, 23, 27, 35, 38, 49, 59, 63, 71, 74, 86, 88, 89], "enhanc": [9, 14, 23, 54, 63, 65, 71, 86], "height": [10, 35, 38, 39, 40, 42, 43, 44, 45, 46, 47, 69], "There": [10, 25, 28, 45, 50, 51, 56, 64, 65, 67, 68, 69, 72, 77, 86], "vertic": [10, 14, 52, 61, 64, 66, 67, 72], "drawn": 10, "endpoint": 10, "th": [10, 15, 30, 43, 55, 75], "x": [10, 45, 52, 55, 61, 66, 69, 72, 86], "axi": 10, "notic": [10, 12, 24], "slant": 10, "maxarea": 10, "area": [10, 14], "max_area": 10, "pointer": [10, 11, 12, 18, 24, 25, 28, 29, 30, 31, 32, 33, 35, 84], "until": [10, 11, 12, 17, 24, 25, 29, 30, 31, 32, 33, 37, 39, 41, 43, 44, 50, 55, 66, 67, 72, 88], "h": [10, 38, 40, 42, 43, 44, 45, 52], "min": [10, 15, 16, 56, 57, 69, 72], "width": [10, 41], "distanc": 10, "shorter": 10, "inward": 10, "possibl": [10, 12, 15, 25, 40, 43, 46, 51, 52, 60, 61, 63, 66, 67, 69, 72, 77, 86], "systemat": [10, 63], "hold": [10, 17, 28], "point": [10, 12, 24, 25, 28, 29, 30, 31, 32, 40, 45, 50, 64], "taller": 10, "won": 10, "increas": [10, 12, 44, 78, 84, 87], "decreas": [10, 12, 32], "smaller": [10, 12, 18, 32, 39, 40, 44, 56, 57, 60, 65, 67, 77], "height1": 10, "height2": 10, "height3": 10, "45": [10, 77], "like": [10, 12, 17, 31, 40, 41, 42, 44, 45, 47, 50, 52, 65, 66, 85, 89], "whose": 10, "consumpt": 10, "variant": [10, 41], "gap": [10, 89], "volum": 10, "uppercas": [11, 17, 18, 61], "non": [11, 32, 40, 42, 72, 81, 84, 87], "alphanumer": 11, "read": [11, 59, 84, 89], "forward": [11, 86], "backward": 11, "printabl": 11, "ispalindrom": 11, "lower": [11, 15], "cleaned_": 11, "join": [11, 45, 59, 72], "isalnum": 11, "clean": [11, 23, 27, 39], "don": [11, 25, 33, 45, 50], "insensit": 11, "digit": [11, 16], "punctuat": 11, "closer": [11, 42, 43], "decrement": [11, 12, 17, 18, 43, 67], "mismatch": [11, 46], "man": 11, "plan": [11, 75, 76], "canal": 11, "panama": 11, "output1": 11, "race": [11, 85], "output2": 11, "input3": 11, "output3": 11, "comprehens": [11, 54, 89], "half": [11, 24, 25, 29, 56], "still": [11, 17, 18, 24, 28, 37, 43, 47, 51, 57, 61], "primarili": [11, 17, 29, 46, 47, 51, 55, 76], "version": [11, 66, 85], "larg": [11, 44, 46, 55, 57], "addition": [11, 16, 30, 39, 40, 46, 61, 66, 68, 72], "babad": 11, "bab": 11, "aba": 11, "limit": [11, 52, 77], "allowed_char": 11, "ignor": [11, 21], "triplet": [12, 14], "j": [12, 33, 52, 61, 64, 66, 69, 72], "3000": [12, 39], "threesum": 12, "ascend": [12, 24, 25, 33, 43, 89], "skip": [12, 30, 60, 75, 76], "elif": [12, 21, 25, 44, 51, 56, 76], "neg": [12, 40, 47, 60], "Then": [12, 15, 56, 69, 72], "larger": [12, 18, 44, 56, 61], "been": [12, 18, 21, 29, 37, 39, 52, 64, 65, 67, 68, 88], "contribut": [12, 25, 29, 41, 55, 61, 65, 77, 85], "roughli": [12, 41], "mani": [12, 18, 51, 60, 66, 68, 72, 77, 89], "criteria": 12, "rather": [12, 24, 64], "gener": [12, 18, 21, 29, 56, 60, 61, 63, 77, 86, 89], "satisfi": [12, 47, 63, 65, 69, 72, 84], "themselv": [12, 29], "some": [12, 17, 21, 28, 44, 45], "125": [13, 14, 84, 87], "15": [13, 14, 24, 37, 38, 39, 40, 61, 84, 87], "3sum": [13, 14, 84, 87], "11": [13, 14, 24, 84, 87], "With": [13, 14, 15, 84, 87], "water": [13, 14, 66, 70, 71, 84, 87], "solut": [13, 14, 19, 20, 22, 23, 26, 28, 32, 33, 34, 42, 46, 48, 49, 53, 54, 58, 59, 62, 63, 70, 71, 73, 74, 78, 79, 80, 81, 82, 83, 84, 86, 89], "strategi": [14, 20, 33, 74, 75, 76, 89], "abil": [14, 27, 28, 35, 50, 63, 65, 89], "maintain": [14, 20, 25, 47, 50, 52, 55, 56, 57, 59, 64, 69, 75], "often": [14, 42, 43, 47, 60, 66, 67], "known": [14, 50, 54, 75, 79, 80, 81, 82, 83, 87, 89], "time": [14, 17, 18, 19, 20, 21, 24, 25, 28, 29, 30, 31, 32, 33, 38, 42, 46, 59, 84, 85, 86, 87], "space": [14, 17, 18, 20, 21, 24, 25, 28, 29, 30, 31, 32, 33, 38, 42, 46, 85, 86], "discov": [14, 20, 23, 49, 59], "leverag": [14, 23, 27, 49, 54, 59, 63, 74, 89], "address": [14, 71, 85, 88], "immers": [14, 27, 89], "yourself": [14, 27, 41, 45, 89], "maxim": [14, 15, 20, 75], "histogram": 14, "showcas": [14, 20, 23, 27, 35, 74], "realm": [14, 49, 59, 63, 74, 89], "learn": [14, 27, 59, 85, 89], "emploi": [14, 24, 63], "three": [14, 38, 39, 44, 45, 60, 61, 76, 77], "embrac": [14, 74, 89], "readabl": [14, 27, 59, 74], "practic": [14, 27, 42, 43, 44, 46, 47, 50, 51, 52, 60, 65, 66, 67, 71, 89], "detail": [14, 29, 60, 68, 69, 72, 86, 88], "explan": [14, 85, 86], "gain": [14, 23, 27, 49, 54], "deep": [14, 27, 38, 39, 64, 66, 69, 74, 89], "prepar": [14, 23, 35, 74, 79, 80, 81, 82, 83, 87, 89], "think": [14, 35], "divers": [14, 27, 49, 54, 85], "commenc": [14, 27, 54, 89], "price": [15, 20], "dai": [15, 88], "profit": [15, 20], "choos": [15, 17, 40, 55, 76], "futur": [15, 84], "transact": 15, "maxprofit": 15, "min_pric": 15, "max_profit": 15, "encount": [15, 16, 21, 23, 27, 40, 65, 66, 74, 86, 89], "made": [15, 32, 33, 39, 46, 50, 51, 56, 57, 60], "seen": [15, 16, 57], "far": [15, 16, 18, 43, 49, 56, 57], "haven": [15, 43, 52], "lowest": [15, 48, 49, 87], "sold": 15, "assum": [15, 39, 51, 64, 65, 66, 68, 72], "bought": 15, "subtract": [15, 60], "previous": [15, 50, 51], "prices1": 15, "prices2": 15, "examin": [15, 49], "rel": [15, 33, 44, 69], "regardless": [15, 17, 28, 29, 31, 32, 44, 75, 76, 85], "memori": [15, 28, 31, 32, 33, 44, 45, 51, 61, 64, 66], "usag": [15, 28, 30, 39, 40, 43, 45, 47, 50, 51, 55, 56, 57, 61, 69, 75], "again": [15, 17, 25, 28, 50], "fee": 15, "symbol": 16, "length_of_longest_substr": 16, "last": [16, 23, 25, 28, 29, 31, 37, 50, 58, 59, 75, 76, 84], "char_index": 16, "necessari": [16, 51, 56, 57, 61], "wa": [16, 18, 24, 31, 50, 72, 85], "new": [16, 28, 29, 30, 31, 32, 33, 35, 37, 39, 40, 41, 44, 45, 50, 51, 52, 54, 55, 57, 64, 68, 72, 75, 84, 86], "fact": 16, "abcabcbb": 16, "bbbbb": 16, "pwwkew": 16, "lookup": 16, "arithmet": [16, 21], "abca": 16, "abcd": 16, "chang": [17, 29, 31, 49, 52, 57, 66, 78, 84, 86, 87], "characterreplac": 17, "max_count": 17, "slide": [17, 18, 56, 84], "window": [17, 56, 84], "char_count": 17, "plu": [17, 40], "minu": [17, 61], "exceed": 17, "shrink": [17, 18], "side": [17, 42, 49, 89], "alwai": [17, 57, 64], "s1": 17, "abab": 17, "k1": [17, 43], "s2": 17, "aababba": 17, "k2": [17, 43], "sever": [17, 45, 86], "inform": [17, 31, 65, 72, 85, 87, 88], "variat": [17, 18, 24, 25, 55, 75, 76], "testcas": 18, "min_window_substr": 18, "t_dict": 18, "current_window_dict": 18, "popul": [18, 57, 67, 68], "char": [18, 21, 50, 51, 52, 72], "min_len": 18, "float": [18, 40, 47, 56], "inf": [18, 40, 47], "min_window": 18, "required_char": 18, "found_char": 18, "window_s": 18, "left_char": 18, "further": [18, 31, 66, 72, 86, 88], "adobecodebanc": 18, "abc": [18, 61], "banc": 18, "aa": 18, "smallest": [18, 24, 25, 48, 49, 56, 57, 72, 87], "No": [18, 85], "121": [19, 20, 84, 87], "best": [19, 20, 38, 40, 43, 84, 85, 87], "bui": [19, 20, 84, 87], "And": [19, 48, 53, 87], "sell": [19, 20, 84, 87], "stock": [19, 20, 84, 87], "424": [19, 20, 84, 87], "76": [19, 20, 84, 87], "chapter": [20, 59, 89], "dynam": [20, 35, 54, 57, 59, 71, 75, 76, 77, 89], "particularli": [20, 28, 50, 54], "subarrai": [20, 78, 80, 87], "wit": [20, 23, 35], "invalu": [20, 54, 89], "tool": [20, 23, 54, 59, 84], "arsen": [20, 23, 54], "relat": [20, 50, 63, 71], "grasp": [20, 59], "manag": [20, 32], "bracket": [21, 23], "close": [21, 23, 43, 47, 65], 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